What Time Did Clock A Measure During Clock B's 4-Minute Journey?

[latentcorpse]

Consider the spacetime metric

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

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

Consider an observer whose worldline is r=0. 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 r=0 the mteric simplifies to

ds^2=-dt^2+dr^2

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 g_{ab}u^au^b=-1 for timelike geodesics we get

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

Now I'm stuck. We know A is measuring proper time I think and so I imagine we want to solve this equation for \frac{dt}{d \tau} 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 g_{ab}u^au^b=-1 true for any timelike curve or just for timelike geodesics, and if so, why?

Thanks a lot.

 

[betel]

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

 

[latentcorpse]

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

How do I go about solving this?

 

[betel]

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.

 

[latentcorpse]

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:

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

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

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

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

\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}}

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

Thanks!

 

[betel]

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.

 

[latentcorpse]

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

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

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

So if I pick x^\mu=r then

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

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

Now if we take \mu=r then the only non zero component is

\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}

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

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


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


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


Thanks again!

 

[betel]

Up to the definition of the Christoffel symbol you are correct. Then you calculated \Gamma^r_{rr} wrong. The mistake is in g^{rr}. 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.
\frac{d u_\alpha}{d\tau} - \frac{1}{2}\partial_\alpha g_{\mu\nu}u^\mu u^\nu=0
Then all you have to do is act with \frac{d}{d\tau} on g_{\alpha\beta}u^\alpha u^\beta and reshuffle the derivatives and use the above geodesic equation.

 

[latentcorpse]

Up to the definition of the Christoffel symbol you are correct. Then you calculated \Gamma^r_{rr} wrong. The mistake is in g^{rr}. 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.
\frac{d u_\alpha}{d\tau} - \frac{1}{2}\partial_\alpha g_{\mu\nu}u^\mu u^\nu=0
Then all you have to do is act with \frac{d}{d\tau} on g_{\alpha\beta}u^\alpha u^\beta and reshuffle the derivatives and use the above geodesic equation.

Ok. So I find that

\Gamma^r{}_{rr}=-\frac{1}{2(1+r)}

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

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

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

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

Thanks.

 

[betel]

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.

 

[latentcorpse]

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:

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

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.

 

[betel]

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.
\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 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.

 

[betel]

Btw. in your first formula for the geodesic equation you accidentially wrote x^\nu x^\rho instead of u^\nu u^\rho but correctly used the u later on.

 

[latentcorpse]

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.
\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 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 \nu,\rho indices are summed over, no?

 

[betel]

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?

 

[latentcorpse]

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

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

So surely the free index is \mu. Now we have picked \mu=r but the \nu, \rho indices are dummy (i.e. summed over) so surely we would have
\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
No?

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

\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
\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

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

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

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?

 

[betel]

We have

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

So surely the free index is \mu. Now we have picked \mu=r but the \nu, \rho indices are dummy (i.e. summed over) so surely we would have
\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

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 g_{\mu \lambda} to get:

\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

Careful: You cannot pull the metric through \frac{d}{d \tau} 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.
\frac{d}{d\tau}(g_{\alpha\beta}u^{\beta}= \ldots
Then using writing \frac{d}{d\tau}=u^\alpha\frac{\partial}{\partial x^\alpha} 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.

 

[latentcorpse]

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 \mu=r:

<br /> \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<br />

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

<br /> \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<br />

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

\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

(Hopefully my Christoffel symbols are correct here!)

Anyway, I'm a bit concerned about the \frac{dr}{dt} 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?

Careful: You cannot pull the metric through \frac{d}{d \tau} 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.
\frac{d}{d\tau}(g_{\alpha\beta}u^{\beta}= \ldots
Then using writing \frac{d}{d\tau}=u^\alpha\frac{\partial}{\partial x^\alpha} you should be able to make the calculation.

Well is it like this:

\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}
But since \delta^\beta{}_\rho is constant, the last term vanishes and so we can rearrange to get
\frac{d^2 x_{\alpha}}{d \tau^2} - \partial_\rho g_{\alpha \beta} u^\rho u^\beta=0.

