My problem with the relativity representation on gravity.

[A.T.]

Isn't it a bit misleading for all of the video's captions to refer to "space" as being curved by gravity? After all, it is not objects which are distorted or bent by the presence of the planet, but space-TIME which is curved, inclining freely *moving* objects to veer toward the Earth unless held aloft.


"...space all around an object is bent...space is bent toward the objects..."

The captions are correct. And that's why this pictures are just as useless as the rubber sheet: it shows ony spatial distortion. Making it 3D doesn't make it better. Eventually worse. I doubt that those distorted 3D grids have anything to do with the spatial Schwarshild metric. Probably just some artists vague idea of distortion.

But the main problem is, they omit the time dimension.

 

[sophiecentaur]

The captions are correct. And that's why this pictures are just as useless as the rubber sheet: it shows ony spatial distortion. Making it 3D doesn't make it better. Eventually worse. I doubt that those distorted 3D grids have anything to do with the spatial Schwarshild metric. Probably just some artists vague idea of distortion.

But the main problem is, they omit the time dimension.

Perhaps a two dimensional projection of a 2D space plus time? Can't get my brain around what it could look like but we can handle 2D presentation of 3D images fairly well.

 

[A.T.]

Perhaps a two dimensional projection of a 2D space plus time? Can't get my brain around what it could look like but we can handle 2D presentation of 3D images fairly well.

For radial fall you just need 1 spatial dimension. That is what is used in the links i posted before. If you want 2+1 curved space time, you need two diagrams.

 

[ikjyotsingh]

I just want to know why everyone here is interested in visualizing spacetime?

 

[Dale]

Spacetime is fairly important in GR.

 

[ikjyotsingh]

Umm yeah

Spacetime is fairly important in GR.

Spacetime is fairly important in GR.

Umm yeah, but visualizing it is such a feeble task. You can only visualize it where an appropriate embedding exists, which according to my memory only exists for Bianchi I, V, and IX models, in addition to Schwarzschild family of metrics.

Why don't you talk about visualizing it in terms of how a pseudo-Riemannian manifold is constructed as a Hausdorff atlas of charts rather than all this diagram nonsense. At least the atlas of charts is correct both mathematically and physically, and provides an interpretation for which there can be no confusion.

 

[1977ub]

The captions are correct.

Are you saying that "space is bent" near the planet?

 

[WannabeNewton]

Why don't you talk about visualizing it in terms of how a pseudo-Riemannian manifold is constructed as a Hausdorff atlas of charts rather than all this diagram nonsense. At least the atlas of charts is correct both mathematically and physically, and provides an interpretation for which there can be no confusion.

This doesn't help you visualize curvature, it is a standard textbook depiction of the requirement of diffeomorphic transitions between overlapping coordinate charts of a maximal atlas. It doesn't help one visualize the physics of relativity in curved space - time in any way. Note however that I'm not saying one MUST have a way of visualizing the physics in GR; it is silly to ask for such visualizations in full generality because we are talking about pseudo - riemannian 4 - manifolds and of course we can't visualize such things innately.

 

[A.T.]

Are you saying that "space is bent" near the planet?

http://en.wikipedia.org/wiki/Schwarzschild_metric#Flamm.27s_paraboloid

But this is not what describes gravity. It describes some of the differences between Newton and GR. To understand how Newtons gravity is modeled in GR you have to include the time dimension. See the links I posted on page 1.

 

[wacki]

The best visualisation of curved space-time I came across are the three papers in

https://www.physicsforums.com/showthread.php?t=381683

see post #7 (from A.T.)

they are mathematically sound (at least for me) and it gave me a bit of intuition.
I think one can't hope for more than what's presented in those papers.

 

[Passionflower]

The spatial curvature is irrelevant to objects at rest in space, like an apple that starts to fall.

What do you mean by 'an object at rest in space'?

 

[pervect]

What do you mean by 'an object at rest in space'?

Wald has a definition on pg 288 of "General Relativity"

In a static spacetime the notion of "staying in place" is well defined; it means following an orbit of the Killing field \xi^a[\itex].<br />

<br /> <br /> It might be a good idea to add that the Killing field is hypersurface orthogonal, but I didn&#039;t see Wald mention that explicitly

 

[WannabeNewton]

It might be a good idea to add that the Killing field is hypersurface orthogonal, but I didn't see Wald mention that explicitly

Page 119. To paraphrase, a space - time $(M,g_{ab})$ is stationary if there exists a one parameter group of isometries on the space - time such that the orbits of the group action are time - like curves. Furthermore, the space - time is static if there exists a space - like hypersurface orthogonal to the orbits. By Frobenius' theorem this is equivalent to the condition that $\xi _{[c}\triangledown _{b}\xi _{a}] = 0$ where $\xi ^{a}$ is the time - like killing vector field. Of course if we had coordinates $(t,x^1,x^2,x^3)$ on some region of this static space - time, with the coordinate vector field of the first coordinate having the usual interpretation as a "time direction" of this coordinate system, then the more intuitive notion of hypersurface orthogonal would be that $\xi ^{a}\propto \triangledown ^{a}t$ which leads to the other condition anyways via a computation.

