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Thread: The God's eye global gist of general relativity

  1. #101  
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    Quote Originally Posted by Farsight View Post
    Fair enough. You could say much the same for the room you're in. Regardless of or shells or balls, if you can't detect any tidal force or any difference in field strength between the floor and ceiling, you might think the gravitational field was uniform.
    The room I am in is different to the underground cave. The inverse square law applies so the field is stronger near the floor then up near the ceiling.
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  2. #102  
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    Quote Originally Posted by Jilan
    You misunderstand me, the cave is not at the centre of the earth, it's just a little way underground.
    You are now trying to cover up your mistake, you have claimed a cave , off-center, under the Earth. Now you are changing your story. Not only that you are wrong, you are dishonest as well.
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  3. #103  
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    I never changed anything. You need to reread my posts carefully.

    Edit: If you want to be treated with respect you need to take up my challenge and put your maths where your mouth is.
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  4. #104  
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    Quote Originally Posted by Jilan
    I don't think you are quite getting it. I am talking about a hollow spherical cave inside a solid spherical body.
    The shape of the cave is irrelevant. You still have it wrong.
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  5. #105  
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    Quote Originally Posted by Farsight
    Then you've got a uniform gravitational field, which means you've got no Riemann curvature, so you haven't got a gravitational field. The moot point is that standing on the surface of the Earth is like accelerating through space, but you aren't actually accelerating. So it isn't actually equivalent.
    You're quite wrong. In fact only amateurs make global statements like that in general relativity, certainly not experts and certainly not Eintein because he made it clear that you can create a gravitational field by changing spacetime coordinates, i.e. chaning your frame of reference. You can't do that with curvature.

    In fact Einstein explicitly stated that the non-vanishing of the Riemann tensor is not the criteria for the non-vanishing of a gravitational field. Its the non-vanishing of the Christoffel sybmbols. Einstein protested when Max von Laue stated that it was the Riemann tensor that was the only representation of the gravitational field and wrote back to him saying
    ... what characterizes the existence of a gravitational field from the empirical standpoint is the non-vanishing of the components of the affine connection (i.e. the Christoffel symbols), not the vanishing of the [components of the Riemann tensor]. If one does not think in such intuitive ways, one cannot grasp why something like curvature should have anything at all to do with gravitation. In any case, no rational person would have hit upon anything otherwise. The key to the understanding of the equality of gravitational mass and inertial mass would have been missing.
    That this is true and current can also be found in the American Journal of Physics in the article

    Cosmic strings: Gravitation without local curvature by T.M. Helliwell, Am. J. Phys., 55(5), May (1987)
    Cosmic strings: Gravitation without local curvature
    Cosmic strings are very long, thin structures which might stretch over vast reaches of the universe. If they exist, they would have been formed during phase transitions in the very early universe. The space‐time surrounding a straight cosmic string is flat but nontrivial: A two‐dimensional spatial section is a cone rather than a plane. This feature leads to unique gravitational effects. The flatness of the cone means that many of the gravitational effects can be understood with no mathematics beyond trigonometry. This includes the observational predictions of the double imaging of quasars and the truncation of the images of galaxies.
    The exact same thing is true for a vacuum domain wall. The gravitational field of such a wall is a uniform gravitational field. But physicists still think of it as a gravtitational field, as real as any other field
    An exact solution of Einstein's equations is found describing the gravitational field of a plane vacuum domain wall. The metric has event horizons, has no true geometric singularities, and is locally flat everywhere except on the wall itself.
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  6. #106  
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    x0x - It’s clear to both Jilan and myself that that the physics is quite correct and how Jilan described it is quite accurate. In fact what Jilan described is a well known fact in physics. It works in both gravity and electrostatics. In fact I had to prove this on a test in electrodynamics that I had to take in graduate school. It’s very easy to show that it’s true in fact. If you had the skills and were actually willing to learn rather than merely make claims that Jilan is wrong - which, by the way, is dead giveaway that you don’t know how to justify/prove your claim. It’s very clear that you’re basing this on some flawed assumptions you picked up the past. Actually if you picked up Gravitation and Spacetime – Third Edition by Ohanian and Ruffini and tried to solve problem 1.1 which reads
    Consider the gravitational field produced by a sphere with a uniform mass density. Show that if you evacuate a concentric spherical cavity in this sphere, the gravitational field will be zero within this cavity (that is g = -grad Phi = 0). Show that if you excavate a spherical cavity not concentric with the original sphere, the gravitational field will be constant (that is g = -grad Phi = constant).
    This is the first problem because it’s a very easy problem to solve but a very important one. I hope you’re not going to claim that you’re right, it’s the world that’s wrong, are you?
    Last edited by Physicist; 06-29-2014 at 12:41 PM.
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  7. #107  
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    Quote Originally Posted by x0x
    the cave is not at the centre of the earth, it's just a little way underground.
    How on Earth could you ever conclude such a thing. Jilan wrote
    the cave is not at the center, it's just a little way underground.
    That's what it means for a cave to be off center. You were thinking that the center of the cave was near the center of the earth and Jilan was thinking that it was near the surface. I don't know about you but I know how a cave could be constructed near the center of the Earth. Caves are near the surface of the Earth. Especially the man made ones.

    But it doesn't matter where the center is, so long as its not at the center of the Earth. The field will be uniform so long as its not centered at the Earth's center and is completely contained within the Earth.
    Last edited by Physicist; 06-30-2014 at 03:57 AM.
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  8. #108  
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    Moderator note: ALL members will desist from using ad-hominems. Personal attacks are not allowed here - attack the words a user posts, but DO NOT attack the user who posts them.
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  9. #109  
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    Quote Originally Posted by Physicists
    That's what it means for a cave to be off center. You were thinking that the center of the cave was near the center of the earth and she was thinking that it was near the surface.
    If the cave is off center, then the gravitational field is NOT uniform.

    I don't know about you but I know how a cave could be constructed near the center of the Earth. Caves are near the surface of the Earth.
    If the cave is near the surface of the Earth , the gravitational field inside it will NOT be uniform, as I already explained about three times.
    The field will be uniform so long as its not centered at the Earth's center and is completely contained within the Earth.
    This is false. The reason is that the force exerted on a test probe is non-null. Since it follows that is not constant.
    Last edited by x0x; 06-29-2014 at 02:36 PM.
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  10. #110  
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    Quote Originally Posted by x0x View Post
    If the cave is off center, then the gravitational field is NOT uniform.



    If the cave is near the surface of the Earth , the gravitational field inside it will NOT be uniform, as I already explained about three times.


    This is false. The reason is that the force exerted on a test probe is non-null. Since it follows that is not constant.
    You seem to be confusing the gravitational potential with the field. The potential changes but it's gradient (the field strength) doesn't. No one said that a test probe would be weightless!
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  11. #111  
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    Quote Originally Posted by Jilan
    You seem to be confusing the gravitational potential with the field. No one said that a test probe would be weightless!
    The measurable entity is force. Force , as I have already shown is NOT constant and is NOT null.
    The gravitational potential is VARIABLE, as such, contrary to your oft repeated claims, the field is NOT "uniform".

    The potential changes but it's gradient (the field strength) doesn't.
    The potential is . Please prove that "it's (sic!) gradient doesn't change". Use math.
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  12. #112  
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    Thankyou SpeedFreak. I don't know about anybody else, but I'm getting rather tired of x0x's ignorance and abuse. If he's the same x0x as per TSR he's just a student who obviously doesn't know much physics. And yet what he's effectively saying is if you're talking about something that I don't know about, you must be a quack.

    Quote Originally Posted by Jilan
    The room I am in is different to the underground cave. The inverse square law applies so the field is stronger near the floor then up near the ceiling.
    No problem. I was trying to make it moot. You can't readily detect the tidal force in the room you're in. But you can readily detect the force of gravity.

    x0x: see gravitational potential on Wikipedia. Note the plot:


    CCSA image by AllenMcC

    The force of gravity at some location relates to the slope of the plot at that location. That's the gradient in potential. Or the derivative of potential if you prefer. The second derivative gives you the tidal force. This relates to the difference in the force of gravity at the floor and ceiling, and to "Riemann curvature". Read the OP to see how you could plot this with light clocks.
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  13. #113  
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    On this cave business...

    If you have a ball of uniform density and radius , and create a ball-shaped cavity within it (anywhere within it) with radius , then in Newtonian gravity the gravitational field strength will be constant throughout the cavity (i.e. the potential will have a constant gradient).

    This is pretty easy to prove. The field of the source is equal to the field of a ball of radius and uniform density (without any cavity) minus the field of a ball of equal density of radius , centered at the appropriate location. So, if you consider any point within the cavity, by the shell theorem the field strength is


    where is the vector position of the point relative to the centre of the ball of radius , and is the vector position of the point relative to the centre of the cavity. However, by basic geometry, the vector is the vector position of the centre of the cavity relative to the centre of the enclosing ball; therefore, it has the same value throughout the cavity, and the field is perfectly uniform (to the extent that Newtonian gravity is valid).

