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 ops: Objection withdrawn.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)

"...because the logic of the lines traced from reality is as poor of aesthetic value as it is strict in consistency. " - Paolo Bozzi (Naive Physics - free translation)

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)

"...because the logic of the lines traced from reality is as poor of aesthetic value as it is strict in consistency. " - Paolo Bozzi (Naive Physics - free translation)

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:

The crew’s calculation of the length of the gantry’s space-time interval, in one year on the crew’s clock, is: s^2 = (vT)^2 - (cT)^2 = (0.7071 * 1 yr)^2 – (1 * 1)^2 = -0.5.

The gantry’s calculation of the length of the rocket’s space-time interval is: s^2 = (0)^2 - (ct)^2 = 0 – (1 * t)^2 = -t^2 = -0.5. Then t^2 = 0.5, and t = 0.7071. In the crew’s frame, in one year on their clock (T = 1), the gantry’s clock elapses 0.7071 years (t = 0.7071). That is correct. This example shows that the gantry’s clock runs slow in the rocket’s frame, and vice versa. It shows that the gantry's rods are length-contracted in the rocket's frame, and vice versa. This spacetime interval example is also compatible with the example in section 2.

The paper defines and uses inertial frames properly. Can you show where it does not?

Those are two different types of velocities. One is an escape velocity. The other is a directly-measured velocity. Above an event horizon, those can be the same. At and below an event horizon, they cannot be the same. Let a test particle free-fall from rest at infinity toward a black hole. Above the event horizon, it falls past a platform that is fixed at some altitude at the escape velocity at that altitude. At and below an event horizon, it does not, because no such platform can exist there; all material objects must fall there. That does not mean that the particle’s escape velocity is less than c there. Its escape velocity is still given by the GR escape velocity equation, which returns a value >= c there.

The reason that no velocity >=c can be directly measured at or below an event horizon is because GR’s escape velocity equation returns a velocity >= c there. The equation shows that, below the horizon, even light must fall. Then everything must fall there, and fall in such a way as to prevent a directly-measured velocity >= c. But (in Schwarzschild geometry) if the escape velocity equation is validly one that returns always less than c, as the paper shows, then no event horizon exists, because an event horizon is by definition where the escape velocity = c.

Thanks!

Yep, that’s me. I came up with the pinwheel paradox. I’m biased of course, but I think it’s a great story problem for learning the concept of excess radius in GR. The resolution of it is (spoiler alert) that the pinwheel gains an excess circumference as it accelerates. This is analogous to the excess radius in GR. Both excess circumference and excess radius are excess lengths or distances. The thread shows that the pinwheel paradox is Bell’s spaceship paradox at every small circumferential segment of the pinwheel. Section 4 of the paper explains excess radius using a free-falling test particle.

The black hole riddle was not a good riddle. It sounds like a riddle but there’s really no riddle there, as shown in the thread. So nothing insightful there.

And doesn't it seem pointlessly redundant, since we have already the first postulate?

The first postulate already says that the laws of physics have the same form, whatever the frame of reference, how does the second postulate imply that "the laws of motion for a uniform gravitational field are the same as the laws of motion for a relativistic rocket"?
Or do you mean "a uniformly accelerating frame of reference is equivalent to a frame of reference with a uniform gravitational field"?


So, you take the distance as measured in the frame of reference of the gantry, and divide it by the travel time as measured in the frame of reference of the crew.
How is that a physically significant quantity? You are using measurements from two different frames.

You set T = 1 y.
I set x = 1 ly, which was what you used for the length contraction:We are not comparing the same intervals.

I was using the interval between two events:
1. Rocket is leaving gantry;
2. Rocket passes the 1 ly signpost (distance measured in the gantry's frame).

Gantry's frame: distance = 1 ly, rocket's velocity = 0.7071c, travel time = 1.4142 y.

Rocket's frame: distance = 0.7071 ly, gantry's velocity = -0.7071c, travel time = 1.225 y (determined by space-time interval invariance).

__________________
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)

"...because the logic of the lines traced from reality is as poor of aesthetic value as it is strict in consistency. " - Paolo Bozzi (Naive Physics - free translation)

I don’t know about that. If it was redundant, it wouldn’t affect the paper.

The way the second postulate implies that is explained on tons of sites. The equivalence principle implies two key things:

- A local frame fixed to the surface of a mass is equivalent to a noninertial frame uniformly accelerating in flat spacetime (e.g. a relativistic rocket).
- A free-falling local frame is equivalent to an inertial frame in flat spacetime.

The two key implications can probably be restated in a thousand different yet valid ways, which is why you see so many different statements of the implications of the equivalence principle on the web and in books. The first implication is restated as the first sentence in section 1 of the paper (what you quoted above). The second implication is incorporated into the paper’s definition for an inertial frame.

Please don’t take this personally, but I don’t want to debate the validity of the sentence quoted above. Since there are so many sources that support it (like the link I gave you), if you still disagree, I’d rather agree to disagree.

It’s okay to do that. Look at eq. 8.7 in the paper, T = asinh(a * t) / a. If this equation is rearranged to become a = asinh(a * t) / T, it's still valid even though two different frames are represented on the right-hand side, with t and T, right? The formula for effective velocity is likewise used correctly as allowed by algebra.

This is correct.

This is incorrect. A distance of 0.7071 ly takes one year to traverse at 0.7071c. Doesn’t that make sense? For both the gantry and rocket, the travel time is proper distance / velocity. Something must be wrong with your calculations. If you can show me where you’re getting your formula from (I don’t recognize it), I’ll take a closer look.

Motion and laws of motion are technical terms in physics, and are not interchangeable.

The statement "the laws of motion for a uniform gravitational field are the same as the laws of motion for a relativistic rocket" is a consequence of the Principle of Relativity.
The statement "the motion of a mass in a uniform gravitational field is the same as in a uniformly accelerated frame" is a consequence of the Principle of Equivalence.

So, which is it?
I already asked you this:
So, you did not mean "laws of motion", but "motion".

If you don't want to discuss the validity of the statements in your paper, why did you start this thread?

Why? They are quantities measured in different frames of reference.

I was talking about the physical meaning of the quantity.

From what you say:
Gantry's frame: distance = 1 ly, rocket's velocity = 0.7071c, travel time = 1.4142 y;
Rocket's frame: distance = 0.7071 ly, gantry's velocity = -0.7071c, travel time = 1 y.

Testing the space-time interval invariance:
rhs: (1 ly)^2 - (c * 1.4142 y)^2 = - 1 ly^2;
lhs: (0.7071 ly)^2 - (c * 1 y)^2 = -0.5 ly^2.

So, where is the problem?

__________________
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)

"...because the logic of the lines traced from reality is as poor of aesthetic value as it is strict in consistency. " - Paolo Bozzi (Naive Physics - free translation)

From Wikipedia, on the topic “general relativity”: “The equivalence principle, which was the starting point for the development of general relativity, ended up being a consequence of the general principle of relativity and the principle that inertial motion is geodesic motion.)”

Then the statement “"the laws of motion for a uniform gravitational field are the same as the laws of motion for a relativistic rocket” is implied by both the equivalence principle and its superset, the general principle of relativity. Then the first sentence in the paper is fine. And you are right, the equivalence principle is redundant in the way you said.

That the motion is the same implies that the laws that predict that motion must be the same. Do you disagree?

To debate the validity of the new claims in the paper, not statements of existing theory that are there for reference. Those are debatable if they can’t be found in the texts, either explicitly or strongly implied. But IMO that statement is strongly implied by many sources.

A valid equation containing measurements from different frames can still be useful. Like how the equation in question was used in the paper to come to other conclusions.

I don’t recognize your formulas for space-time interval invariance. It differs from what I used from Wikipedia above, where I got the values I expected. From where are you getting those formulas?

Please see my next post, which affects our debate here regarding the “effective velocity” in the paper.

I’ve gotten a lot of flak for the section in the paper on “effective velocity”. Even though “effective velocity” is not an actual velocity, and no problem has been found with the related math, that section raises red flags. To forestall the issues from coming up again, I removed that section from the paper and reworked another section. The math is equivalent, save one less derivation. The conclusions are the same. Hopefully I’ve made the paper more convincing; time will tell. Here is the paper.

First, I would not use Wikipedia as a first reference.

Then, if we use the Wikipedia quote, saying that the Equivalence Principle implies (i.e. has as a consequence) that "the laws of motion for a uniform gravitational field are the same as the laws of motion for a relativistic rocket", would be still wrong, because the Equivalence Principle is the consequence and is implied by the statement.

No. The logical order in your paper is reversed.
The Equivalence Principle is a consequence of the statement.
And it is the statement which is redundant, because it is a special case of the Principle of Relativity.

If you want a statement which is a consequence of the Equivalence Principle, than you should use "motion of a mass".


You are wrong.
Newton's second law gives us the law of motion of a mass, but the motion of the mass depends on the force applied to it.
The motion without any force applied is different from the motion when a constant force is applied, and different if a non-constant force is applied.
In all cases the laws of motion is the same, but the motion differs.

This is the reason why "the laws of motion for a uniform gravitational field are the same as the laws of motion for a relativistic rocket" is not the same as "the motion of a mass in a uniform gravitational field is the same as in a uniformly accelerated frame".


And I showed you that your opinion is incorrect.
The statement you put in your paper is not implied by many sources.


I already asked you this: what is the physical meaning of that equation?
This is important especially within General Relativity.


You do not recognize the space-time interval:
(ds)^2 = (dx)^2 - (c*dT)^2 ?
It is invariant for inertial frames of reference (as your example).