 

[betel]

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

<br /> \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<br />


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

<br /> \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<br />

This is almost correct. The mistake is mine, because I missed the sign error on your \Gamma_{tt}^r. 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 \mu=t. Again we can get rid of the last two terms because the clock is traveling radially. This gives:

\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

(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:

\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}

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.

 

[latentcorpse]

This is almost correct. The mistake is mine, because I missed the sign error on your \Gamma_{tt}^r. 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

\Gamma^r{}_{tt}=\frac{1}{2}g^{rr}(-g_{tt,r})=\frac{1}{2}(-(1+r))(-(-1))=-\frac{1}{2}(1+r)
which is what I have.

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

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

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.

<br /> \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}<br />

=\partial_\rho g_{\alpha \beta} u^\rho u^\beta -g_{\alpha \beta} \Gamma^\beta{}_{\lambda \tau} u^\lambda u^\tau
=\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})
=\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})
=\partial_\rho g_{\alpha \beta} u^\rho u^\beta -\frac{1}{2} ( g_{\lambda \alpha, \tau} + g_{\alpha \tau, \lambda} - g_{\lambda \tau, \alpha})
=\partial_\rho g_{\alpha \beta} u^\rho u^\beta -\frac{1}{2} ( g_{\lambda \alpha, \tau} + g_{\alpha \tau, \lambda} - g_{\lambda \tau, \alpha})
=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
=\frac{1}{2} g_{\beta \rho,\alpha} u^\rho u^\beta
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:

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

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

Thanks.

 

[betel]

I can't see the problem
\Gamma^r{}_{tt}=\frac{1}{2}g^{rr}(-g_{tt,r})=\frac{1}{2}(-(1+r))(-(-1))=-\frac{1}{2}(1+r)
which is what I have.

g^{rr}=1+r\neq -(1+r)

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

If you continue correctly yes.

<br /> \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}<br />

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

Correct. The start should be
\frac{d}{d \tau}u_\alpha\neq\frac{d^2}{d\tau^2}x_\alpha though.

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

 

[latentcorpse]

g^{rr}=1+r\neq -(1+r)

Ok. This is my mistake. I had written down that it had a minus in it by accident.

So we have
<br /> \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 <br />

If you continue correctly yes.

So it's just

<br /> \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<br />

Correct. The start should be
\frac{d}{d \tau}u_\alpha\neq\frac{d^2}{d\tau^2}x_\alpha though.

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

Doesn't \frac{d}{d \tau}u_\alpha=\frac{d^2}{d\tau^2}x_\alpha since u_\alpha=\frac{d x_\alpha}{d \tau}? 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?

 

[betel]

So we have
<br /> \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 <br />

Correct.

So it's just

<br /> \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<br />

So what is \frac{\partial r}{\partial t} ?

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

Lowering inidices makes a big difference.
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}

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
\frac{d u_\alpha}{d\tau}-\frac{1}{2}\frac{\partial g_{\mu\nu}{\partial x^\alpha} u^\mu u^\nu=0
Now we want to prove
\frac{d}{d\tau}(g^{\alpha\beta}u_\alpha u_\beta)=0
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.

 

[betel]

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.

 

[latentcorpse]

So what is \frac{\partial r}{\partial t} ?

So from the norm equation I have

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

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

You didn't use this relation, but wrote it at the start and so got the wrong complete relation. The correct is
\frac{d u_\alpha}{d\tau}-\frac{1}{2}\frac{\partial g_{\mu\nu}{\partial x^\alpha} u^\mu u^\nu=0
Now we want to prove
\frac{d}{d\tau}(g^{\alpha\beta}u_\alpha u_\beta)=0
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:

\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 )
=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)

 

[betel]

So from the norm equation I have

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

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 \Gamma^t_{tr}

Can't quite get it to work:

\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 )
=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)

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

 

[latentcorpse]

You didn't really get one wrong, at least not yet. But you missed \Gamma^t_{tr}

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

<br /> <br /> \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<br /> <br />

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

<br /> <br /> \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<br /> <br />

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

<br /> \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<br />

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



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

Ok. I tried writing

\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}

But then I couldn't simplify it any further...

