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#1
Jan911, 05:44 AM

P: 1,441

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 [itex]ds^2=dt^2+dr^2[/itex] 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
Jan1011, 08:16 AM

P: 319

You cannot set r=0 for clock B which is travelling.It will travel on a radial trajectory, so theta and phi are constant.



#3
Jan1011, 08:35 AM

P: 1,441

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? 


#4
Jan1011, 09:33 AM

P: 319

Time Delay
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. 


#5
Jan1011, 11:54 AM

P: 1,441

[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? Thanks! 


#6
Jan1011, 02:40 PM

P: 319

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. 


#7
Jan1011, 06:32 PM

P: 1,441

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! 


#8
Jan1111, 02:24 AM

P: 319

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. 


#9
Jan1111, 06:06 AM

P: 1,441

[itex]\Gamma^r{}_{rr}=\frac{1}{2(1+r)}[/itex] However, from the definition [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] 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 [/tex] What happened to the 1st and second terms from the Christoffel symbol? Thanks. 


#10
Jan1111, 07:06 AM

P: 319

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. 


#11
Jan1111, 07:47 AM

P: 1,441

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!!! 


#12
Jan1111, 07:58 AM

P: 319

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. 


#13
Jan1111, 07:59 AM

P: 319

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.



#14
Jan1111, 08:01 AM

P: 1,441




#15
Jan1111, 08:02 AM

P: 319

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? 


#16
Jan1111, 08:13 AM

P: 1,441

We have [itex] \frac{d^2 x^\mu}{d \tau^2} + \Gamma^\mu{}_{\nu \rho} x^\nu x^\rho=0 [/itex] 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] No? 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? 


#17
Jan1111, 08:22 AM

P: 319

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. 


#18
Jan1111, 11:34 AM

P: 1,441

[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 can now get rid of the last two terms since [itex]\theta, \phi[/itex] are constant. This gives: [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=0 [/itex] 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? [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]. 


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