Time dilation: Gravity force vs Gravitational potential

[Q-reeus]

Good stuff Schmelzer.

Yet you go on to contradict what you seemingly here approve of!

You are plotting the speed of light, which you typically refer to as gravitational potential. The force of gravity at some location depends on the gradient in the speed of light, which relates to the slope of the plot at that location. The curvature you can see in your plot relates to the tidal force, and to the Riemann curvature tensor. This is the "defining feature" of a gravitational field because without this curvature, your plot is flat and level, the speed of light is constant, and therefore light doesn't curve and your pencil doesn't fall down.

Ignoring a few fine points there, above need have nothing or only indirectly anything to do with relative time dilation at a given location. The definition of time dilation is equivalent, as explained in earlier posts, to the redshift. Either 'absolute' - as referring to observer at infinite distance, or relative, as between two different elevations. In either case, a function of the Newtonian gravitational potential, with no gradients involved whatsoever. As given e.g. in #9:
https://en.wikipedia.org/wiki/Gravi...rtant_features_of_gravitational_time_dilation

It isn't. The difference in the metric field at different spacetime points gives you the gravitational gradient.

Wrong. His statement you were there quoting was correct. The radial spatial gradient in the spatial component of metric field gives you approx minus twice the 'g' field. You have some personal interpretation of what a metric field is.

No problem with that.

Which therefore becomes your problem since what you quoted there was wrong. Flat spacetime is not always synonymous with zero redshift i.e. time dilation. Interior of hollow shell as chief example - already covered in earlier posts, but here's another article (See first comment by Richard Muller below main article):
https://www.quora.com/Why-is-gravity-inside-a-spherical-shell-considered-to-be-zero

Your last mixed quote, with you claiming time dilation and 'spatial energy density' are linked is just nonsense. In flat vacuum region, as in hollow interior of a matter shell, there is no 'energy density' owing to any influence of the surrounding shell. Just a depressed potential and in GR (standard Schwarzschild metric) just a depressed temporal metric component, with spatial metric completely undepressed. The latter a strange consequence of transverse spatial metric having no dependence on potential anywhere. But pursuing that can of worms is another topic.

 

[Schneibster]

How do you separate the two?

Gravitational potential is potential energy due to the gravity field of a large object nearby (Earth), only. It's a classical (non-relativistic) way of looking at things, that doesn't take into account other gravity fields (the Sun, the Moon, Jupiter, the cosmological constant, etc.) that might influence the curvature at that specific point (albeit only a little bit, but we're talking theory here). If that's not what Janus meant, it wasn't clear from the context. If the full curvature, not just the curvature due to Earth, what Janus meant, then you're correct, there is no separating them. But I didn't interpret it that way.

I do not see the point of difference between you two. Am I missing something?

Only Janus can answer that. Perhaps I interpreted it incorrectly.

 

[Schneibster]

If we have a hole at the centre of the Earth and the centre of the Moon (ignoring all rotations) with clocks in both. Both clocks in flat spacetime but not at the same potential.

Well, they're not really quite at the same potential. There's the Moon's influence on the Earth, the Earth's influence on the Moon, the Sun's influence on both (their distances from the Sun might be different), Jupiter's influence, the other planets, and the cosmological constant (which because of slight anisotropies might be different at the Earth's center and the Moon's center). All of these influences must be taken into account. In practical terms, we might not be able to measure them, but we're likelier to be able to at the centers of the Moon and Earth than elsewhere, since those locations are pretty flat. Remember that gravity cannot be shielded; it goes right through matter.

Also, it's important to remember that flat means flat; under SRT all inertial (i.e. flat) spacetime locations are, for the purposes of siting a frame there, equivalent and will show the same mechanics, i.e. clocks in flat spacetime will run the same, as long as it's really flat, and as long as they aren't moving relative to one another (you said "ignoring all rotations").

Now, in the real world, from each others' points of view, the two clocks would be moving; but they would do so in a periodic way, with equal and opposite accelerations as I pointed out above in the centrifuge experiments. So from the clock at the center of the Earth's frame, at one point on the Moon's orbit the clock at the center of the Moon would be accelerating in one fashion, and at the opposite end it would be accelerating in an equal but opposite direction, and this would be true all around the orbit, so it would cancel; and vice versa from the clock at the center of the Moon's frame. Given a whole number of rotations, there would ultimately be no net time dilation between them (except for the Sun's, Earth's, Moon's, planets', cosmological constant's, etc.'s influences).

