Yes, I know that anyone who challenges GR is by definition a crackpot. I also know that a paper showing a flaw of GR is uninteresting even if valid, and that I need to fix the Kerr metric and predict gravitational waves too or else it's junk. (Just trying to forestall some typical comments.)

A Flaw of General Relativity, a Fix, and Cosmological Implications

Abstract: A flaw of general relativity is exposed and is shown to source from a misapplication of the equivalence principle, the theory’s core postulate. A replacement for the Schwarzschild metric is simply derived. (The vast majority of experimental tests of general relativity have been tests of the Schwarzschild metric.) The new metric is shown to be confirmed by experiments of the four classical tests of general relativity. The predictions of the new metric are shown to diverge from those of the Schwarzschild metric as gravity strengthens. The cosmological implications explain some observations simpler than do alternative explanations.

I have looked at your webpage and have found one significant error. You are missing several factors of two in the Schwarzschild metric. The Schwarzschild metric (in your notation) is:

dT^2 = (((r - 2* R) / r) * dt^2) - (dr^2 / ((r - 2 * R) / r)) - (r^2 * do^2).

Somehow you would need to replace (r - 2 * R) / r with r / (r + 2 * R) in order for your theory to give results similar to Einstein's in the first post-Newtonian order.

__________________
Microsoft is over if you want it.

The bar has been lowered for the promotion of ATM ideas; the bar for the acceptance of ATM ideas must remain high.

Thanks for your input. Eqs. 5.3 and 5.4 are the new metric for Schwarzschild geometry. In the Schwarzschild metric (the equations for which are not in the paper) the curvature factor is 1 – (R / r), like eq. 5.1, and as it is at Schwarzschild Geometry. Note that, in the Conventions Herein section, R = 2 * M.

I didn't see that! Sorry! ops: Objection withdrawn.

Tonight I will pull out my trusty copy of Theory and Experiment by Clifford M. Will and see what the PPN (parametrized post-Newtonian) parameters of your metric are. If your metric has values of the PPN parameters beta and gamma that are close to the GR values of beta=1 and gamma=1 then your "new" metric is viable.

__________________
Microsoft is over if you want it.

The bar has been lowered for the promotion of ATM ideas; the bar for the acceptance of ATM ideas must remain high.

I profess that I don't understand that, but it sounds like a good test. Thanks in advance for that. If I understand the gist of it right, your test verifies that the new metric approximates the Schwarzschild metric. Of course it need only approximate it in weak gravity, the only case where the Schwarzschild metric has been experimentally tested. As shown in section 6, the new metric matches all experimental tests of the Schwarzschild metric to date to all significant digits. So it should pass such a test.

Can you show us how this implication works?

Why?

I thought time was the problem.
Isn't it sufficient that the relativistic rocket's speed is close enough to c, so that the travel time measured by the crew is less than their lifetimes?

What happens to the time?

0.7071 ly is the distance as measured by the crew.
The speed of the rocket as measured by the crew is zero.
How do you get one year travel time, and in what frame of reference is this time measured?

In the frame of reference of the gantry, wouldn't they take ~1.4142 years to cover 1 ly at 0.7071c speed?
Wouldn't it take ~0.293 years from the point of view of the crew?
(Of course I could have made mistakes in my calculations.)

How did you obtain those numbers?

__________________
papageno


"Why waste time learning, when ignorance is instantaneous?" - Hobbes (Calvin and Hobbes)

"It's all about context!" - Vince Noir (The Mighty Boosh)

"I've never heard of such a brutal and shocking injustice that I cared so little about!" - Zapp Brannigan (Futurama)

I'm trying to understand what you believe the flaw to be. As far as I can make out, it seems to be the notion that an object free-falling in a sufficiently high gravitational field will reach a relative velocity (relative to where it started) greater than c. Is this correct?

There’s a ton of explanations for this on the web. Just google for “equivalence principle” or “principle of equivalence”. For example, this site has some mpeg movies showing what you quoted from the paper.

Because another galaxy is further away in light years than the maximum years in a human lifetime, and the crew cannot attain or exceed the speed of light.

Length contraction is what reduces their proper time to another galaxy, as described in section 2. Instead of traversing 2 million light years, say, to get to another galaxy, they might traverse only 20 light years. The 2 million light years reduced to 20 light years thanks to length contraction.

The crew also observes time dilation, to the same percentage. Passing clocks run at about 70.71% of the rate of the ship’s clocks, as the crew measures. Length contraction and time dilation go hand-in-hand in relativity.

The “as they measured” indicates the crew’s frame. The velocity of the rocket relative to the crew is zero, yes, but the velocity that’s being used here is the rocket’s (or the crew’s) velocity relative to the galaxies in question. That velocity is 0.7071c. At 0.7071c they traverse every formerly-measured light year, now length-contracted to 0.7071 light years, every proper year (a year on their clock).

