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Was Wald wrong? A question with the problem 4 in Chapter 6

  1. Aug 10, 2013 #1
    Hi all,

    I am working on the problems of Wald's General Relativity. I came across this difficult problem: the 4th problem in Chapter 6.

    Part (a) is easy, but I cannot figure out how to prove part (b). Prof. Wald suggested to use the so-called conservation of energy, but [itex]u^{a}[/itex] isn't a geodesic tangent vector. So I feel confused.

    I need your help! Thank you!

    qinglong.1397
     

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  3. Aug 10, 2013 #2

    WannabeNewton

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  4. Aug 13, 2013 #3
    Thanks, WnnabeNewton!

    I read some of the posts in that link and #65 and got inspired by your discussion. I think I can explain the meaning of conservation-of-energy argument now. I think, the correct the wording should be conservation-of-mass.

    Mass of a particle is a constant. But how to measure it is a trick task. The correct way to do so is to set up an pseudo-orthonormal coordinate system. At infinity, the spacetime is asymptotically flat, so the coordinate system (t, x, y, z) given by Wald is naturally pseudo-orthonormal, while at some finite place, [itex]g_{00}=-V^2[/itex]. So the local coordinate system should have [itex]\varepsilon^a/V[/itex] as one of its basic vectors in the time direction. With this coordinate system, the local stationary observer measures the same mass.

    But the observer at the infinity already sets up his own coordinate system, so he insists on using [itex](\partial_t)^a=\varepsilon^a[/itex] as his basic time-like vector, then the mass he measures is mV which is not m! This deviation tells our observer that the mass of the particle has been redshifted.

    The next step is just to virtual work to calculate the forces.
     
  5. Aug 13, 2013 #4

    WannabeNewton

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    Cool! By the way, I don't think Wald was referring to the energy measured at infinity ##E## when he talked about conservation of energy although I don't know why he mentions ##E## when he says to use conservation of energy arguments. Anyways consider the following conservation of energy argument from Poisson's text, although I'm not particularly fond of it because it doesn't really seem rigorous at all: http://postimg.org/image/4fny3mut1/
     
  6. Aug 14, 2013 #5

    pervect

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    I interpreted a "massless string" as a stress-energy tensor that has ##T^{00}= 0##, and ##T^{rr}= P(r)## (in Schwarzschild coordinates). If you consider that ##\nabla_a T^{ab} = 0## (which is a form of local energy conservation, I believe Wald even mentions this), you get a differential equation for P(r).

    In the standard spherical Schwarzschild coordinates (r, theta, phi), r^2 P(r) is the total tension in the string as the string isn't a uniform thickness, the thickness goes up as r^2. You can solve for the Schwarzschild metric to find that r^2 P(r) does indeed vary as the redshift factor. This doesn't quite show what Wald wants though, because I don't see how to do the covariant derivative without the whole metric, and he gives only the Killing vector.
     
  7. Aug 15, 2013 #6
    The argument in Poisson's text is interesting. Sadly, it is for static case and seems not general.

    I think the energy Wald referred should be the one measured by the observer sitting next to the particle. This makes sense because in GR, you can't compare two vectors at different events.
     
  8. Aug 16, 2013 #7
    Hi, Pervect. Thanks for sharing your idea! I'd like also provide a different point of view which sound trivial from a guy in a different forum. She simply interprets the "massless string" as something which doesn't consume any work done by the observer. That's how she understands conservation-of-energy.
     
  9. Aug 16, 2013 #8

    WannabeNewton

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    Is a conservation of energy argument strictly necessary? Was there something wrong with the argument I gave in the thread linked in post #2? In retrospect, it still uses conservation of energy but maybe not as prominently as you would have hoped?

    I'll repost it here: It seems that the solution comes from the fact that for stationary, asymptotically flat space-times we can define a gravitational potential (analogous to the Newtonian potential) by ##\varphi = \frac{1}{2}\ln (-\xi_{a}\xi^{a}) = \frac{1}{2}\ln V^{2} = \ln V##. But note that ##\nabla^{b} \varphi ## will only be the local change in the potential; the change in the potential as measured at infinity will get red-shifted so ##(\nabla^{b}\varphi)_{\infty} = V\nabla^{b}\varphi##. Now the net virtual work done on the particle, as measured at infinity, for a virtual displacement ##\delta s^{b}##, will be ##\delta W_{\infty} = (F_b)_{\infty}\delta s^{b} + m(\nabla_{b}\varphi)_{\infty} \delta s^{b}## i.e. the work done by the observer at infinity plus the work done by the gravitational field as measured at infinity for this arbitrary virtual displacement. But the principle of virtual work says that for stationary particles, ##(F_b)_{\infty}\delta s^{b} + m(\nabla_{b}\varphi)_{\infty} \delta s^{b} = 0## for any virtual displacement ##\delta s^{b}## so this implies that ##(F_b)_{\infty} + m(\nabla_{b}\varphi)_{\infty} = 0## i.e. ##(F_b)_{\infty} = -mV(\nabla_{b}\varphi) = -m\nabla_{b}V = -\nabla_{b}E## because ##E = -m\xi_{a}u^{a} = -mV^{-1}\xi_{a}\xi^{a} = mV## for the stationary particle.

    From this we can see that ##F_{\infty} = m(\nabla_a V \nabla^a V)^{1/2} = VF##.

    Peter helped me significantly with that problem, as you can see from the linked thread aforementioned, so hopefully he sees this thread so as to offer further comments :)
     
    Last edited: Aug 16, 2013
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