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General relativity is self-inconsistent


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#188 andrewgray

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Posted 04 October 2007 - 04:59 PM

I see that in the diagram. What is the significance? What about it would change the fact that the light pulse never reaches the hoverer (thus the definition of "black hole" is satisfied)?


Well, the r=2M coordinate trajectory is the limit of a sequence of constant accelerations, experienced by the curves R=constant or r=constant, which are seen in the Rindler diagram. In this particular diagram, the constant r trajectory sequence is {r=2.0000053M,r=2.0000032M,r=2.0000017.0025MM,r=2.0000006M} These have a constant acceleration sequence of {a=1/.0025M,a=1/.002M,1/.0015M,1/.001M,1/.0005M}.

So the r=2M trajectory is not a light geodesic. It is the limit of a sequence of constant accelerations. This r=2M sequence limit is shown in the Rindler diagram as a wedge shape trajectory. It looks like a less-than sign < .

Agreed?

And finally, since both the Schwarzschild and Rindler metrics are independent of time, we can make t=0 or T=0 be any arbitrary start time that we want.

Agreed?


Andrew

OK, great. Z. Ket it will be. Do you have an institution or company name that you want associated with you?

#189 Guest_Zanket_*

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Posted 04 October 2007 - 08:25 PM

So the r=2M trajectory is not a light geodesic. It is the limit of a sequence of constant accelerations. This r=2M sequence limit is shown in the Rindler diagram as a wedge shape trajectory. It looks like a less-than sign < .

Agreed?

I can't agree that the r=2M trajectory is not a light geodesic, because it can be shown (as I did above) that a photon could remain there indefinitely.

Why can't the r=2M trajectory be both a light geodesic and a limit of a sequence of constant accelerations?

And finally, since both the Schwarzschild and Rindler metrics are independent of time, we can make t=0 or T=0 be any arbitrary start time that we want.

Agreed?

I agree. It's like starting a stopwatch before the race starts.

Do you have an institution or company name that you want associated with you?

No.

#190 andrewgray

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Posted 04 October 2007 - 10:46 PM

Why can't the r=2M trajectory be both a light geodesic and a limit of a sequence of constant accelerations?


Well, the easiest way to understand this is in flat spacetime. Here, the following is universally true:

A constant acceleration frame changes direction in all inertial frames.


Are you with me?

Anyhow, look at the Rindler diagram. You said that you agree with it. The constant r (and constant R) coordinate trajectories all change direction at T=0. The r=2M trajectory is the limit of the the sequence of accelerations shown. So the r=2M trajectory changes direction as well. You either agree or disagree with the Rindler diagram. The Rindler diagram shows this clearly.


Andrew A. Gray

PS. I have your contact info. You probably should edit it out of the post.

#191 Qfwfq

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Posted 05 October 2007 - 02:32 AM

You’re arguing against yourself here, not me. Your contention has been that because SR requires infinitely small frames, SR does not apply in X.

Whaaaaaaaaaaaa?

Now you’re suggesting otherwise. Either SR cannot be confirmed in X, or it can. Which is it?

Goodness gracious, I even said it would be less simple to do so, where the higher order terms are more significant at shorter distances compared to our stomping grounds. If you think that means I said it can't be confirmed where gravitation is strong, learn English and get yourself a good calculus course before getting back to me.

Confirming SR at each point-event of a space-time region X does not remove the fact that, for the tangent spaces at each of two distinct point-events of X, the correspondence between vectors cannot have a spacelike one parallel with a timelike one. This is the main basis of why your paradox doesn't even make sense; it is based on misuse of the equivalence principle.

Seeing as how both of these predictions are those of GR, it would be GR’s contradiction, not mine. The contradiction is explained away in post #185 at “think about this”.

I notice now that you said "in the dropped object’s frame", which means not according to Schwarzschild coordinates. This is what removes the contradiction, the simple fact that the distance you mean is not a but instead decreased due to Lorentz transform. Big deal; you may as well say that, according to the dropped object, the EH and the hoverer approach each other. The same obviously holds for the light pulse as for the EH, given the one being on the other.