In the gantry’s frame, yes, it’ll take the rocket (1 light year / 0.7071c) = ~1.4142 years to traverse a light year. How did you get ~0.293 years for the crew’s frame?

In the crew’s frame, moving at 0.7071c relative to the galaxies in question, one formerly-measured light year (that is, before they began accelerating, or in the gantry’s frame) is contracted to a percentage given by eq. 8.13, = sqrt(1 – (0.7071c)^2) = ~0.7071. 1 formerly-measured light year * 0.7071 = 0.7071 light years. The crew traverse 0.7071 light years in (0.7071 light years / 0.7071c) = 1 proper year (a year on their clock).

Yes. The escape velocity equation in GR is the same as Newton’s. Escape velocity at a given altitude is the same as the free-fall velocity from rest at infinity there. Newton/GR’s escape velocity equation allows a velocity greater than c. Section 1 in the paper shows that escape velocity should always be less than c, since free-fall velocity in a uniform gravitational field approaches a limit of c, and any gravitational field is locally uniform (a nonuniform gravitational field comprises uniform gravitational fields). (GR does not allow a directly-measured velocity to be c or greater. At the event horizon of a black hole, where the escape velocity is c, everything must be falling, preventing a directly-measured velocity to be c or greater. But that doesn’t change the fact that escape velocity should approach a limit of c.) The paper has proven the flaw of GR by this sentence in section 4: “Eq. 4.2 is invalid because it is derived from Newton’s invalid escape velocity equation (eq. 3.1), which general relativity shares.” Eq. 4.2 is Einstein’s.

Those two statements seem contradictory. What am I missing?

The Theory of General Relativity is based on two postulates:
1. General principle of Relativity;
2. Principle of Equivalence,

Now, "the laws of motion for a uniform gravitational field are the same as the laws of motion for a relativistic rocket" looks like a consequence of the first postulate.
The link you gave talks about the consequence of the second postulate.

Can you explain how the Equivalence Principle implies that "the laws of motion for a uniform gravitational field are the same as the laws of motion for a relativistic rocket"?


Why does it have to exceed the speed of light? (See below.)

So, you are looking at it from the point of view of the crew.
Again, why does the crew need to exceed the speed of light?

So, how do yo exceed the speed of light?

I got confused in the calculations.
I should be more careful.

Hang on.
Shouldn't you get:
[x^2 - (ct)^2](gantry's frame) = [x'^2 - (ct')^2](rocket's frame) ?

left: (1 ly)^2 - (c * x/v)^2 = {1 - (1/0.7071)^2 } ly^2 = 1 ly^2,

right: (0.7071 ly)^2 - (c * 1 y)^2 = 0.5 ly^2;

I don't think you are comparing the correct intervals.

If we assume that left equals right, we obtain t' = 1.225 y, the time interval to pass each signpost at one lightyear separation (in gantry's frame).

I hope posters with more experience in Relativity can help.

__________________
papageno


"Why waste time learning, when ignorance is instantaneous?" - Hobbes (Calvin and Hobbes)

"It's all about context!" - Vince Noir (The Mighty Boosh)

"I've never heard of such a brutal and shocking injustice that I cared so little about!" - Zapp Brannigan (Futurama)

I think that the problem here is that in general in GR, you can't define a global inertial reference frame. I suspect that it is this that is leading to what Zanket considers to be a flaw.

You are missing that, at and below an event horizon, the escape velocity is c or greater (even if that cannot be directly measured). Section 1 shows that the escape velocity should be less than c everywhere. When escape velocity is less than c everywhere, there is no event horizon.

No, I'm talking about the contradiction in your two statements that I've now highlighted in red. In one you say allow, the other not allow.

Welcome to BadAstronomy Zanket!

I'm curious about the history of your idea; would you care to comment on The Pinwheel Paradox and black hole riddle, which are two threads in Physics Forum's Special & General Relativity section, started by a member with the name "Zanket"?

The statement from the paper is a consequence of the second postulate.

You watched the movies? The second one shows that the motion (path) of the light beam in a uniform gravitational field is the same as the motion of the light beam in the uniformly noninertially accelerating rocket (a relativistic rocket).

The crew has to effectively exceed the speed of light. In principle they could traverse between the Milky Way and the Andromeda galaxy in 5 minutes on their clock. That galaxy is 2 million light years away as we measure it. Light takes light 2 million years on our clocks to cross that gulf. They can’t get there in 5 minutes without effectively exceeding the speed of light. The word “effectively” is being properly used here. If they get there in 5 minutes on their clock, did they exceed the speed of light? No. Did they effectively exceed it? Yes.

That looks strange to me. Wikipedia has: