Exploring Spacetime Metric for Clock A's Time Interval

In summary: You can use the Euler Lagrange equation for a geodesic to get rid of one equation.You can use the Euler Lagrange equation for a geodesic to get rid of one equation.
  • #1
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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.
 
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  • #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.
 
  • #3
betel said:
You cannot set r=0 for clock B which is travelling.It will travel on a radial trajectory, so theta and phi are constant.

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?
 
  • #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.
 
  • #5
betel said:
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.

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?

Thanks!
 
  • #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.
 
  • #7
betel said:
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.

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!
 
  • #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.
 
  • #9
betel said:
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.

Ok. So I find that

[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
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
betel said:
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.[/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!

betel said:
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.
Sorry but I don't understand what you mean here.
 
  • #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.
 
  • #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.
 
  • #14
betel said:
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.
Thanks but why is it four separate 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?
 
  • #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?
 
  • #16
betel said:
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?

I don't get it.

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 I am finding it pretty tough. What about you, where do you study/work?
 
  • #17
latentcorpse said:
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]
Yes. But you could equally pick mu=t. This would be the second equation.

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]

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.


And I'm studying at Cambridge but as you can probably tell I am finding it pretty tough. What about you, where do you study/work?

I'm in Munich doing a Ph.D. in Cosmoloy.
 
  • #18
betel said:
Yes. But you could equally pick mu=t. This would be the second equation.

Ok. Well, if I go back to my first equation where I picked [itex]\mu=r[/itex]:

[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 traveling 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?
betel said:
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.

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].
 
  • #19
latentcorpse said:
Ok. Well, if I go back to my first equation where I picked [itex]\mu=r[/itex]:

[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]
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.

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

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?
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.

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]
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.
 
  • #20
betel said:
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.

I can't see the problem

[itex]\Gamma^r{}_{tt}=\frac{1}{2}g^{rr}(-g_{tt,r})=\frac{1}{2}(-(1+r))(-(-1))=-\frac{1}{2}(1+r)[/itex]
which is what I have.
betel said:
No. Be careful. In the Christoffelsymbols only partial derivatives appear, whereas you use total derivatives w.r.t. t.

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

betel said:
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.
[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} \frac{du^\beta}{d \tau}
[/itex]

[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?

Thanks.
 
  • #21
latentcorpse said:
I can't see the problem
[itex]\Gamma^r{}_{tt}=\frac{1}{2}g^{rr}(-g_{tt,r})=\frac{1}{2}(-(1+r))(-(-1))=-\frac{1}{2}(1+r)[/itex]
which is what I have.
[tex]g^{rr}=1+r\neq -(1+r)[/tex]

So if I change [itex]\frac{dr}{dt} \rightarrow \frac{\partial r}{\partial t}[/itex] will this be alright?
If you continue correctly yes.



[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} \frac{du^\beta}{d \tau}
[/itex]

[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?
Correct. The start should be
[tex]\frac{d}{d \tau}u_\alpha\neq\frac{d^2}{d\tau^2}x_\alpha[/tex] though.

Here you should stop. This is the relation we wanted and which is easier to use than the usual geodesic equation.
 
  • #22
betel said:
[tex]g^{rr}=1+r\neq -(1+r)[/tex]
Ok. This is my mistake. I had written down that it had a minus in it by accident.

So we have
[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]


betel said:
If you continue correctly yes.

So it's just

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



betel said:
Correct. The start should be
[tex]\frac{d}{d \tau}u_\alpha\neq\frac{d^2}{d\tau^2}x_\alpha[/tex] though.

Here you should stop. This is the relation we wanted and which is easier to use than the usual geodesic equation.

Doesn't [tex]\frac{d}{d \tau}u_\alpha=\frac{d^2}{d\tau^2}x_\alpha[/tex] since [itex]u_\alpha=\frac{d x_\alpha}{d \tau}[/itex]? Or is it different because we've lowered indices?

Also, I don't think we've used this anywhere above, have we? You just introduced this when I asked how we deduce that the norm of the velocity is constant. We haven't yet shown this so surely we still have some work to do?
 
  • #23
latentcorpse said:
So we have
[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]
Correct.
So it's just

[itex]
\frac{d^2t}{d \tau^2} + \frac{1}{1+r} \frac{\partial r}{\partial t} \left( \frac{dt}{d \tau} \right)^2 + \frac{1}{2(1+r)^3} \frac{\partial r}{\partial t} \left( \frac{dr}{d \tau} \right)^2=0
[/itex]
So what is [tex]\frac{\partial r}{\partial t}[/tex] ?
Doesn't [tex]\frac{d}{d \tau}u_\alpha=\frac{d^2}{d\tau^2}x_\alpha[/tex] since [itex]u_\alpha=\frac{d x_\alpha}{d \tau}[/itex]? Or is it different because we've lowered indices?
Lowering inidices makes a big difference.
[tex]u^\alpha:= \frac{d x^\alpha}{d\tau}\Rightarrow u_\alpha = g_{\alpha\beta}\frac{d x^\beta}{d\tau }\neq\frac{d x_\alpha}{d\tau}[/tex]

Also, I don't think we've used this anywhere above, have we? You just introduced this when I asked how we deduce that the norm of the velocity is constant. We haven't yet shown this so surely we still have some work to do?
[/QUOTE]
You didn't use this relation, but wrote it at the start and so got the wrong complete relation. The correct is
[tex]\frac{d u_\alpha}{d\tau}-\frac{1}{2}\frac{\partial g_{\mu\nu}{\partial x^\alpha} u^\mu u^\nu=0[/tex]
Now we want to prove
[tex]\frac{d}{d\tau}(g^{\alpha\beta}u_\alpha u_\beta)=0[/tex]
The derivatives acting on the u-s can be done with the relation we just derived. To get the one on g^(-1) you should find a way to express it in terms of derivatives of g.
 
  • #24
Wait a second. Before I was just focusing on the dr/dt part which should not be in the geodesic equation. But I didn't check, whether you calculated the Christoffelsymbols correctly, which you did not, at least not all, you missed some.

You could recalculate the Christoffelsymbols again or use the geodesic equation we just derived.
 
  • #25
betel said:
So what is [tex]\frac{\partial r}{\partial t}[/tex] ?

So from the norm equation I have

[itex]\frac{\partial r}{\partial \tau} = \sqrt{-1 - \left( \frac{\partial t}{\partial \tau} \right)^2}[/itex]

So i substitute this into my t equation above. Which Christoffel symbols did I get wrong?



betel said:
You didn't use this relation, but wrote it at the start and so got the wrong complete relation. The correct is
[tex]\frac{d u_\alpha}{d\tau}-\frac{1}{2}\frac{\partial g_{\mu\nu}{\partial x^\alpha} u^\mu u^\nu=0[/tex]
Now we want to prove
[tex]\frac{d}{d\tau}(g^{\alpha\beta}u_\alpha u_\beta)=0[/tex]
The derivatives acting on the u-s can be done with the relation we just derived. To get the one on g^(-1) you should find a way to express it in terms of derivatives of g.
Can't quite get it to work:

[itex]\frac{d}{d \tau} ( g^{\alpha \beta} u_\alpha u_\beta)=\frac{d x^\rho}{d \tau} \frac{\partial}{\partial x^\rho} g^{\alpha \beta} u_\alpha u_\beta + g^{\alpha \beta} ( \frac{d}{d \tau} u_\alpha ) u_\beta + g^{\alpha \beta} u_\alpha ( \frac{d}{d \tau} u_\beta )[/itex]
[itex]=g^{\alpha \beta}{}_{, \rho} u^\rho u_\alpha u_\beta + g^{\alpha \beta} ( \frac{1}{2} g_{\lambda \rho, \alpha} u^\rho u^\lambda) u_\beta + g^{\alpha \beta} u_\alpha ( \frac{1}{2} g_{\lambda \rho, \beta} u^\lambda u^\rho)[/itex]
 
  • #26
latentcorpse said:
So from the norm equation I have

[itex]\frac{\partial r}{\partial \tau} = \sqrt{-1 - \left( \frac{\partial t}{\partial \tau} \right)^2}[/itex]

So i substitute this into my t equation above. Which Christoffel symbols did I get wrong?
You didn't really get one wrong, at least not yet. But you missed [tex]\Gamma^t_{tr}[/tex]
Can't quite get it to work:

[itex]\frac{d}{d \tau} ( g^{\alpha \beta} u_\alpha u_\beta)=\frac{d x^\rho}{d \tau} \frac{\partial}{\partial x^\rho} g^{\alpha \beta} u_\alpha u_\beta + g^{\alpha \beta} ( \frac{d}{d \tau} u_\alpha ) u_\beta + g^{\alpha \beta} u_\alpha ( \frac{d}{d \tau} u_\beta )[/itex]
[itex]=g^{\alpha \beta}{}_{, \rho} u^\rho u_\alpha u_\beta + g^{\alpha \beta} ( \frac{1}{2} g_{\lambda \rho, \alpha} u^\rho u^\lambda) u_\beta + g^{\alpha \beta} u_\alpha ( \frac{1}{2} g_{\lambda \rho, \beta} u^\lambda u^\rho)[/itex]

You somehow have to rewrite [tex]g^{\alpha\beta}{}_{,\rho}[/tex] in terms of [tex]g_{\mu\nu,\sigma}[/tex] Try to do this using [tex]g^{\alpha\beta}g_{\beta\rho}=\delta^{\alpha}_\rho[/tex]
 
Last edited:
  • #27
betel said:
You didn't really get one wrong, at least not yet. But you missed [tex]\Gamma^t_{tr}[/tex]

My r equation is (notice I have corrected the signs from earlier because of that minus I had in [itex]g_{rr}[/itex] that wasn't meant to be there!

[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]

which becomes (after using the fact it's radial):

[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]

My t equation is (again I corrected one of the signs and I have also included the new term that I missed)

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

Now is the next step to substitute
[itex]
\frac{\partial r}{\partial \tau} = \sqrt{-1 - \left( \frac{\partial t}{\partial \tau} \right)^2}
[/itex]
into my t equation?



You somehow have to rewrite [tex]g^{\alpha\beta}{}_{,\rho}[/tex] in terms of [tex]g_{\mu\nu,\sigma}[/tex] Try to do this using [tex]g^{\alpha\beta}g_{\beta\rho}=\delta^{\alpha}_\rho[/tex][/QUOTE]

Ok. I tried writing

[itex]\frac{\partial}{\partial x^\rho} g^{\alpha \beta} = \frac{\partial}{\partial x^\mu} \delta^\rho{}_\mu g^{\alpha \beta} - \frac{\partial}{\partial x^\mu} g^{\rho \lambda}g_{\lambda \mu} g^{\alpha \beta}[/itex]

But then I couldn't simplify it any further...
 
  • #28
latentcorpse said:
My r equation is (notice I have corrected the signs from earlier because of that minus I had in [itex]g_{rr}[/itex] that wasn't meant to be there!
which becomes (after using the fact it's radial):

[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]
That's correct.

My t equation is (again I corrected one of the signs and I have also included the new term that I missed)

[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 + \frac{1}{2(1+r)} \frac{dt}{d \tau} \frac{dr}{d \tau}=0
[/itex]
The new term should be double. Check the summation again and remember the Christoffelsymbolds are symmetric. The second and third term are not completely correct. There should be [tex]\frac{\partial r}{\partial t}[/tex] and you should further simplify this.

Now is the next step to substitute
[itex]
\frac{\partial r}{\partial \tau} = \sqrt{-1 - \left( \frac{\partial t}{\partial \tau} \right)^2}
[/itex]
into my t equation?
You can do this, but you have to be careful when taking the root. The particle will come back, so the speed will become negative.

I personally found it easier to first solve for r in terms of tau. This is a very easy equation once you notice that the last part in the geodesic is exactly the relation for the norm of the velocity.

Ok. I tried writing

[itex]\frac{\partial}{\partial x^\rho} g^{\alpha \beta} = \frac{\partial}{\partial x^\mu} \delta^\rho{}_\mu g^{\alpha \beta} - \frac{\partial}{\partial x^\mu} g^{\rho \lambda}g_{\lambda \mu} g^{\alpha \beta}[/itex]

But then I couldn't simplify it any further...
This is wrong. In the first equality you moved rho from a lower to an upper index and now have two lower mu. Try hitting with d/dtau on
[tex]
g^{\alpha\beta}g_{\beta\rho}=\delta^{\alpha}_\rho
[/tex]
 
  • #29
If you want to substitute your square root, I should also notice, that you missed a couple of prefactors and signs.
 
  • #30
betel said:
That's correct.
Grand.

betel said:
The new term should be double. Check the summation again and remember the Christoffelsymbolds are symmetric. The second and third term are not completely correct. There should be [tex]\frac{\partial r}{\partial t}[/tex] and you should further simplify this.
Ok. So the t equation is:

[itex]

\frac{d^2t}{d \tau^2} + \frac{1}{1+r} \frac{\partial r}{\partial t} \left( \frac{dt}{d \tau} \right)^2 - \frac{1}{2(1+r)^3} \frac{\partial r}{\partial t} \left( \frac{dr}{d \tau} \right)^2 + \frac{1}{(1+r)} \frac{dt}{d \tau} \frac{dr}{d \tau}=0

[/itex]

betel said:
You can do this, but you have to be careful when taking the root. The particle will come back, so the speed will become negative.
I personally found it easier to first solve for r in terms of tau. This is a very easy equation once you notice that the last part in the geodesic is exactly the relation for the norm of the velocity.

Well. I think I should probably do it the easy way!But I don't know how to use these two equations to solve for [itex]r(\tau)[/itex]?

I find [itex]g^{\mu \nu} u_\mu u_\nu = - \frac{1}{1+r} \left( \frac{\partial t}{\partial \tau} \right)^2 + (1+r) \left( \frac{dr}{d \tau} \right)^2[/itex] and so I would be able to substitue this into the last part of my r equation but it appears that the terms don't match up! Have I made a mistake? I see what you mean by this making it easy because then I would just get

[itex]\frac{d^2r}{dt^2} + g^{\mu \nu}u_\mu u_\nu=0 \Rightarrow \frac{d^2r}{dt^2} - \sigma =0 \Rightarrow \frac{d^2r}{dt^2}=1[/itex] since the clock follows a timelike geodesic so [itex]\sigma=1[/itex].

Then [itex]\frac{dr}{d \tau} = \tau + k_1 \Rightarrow r(\tau)=\tau^2+k_1 \tau+k_2[/itex]

How is that?
betel said:
This is wrong. In the first equality you moved rho from a lower to an upper index and now have two lower mu. Try hitting with d/dtau on
[tex]
g^{\alpha\beta}g_{\beta\rho}=\delta^{\alpha}_\rho
[/tex]
So this reduces to [itex]g^{\alpha \beta}{}_{, \lambda} u^\lambda g_{\beta \rho} = - g^\alpha \beta} g_{\beta \rho, lambda} u^\lambda[/itex]
[itex]g^{\alpha \kappa}{}_{, \lambda u^\lambda} = - g^{\alpha \beta} g^{\rho \kappa} g_{\alpha \beta, \lambda} u^\lambda[/itex]
And then with some relabelling this can be used to cancel the other two terms as required! Great!
 
  • #31
latentcorpse said:
Grand.
Ok. So the t equation is:

[itex]

\frac{d^2t}{d \tau^2} + \frac{1}{1+r} \frac{\partial r}{\partial t} \left( \frac{dt}{d \tau} \right)^2 - \frac{1}{2(1+r)^3} \frac{\partial r}{\partial t} \left( \frac{dr}{d \tau} \right)^2 + \frac{1}{(1+r)} \frac{dt}{d \tau} \frac{dr}{d \tau}=0

[/itex]
Remember: r and t are independent coordinates. You will smack your head after you got this.


Well. I think I should probably do it the easy way!But I don't know how to use these two equations to solve for [itex]r(\tau)[/itex]?

I find [itex]g^{\mu \nu} u_\mu u_\nu = - \frac{1}{1+r} \left( \frac{\partial t}{\partial \tau} \right)^2 + (1+r) \left( \frac{dr}{d \tau} \right)^2[/itex]
No. Watch the definition of u. You mixed contra and covariant u. Also there should be only total derivatives.

[itex]\frac{d^2r}{dt^2} + g^{\mu \nu}u_\mu u_\nu=0 \Rightarrow \frac{d^2r}{dt^2} - \sigma =0 \Rightarrow \frac{d^2r}{dt^2}=1[/itex] since the clock follows a timelike geodesic so [itex]\sigma=1[/itex].

Then [itex]\frac{dr}{d \tau} = \tau + k_1 \Rightarrow r(\tau)=\tau^2+k_1 \tau+k_2[/itex]

How is that?
In principle yes. Up to some factors. And sigma=-1 for timelike geodesic.

So this reduces to [itex]g^{\alpha \beta}{}_{, \lambda} u^\lambda g_{\beta \rho} = - g^\alpha \beta} g_{\beta \rho, lambda} u^\lambda[/itex]
[itex]g^{\alpha \kappa}{}_{, \lambda u^\lambda} = - g^{\alpha \beta} g^{\rho \kappa} g_{\alpha \beta, \lambda} u^\lambda[/itex]
And then with some relabelling this can be used to cancel the other two terms as required! Great!
Good.
 
  • #32
betel said:
Remember: r and t are independent coordinates. You will smack your head after you got this.
So [itex]\frac{\partial r}{\partial t}=\frac{dr}{d \tau} \frac{d \tau}{dt}[/itex]? That would give

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

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

Is this correct?

betel said:
No. Watch the definition of u. You mixed contra and covariant u. Also there should be only total derivatives.

Well I find [itex]u_\mu = \frac{d}{d \tau} (g_{\mu \nu} x^\nu)=g_{\mu \nu , \lambda} u^\lambda x^\nu + g_{\mu \nu} u^\nu[/itex]
But this seems like we're going to get two too many terms when we work out [itex]g^{\mu \nu} u_\mu u_\nu[/itex]? And so the next bit that you said was (in principle) correct won't work!
 
  • #33
latentcorpse said:
So [itex]\frac{\partial r}{\partial t}=0[/itex]? I don't see how that can be? Even if they are independent, it's still possible for a particle to follow a worldline along which both r and t vary, isn't it?
That is the big difference between total and partial derivative. The total takes exactly those dependencies into account as you mentioned above. The partial only cares for explicit dependence on a coordinate. Compare e.g. to the Lagrange equations. There you also have partial derivatives w.r.t to x and xdot. Although as x changes usually xdot changes they are independent. Only when you consider the implicit dependence of t you have to use total derivatives.

Well I find [itex]u_\mu = \frac{d}{d \tau} (g_{\mu \nu} x^\nu)=g_{\mu \nu , \lambda} u^\lambda x^\nu + g_{\mu \nu} u^\nu[/itex]
But this seems like we're going to get two too many terms when we work out [itex]g^{\mu \nu} u_\mu u_\nu[/itex]? And so the next bit that you said was (in principle) correct won't work!
That is not the definition of [tex]u_\alpha[/tex]. The velocity is defined as
[tex]u^\alpha=\frac{d}{d\tau} x^\alpha[/tex] and the covariant velocity is [tex]u_\alpha=g_{\alpha\beta}u^\beta[/tex]. All you have to do is swap the indices in lower and upper position.
[tex]g^{\alpha\beta}u_\alpha u_\beta = g_{\alpha\beta}u^\alpha u^\beta[/tex]
 
  • #34
betel said:
That is the big difference between total and partial derivative. The total takes exactly those dependencies into account as you mentioned above. The partial only cares for explicit dependence on a coordinate. Compare e.g. to the Lagrange equations. There you also have partial derivatives w.r.t to x and xdot. Although as x changes usually xdot changes they are independent. Only when you consider the implicit dependence of t you have to use total derivatives.

So my r equation is:

[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]

and the t equation is (getting rid of [itex]\frac{\partial r}{\partial t}[/itex] terms:

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


betel said:
That is not the definition of [tex]u_\alpha[/tex]. The velocity is defined as
[tex]u^\alpha=\frac{d}{d\tau} x^\alpha[/tex] and the covariant velocity is [tex]u_\alpha=g_{\alpha\beta}u^\beta[/tex]. All you have to do is swap the indices in lower and upper position.
[tex]g^{\alpha\beta}u_\alpha u_\beta = g_{\alpha\beta}u^\alpha u^\beta[/tex]

Ok so we have [itex]g_{\alpha \beta}u^\alpha u^\beta = - (1+r) \left( \frac{dt}{d \tau} \right)^2 + \frac{1}{1+r} \left( \frac{dr}{d \tau} \right)^2[/itex]
And so the r equation implies

[itex]\frac{d^2r}{d \tau^2} - \frac{1}{2} g_{\alpha \beta} U^\alpha u^\beta=0[/itex]
[itex]\frac{d^2r}{d \tau^2} +\frac{1}{2}=0[/itex] since we are on a timelike geodesic.
The solution of which is
[itex]r(\tau)=-\frac{1}{2} \tau^2 + k_1 \tau + k_2[/itex]
But since the observer throws the clock from [itex]r=0[/itex] at what we can assume is [itex]\tau=0[/itex], we conclude that [itex]k_2=0[/itex] and so
[itex]r(\tau)=-\frac{1}{2} \tau^2 + k_1 \tau[/itex]

Now, how do we find the value of [itex]k_1[/itex]?

And I tried substituting this into my t equaiton and solving for t as a function of tau - do i do this using an auxiliary equation?

Thanks.
 
  • #35
latentcorpse said:
So my r equation is:

[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]

and the t equation is (getting rid of [itex]\frac{\partial r}{\partial t}[/itex] terms:

[itex]
\frac{d^2t}{d \tau^2} + \frac{1}{2(1+r)} \frac{dt}{d \tau} \frac{dr}{d \tau}=0
[/itex]
The second term has two be without the 1/2 because you get twice the same contribution from the t-r and r-t.
Then you can combine the two.


Ok so we have [itex]g_{\alpha \beta}u^\alpha u^\beta = - (1+r) \left( \frac{dt}{d \tau} \right)^2 + \frac{1}{1+r} \left( \frac{dr}{d \tau} \right)^2[/itex]
And so the r equation implies

[itex]\frac{d^2r}{d \tau^2} - \frac{1}{2} g_{\alpha \beta} U^\alpha u^\beta=0[/itex]
[itex]\frac{d^2r}{d \tau^2} +\frac{1}{2}=0[/itex] since we are on a timelike geodesic.
The solution of which is
[itex]r(\tau)=-\frac{1}{2} \tau^2 + k_1 \tau + k_2[/itex]
But since the observer throws the clock from [itex]r=0[/itex] at what we can assume is [itex]\tau=0[/itex], we conclude that [itex]k_2=0[/itex] and so
[itex]r(\tau)=-\frac{1}{2} \tau^2 + k_1 \tau[/itex]

Now, how do we find the value of [itex]k_1[/itex]?
You also know, that it is back at r=0 at tau=4. That will be enough.

And I tried substituting this into my t equaiton and solving for t as a function of tau - do i do this using an auxiliary equation?
Thanks.
If you take the correct equation for t it will be easier.
 

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