    Note that for a ball of non-uniform density, the result does not necessarily hold (even if the density is spherically symmetric, aside from the cavity).
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  14. #114  
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    Here you are. But you should have taken up the challenge yourself!
    Physics by Example: 200 Problems and Solutions - W. G. Rees - Google Books
    See problem 44.
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  15. #115  
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    Quote Originally Posted by btr
    Note that for a ball of non-uniform density, the result does not necessarily hold (even if the density is spherically symmetric, aside from the cavity).
    You are absolutely correct. The potential varies linearly with "r" inside a sphere. So, IF the cavity is spherical, then . If the cavity is anything but spherical, then the field is not uniform, because the strength due to the subtracted cavity DOES NOT vary linearly with "r", so the formula does not apply.
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  16. #116  
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    Quote Originally Posted by Jilan
    Here you are.
    Try with a non-spherical cavity, one that has ceiling and floor.
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  17. #117  
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    If I cast my mind back far enough (to post #74), I remember that Jilan wanted to demonstrate the possibility of creating a uniform gravitational field by methods other than accelerating rocket ships. Maybe we can at least all agree that this has been demonstrated?
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  18. #118  
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    Quote Originally Posted by x0x View Post
    Try with a non-spherical cavity, one that has ceiling and floor.
    I believe in post#74 I said that a spherical underground cave would do the job nicely. If this is as close to an apology as I am getting I will accept it. No hard feelings.
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  19. #119  
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    Jilan,

    My correction came to the error in your post 83, not 74:

    Quote Originally Posted by Jilan
    You misunderstand me, the cave is not at the centre of the earth, it's just a little way underground. The acceleration due to gravity within the cave is constant between the bottom and the top of the cave.
    I was very clear in saying that the field is not uniform between the ceiling and the floor of the cave. So, once again, try the exercise in your book with a box-shaped cave. I challenge you.
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  20. #120  
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    Quote Originally Posted by x0x View Post
    Jilan,

    My correction came to the error in your post 83:



    I was very clear in saying that the field is not uniform between the ceiling and the floor of the cave. So, once again, try the exercise in your book with a box-shaped cave. I challenge you.
    Why would I want to do that, we were never discussing box shaped caves, lol!
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  21. #121  
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    Why would I want to do that, we were never discussing box shaped caves, lol!
    Because this is the type of cavity that has "top" and "bottom". What I was correcting was your error from post 83, not 74 (I have not even seen post 74). So, what is the answer, Jilan? Still claiming uniform gravitational field or do you want to retract that claim?
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  22. #122  
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    Lol, my sides are splitting.
    Why do you think that a spherical cave doesn't have a top or a bottom?
    I should imagine there are better things you could be doing right now. Perhaps reading the book that Physicist mentioned?
    I wish you well, but I am not getting dragged into a crazy argument. Go learn some stuff, we may speak again, hopefully.
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  23. #123  
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    Quote Originally Posted by Jilan
    Why do you think that a spherical cave doesn't have a top or a bottom?
    Because it doesn't. Last I checked , it has full spherical symmetry, so there is not "top" and there is no "bottom". At this point you are desperately trying to cover your errors from post 83 and 110. This one is the most hilarious:

    Quote Originally Posted by Jilan
    The potential changes but its gradient doesn't
    How is your calculus knowledge? eh?
    Initially, I thought that your support for Farsight crackpoterry was accidental, now I don't think that anymore.
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  24. #124  
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    I promised myself, but.....
    If I had a spherical room in my house would it have a top and a bottom? So with an underground cave. A mass will fall from the top to the bottom. My calculus is just fine thank you. Post #110, really? Please you must stop soon or my sides will split!
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  25. #125  
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    x0x, I think you may have missed some important context, such as post #74 and the discussion leading up to it. With that context, it was pretty obvious that Jilan was referring to a spherical cavity rather than some generic one, and that "top" and "bottom" referred to the points respectively furthest from and nearest to the centre of the enclosing ball of matter.

    As for Jilan's statement that "the potential changes but the gradient doesn't," I don't see a problem with that. We've seen that the field strength (and therefore the potential gradient) is constant and non-zero in the cavity, and so the potential must vary. In fact, in the cavity is given by


    in a Cartesian coordinate system with its origin at the cavity's centre, and with the axis parallel to the line joining the cavity's centre to the enclosing ball's centre; the constant is the distance from the cavity's centre to the centre of the enclosing ball of matter.
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    Yeah, I have to "side" with a spherical void having a top and bottom, even if chosen arbitrarily.
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    Apologies to Farsight, for thread hijacking, it wasn't intentional. Can we carry on?
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    Yes. But please don't feed the troll.

    So, do you understand the OP? Do you have any problems with it? Can you find any flaws?
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    The obvious flaw is that there isn't any way presented to do physics with the "interpretation" presented.
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  30. #130  
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    As for Jilan's statement that "the potential changes but the gradient doesn't," I don't see a problem with that.
    Really? You too? =>

    x0x, I think you may have missed some important context, such as post #74
    I started criticizing from post 83. As it stands , by itself, post 83 is in error, arbitrary cavities inside the Earth do not exhibit uniform gravitational field. .
    Last edited by x0x; 06-30-2014 at 03:00 AM.
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  31. #131  
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    Quote Originally Posted by x0x View Post
    Really? You too? =>
    This will be my last post on this derail.

    x0x, as an exercise, compute the gradient of , where and are constants. Feel free to start another thread if you want any help.

    Farsight and everyone else, my apologies.
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  32. #132  
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    I knew that you'll come up with this example. For the particular case that you posted, yes. For the general case, obviously, no. Math is not done by enumeration (actually, science in general is not done by enumeration).
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  33. #133  
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    Quote Originally Posted by x0x View Post
    If the cave is off center, then the gravitational field is NOT uniform.
    That is incorrect.

    Quote Originally Posted by x0x View Post
    If the cave is near the surface of the Earth , the gravitational field inside it will NOT be uniform, as I already explained about three times.
    And in all three times you've been incorrect. Repeating a mistake doesn't make it right you know.

    Quote Originally Posted by x0x View Post
    This is false. The reason is that the force exerted on a test probe is non-null. Since it follows that is not constant.
    [/quote]
    That too is incorrect. I've derived the proof of my assertion many times in the past including on a test during an exam. I got it right there too.

    The solution can be found on the internet and physics textbooks. Use Google to search using the terms "uniform gravitational field" + "spherical cavity" . When the list comes up choose the one that says
    Physics by Example: 200 Problems and Solutions - Google Books Result
    books.google.com/books?isbn=0521449758
    W. G. Rees - ‎1994 - 374 pages
    ... 47 Now for a body falling from rest in a uniform gravitational field of strength g,
    ... A sphere of uniform density p has within it a spherical cavity whose centre is a
    ...
    Read problem 44. The book is called Physics by Example: 200 Problems and Solutions by W. Rees. You can download the entire text at bookos-z1

    From now on you'd be wise to check your facts before taking a firm stand on something. I can't speak for anybody else but when I make a statement I do so because I know it to be a fact. That means that I've either calculated it for myself or I know several standard well respected textbooks in which the proof can be found or where an axiom/principle/law of physics can be found or where I can find the experimental evidence of what I state as true.
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  34. #134  
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    Quote Originally Posted by Physicist
    That is incorrect.
    Actually, it is correct. Only spherical cavities produce a uniform field when positioned off-center. Generic shape cavities don't.
    That too is incorrect. I've derived the proof of my assertion many times in the past including on a test during an exam. I got it right there too.
    One more thing: non-null force means non-null gradient for . Non-null gradient means that is not constant. The fact that you got it right for spherical cavities (congratulations!), doesn't mean that you have it right for generic ones (you don't, see above).

    Read problem 44. The book is called Physics by Example: 200 Problems and Solutions by W. Rees. You can download the entire text at bookos-z1
    Getting the answer right for a specific problem in a textbook doesn't make you a physicist. You need to learn how to think about the general case. This is why I challenged "Jilan" to solve the case of a box-like cavity.
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  35. #135  
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    Quote Originally Posted by Farsight View Post
    Yes. But please don't feed the troll.

    So, do you understand the OP? Do you have any problems with it? Can you find any flaws?
    Yes I understand it. Path deflection on the horizontal part of the orbit. Whether it's EM or something else, it's an interesting concept.
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  36. #136  
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    Quote Originally Posted by Farsight View Post
    I don't know what the other side of the globe has got to do with spacetime curvature.
    Geodesic deviation associated with the spacetime curvature due to earth's gravity has changed the direction of the acceleration on the other side of the globe. But the equivalence principle is about local physics such that the change in acceleration is insignificant. Under these conditions, we can treat the acceleration as in a locally Minkowskian spacetime.


    Quote Originally Posted by Farsight View Post
    Standing on the ground can be likened to accelerating through space, but you aren't actually accelerating, nor is the ground, nor are the objects near you.

    Let's say it's a rocket that has enough exhaust to keep it hovering. A force is being exerted. Energy is being expended. But because there's a gravitational field present the rocket isn't accelerating. It's like it's accelerating through space when there's no gravitational field present. But the two situations are not the same.
    Well then, what do think acceleration is? How can something be like acceleration but not acceleration? I should point out that if we accept that light slows down in a gravitational field, then it also slows down in an accelerated frame of reference, so you cannot use the speed of light to distinguish between gravity and acceleration.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
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    Quote Originally Posted by KJW View Post
    Well then, what do think acceleration is? How can something be like acceleration but not acceleration? I should point out that if we accept that light slows down in a gravitational field, then it also slows down in an accelerated frame of reference, so you cannot use the speed of light to distinguish between gravity and acceleration.
    Predicted answer: frames of reference don't really exist.
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  38. #138  
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    Quote Originally Posted by x0x
    Actually, it is correct.
    Wrong again.

    Quote Originally Posted by x0x
    Only spherical cavities produce a uniform field when positioned off-center.
    You need to pay much closer attention to the conversation because we’re talking about that kind of cavity, nothing more and nothing less.

    Quote Originally Posted by x0x
    Generic shape cavities don't.
    So what? Nobody is talking about general distributions of matter. We’re talking about one and only one distribution, i.e. the one Jilan and I told you about.

    Quote Originally Posted by x0x
    Getting the answer right for a specific problem in a textbook doesn't make you a physicist.
    No. It doesn’t. Going to undergraduate and graduate school in physics and getting a degree in physics and working as a physicist is what makes me a physicist.

    Sorry x0x but since you’re not reading and responding directly to what Jilan and I are talking about and are making things up as you go along I will not respond to anymore of your posts because all of them so far contain false assumptions and I see no hope of that changing now or in the future. Good bye.
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  39. #139  
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    Quote Originally Posted by Physicist
    You need to pay much closer attention to the conversation because we’re talking about that kind of cavity, nothing more and nothing less.
    I did, I picked up the "conversation" at post 83. As it stands, post 83 is wrong due to sloppy language: an arbitrary cavity inside the Earth DOES NOT exhibit uniform gravitational field.

    No. It doesn’t. Going to undergraduate and graduate school in physics and getting a degree in physics and working as a physicist is what makes me a physicist.
    Not if you are sloppy.
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    Quote Originally Posted by PhysBang
    Predicted answer: frames of reference don't really exist.
    Who was it that said or do you predict will say this? A frame of reference in SR is defined as a coordinate system along with a system of clocks and rods used to record events. In practice there are much less clocks are rods than in principle. In practice they consist of measurement apparatus.
    Frame of reference - Wikipedia, the free encyclopedia

    In reality a frame of reference is something such as the laboratory at a particle accelerator lab such as CERN.