From my textbooks (e.g. Eisberg & Resnick, Quantum Physics, which has a chapter on Special Relativity), and from here.


Our discussion is about more basic concepts than your "effective velocity".

__________________
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)

"...because the logic of the lines traced from reality is as poor of aesthetic value as it is strict in consistency. " - Paolo Bozzi (Naive Physics - free translation)

When the equivalence principle is the consequence of the general principle of relativity, the statement “the laws of motion for a uniform gravitational field...” is still implied by the equivalence principle. That the equivalence principle is a consequence of the general principle of relativity means that I can optionally replace “equivalence principle” with “general principle of relativity” in the statement.

That’s because those are three different experiments, so that’s beside the point. The point is, where the motion in any two identical experiments is the same, the laws that predict that motion are the same. Any experiment of motion performed in a uniform gravitational field will have the same result as when performed in a relativistic rocket having the same acceleration as the field. Then the laws of motion are the same in both situations. In either situation, three different experiments can yield three different results; that does not imply that the laws of motions are not the same in both situations.

My example was a rocket crossing 2 million light years in the gantry’s frame, in 5 minutes as the crew measures. So the equation in question is:

effective velocity = veff = 2 million light years / 5 minutes in years

I can’t think of any physical meaning it has. Doesn’t matter though, because it’s equivalent to:

veff = v * γ

And I can rearrange the relativistic rocket equations to get v * γ by itself on one side, so if v * γ is not physically meaningful then neither is it in SR, hence not in GR either, and then whether it’s physically meaningful is irrelevant. Algebra stills allows me to substitute veff for v * γ in the rearrangement.

No, I didn’t recognize it. As Wikipedia says, “Space-time intervals are difficult to imagine.” I ignore the spacetime interval and instead imagine what’s going on in SR situations using the far simpler equation, the Lorentz factor, that is derivable from the spacetime interval. I reviewed the spacetime interval. The example I gave above, using the spacetime interval from Wikipedia, is correct. (Your x is vT in that example.) You seem to be using it wrong; you are using it differently than Wikipedia does. Let’s go back to what you said before:

In Wikipedia’s usage of the spacetime interval, either the gantry or the rocket is taken to be stationary. Let the gantry be stationary. Set the values of the variables:

c = 1
x = 0 ly (the gantry is stationary)
t = 0.7071 yr (in the rocket's frame the gantry's clock runs at 70.71% of the normal rate)
x' = 0.7071 ly
t' = 1 yr

Then:

left: 0^2 – 0.7071^2 = -0.5
right: 0.7071^2 – 1^2 = -0.5

Then left = right.

No.
The statement with the laws of motion is a consequence of the Principle of Relativity.
The Equivalence Principle implies the satement with the motion of a mass.

No.
You forgot the postulate "that inertial motion is geodesic motion.": you quoted it from Wikipedia.


They relate to the Equivalence Principle: motion with no forces in a uniformly accelerated frame, is the same as in a stationary frame in a uniform gravitational field.

And this is the consequence of the Equivalence Principle.


Wrong.
In the first you have a force, in the second you have not, hence F=ma is different.
In the first case the acceleration is blamed on a force; in the second the acceleration is blamed on the frame.

If the motion is different, why would they have the laws of motion (within the same frame)?

If veff has no physical meaning, what physical meaning has an equation involving it?

Algebra is not physics.
Algebraic gymnastic does not necessarily give physically meaningful results.


I was talking about the formula (the algebra).


So you ignore the fundamental concept of space-time interval and its invariance in SR, in favor of your imagination.
Why should we take your algebraic gymnastic seriously?


As I suspected: you use a different space-time interval.
Now, for t = 0.7071 yr, the distance rocket-gantry in the gantry's frame is Dx = (0.7071c)*t = 0.5 ly; in the rocket's frame it is Dx' = gamma*Dx = 0.7071*0.5 ly = 0.3536 ly which is not the distance you got: x' = 0.7071 ly.
So, you are using positions and times for different events: no wonder there problems wih the invariance of space-time interavls.


So far the only flaws you have shown is in your understanding of basic concepts in Relativity.

__________________
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)

"...because the logic of the lines traced from reality is as poor of aesthetic value as it is strict in consistency. " - Paolo Bozzi (Naive Physics - free translation)

I leave the discussion at rudeness. You have shown no flaw in my understanding.

At the beginning of your web page you write:
The discussion in this thread brought to light your confusion about laws of motion and motion, and how these relate to the postulates of General Relativity.
Showing that you are wrong is not being rude.

__________________
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)

"...because the logic of the lines traced from reality is as poor of aesthetic value as it is strict in consistency. " - Paolo Bozzi (Naive Physics - free translation)

It is implicit in your assumption that you cannot exceed c. The constancy of the speed of light is a property of inertial frames of reference, i.e. the tangent spaces where space-time is locally flat. Ask yourself what kind of frame the free-falling observer is in.