
You are now trying to cover up your mistake, you have claimed a cave , offcenter, under the Earth. Now you are changing your story. Not only that you are wrong, you are dishonest as well.Originally Posted by Jilan
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.
The shape of the cave is irrelevant. You still have it wrong.Originally Posted by Jilan
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.Originally Posted by Farsight
In fact Einstein explicitly stated that the nonvanishing of the Riemann tensor is not the criteria for the nonvanishing of a gravitational field. Its the nonvanishing 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
That this is true and current can also be found in the American Journal of Physics in the article... what characterizes the existence of a gravitational field from the empirical standpoint is the nonvanishing 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.
Cosmic strings: Gravitation without local curvature by T.M. Helliwell, Am. J. Phys., 55(5), May (1987)
Cosmic strings: Gravitation without local curvature
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 fieldCosmic 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.
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.
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
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?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).
Last edited by Physicist; 06292014 at 12:41 PM.
How on Earth could you ever conclude such a thing. Jilan wroteOriginally Posted by x0x
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.the cave is not at the center, it's just a little way underground.
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; 06302014 at 03:57 AM.
Moderator note: ALL members will desist from using adhominems. Personal attacks are not allowed here  attack the words a user posts, but DO NOT attack the user who posts them.
If the cave is off center, then the gravitational field is NOT uniform.Originally Posted by Physicists
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.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.
This is false. The reason is that the force exerted on a test probe is nonnull. Since it follows that is not constant.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 x0x; 06292014 at 02:36 PM.
The measurable entity is force. Force , as I have already shown is NOT constant and is NOT null.Originally Posted by Jilan
The gravitational potential is VARIABLE, as such, contrary to your oft repeated claims, the field is NOT "uniform".
The potential is . Please prove that "it's (sic!) gradient doesn't change". Use math.The potential changes but it's gradient (the field strength) doesn't.
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.
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.Originally Posted by Jilan
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.
On this cave business...
If you have a ball of uniform density and radius , and create a ballshaped 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 nonuniform density, the result does not necessarily hold (even if the density is spherically symmetric, aside from the cavity).
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.
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.Originally Posted by btr
Try with a nonspherical cavity, one that has ceiling and floor.Originally Posted by Jilan
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?
Jilan,
My correction came to the error in your post 83, not 74:
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 boxshaped cave. I challenge you.Originally Posted by Jilan
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?Why would I want to do that, we were never discussing box shaped caves, lol!
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.
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:Originally Posted by Jilan
How is your calculus knowledge? eh?Originally Posted by Jilan
Initially, I thought that your support for Farsight crackpoterry was accidental, now I don't think that anymore.
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!
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 nonzero 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.
Yeah, I have to "side" with a spherical void having a top and bottom, even if chosen arbitrarily.
Apologies to Farsight, for thread hijacking, it wasn't intentional. Can we carry on?
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?
The obvious flaw is that there isn't any way presented to do physics with the "interpretation" presented.
Really? You too? =>As for Jilan's statement that "the potential changes but the gradient doesn't," I don't see a problem with that.
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. .x0x, I think you may have missed some important context, such as post #74
Last edited by x0x; 06302014 at 03:00 AM.
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).
That is incorrect.
And in all three times you've been incorrect. Repeating a mistake doesn't make it right you know.
[/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
Read problem 44. The book is called Physics by Example: 200 Problems and Solutions by W. Rees. You can download the entire text at bookosz1Physics 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
...
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.
Actually, it is correct. Only spherical cavities produce a uniform field when positioned offcenter. Generic shape cavities don't.Originally Posted by Physicist
One more thing: nonnull force means nonnull gradient for . Nonnull 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).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.
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 boxlike cavity.Read problem 44. The book is called Physics by Example: 200 Problems and Solutions by W. Rees. You can download the entire text at bookosz1
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.
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.
Wrong again.Originally Posted by x0x
You need to pay much closer attention to the conversation because we’re talking about that kind of cavity, nothing more and nothing less.Originally Posted by x0x
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.Originally Posted by x0x
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.Originally Posted by x0x
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.
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.Originally Posted by Physicist
Not if you are sloppy.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.
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.Originally Posted by PhysBang
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.
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?
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:
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.
I guess it fine to point out the error of their ways with recourse to the evidence without calling them a liar.
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.
A change in speed or direction. When you're standing on the ground, it isn't happening to you.
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.Originally Posted by KJW
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.
Agreed. See Don Koks' article, and in particular the section entitled The Speed of Light as Measured by NonInertial 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.Originally Posted by KJW
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.Originally Posted by Farsight
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
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 writeOriginally Posted by Farsight

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.
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.... what characterizes the existence of a gravitational field from the empirical standpoint is the nonvanishing 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.
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)
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.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 coordinates. 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.
I obviously don't think I have, but let's look at this more closely.
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.
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 referencebody 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".
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.Originally Posted by Physicist
As above.Originally Posted by Physicist
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.Originally Posted by Physicist
Duffield,Originally Posted by Farsight
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?
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.
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.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.
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?Originally Posted by PhysBang
Peter M. Brown is a well known fringe person who fashions himself as a "physicist"
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.
Who said anything about flat meaning horizontal?
What does "change in speed or direction" actually mean? For example, consider an observer at rest on the ground and another observer in freefall 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 freefall above the ground is inertial (not accelerating).
It's for exactly the same reason.
But it is still the upward acceleration that keeps us on the ground.
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?