 

[betel]

My r equation is (notice I have corrected the signs from earlier because of that minus I had in g_{rr} that wasn't meant to be there!
which becomes (after using the fact it's radial):

<br /> <br /> \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<br /> <br />

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)

<br /> \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<br />

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 \frac{\partial r}{\partial t} and you should further simplify this.

Now is the next step to substitute
<br /> \frac{\partial r}{\partial \tau} = \sqrt{-1 - \left( \frac{\partial t}{\partial \tau} \right)^2}<br />
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

\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}

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
<br /> g^{\alpha\beta}g_{\beta\rho}=\delta^{\alpha}_\rho<br />

 

[betel]

If you want to substitute your square root, I should also notice, that you missed a couple of prefactors and signs.

 

[latentcorpse]

That's correct.

Grand.

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 \frac{\partial r}{\partial t} and you should further simplify this.

Ok. So the t equation is:

<br /> <br /> \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<br /> <br />

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 r(\tau)?

I find 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 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

\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 since the clock follows a timelike geodesic so \sigma=1.

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

How is that?

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
<br /> g^{\alpha\beta}g_{\beta\rho}=\delta^{\alpha}_\rho<br />

So this reduces to g^{\alpha \beta}{}_{, \lambda} u^\lambda g_{\beta \rho} = - g^\alpha \beta} g_{\beta \rho, lambda} u^\lambda
g^{\alpha \kappa}{}_{, \lambda u^\lambda} = - g^{\alpha \beta} g^{\rho \kappa} g_{\alpha \beta, \lambda} u^\lambda
And then with some relabelling this can be used to cancel the other two terms as required! Great!

 

[betel]

Grand.
Ok. So the t equation is:

<br /> <br /> \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<br /> <br />

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 r(\tau)?

I find 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

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

\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 since the clock follows a timelike geodesic so \sigma=1.

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

How is that?

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

So this reduces to g^{\alpha \beta}{}_{, \lambda} u^\lambda g_{\beta \rho} = - g^\alpha \beta} g_{\beta \rho, lambda} u^\lambda
g^{\alpha \kappa}{}_{, \lambda u^\lambda} = - g^{\alpha \beta} g^{\rho \kappa} g_{\alpha \beta, \lambda} u^\lambda
And then with some relabelling this can be used to cancel the other two terms as required! Great!

Good.

 

[latentcorpse]

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

So \frac{\partial r}{\partial t}=\frac{dr}{d \tau} \frac{d \tau}{dt}? That would give

\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<br />

And so,
<br /> \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<br />

Is this correct?

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

Well I find u_\mu = \frac{d}{d \tau} (g_{\mu \nu} x^\nu)=g_{\mu \nu , \lambda} u^\lambda x^\nu + g_{\mu \nu} u^\nu
But this seems like we're going to get two too many terms when we work out g^{\mu \nu} u_\mu u_\nu? And so the next bit that you said was (in principle) correct won't work!

 

[betel]

So \frac{\partial r}{\partial t}=0? 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 u_\mu = \frac{d}{d \tau} (g_{\mu \nu} x^\nu)=g_{\mu \nu , \lambda} u^\lambda x^\nu + g_{\mu \nu} u^\nu
But this seems like we're going to get two too many terms when we work out g^{\mu \nu} u_\mu u_\nu? And so the next bit that you said was (in principle) correct won't work!

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

 

[latentcorpse]

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:

<br /> \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<br />

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

<br /> \frac{d^2t}{d \tau^2} + \frac{1}{2(1+r)} \frac{dt}{d \tau} \frac{dr}{d \tau}=0 <br />

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

Ok so we have 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
And so the r equation implies

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

Now, how do we find the value of k_1?

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.

