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Thread: How come Light can not escape a Black hole

  1. #1 How come Light can not escape a Black hole 
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    Hello

    From my understanding, black holes are just ultra dense matter and due to its denseness gravity is so high that the escape velocity is faster than the speed of light.
    So gravity attracts mass?

    Photons have no mass.
    How come gravity attracts something without mass?
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  2. #2  
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    So gravity attracts mass?
    How come gravity attracts something without mass?
    Gravity is a geometric property of space-time, and as such it affects everything, not just mass. Photons are null geodesics in space-time, so these world lines are affected just like any other object. The only difference is in the geometry of the world line in question.
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    So how is the photon's world line affected if you stand on the surface of the Earth and shine a laser beam straight up?

    And how is it affected if you stand on a more massive body and shine a laser beam straight up?

    And on an even more massive body?
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    So how is the photon's world line affected if you stand on the surface of the Earth and shine a laser beam straight up?
    And how is it affected if you stand on a more massive body and shine a laser beam straight up?
    And on an even more massive body?
    You know the answer to this question, Farsight. The world line which represents the photon becomes increasingly "bent" in space-time as you increase the mass of the central body. Since tidal forces do not play a role here, the entire gravitational effect is found along the time-like direction, so the projection of the world line onto 3-space yields a straight radial line. If the central body undergoes gravitational collapse and the photon emitter happens to be behind or on the event horizon, then the world line will curve towards the centre of mass, and never escape to infinity. Note that such an emitter can not be stationary. And please do not respond with the usual "world lines / space-time are just mathematical abstractions" comment, because this is not your thread, so the OP should hear the accepted explanation of this phenomenon.

    Here's a very good visualisation, you can play around with the parameters a little to get a feel for the relationship between coordinate and proper time, and how mass affects this :

    http://www.adamtoons.de/physics/gravitation.swf

    Note how the graph on the left gets shorter and shorter and more and more "bent" as you increase the central mass, which is what answers your question.
    x0x and Glitch like this.
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    I think I know the answer, and that therefore the "accepted explanation" is inadequate. Your photon is emitted on a straight radial line. It's path isn't bent. It doesn't slow down, and it doesn't fall back. So how come the light doesn't get out? Your answer conflates space with spacetime, suggesting that the photon's path is curved. It isn't curved, it's emitted on a straight radial line. By the way, the http://www.adamtoons.de/physics/gravitation.swf animation isn't particularly useful. Yes, it allows me to select an initial speed of c to emulate a photon, and on the left it shows this going slower as it traverses the green spherical mass. That's good. But the maximum mass setting is 0.75 that of a black hole. It dodges the bullet. If you'd rather talk about this on another thread no problem.
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    Thanks for the answer, i have no idea what world lines are of what null geodesics means, googling it didnt help much
    But i was wrong in assuming gravity only attracts mass, i got that far thanks

    Should gravity also affect magnetic fields or xrays? (is that question stupid?)

    How come hawking radiation can escape a black hole?
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    How I look at this problem is that time around a massive object gets slowed down... inside a black hole time slows to a (near?) stop.

    So it allows light to curl up upon itself end finally fall into the black hole. since light moves at 300 000km/s but time does not exist it moves at 300 000 km/s but a second takes forever, from the onlookers point of view it just stops and light can never escape.

    As far as I know X-rays should be considered light.

    About Magnetic fields and (as far as I know unproven) hawking radiation I have no clue what to tell you.
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  8. #8  
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    Should gravity also affect magnetic fields or xrays? (is that question stupid?)
    Yes, it affects both.

    How come hawking radiation can escape a black hole?
    It originates from a region just outside the event horizon.
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    I think I know the answer, and that therefore the "accepted explanation" is inadequate.
    "I think I know the answer, therefore everyone else is wrong."

    I don't know what kind of logic this is, but it certainly has very little to do with science !
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    Quote Originally Posted by hagara View Post
    Thanks for the answer, i have no idea what world lines are of what null geodesics means, googling it didnt help much
    But i was wrong in assuming gravity only attracts mass, i got that far thanks

    Should gravity also affect magnetic fields or xrays? (is that question stupid?)

    How come hawking radiation can escape a black hole?
    Consider how Hawking radiation is created. The following is the plausibility argument used by Schutz; According to quantum field theory the vacuum is filled with vacuum fluctuations which consists of pairs of photons. One of the photons has +E and the other -E. If these pairs are produced near the event horizon then its possible that the -E photon will be captured by the event horizon and the other escapes to infinity. This is why what Markus said is true, that the photons are created outside the event horizon. Since the black hole is getting -E photons the mass if the black hole is decreasing.
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    With respect, Physicist, there are no negative-energy photons. Again the "accepted explanation" is inadequate.


    Quote Originally Posted by Markus Hanke
    I don't know what kind of logic this is, but it certainly has very little to do with science!
    It's got everything to do with science, Markus. Accepting a non-explanation that simply doesn't hold water has very little to do with science. Do you know of any negative-energy particles? No. So you must surely find the given explanation for Hawking radiation to be inadequate.
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    Quote Originally Posted by Farsight View Post
    With respect, Physicist, there are no negative-energy photons. Again the "accepted explanation" is inadequate.


    It's got everything to do with science, Markus. Accepting a non-explanation that simply doesn't hold water has very little to do with science. Do you know of any negative-energy particles? No. So you must surely find the given explanation for Hawking radiation to be inadequate.
    This is a non-answer that avoids the question!

    As a guiding principle for physics, I suppose that one should adopt the theory that allows us to produce accurate, agreeing measurements of physical phenomena from those phenomena as we practice physics according to that theory over time. Practicing physics means interacting with physical systems, so we must of necessity make predictions, if only with our applications, and these predictions and our applications involve generating measurements of these systems.

    The current explanation of black hole physics relies on a theoretical framework built on theories tested in this manner. I await the day when Farsight-Relativity is able to be compared to physical phenomena in any way.
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    Do you know of any negative-energy particles?

    And when did you last see any Hawking radiation?
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    Quote Originally Posted by Farsight View Post
    With respect, Physicist, there are no negative-energy photons.
    Prove it. And it's not as if you have a great reputation for being flawless in your ideas in physics my friend. It's not as if your knowledge of physics is better than Hawking's is.

    Quote Originally Posted by Farsight View Post
    Again the "accepted explanation" is inadequate.
    Just because you don't accept it doesn't mean that mainstream cosmologists don't.
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    OK, let me rephrase the question: give me a reference to some evidence that demonstrates the existence of negative-energy particles.
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    Quote Originally Posted by Farsight View Post
    OK, let me rephrase the question: give me a reference to some evidence that demonstrates the existence of negative-energy particles.
    There are two major problems with that request.

    Problem #1) It's a logical fallacy. Lack of being able to provide evidence is not proof of your claim that there are no negative-energy photons. For example; 10 years ago some physicists claimed that neutrinos had no mass. If someone disagreed and said it hadn't been proved yet and they said give me a reference to some evidence that demonstrates the neutrino has mass then that too would be a logical fallacy. In that case all it took was a few years for it to be proven experimentally. Right now they exist in theory. Perhaps you just need to learn more about virtual particles since you don't seem to be aware of the notion that they can have negative energy.

    Problem #2) Such photons cannot be observed because they're either virtual particles or when they become real they've already passed into the event horizon of the black hole and therefore cannot be observed.
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    It's no fallacy. As you know we he have never ever detected any negative-energy particles. So a hypothesis that demands the existence of negative-energy particles is suspect. Just like a hypothesis that demands the existence of fairies and unicorns.

    Neutrino mass is not relevant to this. Particularly since the (rest) mass of a body is a measure of energy-content, and a photon has a non-zero inertial mass. Which is the same as its active gravitational mass, and is positive. You will also be aware that the photon has a non-zero effective mass when you slow it down in some medium to a speed less than c.

    Quote Originally Posted by physicists
    Perhaps you just need to learn more about virtual particles since you don't seem to be aware of the notion that they can have negative energy.
    With respect, my knowledge of virtual particles is fine thanks. They aren't short-lived real particles that pop in and out of existence spontaneously. Like worms from mud. Like magic. They aren't short-lived real particles which can have negative energy and go faster than light to boot. The reality that underlies virtual photons is the evanescent wave. See for example Phys. Rev. D 7, 1668 (1973) - Evanescent Waves in Quantum Electrodynamics with Unquantized Sources and do your own research.

