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Time Delay

  1. Jan 9, 2011 #1
    Consider the spacetime metric

    [itex]ds^2=-(1+r)dt^2+\frac{dr^2}{(1+r)} + r^2 ( d \theta^2 + \sin^2{\theta} d \phi^2)[/itex]

    where [itex]\theta, \phi[/itex] are polar coordinates on the sphere and [itex]r \geq 0[/itex].

    Consider an observer whose worldline is [itex]r=0[/itex]. He has two identical clocks, A and B. He keeps clock A with himself and throws clock B away which returns to him after an interval of 4 minutes according to clock B. What time interval has elapsed on clock A?

    So by setting [itex]r=0[/itex] the mteric simplifies to


    Now I said that we can assume that the clock will travel on a timelike geodesic (since it is essentially a massive particle). And so using [itex]g_{ab}u^au^b=-1[/itex] for timelike geodesics we get

    [itex]-1= \left( \frac{dt}{d \tau} \right)^2 + \left( \frac{dr}{d \tau} \right)^2[/itex].

    Now I'm stuck. We know A is measuring proper time I think and so I imagine we want to solve this equation for [itex]\frac{dt}{d \tau}[/itex] and then use that to get an equation for t in terms of tau and then solve for tau when t is equal to 4. Am I right?

    Also, is [itex]g_{ab}u^au^b=-1[/itex] true for any timelike curve or just for timelike geodesics, and if so, why?

    Thanks a lot.
  2. jcsd
  3. Jan 10, 2011 #2
    You cannot set r=0 for clock B which is travelling.It will travel on a radial trajectory, so theta and phi are constant.
  4. Jan 10, 2011 #3
    Will it be timelike?

    So do I get [itex]-1=-(1+r) (\frac{dt}{d \tau})^2 + \frac{1}{1+r} ( \frac{dr}{d \tau} )^2[/itex]?

    How do I go about solving this?
  5. Jan 10, 2011 #4
    Yes it is a normal massive particle, so it has to be timelike.
    So far you only have one equation for two unknown functions. You have to use the e.o.m. for a geodesic to eliminate one of them.
  6. Jan 10, 2011 #5
    Well how is this:

    [itex]L=-(1+r) \dot{t}^2 + \frac{1}{1+r} \dot{r}^2 + r^2 \dot{\theta}^2 + r^2 \sin^2{\theta{ \dot{\phi}^2[/itex]

    So it seems like this is going to be simplest to use Euler Lagrange on the t coordinate so we get:

    [itex]\frac{\partial L}{\partial x^\mu} = \frac{d}{d \tau} \left( \frac{\partial L}{\partial \dot{x}^\mu} \right)[/itex]
    [itex]\Rightarrow \frac{d}{d \tau} \left( -2 (1+r) \dot{t} \right)=0[/itex]

    Now by cancelling the -2 and then by product rule we get

    [itex] \dot{(1+r)} \dot{t} + (1+r) \ddot{t}=0 \Rightarrow \dot{r} \dot{t} + ( 1+r) \ddot{t}=0 \Rightarrow \dot{r} = - \frac{(1+r) \ddot{t}}{\dot{t}}[/itex]

    Is this correct? Should I go ahead and substitute this back in?

  7. Jan 10, 2011 #6
    How did you get the idea to take the norm of the velocity as Lagrange function?
    Somewhere in your lecture you should have derived something called geodesic equation or similar. It should be a differential equation defining the motion of a particle on a geodesic. The equation with the -1 in your first post follows from this differential equations.
    So you need one of these explicit equations to help you eliminate one function.
  8. Jan 10, 2011 #7
    Oops. Kind of went of at a tangent there!

    So the geodesic equation is

    [itex]\frac{d^2 x^\mu}{d \tau^2} + \Gamma^\mu{}_{\nu \rho} x^\nu x^\rho=0[/itex]

    So I assume I want to get rid of [itex]\frac{dr}{d \tau}[/itex] since we are interested in how [itex]t[/itex] varies with [itex]\tau[/itex].

