# Thread: A Refutation of General Relativity

1. A refutation of general relativity (hereafter "GR"): At A, let a test particle be in free fall near a star. (All drawings are from the particle’s perspective.) At B, let the star be hurled into the particle at an arbitrary velocity v. Special relativity (hereafter "SR") applies to a direct measurement, so the star is length-contracted in the particle’s frame. At C, let the particle instead free-fall to hit the star at v. According to GR, a superset of SR, the star is not length-contracted in the particle’s frame. But B and C are identical situations at impact—GR cannot have it both ways. Then GR is refuted as inconsistent.

Drawing B is validated by the muon experiment, a confirmation of SR, in which the whole Earth is length-contracted in the frame of a muon free-falling toward it.

Drawing C is validated by the book Exploring Black Holes, pg. 3-22: The general relativistic equation that returns the proper time elapsed of an object free-falling from rest at infinity (or a great distance) toward a Schwarzschild mass (i.e., a spherically symmetric, uncharged, nonrotating mass), is an integration, from a higher-altitude r-coordinate to a lower-altitude r-coordinate, of the equation -sqrt(r / 2M) * dr, where r is the r-coordinate (Euclidian radius) of the object from the center of the mass, M is the mass in geometric units (same units as r), and dr is the increment of r-coordinate. The proper time elapsed is just the sum of the proper time for the object to traverse each increment of the r-coordinate at the escape velocity at that r-coordinate. The same integration applies to Newtonian mechanics (GR shares Newton’s escape velocity equation; only the interpretations differ). Then in GR a Schwarzschild mass is not length-contracted in the frame of an object free-falling toward it. When the integration is from r = 2M (horizon) to r = 0 (singularity), the resulting equation is equivalent to 6.57 * 10<sup>-6</sup> (M / M<sub>Sun</sub>) seconds, where M is the mass of the black hole and M<sub>Sun</sub> is the mass of the Sun.

Let a particle free-fall from rest at infinity (or a great distance) to a star whose escape velocity at its surface is almost c. The particle reaches the surface at almost c. Drawing B shows that the star will be length-contracted in the particle’s frame to almost zero length. Were the particle able to pass unimpeded through the star, it would do so almost instantly in its frame (by its clock). (The particle’s minimum velocity while traversing the star is at the star’s surface).

GR, however, says that the proper time for the particle to pass through this star is dependent on the star’s mass. The more mass, the more time required, with no upper time limit. This dependency is not because of some black hole strangeness—the derivation is not dependent on any feature of a black hole. The reason is because, in GR, the star is not length-contracted in the particle’s frame. The more massive the star, the greater the r-coordinate at which the escape velocity is almost c. Then the more massive the star, the greater the (uncontracted) distance the particle must traverse to pass through the star.

GR is a superset of SR. Then every SR prediction must be equivalently predicted by GR. Because GR disagrees with SR for the same scenario, as shown above, GR is refuted as inconsistent.

2.

3. Hi Zanket. How's it going? I'd still love to see the GR field equation solutions for this problem.

The GR field equations are above my head, but according to the book cited above, Exploring Black Holes, all of the derivations within it are based on metrics that are solutions of the field equations. The derivation I cited above is based on the Schwarzschild metric, Schwarzschild's solution for a spherically symmetric, uncharged, nonrotating mass. (This is a great book! It covers a multitude of scenarios related to Schwarzschild black holes, while requiring only basic calculus. One of its authors, John Archibald Wheeler, co-authored the definitive tome Gravitation.)

5. James is registered on this site and has posted a time or two, but hasn't offered himself in any of the intellectual threads, merely talking about the other place. It'd be great if you could get him to contribute though. There is a lack of experienced maths and physics peoples here. James could fill a large hole and start the ball rolling

6. Zanket,

Looks like I need to make a trip to my local Borders bookstore...

7. I found a relevant derivation online. See eq. 51 on pg. 9 of this PDF (Adobe Acrobat) file. Plug in 2m (horizon) for r<sub>2</sub> and 0 (singularity) for r<sub>1</sub>. This results in 4/3 * m, where m is in geometric units where c = 1. The mass of the Sun is 1477 meters, which is ((1477 meters) / (299792458 meters / second)) = 4.927 * 10<sup>-6</sup> seconds of light-travel time. Multiply that by 4/3 to get 6.57 * 10<sup>-6</sup> seconds for the proper time to fall from horizon to singularity of a point mass equal to the Sun’s mass. This matches the result of the derivation cited above, from the book Exploring Black Holes.

8. SR is a subset of GR but is only applies to things in constant motion, so no accelerations or changes in direction. A particle in freefall is accelerating, therefore SR is not applicable.

9. Welcome to the discussion, Destruct.

In that case, the muon experiment, in which the muon is free-falling, could not be the confirmation of special relativity that it is.

The equivalence principle of GR says that SR applies in the local frame (generally, the immediate vicinity) of an object in free fall, despite its gravitational acceleration. In the case of drawing B, the velocity is directly measured at the moment of impact. A direct measurement is as local as it gets. Then SR applies.

10. Thanks Zaket.

I don't know about the muon experiement, but SR only applies when the observers involved are moving in straight lines at constant speeds.

From Wikipedia, http://en.wikipedia.org/wiki/Special_relativity

The theory is called "special" because it is a special case of Einstein's theory of relativity where the effects of gravity can be ignored. Ten years later, Einstein published the theory of general relativity which incorporates gravitation.
...
Special relativity is only accurate when gravitational effects are negligible or very weak, otherwise it must be replaced by general relativity.
...
Special Relativity rejects the idea of any absolute ('unique' or 'special') frame of reference; rather physics must look the same to all observers travelling at a constant velocity (inertial frame).
Gravity and acceleration invalidate the use of Special Relativity.

11. Originally Posted by Destruct

I don't know about the muon experiement, but SR only applies when the observers involved are moving in straight lines at constant speeds.
This is true in general, but I'm sorry to say that Zanket is quite right.
It is customary and quite legitimate to describe curved space-time as locally Minkowskian i.e flat. In fact, the whole study of manifolds makes heavy use of the homomorphism between the R<sup>n</sup> manifold and its R<sup>n - m</sup> chart. In fact, I might go so far as to say that's what distiguishes a manifold from a topological space. Yes, here it is: a manifold is a topological space that can be covered by appropriate coordinate neighbourhoods.

Why am I "sorry to say"? Because, although I can't see the answer to his puzzle, I'm pretty sure he is premature in calling it the death of GR.

Hey! That's rude of me! Sorry Zanket, let me talk directly to you now.

In your picture "B", are you assuming the presence of a field or not? Clearly you are in "A" and "C". My instinct, for what it's worth - nothing, probably - would be that unless you are, you're not comparing like with like. And if you are assuming the presence of a field, you are still not comparing like with like, for then in "B" you will have the combined effects of the field and linear transformation.

Oh well.

12. Originally Posted by Destruct
From Wikipedia, http://en.wikipedia.org/wiki/Special_relativity

The theory is called "special" because it is a special case of Einstein's theory of relativity where the effects of gravity can be ignored.
To add to what Guitarist said, note that the effects of gravity can be ignored in the immediate vicinity of a free-falling object, as implied by GR's equivalence principle. That is how SR applies in a universe that is nowhere free of gravitational influence. Check out the explanation and use (by a free-falling observer) of the equivalence principle at A Non-Mathematical Proof of Gravitational Time Dilation.

13. Guitarist, Hi.

I can't claim to be that knowledgable on the technical aspects of GR and SR, but I was wondering about this:
It is customary and quite legitimate to describe curved space-time as locally Minkowskian i.e flat.
Wouldn't that be invalidated by this?
Let a particle free-fall from rest at infinity (or a great distance) to a star whose escape velocity at its surface is almost c. The particle reaches the surface at almost c.
I thought maybe the fact that the space-time at the place in question was warped almost to the point of being a black hole makes local considerations highly unlike a flat space-time. It would be made even worse by the particle moving at close to c when it hits, amplifiying the intensity of the warp from the POV of the particle, resulting in situation where, even locally, things are way too distorted to be analyzed with SR.

[Edit: I also wonder if perhaps approximating locally as if SR is applicable is fine usually, but not in extreme situations such as gravitational warps close to black-hole level being hit by particles moving almost the speed of light. The would normally be slight discrepancies between a situation described first by a locally based SR and secondly by GR, I'd expect GR to be more accurate. And in some situations I'd expect the discrepency to be large.]