    Edit:

    Quote Originally Posted by Physicist
    Problem #2) Such photons cannot be observed because they're either virtual particles or when they become real they've already passed into the event horizon of the black hole and therefore cannot be observed.
    Such photons cannot be observed because there are no such photons. Virtual photons are not real photons. Hydrogen atoms don't twinkle, and magnets don't shine. And there are no photons with negative energy. So I'm afraid the given explanation for Hawking radiation is woefully inadequate.
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    Quote Originally Posted by Farsight
    It's no fallacy.
    It most certainly is, undeniably so in fact. It’s called The Fallacy of the Argument from Ignorance. Basically the argument-from-ignorance fallacy is an argument that confuses a lack of proof (ignorance) with a refutation.
    I explained what’s wrong with that kind of reasoning by using an example. You simply can’t correctly/logically argue that because someone can’t provide evidence that he’s right then he must therefore be wrong. Einstein’s theory of relativity wasn’t fully accepted until evidence of his predictions were observed in nature. If someone claimed he was wrong before that because he didn’t yet have proof then that’d obviously be an illogical deduction. You can learn much much about it the following links –
    Argument from ignorance - Wikipedia, the free encyclopedia
    Argumentum ad Ignorantiam
    Logical Fallacy: Appeal to Ignorance
    http://www.logicalfallacies.info/pre...from-ignorance
    Argument from Ignorance
    etc. That will give you plenty of information for you to understand the nature of this kind of logical fallacy.
    Quote Originally Posted by Farsight
    As you know we he have never ever detected any negative-energy particles.
    There you go. Using that same fallacy again. I’ve given you the link that explains what’s wrong with that type of reasoning sufficiently now so I won’t respond to that kind of argument again. As I said, something’s can’t be observed in nature, virtual particles being examples of that.
    Quote Originally Posted by Farsight
    So a hypothesis that demands the existence of negative-energy particles is suspect. Just like a hypothesis that demands the existence of fairies and unicorns.
    You didn’t read closely enough because I wrote The following is the plausibility argument used by Schutz. I don’t know whether Hawking himself used that as proof. I doubt it.
    Quote Originally Posted by Farsight
    Neutrino mass is not relevant to this.
    That’s enough of this. Nothing personal my friend but that’s quite an ignorant comment. I used neutrino mass as in an analogy. You keep making serious logical mistakes and I’m getting frustrated correcting them so from now on I won’t respond to your posts.
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    It isn't a logical fallacy to point out that we have no evidence and no theory that supports the existence of negative-energy particles. Photons are comprised of positive energy, so are neutrinos, so are electrons, so are positrons, so are protons, and so is every other particle that has ever been observed. And it definitely not some argument from ignorance to point this out. It's an argument from knowledge, one that refuses to accept the specious garbage that describes black holes evaporating due to the spontaneous creation of particle pairs, one of which has positive energy, the other negative energy. Wherein it's always the negative-energy particle that falls into the black hole. Never the other way round, and no mention of any 50:50 chance. The given explanation doesn't hold water for a moment.

    You will refuse to respond to my posts because you cannot refute what I've said. That's why you have resorted to specious accusations such as logical fallacy and argument from ignorance, and that's why you have cut and run.
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    Hey guys come on. We know that energy is a relative concept. When we talk about negative energy it is relative to a baseline. So... What is the baseline and how would it differ at the event horizon to free space?
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    It's not just a relative concept, Jilan. Matter is made of it. E=mc² and all that. Or for photons, E=hf, h being Planck's constant and f being frequency.

    Do you know of any particles with negative mass? Do you know of any photons with negative energy? Do you know of any negative-frequency photons?

    The answer is no. The baseline is zero, and there's no going below it.
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    So are you unfamiliar with the concept of gravitational energy being negative?
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    Not at all. And it isn't really negative. Gravitational potential energy is only negative by convention, because the zero is set at infinity, see hyperphysics.

    When two bodies fall together, some of their mass-energy is converted into kinetic energy. When they collide this kinetic energy is radiated away into space. The mass of the consolidated single body is then less than the original mass of the two bodies, and there's a mass deficit. We then talk of binding energy as negative energy, but there isn't any actual negative energy anywhere, just less positive energy. And to separate the two masses you have to add more positive energy to get back to where you started.
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    Farsight, you do seem to understand the zero energy level is a matter of convention, which is encouraging. However when two bodies "fall together" their mass energy is not converted into kinetic energy, their gravitational energy is.

    If I am at the event horizon I can call my energy zero (by my convention). I could hold a ball out in each hand either side of the event horizon and declare that one had positive energy and the other one had negative energy.
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    Quote Originally Posted by Physicist
    You keep making serious logical mistakes and I’m getting frustrated correcting them so from now on I won’t respond to your posts.
    Sorry, John. I was irritated by your last few posts and got carried away. I'm trying to stay away from long drawn out discussions about things I know to be logically correct and I won't continue proving something I've already proven. So I'm trying to take you out of my ignore list. However something's wrong with the software. It's not letting me do it. So for a while I won't be able to read your posts.
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    Quote Originally Posted by Jilan View Post
    Farsight, you do seem to understand the zero energy level is a matter of convention, which is encouraging. However when two bodies "fall together" their mass energy is not converted into kinetic energy, their gravitational energy is.
    This gravitational potential energy is mass-energy. When you lift a brick you do work on it, you give it energy, and you increase its mass. The mass of a body is a measure of its energy-content. When you drop the brick some of its mass-energy is converted into kinetic energy. After you dissipate this kinetic energy, you'll find that the brick's mass has reduced to its original value before you picked it up. Note though that the mass change is very slight, and you can't practically measure it. Especially using a balance scale. Because the mass of your counterweight has changed too.

    Quote Originally Posted by Jilan
    If I am at the event horizon I can call my energy zero (by my convention). I could hold a ball out in each hand either side of the event horizon and declare that one had positive energy and the other one had negative energy.
    Let's forget about the event horizon and just say a higher ball has more energy, and a lower ball has less. Both balls comprise positive energy. There isn't any actual negative energy anywhere.

    By the way, we're off topic with this, it's rather impolite to hijack a thread so. So if you want to discuss it further, please can you start another thread?

    Physicist: noted.
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    Quote Originally Posted by Farsight View Post
    Not at all. And it isn't really negative. Gravitational potential energy is only negative by convention, because the zero is set at infinity...
    Not true. That link says that we set the zero potential to be zero at infinity. That doesn't mean that if we didn't set it to zero at infinity that the gravitational potential would be positive everywhere. In fact it makes no difference where you set your zero potential to. In Newtonian gravity for a point particle there will always be regions of space where the potential is negative.

    That doesn't apply to energy density though. E.g.
    Negative mass - Wikipedia, the free encyclopedia
    In electromagnetism one can derive the energy density of a field from Gauss's law, assuming the curl of the field is 0. Performing the same calculation using Gauss's law for gravity produces a negative energy density for a gravitational field.
    Quote Originally Posted by Farsight View Post
    When two bodies fall together, some of their mass-energy is converted into kinetic energy.
    What? Why do you say that? And what do you mean by "fall together"? Do you mean if they fall side by side so that there's no relative motion between the two bodies?

    Quote Originally Posted by Farsight View Post
    When they collide this kinetic energy is radiated away into space.
    How can they collide if they fall side by side?

    This post doesn't make a lot of sense to me.
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    Quote Originally Posted by Farsight View Post

    Let's forget about the event horizon and just say a higher ball has more energy, and a lower ball has less. Both balls comprise positive energy. There isn't any actual negative energy anywhere.
    So far so good, but you cannot dismiss the relative frames of reference as energy is frame dependent, Now take yourself far away for the event horizon, from that frame it is possible that a particle has negative energy. That is how you need to read that it has ""negative energy"

    By the way, we're off topic with this, it's rather impolite to hijack a thread so. So if you want to discuss it further, please can you start another thread?
    I don't approve of thread-jacking, but this is well on-thread with the OP (How come light can't escape a black hole?)
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    One of the photons has +E and the other -E. If these pairs are produced near the event horizon then its possible that the -E photon will be captured by the event horizon and the other escapes to infinity...

    The explanation I got some while ago was different, because I could not accept (meaning with logic) this one. Like Farsight said the probability to receive +E et -E is equal.

    What I was told is that those quantum fluctuation was indeed real-like particles pair(P and anti-P) and not virtual-like pair (P and nega-P). Those 2 real-thing of positive energy P or -P was borrowed (the energy == 2*E) from the gravitation field energy (or was it a Mexican hat?), actually puncturing *two* virtual holes in space (== 2*-E). Then out of these 4 things only 1 escape (the P or -P) == E and the 3 others enter the horizon (of a total of -E).

    That I found logical enough.
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    Yep, good enough. The gravitational field makes it possible.
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    Quote Originally Posted by Jilan View Post
    So far so good, but you cannot dismiss the relative frames of reference as energy is frame dependent, Now take yourself far away for the event horizon, from that frame it is possible that a particle has negative energy.
    A brick that's lower than you doesn't comprise negative energy. Nor does an electron, or a photon. Even when it's a lot lower than you. An electron at rest might comprise less energy than an electron at rest at your altitude, because mass-energy was converted into kinetic energy then shed. But not zero energy, and not less than zero energy. The photon doesn't lose any energy as it descends, because it's all kinetic energy anyway.

    Quote Originally Posted by Jilan
    That is how you need to read that it has ""negative energy"
    Jilan, a negative-energy particle is a myth. There aren't any. Not from any frame.

    Physicist: the two bodies fall towards each other. And the energy density of a gravitational field is not negative. The energy of the gravitational field shall act gravitatively in the same way as any other kind of energy.

    Boing: what isn't logical is that the energy could be "borrowed" from the space near the black hole, then all 4 things fall in. The black hole would consume vacuum energy and grow as a result. It just doesn't add up.
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    In reply to Farsight, re: your #23 post.

    "Kinetic energy is radiated away into space?" How so? By what mechanism is mechanical-force being conducted? (this is the first time I have ever heard of this!)

    (Thanks for reading!)
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    what isn't logical is that the energy could be "borrowed" from the space near the black hole, then all 4 things fall in
    Wrong and wrong. It would be utterly illogical to create something from nothing. That's called magic.
    Logic, on the other side, implies that things are conserved, from sentences to sentences, from cause to effect, from sides of equations.