    So if I pick [itex]x^\mu=r[/itex] then

    [itex]\frac{d^2r}{d \tau^2} + \Gamma^r{}_{\nu \rho} x^\nu x^rho=0[/itex]

    Now [itex]\Gamma^\mu{}_{\nu \rho} = \frac{1}{2} g^{\mu \sigma} \left( g_{\nu \sigma, \rho} + g_{\sigma \rho, \nu} - g_{\nu \rho, \sigma} \right)[/itex]

    Now if we take [itex]\mu=r[/itex] then the only non zero component is

    [itex]\Gamma^r{}_{rr} = \frac{1}{2} g^{rr} g_{rr,r} = \frac{1}{2} \left( - \frac{1}{1+r} \right) \left( \frac{1}{(1+r)^2} \right) = - \frac{1}{(1+r)^3}[/itex]

    So this would go back into the geodesic equation to give
    [itex]\frac{d^2r}{d \tau^2} - \frac{1}{(1+r)^3} \left( \frac{dr}{d \tau} \right)[/itex]

    And hence [itex]\left( \frac{dr}{d \tau} \right)^2 = - (1+r)^3 \frac{d^2r}{d \tau^2}[/itex]

    So should I substitute this back in? How would I get rid of the [itex]\frac{d^2r}{d \tau^2}[/itex] term?

    Also, can you remind me how we derive the equation [itex]g_{ab}u^au^b=-\sigma[/itex] please?

    Thanks again!
  9. Jan 11, 2011 #8
    Up to the definition of the Christoffel symbol you are correct. Then you calculated [tex]\Gamma^r_{rr}[/tex] wrong. The mistake is in [tex]g^{rr}[/tex]. And the 1/2 disappeared. But this is not the only nonzero component.

    To dervie that the norm is constant it is convenient to first rewrite the geodsic equation in the following form.
    [tex]\frac{d u_\alpha}{d\tau} - \frac{1}{2}\partial_\alpha g_{\mu\nu}u^\mu u^\nu=0[/tex]
    Then all you have to do is act with [tex]\frac{d}{d\tau}[/tex] on [tex]g_{\alpha\beta}u^\alpha u^\beta[/tex] and reshuffle the derivatives and use the above geodesic equation.
  10. Jan 11, 2011 #9
    Ok. So I find that


    However, from the definition
    \Gamma^\mu{}_{\nu \rho} = \frac{1}{2} g^{\mu \sigma} \left( g_{\nu \sigma, \rho} + g_{\sigma \rho, \nu} - g_{\nu \rho, \sigma} \right)
    We see that having picked [itex]\mu=r[/itex], we must take [itex]\sigma=r[/itex] but we can get a contribution from the third term in the definition of the Christoffel symbols when [itex]\nu=\rho[/itex] also,

    So [itex]\Gamma^r{}_{tt}=-\frac{1}{2}(1+r)[/itex]
    [itex]\Gamma^r{}_{\theta \theta}=r(1+r)[/itex]
    [itex]\Gamma^r{}_\phi \phi} = r(1+r) \sin^2{\theta}[/itex]

    So are all these correct now? What's next? Plug them back into
    [itex]\left( \frac{dr}{d \tau} \right)^2 + \left( \frac{d t }{d \tau} \right)^2=-1[/itex]?

    And secondly, you wrote [tex]
    \frac{d u_\alpha}{d\tau} - \frac{1}{2}\partial_\alpha g_{\mu\nu}u^\mu u^\nu=0
    What happened to the 1st and second terms from the Christoffel symbol?

  11. Jan 11, 2011 #10
    The Christoffel symbols now are correct. You should now write the relevant geodesic equations and try to find out which ones to use to solve for r(tau) and t(tau).