    But they're useful ideas just as numbers are useful ideas. Do numbers exist? What they describe certainly does.
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  41. #141  
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    Quote Originally Posted by Physicist View Post
    Who was it that said or do you predict will say this?
    I believe PhysBang is predicting this to be Farsight's response to my post.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
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  42. #142  
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    Quote Originally Posted by x0x View Post
    I did, I picked up the "conversation" at post 83. As it stands, post 83 is wrong due to sloppy language: an arbitrary cavity inside the Earth DOES NOT exhibit uniform gravitational field.

    Not if you are sloppy.
    x0x, I can see what happened now, you jumped in a post #60 Then you ignore the thread again until post #83 without catching up on the discussion in the meantime. Don't you think it would be a little unrealistic for every post to bring you up to speed with the entire contents so the thread so far? If you having a conversation with someone about the best way to grow apples and then someone arrives and says "no that is not the best way to grow pears" what would you think?
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  43. #143  
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    Quote Originally Posted by x0x View Post

    Quote Originally Posted by Physicist
    You need to pay much closer attention to the conversation because we’re talking about that kind of cavity, nothing more and nothing less.
    I did, I picked up the "conversation" at post 83. As it stands, post 83 is wrong due to sloppy language: an arbitrary cavity inside the Earth DOES NOT exhibit uniform gravitational field.

    No. It doesn’t. Going to undergraduate and graduate school in physics and getting a degree in physics and working as a physicist is what makes me a physicist.
    Not if you are sloppy.
    Well, it was remiss of YOU not to track the conversation back to where it started. Post 83 is quoting only part of post 78, whilst post 78 itself quotes the original statement made by Jilan in post 74, and it is the original statement made there that was subsequently being discussed. Here is that statement:

    Quote Originally Posted by Jilan View Post
    I guess it all comes down to that inverse square law resulting in acceleration is not uniform. To model the real equivalence you would need a constant field strength. A spherical underground cave should do the job nicely.
    So, the discussion you jumped in on in post 83 was about a spherical cave.

    Now then, to put on my admin hat...

    You, x0x, will NOT call people sloppy. That is an ad hominem and I have already posted a note about this. I would also advise you to lose the attitude. Consider this a warning.

    TO EVERYONE: I will not have people called liars, trolls, ignorant or any other ad hominems. The next person who posts an ad hom will be suspended. Please learn to play together nicely.
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  44. #144  
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    Quote Originally Posted by SpeedFreek View Post
    TO EVERYONE: I will not have people called liars, trolls, ignorant or any other ad hominems. The next person who posts an ad hom will be suspended. Please learn to play together nicely.[/COLOR]
    So if a poster here tells an obvious lie, what are we to do about it, pretend that it doesn't exist?
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    I guess it fine to point out the error of their ways with recourse to the evidence without calling them a liar.
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    Quote Originally Posted by KJW View Post
    Geodesic deviation associated with the spacetime curvature due to earth's gravity has changed the direction of the acceleration on the other side of the globe. But the equivalence principle is about local physics such that the change in acceleration is insignificant. Under these conditions, we can treat the acceleration as in a locally Minkowskian spacetime.
    OK, but let's not get confused between flat as in horizontal and flat as in not curved. The "tilt" in the plot of gravitational potential relates to the force of gravity. The more tilted it is, the stronger the force of gravity at that location.

    Quote Originally Posted by KJW View Post
    Well then, what do think acceleration is?
    A change in speed or direction. When you're standing on the ground, it isn't happening to you.

    Quote Originally Posted by KJW
    How can something be like acceleration but not [be] acceleration?
    Because you feel a force like you would if you were accelerating, even though you aren't. Standing on the ground feels like accelerating through space. But you aren't accelerating through space.

    When you're accelerating through space you see a light beam curve and feel a force on your feet. When you're standing on the ground you see a light beam curve and feel a force on your feet. But not for the same reason. The light curves and you feel that force because space is inhomogeneous.

    Quote Originally Posted by KJW
    I should point out that if we accept that light slows down in a gravitational field, then it also slows down in an accelerated frame of reference, so you cannot use the speed of light to distinguish between gravity and acceleration.
    Agreed. See Don Koks' article, and in particular the section entitled The Speed of Light as Measured by Non-Inertial Observers. It's the "Riemann curvature" that distinguishes gravity from acceleration. You can't transform it away. It relates to the curvature in the plot of gravitational potential. Without it, the plot can't get off the flat and level.
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    Quote Originally Posted by PhysBang View Post
    So if a poster here tells an obvious lie, what are we to do about it, pretend that it doesn't exist?
    Correct them. Tell them they are mistaken. Point out the inconsistency in what they have posted. Use logic. But do not question their character.
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    Quote Originally Posted by Farsight
    Then you've got a uniform gravitational field, which means you've got no Riemann curvature, so you haven't got a gravitational field. The moot point is that standing on the surface of the Earth is like accelerating through space, but you aren't actually accelerating. So it isn't actually equivalent.
    Perhaps in Farsight’s theory of relativity, but certainly not in Einstein’s theory of relativity in which case you’ve got it all wrong. That means that you’ve got the definitions all wrong.

    In Einstein’s general theory of relativity, as defined and interpreted by Einstein (not Farsight or anybody else), Jilan is 100% correct. In fact a uniform gravitational field was the first gravitational field that Einstein studied.

    Regarding your comment
    Quote Originally Posted by Farsight
    …which means you've got no Riemann curvature, so you haven't got a gravitational field.
    It's incorrect to assert that spacetime curvature, i.e. the Riemann tensor, is that which determines the presence of a gravitational field. Its the Christoffel symbols, G^a_mn (G = Christoffel symbol = capital gamma), which are responsible. Or if you have Gravitation by Misner, Thorne, and Wheeler turn to page 467 where the authors write
    -------------------------------------------------------
    One can always find in any given locality a frame of reference in which all local "gravitational fields (all Christoffel symbols: all G^a_mn) disappear. No G's means no "gravitational field....
    -------------------------------------------------------

    In a letter to Max von Laue, Einstein wrote the following in response to Laue's assertion that gravity exists only where spacetime curvature exists.
    ... what characterizes the existence of a gravitational field from the empirical standpoint is the non-vanishing of the components of the affine connection] not the vanishing of the [components of the Riemann tensor]. If one does not think in such intuitive (anschaulich) ways, one cannot grasp why something like curvature should have anything at all to do with gravitation. In any case, no rational person would have hit upon anything otherwise. The key to the understanding of the equality of gravitational mass and inertial mass would have been missing.
    You’ve been interpreting all of this backwards. You think that since a uniform gravitational field is identical to a uniform gravitational field then since there’s no gravitational field just a transformation to a acceleration then there’s no “real” gravitational field whereas Einstein saw it the other way around. I.e. since they’re identical the accelerating frame in flat spacetime produces a “real” gravitational field merely by changing frames to an accelerating one.
    Recall how Einstein phrased this in his 1916 review paper, i.e. The Foundation of the General Theory of Relativity by Albert Einstein, Annalen der Physik, 49. It’s online at The Foundation of the General Theory ... (1 to 3, of 22, 1916)
    It will be seen from these reflexions that in pursuing the general theory of relativity we shall be led to a theory of gravitation, since we are able to "produce" a gravitational field merely by changing the system of co-ordinates. It will also be obvious that the principle of the constancy of the velocity of light in vacuo must be modified, since we easily recognize that the path of a ray of light with respect to K' must in general be curvilinear, if with respect to K light is propagated in a straight line with a definite constant velocity.
    So you see, Farsight? You’ve got it all backwards. It’s simply an historical fact that you’re (unsubstantiated) claim Then you've got a uniform gravitational field, which means you've got no Riemann curvature, so you haven't got a gravitational field. is quite wrong.
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    Quote Originally Posted by Physicist View Post
    Perhaps in Farsight’s theory of relativity, but certainly not in Einstein’s theory of relativity in which case you’ve got it all wrong. That means that you’ve got the definitions all wrong.
    I obviously don't think I have, but let's look at this more closely.

    Quote Originally Posted by Physicist View Post
    In Einstein’s general theory of relativity, as defined and interpreted by Einstein (not Farsight or anybody else), Jilan is 100% correct. In fact a uniform gravitational field was the first gravitational field that Einstein studied.
    Agreed. I've said the principle of equivalence was an "enabling" principle. Einstein's happiest thought was of the falling man. But there really is no escaping the issue that the principle of equivalence applies to an infinitesimal region. A region of zero extent. No region.

    Quote Originally Posted by Physicist View Post
    Regarding your comment

    …which means you've got no Riemann curvature, so you haven't got a gravitational field.

    It's incorrect to assert that spacetime curvature, i.e. the Riemann tensor, is that which determines the presence of a gravitational field. Its the Christoffel symbols, G^a_mn (G = Christoffel symbol = capital gamma), which are responsible. Or if you have Gravitation by Misner, Thorne, and Wheeler turn to page 467 where the authors write
    -------------------------------------------------------
    One can always find in any given locality a frame of reference in which all local "gravitational fields (all Christoffel symbols: all G^a_mn) disappear. No G's means no "gravitational field....
    -------------------------------------------------------
    I'm not the first person to say Riemann curvature is the defining feature of a gravitational field. Maybe there's some confusion here in that Einstein referred to a "special" gravitational field that you can't transform away. I tend to call this a "real" gravitational field, or even just a gravitational field. That's what I was talking about here. See section 20 of Relativity: the special and general theory where you can read this:

    "We might also think that, regardless of the kind of gravitational field which may be present, we could always choose another reference-body such that no gravitational field exists with reference to it. This is by no means true for all gravitational fields, but only for those of quite special form. It is, for instance, impossible to choose a body of reference such that, as judged from it, the gravitational field of the earth (in its entirety) vanishes".

    MTW arguably contradicts Einstein on this point. Have a read of this essay on arXiv about some of the differences between what Peter M Brown calls "Einstein's interpretation" and "the modern interpretation".

    Quote Originally Posted by Physicist
    In a letter to Max von Laue, Einstein wrote the following in response to Laue's assertion that gravity exists only where spacetime curvature exists.
    Noted. We have said somewhere that things don't actually fall down because spacetime is curved. Once you appreciate that Einstein talked of a "special" gravitational field when I just say gravitational field I hope you can understand why there isn't an issue here.