14. Originally Posted by Zanket
In that case, the muon experiment, in which the muon is free-falling, could not be the confirmation of special relativity that it is.
The muons are already travelling at a speed close to c. The speed increase caused by Earth's gravity over the distance of the experiment is so utterly small that it's probably okay to consider the muons to be in uniform motion.

15. Originally Posted by Guitarist
In your picture "B", are you assuming the presence of a field or not?
If you mean a gravitational field, then yes of course, since all stars have one.

And if you are assuming the presence of a field, you are still not comparing like with like, for then in "B" you will have the combined effects of the field and linear transformation.
If I understand your point… The Lorentz transformation applies to the whole star to the same degree regardless of the curvature of spacetime around the star; i.e. regardless of the star’s gravitational field. This behavior is validated by the muon experiment as noted above, and by numerous books on SR: trans-galaxy rocket travel is possible in the crew’s lifetimes only if the whole galaxy is length-contracted in their frame. Then so must be the stars composing the galaxy. The Lorentz transformation is dependent only on the relative velocity of an object, and not on the curvature of spacetime around it.

16. To counter this, I guess you just keep defining "local" as a smaller and smaller region, so provided the gravitational field is not warped to infinity and the particle is not quite moving at C you can always set up a situation where "local" is arbitrarily small and so reclaim an almost flat space-time.

In this case though, are the regions being traversed all contained within this tiny definition of "local"? Is it still valid to talk about covering distances if they are larger than what can safely be considered a flat space? For instance, from one side of the length contracted star to the other, can this be considered a region of flat space? I know it's a small span from the particle's pov, but is it small enough?

17. Originally Posted by Destruct
I thought maybe the fact that the space-time at the place in question was warped almost to the point of being a black hole makes local considerations highly unlike a flat space-time. It would be made even worse by the particle moving at close to c when it hits, amplifiying the intensity of the warp from the POV of the particle, resulting in situation where, even locally, things are way too distorted to be analyzed with SR.
Well, it all depends what's meant by "local". Clearly, the greater the curvature of space-time, the more local is "local", if you follow me!

The would normally be slight discrepancies between a situation described first by a locally based SR and secondly by GR, I'd expect GR to be more accurate. And in some situations I'd expect the discrepency to be large.
Not quite sure I follow you here. Locally, SR can always be applied, globally you need GR. There is no "discrepency". SR applies within GR exactly as first formulated, but only locally.

Originally Posted by Zanket
Were the particle able to pass unimpeded through the star, it would do so almost instantly in its frame (by its clock). (The particle’s minimum velocity while traversing the star is at the star’s surface).

GR, however, says that the proper time for the particle to pass through this star is dependent on the star’s mass. The more mass, the more time required, with no upper time limit.
Hm. Are you suggesting here that mass Lorentz transforms? I don't think that's right.

Good post, though.

<EDIT by HomoUniversalis: Compiled the two posts into one>

18. Not quite sure I follow you here. Locally, SR can always be applied, globally you need GR. There is no "discrepency". SR applies within GR exactly as first formulated, but only locally.
I'm guessing that the bigger you make "local", the greater the discrepancy between a GR description and a SR description, and conversely the smaller we consider local the smaller the discrepancy. But isn't there always going to be some tiny difference, maybe at the billionth decimal place, except maybe at some infinitesimel limit of local?
[Edit: At least in cases involving gravity and acceleration.]

19. Originally Posted by DEChengst
The muons are already travelling at a speed close to c. The speed increase caused by Earth's gravity over the distance of the experiment is so utterly small, that it's probably okay to consider the muons to be in uniform motion.
The increase in a muon’s velocity relative to the Earth is ignored as negligible in the experiment, but it need not be. As the muon’s velocity slightly increases, the whole Earth becomes slightly more length-contracted in its frame, and the muon’s predicted remaining proper time elapsed to reach the ground is slightly reduced. Are you suggesting that your point invalidates drawing B?

Originally Posted by Guitarist
Originally Posted by Zanket
Were the particle able to pass unimpeded through the star, it would do so almost instantly in its frame (by its clock). (The particle’s minimum velocity while traversing the star is at the star’s surface).

GR, however, says that the proper time for the particle to pass through this star is dependent on the star’s mass. The more mass, the more time required, with no upper time limit.
Hm. Are you suggesting here that mass Lorentz transforms? I don't think that's right.
If by “here” you refer to the first sentence you quoted, then yes. The objects that length-contract in SR are masses.

Originally Posted by Destruct
I'm guessing that the bigger you make "local", the greater the discrepancy between a GR description and a SR description, and conversely the smaller we consider local the smaller the discrepancy. But isn't there always going to be some tiny difference, maybe at the billionth decimal place, except maybe at some infinitesimel limit of local?
[Edit: At least in cases involving gravity and acceleration.]
Yes, even when infinitesimally close. There is no difference, however, for a direct measurement.

Originally Posted by Destruct
In this case though, are the regions being traversed all contained within this tiny definition of "local"? Is it still valid to talk about covering distances if they are larger than what can safely be considered a flat space? For instance, from one side of the length contracted star to the other, can this be considered a region of flat space? I know it's a small span from the particle's pov, but is it small enough?
(I’m not sure if you’re addressing me…) The region surrounding the star need not be considered flat in the particle’s frame. See my “Sun Jul 03, 2005 2:38 am” post to Guitarist above.

<EDIT by HomoUniversalis: Compiled the four posts into one>

20. Originally Posted by Zanket
Are you suggesting that your point invalidates drawing B?
Nope. I'm saying that the muons are close enough to uniform motion to use it as proof for SR.

21. Originally Posted by Zanket
If by “here” you refer to the first sentence you quoted, then yes. The objects that length-contract in SR are masses.
Then this is where we part company. Mass is a Lorentz invariant. The "length-contraction of mass" makes no sort of sense.

Zanket: Here's your solution (I think). You won't like it as it's pretty trivial.

Although you rightly stress that SR is a subset of GR, Einstein explicitly states (in his 1916 synopsis) that SR only applies over infinitely small distances and/or times. Hence, as Destruct clearly saw, for high escape velocities, locally Minkowskian means pretty damn local.

Let's consider the regime where a = v (a is acceleration in the field, v is linear velocity). In this regime, let SR apply. Then your test particle and star have an equally valid claim to consider themselves at rest, any other scenario isn't SR and doesn't count here. So, if the star is length-contracted in your figure B, then it must be in your figure A.

Once we move out of the a = v regime SR no longer applies, by the definition of SR. When SR no longer applies, GR does, and explicitly refers to different rates of clocks at different points in the gravitational field, but, as I recall, makes no explicit reference to length contraction.

And for the reasons I hinted at earlier, even if it did, length contraction in no way implies change of mass. So the "transit time" argument is really a red herring.
Or have I totally missed your point?

<EDIT by HomoUniversalis: Compiled the two posts into one>

22. Zanket

Diagram B introduces length contraction for both objects yet diagram C introduces only one object length contracted and the other not.

23. Using the

Code:
```

<Quote goes here>

```
tags, you can quote multiple posts in one post. Please try to keep double posting to a minimum. The EDIT button is your friend .

Mr U

24. Originally Posted by HomoUniversalis
Please try to keep double posting to a minimum. The EDIT button is your friend ;).

Mr U
Ha! We're not allowed to think between posts?

25. Hehe. Silly rabbit. If you think of something else, and no one has posted since, use the edit button. I don't really mind an occasional double post if a thread has been forgotten, and you think of something new, but four posts in a row is a bit over the top.

Mr U

26. Originally Posted by HomoUniversalis
Hehe. Silly rabbit. If you think of something else, and no one has posted since, use the edit button. I don't really mind an occasional double post if a thread has been forgotten, and you think of something new, but four posts in a row is a bit over the top.

Mr U
Right, point taken, my friend. BUT: let's say I see an interesting new thread, and post back saying "interesting, lemme think about it". And then come up with a brilliant reply (my speciality!!??) I don't expect the originator to re-click on my post, if it's not showing a "new post" flag.

Maybe I'm wrong. I certainly don't open subs unless they have a new post flag.

Don't you just hate rainy Sundays?

27. Originally Posted by DEChengst
Nope. I'm saying that the muons are close enough to uniform motion to use it as proof for SR.
OK.