    I don't know if nature does such a thing. What I know is if it does, it does it pretty logically. My banker do create money from thin air, so why mother nature could not do that too ?

    If you'd care to read what I said, out of those 4 things, only 1 will escape, P or -P. The 2 "borrows" may have no momentum, for all I know. The only trouble with that picture (I can play devil advocate with my own thoughts) is that those 2 holes being "negative energy", may as well be repelled by the BH gravitation.
    But those two holes are probably the Planck bubble where the particles appear (and disappear, if not EH was there to bug the process).

    I don't say it is happening, I say it is a logical description of a process.
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    Quote Originally Posted by Gerry Nightingale View Post
    "Kinetic energy is radiated away into space?" How so?
    Two planets collide and there's heat and light, and the combined planet is red hot then it radiates and cools down.

    Quote Originally Posted by Boing3000
    Wrong and wrong. It would be utterly illogical to create something from nothing. That's called magic.
    Hawking radiation is said to be due to particles popping into existence above the event horizon. Their energy is said to have been borrowed from the black hole. It could equally be borrowed from vacuum energy.

    Quote Originally Posted by Boing3000
    My banker do create money from thin air, so why mother nature could not do that too?
    Conservation of energy. There are no perpetual motio machines.

    Quote Originally Posted by Boing3000
    If you'd care to read what I said, out of those 4 things, only 1 will escape, P or -P. The 2 "borrows" may have no momentum, for all I know. The only trouble with that picture (I can play devil advocate with my own thoughts) is that those 2 holes being "negative energy", may as well be repelled by the BH gravitation.
    But those two holes are probably the Planck bubble where the particles appear (and disappear, if not EH was there to bug the process). I don't say it is happening, I say it is a logical description of a process.
    Only it isn't logical. Because there are no negative-energy particles.
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    Their energy is said to have been borrowed from the black hole. It could equally be borrowed from vacuum energy.

    From both. First from the fluctuation of vacuum account, and because that Planck bubble is crossing the horizon, the merging and acquisition process of that bubble transfer the negative energy 2*-E + 1*E onto the BH account. What's left outside is 1*E.
    Total ZERO. Lawyers are happy, and Gandalf not.

    Conservation of energy. There are no perpetual motio machines.

    That's my point, exactly. I go further, it is because of the conservation of logic. Any non-magical environment implies conservation of energy. I bet that Hawkins would agree, and he have a sense of humor
    You do agree, so what is your point exactly ?

    Only it isn't logical. Because there are no negative-energy particles.

    Those two phrases have no logical connection whatsoever.
    I doubt that the 2 holes/borrow are actual particle in the theory. I think the energy is borrowed from the gravitational field or space time itself. Nor you and I can be bothered by complicate math. So let's stick with simple notion. My banker do believe in money, so should you. Guess what, the total value is ZERO.

    Now, beside that nobody talk about negative particle but you, the second phrase is just your opinion. Not a fact. And it is a negative, very very hard to prove (in my logical book impossible).

    If one calm day on the side of a lake (flat, zero energy level), you see a flying fish pop out into existence (or two), even you will agree that there is a temporary negative hole waiting for the fish to land.
    Do you ?



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    Thanks for the explanations!


    p.s. didnt wanna start a war
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    Quote Originally Posted by Farsight
    With respect, my knowledge of virtual particles is fine thanks.
    No, it isn't. Every single comment that you've made about them is contrary to what virtual particles are.

    Quote Originally Posted by Farsight
    They aren't short-lived real particles that pop in and out of existence spontaneously. Like worms from mud. Like magic. They aren't short-lived real particles which can have negative energy and go faster than light to boot.
    Magic my butt. If it occurs in nature then, by definition, its not magic. You have a strong tendency to dump on things you don't understand and you then claim you understand it and its everyone else that doesn't. That got very old John and I'm tired or it. Your claims are all incorrect. Any particle physicist will tell you that your claims here are all nonsense boarding on crackpot claims.

    Quote Originally Posted by Farsight
    The reality that underlies virtual photons is the evanescent wave. See for example Phys. Rev. D 7, 1668 (1973) - Evanescent Waves in Quantum Electrodynamics with Unquantized Sources and do your own research.
    That article has nothing to do with your claims. Why you think it is baffles me. That in no way implies or has anything to do with virtual particles having negative energy. You appear to be using your own theory of physics again. For example; download Introduction to Elementary Particles by David Griffiths at
    Introduction to Elementary Particles | David Griffiths | digital library BookOS
    and show me where in that text that what claim is true. Or, unless you're too scared, you can contact the author himself. I have his e-mail address if you have the courage?

    Quote Originally Posted by Farsight
    Such photons cannot be observed because there are no such photons. Virtual photons are not real photons.
    You weren't paying close attention. I wrote Since the black hole is getting -E photons the mass of the black hole is decreasing. If you knew about this theory then you'd know that when this happens the photon becomes real. That's how the mass of the photon decreases the mass of a black hole.

    I have an idea. Why don't you contact Steven Hawking and prove how wrong he is. I'm sure he'll be happy that you found the flaw in his theory. Or perhaps you don't believe yourself enough to actually do something useful with your claim? Or are you afraid that Hawking will explain your mistake? I give up. Exactly why wouldn't you attempt to publish your claim? My guess is that you don't know your claim well enough to prove it?

    Quote Originally Posted by Farsight
    Hydrogen atoms don't twinkle, and magnets don't shine. And there are no photons with negative energy. So I'm afraid the given explanation for Hawking radiation is woefully inadequate.
    Every time you make comments like this you come across more an more irrational to me.
    Last edited by Physicist; 08-03-2014 at 11:20 AM.
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    In reply to Physicist, re: your #37 post.

    Cheers!

    In regard to Hawking, he has "modified" his BH gravitational field theory somewhat (published around early Feb. of this year) I would imagine he's hesitant to go any further for fear

    of losing control of what the implications may imply.

    ......

    Hawking's "modifications" seem to vindicate GR conditions somewhat (at least in my view) in that the relationship of matter-density and gravimetric-fields may not be "all that it seems".

    ( I personally feel that the relationship of matter-gravity is a proper "one-to-one" GR value)


    (Thanks for reading!)
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    In reply to Farsight, re: your #34 post.

    There must exist a media for the conduction of Kinetic energy...such as matter to matter, mass to mass. (kinesis represents movement of a real "thing" of matter/mass, not quantum

    energies...light has no initial inertia as a result of having no matter, and therefore cannot "impart" kinetic-force to anything)


    (Thanks for reading!)
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    Quote Originally Posted by Gerry Nightingale View Post
    I.light has no initial inertia as a result of having no matter, and therefore cannot "impart" kinetic-force to anything)
    That's not correct, Gerry. Light most definitely has momentum, and it can -- and does -- transfer that momentum as anything else that possesses momentum. This momentum has been measured, and it has real effects. The "Pioneer 10 anomaly" was finally explained as due to precisely this momentum (which, by the way, is h/lambda). The Compton Effect is another famous example, and there are infinitely many more.
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    Quote Originally Posted by Physicist View Post
    Magic my butt. If it occurs in nature then, by definition, its not magic.
    Have you ever actually see Hawking radiation? Or a negative-energy particle? Has anybody? No. So please spare me the occurs in nature, and the insults. If you'd like to quote something from some textbook at me that I think is wrong, I'll happily email the author provided he's hale and hearty.

    Quote Originally Posted by Physicist View Post
    You weren't paying close attention. I wrote Since the black hole is getting -E photons the mass if the black hole is decreasing. If you knew about this theory then you'd know that when this happens the photon becomes real. That's how the mass of the photon decreases.
    The mass of the photon decreases? It's you who isn't paying attention. Now start paying attention: there are no negative energy photons. And do note that the given explanation talks about pair production that creates two photons. Not an electron and a positron. The whole thing is just popscience for kids every step of the way.

    Quote Originally Posted by Physicist View Post
    I have an idea. Why don't you contact Steven Hawking and prove how wrong he is...
    Because he's old and unwell. It might harm his physical and/or mental health.
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    And do note that
    the given explanation talks about pair production that creates two photons.

    Whose explanation ? Even this vague summary "for kids" only mention particles...
    Would you mind backing up your claims by quotations ?

    Not an electron and a positron.

    Not only an electron ... you mean.
    A whole bestiary of them, some kind of quantum soup, if I've learned my pop-sci science correctly

    The whole thing is just popscience for kids every step of the way.

    That much I would agree on, like most of the science I know, which is the same science you know.
    Our science horizon begins when we have no more time to spent digging further.
    So science "pop" some kind of temperature, which would warm our collective appetite of being "in charge". Somehow, we all have some kind a black spot of infinite ignorance. Will it finally evaporate to reveal nothing is left ?

    aahhh sweet sweet analogies...

    Maybe that's why photon does not escape black hole, they have given up the hope to enlighten anyone anymore.

    R.I.P.
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    Quote Originally Posted by Farsight View Post
    Have you ever actually see Hawking radiation? Or a negative-energy particle? Has anybody?
    Why do you keep using logical fallacies, i.e. the logical fallacy of the argument from ignorance? It sure is no way to present a logical argument, that's for sure. You constantly imply that what hasn't been observed yet doesn't exist

    Quote Originally Posted by Farsight View Post
    please spare me the occurs in nature, and the insults.
    Insults? WTH? I never insulted you so please knock that kind of comment off. It's the truth. For you the truth hurts but I'm not insulting you. What I said are facts.