All of this is of course complicated by the fact that the Moon's orbit is not a circle but an ellipse; but Kepler's laws of orbital mechanics show that this is immaterial because the speed of the Moon at perigee and apogee varies just enough to account for the difference in the Earth clock's frame, and vice versa for the perilune and apolune from the Moon clock's frame.

The net net of all of this is that they'd run the same (ignoring all those other minor influences).

 

[QuarkHead]

Exactly. It's hard to see how Quarkhead, last line in #16, could NOT see the obvious difference.............Yet there is no concession of error(s) from that poser - as expected.

If the "poser" you refer to is me, then I would ask you to desist from such remarks. If not, I would ask you to desist anyway.

But look, in your hurry to tell me I am wrong, you guys seem not to have read my post sufficiently carefully.

In it (my post, that is) I referred to a region of spacetime with a constant metric field which mandates a flat spacetime in that region and therefore constant clock rates again in that region. I explicitly used these terms in my post.

I never claimed that any 2 (or more) disconnected spacetime regions where the metric field is constant necessarily agreed on clock rates. Indeed, any thinking reader would assume that the term "constant" meant just that - the same throughout the region, but possibly different from the constant metric field throughout any other disconnected region

 

[Q-reeus]

Some (NOT referring to Quarkhead btw) just don't know when to quit. Throwing up chaff in the form of bringing in utterly extraneous issues like 'effect of the rest of the solar system' is nothing short of desperation tending to thread derailing. I'll leave the specific mopping-up operation to one or two others.

 

[Q-reeus]

If the "poser" you refer to is me, then I would ask you to desist from such remarks. If not, I would ask you to desist anyway.

Not you - I'd had expected the context of relevant passage in #18 would have made that clear, but understand a possible confusion since I deliberately avoided naming.

But look, in your hurry to tell me I am wrong, you guys seem not to have read my post sufficiently carefully.

In it (my post, that is) I referred to a region of spacetime with a constant metric field which mandates a flat spacetime in that region and therefore constant clock rates again in that region. I explicitly used these terms in my post.

No problem to that point - which is something I referred to specifically in #21 - although it was via quoted text by Farsight.

I never claimed that any 2 (or more) disconnected spacetime regions where the metric field is constant necessarily agreed on clock rates. Indeed, any thinking reader would assume that the term "constant" meant just that - the same throughout the region, but possibly different from the constant metric field throughout any other disconnected region

I will allow that's what you actually meant Quarkhead, but what I and others read in #16 had a different translation. Anyway, main thing is we seem to be on the same page. Not everyone obviously, but at least 4 posters and maybe one or two more. You are ok with that much?

 

[Schmelzer]

Gravitational potential is potential energy due to the gravity field of a large object nearby (Earth), only. It's a classical (non-relativistic) way of looking at things, that doesn't take into account other gravity fields (the Sun, the Moon, Jupiter, the cosmological constant, etc.) that might influence the curvature at that specific point (albeit only a little bit, but we're talking theory here).

Wrong. The Newtonian potential takes into account the gravity field of other bodies, like the Sun, the Moon, Jupiter and so on. This follows from the classical Newtonian formula for the potential

$$\Phi(x) = - \int G \frac{\rho(x')}{|x-x'|} d^3 x'$$

Of course, in my consideration with the hole in the center of the Earth, I have not considered these distortion, for simplicity. But the Newtonian potential takes them into account. As a consequence, the Newtonian approximation of GR, which simply uses the Newtonian potential to define the GR metric as

$$ds^2 = (1+2\Phi) dt^2 - (1-2\Phi)(ds^2+dy^2+dz^2)$$

also takes into account Sun, Jupiter and so on. It does not take into account the cosmological constant, but for the Solar system it is far too small to become relevant. So, the problem is not about classical vs. special relativistic vs. general relativistic. The Newtonian approximation of GR (which is not Newtonian theory, note, because in Newtonian theory there is no time dilation at all, in the Newtonian approximation of GR we have one) is, at least for the purposes discussed here in the forum, completely sufficient.

