My problem with the relativity representation on gravity.

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The forum discussion critiques common analogies used to explain gravity in the context of general relativity, particularly the rubber sheet model and funnel graphics. Participants argue that these representations fail to accurately convey the mechanics of gravity, often leading to misconceptions. They emphasize the necessity of understanding the mathematics behind general relativity for a comprehensive grasp of the concepts. Notably, the Schwarzschild geometry is mentioned as a more accurate framework for understanding gravitational effects.

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  • #61
A.T. said:
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'?
 
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  • #62
Passionflower said:
What do you mean by 'an object at rest in space'?

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

Wald said:
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
 
  • #63
pervect said:
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.
 
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  • #64
pervect said:
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.
 
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  • #65
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.
 
  • #66
WannabeNewton said:
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.
 
  • #67
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.
 
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  • #68
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'.
 
  • #69
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?
 
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  • #70
WannabeNewton said:
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.
 
  • #71
Passionflower said:
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?
 

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