 

[betel]

So my r equation is:

<br /> \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<br />

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

<br /> \frac{d^2t}{d \tau^2} + \frac{1}{2(1+r)} \frac{dt}{d \tau} \frac{dr}{d \tau}=0 <br />

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 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
And so the r equation implies

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

Now, how do we find the value of k_1?

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.

 

[latentcorpse]

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.
You also know, that it is back at r=0 at tau=4. That will be enough.If you take the correct equation for t it will be easier.

So I find r(\tau)=-\frac{1}{2} \tau^2+2 \tau

and then the t equation becomes

<br /> \frac{d^2t}{d \tau^2} + \frac{1}{(1+r)} \frac{dt}{d \tau} \frac{dr}{d \tau}=0 <br />

What do you mean by now I can combine the two? Do you mean substitute the for r into the t equation or do you mean do some further simplification? Presumably further simplification because substitution yields:

\frac{d^2t}{d \tau^2} + \frac{- \tau + 2}{ -\frac{1}{2} \tau^2 + 2 \tau + 1} \frac{dt}{d \tau}=0
which doesn't look very promising!

 

[betel]

So I find r(\tau)=-\frac{1}{2} \tau^2+2 \tau

Dammit again, I only checked you had tau^2, but didn't check your factor which is wrong. You made a mistake when you integrated. Otherwise correct.

and then the t equation becomes

<br /> \frac{d^2t}{d \tau^2} + \frac{1}{(1+r)} \frac{dt}{d \tau} \frac{dr}{d \tau}=0 <br />

What do you mean by now I can combine the two? Do you mean substitute the for r into the t equation or do you mean do some further simplification? Presumably further simplification because substitution yields:

\frac{d^2t}{d \tau^2} + \frac{- \tau + 2}{ -\frac{1}{2} \tau^2 + 2 \tau + 1} \frac{dt}{d \tau}=0
which doesn't look very promising!

Well, that's possible to solve. But it is easier if you look a the bare t equations. Try to write it as one total derivative. Then you are down to only a first order equation.

 

[latentcorpse]

Dammit again, I only checked you had tau^2, but didn't check your factor which is wrong. You made a mistake when you integrated. Otherwise correct.


Well, that's possible to solve. But it is easier if you look a the bare t equations. Try to write it as one total derivative. Then you are down to only a first order equation.

So do we agree that the equation to solve is

<br /> \frac{d^2r}{d \tau^2} - \frac{1}{2} g_{\alpha \beta} u^\alpha u^\beta=0<br />

But then using g_{\alpha \beta}u^\alpha u^\beta = g^{\alpha \beta} u_\alpha u_\beta=-1 for timelike curves we get

\frac{d^2r}{d \tau^2} +\frac{1}{2}=0

As far as I can see integrating this straight up gives the same as last time?
Although I notice that if I do this with an auxiliary equation I get something with sine and cosine in it - this is definitely wrong so what is the reason why we can't use an auxiliary equation here?

 

[betel]

So do we agree that the equation to solve is

<br /> \frac{d^2r}{d \tau^2} - \frac{1}{2} g_{\alpha \beta} u^\alpha u^\beta=0<br />

But then using g_{\alpha \beta}u^\alpha u^\beta = g^{\alpha \beta} u_\alpha u_\beta=-1 for timelike curves we get

\frac{d^2r}{d \tau^2} +\frac{1}{2}=0

As far as I can see integrating this straight up gives the same as last time?
Although I notice that if I do this with an auxiliary equation I get something with sine and cosine in it - this is definitely wrong so what is the reason why we can't use an auxiliary equation here?

You are fine up to here. But upon integrating you forgot the factor. Just reverse check from your solution and you will see.

 

[latentcorpse]

You are fine up to here. But upon integrating you forgot the factor. Just reverse check from your solution and you will see.

Oh yeah...it appears I forgot the basic rules of integration momentarily!

Ok so r(\tau)=-\frac{1}{4} \tau^2 + \tauNow, as for this t equation, I really have no idea how to rewrite this. I assume we do something fancy with the \frac{1}{1+r} and \frac{dr}{d \tau}?