    You have to notice that compared to the original geodesic equation this one is now for the kovariant velocity. You should try to derive my expression from the usual one, but it is straight forward.
  12. Jan 11, 2011 #11
    You have to notice that compared to the original geodesic equation this one is now for the kovariant velocity. You should try to derive my expression from the usual one, but it is straight forward.[/QUOTE]

    Surely there is only one geodesic equation, namely:

    [itex]\frac{d^2r}{d \tau^2} - \frac{1}{2(1+r)} \left( \frac{dr}{d \tau} \right)^2 - \frac{1}{2} ( 1+r) \left( \frac{dt}{d \tau} \right)^2 + r (1+r) \left( \frac{d \theta}{d \tau} \right)^2 + r(1+r) \sin^2{\theta} \left( \frac{d \phi}{d \tau} \right)^2=0[/itex]

    I don't see how I can solve this for r(tau) or t(tau) since I now have one equation and 5 unknowns!!!

    Sorry but I don't understand what you mean here.
  13. Jan 11, 2011 #12
    No. You have for equations. One for each t,r,theta,phi. And two of the functions are known. The observer throws the clock on a radial trajectory, so theta=const and phi=const.

    So you have two differential equations for two unknown function which is enough to solve the problem. Or you can use one of the DE and the relation for the norm of the velocity, which will give the same result.

    On the expression for the geodesic equation: What I meant is that both expressions are equivalent.
    [tex] \frac{d u_\alpha}{d\tau} - \frac{1}{2}\partial_\alpha g_{\mu\nu}u^\mu u^\nu=0\Leftrightarrow \frac{d^2 x^\mu}{d \tau^2} + \Gamma^\mu{}_{\nu \rho} u^\nu u^\rho=0\Leftrightarrow \frac{d^2 x_\mu}{d \tau^2} - \Gamma^\nu{}_{\mu \rho} u_\nu u^\rho=0[/tex] and you can use whichever one is more convenient to you.
    The derivation of this relation will be about four lines, so you should try to prove it.
  14. Jan 11, 2011 #13
    Btw. in your first formula for the geodesic equation you accidentially wrote [tex]x^\nu x^\rho[/tex] instead of [tex]u^\nu u^\rho[/tex] but correctly used the u later on.
  15. Jan 11, 2011 #14
    Thanks but why is it four seperate equations. Surely in the definition of the Christoffel symbols, we are using Einstein summation convention and so the [itex]\nu,\rho[/itex] indices are summed over, no?
  16. Jan 11, 2011 #15
    Yes, but you have on free index, alpha.

    I just realized, that it is you again latentcorpse :)
    Seems I always choose to answer your questions. Where are studying?
  17. Jan 11, 2011 #16
    I don't get it.

    We have

    \frac{d^2 x^\mu}{d \tau^2} + \Gamma^\mu{}_{\nu \rho} x^\nu x^\rho=0

    So surely the free index is [itex]\mu[/itex]. Now we have picked [itex]\mu=r[/itex] but the [itex]\nu, \rho[/itex] indices are dummy (i.e. summed over) so surely we would have
    [itex]\frac{d^2 r}{d \tau^2} + \Gamma^r{}_{tt} u^tu^t +\Gamma^r{}_{rr} u^ru^r + \Gamma^r{}_{\theta \theta} u^\thetau^\theta + \Gamma^r{}_{\phi \phi} u^\phi u^\phi=0[/itex]

    And for the derivation of the norm of the velocity equation I multiplied the whole thing through by [itex]g_{\mu \lambda}[/itex] to get:

    [itex]\frac{d^2 x_\lambda}{d \tau^2} + \frac{1}{2} g_{\mu \lambda} g^{\mu \sigma} ( g_{\nu \sigma, \rho + g_{\sigma \rho, \nu} - g_{\nu \rho, \sigma}) u^\nu u^\rho=0[/itex]
    [itex]\frac{d^2 x_\lambda}{d \tau^2}+\frac{1}{2} ( g_{\nu \lambda, \rho} + g_{\lambda \rho, \nu} - g_{\nu \rho, \lambda})u^\nu u^\rho=0[/itex]

    And then if we relabel [itex]\lambda \rightarrow \mu[/itex] and use the symmetry fo the metric and the u terms, we can rewrite it as

    [itex]\frac{d^2x_\mu}{d \tau^2} + \frac{1}{2} ( 2g_{\nu \mu,\rho} - g_{\nu \rho,\mu})u^\nu u^\rho=0[/itex]

    So it appears I have an extra term that you don't have?

