A simply-put flaw of general relativity

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zanket

10-04-06, 09:55 PM

superluminal

10-04-06, 10:06 PM

Fig. 1. An inconsistency of general relativity. The drawing at left represents a ball free to move within a relativistic rocket. The drawing at right represents a ball in free fall within a uniform gravitational field above a planet. Let the crew of the rocket traverse a chord of a galaxy, starting and finishing at rest with respect to it, in an arbitrarily short time on their clock (as allowed by the relativistic rocket equations). Then the ball, a loose object within the rocket, traverses it in an arbitrarily short time in their frame. According to general relativity’s equivalence principle, the crew experiences the equivalent of a uniform gravitational field. But also according to general relativity, the ball at right cannot traverse the uniform gravitational field (the box) in an arbitrarily short time in the tree’s frame, even in principle.* Then the rocket’s crew does not experience the equivalent of a uniform gravitational field, and the theory contradicts itself. It misapplies the equivalence principle.

The ball at left traverses the "box" in some arbitrarily short time. This is the same as saying the equivalent gravitational field can be arbitrarily strong, thus accelerating the ball on the right (i.e. traverse the "box") in an arbitrarily short time also.

Do the thought experiment at the event horizon of a black hole. This will be similar to the rocket accelerating at extreme g's.

zanket

10-05-06, 01:28 AM

The ball at left traverses the "box" in some arbitrarily short time. This is the same as saying the equivalent gravitational field can be arbitrarily strong, thus accelerating the ball on the right (i.e. traverse the "box") in an arbitrarily short time also.
I agree. The problem is, as the paper says, “But also according to general relativity, the ball at right cannot traverse the uniform gravitational field (the box) in an arbitrarily short time in the tree’s frame, even in principle.” In which case GR contradicts itself. Fig. 1 includes reasoning to show that given what you say here, the observable universe can be arbitrarily large at any given moment, a prediction GR precludes. Can you refute the reasoning?

superluminal

10-05-06, 07:35 PM

“But also according to general relativity, the ball at right cannot traverse the uniform gravitational field (the box) in an arbitrarily short time in the tree’s frame, even in principle.”
And why are the two examples not the same?

1) The ship can accelerate at a rate sufficient to traverse the box, not in an arbitrarily short time (I think you're mistaken about that) , but a time determined by the acceleration (uniform g field) and limited to the speed of light, in the box's FOR.

2) In the tree's FOR the ball can fall (traverse) under the same conditions as above, with the spacecraft acceleration replaced by the gravitational field.

Please explain where I have gone wrong.

zanket

10-06-06, 03:56 PM

1) The ship can accelerate at a rate sufficient to traverse the box, ...
I assume you mean the ball can do that, since the box on the left is the ship.

...not in an arbitrarily short time (I think you're mistaken about that) , but a time determined by the acceleration (uniform g field) and limited to the speed of light, in the box's FOR.
The greater the crew’s acceleration, the less time required in their frame for them to traverse a galaxy, say, according to the equations here ( http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html). In principle there’s no limit on the degree of acceleration, so the time can be arbitrarily short in their frame. Then the time for the ball to traverse the rocket (the box) can be arbitrarily short in their frame.

2) In the tree's FOR the ball can fall (traverse) under the same conditions as above, with the spacecraft acceleration replaced by the gravitational field.
Yes, according to the equivalence principle, which the paper does not dispute.

Please explain where I have gone wrong.
The paper shows that GR misapplies the equivalence principle. If what you say in (2) were valid in GR, then the last paragraph in fig. 1 shows that it is deducible that the observable universe can be arbitrarily large at any given moment. But GR precludes that prediction. Then it does not allow (2) to occur, even in principle.

Ophiolite

10-07-06, 07:45 AM

Zanket, my grasp of the logic and the mathematics to critique your hypothesis is inadequate. (This is why I never buy bridges from physicists.) However, on a human interest note, how high do you personally rate the odds that you are right and thousands of physicists and mathematicians are wrong?

zanket

10-07-06, 04:18 PM

Never got that question before! I’d say about 90% at this point. I’ve been arguing the full paper for a year with lots of people knowledgeable about the subject matter. The odds are helped by the fact that the inconsistency is subtle.