Of course, introducing some minor asymmetries would destroy the point of my thought experiment with a hole inside the Earth, because with such minor distortions there would be some minor amount of curvature even inside the hole. Nonetheless, this does not change the main point that curvature is irrelevant for time dilation. What defines time dilation is proper time of a clock

$$ \tau = \int \sqrt{g_{\mu\nu}(x) dx^\mu dx^\nu}$$

which for a stationary clock reduces to $\tau = \int g_{00} dt$ and in no way depends on second derivatives of the metric, which define curvature, but on the metric itself.

 

[Schneibster]

If the "poser" you refer to is me, then I would ask you to desist from such remarks. If not, I would ask you to desist anyway.

Personally I take such remarks as "going outside the box" because the poster doesn't have any answers for the points you made. YMMV.

 

[Layman]

General Relativity assumes an equivalence between gravity and acceleration, so being at the center of the Earth should have the same effect of someone spinning around in a circle. Then the person would only experience being pulled 1/2 of a g at the center of the Earth in all directions. Then time would most likely go by slower for someone on the surface.

 

[Schmelzer]

General Relativity assumes an equivalence between gravity and acceleration, so ....

This idea was what Einstein has named the best idea he has ever had. With some justification, because it has lead him to the discovery of GR, which was his greatest scientific success.

Nonetheless, in the theory which was the result of this idea, there is no such equivalence - or, more accurate, it is only an approximate one. An approximation which is correct only if one ignores curvature. Which is, of course, not really a good idea, given the role curvature plays in the Einstein equations of GR.

So it is not really a good idea, in particular for a layman, to derive something about GR from this idea.

 

[Farsight]

Hence Pete's midwife quote by Synge in in http://arxiv.org/abs/physics/0204044 :

"I have never been able to understand this principle…Does it mean that the
effects of a gravitational field are indistinguishable from the effects of an
observer’s acceleration? If so, it is false. In Einstein’s theory, either there is
a gravitational field or there is none, according as the Riemann tensor does
not or does vanish. This is an absolute property; it has nothing to do with any
observers world line … The Principle of Equivalence performed the essential
office of midwife at the birth of general relativity, but, as Einstein remarked,
the infant would never have gone beyond its long clothes had it not been for
Minkowski’s concept [of space-time geometry]. I suggest that the midwife be
buried with appropriate honours and the facts of absolute space-time faced."

 

[Layman]

This idea was what Einstein has named the best idea he has ever had. With some justification, because it has lead him to the discovery of GR, which was his greatest scientific success.

Nonetheless, in the theory which was the result of this idea, there is no such equivalence - or, more accurate, it is only an approximate one. An approximation which is correct only if one ignores curvature. Which is, of course, not really a good idea, given the role curvature plays in the Einstein equations of GR.

So it is not really a good idea, in particular for a layman, to derive something about GR from this idea.

The equivalence principal is said to be completely accurate in every right, in books designed for laymen if they mention it. Einstien uses it and mentions it here in this paper; http://www.nevis.columbia.edu/~zajc/acad/W3072/EinsteinWormhole.pdf , as equation 1.

Then the curvature at the center of the mass is almost zero, according the diagram which was posted earlier.

 

[Layman]

Even though this diagram of the curvature doesn't show it, a person in the center would be pulled outwards. Then it is in higher dimensional space, so it could be impossible to picture that. If someone spun around in a circle and their arms were lifted up from it, their fingers would be pulled more than their hands. It would be the same as in a center of a mass. The outside of their body would be pulled more. Then that metric would have to be considered, like in Einstein-Rosen's paper where they used this new metric to eliminate the singularity.

 

[Layman]

You would need to consider the tensor T_ik, which would be the fictitious force like Einstein used in equation 4 of the Einstein-Rosen paper. I think it would be about half a g.

 

[Schmelzer]

The equivalence principal is said to be completely accurate in every right in books designed for laymen if they mention it.

Yes. And there are several formulations of the principle which are indeed completely accurate. Note: Several. There is, in particular, the Weak Equivalence Principle (WEP), the Einstein Equivalence Principle (EEP) and the Strong Equivalence Principle (SEP).

Notably, these principles are different. And it may happen that for one theory, say, the EEP holds but the SEP not. My ether theory is an example of such a theory. I have no nice example to show the difference between WEP and EEP, nor do I have a simple explanation what makes the difference. For the difference between EEP and SEP the situation is easier: Here the difference is if only the equations for matter fields (EEP) or all equations (SEP) are independent of the choice of preferred coordinates. In GR, the SEP holds, together with the weaker EEP and WEP, in my ether theory only the EEP and the WEP holds, but not the SEP.