    Quote Originally Posted by Physicist
    You’ve been interpreting all of this backwards. You think that since a uniform gravitational field is identical to a uniform gravitational field then since there’s no gravitational field just a transformation to a acceleration then there’s no “real” gravitational field whereas Einstein saw it the other way around. I.e. since they’re identical the accelerating frame in flat spacetime produces a “real” gravitational field merely by changing frames to an accelerating one.
    As above.

    Quote Originally Posted by Physicist
    quote=Recall how Einstein phrased this in his 1916 review paper, i.e. The Foundation of the General Theory of Relativity by Albert Einstein, Annalen der Physik, 49. It’s online at The Foundation of the General Theory ... (1 to 3, of 22, 1916)

    It will be seen from these reflexions that in pursuing the general theory of relativity we shall be led to a theory of gravitation, since we are able to "produce" a gravitational field merely by changing the system of co-ordinates. It will also be obvious that the principle of the constancy of the velocity of light in vacuo must be modified, since we easily recognize that the path of a ray of light with respect to K' must in general be curvilinear, if with respect to K light is propagated in a straight line with a definite constant velocity.

    So you see, Farsight? You’ve got it all backwards. It’s simply an historical fact that you’re (unsubstantiated) claim Then you've got a uniform gravitational field, which means you've got no Riemann curvature, so you haven't got a gravitational field. is quite wrong.
    With respect Physicist, just insert the word "special" or "real" and the problem goes away. There are no real gravitational fields where the gravitational potential can be depicted as an inverted cone. You need that Riemann curvature to get your plot off the flat and level. The Wikipedia article on gravitational potential is rather Newtonian, but the plot is adequate. Incidentally, re the cavity, there's no gradient in potential if you're right in the middle of the Earth. However when you move away from the middle there is. That means the gradient can't be absolutely uniform as you go from one situation to the other.
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  50. #150  
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    Quote Originally Posted by Farsight
    I'm not the first person to say Riemann curvature is the defining feature of a gravitational field.
    Duffield,

    This is what happens when you try to do physics by cherry picking quotes. The Riemann tensor is NOT "the defining feature of a gravitational field". Another tensor IS. Which one is it, John?
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    Quote Originally Posted by Farsight View Post
    MTW arguably contradicts Einstein on this point. Have a read of this essay on arXiv about some of the differences between what Peter M Brown calls "Einstein's interpretation" and "the modern interpretation".
    This is another example of you treating Einstein's work as the holy word. Einstein made mistakes. The way to determine whether or not Einstein was correct about something is not to reference unpublished crank papers that made it on to the archiv, the way is to actually learn the relevant details, in this case many are mathematical, and work through the supposed differences.
    Noted. We have said somewhere that things don't actually fall down because spacetime is curved. Once you appreciate that Einstein talked of a "special" gravitational field when I just say gravitational field I hope you can understand why there isn't an issue here.
    All you have to do to convince people is show a physics example that highlights the differences. Show the physics, not a word salad. Until you do this, it merely appears that you are mistaken in a way that does not stain your character.
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  52. #152  
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    Quote Originally Posted by PhysBang
    Einstein made mistakes. The way to determine whether or not Einstein was correct about something is not to reference unpublished crank papers that made it on to the archiv...
    Help me understand something. By this comment do you mean that you read it and know that it's all wrong and that's why the author is a crank? Or is it because the paper is unpublished and therefore it could have been written by a crank? I ask because I want to know whether you think its worth reading or not?
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  53. #153  
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    Peter M. Brown is a well known fringe person who fashions himself as a "physicist"
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  54. #154  
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    Quote Originally Posted by x0x View Post
    Duffield,

    This is what happens when you try to do physics by cherry picking quotes. The Riemann tensor is NOT "the defining feature of a gravitational field". Another tensor IS. Which one is it, John?
    Is it called the Einstein tensor?
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    Quote Originally Posted by Physicist View Post
    Help me understand something. By this comment do you mean that you read it and know that it's all wrong and that's why the author is a crank? Or is it because the paper is unpublished and therefore it could have been written by a crank? I ask because I want to know whether you think its worth reading or not?
    Sadly, I did read the paper, years ago. I discovered this because when I went to download the paper, I found I already had it in my "crank" folder.

    I merely point out that the paper is written, in my opinion, by a crank in the sense that the person is someone who is trying to make unjustified physics claims. I also believe that the paper is not worth reading in the context of this discussion: there are more direct routes to solving this dilemma that turning to some dubious secondary source. Ironically, Brown in this case provides a guide to what Farsight should be doing: trying to work through the relevant details of the physics.
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  56. #156  
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    Quote Originally Posted by Farsight View Post
    OK, but let's not get confused between flat as in horizontal and flat as in not curved.
    Who said anything about flat meaning horizontal?


    Quote Originally Posted by Farsight View Post
    Quote Originally Posted by KJW View Post
    Well then, what do think acceleration is?
    A change in speed or direction. When you're standing on the ground, it isn't happening to you.
    What does "change in speed or direction" actually mean? For example, consider an observer at rest on the ground and another observer in free-fall above the ground. Each observer can naively claim that the other observer is changing in speed and/or direction, but only one of them is correct. How do we decide which one is correct? It is essential to note that the decision methodology must be universally applicable. The methodology used by standard general relativity is that a straight line in a locally Minkowskian coordinate system is a universally straight line. Mathematically, this corresponds to the formula:



    When applied to the observers in gravity, it is the observer at rest on the ground that is accelerating, and the observer in free-fall above the ground is inertial (not accelerating).


    Quote Originally Posted by Farsight View Post
    Quote Originally Posted by KJW View Post
    How can something be like acceleration but not [be] acceleration?
    Because you feel a force like you would if you were accelerating, even though you aren't. Standing on the ground feels like accelerating through space. But you aren't accelerating through space.

    When you're accelerating through space you see a light beam curve and feel a force on your feet. When you're standing on the ground you see a light beam curve and feel a force on your feet. But not for the same reason.
    It's for exactly the same reason.


    Quote Originally Posted by Farsight View Post
    It's the "Riemann curvature" that distinguishes gravity from acceleration. You can't transform it away.
    But it is still the upward acceleration that keeps us on the ground.
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    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
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  57. #157  
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    btr - I noticed that you didn't comment on a large number of x0x's errors. Was that intentional or did you not notice them all?
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  58. #158  
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    Quote Originally Posted by Physicist View Post
    btr - I noticed that you didn't comment on a large number of x0x's mistakes. Was the intentional of did you not notice them all?
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    Bickering and snide posts are unhelpful.

    There is not just or two guilty.
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    btr - I don't understand your response. Are you trying to tell me that you don't want to talk about it? If so then that's perfectly fine with me.
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    Beer w/Straw: I don't understand your last response. What does
    There is not just or two guilty.
    What does that supposed to mean?
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  62. #162  
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    Quote Originally Posted by Physicist View Post
    btr - I don't understand your response. Are you trying to tell me that you don't want to talk about it? If so then that's perfectly fine with me.
    No, I presumed you were joking about my habit of responding, somewhat fruitlessly, to errors elsewhere. Apologies for being dense (it is getting late here).

    In this thread, my aim at the time was to try to move the discussion past the unimportant (in the larger context) details of the spherical cavity, and back to the argument I thought Jilan was going to make involving uniform fields (starting at post #74). I didn't really want to get too involved beyond that.
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    Okay. Thanks btr. I greatly appreciate you're civil attitude.
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  64. #164  
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    And likewise. I've enjoyed reading your posts.
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  65. #165  
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    Quote Originally Posted by KJW View Post
    Who said anything about flat meaning horizontal?
    Nobody. But in my experience people do confuse the two.

    Quote Originally Posted by KJW
    What does "change in speed or direction" actually mean?
    What it says. For example Compton scattering accelerates a photon. It doesn't slow it down, it changes its direction.

    Quote Originally Posted by KJW
    For example, consider an observer at rest on the ground and another observer in free-fall above the ground. Each observer can naively claim that the other observer is changing in speed and/or direction, but only one of them is correct. How do we decide which one is correct?
    By looking at the surroundings. Their context. We'd say the observer on the ground isn't moving or accelerating, whilst we'd say the falling observer is accelerating at 9.8 m/s/s. We know that there's no force acting upon him, and we know that the guy on the ground feels a force on his feet like an observer in a rocket, but we wouldn't say the falling observer wasn't moving.

    Quote Originally Posted by KJW
    It is essential to note that the decision methodology must be universally applicable. The methodology used by standard general relativity is that a straight line in a locally Minkowskian coordinate system is a universally straight line. Mathematically, this corresponds to the formula:



    When applied to the observers in gravity, it is the observer at rest on the ground that is accelerating, and the observer in free-fall above the ground is inertial (not accelerating).
    But he isn't actually accelerating. He feels a force like the guy in the rocket, but he isn't in the same situation.

    Quote Originally Posted by KJW
    It's for exactly the same reason.
    No it isn't exactly the same. The horizontal light beam bends downward because when the space you're in is inhomogeneous. Or when you're accelerating through homogeneous space.

    Quote Originally Posted by KJW
    But it is still the upward acceleration that keeps us on the ground.
    It's the upward force on your feet that does it. If it wasn't there you'd be falling.
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    Quote Originally Posted by btr
    And likewise. I've enjoyed reading your posts.
    Thanks. The powers that be recently helped me figure out how to put people in my ignore list. Now that I've done that I can't see any posts by the forum crackpots anymore. Have ever considered doing the same or do you like having sport with them?
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  67. #167  
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    Quote Originally Posted by Physicist View Post
    Thanks. The powers that be recently helped me figure out how to put people in my ignore list. Now that I've done that I can't see any posts by the forum crackpots anymore. Have ever considered doing the same or do you like having sport with them?
    I've not yet used the ignore function built into the forum software, but I do sometimes use the one built into my mouse (i.e. the scroll wheel). The main reasons are (a) I can't silence my inner optimist, who keeps telling me that such-and-such might turn a corner (despite ever-mounting evidence to the contrary); and (b) sometimes a bit of debunking is called for, mostly for (what I see as) good reasons like the benefit of people who come here looking for reasonable answers, but also partly for the less noble reason that I find it entertaining.
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  68. #168  
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    Quote Originally Posted by Physicist
    Now that I've done that I can't see any posts by the forum crackpots anymore.
    This must be the reason you don't see the errors in this post.
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  69. #169  
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    Quote Originally Posted by Farsight View Post
    Quote Originally Posted by KJW
    What does "change in speed or direction" actually mean?
    What it says.
    It is not as straightforward as that. For example, if I'm in an accelerated spaceship and there is an object at rest on the floor, then for me, the object is at rest and is not changing in speed and/or direction, yet for a nearby inertial observer, the object is accelerating and changing in speed and/or direction. It's the same object, but different observers disagree about its motion. Only one of the observers is right and there has to be a universally applicable method to determine this unequivocally.