    Quote Originally Posted by Farsight View Post
    If you'd like to quote something from some textbook at me that I think is wrong, I'll happily email the author provided he's hale and hearty.
    Why on earth would you ever think that something that takes a book to read and understand can be relayed in a simple quote? Besides, I already did. You just weren't paying attention. The argument I explained to you that the argument using negative energy photons comes from a plausibility argument used by Schutz. It's in A First Course in General Relativity - Second Edition page 323
    It can be downloaded here A First Course in General Relativity | Bernard Schutz | digital library BookOS
    Hawking’s calculation (Hawking 1975) uses the techniques of quantum field theory, but we can derive its main prediction very simply from elementary considerations. What follows, therefore, is a ‘plausibility argument’, not a rigorous discussion of the effect. One
    form of the uncertainty principle is dEdt ≥ /2, where E is the minimum uncertainty in a particle’s energy which resides in a quantum mechanical state for a time t. According to quantum field theory, ordinary space is filled with ‘vacuum fluctuations’ in electromagnetic
    fields, which consist of pairs of photons being produced at one event and recombining at another. Such pairs violate conservation of energy, but if they last less thant = /2E, where E is the amount of violation, they violate no physical law. Thus, in the largescale, energy conservation holds rigorously, while, on a small scale, it is always being violated. Now, as we have emphasized before, spacetime near the horizon of a black hole is perfectly ordinary and, in particular, locally flat. Therefore these fluctuations will also be happening
    there. Consider a fluctuation which produces two photons, one of energy E and the other with energy −E. In flat spacetime the negative-energy photon would not be able to propagate freely, so it would necessarily recombine with the positive-energy one within
    a time /2E. But if produced just outside the horizon, it has a chance of crossing the horizon before the time /2E elapses; once inside the horizon it can propagate freely, as we shall now show.
    To read how that's calculated download the text. When will you be writing to Schutz to explain to him the nature of his mistake? When will you be publishing your corrections to the theory?

    Quote Originally Posted by Farsight View Post
    The mass of the photon decreases? It's you who isn't paying attention.
    I made the mistake of not finishing the sentence which I did now. It should have read That's how the mass of the photon decreases the mass of a black hole. - referring to negative mass photons

    Quote Originally Posted by Farsight View Post
    Now start paying attention:..
    Not to someone who doesn't understand the nature of virtual photons.

    are no negative energy photons[/I]. And do note that the given explanation talks about pair production that creates two photons. Not an electron and a positron. The whole thing is just popscience for kids every step of the way.

    Quote Originally Posted by Farsight View Post
    Because he's old and unwell. It might harm his physical and/or mental health.
    Oh, ya! Like that'd actually happen. ROTLF!!!!

    Sorry John but I'll never have anymore time to read your bogus claims or listen to your exceptionally poor understanding of physics ever again. It was clearly a bad idea to take you off my ignore list. Not when you can't fathom what a logical fallacy of the argument from ignorance is and you keep using it. Sheesh!
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    Quote Originally Posted by Gerry Nightingale View Post
    In reply to Farsight, re: your #34 post.

    There must exist a media for the conduction of Kinetic energy...such as matter to matter, mass to mass. (kinesis represents movement of a real "thing" of matter/mass, not quantum

    energies...light has no initial inertia as a result of having no matter, and therefore cannot "impart" kinetic-force to anything)


    (Thanks for reading!)
    That's not true. Light does have mass. I.e. it has what physicists call inertial mass aka relativistic mass defined as m = p/v. That means that whatever has momentum, such as a photon where p = E/c, has mass. The energy of a photon is all kinetic energy and since it can interact with charge particles such as in Compton scattering it can transfer some of that kinetic energy to charged particles.
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    You have a strong tendency to dump on things you don't understand and you then claim you understand it and its everyone else that doesn't.
    Nailed it.
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    Quote Originally Posted by Physicist View Post
    ...The argument I explained to you that the argument using negative energy photons comes from a plausibility argument used by Schutz. It's in A First Course in General Relativity - Second Edition page 323. It can be downloaded here A First Course in General Relativity | Bernard Schutz | digital library BookOS

    Hawking’s calculation (Hawking 1975) uses the techniques of quantum field theory, but we can derive its main prediction very simply from elementary considerations. What follows, therefore, is a ‘plausibility argument’, not a rigorous discussion of the effect. One form of the uncertainty principle is dEdt ≥ /2, where E is the minimum uncertainty in a particle’s energy which resides in a quantum mechanical state for a time t. According to quantum field theory, ordinary space is filled with ‘vacuum fluctuations’ in electromagnetic fields, which consist of pairs of photons being produced at one event and recombining at another. Such pairs violate conservation of energy, but if they last less than t = /2E, where E is the amount of violation, they violate no physical law. Thus, in the largescale, energy conservation holds rigorously, while, on a small scale, it is always being violated. Now, as we have emphasized before, spacetime near the horizon of a black hole is perfectly ordinary and, in particular, locally flat. Therefore these fluctuations will also be happening there. Consider a fluctuation which produces two photons, one of energy E and the other with energy −E. In flat spacetime the negative-energy photon would not be able to propagate freely, so it would necessarily recombine with the positive-energy one within a time /2E. But if produced just outside the horizon, it has a chance of crossing the horizon before the time /2E elapses; once inside the horizon it can propagate freely, as we shall now show.
    I'm afraid it's wrong on a number of counts. Vacuum fluctuations aren't virtual particles. Pairs of photons aren't literally being created and destroyed. Energy conservation is not violated, there are no negative-energy photons, if it can't propagate freely it can't cross the horizon, and if it can the positive energy-photon could, so the black hole could consume vacuum energy and grow. No, what Schulz wrote wasn't a rigorous discussion. And as a 'plausibilty argument' it was utterly unconvincing. It isn't plausible. Instead it's pseudoscience junk. No way should this be in a general relativity textbook. You want me to email Schutz and tell him? My pleasure. I'll let you know if he responds, but don't hold your breath.

    Quote Originally Posted by Physicist
    I made the mistake of not finishing the sentence which I did now. It should have read That's how the mass of the photon decreases the mass of a black hole. - referring to negative mass photons
    Groan. There ARE no negative-mass photons. There ARE no negative-energy particles. There ARE no unicorns. There ARE no fairies.

    Quote Originally Posted by Physicist
    Sorry John but I'll never have anymore time to read your bogus claims or listen to your exceptionally poor understanding of physics
    I understand it far better than you. And as for bogus claims, when you see a negative-energy photon, you can tell me about it.


    Quote Originally Posted by Markus Hanke
    Nailed it.
    What was it you said Markus? Let's see now. I said For example Hawking radiation is mainstream. And you said this:

    "Yes, one can probably consider it mainstream at this point, but I agree with you that it is somewhat problematic. Granted, the case for its existence is a good and strong one, yet as I have mentioned previously on several occasions, I have certain lingering doubts about QFT in general and QFT/CST ( in curved space-time ) in particular. I might very well be wrong though, I don't know, it's just a gut feeling at this stage."

    That would be a gut feeling about negative-energy photons magically springing into existence and conveniently falling into the black hole so we never see 'em.
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    Quote Originally Posted by Physicist
    Light does have mass. I.e. it has what physicists call inertial mass aka relativistic mass defined as m = p/v. That means that whatever has momentum, such as a photon where p = E/c, has mass. The energy of a photon is all kinetic energy and since it can interact with charge particles such as in Compton scattering it can transfer some of that kinetic energy to charged particles.
    That may have been correct 40 years ago when you got your BA, Peter. It is no longer correct in contemporary physics, you may want to take a class.
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    x0x, how would you put it alternatively?
    Last edited by Jilan; 08-03-2014 at 08:18 PM.
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    OK, so this thread is became a generic about "how come light can not ...

    -Have a mass.... Easy , if it had any, moving trough space would meant create a tiny tiny gravitational wave. Enough to suck all the energy out of the beast. Any experiment possible within our reach ? (proving I was wrong to try to create small BH by stuffing enough photon in one place)

    -Be of negative energy... Physicist found some article, but it also contains "plausibility argument’, not a rigorous discussion of the effect." Markus, maybe you can confirm (or not) to the absolute layman that I am, that some accepted theories model some particles which are not "anti-" but "nega-" ? I was stuck with the explanation (a Feynam lecture perhaps) that anti-particle can be seen as themselves but moving backward in time. A easy picture for me, especially that both energy are identical(ly positive), only all other attributes (charge, momentum, whatever) being reverse in the "mirror".
    Also an easy picture, for quantum fluctuation, the partner being the future me coming backward from the future, meet me, and cancel me (me being a photon, I can see how two identically reverse wave would cancel).
    But like Farsight said, no energy from vacuum has been observed, beside the dark one, which correspond to hypothesis 1 in this resume. Do Hawkins radiations been observed ? Is it even possible ?

    -Escape black hole ... Well, Markus explains that there is simply not 4D-path leading from inside to the outside. I can imagine that. Nothing can escape a BH, kind of by-definition. Except for the BH gravity, its only hair.
    After all we still get this information out of the BH "I am here". Can I ask someone if that mean that graviton do not exist or behave like tachyon inside the BH ?

    -Be the symbol of peace and harmony ... the end of war of words, if not war of world ?
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    In reply to tk421, re: your #40 post.




    No...I think "light momentum" does not exist! The impartation of energy is not the transfer of kinetic motion from one "factor" to another.