Originally Posted by Guitarist
Then this is where we part company. Mass is a Lorentz invariant. The "length-contraction of mass" makes no sort of sense.
That isn’t what I meant; I meant only that masses (objects) length-contract in SR, not that the object’s mass changes. Maybe rephrase your question.

Originally Posted by Guitarist
Although you rightly stress that SR is a subset of GR, Einstein explicitly states (in his 1916 synopsis) that SR only applies over infinitely small distances and/or times. Hence, as Destruct clearly saw, for high escape velocities, locally Minkowskian means pretty damn local.
Yes, what is local is arbitrary. If SR applied only in infinitesimal regions then the relativistic rocket equations, for example, would be invalid. In practice SR applies within a range throughout which the tidal force is negligible. As far as this refutation of general relativity goes, the star's velocity relative to the particle is directly measured. SR always applies then.

Originally Posted by Guitarist
Let's consider the regime where a = v (a is acceleration in the field, v is linear velocity). In this regime, let SR apply. Then your test particle and star have an equally valid claim to consider themselves at rest, any other scenario isn't SR and doesn't count here. So, if the star is length-contracted in your figure B, then it must be in your figure A.

Once we move out of the a = v regime SR no longer applies, by the definition of SR. When SR no longer applies, GR does, and explicitly refers to different rates of clocks at different points in the gravitational field, but, as I recall, makes no explicit reference to length contraction.

And for the reasons I hinted at earlier, even if it did, length contraction in no way implies change of mass. So the "transit time" argument is really a red herring.
Or have I totally missed your point?
Sorry, I don’t understand what you’re getting at here. Setting acceleration = velocity, and “move out of the a = v regime”, makes no sense to me. Maybe you can rephrase.

In A, the star is not length-contracted because the star is negligibly moving relative to the particle. In B the star is appreciably moving relative to the particle, so the star is drawn as length-contracted. Gravitational time dilation is not a factor, because the star’s velocity relative to the particle is directly measured.

Originally Posted by (Q)
Diagram B introduces length contraction for both objects yet diagram C introduces only one object length contracted and the other not.

Hi, (Q).

All drawings are from the particle’s perspective; i.e. if the star is length-contracted in the particle’s frame then the star is drawn as length-contracted. So in none of the drawings is the particle length-contracted. In drawing B the star is length-contracted, as determined by SR. In drawing C the star is not length-contracted, as determined by GR. If I missed your point, please rephrase.

28. Originally Posted by Zanket
The equivalence principle says that SR applies when the measurement of velocity is taken locally.
No. The equivalence principle, as usually understood, syas that the experience of acceleration and of gravity are indistinguishable. In either case, except in the infinitly small, SR most emphatically does not apply.
"It seems that the equivalence principle holds only for infinitely small fields..." (Annalen dur Physik, 38; 443 (1912). Equivalence has nothing to do with the equivalence of inertial and non-inertial frames, as you seem to suggest.
Originally Posted by Guitarist
Then this is where we part company. Mass is a Lorentz invariant. The "length-contraction of mass" makes no sort of sense.
That isn’t what I meant; I meant only that masses (objects) length-contract in SR, not that the object’s mass changes.
Ah! Mass is a measurement, a scalar. Have you been confused by your reading things like ".......let a mass be suspended, blah, blah" Truly, this is sloppy physicists speak. The apparent interchangeability of "mass" and " object", is only apparent. What they really mean is either a massive object, or an object with a defined mass m.

But when GR asserts that transit through a massive object is a function of its mass, it means a function of the Lorentz invariant scalar m.

Yes, what is local is arbitrary. If SR applied only in infinitesimal regions ...
Have I wasted my time? On a Reimannian manifold, if you want to be posh, Lorentz (i.e. linear) coordinate transformation is a strictly local effect. You simply cannot globalize it. Why do you think that GR was born?

the relativistic rocket equations, for example, would be invalid.
Why? Sufficiently distant from a field source, the energy-momentum tensor approaches zero, then, by the field equations, so does the curvature tensor and space-time by the definition is "flat-ish, and SR can approximate.
In practice SR applies within a range throughout which the tidal force is negligible.
Which is just what I very pretentiousy said!
Originally Posted by Guitarist
Once we move out of the a = v regime SR no longer applies, by the definition of SR.
Sorry, I don’t understand what you’re getting at here. Setting acceleration = velocity, and “move out of the a = v regime”, makes no sense to me. Maybe you can rephrase.
Not sure....um... when a noteq v then SR doesn't apply? OK. Set v = ds/dt, and a = d<sup>2</sup>s/dt<sup>2</sup>. Can you now see why a = v only in the infinitely small?

In A, the star is not length-contracted because the star is negligibly moving relative to the particle. In B the star is appreciably moving relative to the particle, so the star is drawn as length-contracted.
Now you're contradicting youself. If the particle is in free-fall in the neighbourhood of a star with an escape velocity near c, the star is moving relative to your test particle at precisely the same velocity as vice versa.
That's SR!

Look Z. Let's not fall out over this. It's possible we are talking past each other, but frankly I don't quite see your problem.
Gimme math if you want, that's cool with me.

29. Originally Posted by Guitarist
"It seems that the equivalence principle holds only for infinitely small fields..." (Annalen dur Physik, 38; 443 (1912).
Regarding this and other info you gave about the equivalence principle, to avoid going off on a tangent here, let me just say that the measurement of velocity in the refutation is taken directly. Then the equivalence principle allows SR to apply to this measurement.

Originally Posted by Guitarist
On a Reimannian manifold, if you want to be posh, Lorentz (i.e. linear) coordinate transformation is a strictly local effect. You simply cannot globalize it. Why do you think that GR was born?
I assume here that you disagree that the whole star is length-contracted in the particle’s frame in SR (drawing B). If so, then how do you explain that the whole Earth is length-contracted in the frame of a muon free-falling toward it, in the muon experiment? And how do you explain trans-galaxy travel in the lifetimes of a rocket’s crew, if the whole galaxy is not length-contracted in their frame? And if an object moving relative to you is only locally length-contracted, then, tell me, at what precise limit of range does the object cease being length-contracted? A measurement of velocity relative to a system (e.g. a star) must be done locally (as implied by the equivalence principle) so that the tidal force does not sully the measurement, but the resulting Lorentz coordinate transformation applies to the whole system regardless of the curvature of spacetime around it.

Originally Posted by Guitarist
Ah! Mass is a measurement, a scalar. Have you been confused by your reading things like ".......let a mass be suspended, blah, blah" Truly, this is sloppy physicists speak. The apparent interchangeability of "mass" and " object", is only apparent. What they really mean is either a massive object, or an object with a defined mass m.
I use the dictionary definition of mass also used by some of my relativity books. But that's a good point.

Originally Posted by Guitarist
Now you're contradicting youself. If the particle is in free-fall in the neighbourhood of a star with an escape velocity near c, the star is moving relative to your test particle at precisely the same velocity as vice versa.
That's SR!
When I say, “Let a particle free-fall from rest at infinity (or a great distance) to a star whose escape velocity at its surface is almost c”, there is no reference to drawing A. I am starting a new example there.

Originally Posted by Guitarist
Look Z. Let's not fall out over this. It's possible we are talking past each other, but frankly I don't quite see your problem.
Gimme math if you want, that's cool with me.
I think we are talking past each other a bit. I’m trying to grasp what part of the refutation you still disagree with. Do you disagree that the whole star is length-contracted in the particle’s frame in SR (drawing B)—and if so, is that why you are bringing up the equivalence principle? I gave validation for both drawings and showed that they should match. For SR (drawing B) the math is implied. For GR (drawing C) the validation includes the math. Anybody who will refute the refutation needs to show either that I drew them wrong (i.e. validation is faulty) or that they should not match. What I’d like to see is something like this: “Your validation of drawing C is wrong because…” or “B and C should not match because…”

30. I'll watch you guys duke out the math, it's not my forte, I'm more a broad principle kinda person.

The lesson I am taking away from this, so far, is that approximating GR with SR is not such a good idea when the space involved is quite obviously not very flat.

On a brief tangent, I've been thinking about this in relation to the twins paradox, Twin A flies away fast, turns around, comes back to Twin B, SR says they should both be older than each other from their respective POVs.

My usual mental solution is that because twin A turns around, even if we imagine he somehow otherwise always travels at a constant velocity this turnaround still shatters the symmetry and so SR cannot be applied. GR says Twin A is the one who experiences time dilation and both twins agree B is now older.