    And I'm studying at Cambridge but as you can probably tell Im finding it pretty tough. What about you, where do you study/work?
  18. Jan 11, 2011 #17
    Yes. But you could equally pick mu=t. This would be the second equation.

    Careful: You cannot pull the metric through [tex]\frac{d}{d \tau}[/tex] This is not a covariant derivative.
    I think it is easier if you start with the metric inside and then pull it out step by step.
    [tex]\frac{d}{d\tau}(g_{\alpha\beta}u^{\beta}= \ldots[/tex]
    Then using writing [tex]\frac{d}{d\tau}=u^\alpha\frac{\partial}{\partial x^\alpha}[/tex] you should be able to make the calculation.

    I'm in Munich doing a Ph.D. in Cosmoloy.
  19. Jan 11, 2011 #18
    Ok. Well, if I go back to my first equation where I picked [itex]\mu=r[/itex]:

    \frac{d^2r}{d \tau^2} - \frac{1}{2(1+r)} \left( \frac{dr}{d \tau} \right)^2 - \frac{1}{2} ( 1+r) \left( \frac{dt}{d \tau} \right)^2 + r (1+r) \left( \frac{d \theta}{d \tau} \right)^2 + r(1+r) \sin^2{\theta} \left( \frac{d \phi}{d \tau} \right)^2=0

    I can now get rid of the last two terms since [itex]\theta, \phi[/itex] are constant. This gives:

    \frac{d^2r}{d \tau^2} - \frac{1}{2(1+r)} \left( \frac{dr}{d \tau} \right)^2 - \frac{1}{2} ( 1+r) \left( \frac{dt}{d \tau} \right)^2=0

    Now I can get the other equation by setting [itex]\mu=t[/itex]. Again we can get rid of the last two terms because the clock is travelling radially. This gives:

    [itex]\frac{d^2t}{d \tau^2} + \frac{1}{1+r} \frac{dr}{dt} \left( \frac{dt}{d \tau} \right)^2 + \frac{1}{2(1+r)^3} \frac{dr}{dt} \left( \frac{dr}{d \tau} \right)^2=0[/itex]

    (Hopefully my Christoffel symbols are correct here!)

    Anyway, I'm a bit concerned about the [itex]\frac{dr}{dt}[/itex] terms in the second equation. How do I get rid of them?

    Also, we know the clock will travel radially since our observer is at r=0 and so no matter where he throws it is going to be radial with respect to our coordinate system, correct?
    We know it's timelike since it's a massive particle, correct?
    But how do we know it will travel on a geodesic and not just a timelike curve? Is this because it is a free particle? If so, what sort of particle would travel on a curve that isn't a geodesic - something like a particle that is in a potential?

    Well is it like this:

    [itex]\frac{d^2 x_\alpha}{d \tau^2}=\frac{d}{d \tau} ( g_{\alpha \beta} u^\beta) = u^\rho \frac{\partial}{\partial x^\rho} ( g_{\alpha \beta} u^\beta )=u^\rho g_{\alpha \beta, \rho} u^\beta + g_{\alpha \beta} u^\rho \frac{\partial u^\beta}{\partial x^\rho} = \partial_\rho g_{\alpha \beta} u^\rho u^\beta + g_{\alpha \beta} u^\rho \frac{d}{d \tau} \delta^\beta{}_{\rho}[/itex]
    But since [itex]\delta^\beta{}_\rho[/itex] is constant, the last term vanishes and so we can rearrange to get
    [itex]\frac{d^2 x_{\alpha}}{d \tau^2} - \partial_\rho g_{\alpha \beta} u^\rho u^\beta=0[/itex].
  20. Jan 11, 2011 #19
    This is almost correct. The mistake is mine, because I missed the sign error on your [tex]\Gamma_{tt}^r[/tex]. Otherwise that's fine. Now you could also write the equation for the norm of the velocity and see if you can simplify it that way. You can also do it with the next DE but this way it is easier.

    No. Be careful. In the Christoffelsymbols only partial derivatives appear, whereas you use total derivatives w.r.t. t.

    Any particle whose motion is give solely by the metric and not some other external (i.e. nongravitational force) will follow a geodesic. So potential would not be good idea, because usually you would have a gravitational potential. Examples of nongeodesic motion are e.g. obeservers at fixed points in a given gravitational field, e.g. a black hole. If following a geodesic they would fall into the BH, but we usually put them at a fixed radius.

    Until the last step yes. But then you mixed it up a bit. All you have to do now is revert to d/d tau again in the last term and use the geodesic equation. Then you have to use the expression for the Christoffelsymbols in terms of the metric and you are done.
  21. Jan 11, 2011 #20
    I can't see the problem

    which is what I have.

    So if I change [itex]\frac{dr}{dt} \rightarrow \frac{\partial r}{\partial t}[/itex] will this be alright?

    \frac{d^2 x_\alpha}{d \tau^2}=\frac{d}{d \tau} ( g_{\alpha \beta} u^\beta) = u^\rho \frac{\partial}{\partial x^\rho} ( g_{\alpha \beta} u^\beta )=u^\rho g_{\alpha \beta, \rho} u^\beta + g_{\alpha \beta} u^\rho \frac{\partial u^\beta}{\partial x^\rho} = \partial_\rho g_{\alpha \beta} u^\rho u^\beta + g_{\alpha \beta} \frac{du^\beta}{d \tau}

    [itex]=\partial_\rho g_{\alpha \beta} u^\rho u^\beta -g_{\alpha \beta} \Gamma^\beta{}_{\lambda \tau} u^\lambda u^\tau[/itex]
    [itex]=\partial_\rho g_{\alpha \beta} u^\rho u^\beta -\frac{1}{2}g_{\alpha \beta}g^{\beta \sigma} ( g_{\lambda \sigma, \tau} + g_{\sigma \tau, \lambda} - g_{\lambda \tau, \sigma})[/itex]
    [itex]=\partial_\rho g_{\alpha \beta} u^\rho u^\beta -\frac{1}{2}\delta^\sigma{}_\alpha ( g_{\lambda \sigma, \tau} + g_{\sigma \tau, \lambda} - g_{\lambda \tau, \sigma})[/itex]
    [itex]=\partial_\rho g_{\alpha \beta} u^\rho u^\beta -\frac{1}{2} ( g_{\lambda \alpha, \tau} + g_{\alpha \tau, \lambda} - g_{\lambda \tau, \alpha})[/itex]
    [itex]=\partial_\rho g_{\alpha \beta} u^\rho u^\beta -\frac{1}{2} ( g_{\lambda \alpha, \tau} + g_{\alpha \tau, \lambda} - g_{\lambda \tau, \alpha})[/itex]
    [itex]=g_{\alpha \beta, \rho} u^\rho u^\beta - \frac{1}{2} g_{\beta \alpha, \rho} u^\rho u^\beta - \frac{1}{2} g_{\alpha \beta, \rho} u^\rho u^\beta + \frac{1}{2} g_{\beta \rho, \alpha} u^\rho u^\beta[/itex]
    [itex]=\frac{1}{2} g_{\beta \rho,\alpha} u^\rho u^\beta[/itex]
    which I think is what we wanted, no?

    So you now said I should take the derivative wrt tau on this whole thing. That gives:

    [itex]\frac{d^2 x_\alpha}{d \tau^2} - \frac{1}{2} \frac{d}{d \tau} g_{\beta \rho, \alpha} u^\rho u^\beta=0[/itex]

    Now is the first term zero? If so, why? And then how do we get rid of the [itex]\frac{\partial}{\partial x^\alpha}[/itex] in the second term?

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