Bickering and snide posts are unhelpful.
There is not just or two guilty.
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.
Beer w/Straw: I don't understand your last response. What does
What does that supposed to mean?There is not just or two guilty.
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.
Okay. Thanks btr. I greatly appreciate you're civil attitude.
And likewise. I've enjoyed reading your posts.
Nobody. But in my experience people do confuse the two.
What it says. For example Compton scattering accelerates a photon. It doesn't slow it down, it changes its direction.Originally Posted by KJW
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.Originally Posted by KJW
But he isn't actually accelerating. He feels a force like the guy in the rocket, but he isn't in the same situation.Originally Posted by KJW
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.Originally Posted by KJW
It's the upward force on your feet that does it. If it wasn't there you'd be falling.Originally Posted by KJW
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?Originally Posted by btr
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 suchandsuch might turn a corner (despite evermounting 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.
This must be the reason you don't see the errors in this post.Originally Posted by Physicist
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.
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?
Why? By what criteria can we say this?
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
The above mathematics says you're wrong.
To the first order change in the metric, which determines the acceleration, they are exactly the same.
People talk about using the CMB reference frame to gauge your motion through the universe.
See if it changes position relative to other objects.Originally Posted by KJW
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.Originally Posted by KJW
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.Originally Posted by KJW
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.Originally Posted by KJW
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.Originally Posted by KJW
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.
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.
Because there is no force on the falling observer, he is not accelerating (Newton's first law of motion).
No, I'll focus on the mathematics because that is where the logic of physics resides.
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.
How does that change things.
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.
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?
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?See if it changes position relative to other objects.
[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.
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?Focus on the physics.
Please show us a toy example with (in)homogeneous space.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.
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°...
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.Originally Posted by KJW
No they aren't the same thing. They feel like the same thing, but they aren't the same thing.Originally Posted by KJW
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.Originally Posted by KJW
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.Originally Posted by KJW
I'm not going to. The CMB dipole anisotropy is not in doubt.
Huh? See Inhomogeneous Vacuum: An Alternative Interpretation of Curved Spacetime . We've previously said this is somewhat limited and deals only with static situations.
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.
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.Originally Posted by PhysBang
No. I've given enough explanation.Originally Posted by PhysBang
See above and look on arXiv.Originally Posted by PhysBang
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.
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.What are you saying? When two objects collide they collide. You can't adopt a coordinate system that pretends that they don't.
You, however, confuse time coordinates with clocks.It's similar for when a clock stops. It's stopped, and that's it.
To satisfy your own dogma, you certainly have. Yet you have never given us physics. That is a shame, but not ours.No. I've given enough explanation.
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).
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 sayso of some blindfolded guy in a box.
I didn't claim they were the same. The man standing on the Earth feels a force as if he's accelerating through gravityfree 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.Originally Posted by KJW
The reason for choosing the socalled "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.
You claim they have the same acceleration. But their differences ensure they do not have the same acceleration.
Both these conflict with Newton's first and second laws of motion.
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.
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.Originally Posted by KJW
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.Originally Posted by KJW
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.Originally Posted by KJW
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.Originally Posted by KJW
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 offcentre 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.
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.
Nope, wrong again, Duffield. Don't let science get in the way of your spamming your fringe ideas all over the internet.Originally Posted by Farsight
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 energymomentum distribution from an accelerated frame of reference in flat spacetime, is associated with the secondorder change in the metric, whereas the acceleration itself is associated with the firstorder change in the metric. Because the first and secondorder 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.
Regardless of how you say it, general relativity says you are wrong. This is the equivalence principle.
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.
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.
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.
Not for a real gravitational field. You need the Riemann curvature to get that plot off the flat and level.Originally Posted by KJW
I agree that when your pencil falls down, it isn't because of spacetime curvature.Originally Posted by KJW
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.Originally Posted by KJW
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.Originally Posted by KJW
It applies to an infinitesimal region only. So your pencil in your room isn't exactly equivalent to a pencil in an accelerating rocket.Originally Posted by KJW
He experiences an upward force on his feet, but he isn't accelerated upward.Originally Posted by KJW
He experiences no force, but he's accelerating at 9.8m/s/s.Originally Posted by KJW
I'm sorry KJW, but when you're standing still in your room, you just aren't accelerating.Originally Posted by KJW
Repeating the same falsity over and over doesn't make it true, John.Originally Posted by Duffield
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.
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?
In conflict with the Newton's first and second laws of motion.
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".
The accumulation of secondorder changes over distance is a firstorder change. By means of geodesic deviation, curvature does produce acceleration.
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/s². In the other the room is motionless and the pencil accelerates downwards at 9.8m/s².
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.
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.Originally Posted by KJW
That sounds similar to what I said about needing the Riemann curvature to get your plot off the flat and level.Originally Posted by KJW
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.
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.
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.
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 everfaster through homogeneous space. Again see Einstein's 1920 Leyden Address where he described a gravitational field as inhomogeneous space:
"This spacetime 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."
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.Originally Posted by KJW
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.Originally Posted by KJW
Again you cherrypick 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.
Book thumping and quote mining.
I think that Albert Einstein was being imprecise here, because in general relativity, it's *spacetime* that's curved, and not just 3space. 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 spacetime 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.
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 nomotion in inhomogeneous space. See above. Einstein described a gravitational field as a place where space is neither homogeneous nor isotropic.
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 Einsteinde 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.
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.Originally Posted by PhysBang
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.
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.
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