 

[betel]

Oh yeah...it appears I forgot the basic rules of integration momentarily!

Ok so r(\tau)=-\frac{1}{4} \tau^2 + \tau

Correct.

Now, as for this t equation, I really have no idea how to rewrite this. I assume we do something fancy with the \frac{1}{1+r} and \frac{dr}{d \tau}?

Maybe it will help if you multiply the equation through by 1+r, then it is easier to see. You will get it at once if you use the alternative version of the geodesic equation we derived before.
You can also insert the solution for r in the equation for the norm of the velocity which will give the same equation for t.

 

[latentcorpse]

Correct.


Maybe it will help if you multiply the equation through by 1+r, then it is easier to see. You will get it at once if you use the alternative version of the geodesic equation we derived before.
You can also insert the solution for r in the equation for the norm of the velocity which will give the same equation for t.

Aaarghh! I'm getting confused.

We had \frac{d u_\alpha}{d \tau} - \frac{1}{2} \partial_\alpha g_{\mu \nu} u^\mu u^\nu=0

so to match up with our t equation we take \alpha=t,\mu=t,\nu=r
but then we have a g_{rt}=0 and I get confused because this removes all r dependence from the t equation so i have clearly gone wrong somewhere!

 

[betel]

Aaarghh! I'm getting confused.

We had \frac{d u_\alpha}{d \tau} - \frac{1}{2} \partial_\alpha g_{\mu \nu} u^\mu u^\nu=0

so to match up with our t equation we take \alpha=t,\mu=t,\nu=r
but then we have a g_{rt}=0 and I get confused because this removes all r dependence from the t equation so i have clearly gone wrong somewhere!

No, everything fine. just remember the difference between co and contravariant velocity.

 

[betel]

You cannot really choose mu and nu, But by choosing alpha=t and as no metric component depends on t the last some will be zero.

 

[latentcorpse]

You cannot really choose mu and nu, But by choosing alpha=t and as no metric component depends on t the last some will be zero.

So this tells us that \frac{du_t}{d \tau}=0

But u_\alpha = g_{\alpha \beta} u^\beta = g_{\alpha \beta} \frac{d x^\beta}{d \tau}

And so \frac{d}{d \tau} ( g_{tt} \frac{dt}{d \tau})=0

But now what?

We know g_{tt}=-(1+r)

So expanding gives -\frac{dr}{d \tau} \frac{dt}{d \tau} - (1+r) \frac{d^2t}{d \tau^2}=0

But I thought we just showed that \frac{d^2t}{d \tau^2}=0?

I appear to be going round in circles!

 

[betel]

Take a break. You have shown
<br /> \frac{d}{d\tau} \left((1+r)\frac{dt}{d\tau}\right)=0<br />
which is correct. No need to expand. You can directly conclude
<br /> \frac{dt}{d\tau}=\frac{const}{1+r}<br />
The constant still has to be determined but this can be done from the norm of the velocity.
Now just insert your solution for r and integrate.

 

[latentcorpse]

Take a break. You have shown
<br /> \frac{d}{d\tau} \left((1+r)\frac{dt}{d\tau}\right)=0<br />
which is correct. No need to expand. You can directly conclude
<br /> \frac{dt}{d\tau}=\frac{const}{1+r}<br />
The constant still has to be determined but this can be done from the norm of the velocity.
Now just insert your solution for r and integrate.

The norm of the velocity for a timelike curve is g_{\alpha \beta} u^\alpha u^\beta = -1

so -(1+r) \frac{k^2}{(1+r)^2} + \frac{1}{1+r} ( - \frac{1}{2} \tau + 1 )^2 =-1

since \frac{dr}{d \tau}=-\frac{1}{2} \tau +1
so -k^2 + (1-\frac{1}{2} \tau)^2 = - (1+r) \Rightarrow k^2=1+r + (1-\frac{1}{2} \tau)^2
k^2=1+ \tau -\frac{1}{4} \tau^2 + 1 - \tau + \frac{1}{4} \tau^2=2 \Rightarrow k = \sqrt{2}
I took the positive root but presumably the negative is also valid or is there a reason for chosing one or the other?