    Quote Originally Posted by Farsight View Post
    By looking at the surroundings.
    Looking at the surroundings for what? This answer carries with it the presumption that (1) there is some object in the surroundings, and (2) the motion of the object in the surrounding is known. Given that we are trying to determine the motion of a given object, how do we determine the motion of an object in the surrounding?


    Quote Originally Posted by Farsight View Post
    We'd say the observer on the ground isn't moving or accelerating, whilst we'd say the falling observer is accelerating at 9.8 m/s/s.
    Why? By what criteria can we say this?


    Quote Originally Posted by Farsight View Post
    ... we'd say the falling observer is accelerating at 9.8 m/s/s. We know that there's no force acting upon him, and we know that the guy on the ground feels a force on his feet like an observer in a rocket, but we wouldn't say the falling observer wasn't moving... But he isn't actually accelerating. He feels a force like the guy in the rocket...
    Aren't you forgetting Newton's first and second laws of motion? If there is an upward force on the feet of a person on the ground (and no other force), then that person will be accelerated upward in accordance with


    Quote Originally Posted by Farsight View Post
    Quote Originally Posted by KJW
    It is essential to note that the decision methodology must be universally applicable. The methodology used by standard general relativity is that a straight line in a locally Minkowskian coordinate system is a universally straight line. Mathematically, this corresponds to the formula:



    When applied to the observers in gravity, it is the observer at rest on the ground that is accelerating, and the observer in free-fall above the ground is inertial (not accelerating).
    But he isn't actually accelerating. He feels a force like the guy in the rocket, but he isn't in the same situation.
    The above mathematics says you're wrong.


    Quote Originally Posted by Farsight View Post
    Quote Originally Posted by KJW
    It's for exactly the same reason.
    No it isn't exactly the same.
    To the first order change in the metric, which determines the acceleration, they are exactly the same.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
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    Quote Originally Posted by KJW View Post
    It is not as straightforward as that. For example, if I'm in an accelerated spaceship and there is an object at rest on the floor, then for me, the object is at rest and is not changing in speed and/or direction, yet for a nearby inertial observer, the object is accelerating and changing in speed and/or direction. It's the same object, but different observers disagree about its motion. Only one of the observers is right and there has to be a universally applicable method to determine this unequivocally.
    People talk about using the CMB reference frame to gauge your motion through the universe.

    Quote Originally Posted by KJW
    Looking at the surroundings for what? This answer carries with it the presumption that (1) there is some object in the surroundings, and (2) the motion of the object in the surrounding is known. Given that we are trying to determine the motion of a given object, how do we determine the motion of an object in the surrounding?
    See if it changes position relative to other objects.

    Quote Originally Posted by KJW
    Why? By what criteria can we say this?
    We think of the Earth as our frame of reference. It's spinning and it's orbiting the Sun, which is orbiting the galaxy which is in itself moving. But we discount all these motions.

    Quote Originally Posted by KJW
    Aren't you forgetting Newton's first and second laws of motion? If there is an upward force on the feet of a person on the ground (and no other force), then that person will be accelerated upward in accordance with
    Sorry, there's no force in the Newtonian sense acting on the falling observer. He falls faster and faster but he isn't acquiring energy like he would be if you accelerated him in a rocket.

    Quote Originally Posted by KJW
    The above mathematics says you're wrong.
    Focus on the physics. An observer in an accelerating rocket is not in the same situation as an observer standing on the ground. It might feel similar, and the same mathematics can apply. But those two situations are not the same. One observer is moving faster and faster through homogeneous space, the other is standing motionless on the surface of a planet in inhomogeneous space.

    Quote Originally Posted by KJW
    To the first order change in the metric, which determines the acceleration, they are exactly the same.
    And again, they aren't exactly the same. What's the metric? It's an abstract thing related to what you measure. And the two observers' measurements are the same. They're each in a box, they can't detect any tidal force. They can't tell whether they're accelerating through space or standing on the surface of the Earth. Until they lift the blinds and look out of the window.
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  71. #171  
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    Quote Originally Posted by Farsight View Post
    Quote Originally Posted by KJW View Post
    It is not as straightforward as that. For example, if I'm in an accelerated spaceship and there is an object at rest on the floor, then for me, the object is at rest and is not changing in speed and/or direction, yet for a nearby inertial observer, the object is accelerating and changing in speed and/or direction. It's the same object, but different observers disagree about its motion. Only one of the observers is right and there has to be a universally applicable method to determine this unequivocally.
    People talk about using the CMB reference frame to gauge your motion through the universe.


    Quote Originally Posted by KJW
    Looking at the surroundings for what? This answer carries with it the presumption that (1) there is some object in the surroundings, and (2) the motion of the object in the surrounding is known. Given that we are trying to determine the motion of a given object, how do we determine the motion of an object in the surrounding?
    See if it changes position relative to other objects.
    If you are going to determine the motion of objects by determining the motion relative to other objects, then you need at least one object for which the motion is known. You appear to have chosen the CMB for that object but you actually don't know the motion of the CMB, so it can't be used as the reference object. Also, the CMB is a distant object and a lot can happen with the intervening spacetime between the CMB and the object whose acceleration you are determining. If you are going to use a reference object, then it needs to be a local object.


    Quote Originally Posted by Farsight View Post
    Quote Originally Posted by KJW
    Why? By what criteria can we say this?
    We think of the Earth as our frame of reference. It's spinning and it's orbiting the Sun, which is orbiting the galaxy which is in itself moving. But we discount all these motions.
    It's not enough to have a frame of reference (coordinate system). One needs a connection over that frame of reference (coordinate system). The connection is what characterises the spacetime.


    Quote Originally Posted by Farsight View Post
    Quote Originally Posted by KJW
    Aren't you forgetting Newton's first and second laws of motion? If there is an upward force on the feet of a person on the ground (and no other force), then that person will be accelerated upward in accordance with
    Sorry, there's no force in the Newtonian sense acting on the falling observer. He falls faster and faster but he isn't acquiring energy like he would be if you accelerated him in a rocket.
    Because there is no force on the falling observer, he is not accelerating (Newton's first law of motion).


    Quote Originally Posted by Farsight View Post
    Quote Originally Posted by KJW
    The above mathematics says you're wrong.
    Focus on the physics.
    No, I'll focus on the mathematics because that is where the logic of physics resides.


    Quote Originally Posted by Farsight View Post
    They can't tell whether they're accelerating through space or standing on the surface of the Earth.
    That's the point. If they can't tell the difference between accelerating through space and standing on the surface of the Earth then they are the same thing.


    Quote Originally Posted by Farsight View Post
    Until they lift the blinds and look out of the window.
    How does that change things.


    Quote Originally Posted by Farsight View Post
    An observer in an accelerating rocket is not in the same situation as an observer standing on the ground. It might feel similar, and the same mathematics can apply. But those two situations are not the same. One observer is moving faster and faster through homogeneous space, the other is standing motionless on the surface of a planet in inhomogeneous space.

    Quote Originally Posted by KJW
    To the first order change in the metric, which determines the acceleration, they are exactly the same.
    And again, they aren't exactly the same. What's the metric? It's an abstract thing related to what you measure. And the two observers' measurements are the same. They're each in a box, they can't detect any tidal force. They can't tell whether they're accelerating through space or standing on the surface of the Earth. Until they lift the blinds and look out of the window.
    Your viewpoint not only violates the principle that indistinguishable things are really the same, but also the principle that distinguishable things are really different. This is because you ignore the metric (or connection) that determines acceleration, but include objects such as the earth or the CMB that doesn't determine acceleration.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
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    Quote Originally Posted by Farsight View Post
    People talk about using the CMB reference frame to gauge your motion through the universe.
    Can you provide a scientific citation for this? Please also show us a citation that does not use a system of coordinates in which space is homogeneous, since you do not think that is valid physics. Can you also explain how to use this reference frame to describe a black hole> Or any system, even a toy system?
    See if it changes position relative to other objects.
    We can always assign a coordinate system that keeps two objects stationary relative to one another. How do we do physics in such a system?
    [quote]We think of the Earth as our frame of reference. It's spinning and it's orbiting the Sun, which is orbiting the galaxy which is in itself moving. But we discount all these motions.
    Focus on the physics.
    Yes, please. I assume that this means you will put some toy examples up or some actual physics descriptions to back up your textual interpretation claims?
    An observer in an accelerating rocket is not in the same situation as an observer standing on the ground. It might feel similar, and the same mathematics can apply. But those two situations are not the same. One observer is moving faster and faster through homogeneous space, the other is standing motionless on the surface of a planet in inhomogeneous space.
    Please show us a toy example with (in)homogeneous space.
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    Quote Originally Posted by KJW View Post
    If you are going to determine the motion of objects by determining the motion relative to other objects, then you need at least one object for which the motion is known. You appear to have chosen the CMB for that object but you actually don't know the motion of the CMB, so it can't be used as the reference object. Also, the CMB is a distant object and a lot can happen with the intervening spacetime between the CMB and the object whose acceleration you are determining. If you are going to use a reference object, then it needs to be a local object.
    See the CMBR dipole anisotropy on Wike where you can read this:

    From the CMB data it is seen that our local group of galaxies (the galactic cluster that includes the Solar System's Milky Way Galaxy) appears to be moving at 369±0.9 km/s relative to the reference frame of the CMB (also called the CMB rest frame, or the frame of reference in which there is no motion through the CMB) in the direction of galactic longitude l = 263.99±0.14°, b = 48.26±0.03°...