    A "packet of energy" has no mass. Photons can transfer (impart) energy to matter, not induce kinesis...a photon is a massless energy state.

    (Yes, I know about the spacecraft "effect"...there are least 3 major camps of theory debating this...and the "momentum" faction is losing)
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    Quote Originally Posted by Jilan
    x0x, how would you put it alternatively?
    The standard way used in contemporary physics: the photon is a massless particle.
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    Quote Originally Posted by x0x View Post
    That may have been correct 40 years ago
    It's correct now. Light has a zero rest mass, but a non-zero inertial mass and a non-zero active gravitational mass. See for example Light is Heavy by van der Mark and 't Hooft. (Not the Nobel 't hooft). Also see Einstein's Does the Inertia of a Body Depend upon its Energy-content? and note the last line:

    "If the theory corresponds to the facts, radiation conveys inertia between the emitting and absorbing bodies."

    Radiation conveys inertia. Light has a non-zero inertial mass.
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    Quote Originally Posted by Duffield
    It's correct now.
    No one takes what you say seriously, Farsight.
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    Quote Originally Posted by Gerry Nightingale
    No...I think "light momentum" does not exist!
    Why? It's not all that hard to prove. All textbooks on electrodynamics prove that to be the case. And it's been thoroughly tested. Light cannot exert pressure on any object unless it has momentum itself. If it didn't then when it causes an object to increase its momentum but it doesn't have momentum itself then there'd be a violation in the law of conservation of momentum. This is not something that's really open for debate. It's a well accepted fact in mainstream physics. So I take it that you disagree with mainstream physics?

    Quote Originally Posted by Gerry Nightingale
    A "packet of energy" has no mass.
    Sure it does, so long as you're referring to what is known in relativity as relativistic mass.

    Quote Originally Posted by Gerry Nightingale
    Photons can transfer (impart) energy to matter,
    And how do you think it does that? And where do you think that the momentum that's imparted to the matter comes from? According to you there is a violation of the law of the conservation of momentum.
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    x0x, how would you put it alternatively?
    Basically what it comes down to is that the term "relativistic mass" has fallen out of favor. Instead, we just cut to the chase and call it 'energy'. 'Mass' means what used to be called "rest mass". The fact that the inertia of a moving object increases with velocity is due to the fact that the energy of the object contributes to its inertia. IOW, momentum and inertia are no longer are tied exclusively to the concept of "mass".

    Since "relativistic mass" was really just the mass equivalence of the object's energy, the new convention merely eliminates the middle-man as it were and cleans up the nomenclature so that we no longer have to deal with two types of "mass".

    Since a photon has energy even though it has zero mass, it still exhibits momentum and inertia.
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    Quote Originally Posted by Janus
    Basically what it comes down to is that the term "relativistic mass" has fallen out of favor.
    Nice response Janus. I'd like to add some more to what you already said.

    When particle physics started talking off and became a large field of study, rest mass became a more useful concept to use, namely because its rest mass that was being studied along with other properties. It's use in particle physics is restricted mainly to closed systems and mostly to particles. For example; nobody measures the mass of a charged capacitor in a particle accelerator lab. If they did then things would become more complicated because then you'd have to take into account Poincare stress.

    See A simple relativistic paradox about electrostatic energy by Wolfgang Rindler and Jack Denur, Am. J. Phys., 56(9), Sep. (1988)
    A simple relativistic paradox about electrostatic energy
    A charged parallel‐plate vacuum capacitor moves uniformly through an inertial frame. Its field energy alone does not transform according to the familiar law ‘‘energy=γ× rest energy.’’ However, when the stresses in the supports are taken into account, the entire system d o e s satisfy this relation.
    See also The inertia of stress by Rodrigo Medina, Am. J. Phys., 74(11), Nov. (2006)
    The inertia of stress
    We present a simple example in which the importance of the inertial effects of stress is evident. The system is an insulating solid narrow disc whose faces are uniformly charged with charges of equal magnitude and opposite signs. The motion of the system in two different directions is considered. It is shown how the contributions to energy and momentum of the stress that develops inside the solid to balance the electrostatic forces have to be added to the electromagnetic contributions to obtain the results predicted by the relativistic equivalence of mass and energy.
    When it comes to other fields such as the early universe it comes in handy to think in those terms. I'm sure you've heard of Alan Guth, correct? In his course notes in the early universe he wrote
    We are perhaps not used to thinking of electromagnetic radiation as having mass, but it is well-known that radiation has an energy density. If the energy density is denoted by [I]u[/u], then special relativity implies that the electromagnetic radiation has a mass density given by

    (7.3)

    To my knowledge nobody has ever actually "weighed" electromagnetic radiation in any way, but the theoretical evidence in favor of Eq. (7.3) is overwhelming. - light does have mass. (Nonetheless, the photon has zero rest mass, meaning that it cannot be brought to rest ....)
    So it is used. Other textbooks speak of the mass density of radiation such as Gravitation and Spacetime - Third Edition by Ohanian and Ruffini.

    For more see Relativistic mass

    A study was done not too long ago by Gary Oas. He did a survey of some 300 books on relativity textbooks meant for physics students and found that about 40% of them used relativistic mass. And those were texts published not too long ago. It's far from being out of use. In fact it's still in use even in a few GR textbooks I've seen published within the last year or two.

    Quote Originally Posted by Janus
    However when it comes to other fields its still a useful concept.
    Exactly. But all that can be said is that since less people work outside of particle physics less people use it because its not what they're studying. It's very misleading to say that its fallen out of favor.

    Quote Originally Posted by Janus
    Instead, we just cut to the chase and call it 'energy'.
    Many people have made the mistake of thinking that just because E = mc2 that mass is the same as energy. However that is a serious error. They have very different meanings. All that can be said is that there is a numerical relationship between the two just like E = hf. This equation most certainly doesn't say that energy and frequency are the same thing. Mass is what gives objects momentum and energy is a quantity that is conserved. Those are entirely different concepts. When energy leaves a body then the mass of the body decreases according to E = mc2. That's what it means.
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    Quote Originally Posted by Janus View Post
    Basically what it comes down to is that the term "relativistic mass" has fallen out of favor. Instead, we just cut to the chase and call it 'energy'. 'Mass' means what used to be called "rest mass". The fact that the inertia of a moving object increases with velocity is due to the fact that the energy of the object contributes to its inertia. IOW, momentum and inertia are no longer are tied exclusively to the concept of "mass".

    Since "relativistic mass" was really just the mass equivalence of the object's energy, the new convention merely eliminates the middle-man as it were and cleans up the nomenclature so that we no longer have to deal with two types of "mass". Since a photon has energy even though it has zero mass, it still exhibits momentum and inertia.
    Good stuff, Janus. Have we met? Note that you can't slow down a photon in vacuo like you can slow down a fast-moving electron. But you can decelerate it. Like this. And you can slow it down in glass, see photon effective mass. But if you shine your torch straight up, the gravitational field doesn't slow down a photon. So to get us back on topic:

    How come light cannot escape a black hole?
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    Quote Originally Posted by Farsight View Post
    But if you shine your torch straight up, the gravitational field doesn't slow down a photon.
    In what metric are you saying that the speed does not slow down. In relativity theory, the speed of light is either constant at the level of the points of the manifold or dependent on the details of the metric.

    In the future, please indicate the metric you are using to make claims about the speed of light so that we can be clear about what you mean.
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    There is no "metric in which the speed does not slow down". A metric is an abstract thing, related to "what you measure". And you don't measure a vertical light beam slowing down as it ascends. So again, we have a vertical light beam, and it doesn't slow down. So:

    How come light cannot escape a black hole?
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    Quote Originally Posted by Farsight View Post
    There is no "metric in which the speed does not slow down". A metric is an abstract thing, related to "what you measure". And you don't measure a vertical light beam slowing down as it ascends. So again, we have a vertical light beam, and it doesn't slow down. So:

    How come light cannot escape a black hole?
    Because the velocity (the vector that you do not understand, despite years of explanations), changes direction. The motion is not radial, it tends asymptotically to the EH.
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    No it doesn't.

    If you stand on a planet and shine your laser beam straight up, it goes straight up. The light doesn't curve round, it doesn't slow down, and it doesn't fall back down.

    If we make the planet ten times as massive, the light still goes straight up. The light doesn't curve round, it doesn't slow down, and it doesn't fall back down.

    If we make the planet a million times as massive, the light still goes straight up. The light doesn't curve round, it doesn't slow down, and it doesn't fall back down.

    There is no point at which the light starts curving round or slowing down or falling down. Even when we take it to the limit and make our planet a black hole. So:

    How come light cannot escape a black hole?
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    Quote Originally Posted by Farsight View Post
    No it doesn't.

    I didn't expect you to understand, Duffield. The explanation was for the benefit of the others reading the thread, you are beyond any redemption, too deep in your bombastic ignorance.

    Quote Originally Posted by Duffield
    If we make the planet a million times as massive, the light still goes straight up.
    Nope, you are the same old crank as usual.