But if we start insisting that we can always approximate GR with SR and that all the entities and measures involved don't need to fit in some tiny span considered "local" (in space and time) then the twin paradox would be reborn, wouldn't it?

31. Originally Posted by Destruct
My usual mental solution is that because twin A turns around, even if we imagine he somehow otherwise always travels at a constant velocity this turnaround still shatters the symmetry and so SR cannot be applied. GR says Twin A is the one who experiences time dilation and both twins agree B is now older.
Although I don't normally debate the special theory, because you had some interesting thoughts on Zanket's nice little puzzle, here's a tip. First, the twin paradox is explicable in the special theory, and if you don't care for math, try this trick (it takes a while to explain, but it's easy to do:

Draw a pair of perpendicular coordinates, label the vertical axis "time", and the horizontal on "space". Call it reference frame A. Draw B moving in A (a line sloping the right). Now any theory of relativity asserts that B has a right to consider himeself at rest in his own frame. Superimpose that on A (it will be a pair of coordinates tilted relative to A).

OK? Now start again, but this time draw the so-called "world-line" of B (in A) going out and coming back. A dog-leg, right? There's nothing, nothing at all you can draw which will provide B with a coordinate set, a reference frame, which gives him the right to consider himself at rest during the journey i.e. the time dilation this time is a one-way street.

Happy drawing!

32. Guitarist:

Doesn't SR forbid changes in direction? [For fear of SR no longer being applicable.]

Draw a pair of perpendicular coordinates, label the vertical axis "time", and the horizontal on "space". Call it reference frame A. Draw B moving in A (a line sloping the right). Now any theory of relativity asserts that B has a right to consider himeself at rest in his own frame. Superimpose that on A (it will be a pair of coordinates tilted relative to A).

OK? Now start again, but this time draw the so-called "world-line" of B (in A) going out and coming back. A dog-leg, right? There's nothing, nothing at all you can draw which will provide B with a coordinate set, a reference frame, which gives him the right to consider himself at rest during the journey i.e. the time dilation this time is a one-way street.
If I replace A with B in the above description (as all observers have equivalent descriptions of physics) wouldn't this still lead to the classic twins paradox?

Not that I believe the twins paradox, but what makes you so sure that A was the one moving and not B? I'm saying that it is because A experienced an acceleration (actually three, the take off, the change in direction, and the return landing) and that this invalidates an SR description.

There are two observers, A and B, both percieve the other to suddenly move away, change direction, and come back. One will actually experience moments of acceleration at the take off, turn around point and landing, one will not. SR does not cover situations where observers experience acceleration. Therefore GR distinguishes either A or B as accelerating, whereas SR does not and is not valid as a description.

So if I don't say whether A is the twin on Earth, and B is the twin in the rocket (maybe A is in the rocket and B is on earth, maybe not) who would you say is older when they meet again?

To further clarify, how do you know that the rocket flies away and back to the Earth, and not that the Earth flies away and back to the rocket? I am saying it is because the observer in the rocket experiences acceleration. [Ignoring negligible orbital accelartion on the Earth. We can use satellites if this is a sticking point.]

33. Destruct:

You can use that argument if you must, I was simply trying to show that you don't need to.
Remember the axis tilting trick I showed? (you did try it, didn't you?). Let's make ourselves sound clever and give it fancy name. How d'you like "linear coordinate transformation"? Then, at the risk of slightly overstating my case, let me be dogmatic and say:

If there is a transformation that will bring B to rest in his own frame, A and B are never going to agree on who is moving, and all the usual guff about mutual time dilation, length contraction etc applies.

If there is no such single transformation, they must agree, and time dilation isn't mutual.

Any good?

34. You can use that argument if you must, I was simply trying to show that you don't need to.
As long as what I said isn't actually incorrect. My consideration was to why "there is no such single transformation" in the Twins paradox. I'm saying it's because the change in direction halfway through the expriment breaks the symmetry that SR requires.

35. Destruct: No, it's not wrong, it's just a more difficult argument to sustain. For, unless you're very careful you might find yourself implyng something like: the special theory requires mutual time dilation, the twin journey doesn't involve mutual time dilation and therefore is not in the special theory. Whish is at best circular, at worst nonsense.

There is (I think) a valid argument that says, as long as the total distance travelled is the same, it makes no odds what trajectory you take to get back to your starting point, so I am free to choose the circular orbit which is accelerated motion which, by the definition is not part of the special theory. And, in purely kinematic terms, anything I say about the circular orbit of necessity applies to any other equal length trajectory that brings you back.

36. Hey Zanket: Where are you?
Anyway. Here's your solution, and you'll like it even less than my last moronic attempt.

Leaving aside your drawings, let's state the general principle.

Any massive object, with its associated G-field, "carries" with it its own space-time geometry. In other words, there is no meaningful sense in which a G-source can be said to be moving at uniform relative velocity to a test particle in free fall in the G-field.
It is therefore meaningless to talk, at one and the same time, as a particle in free fall in a G-field and the relative inertial velocities of the two.

In short (I'm sorry to say, as you seem like a decent chap) the question was wrongly posed. There is no contradiction.

I hope this is clear, as I'm rushing off out.

37. Originally Posted by Guitarist
It is therefore meaningless to talk, at one and the same time, as a particle in free fall in a G-field and the relative inertial velocities of the two.
That denies the equivalence principle, and hence, if that were true, then all SR experiments would be invalidated, since SR would not be testable (a gravitational field exists everywhere). For example, the muon experiment would be worthless rather than the validation of special relativity that it is. The muon is free-falling in the experiment. Yet its velocity relative to the Earth is measured and SR is applied to the situation just fine. No, the equivalence principle implies that a directly-measured velocity can always be plugged in to SR, and the result is accurate (meaningful) at that moment.

(The result given by SR in the muon experiment, for the muon’s expected proper time to traverse between the Earth’s upper atmosphere and the ground, is an approximation, because the muon continues to gravitationally accelerate after its velocity is measured, but the approximation is accurate as a maximum time.)

38. Originally Posted by Zanket
That denies the equivalence principle, and hence, if that were true, then all SR experiments would be invalidated, since SR would not be testable (a gravitational field exists everywhere).
I had rather hoped we could agree to disagree what the equivalence principle is. But as it now seems to be your major tool, let me insist:

The equivalence principle, as enunciated by Einstein states simply that the experience of being in a gravitational field is indistinguishable from that of being accelerated in a field-free environment. It has nothing at all to do with the equivalence of inertial and non-inertial frames.

So leave that. If you want to discuss the regime in which the special and general theories meet, let's do it. But I have a sinking feeling that what you really want to do is tear down accepted science.

39. Originally Posted by Guitarist
[The equivalence principle] has nothing at all to do with the equivalence of inertial and non-inertial frames.
I have neither said nor implied that. The test particle’s frame is inertial, as is the frame of the muon in the muon experiment. Do you disagree that a free-falling test particle’s frame is inertial? That’s what the equivalence principle implies (it implies more than it says, as most any book on GR explains).

If you want to discuss the regime in which the special and general theories meet, let's do it. But I have a sinking feeling that what you really want to do is tear down accepted science.
You haven’t shown where I have misused either theory. I ask you to go back to my previous post and tell me how the muon experiment can be validly conducted if it is meaningless to talk about the muon’s velocity relative to the Earth, as you implied in your previous post.

40. Originally Posted by Zanket
Do you disagree that a free-falling test particle’s frame is inertial? That’s what the equivalence principle implies (it implies more than it says, as most any book on GR explains).
Ok, I see now where we are at cross-purpose. So, let's leave the muon out of it for now, and agree that one formulation of the equivalence principle states that a person in free-fall in the presence of a G-field has every right to consider himself and everything also in free-fall in a field-free environment. He can therefore claim that he is in Minkowski space-time.

Therefore he can perform experiments to prove that the special theory applies, provided, and only provided that his experiments are also in free-fall. Agreed?

Now let him look at the gravitational source. He sees it accelerating towards him, and therefore the special theory does not apply with respect to his relation to the source - it is not length-contracted.

Now to your figure "B". If your test particle is in free-fall in the presence of the star's field, it makes no sense to talk of the star "moving relative to the particle" The only relative motion allowed in that situation is that due to the field. (All gravitational sources are moving at high velocity anyway!).

Now to the muon (why do you call it an experiment, by the way? Surely it's merely an observation of a natural process - or is there something I don't know?).