    Quote Originally Posted by KJW
    Because there is no force on the falling observer, he is not accelerating (Newton's first law of motion).
    He's accelerating at 9.8 m/s/s in a downward direction. And then he smashes into the ground, decelerating rather rapidly. The guy standing on the ground feels a force like the guy in the SR rocket, but he isn't actually accelerating. It's like he's accelerating, but he isn't actually accelerating.

    Quote Originally Posted by KJW
    That's the point. If they can't tell the difference between accelerating through space and standing on the surface of the Earth then they are the same thing.
    No they aren't the same thing. They feel like the same thing, but they aren't the same thing.

    Quote Originally Posted by KJW
    How does that change things.
    When they look out of the window they can see whether they're on the ground or accelerating through space. They can tell the difference between the two situations. Keep on accelerating through space, and you can use the CMB to gauge your changing speed.

    Quote Originally Posted by KJW
    Your viewpoint not only violates the principle that indistinguishable things are really the same, but also the principle that distinguishable things are really different.
    I reject such principles. I have said that sometimes something changes, but you change too, so you don't know it changes. And that sometimes something doesn't change but you do, so you think it has changed when it hasn't. You cannot merely take your measurements at face value. When you can't distinguish between two situations, you take more measurements. You open that window.
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    Quote Originally Posted by Farsight View Post
    See the CMBR dipole anisotropy on Wike where you can read this:

    From the CMB data it is seen that our local group of galaxies (the galactic cluster that includes the Solar System's Milky Way Galaxy) appears to be moving at 369±0.9 km/s relative to the reference frame of the CMB (also called the CMB rest frame, or the frame of reference in which there is no motion through the CMB) in the direction of galactic longitude l = 263.99±0.14°, b = 48.26±0.03°...
    Given that this determination relies on an idea of a flat spacetime with gravitational effects, this citation is inconsistent with your theory and thus cannot be considered an acceptable citation to support your claims.
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    Quote Originally Posted by PhysBang View Post
    Can you provide a scientific citation for this?
    I'm not going to. The CMB dipole anisotropy is not in doubt.

    Quote Originally Posted by PhysBang View Post
    Please also show us a citation that does not use a system of coordinates in which space is homogeneous, since you do not think that is valid physics.
    Huh? See Inhomogeneous Vacuum: An Alternative Interpretation of Curved Spacetime . We've previously said this is somewhat limited and deals only with static situations.

    Quote Originally Posted by PhysBang View Post
    Can you also explain how to use this reference frame to describe a black hole> Or any system, even a toy system?
    Quoi? See post 17 on the black hole thread. The typical plot of gravitational potential relates to the "coordinate" speed of light, only the plot ends at the event horizon where the coordinate speed of light goes to zero and can't go any lower.

    Quote Originally Posted by PhysBang
    We can always assign a coordinate system that keeps two objects stationary relative to one another. How do we do physics in such a system?
    What are you saying? When two objects collide they collide. You can't adopt a coordinate system that pretends that they don't. It's similar for when a clock stops. It's stopped, and that's it.

    Quote Originally Posted by PhysBang
    Yes, please. I assume that this means you will put some toy examples up or some actual physics descriptions to back up your textual interpretation claims?
    No. I've given enough explanation.

    Quote Originally Posted by PhysBang
    Please show us a toy example with (in)homogeneous space.
    See above and look on arXiv.
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    Quote Originally Posted by Farsight View Post
    Quoi? See post 17 on the black hole thread. The typical plot of gravitational potential relates to the "coordinate" speed of light, only the plot ends at the event horizon where the coordinate speed of light goes to zero and can't go any lower.
    That response, like all your responses to requests for details, is a mistake that does not stain your character.

    You have never, ever, tried using your theory to describe a physics scenario. Until you do, you have no details. Only mistakes that don't stain your character.
    What are you saying? When two objects collide they collide. You can't adopt a coordinate system that pretends that they don't.
    Indeed you can: you can create a coordinate system that assigns the same spatial coordinate system to the objects at all times. You could even adjust the time coordinate so that clocks stop before any collision.
    It's similar for when a clock stops. It's stopped, and that's it.
    You, however, confuse time coordinates with clocks.

    No. I've given enough explanation.
    To satisfy your own dogma, you certainly have. Yet you have never given us physics. That is a shame, but not ours.
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  77. #177  
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    Quote Originally Posted by Farsight View Post
    You cannot merely take your measurements at face value. When you can't distinguish between two situations, you take more measurements.
    And if by taking more measurements, you still can't distinguish between two situations, what then? When do you accept that they are really the same?

    However, you also fail to distinguish between situations that are really different. For example, the person standing on the surface of the earth is distinguishable from an observer who is not accelerating, yet you claim they are the same (that the person standing on the surface of the earth is not accelerating).
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
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    Quote Originally Posted by KJW View Post
    And if by taking more measurements, you still can't distinguish between two situations, what then? When do you accept that they are really the same?
    When your logic works and you feel you understand both situations absolutely. What you don't do is ignore the patent differences between them on the say-so of some blindfolded guy in a box.

    Quote Originally Posted by KJW
    However, you also fail to distinguish between situations that are really different. For example, the person standing on the surface of the earth is distinguishable from an observer who is not accelerating, yet you claim they are the same (that the person standing on the surface of the earth is not accelerating).
    I didn't claim they were the same. The man standing on the Earth feels a force as if he's accelerating through gravity-free homogeneous space, but he isn't accelerating, and he isn't in homogeneous space. A man in a box floating in space above the Earth isn't accelerating, and he doesn't feel any force. If he starts falling towards Earth he's accelerating at say 9.8 m/s/s but he still doesn't feel any force. Until he hits the ground. Blam. Then he feels a marked force whilst he's accelerating. Deceleration being acceleration in the wider sense.
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    Quote Originally Posted by Farsight View Post
    What you don't do is ignore the patent differences between them on the say-so of some blindfolded guy in a box.
    The reason for choosing the so-called "blindfolded guy in a box" is that it is the immediate environment of the observer that determines his acceleration. After all, it is the immediate environment that has any effect on the observer. The distant CMB doesn't have any effect on the observer, so its motion is irrelevant to the acceleration of the observer.


    Quote Originally Posted by Farsight View Post
    I didn't claim they were the same.
    You claim they have the same acceleration. But their differences ensure they do not have the same acceleration.


    Quote Originally Posted by Farsight View Post
    The man standing on the Earth feels a force as if he's accelerating ... but he isn't accelerating

    ...

    If he starts falling towards Earth he's accelerating at say 9.8 m/s/s but he still doesn't feel any force
    Both these conflict with Newton's first and second laws of motion.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
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  80. #180  
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    Quote Originally Posted by Farsight View Post
    Quote Originally Posted by KJW View Post
    If you are going to determine the motion of objects by determining the motion relative to other objects, then you need at least one object for which the motion is known. You appear to have chosen the CMB for that object but you actually don't know the motion of the CMB, so it can't be used as the reference object. Also, the CMB is a distant object and a lot can happen with the intervening spacetime between the CMB and the object whose acceleration you are determining. If you are going to use a reference object, then it needs to be a local object.
    See the CMBR dipole anisotropy on Wike where you can read this:

    From the CMB data it is seen that our local group of galaxies (the galactic cluster that includes the Solar System's Milky Way Galaxy) appears to be moving at 369±0.9 km/s relative to the reference frame of the CMB (also called the CMB rest frame, or the frame of reference in which there is no motion through the CMB) in the direction of galactic longitude l = 263.99±0.14°, b = 48.26±0.03°...
    Your reply doesn't address my statement. You gave a relative velocity whereas I'm talking about an absolute acceleration.


    Quote Originally Posted by Farsight View Post
    Quote Originally Posted by KJW
    How does that change things.
    When they look out of the window they can see whether they're on the ground or accelerating through space. They can tell the difference between the two situations. Keep on accelerating through space, and you can use the CMB to gauge your changing speed.
    This conflicts with the local connection that actually determines the absolute acceleration.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
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    Quote Originally Posted by KJW View Post
    The reason for choosing the so-called "blindfolded guy in a box" is that it is the immediate environment of the observer that determines his acceleration.
    I reject that. You've been talking about two situations. Let's say one is where an electron falls down inside the box, because it's in a gravitational field. The other where the electron is stationary and the box rises up because it's propelled by rockets. From inside the box the two situations look similar, but they just aren't the same. Einstein spoke of a "special" gravitational field that can't be transformed away. We'd call this a real gravitational field.

    Quote Originally Posted by KJW
    After all, it is the immediate environment that has any effect on the observer. The distant CMB doesn't have any effect on the observer, so its motion is irrelevant to the acceleration of the observer.
    Agreed. The observer falls down because the immediate environment has an effect on him. He doesn't fall down because of some magical mysterious action at a distance. He falls down because the space he's in is inhomogeneous, and he's a more complex version of the electron.

    Quote Originally Posted by KJW
    You claim they have the same acceleration. But their differences ensure they do not have the same acceleration.
    I can't say it any other way: the observer standing on the surface of the Earth just isn't accelerating. He might feel like he's accelerating. But he isn't. In similar vein the observer in free fall might feel like he isn't accelerating, but he is.

    Quote Originally Posted by KJW
    Both these conflict with Newton's first and second laws of motion.
    The man standing on the surface of the Earth isn't moving. The falling man is moving faster and faster and faster. There's no conflict.

    Quote Originally Posted by KJW
    Your reply doesn't address my statement. You gave a relative velocity whereas I'm talking about an absolute acceleration.

    This conflicts with the local connection that actually determines the absolute acceleration.
    That doesn't seem to square with If you are going to use a reference object, then it needs to be a local object. Please restate the issue.
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    Farsight, you have the first point backwards I think. It's only a special type of gravitational field that can be transformed away. Remember the off-centre spherical cave? That is of the special kind. Most others cannot due to the inverse square law.
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    Quote Originally Posted by Jilan View Post
    Farsight, you have the first point backwards I think. It's only a special type of gravitational field that can be transformed away. Remember the off-centre spherical cave? That is of the special kind. Most others cannot due to the inverse square law.
    It's not due only to the inverse square law. Consider an infinitely long rod. Near the rod the field varies as 1/r instead of 1/r^2. Inside a sphere with uniform mass density the field varies directly proportional to r. To a first approximation the gravitational field of the galaxy can be modeled as a disk and that too will create spacetime curvature. And actually near the plane of the disk the field is uniform. Misner, Thorne and Wheeler in their famous text Gravitation use this model to describe a star oscillating through the plane of the galaxy.