    The light doesn't curve round
    Actually, it DOES. Out of defiance for the old crank John Duffield.
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    Quote Originally Posted by Farsight View Post
    There is no "metric in which the speed does not slow down". A metric is an abstract thing, related to "what you measure". And you don't measure a vertical light beam slowing down as it ascends. So again, we have a vertical light beam, and it doesn't slow down.
    It is trivial to create a system of coordinates in which the light does not slow down. And a metric is an abstract thing required for attributions of measure. So, since you are making claims about measurements, I again ask you to be clear what metric you are using and what system of coordinates.
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    Farsight, if I am just inside the event horizon (as deemed by you who is a great distance from it) and I shine my torch outwards it will appear to me that the light will go up and keep going. However as far as you concerned it gets stuck at the horizon. This is because we are looking at infinite time dilation at the horizon. The light is moving infinitely slowly from your viewpoint so the light never makes any progress out of the black hole. It is worth bearing in my mind that from my point of view the event horizon is somewhat closer to the black hole than I am. This is always the case. The position of the event horizon is frame dependent, namely it depends on the gravitational potential of the observer with respect to it.
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    Quote Originally Posted by Jilan View Post
    Farsight, if I am just inside the event horizon (as deemed by you who is a great distance from it) and I shine my torch outwards it will appear to me that the light will go up and keep going.
    This is incorrect, the light doesn't "go up", it doesn't follow the radial direction. The light will "turn" such that it "boomerangs" to the singularity. The spacetime inside the EH is curved in such a way that all null geodesics (paths followed by light) lead to . Here is an excellent treatment of the subject.
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    Thanks xox. All the diagrams in the excellent treatment are drawn from the viewpoint of the distant observer, so how are they relevant?

    To quote wiki.
    "A misconception concerning event horizons, especially black hole event horizons, is that they represent an immutable surface that destroys objects that approach them. In practice, all event horizons appear to be some distance away from any observer, and objects sent towards an event horizon never appear to cross it from the sending observer's point of view (as the horizon-crossing event's light cone never intersects the observer's world line)."
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    If you stand on a planet and shine your laser beam straight up
    You can stand stationary on a planet, but you cannot stand stationary at a point below the EH; that's physically impossible, and kind of the whole point in having an EH. Furthermore, as has been pointed out by several people here already, a beam of light fired anywhere in a curved space-time does not undergo straight radial motion, so the arguments about speeds and velocity and directions are nonsensical, and completely miss the point. Null geodesics in Schwarzschild space-time aren't straight lines ! As for what you "see" - you see each photon hitting your eye, no more and no less.

    Here is how light moves in Schwarzschild space-time ( set K=0 for null geodesics ) :

    Wolfram Demonstrations Project

    This can be extended below the horizon by choosing an appropriate system of coordinates.
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    I think it's high time we do this explicitly and calculate the actual trajectory of our light ray, instead of arguing for pages and pages about speeds and times and velocity. Light satisfies the geodesic equation



    with some affine parameter . Because for light we have ds=0, it also must satisfy the additional parametrisation condition



    Starting with the Schwarzschild metric, we can now calculate the connection coefficients, insert these into the above conditions, and solve the resulting set of partial differential equations. Typing all of that out is way too tedious, so I will skip that and simply state the end result ( the dot denotes differentiation with respect to ) :









    Herein , and the constants of integration have to be determined from the initial conditions of the trajectory. Having said that, due to the spherical symmetry of Schwarzschild space-time, we are free to fix one more constant at random; if we pick F=1, and fix E=0 for light, the equations of motion are solved by



    wherein A and B are the coefficients of the Schwarzschild metric. As is immediately obvious without even having to evaluate the integral, the trajectory of light in Schwarzschild space-time is not a straight radial line under any circumstance. If you shine a beam of light "straight up", regardless of where you are, the trajectory of the beam is given by the above integral, and it is not a straight line.
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    Quote Originally Posted by Jilan View Post
    Farsight, if I am just inside the event horizon (as deemed by you who is a great distance from it) and I shine my torch outwards it will appear to me that the light will go up and keep going. However as far as you concerned it gets stuck at the horizon. This is because we are looking at infinite time dilation at the horizon. The light is moving infinitely slowly from your viewpoint so the light never makes any progress out of the black hole.
    I agree with this, Jilan. Others don't. See Markus's posts. He goes all round the houses and gets lost in abstraction without defining his terms and whilst totally evading the central issue: as you increase the mass of the gedanken planet, there is no point when the radial light beam starts curving round.

    Markus: your "calculation" is misleading because your are confusing space and spacetime. The spacetime interval ds=0 is not a reflection of reality, light takes a non-zero time to move a non-zero distance. And whilst the trajectory of light in Schwarzschild space-time is not a straight radial line, the trajectory of light in space is.
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    Quote Originally Posted by Jilan View Post
    Thanks xox. All the diagrams in the excellent treatment are drawn from the viewpoint of the distant observer, so how are they relevant?
    The SHAPE of the trajectory (non-radial, see Markus' excellent post) is observer invariant, all observers measure the trajectory as being curved, not radial (as the John Duffield crank keeps insisting). The "boomeranging" effect (the ray returning to the singularity) is also observer invariant.

    To quote wiki.
    "A misconception concerning event horizons, especially black hole event horizons, is that they represent an immutable surface that destroys objects that approach them. In practice, all event horizons appear to be some distance away from any observer, and objects sent towards an event horizon never appear to cross it from the sending observer's point of view (as the horizon-crossing event's light cone never intersects the observer's world line)."
    What does the above have to do with what has been explained to you?
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    Quote Originally Posted by Farsight View Post
    I agree with this, Jilan. Others don't. See Markus's posts. He goes all round the houses and gets lost in abstraction without defining his terms and whilst totally evading the central issue: as you increase the mass of the gedanken planet, there is no point when the radial light beam starts curving round.

    Markus: your "calculation" is misleading because your are confusing space and spacetime. The spacetime interval ds=0 is not a reflection of reality, light takes a non-zero time to move a non-zero distance. And whilst the trajectory of light in Schwarzschild space-time is not a straight radial line, the trajectory of light in space is.
    Duffield,

    Keep digging yourself in your ridiculous posturing.
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    Quote Originally Posted by x0x View Post
    This is incorrect, the light doesn't "go up", it doesn't follow the radial direction. The light will "turn" such that it "boomerangs" to the singularity. The spacetime inside the EH is curved in such a way that all null geodesics (paths followed by light) lead to . Here is an excellent treatment of the subject.
    It falls at the first hurdle by saying this:

    • Space becomes so curved that there are no paths for light to follow from an interior to exterior region.

    A gravitational field is described by curved spacetime, not curved space. Space is inhomogeneous, not curved. And note this:

    As viewed from a distance outside rs, the spacecraft appears to slow as it approaches rs, and eventually stops as r→ rs.

    Light appears to stop. Light clocks appear to stop, because they have stopped. The coordinate speed of light is zero at the event horizon. And yet:

    The occupants will use their own clocks (measuring proper time) to gauge the passage of time.

    Light has stopped, light clocks have stopped, but occupants allegedly see their own clocks ticking normally. Even though they're stopped too. Their finite proper time to hit the point-singularity takes infinite coordinate time. It hasn't happened yet, and it never ever will. It's a fairy tale.

    Quote Originally Posted by x0x
    The SHAPE of the trajectory (non-radial, see Markus' excellent post) is observer invariant, all observers measure the trajectory as being curved, not radial (as the John Duffield crank keeps insisting). The "boomeranging" effect (the ray returning to the singularity) is also observer invariant.
    Shine that radial laser beam straight up from a planet, and all observers will agree it's straight. Even an accelerating observer. He knows he's accelerating. When we make the planet more massive, it's still straight. More massive still? Still straight. Even more massive? Still straight. So at what point does the laser beam stop being straight? Answer: at no point.
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    Quote Originally Posted by Farsight View Post
    It falls at the first hurdle by saying this:
    No one cares about your crank opinions, Duffield. Live with it.
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    It fell at the first hurdle by confusing curved space with curved spacetime. See this Baez article:

    "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."

    What I say isn't some crank opinion, instead it's good physics. And people do care for that. What they don't care for is abuse from an ignorant troll.
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    Quote Originally Posted by Farsight View Post
    It falls at the first hurdle by saying this:

    • Space becomes so curved that there are no paths for light to follow from an interior to exterior region.

    A gravitational field is described by curved spacetime, not curved space. Space is inhomogeneous, not curved.
    OK, so rather than address the physics, you are going to address a word choice.

    Not a surprising move for a coward who would rather attack scientists than learn science.

    Look, we all know that you are a pathetic human being, getting banned from message board after message board because you insult scientists and try to substitute cheap and incorrect critical theory for actual science.

    You yammer on about "inhomogeneous space", but the embarrassing fact is that you can't use this supposed concept to describe a pencil falling, let alone a black hole.


    And note this:

    As viewed from a distance outside rs, the spacecraft appears to slow as it approaches rs, and eventually stops as r→ rs.

    Light appears to stop. Light clocks appear to stop, because they have stopped. The coordinate speed of light is zero at the event horizon.
    Yeah, the coordinate speed in one set of coordinates. When someone writes "viewed from a distance" in this context, the really mean (or should mean) "assigned coordinates such that".

    You have the chance to describe the scenario using your inhomogeneous space if you want.

    Oh, wait, I forgot: you can't. You are talking out of your ass.

    Shine that radial laser beam straight up from a planet, and all observers will agree it's straight.
    What is an observer? It is trivial to assign coordinates such that light does not travel in a straight line. If you want to be correct, tell us the coordinate system. Tell us the metric.

    But you don't want to be correct.