And you are completely wrong about equation (1) of the Einstein paper. It tells us that the principle does not hold exactly. Instead, it tells us that an accelerated system of coordinates gives (exactly) the metric (1). The next sentence tells us that in this case holds (2). Exactly. The curvature tensor would have to be zero, exactly. Which is possible only if there is a vacuum. (And, as shown by the Schwarzschild solution, it does not hold always even in the vacuum).

So, even the SEP (which holds in GR) does not mean that a gravitational field is exactly equivalent to acceleration.

Then the curvature at the center of the mass is almost zero, according the diagram which was posted earlier.

No. This is nothing you can reasonably see in such a picture. What you can see there is that in the center, the first derivative is zero. Which defines, in the Newtonian case, the gravitational force.

The curvature is essentially the second derivative. It can be nonzero at a minimum. And, if there is matter in the center, the Einstein equation $G_{mn} = T_{mn}$ tells us that, oncve $T_{mn}\neq 0 $ (there is matter), we have also $G_{mn} \neq 0$, thus, the curvature is nontrivial.

 

[Layman]

And you are completely wrong about equation (1) of the Einstein paper. It tells us that the principle does not hold exactly. Instead, it tells us that an accelerated system of coordinates gives (exactly) the metric (1). The next sentence tells us that in this case holds (2). Exactly. The curvature tensor would have to be zero, exactly. Which is possible only if there is a vacuum. (And, as shown by the Schwarzschild solution, it does not hold always even in the vacuum).

Equation 1 didn't work, because in that particular situation the gravitational pull at the center would be zero. That gives a zero in the denominator, which would result into infinities. Then that is not actually what would happen at the center of a mass. Then he modified the equation into equation 1a, on the same page of the paper. That equation only takes into account the centrifugal force. (The fictitious force that he mentions). Then the gravitational force actually isn't zero. Since acceleration and gravitation are equivalent, it makes it non-zero by adding the centrifugal force to it. Then the calculation can be done without getting division by zero.

Like the OP originally stated, the gravitational force would be zero at the center of the Earth. Then the problem wouldn't be able to be solved, because you would end up getting division by zero. Then they would feel like they are being pulled outwards, like the centrifugal force, so it actually isn't zero. It would actually be half of a g. Then you would need to solve it with equation 1a, from it having this similar problem of division by zero. Basically, this paper is saying that the centrifugal force is equivalent to gravity as well...

 

[Layman]

That paper is basically the reason why pop physics says you can create a wormhole out of a black hole from it having a large enough amount of spin. If the spin creates a force which is the same as the gravitational pull of the black hole, then it cancels out the force of gravity in the equations, and then it would create an open singularity. This could easily happen in nature due to the collisions of black holes. They rotate faster and faster around each other, and then it would add to the amount of spin it has after the collision. Then it would create a white hole, no anti-gravity required. That is just what Einsteins equations predict if you follow it to the letter.

 

[Layman]

That paper is also the reason why people have made predictions that it would take as much energy contained in the sun to create an electrical charge, which could create a wormhole or turn a black hole into a wormhole/white hole. That is what you should find if you do the maths from the paper where it also unifies electricity with the force of gravity.

 

[Q-reeus]

Personally I take such remarks as "going outside the box" because the poster doesn't have any answers for the points you made. YMMV.

Making up conveniently undefined 'points' I supposedly had 'no answers' for is both hilarious and THAT is truly "going outside the box". And a nakedly obvious cheap attempt at 'winning over' someone shrewdly deemed to be a potentially valuable 'ally'.With any luck, the shameless poser will now just exit the scene completely.

 

[QuarkHead]

Can someone please explain what YMMV means? Is it offensive? (yes, I know - I have led a sheltered life. For years I thought that Paris Hilton was a French hotel!)

Seriously though. Let's cut out the personal remarks and fighting - no purpose is served thereby, it wastes the site's bandwidth allowance and does not make interesting reading. While we're at it, let's keep links (and verbatim quotes therefrom) to a minimum - we all have internet access and a search engine (incidentally, I use StartPage which is free, cross-platform and as secure as you can get these days).

What do you say?