Ophiolite

10-08-06, 11:38 AM

Thank you for your reply. The inconsistency is too subtle for my mind. I wish you luck with it. I suspect relativity, general and special, will be overturned one day, just as Newton was. It is the nature of science that theories become discarded. Perhaps you may be the one to do it.
However, my natural scepticism leads me to believe that it is very unlikely that I have exchanged views on the internet with the person who future generations will know as the one who overthrew Einstein. For the moment, I'll stick with the thousands of scientists who seem happy with the current reading of things.

CANGAS

10-20-06, 02:46 AM

Sometimes, just for the novelty of it, a person could think things things through for themself, rather than gaze blankly and agreeably upon the herds of people who bleat out their mocking mimicry of each other's learned responses to the herd standard and respond innocently in agreement.

P.S. I have not, to my memory of the situation, heard O provide a view other than a very general disdain for anything other than the herd bleat.

So, how could O have had a chance to EXCHANGE views with the one who would provide correction to Einstein? EXCHANGE is a two way process.

Trilairian

10-30-06, 01:32 PM

This paper (http://zanket.home.att.net/intro.htm) is my latest attempt to show a flaw of general relativity (GR) in a simple way. Polite discussion / criticism / attempts to refute appreciated.

(The paper is an excerpt of another paper (http://zanket.home.att.net/) open for discussion in this thread (http://www.sciforums.com/showthread.php?t=47434), but not here please.)

There isn't a mistake in general relativity, there is only a bad terminology leading to your missunderstanding. The bad terminology isn't your fault, but your assumption that you are showing a mistake in GR without even looking at the actual math is your fault. The Riemannian spacetime curvature is expressed by the Riemann tensor. 
Rlmrn = Glmn,r - Glmr,n + GlsrGsmn - GlsnGsmr 
On the right appear products of the affine connection or Christoffel symbols
Glmn and the gradients of the affine connections Glmn,r.
The affine connections are what yield gravitational acceleration through the geodesic equation
d2xl/dt2 = - Glmn(dxm/dt)(dxn/dt)
They are therefor the analog of the Newtonian gravitational acceleration field. Tidal gravity then in a literal analogy would be the gradients Glmn,r.
Now consider going to a local free fall frame. The affine connections will vanish. But when there is Riemannian spacetime curvature the Riemann tensor isn't zero and being a tensor it can not be made to vanish, so the gradient terms corresponding to tidal gravity won't vanish either. So it isn't that spacetime curvature is equivalent to tidal gravity, bad terminology. It is that spacetime curvature is equivalent to tidal gravity IN FREE FALL. Now you go on to talk about an accelerating rocket in flat spacetime and how it experiences tides, but so what? You didn't bother to look at the math behind what he said to see that it would be irrelevent because it is not in free fall. In fact the math behind what he said says that there has to be tides for the ship because in flat spacetime the terms adding up to the Riemann tensor have to add up to zero, but its accelerated frame yield nonzero affine connetions trivially. 
To sum it up if your in curved spacetime *when you are in free fall* you will have tides and when you are in flat spacetime, whether you have tides depends directly on whether you are in free fall, and so it would have been correct for him to say spacetime curvature is equivalent to tidal gravity in free fall, but was wrong for you to assume something was wrong with GR without looking at the math behind what he said.

cosmodel

10-30-06, 03:44 PM

Again you present math formulas. I just talk about your second formula.
According to your understanding, on the left hand is acceleration. On the right hand is the constant connecttion at the spacetime point, times the generalized velocity. THEREFORE, ACCELERATION DEPENDS ON TEST PARTICLES` VELOCITY LOCALLY. Locally particles do not share common acceleration. No freely-falling frame locally which cancels gravity.
You used freely-falling frames frequently!!

GENERALLY IS UTTERLY NONESENCE!!!!!

Trilairian

10-30-06, 03:53 PM

THEREFORE, ACCELERATION DEPENDS ON TEST PARTICLES` VELOCITY LOCALLY.
Yes, so?
No freely-falling frame locally which cancels gravityIn local free fall all Glmn = 0 so the right hand side is zero no matter what the test particle's velocity is.
GENERALLY IS UTTERLY NONESENCE!!!!!
No, your statement without the support of a sound arguement is nonsence.

cosmodel

10-30-06, 04:11 PM

You do not know that the formula of accerelation given by yourself is 4-dimmensinal, and you do not know zero common velocity does not mean zero generalized velocity.