    Even an accelerating observer. He knows he's accelerating.
    Really? Every observer is accelerating relative to some standard. How can you say that someone knows that they are accelerating in a non-trivial way?

    When we make the planet more massive, it's still straight. More massive still? Still straight. Even more massive? Still straight. So at what point does the laser beam stop being straight? Answer: at no point.
    OK, so show us what happens in your inhomogeneous space theory, with numbers.

    Oh, wait, I forgot: you can't. You are talking out of your ass.
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    Quote Originally Posted by PhysBang View Post

    Oh, wait, I forgot: you can't. You are talking out of your ass.
    yep



    Oh, wait, I forgot: you can't. You are talking out of your ass.
    yep
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    He goes all round the houses and gets lost in abstraction without defining his terms and whilst totally evading the central issue
    The central issue is that the trajectory of a ray of light emitted by an observer "straight up" is not a straight radial line, as I have demonstrated. This has nothing to do with space and space-time, the geodesic is a simple function of and , all in a fixed plane.
    So I am now giving you three options here :

    1. If you think there is an error in these maths, please point it out
    2. Accept the fact that null geodesics in Schwarzschild space-time are not straight lines
    3. Refuse to accept (2) while being unable to do (1)

    The trajectory of a ray of light is not any more of an "abstraction" than the trajectory of a ball fired out of a canon. Both are quite real.

    Shine that radial laser beam straight up from a planet, and all observers will agree it's straight.
    The maths are clear and they disagree with you - see post 68. Your words are meaningless unless you can substantiate them with a mathematical proof of your claim.

    What I say isn't some crank opinion, instead it's good physics. And people do care for that. What they don't care for is abuse from an ignorant troll.
    Actually what everyone here would care for is you demonstrating to us mathematically that the ray of light will be straight, by solving the geodesic equation and explicitly showing us how you obtain a straight line. The problem with this is that you can't do it, and you can't do it because it simply isn't true. You might as well argue that 1+1=3.
    But go ahead and give it a shot like I did, perhaps you can surprise us
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    Quote Originally Posted by Markus Hanke View Post
    The central issue is that the trajectory of a ray of light emitted by an observer "straight up" is not a straight radial line, as I have demonstrated.
    You haven't demonstrated that at all. All you've done is given some smoke and mirrors and produced a rabbit out the hat. Voila! The trajectory is curved!

    Quote Originally Posted by Markus Hanke View Post
    I am now giving you three options here :

    1. If you think there is an error in these maths, please point it out
    2. Accept the fact that null geodesics in Schwarzschild space-time are not straight lines
    3. Refuse to accept (2) while being unable to do (1)
    I already pointed out your error, which was ds=0. It has no connection to reality. And that you have confused a curved trajectory in spacetime with a straight-line motion through space. You did something similar when we were talking about a Shapiro delay for a light-beam moving between two close stars. If you want more, define your terms, because that's where your real error lies. Your terms do not reflect reality. And the reality is that the "coordinate" speed of light is zero at the event horizon. Light doesn't just appear to stop, it stops. Your maths is blind to this, and invents a fictional curved trajectory instead. And stopped observers who see their stopped clocks ticking in a neverneverland beyond the end of time.

    Quote Originally Posted by Markus Hanke View Post
    The trajectory of a ray of light is not any more of an "abstraction" than the trajectory of a ball fired out of a canon. Both are quite real.
    Yes, they are real. Now imagine the Earth is below us and we have no motion relative to it. And we drop our cannonball. It falls straight down. We shine a line beam to illuminate it. That light beam goes straight too. It isn't curved at all. You know what's next.

    Quote Originally Posted by Markus Hanke View Post
    The maths are clear and they disagree with you - see post 68. Your words are meaningless unless you can substantiate them with a mathematical proof of your claim.
    No, your math is meaningless unless you can substantiate it with a clear explanation of what happens in reality as you increase the mass of the gedanken planet. You can't, because it's mathematical fiction that gives rise to science fiction. Einstein would never have agreed with point singularities.

    Quote Originally Posted by Markus Hanke View Post
    Actually what everyone here would care for is you demonstrating to us mathematically that the ray of light will be straight, by solving the geodesic equation and explicitly showing us how you obtain a straight line. The problem with this is that you can't do it, and you can't do it because it simply isn't true. You might as well argue that 1+1=3. But go ahead and give it a shot like I did, perhaps you can surprise us
    Our cannonball falls straight down, our light beam goes straight down. And the light beam shone back up at us goes straight up. And it doesn't curve round, and it doesn't slow down, and it doesn't fall back. And again, as we increase the mass of the planet, there is no point at which any of the above begins to occur. So again, how come light cannot escape a black hole? Explain it. To your grandmother. And if you can't, face up to it.

    Oh, and do something about the trolls and abuse too, because you are a moderator, and they do not reflect well upon you, or this forum.
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    Quote Originally Posted by Farsight View Post
    I already pointed out your error, which was ds=0. It has no connection to reality.
    It is called the condition of null geodesic, pretender.

    Your maths is blind to this, and invents a fictional curved trajectory instead. And stopped observers who see their stopped clocks ticking in a neverneverland beyond the end of time.
    LOL.

    Oh, and do something about the trolls and abuse too, because you are a moderator, and they do not reflect well upon you, or this forum.
    Precisely, time to shoot down the John Duffield drone.
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    I already pointed out your error, which was ds=0.
    That's not an error, it's a basic feature of relativity. I thought you agreed that Einstein's relativity is a good model; in relativity, light is a null geodesic, and ds=0 is the condition for null geodesics - this is what distinguishes particles without rest mass from massive particles.
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    It has no connection to reality.
    Then neither has coordinate time, since both are based on the same metric. You are just undermining your own arguments here.
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    You haven't demonstrated that at all. All you've done is given some smoke and mirrors and produced a rabbit out the hat. Voila! The trajectory is curved!
    So you are openly denying now that freely falling objects trace out geodesics ?
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    Quote Originally Posted by Markus Hanke View Post
    So you are openly denying now that freely falling objects trace out geodesics ?
    He's been trolling you (and the forum). He has no place in this forum.
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    Quote Originally Posted by Markus Hanke View Post
    I think it's high time we do this explicitly and calculate the actual trajectory of our light ray, instead of arguing for pages and pages about speeds and times and velocity. Light satisfies the geodesic equation



    with some affine parameter . Because for light we have ds=0, it also must satisfy the additional parametrisation condition



    Starting with the Schwarzschild metric, we can now calculate the connection coefficients, insert these into the above conditions, and solve the resulting set of partial differential equations. Typing all of that out is way too tedious, so I will skip that and simply state the end result ( the dot denotes differentiation with respect to ) :









    Herein , and the constants of integration have to be determined from the initial conditions of the trajectory. Having said that, due to the spherical symmetry of Schwarzschild space-time, we are free to fix one more constant at random; if we pick F=1, and fix E=0 for light, the equations of motion are solved by



    wherein A and B are the coefficients of the Schwarzschild metric. As is immediately obvious without even having to evaluate the integral, the trajectory of light in Schwarzschild space-time is not a straight radial line under any circumstance. If you shine a beam of light "straight up", regardless of where you are, the trajectory of the beam is given by the above integral, and it is not a straight line.
    Markus, I am struggling with this big style. The issue I am having with it is that the "t" in the Schwarzschild metric is the t for a distant observer. The question is from a different frame of reference, one inside the event horizon, am I missing something obvious?
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    MODERATOR NOTE : Right Farsight, I think this has gone far enough. I don't know if you really thought that you were going to fool anyone here into believing that any of what you present has any relation to Einstein's relativity; I allowed you to post on this forum because I was hoping you were prepared to discuss this in earnest, and perhaps be man enough to see sense and admit you are wrong on at least some of these very basic points. As it turns out I was misguided - it started off well enough, but we are now just going around in circles, which is meaningless and a waste of time for everyone concerned, including yourself. My bad for giving you the benefit of the doubt. Let me use the example of our latest exchange on this thread to demonstrate just how far off the mark you really are. From Einstein's very own text "The meaning of relativity" :

    (1) The condition ds=0, which you said has no connection to reality, is formula (22) on page 30, where Einstein discusses the propagation of light. You told me I used that condition in error - well then, so did Einstein.
    (2) The geodesic equations are formula (90) on page 84, together with the discussion as to how and why these describe the trajectory of freely falling particles. You called this smoke and mirrors, and a rabbit out of a hat - well, Einstein was quite a magician, then.
    (3) You keep going on about variable speeds of light being the meaning of GR - there is no mention of this whatsoever anywhere in Einstein's above referenced text. Funny, given the title of Einstein's book.

    When you first started posting here again a few months back, I explicitly reiterated to you that this is a mainstream forum, and to please respect that. You didn't. Not only are you rejecting basic mathematics, you are now rejecting Einstein's very own words, distorting it to fit your personal ideas about what GR should be, not what it is. Heck, you are even rejecting basic calculus, such as the covariant derivative ! There can be no discussion about mainstream physics on that basis; there simply is no common ground here, if you just decide anything and everything we calculate for you to be meaningless. That's not physics.

    So here is how it is going to be - we will be going back to the previous arrangement, and you will henceforth restrict yourself to posting exclusively in the "Personal Theories" section, because that is what your ideas are : personal theories. They have nothing to do with Einstein's relativity, or mainstream physics. In particular, you will not respond to questions posted in the main forums with your personal ideas - such posts will be moved immediately. To be perfectly honest, I see no future here for you - this forum is just not for you, it is for mainstream science. You are free to disagree and reject the mainstream and have your own opinions, but please don't expect anyone to take those seriously, and this forum is most certainly not the right platform to try and proliferate such ideas.