The speed of the muon approaching Earth is sufficiently close to c and it's proper half-life so short that we can account for its apparent longer half-life in our frame using the special theory as a fair approximation. Just as, in our everyday life, we use Newtonian kinematics and dynamics as a sufficiently close approximation for most purposes.

So finally, I think the apparent incompatability between your originial figres lies in the fact fact that "B" is wrong.

41. Zanket

Einstein's equivalence princple does not hold good near self-gravitaing bodies like massive stars and black holes. Such bodies are associated with gravitational tidal forces that localize the equivalence principle over very small distances. This strong equivalence principle is different from Einstein's equivalence principle.

In short, the frame that is free falling on a BH would attain relativistic velocity near the BH and tidal forces (say if you free fall on the BH head down and as you approach nearer your head will have more G-force acting on it than your on your feet) come into play and space-time curvature is so sharp that SR would not hold good. The force gradient due to gravitational tidal force make the free falling body's frame non-inertial. SR gives way to GR.

42. I am not, course, responding on Zanket's behalf. But....
Originally Posted by everneo
Einstein's equivalence princple does not hold good near self-gravitaing bodies like massive stars and black holes.
As opposed, for example, to non self gravitating bodies?
.... that localize the equivalence principle over very small distances.
A "principle" is still a principle, even in the nanometer range. It has already been agreed in this thread that the greater the curvature of space-time, the smaller interval over which the equivalence principle holds.

SR gives way to GR.
Oh dear. This characterises a woeful misconcenception that pervades this forum. The special theory, as enunciated in 1905 effectively became redundant in 1915, when the general theory was born. "SR" from then on became a shorthand for the situation in the general theory where space-time is Minkowskian i.e "flat". Einstein showed in 1916 how to derive all the conclusions of the classic special theory from the field equations, under the assumption that space-time is Minkowskian.
In other words, is just a matter of what values one plugs into the GR metric, not "SR vs GR"

43. Originally Posted by Guitarist
A "principle" is still a principle, even in the nanometer range. It has already been agreed in this thread that the greater the curvature of space-time, the smaller interval over which the equivalence principle holds.
The scope of the particle's frame as being inertial should stop at that 'nano meter' level; talking about length contraction of BH that is beyond the 'horizon' of applicablity of einstein's equivalence principle in a steeply curved space-time is non-sense. right?

Originally Posted by Guitarist
Originally Posted by everneo
SR gives way to GR.
Oh dear. This characterises a woeful misconcenception that pervades this forum. The special theory, as enunciated in 1905 effectively became redundant in 1915, when the general theory was born. "SR" from then on became a shorthand for the situation in the general theory where space-time is Minkowskian i.e "flat". Einstein showed in 1916 how to derive all the conclusions of the classic special theory from the field equations, under the assumption that space-time is Minkowskian.
In other words, is just a matter of what values one plugs into the GR metric, not "SR vs GR"
I think you misunderstand. By 'SR gives way to GR', i meant the current problem falls under GR's scope as SR is not applicable here.
There is no inertial frames involved here that warrant SR.

44. Originally Posted by everneo
The scope of the particle's frame as being inertial should stop at that 'nano meter' level; talking about length contraction of BH that is beyond the 'horizon' of applicablity of einstein's equivalence principle in a steeply curved space-time is non-sense.
Er... I'm tempted to say yes, no, dunno. I don't understand this, at all. Duh

I think you misunderstand. By 'SR gives way to GR', i meant the current problem falls under GR's scope as SR is not applicable here.
There is no inertial frames involved here that warrant SR.
But you seem to be implying there are two different theories. There are not!
No. What I was trying to say, in my inarticulate way, is that the handle SR can be jettisoned (is that how you spell it?). Take the field equations of GR, insist that the metric is Minkowskian, and "pop" there's the conclusions of SR. No Lorentz transform, no quibbling about light speed, who sees what when and where. It just drops out so perfectly.

When Z. or anybody else says that that SR is a subset of GR, what they mean (I earnestly hope) is that the conclusions of the 1905 SR theory can be obtained in a staightforward way from the field equations of the 1915 GR theory

45. I have no objection if you call SR as GR in minkowski space. Agree? disagree? or agree&disagree???

46. Originally Posted by everneo
I have no objection if you call SR as GR in minkowski space. Agree? disagree? or agree&disagree???
Agree. Just don't give it a special name.

47. Originally Posted by Guitarist
Therefore he can perform experiments to prove that the special theory applies, provided, and only provided that his experiments are also in free-fall. Agreed?
If you replace “can perform experiments to prove that the special theory applies” with “can perform valid SR experiments”, and replace “his experiments are” with “his measuring equipment is”, then I agree.

Originally Posted by Guitarist
Now let him look at the gravitational source. He sees it accelerating towards him, and therefore the special theory does not apply with respect to his relation to the source - it is not length-contracted.
It is length-contracted, as it is in the muon experiment. The muon is analogous to your free-falling observer, and the length-contraction in the muon’s frame reduces the muon’s proper time to traverse the Earth’s atmosphere. The whole Earth is length-contracted in the muon’s frame. This experiment cannot be ignored. It is a confirmation of SR, and it refutes what you are saying.

This also doubles back to a previous post I made to you, on Mon Jul 04, 2005 1:08 am. I asked you questions that are tough to answer if what you say is true. You didn’t answer those questions.

In the frame of a free-falling observer, SR applies in a single moment for a locally measured (e.g. directly measured) velocity. In the muon’s case, the whole Earth is length-contracted in the muon’s frame to, say, x% of its proper length, and in the next moment the muon’s velocity relative to the Earth is increased and in that moment the Earth is length-contracted to less than x%. In each moment SR is accurate. What SR cannot do is give an exact value when a scenario involves gravitational acceleration over time. But for simple scenarios, such as for a free-falling observer, SR can be used to make an accurate prediction within a range.

Originally Posted by Guitarist
Now to your figure "B". If your test particle is in free-fall in the presence of the star's field, it makes no sense to talk of the star "moving relative to the particle" The only relative motion allowed in that situation is that due to the field. (All gravitational sources are moving at high velocity anyway!).
This denies SR, and the essence of relativity theory. If a star is hurled into you, you are surely going to feel its relative movement, the same as you would if you free-fell into it at the same impact velocity. It does not matter how the star got to you.

In the muon experiment, the muon hurls into the Earth. SR applies just fine.

Originally Posted by Guitarist
Now to the muon (why do you call it an experiment, by the way? Surely it's merely an observation of a natural process - or is there something I don't know?).
Google for “muon experiment”. An experiment can be an observation of a natural process.

Originally Posted by Guitarist
The speed of the muon approaching Earth is sufficiently close to c and it's proper half-life so short that we can account for its apparent longer half-life in our frame using the special theory as a fair approximation. Just as, in our everyday life, we use Newtonian kinematics and dynamics as a sufficiently close approximation for most purposes.
Yes, in the experiment SR gives an approximation of what GR should predict. But I showed in my original post, for an example similar to the muon experiment, that GR’s prediction is incompatible with SR’s. Using SR, a range can be obtained for the prediction, and GR’s prediction must fit within that range. But instead GR’s prediction may be greater than the greatest value in SR’s range by a factor up to infinity.

Originally Posted by everneo
In short, the frame that is free falling on a BH would attain relativistic velocity near the BH and tidal forces (say if you free fall on the BH head down and as you approach nearer your head will have more G-force acting on it than your on your feet) come into play and space-time curvature is so sharp that SR would not hold good.
SR applies in the frame of a free-falling observer anywhere above a singularity, so long as the velocity is directly measured, or measured in a region throughout which the tidal force is negligible. The tidal force on you near the horizon of a black hole can be less than the tidal force on you now, if the black hole is sufficiently massive. Note that the example in my original post is for a star, not a black hole.

Originally Posted by everneo
The force gradient due to gravitational tidal force make the free falling body's frame non-inertial. SR gives way to GR.
No, locally it’s always inertial, and SR applies. A test particle always fits in a region small enough that the tidal force within the region is negligible, even right above a singularity.

48. Originally Posted by Zanket
Originally Posted by everneo
In short, the frame that is free falling on a BH would attain relativistic velocity near the BH and tidal forces (say if you free fall on the BH head down and as you approach nearer your head will have more G-force acting on it than your on your feet) come into play and space-time curvature is so sharp that SR would not hold good.
SR applies in the frame of a free-falling observer anywhere above a singularity, so long as the velocity is directly measured, or measured in a region throughout which the tidal force is negligible.
So long as the velocity is directly measured it is also measured to be a increasing relative velocity. Accelerating massive star (from particle's frame), with tidal forces even over short distances around it, does not imply both frames are inertial.