And so t(\tau)=\frac{\sqrt{2} \tau}{1+r}

Now I have to get the substitutions right - do I substitute t=4 or \tau=4.

Well we have derived the geodesic equations for clock B and so clock B will correspond to proper time in this case and t will be the coordinate time (i.e. that of the observer).

So t(4)=\frac{4\sqrt{2}}{1+0}=4 \sqrt{2}

Is this correct?

 

[betel]

The norm of the velocity for a timelike curve is g_{\alpha \beta} u^\alpha u^\beta = -1

so -(1+r) \frac{k^2}{(1+r)^2} + \frac{1}{1+r} ( - \frac{1}{2} \tau + 1 )^2 =-1

since \frac{dr}{d \tau}=-\frac{1}{2} \tau +1
so -k^2 + (1-\frac{1}{2} \tau)^2 = - (1+r) \Rightarrow k^2=1+r + (1-\frac{1}{2} \tau)^2

k^2=1+ \tau -\frac{1}{4} \tau^2 + 1 - \tau + \frac{1}{4} \tau^2=2 \Rightarrow k = \sqrt{2}
I took the positive root but presumably the negative is also valid or is there a reason for chosing one or the other?

Fine up to here. Either one is good, as time should run strictly monotoneous. Usually we choose time to be increasing

And so t(\tau)=\frac{\sqrt{2} \tau}{1+r}

You have to integrate the derivative of the time and remember that r depens on tau too.

Now I have to get the substitutions right - do I substitute t=4 or \tau=4.
Well we have derived the geodesic equations for clock B and so clock B will correspond to proper time in this case and t will be the coordinate time (i.e. that of the observer).

So t(4)=\frac{4\sqrt{2}}{1+0}=4 \sqrt{2}

Is this correct?

The consideration is correct, the value is not.

 

[latentcorpse]

Fine up to here. Either one is good, as time should run strictly monotoneous. Usually we choose time to be increasing

You have to integrate the derivative of the time and remember that r depens on tau too.The consideration is correct, the value is not.

so

t(\tau)=\int \frac{ \sqrt{2} \tau d \tau}{ -\frac{1}{4} \tau^2 + \tau + 1} = \sqrt{2} \frac{1}{ -\frac{1}{4}} \int \frac{ \tau d \tau}{ \tau - 4 \tau - 4} = -4 \sqrt{2} \int \frac{ \tau d \tau}{ ( \tau -2 - 2 \sqrt{2})( \tau -2 + 2 \sqrt{2})}

Now partial fractions gives me

t(\tau)=-4 \sqrt{2} ( \frac{1+ \sqrt{2}}{ 2 \sqrt{2}}) \int \frac{d \tau}{ \tau -2 -2 \sqrt{2}} - 4 \sqrt{2} ( \frac{\sqrt{2}-1}{2 \sqrt{2}}) \int \frac{d \tau}{ \tau -2 + 2 \sqrt{2}}
=( \sqrt{2}-1) \ln{( \tau -2 -2 \sqrt{2})} + (1- \sqrt{2}) \ln{( 2 +2 \sqrt{2})}

And so

t(4)=( \sqrt{2}-1) \ln{(2-2 \sqrt{2})} + (1- \sqrt{2}) \ln{(2+ 2 \sqrt{2})}

How's that?

 

[betel]

so

t(\tau)=\int \frac{ \sqrt{2} \tau d \tau}{ -\frac{1}{4} \tau^2 + \tau + 1} = \sqrt{2} \frac{1}{ -\frac{1}{4}} \int \frac{ \tau d \tau}{ \tau - 4 \tau - 4} = -4 \sqrt{2} \int \frac{ \tau d \tau}{ ( \tau -2 - 2 \sqrt{2})( \tau -2 + 2 \sqrt{2})}

One tau to much here.
But partial fractions is a good way to go later.