    So there are a lot of configurations of matter in nature which generate spacetime curvature. Even some that don't, such as cosmic strings and vacuum domain walls.
    Jilan likes this.
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    And going back to the straightforward situation, look at the plot of Newtonian gravitational potential for the Earth:


    CCBYSA image by AllenMcC, see Gravitational potential - Wikipedia, the free encyclopedia

    Right in the very centre of the plot, at the bottom, you're in a spherical cave in the very centre of the Earth. There's no gradient in potential there, so you don't fall down. A third of the way up the plot you're in your off centre spherical cave where the gradient in gravitational potential is uniform. But at location between the two the plot is going from 60 degrees to 0 degrees, so the gradient isn't uniform.

    NB: Einstein said the special gravitational field can't be transformed away. I've referred to that as a real gravitational field. In a nutshell you need the "Riemann curvature" that is associated with a real gravitational field to get your plot off the flat and level. If your plot can't curve upwards, it stays flat as a pancake. See Kevin Brown's mathpages re the gravitational field for an infinite plane. It isn't a realistic situation.
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  85. #185  
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    Quote Originally Posted by Farsight
    A third of the way up the plot you're in your off centre spherical cave where the gradient in gravitational potential is uniform.
    Nope, wrong again, Duffield. Don't let science get in the way of your spamming your fringe ideas all over the internet.
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  86. #186  
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    Quote Originally Posted by Farsight View Post
    Einstein spoke of a "special" gravitational field that can't be transformed away. We'd call this a real gravitational field.
    At this point, I'd like to point out that the discussion has been about why objects fall in a gravitational field, not about the definition of a "gravitational field". The spacetime curvature field "that can't be transformed away", and distinguishes the gravitation external to an energy-momentum distribution from an accelerated frame of reference in flat spacetime, is associated with the second-order change in the metric, whereas the acceleration itself is associated with the first-order change in the metric. Because the first- and second-order changes are independent, each can occur without the other. In other words, the spacetime curvature field "that can't be transformed away" is irrelevant to the reason objects fall in a gravitational field, and objects fall because they are in inertial motion relative to an accelerated frame of reference. Spacetime curvature however does cause the surface of the earth to be accelerating outwards without expansion of this surface.


    Quote Originally Posted by Farsight View Post
    I can't say it any other way: the observer standing on the surface of the Earth just isn't accelerating. He might feel like he's accelerating. But he isn't. In similar vein the observer in free fall might feel like he isn't accelerating, but he is.
    Regardless of how you say it, general relativity says you are wrong. This is the equivalence principle.


    Quote Originally Posted by Farsight View Post
    Quote Originally Posted by KJW
    Both these conflict with Newton's first and second laws of motion.
    The man standing on the surface of the Earth isn't moving. The falling man is moving faster and faster and faster. There's no conflict.
    The man standing on the surface of the Earth experiences and upward force on his feet - he is accelerated upward - conflict with Newton's first and second laws of motion.

    The falling man is experiencing no force upon him - no acceleration - conflict with Newton's first and second laws of motion.


    Quote Originally Posted by Farsight View Post
    Quote Originally Posted by KJW
    Your reply doesn't address my statement. You gave a relative velocity whereas I'm talking about an absolute acceleration.

    This conflicts with the local connection that actually determines the absolute acceleration.
    That doesn't seem to square with If you are going to use a reference object, then it needs to be a local object. Please restate the issue.
    Note that unlike velocity, acceleration is absolute in that it can be determined without being relative to any other object. But acceleration cannot be determined from the trajectory alone. It requires the connection local to the trajectory. This is the in the formula:



    It's worthwhile noting that the connection is required for the acceleration to be covariant.

    Thus, your claim that acceleration of an object can be determined by its motion relative to some other object is misguided.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
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    Quote Originally Posted by KJW View Post
    At this point, I'd like to point out that the discussion has been about why objects fall in a gravitational field, not about the definition of a "gravitational field". The spacetime curvature field "that can't be transformed away", and distinguishes the gravitation external to an energy-momentum distribution from an accelerated frame of reference in flat spacetime, is associated with the second-order change in the metric, whereas the acceleration itself is associated with the first-order change in the metric.
    That's fairly reasonable. In other words: the force of gravity at some location relates to the first derivative of potential, whereas the tidal force which is associated with the "Riemann curvature" relates to the second derivative of potential. But remember that spacetime is an abstract static mathematical model. There is no motion in it. The field is in space rather than spacetime, and you plot your curved spacetime using light clocks in say an equatorial slice of space through and around the Earth.

    Quote Originally Posted by KJW
    Because the first- and second-order changes are independent, each can occur without the other.
    Not for a real gravitational field. You need the Riemann curvature to get that plot off the flat and level.

    Quote Originally Posted by KJW
    In other words, the spacetime curvature field "that can't be transformed away" is irrelevant to the reason objects fall in a gravitational field
    I agree that when your pencil falls down, it isn't because of spacetime curvature.

    Quote Originally Posted by KJW
    objects fall because they are in inertial motion relative to an accelerated frame of reference.
    No! Objects fall because the space they're in is inhomogeneous. A frame of reference is an abstract thing. It isn't something that's actually there in the room you're in. Your pencil is there, and air is there, and space is there. Take one electron from that pencil and remember the wave nature of matter to appreciate why objects fall down.

    Quote Originally Posted by KJW
    Spacetime curvature however does cause the surface of the earth to be accelerating outwards without expansion of this surface.
    If your pencil doesn't fall down because of spacetime curvature, then spacetime curvature isn't accelerating it when it's sitting on your desk.

    Quote Originally Posted by KJW
    Regardless of how you say it, general relativity says you are wrong. This is the equivalence principle.
    It applies to an infinitesimal region only. So your pencil in your room isn't exactly equivalent to a pencil in an accelerating rocket.

    Quote Originally Posted by KJW
    The man standing on the surface of the Earth experiences and upward force on his feet - he is accelerated upward - conflict with Newton's first and second laws of motion.
    He experiences an upward force on his feet, but he isn't accelerated upward.

    Quote Originally Posted by KJW
    The man is experiencing no force upon him - no acceleration - conflict with Newton's first and second laws of motion.
    He experiences no force, but he's accelerating at 9.8m/s/s.

    Quote Originally Posted by KJW
    Note that unlike velocity, acceleration is absolute in that it can be determined without being relative to any other object. But acceleration cannot be determined from the trajectory alone. It requires the connection local to the trajectory. This is the in the formula:



    It's worthwhile noting that the connection is required for the acceleration to be covariant.

    Thus, your claim that acceleration of an object can be determined by its motion relative to some other object is misguided.
    I'm sorry KJW, but when you're standing still in your room, you just aren't accelerating.
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    Quote Originally Posted by Duffield
    I'm sorry KJW, but when you're standing still in your room, you just aren't accelerating
    Repeating the same falsity over and over doesn't make it true, John.

    1. Duffield is standing on the ground holding an accelerometer in his hand. What does the accelerometer show? Zero? Hint: no.
    2. Drop Duffield from a building (preferably, very high). What does the accelerometer show? Hint : zero.
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  89. #189  
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    Quote Originally Posted by Farsight View Post
    Quote Originally Posted by KJW
    Regardless of how you say it, general relativity says you are wrong. This is the equivalence principle.
    It applies to an infinitesimal region only. So your pencil in your room isn't exactly equivalent to a pencil in an accelerating rocket.
    What is the tidal force experienced by the pencil on earth? Do you think this makes a significant difference to the behaviour of the pencil compared to being in an accelerating rocket?


    Quote Originally Posted by Farsight View Post
    Quote Originally Posted by KJW
    The man standing on the surface of the Earth experiences and upward force on his feet - he is accelerated upward - conflict with Newton's first and second laws of motion.
    He experiences an upward force on his feet, but he isn't accelerated upward.

    Quote Originally Posted by KJW
    The man is experiencing no force upon him - no acceleration - conflict with Newton's first and second laws of motion.
    He experiences no force, but he's accelerating at 9.8m/s/s.
    In conflict with the Newton's first and second laws of motion.


    Quote Originally Posted by Farsight View Post
    Quote Originally Posted by KJW
    Note that unlike velocity, acceleration is absolute in that it can be determined without being relative to any other object. But acceleration cannot be determined from the trajectory alone. It requires the connection local to the trajectory. This is the in the formula:



    It's worthwhile noting that the connection is required for the acceleration to be covariant.

    Thus, your claim that acceleration of an object can be determined by its motion relative to some other object is misguided.
    I'm sorry KJW, but when you're standing still in your room, you just aren't accelerating.
    I'm sorry Farsight, but you need to come up with more than just the assertion that "when you're standing still in your room, you just aren't accelerating".


    Quote Originally Posted by Farsight View Post
    Quote Originally Posted by KJW
    Spacetime curvature however does cause the surface of the earth to be accelerating outwards without expansion of this surface.
    If your pencil doesn't fall down because of spacetime curvature, then spacetime curvature isn't accelerating it when it's sitting on your desk.
    The accumulation of second-order changes over distance is a first-order change. By means of geodesic deviation, curvature does produce acceleration.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
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    Quote Originally Posted by KJW View Post
    What is the tidal force experienced by the pencil on earth? Do you think this makes a significant difference to the behaviour of the pencil compared to being in an accelerating rocket?
    The tidal force is very slight. It doesn't make a significant difference to the behaviour of the pencil. But nevertheless the two situations are not identical. They look similar because in the rocket the pencil is motionless and the room accelerates upwards at 9.8m/s˛. In the other the room is motionless and the pencil accelerates downwards at 9.8m/s˛.

    Quote Originally Posted by KJW View Post
    In conflict with the Newton's first and second laws of motion.
    There's no conflict. Gravity isn't a force in the Newtonian sense. There's no force acting on the falling pencil, the force is exerted when it's lying on the ground.

    Quote Originally Posted by KJW
    I'm sorry Farsight, but you need to come up with more than just the assertion that "when you're standing still in your room, you just aren't accelerating".
    See Wikpedia or any other place where you can find a definition. Acceleration is a change in your state of motion. When you're standing in your room, your state of motion isn't changing, so you aren't accelerating. I can't put it any simpler.