 

[Passionflower]

Wald has a definition on pg 288 of "General Relativity"

Where does he talk about something 'at rest in space'?

Wald is talking about an object in a static asymptotically flat vacuum space-time that maintains a fixed coordinate distance from the Schwazschild radius. That is obviously not the same as 'at rest in space'.

In relativity objects simply have no velocity wrt space.

 

[WannabeNewton]

A particle is at rest (stationary) in a static space - time if it follows an orbit of the space - like hypersurface orthogonal time - like killing field i.e. $u^{a} = \frac{\xi ^{a}}{(-\xi^{b} \xi _{b})^{1/2}}$ where as noted above, $\xi ^{a}$ satisfies $\xi _{[c}\triangledown _{b}\xi _{a]} = 0$ and $u^{a}$ is the 4 - velocity of the particle. Simple as that. This is more general than the special case of the Schwarzschild space - time so I have no idea why you are even bringing that up.

 

[Passionflower]

A particle is at rest (stationary)...

Stationary is not the same as 'at rest is space'.

Anyone who thinks that an object can be at rest in space does clearly not understand the principle of relativity.

Space is not something one can compare one's velocity against.

 

[WannabeNewton]

So what is your proof that we cannot define rest with respect to the orbits of the time - like killing vectors of the static space - time? They are certainly a geometric aspect of the space - time.

I can even "abstract" this to stationary axisymmetric space - times. In $(t,\phi,x^2,x^3)$ coordinates I define the family of locally non rotating observers to be the observers who are 'at rest' with respect to the $t = \text{const.}$ hypersurfaces i.e. whose 4 - velocity is given by $u^a = \alpha \triangledown ^{a}t$. The angular momentum is given by $L = g_{ab}u^{a}\psi^{b} = \alpha g_{ab}\triangledown ^{a}t\psi^{b} = \alpha g_{ab}g^{ac}\triangledown _{c}t \psi^{b} = \alpha \delta^{c}_{b}\triangledown _{c}t \psi^{b}$ where $\psi^{a}$ is the space - like killing field whose integral curves are closed. In the coordinate basis this is of course just $L = \alpha \delta^{\mu}_{\nu}\triangledown _{\mu}t \psi^{\nu} = \alpha \delta^{0}_{1} = 0$. Are you telling me it's just a coincidence that my defining the notion of being locally non - rotating as following an orbit of $\triangledown ^{a}t$ actually gave $L = 0$? Pray tell.

 

[Passionflower]

Not sure what you are trying to show with all your math, I am well aware of the meaning of static and stationary spacetimes.

I remain of the opinion that the concept of being "at rest in space" is nonsense in relativity, in fact it is even nonsense in Galilean relativity.

Things can be at rest wrt to other things but not wrt to space.

And one more time: Wald does not use the phrase 'at rest in space'.

 

[WannabeNewton]

The quote by Pervect never made any claim about being at absolute rest in space. Of course that is nonsensical. It just said that given a space - time possessing a certain one parameter isometry group, we can utilize the orbits of the group action of said isometry group to define a notion of rest with respect to the orbits. In the above example of defining locally non rotating observers in stationary axisymmetric space - times notice how I said an observer is 'at rest' with respect to the spatial hypersurfaces $t = \text{const.}$ if his 4 - velocity $u^{a} = \alpha \triangledown^{a} t$ (the proportionality scalar field $\alpha$ is just the normalization); this isn't alien from galilean relativity wherein we describe observers in collinear uniform motion with respect to one another. This example was meant to show that we can deduce physical properties of the observer with precisely that definition of locally non rotating e.g. the fact that the observer's angular momentum vanishes, as shown above.

I don't recall me nor Pervect claiming absolute rest. Do you agree with the above however?

 

[Passionflower]

The quote by Pervect never made any claim about being at absolute rest in space. Of course that is nonsensical. It just said that given a space - time possessing a certain isometry group, we can utilize the orbits of the group action of said isometry group to define a notion of rest with respect to the orbits. I don't recall me nor Pervect claiming absolute rest. Do you agree with the above however?

Certainly things can be at rest wrt to other things in curved spacetime except of course when a spacetime is non-stationary. I also see no issues with considering things at rest wrt certain coordinate values, for instance a Schwarzschild radius or a shell with a given r-value.

 

[WannabeNewton]

Certainly things can be at rest wrt to other things in curved spacetime except of course when a spacetime is non-stationary. I also see no issues with considering things at rest wrt certain coordinate values, for instance a Schwarzschild radius or a shell with a given r-value.

We are in agreement then; I'm not sure anymore what this little quibble was about lol. What exactly is the issue in conclusion?