Originally Posted by Zanket
The tidal force on you near the horizon of a black hole can be less than the tidal force on you now, if the black hole is sufficiently massive. Note that the example in my original post is for a star, not a black hole.
A massive star / BH has more acute curvature of space-time closer to it. What you are saying is in contrary to this.

Originally Posted by Zanket
Originally Posted by everneo
The force gradient due to gravitational tidal force make the free falling body's frame non-inertial. SR gives way to GR.
No, locally it’s always inertial, and SR applies. A test particle always fits in a region small enough that the tidal force within the region is negligible, even right above a singularity.
A test particle can fit in 'planck's length' too. Can you say then also it is inertial?! I find it more prudent to talk about the frame of test particle than the particle itself. In that frame there are umpteen number of 'distances' with different G-forces. You cannot find a 'free particle' moving freely in that frame with no net force acting on it. You can hardly call it an inertial frame.

49. Frankly, I'm starting to get impatient. I gave it my best shot, but let me correct one or two things that might help you.
Originally Posted by Zanket
It is length-contracted, as it is in the muon experiment. The muon is analogous to your free-falling observer,
No! The muon has it's own velocity (via E = mc<sup>2</sup>, the result of atomic disintegration) and is assumed to be insignificantly affected by the field. "Free-fall" refers to an object moving only as a result of the field.
and the length-contraction in the muon’s frame reduces the muon’s proper time to traverse the Earth’s atmosphere.
No! Time dilation and length contraction are not consequences of each other. No way.
In the frame of a free-falling observer, SR applies in a single moment for a locally measured (e.g. directly measured) velocity.
Once again, no. I showed you the correct way to think of it. It's true, I did toy with the idea in an earlier post that one may be able to select a regime where a = v, but I realised that, while logically true, didn't help in the present case.
In the muon’s case, the whole Earth is length-contracted in the muon’s frame to, say, x% of its proper length, and in the next moment the muon’s velocity relative to the Earth is increased and in that moment the Earth is length-contracted to less than x%. In each moment SR is accurate. What SR cannot do is give an exact value when a scenario involves gravitational acceleration over time. But for simple scenarios, such as for a free-falling observer, SR can be used to make an accurate prediction within a range.
I can make no sense of this.

If a star is hurled into you, you are surely going to feel its relative movement, the same as you would if you free-fell into it at the same impact velocity. It does not matter how the star got to you.
But you wanted to be in free-fall (or rather I asked you outright if the star had a field in "B", which amounts to the same thing), so, for the reasons I gave in my last, movement of the star and its associated field is undetectable by an object in free-fall.

. A test particle always fits in a region small enough that the tidal force within the region is negligible, even right above a singularity.
Sigh. Where do tidal forces come into it? You know what they are, I presume?

Look, I've tried to resist the feeling that your agenda is not really comprehension, but it's sneaking up on me. Shame really, as you seem like a decent chap.

Anyway, it's the end of the road for me, I'm afraid.

50. Originally Posted by everneo
So long as the velocity is directly measured it is also measured to be a increasing relative velocity. Accelerating massive star (from particle's frame), with tidal forces even over short distances around it, does not imply both frames are inertial.
SR applies moment by moment. The particle momentarily moves at constant velocity. That is how SR is used to integrate the relativistic rocket equations for a noninertially accelerating rocket.

Originally Posted by everneo
A massive star / BH has more acute curvature of space-time closer to it. What you are saying is in contrary to this.
No contradiction. There is no limit to how mild the tidal force on you can be at the horizon. How else can survival below a horizon be possible, as most any book on GR will tell you?

Originally Posted by everneo
A test particle can fit in 'planck's length' too. Can you say then also it is inertial?!
According to GR it is. That’s all that matters here. I’m arguing against GR, not reality or some other theory.

Originally Posted by everneo
I find it more prudent to talk about the frame of test particle than the particle itself. In that frame there are umpteen number of 'distances' with different G-forces. You cannot find a 'free particle' moving freely in that frame with no net force acting on it. You can hardly call it an inertial frame.
You can call it an inertial frame if it is in free fall and the tidal force throughout it is negligible. There is no upper size limit on such frame. For example, when the relativistic rocket equations are used to calculate the proper time required to traverse a galaxy, the inertial frame encloses the rocket’s path across the galaxy.

Originally Posted by Guitarist
No! The muon has it's own velocity (via E = mc<sup>2</sup>, the result of atomic disintegration) and is assumed to be insignificantly affected by the field.
So when the field gets stronger the length contraction observed by the muon suddenly disappears? You ignored those questions I posed, that would be difficult to answer if you were right.

Originally Posted by Guitarist
"Free-fall" refers to an object moving only as a result of the field.
I did not say or imply to the contrary.

Originally Posted by Guitarist
No! Time dilation and length contraction are not consequences of each other. No way.
I said it reduces its proper time. Less distance to travel at the same velocity, less time required—that’s not time dilation. Would it hurt you to become familiar with the muon experiment? It takes only a few minutes.

Originally Posted by Guitarist
Once again, no. I showed you the correct way to think of it.
Your way is unlike SR in any book I’ve seen. For example, “the result of atomic disintegration” as it relates to relative velocity is not in any book I’ve seen.

Originally Posted by Guitarist
But you wanted to be in free-fall (or rather I asked you outright if the star had a field in "B", which amounts to the same thing), so, for the reasons I gave in my last, movement of the star and its associated field is undetectable by an object in free-fall.
In my example, the movement is directly measured when the particle hits the star. The muon’s movement is relative to Earth’s upper atmosphere. Gravitational acceleration refers to the relative movement of a free-falling object. That movement can be measured. If the movement is measured in an inertial frame, SR applies to the velocity.

When you jump off a cliff, do you move relative to the cliff at undetectable velocity? Of course not. Then it is not meaningless to talk about relative movement in a gravitational field. Such is not implied when GR says that your frame is locally inertial—gravity undetectable within your frame—during your fall.

In drawing B, the star is hurled into the particle, striking it at an arbitrary velocity v, the same velocity at which the particle impacts the star in drawing C, having free-fallen to it. At impact those situations are identical.

Originally Posted by Guitarist
Where do tidal forces come into it?
They come into it as I stated.

Originally Posted by Guitarist
Look, I've tried to resist the feeling that your agenda is not really comprehension, but it's sneaking up on me. Shame really, as you seem like a decent chap.
With all due respect, it’s the other way round.

51. Originally Posted by Zanket
SR applies moment by moment. The particle momentarily moves at constant velocity. That is how SR is used to integrate the relativistic rocket equations for a noninertially accelerating rocket.
From one 'moment' to next 'moment' the velocity vlaue changes. The increase in velocity between such successive 'moments' is calculated as acceleration. (For example in a cicular motion the velocity vector at a moment differes the next moment. It is not inertial.) This resricts the scope of SR to moments. or to short distances. You need more generalized relativity, GR.

Originally Posted by Zanket
Originally Posted by everneo
A massive star / BH has more acute curvature of space-time closer to it. What you are saying is in contrary to this.
No contradiction. There is no limit to how mild the tidal force on you can be at the horizon. How else can survival below a horizon be possible, as most any book on GR will tell you?

Originally Posted by Zanket
Originally Posted by everneo
A test particle can fit in 'planck's length' too. Can you say then also it is inertial?!
According to GR it is. That’s all that matters here. I’m arguing against GR, not reality or some other theory.
Can you tell how it is according to GR?

Originally Posted by Zanket
Originally Posted by everneo
I find it more prudent to talk about the frame of test particle than the particle itself. In that frame there are umpteen number of 'distances' with different G-forces. You cannot find a 'free particle' moving freely in that frame with no net force acting on it. You can hardly call it an inertial frame.
You can call it an inertial frame if it is in free fall and the tidal force throughout it is negligible. There is no upper size limit on such frame. For example, when the relativistic rocket equations are used to calculate the proper time required to traverse a galaxy, the inertial frame encloses the rocket’s path across the galaxy.
For your star in question, you cannot assume the tidal force is negligible.

As for the rocket equations, it is a rocket in motion, its frame is not inertial as they calculate the acceleration due to rocket's motion.