    Quote Originally Posted by KJW
    The accumulation of second-order changes over distance is a first-order change. By means of geodesic deviation, curvature does produce acceleration.
    That sounds similar to what I said about needing the Riemann curvature to get your plot off the flat and level.
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  91. #191  
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    Quote Originally Posted by Farsight View Post
    Quote Originally Posted by KJW View Post
    What is the tidal force experienced by the pencil on earth? Do you think this makes a significant difference to the behaviour of the pencil compared to being in an accelerating rocket?
    The tidal force is very slight. It doesn't make a significant difference to the behaviour of the pencil. But nevertheless the two situations are not identical. They look similar because in the rocket the pencil is motionless and the room accelerates upwards at 9.8m/s˛. In the other the room is motionless and the pencil accelerates downwards at 9.8m/s˛.
    It's true that they are not exactly identical, but they are so similar that the difference in acceleration that you claim cannot be justified by the whether or not there is a tidal force.


    Quote Originally Posted by Farsight View Post
    Quote Originally Posted by KJW View Post
    In conflict with the Newton's first and second laws of motion.
    There's no conflict. Gravity isn't a force in the Newtonian sense. There's no force acting on the falling pencil, the force is exerted when it's lying on the ground.
    If gravity were a force in the Newtonian sense, there would be no conflict with Newton's first and second laws of motion. It is the absence of the gravitational force that makes your belief about the acceleration conflict with Newton's first and second laws of motion. I can't understand why you refuse to see that an upward force on your feet from the ground is accelerating you upward as a result of Newton's first and second laws of motion.


    Quote Originally Posted by Farsight View Post
    Quote Originally Posted by KJW
    I'm sorry Farsight, but you need to come up with more than just the assertion that "when you're standing still in your room, you just aren't accelerating".
    See Wikpedia or any other place where you can find a definition. Acceleration is a change in your state of motion. When you're standing in your room, your state of motion isn't changing, so you aren't accelerating. I can't put it any simpler.
    I presented you with the general relativistic formula for acceleration. This formula accounts for different frames of reference. For example, it may seem as if the person standing on the ground isn't changing in motion, but if viewed by a falling observer, the person standing on the ground does change in motion. The general relativistic formula for acceleration resolves the disagreement and says that the viewpoint of the falling observer is the more "correct" viewpoint and the person standing on the ground is accelerating.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
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    Quote Originally Posted by KJW View Post
    It's true that they are not exactly identical, but they are so similar that the difference in acceleration that you claim cannot be justified by the whether or not there is a tidal force.
    Let me restate: the tidal force has got nothing to do with why the pencil falls down. And the difference between the two situations is the underlying cause. In one situation the pencil falls down because the space its in is inhomogeneous. In the other situation the room moves ever-faster through homogeneous space. Again see Einstein's 1920 Leyden Address where he described a gravitational field as inhomogeneous space:

    "This space-time variability of the reciprocal relations of the standards of space and time, or, perhaps, the recognition of the fact that "empty space" in its physical relation is neither homogeneous nor isotropic, compelling us to describe its state by ten functions (the gravitation potentials gmn), has, I think, finally disposed of the view that space is physically empty."

    Quote Originally Posted by KJW
    If gravity were a force in the Newtonian sense, there would be no conflict with Newton's first and second laws of motion. It is the absence of the gravitational force that makes your belief about the acceleration conflict with Newton's first and second laws of motion. I can't understand why you refuse to see that an upward force on your feet from the ground is accelerating you upward as a result of Newton's first and second laws of motion.
    It's a distinction between force and acceleration. There's an upward force on your feet, but you aren't moving, you aren't actually accelerating. To claim that you are accelerating when you're motionless on the ground and not accelerating when you're falling to Earth is contrary to the definition of what acceleration is.

    Quote Originally Posted by KJW
    I presented you with the general relativistic formula for acceleration. This formula accounts for different frames of reference. For example, it may seem as if the person standing on the ground isn't changing in motion, but if viewed by a falling observer, the person standing on the ground does change in motion. The general relativistic formula for acceleration resolves the disagreement and says that the viewpoint of the falling observer is the more "correct" viewpoint and the person standing on the ground is accelerating.
    I reject that. One of the tenets of general relativity is that "all reference frames are equally valid". And do note that Einstein gave the equations of motion for planets like Mercury and for the falling observer. He didn't give the equations of motion for the guy standing on the ground.
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    Quote Originally Posted by Farsight View Post
    Let me restate: the tidal force has got nothing to do with why the pencil falls down. And the difference between the two situations is the underlying cause. In one situation the pencil falls down because the space its in is inhomogeneous. In the other situation the room moves ever-faster through homogeneous space. Again see Einstein's 1920 Leyden Address where he described a gravitational field as inhomogeneous space:

    "This space-time variability of the reciprocal relations of the standards of space and time, or, perhaps, the recognition of the fact that "empty space" in its physical relation is neither homogeneous nor isotropic, compelling us to describe its state by ten functions (the gravitation potentials gmn), has, I think, finally disposed of the view that space is physically empty."
    Again you cherry-pick this quotation to serve as your answer, but you refuse to demonstrate how "inhomogeneous space" can be used to actually describe the motion of the pencil.

    And, unfortunately, that is a personal failure on your part that does stain your character. You are arguing in bad faith.
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    Quote Originally Posted by Farsight View Post
    Again see Einstein's 1920 Leyden Address where he described a gravitational field as inhomogeneous space:

    "This space-time variability of the reciprocal relations of the standards of space and time, or, perhaps, the recognition of the fact that "empty space" in its physical relation is neither homogeneous nor isotropic, compelling us to describe its state by ten functions (the gravitation potentials gmn), has, I think, finally disposed of the view that space is physically empty."
    Book thumping and quote mining.

    I think that Albert Einstein was being imprecise here, because in general relativity, it's *space-time* that's curved, and not just 3-space. Einstein was attempting to explain general relativity in nontechnical language, thus the imprecision. I also don't see why that statement qualifies as Holy Writ, which is what Farsight has been treating it as.
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  95. #195  
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    Quote Originally Posted by Farsight View Post
    One of the tenets of general relativity is that "all reference frames are equally valid".
    Yes, which is why the person standing on the ground and everyone else agrees with the falling person in saying that the person standing on the ground is accelerating upward.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
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    Quote Originally Posted by lpetrich View Post
    I think that Albert Einstein was being imprecise here, because in general relativity, it's *space-time* that's curved, and not just 3-space. Einstein was attempting to explain general relativity in nontechnical language, thus the imprecision. I also don't see why that statement qualifies as Holy Writ, which is what Farsight has been treating it as.
    He wasn't being imprecise. When Einstein said space-time he meant spacetime, when he said space he meant space. The imprecision comes from people who confuse the two. And you've totally missed the important thing Einstein said, which is that inhomogeneous space is what a gravitational field. See Inhomogeneous Vacuum: An Alternative Interpretation of Curved Spacetime for a modern treatment that relates to static spherically symmetric fields. Then see this Baez article where you can read this:

    "Similarly, in general relativity gravity is not really a 'force', but just a manifestation of the curvature of spacetime. Note: not the curvature of space, but of spacetime. The distinction is crucial."

    Space isn't curved where a gravitational field is. Instead it's inhomogeneous. We model it as curved spacetime. Read the OP.
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    Quote Originally Posted by KJW View Post
    Yes, which is why the person standing on the ground and everyone else agrees with the falling person in saying that the person standing on the ground is accelerating upward.
    You said the viewpoint of the falling observer is the more "correct" viewpoint. It isn't. One of the tenets of general relativity is that "all frames are equal", so the falling observer's frame is not more correct. There's no force on him but his motion is changing because of the wave nature of matter and inhomogeneous space. There is a force on the observer on the ground but his motion isn't changing. It feels like acceleration or "changing motion" in homogeneous space, but it isn't. Instead it's no-motion in inhomogeneous space. See above. Einstein described a gravitational field as a place where space is neither homogeneous nor isotropic.
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    Quote Originally Posted by PhysBang View Post
    Again you cherry-pick this quotation to serve as your answer, but you refuse to demonstrate how "inhomogeneous space" can be used to actually describe the motion of the pencil.
    I described it in the OP. A pencil is made up of electrons and other particles. We make electrons out of light in pair production, and we can diffract electrons. The wave nature of matter is not in doubt. Nor is electron spin or magnetic moment or the Einstein-de Haas effect. So we simply treat the electron as light going round a closed path, then simplify it further to a square, and the horizontals bend downward. The electron falls down, the pencil falls down, and the deflection of matter is half the deflection of light. Simple.



    Quote Originally Posted by PhysBang
    And, unfortunately, that is a personal failure on your part that does stain your character. You are arguing in bad faith.
    I'm not. I'm referring to Einstein. He said what he said, including a curvature of light can only occur when the speed of light varies with position. Au contraire, it's bad faith to dismiss Einstein when discussing relativity.
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    Quote Originally Posted by Farsight View Post
    I described it in the OP.
    That answer is, again, a dodge of the relevant question that is a sign of arguing in bad faith.

    You need to actually describe the fall of a pencil using "inhomogeneous space" using actual details, i.e., precise assignation of numbers. Nobody in the world has yet done so, yet you are claiming that this is the secret of GR as Einstein meant but never wrote down.
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  100. #200  
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    Quote Originally Posted by Farsight View Post
    Quote Originally Posted by KJW View Post
    Yes, which is why the person standing on the ground and everyone else agrees with the falling person in saying that the person standing on the ground is accelerating upward.
    You said the viewpoint of the falling observer is the more "correct" viewpoint. It isn't. One of the tenets of general relativity is that "all frames are equal", so the falling observer's frame is not more correct.
    The definition of acceleration that you are using does not come from general relativity. Therefore, you are not entitled to say that the falling observer's frame is not more correct. The definition of acceleration that you are using assumes preferred frames of reference, and applying the equivalence principle indicates that the falling observer's frame is the more correct frame from which to determine acceleration. If you were to apply the definition of acceleration that comes from general relativity, then it wouldn't matter which frame to consider because they all say that the person standing on the ground is accelerating.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
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