    Don't bother sending me PMs - I can open them, but I cannot reply or compose new messages, due to the forum software issues we are experiencing, so don't waste your time. In any case, the options are clear - keep posting in personal theories, or leave. It's up to you now.
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    Markus, I am struggling with this big style. The issue I am having with it is that the "t" in the Schwarzschild metric is the t for a distant observer. The question is from a different frame of reference, one inside the event horizon, am I missing something obvious?
    I am not using Schwarzschild coordinate time anywhere in this, I am using an affine parameter to parametrise my geodesics ( as has to be the case for light ), and the dot denotes differentiation with respect to this parameter. Time does not appear in the final solution at all, because the relationship between the azimuthal angle and the radial coordinate is constant; remember that all we wanted to obtain was the trajectory in space, i.e. the azimuthal angle as a function of radial coordinate. We don't need to know where the ray of light is at a given instant in time, we only need to know what the trajectory as a whole looks like - in particular, the point of contention was wether or not it was a straight radial line, which of course it is not.

    Btw, you will find the detailed calculation ( in slightly different notation ) in T. Fliessbach, General Relativity, chapter 55. That is my primary reference to support this result.
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    I forgot to mention also that light trajectories in Schwarzschild space-time are never straight lines, except asymptotically at infinity, or in the case of M=0 where the metric reduces to the Minkowski metric.
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    Thanks Markus, I'll keep trying to figure it out. But if you are very distant and I am inside the EH as reckoned by you but not my me, why on earth would the light not go straight up in my frame of reference?
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    Quote Originally Posted by Jilan View Post
    Thanks Markus, I'll keep trying to figure it out. But if you are very distant and I am inside the EH as reckoned by you but not my me, why on earth would the light not go straight up in my frame of reference?
    Because the solution of the differential equation describing the light geodesic is and that is NOT a straight line.
    Nothing to do with any frame of reference. I explained that to you, Markus explained that to you.
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    Ah hmm, In post # 84 is says "starting with the Schwartzchild metric". That's the part I am having problems with. I thought that metric was from the viewpoint of a distant oberver. Are you really following this discussion xox or are you just pretending to?
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    Quote Originally Posted by Jilan View Post
    . I thought that metric was from the viewpoint of a distant oberver.
    It isn't. You are showing your ignorance and your impertinence again. A very toxic combination.

    Are you really following this discussion xox or are you just pretending to?
    I am following. The problem is that you are incapable to comprehend simple answers. Instead of making the effort to comprehend, you are acting petulant. You revert to this behavior every time you can't follow basic science.
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    In reply to Jilan, re: your #90 post

    Go back to A.E.'s emission theories...the answers are there. (and ignore nasty, petulant TROLLS...just as Albert did. Einstein managed to create the modern interpretations of physics

    theory w/ observations and summations...he didn't need 306 "posts of nothing" from poseurs)

    .....

    Einsteins' answers to interpretations of "what he means regarding photons?" "Scoundrels...who THINK they know!"<direct quote.

    Stay w/ a true genius, Jilan...he thought of all this long ago, long before pretenders.

    Thanks for reading!) Cheerio!
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  93. #93  
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    In reply to Physicist, re: your # 44 post

    Was it #44 or #54...whatever, I'm old and I forget things.

    Yes, I strongly disagree w/ "light imparting momentum" but I would have to post "why" on an "alternate theory" thread. I simply don't understand this concept at all.

    Yes, I'm not happy w/ "mainstream" regarding assigning a matter-function to a non-mass entity such as light. (I think if this "kinesis" is true, there is really not much point in studying

    anything...it means the end of Relativity and the "constant" factor of light)

    ....

    Perhaps QM can "explain" such a concept w/ "curves" & "affines" to make it "real", but my mind screams NO in terms of logic and Relativity proportion.


    Of course, it's of no real consequence what I think...still, it IS what I think.


    (Thanks for reading!)
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    Quote Originally Posted by Gerry Nightingale
    In reply to Physicist, re: your # 44 post
    Was it #44 or #54...whatever, I'm old and I forget things.
    Was what #44 or #54?
    Quote Originally Posted by Gerry Nightingale
    Yes, I strongly disagree w/ "light imparting momentum" but I would have to post "why" on an "alternate theory" thread. I simply don't understand this concept at all.
    Are you familiar with what’s called Compton Scattering? If not then please read Compton scattering - Wikipedia, the free encyclopedia

    In Compton scattering a photon scatters off of an electron imparting both energy and momentum to the photon thus decreasing both the energy and momentum of the photon and increasing the energy and momentum of the electron such that both energy and momentum is conserved. Momentum is a vector quantity which means that all components of the momentum are conserved. Therefore the light imparts momentum to the electron. That, my dear friend Gerry, is an experimental fact. Anything else is going to be mere word play and I’m not very interested in word play.

    Quote Originally Posted by Gerry Nightingale
    Yes, I'm not happy w/ "mainstream" regarding assigning a matter-function to a non-mass entity such as light. (I think if this "kinesis" is true, there is really not much point in studying anything...it means the end of Relativity and the "constant" factor of light)
    Sorry then. I don’t have any interest in non-mainstream physics so I won’t be reading the other forums on this subject. Nothing personal, okay my friend?
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    Light can ONLY follow the geodesic, and when you're past the event horizon, all geodesics are bent so that none of them reemerge from the EH.
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    Quote Originally Posted by AlexG View Post
    Light can ONLY follow the geodesic, and when you're past the event horizon, all geodesics are bent so that none of them reemerge from the EH.
    Correct.
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    Ah hmm, In post # 84 is says "starting with the Schwartzchild metric". That's the part I am having problems with. I thought that metric was from the viewpoint of a distant oberver. Are you really following this discussion xox or are you just pretending to?
    The Schwarzschild metric is just a particular choice of observer; the geodesic structure of space-time is the same for all observers, so it does not really matter what coordinate basis you choose to calculate it. The mathematical reason is that curvature and geodesics arise not from the metric, but from the connection - that's why you find the Christoffel symbols in the geodesic deviation equation. I could have chosen a different metric, e.g. Gullstrand-Painleve, or Kruskal-Szekeres, and arrived at the same result - these different metrics are all just different ways to describe the same space-time. This is the core argument that Farsight does not understand - he declares Schwarzschild coordinates to be the absolute physical truth everywhere, without understanding that curvature and geodesic structure do not arise from the metric but the connection - hence all choices of coordinates are equally valid, and none of them is in any way privileged.
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    In reply to Physicist, re: your #94 post.

    Did I mention I'm not a fan of completely unprovable, suppositional "particle" states...I cannot respond to your post properly because of the tenets of the forum (mainstream only protocols)

    Jut forget about me and responding...I guarantee you would find me too "shocking"...and yet, think on this. There has not been a single provable theory put forth since the death

    of Einstein...this was the beginning of the end of pure theory. EVERYTHING is "numbers" now that explain nothing!

    ......

    (If you are fan of quote-mining of suppositional "facts?", this not a bad place to be! In fact, if you can produce enough number theory to verify a supposition...you could win a Nobel just

    from writing here! Why not? Gerardus 't Hooft did...for exactly the same thing!


    Thanks for reading! Bye.
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    Quote Originally Posted by Markus Hanke View Post
    The Schwarzschild metric is just a particular choice of observer; the geodesic structure of space-time is the same for all observers, so it does not really matter what coordinate basis you choose to calculate it. The mathematical reason is that curvature and geodesics arise not from the metric, but from the connection - that's why you find the Christoffel symbols in the geodesic deviation equation. I could have chosen a different metric, e.g. Gullstrand-Painleve, or Kruskal-Szekeres, and arrived at the same result - these different metrics are all just different ways to describe the same space-time. This is the core argument that Farsight does not understand - he declares Schwarzschild coordinates to be the absolute physical truth everywhere, without understanding that curvature and geodesic structure do not arise from the metric but the connection - hence all choices of coordinates are equally valid, and none of them is in any way privileged.
    OK, I think I'm getting there. So am I following this right: If I am standing at the event horizon and shine my torch up the light will initially go straight up and then start veering off following a geodesic. If I am slightly inside the event horizon this geodesic will eventually return to the BH and if I am slightly outside it will still veer off and curve but the geodesic never brings it back? If I am at the event horizon and shine my torch sideways the light achieves orbit at that radius.

    Afterthought: if I stand on the earth and fire my torch upwards it also veers off but only slightly?
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    Quote Originally Posted by Gerry Nightingale View Post
    Did I mention I'm not a fan of completely unprovable, suppositional "particle" states...I cannot respond to your post properly because of the tenets of the forum (mainstream only protocols)
    Yes. I'm aware of that. I'm only interested in what you contribute to discussions on mainstream physics.

    Quote Originally Posted by Gerry Nightingale View Post
    Jut forget about me and responding...I guarantee you would find me too "shocking"...
    I doubt that very much.

    Quote Originally Posted by Gerry Nightingale View Post
    ...and yet, think on this. There has not been a single provable theory put forth since the death of Einstein.
    That's because its not the aim of science to prove theories. Anybody who thoroughly understands physics knows this fact very well. Neither science nor any other human endeavor has or can have that ability.
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