52. Originally Posted by everneo
From one 'moment' to next 'moment' the velocity vlaue changes. The increase in velocity between such successive 'moments' is calculated as acceleration. (For example in a cicular motion the velocity vector at a moment differes the next moment. It is not inertial.) This resricts the scope of SR to moments. or to short distances. You need more generalized relativity, GR.
Then how do you explain that the relativistic rocket equations, for a noninertially accelerating rocket traversing, say, a galaxy, are equations of SR, not GR? Check out that link, including the sublink on accelerating clocks. SR applies in each moment, where velocity is constant; it applies in a momentarily comoving inertial frame. Those moments can be treated as increments in an integration, without invoking GR.

And let’s be clear that this is a side discussion. In the example in my original post, the only SR prediction I use is for a single moment, at the moment of the particle’s impact with the star.

Originally Posted by everneo
The star can be any typical size, or even house-sized, because the test particle is infinitesimally small. It always fits in an inertial frame. Think of it in terms of proportionality. When can you survive crossing a horizon? When the black hole is sufficiently bigger than you are.

Originally Posted by everneo
Can you tell how it is according to GR?
In GR, a test particle always fits in an inertial frame. An inertial frame can exist anywhere except a singularity. The equivalence principle implies that SR applies everywhere in infinitesimal regions in free fall, except a singularity (where not even GR applies).

Originally Posted by everneo
For your star in question, you cannot assume the tidal force is negligible.
I can, as above.

Originally Posted by everneo
As for the rocket equations, it is a rocket in motion, its frame is not inertial as they calculate the acceleration due to rocket's motion.
The frames used are inertial, as seen in the text of the link, and the text of the sublink on accelerating clocks. Successive momentarily comoving inertial frames are used to integrate the rocket’s overall movement, and the rocket’s entire trip is assumed to take place in an inertial frame (otherwise, if the rocket gets too close to a star, say, the curvature of spacetime around the star will throw off the results of the equations).

53. Originally Posted by Zanket

Then how do you explain that the relativistic rocket equations, for a noninertially accelerating rocket traversing, say, a galaxy, are equations of SR, not GR? Check out that link, including the sublink on accelerating clocks. SR applies in each moment, where velocity is constant; it applies in a momentarily comoving inertial frame. Those moments can be treated as increments in an integration, without invoking GR.

And let’s be clear that this is a side discussion. In the example in my original post, the only SR prediction I use is for a single moment, at the moment of the particle’s impact with the star.
You quoted the relativistic rocket equations applying SR in 'moments' in response to my post discussing free falling particle and 'accelerating' star (from particle's frame). SR's application in rocket equations has itw own justification different than that of free falling particle onto a star with tidal force. That led to this side discussion.

Originally Posted by Zanket
Originally Posted by everneo
The star can be any typical size, or even house-sized, because the test particle is infinitesimally small. It always fits in an inertial frame. Think of it in terms of proportionality. When can you survive crossing a horizon? When the black hole is sufficiently bigger than you are.
If a star is not a BH then the question of event horizon does not arise. The particle just heads for the surface of the star and faces more tidal force even within small distances; from the frame of the free falling particle the star is progressively accelerating towards it. the frames invloved are not inertial.

Originally Posted by Zanket

In GR, a test particle always fits in an inertial frame. An inertial frame can exist anywhere except a singularity.
I would like to know whether GR says that a test particle's frame is always inertial?

Originally Posted by Zanket
The frame is inertial, as you will see if you read the text in the link, and the text in the sublink on accelerating clocks. Successive momentarily comoving inertial frames are used to integrate the rocket’s overall movement, and the rocket’s entire trip is assumed to take place in an inertial frame (otherwise, if the rocket gets too close to a star, say, the curvature of spacetime around the star will throw off the results of the equations).
In the link the sentence before equations says "The relativistic equations for a rocket with constant positive acceleration a > 0 are : "

Do we have this advantage of equivalence principle in the case of 'free falling particle' in a changing g-field with, again, tidal forces?

54. Originally Posted by everneo
SR's application in rocket equations has itw own justification different than that of free falling particle onto a star with tidal force. That led to this side discussion.
For a radially free-falling test particle, the effect of the tidal force is just to change the acceleration from moment to moment. If you integrated the sum of the SR predictions for, say, the elapsed proper time it takes the particle to cross each infinitesimal radial segment at the constant velocity there, you should get the GR prediction returned by the equation I cited in my original post. (Not only does that make common sense given that SR is a subset of GR, but also the book Gravitation, section 6.2, which has the derivation of the relativistic rocket equations, states this in so many words.)

Originally Posted by everneo
If a star is not a BH then the question of event horizon does not arise. The particle just heads for the surface of the star and faces more tidal force even within small distances; from the frame of the free falling particle the star is progressively accelerating towards it. the frames invloved are not inertial.
The frame of a test particle is always inertial, whether it falls toward the star or a black hole. This is implied by the equivalence principle.

Originally Posted by everneo
I would like to know whether GR says that a test particle's frame is always inertial?
Yes, it is implied; for confirmation, google on “equivalence principle” or “principle of equivalence”.

Originally Posted by everneo
Do we have this advantage of equivalence principle in the case of 'free falling particle' in a changing g-field with, again, tidal forces?
Yes. The equivalence principle is the core postulate of GR. The principle applies everywhere except at a singularity.

55. Originally Posted by Zanket
Originally Posted by everneo
SR's application in rocket equations has itw own justification different than that of free falling particle onto a star with tidal force. That led to this side discussion.
For a radially free-falling test particle, the effect of the tidal force is just to change the acceleration from moment to moment. If you integrated the sum of the SR predictions for, say, the elapsed proper time it takes the particle to cross each infinitesimal radial segment at the constant velocity there, you should get the GR prediction returned by the equation I cited in my original post. (Not only does that make common sense given that SR is a subset of GR, but also the book Gravitation, section 6.2, which has the derivation of the relativistic rocket equations, states this in so many words.)
individual link cannot be called a chain. ...without links there cannot be a chain..

Originally Posted by Zanket
Originally Posted by everneo
If a star is not a BH then the question of event horizon does not arise. The particle just heads for the surface of the star and faces more tidal force even within small distances; from the frame of the free falling particle the star is progressively accelerating towards it. the frames invloved are not inertial.
The frame of a test particle is always inertial, whether it falls toward the star or a black hole. This is implied by the equivalence principle.

Originally Posted by everneo
I would like to know whether GR says that a test particle's frame is always inertial?
Yes, it is implied; for confirmation, google on “equivalence principle” or “principle of equivalence”.

Originally Posted by everneo
Do we have this advantage of equivalence principle in the case of 'free falling particle' in a changing g-field with, again, tidal forces?
Yes. The equivalence principle is the core postulate of GR. The principle applies everywhere except at a singularity.
the difference is in the scale. you can say that at any given 'moment' the particle and the star does not move at all, hence they are inertial.. As you like.

56. Originally Posted by everneo
individual link cannot be called a chain. ...without links there cannot be a chain..
Each “link” is a moment of constant (negligibly varying) velocity. The relativistic rocket equations, equations of SR, create a “chain” by integrating the links, where the relationship between links is a constant acceleration. GR creates chains by integrating links where the relationship between links is a variable acceleration.

Originally Posted by everneo
the difference is in the scale. you can say that at any given 'moment' the particle and the star does not move at all, hence they are inertial.. As you like.
Each moment is an infinitesimal span of proper time, not a zero span. Within the infinitesimal span, the particle or the star move at constant (negligibly varying) velocity. In each moment an object moves in lockstep with a momentarily comoving inertial frame, whether the object is a noninertially accelerating rocket or a falling particle.

57. Originally Posted by Zanket
Let a particle free-fall from rest at infinity (or a great distance) to a star whose escape velocity at its surface is almost c. The particle reaches the surface at almost c. Drawing B shows that the star will be length-contracted in the particle’s frame to almost zero length. Were the particle able to pass unimpeded through the star, it would do so almost instantly in its frame (by its clock). (The particle’s minimum velocity while traversing the star is at the star’s surface).
IF the particle's velocity is almost c at the surface then not only the star but the whole length in the direction of the particle's path is length-contracted to almost zero - where would the particle go after reaching the star?

58. Originally Posted by everneo
IF the particle's velocity is almost c at the surface then not only the star but the whole length in the direction of the particle's path is length-contracted to almost zero - where would the particle go after reaching the star?
Bingo. Since posting this thread I have learned that only a tiny fraction of people can think intuitively like you do here. Just try and get someone else to understand your question--they will argue essentially that in GR there is no length contraction beyond an infinitesimal region in which the velocity is measured. This can be countered, but it's too much work. For that reason, in a paper of mine, I decided to drop the argument as originally posted in this thread, and just stick with more of a mathematical argument that attacks GR from a different angle. That's in the "Flaw of General Relativity..." thread.