I, therefore, have a serious question for you. Where do you find the `bible` formulas and try hard to serve those sinful relativists, like the preacher Hawking?????

cosmodel

10-30-06, 04:17 PM

Trilairian

10-30-06, 04:25 PM

Sorry, I saw your connection as common velocity mistakenly.

You did not understand my commont. MY COMMON IS THAT NO FREELY-FALLING FRAMES AT ALL. And you say, in a freely-falling frame, acceleration is zero.

You need go to school for common logic!!!!

I aced logic in college. You are insane. If you have a real question ask it. If you want to continue to argue your insane "point" religously then we are done.

zanket

11-12-06, 02:28 AM

So it isn't that spacetime curvature is equivalent to tidal gravity, bad terminology. It is that spacetime curvature is equivalent to tidal gravity IN FREE FALL.
It’s the terminology of Thorne, Taylor, and Wheeler, not me.

Now you go on to talk about an accelerating rocket in flat spacetime and how it experiences tides, but so what?
I don’t mention that. Quote me. This covers the rest of your post.

Trilairian

11-12-06, 09:24 AM

It’s the terminology of Thorne, Taylor, and Wheeler, not me.
I didn't say it was you. Yes, I ment them.
I don’t mention that. Quote me. This covers the rest of your post.Yes you do. Your own figure one is sufficient for a quote.
http://zanket.home.att.net/intro.htm

zanket

11-12-06, 02:34 PM

Yes you do. Your own figure one is sufficient for a quote.
Nowhere in fig. 1 or elsewhere do I talk about how an accelerating rocket experiences a tidal force. Quote me.

Trilairian

11-12-06, 03:04 PM

Nowhere in fig. 1 or elsewhere do I talk about how an accelerating rocket experiences a tidal force. Quote me.A picture is worth a thousand words. If you didn't want the two to be equal you shouldn't have put the sign in between. Since you seem to be ignorant of the meaning of the figure which you cut and pasted to use without apparently meaning to actually convey what it meant because you didn't know what it meant then I will explain this aspect of it to you. I am telling you that the rocket on the left experiences just such a tide just like the earth lab on the right. No matter how uniform you make the field there will be a tide in both scenarios. The most natural frame for an accelerated observer yields the spacetime geometry represented in
ds2 = (1 + a×r/c2)2dct2 - ds 2
a = axex + ayey + azez where ax, ay, az are functions of time, and r = xex + yey + zez and ds 2 = dx2 + dy2 + dz2
This actually allows for arbitrarily time dependent accelerations in arbitrary directions, but as per this discussion take them in the z direction and constant to get
ds2 = (1 + az/c2)2dct2 - dx2 - dy2 - dz2
When you hold a test mass at constant position with respect to these coordinates the force felt by that mass will vary with respect to position along the z axis and will in fact have to go infinite at z = -c2/a where there is an event horizon. In the scenario with a mass's gravitation on the right there will also be tides and given a small enough mass, an event horizon. The difference is that in free fall the rocket frame's tides vanish along with its force as the Riemann tensor was zero the spacetime curvature was zero and so the sums of Christoffel symbols and their derivatives that go into it were always zero, but in the scenario on the right the tides don't vanish in free fall as the Riemann tensor was not zero, the spacetime curvature was not zero and so the Christoffel symbols and their derivatives can not both be made zero at the same time for any frame.
I am telling you that this is the only thing distinguishing the frames and as long as space and time intervals are kept small enough so that the Riemann tensor can not enter explicitly into any dynamical equations that there is no way to experimentally tell the difference, and that is the equivalence.

zanket

11-15-06, 09:48 PM

A picture is worth a thousand words. If you didn't want the two to be equal you shouldn't have put the sign in between.
The two situations are intended to be equivalent, as described in the text (and not by a sign in between, which doesn’t exist). Nothing suggests otherwise.

Since you seem to be ignorant of the meaning of the figure which you cut and pasted ...
No need for rudeness here.

I am telling you that this is the only thing distinguishing the frames and as long as space and time intervals are kept small enough so that the Riemann tensor can not enter explicitly into any dynamical equations that there is no way to experimentally tell the difference, and that is the equivalence.
Yes, according to the equivalence principle, which the paper does not dispute.

The paper shows that it can be inferred by means GR allows (and having nothing to do with the tidal force) that GR disagrees. You haven't addressed that, so you haven't stated a problem with the paper.

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