To answer your question, though, the axis of the particle's motion is contracted to infinitesimal length, but not zero length, so there's still an infinite distance in both directions to go to (consider that 0.000001% of infinity is still infinity). But, at the horizon of a black hole, which the particle would hit at c, the universe would be contracted to zero length.

59. To answer your question, though, the axis of the particle's motion is contracted to infinitesimal length, but not zero length, so there's still an infinite distance in both directions to go to (consider that 0.000001% of infinity is still infinity). But, at the horizon of a black hole, which the particle would hit at c, the universe would be contracted to zero length.
zanket, you choose to explain most gedankens with length contraction.
I prefer time dilation myself. Example: But, at the horizon of a black hole,
which the particle would hit at c, its clock would stop. With no change in temporal coordinates, the spatial coordinates of the particle would be fixed, an absolute frame of rest. Thus, the particle could never exit the event horizon according to a faraway observer.

60. zanket, your description of a particle crossing the event horizon of a black hole created a problem in my mind that I could not construct a solution for. I wondered if anyone else saw it too. I gave a hint in constructing my 'dilation based' explaination of the event, the assertion of an 'absolute' frame of rest. In your scenario, the universe is contracted to
zero length along the particles vector of travel in the particles frame of reference. The particle cannot continue to travel in a universe with zero length in its frame of reference. According to a faraway observer, the particle cannot continue to travel when time has stopped for the particle,
spatial coordinates are fixed with no, or an infintesimal, temporal coordinate. Putting both together, it would seem the particle is in the absolute frame of rest I alluded to from both the particles and the faraway observers frames. But that is not true according to current science. Evidence seems to indicate a black hole in the center of the Milky Way and many other galaxies. All galaxies in our local group are moving at very high rates of velocity towards a point near the center of the Virgo
cluster. The postulated black holes are traveling with the galaxies, are not fixed in spacetime. I said I was unaware of the solution, I did not state there was no solution. Does anyone here know the solution?

61. I know of the solution, but nothing substantive about it.

Apparently the Event Horizon is itself a singularity.

In the very fuzzy way I understand it, there are problems with trying to work GR maths at the event horizon, and you have to approachi it is a special way to get around those problems.

The Event Horizon singularity isn't like the central singularity. With the central singularity, there are always difficulties with applying the GR maths at that place - there are no ways around it. The Event Horizon singularity is removable - there are ways to do it.

Corrections are welcome.

62. 2inquisitive, I'm not sure whether you're talking about GR or some new theory. I don't want to mislead you.

GR says a particle's clock does not stop at the horizon. Nothing strange need happen there. Hypothetically we could have crossed a horizon while reading this (hopefully it would be a real long time before we hit the singularity). It's only from a faraway observer's perspective that the particle's clock gets ever slower as it ever more slowly approaches but never reaches the horizon. If the particle crosses the horizon on its clock at 5pm Sunday, the faraway observer sees the particle's clock ever more closely approach but never reach 5pm Sunday. This effect has to do with the fact that photons (that bring the clock image to the faraway observer) are dragged down by gravity and so move outward very slowly from just above the horizon, and move outward not at all from the horizon (there they stay in place).

63. GR says a particle's clock does not stop at the horizon. Nothing strange need happen there.
Here is my quote:
"But, at the horizon of a black hole,
which the particle would hit at c, its clock would stop. With no change in temporal coordinates, the spatial coordinates of the particle would be fixed, an absolute frame of rest. Thus, the particle could never exit the event horizon according to a faraway observer."

I admit that I was unclear about which frame of reference I was referring to until the last line. I thought you understood that clocks tick slow according to a faraway observer and distances contract according to the particles frame of reference. You do understand that, don't you?

This effect has to do with the fact that photons (that bring the clock image to the faraway observer) are dragged down by gravity and so move outward very slowly from just above the horizon, and move outward not at all from the horizon (there they stay in place).
Now I will have to ask you the same question you asked me: Are you speaking of General Relativity or some new theory? To begin with, I do
agree with what you said, that the photon would move very slowly from just above the horizon,etc. But relativity theory states the speed of the photon is not 'dragged down by gravity'. Relativity theory states the photon will move at c just above the horizon and is only highly redshifted by the gravity from the black hole. GR states the photon would be either
unable to escape the event horizon because the escape velocity would be greater than the speed of light at a certain point on the event horizon, or
if the photon were above this point on the event horizon, it would travel at c. I have stated long ago on sciforums that I believed that the photon would do just as you describe because a light clock would beat ever slower as it approached the event horizon relative to the light clock of a faraway observer, but that is not the way GR describes the situation.

64. Originally Posted by 2inquisitive
You do understand that, don't you?
Yes on both. That's why I mentioned the faraway observer.

Now I will have to ask you the same question you asked me: Are you speaking of General Relativity or some new theory?
All GR.

To begin with, I do
agree with what you said, that the photon would move very slowly from just above the horizon,etc. But relativity theory states the speed of the photon is not 'dragged down by gravity'. Relativity theory states the photon will move at c just above the horizon and is only highly redshifted by the gravity from the black hole. GR states the photon would be either
unable to escape the event horizon because the escape velocity would be greater than the speed of light at a certain point on the event horizon, or
if the photon were above this point on the event horizon, it would travel at c.
“'dragged down by gravity” is just a rough way of putting it. Talking only about outward moving photons here, the photon is only dragged down from the faraway observer’s perspective, in that the photon will take longer to get to the observer, and yes it will be redshifted for that observer. Any observer directly measuring the photon’s speed will measure it as c. Just above the horizon such observer is moving downward at near the speed of light relative to the horizon, whereas the photon is barely moving upward relative to the horizon. Every observer crosses the horizon at exactly c, because the escape velocity there is exactly the speed of light, so they can measure a photon there at c even though the photon does not move from the horizon. When you say “Relativity theory states the photon will move at c just above the horizon”, that’s misleading, because it doesn’t say relative to what. Relativity states that all observers directly measure a photon’s speed as c anywhere.

From Scharzschild Geometry:
Free-fall coordinates reveal that the Schwarzschild geometry looks like ordinary flat space, with the distinctive feature that space itself is flowing radially inwards at the Newtonian escape velocity [formula]. The infall velocity v passes the speed of light c at the horizon.

Picture space as flowing like a river into the black hole. Imagine light rays, photons, as canoes paddling fiercely in the current. Outside the horizon, photon-canoes paddling upstream can make way against the flow. But inside the horizon, the space river is flowing inward so fast that it beats all canoes, carrying them inevitably towards their ultimate fate, the central singularity.

Does the notion that space inside the horizon of a black hole falls faster than the speed of light violate Einstein's law that nothing can move faster than light? No. Einstein's law applies to the velocity of objects moving in spacetime as measured with respect to locally inertial frames. Here it is space itself that is moving.
Originally Posted by 2inquisitive
I have stated long ago on sciforums that I believed that the photon would do just as you describe because a light clock would beat ever slower as it approached the event horizon relative to the light clock of a faraway observer, but that is not the way GR describes the situation.
How do you think GR describes the situation?

65. When you say “Relativity theory states the photon will move at c just above the horizon”, that’s misleading, because it doesn’t say relative to what. Relativity states that all observers directly measure a photon’s speed as c anywhere.
I have seen GR interpreted in different ways, both in forums and at different websites, mostly university websites. There seem to be many
fiercely supportive of SR, and SR states the speed of light is constant for
ALL inertial frames of reference. They usually extend this to inertial frames in freefall and even to non-inertial frames. So when I mentioned a photon will move at c just above the event horizon according to relativity
theory, I was repeating what is often stated by many relativists. You would have to leave out the modifier 'directly' to suit those physicists.
So, above the event horizon, the photon will move at c, regardless if measured by an observer near the horizon or one at a faraway location.
As I said, I don't agree with this position.

 Bookmarks
##### Bookmarks
 Posting Permissions
 You may not post new threads You may not post replies You may not post attachments You may not edit your posts   BB code is On Smilies are On [IMG] code is On [VIDEO] code is On HTML code is Off Trackbacks are Off Pingbacks are Off Refbacks are On Terms of Use Agreement