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

[betel]

Grand.
Ok. So the t equation is:

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

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

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

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

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:

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

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

$\frac{d^2t}{d \tau^2} + \frac{1}{2(1+r)} \frac{dt}{d \tau} \frac{dr}{d \tau}=0$

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:

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

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

$\frac{d^2t}{d \tau^2} + \frac{1}{2(1+r)} \frac{dt}{d \tau} \frac{dr}{d \tau}=0$

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

$\frac{d^2t}{d \tau^2} + \frac{1}{(1+r)} \frac{dt}{d \tau} \frac{dr}{d \tau}=0$

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

$\frac{d^2t}{d \tau^2} + \frac{1}{(1+r)} \frac{dt}{d \tau} \frac{dr}{d \tau}=0$

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

$\frac{d^2r}{d \tau^2} - \frac{1}{2} g_{\alpha \beta} u^\alpha u^\beta=0$

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

$\frac{d^2r}{d \tau^2} - \frac{1}{2} g_{\alpha \beta} u^\alpha u^\beta=0$

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 $r(\tau)=-\frac{1}{4} \tau^2 + \tau$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 $\frac{dr}{d \tau}$?

 

[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
$$\frac{d}{d\tau} \left((1+r)\frac{dt}{d\tau}\right)=0$$
which is correct. No need to expand. You can directly conclude
$$\frac{dt}{d\tau}=\frac{const}{1+r}$$
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
$$\frac{d}{d\tau} \left((1+r)\frac{dt}{d\tau}\right)=0$$
which is correct. No need to expand. You can directly conclude
$$\frac{dt}{d\tau}=\frac{const}{1+r}$$
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.

 

[latentcorpse]

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

Baah! I'm an idiot!

Ok so I get

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

and so

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

?

 

[betel]

Baah! I'm an idiot!

Ok so I get

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

How did you get the 1/4Sqrt(2) prefactor. Without that i agree

$=- \int \frac{d \tau}{\tau -2 -2 \sqrt{2}} + \int \frac{d \tau}{\tau -2 + 2 \sqrt{2}}$
$= \ln{( \tau -2 + 2 \sqrt{2})} - \ln{( \tau -2 -2 \sqrt{2})}$

and so

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

?

I think you missed the lower limit.

 

[betel]

Sorry, i was a bit too quick. I get only a prefactor of 1/2Sqrt(2) and taking the lower limit gives me 4 times your result.

 

[betel]

We covered that point already.

 

[betel]

pfff, not my day. I missed the 2, so your prefactor is all fine. But you still have to include the lower limit.

 

[latentcorpse]

How did you get the 1/4Sqrt(2) prefactor. Without that i agree

I think you missed the lower limit.

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

Now $\frac{1}{( \tau -2 - 2 \sqrt{2})( \tau -2 + 2\sqrt{2})}=\frac{A}{\tau -2 - 2 \sqrt{2}} + \frac{B}{\tau -2 + 2 \sqrt{2}}$

So to get B

$\frac{1}{\tau - 2 - 2 \sqrt{2}} = \frac{A ( \tau - 2 + 2 \sqrt{2})}{ \tau -2 - 2 \sqrt{2}} + B$
set $\tau = 2 - 2 \sqrt{2} \Rightarrow B= -\frac{1}{4 \sqrt{2}}$

and similarly we find $A=\frac{1}{4 \sqrt{2}}$

this means then that

$t (\tau) = -4 \sqrt{2} \frac{1}{4 \sqrt{2}} \int_0^\tau \frac{d \tau}{\tau -2 - 2 \sqrt{2}} - 4 \sqrt{2} (-\frac{1}{4 \sqrt{2}}) \int_0^\tau \frac{d \tau}{\tau -2 + 2 \sqrt{2}}$
$=-\int_0^tau \frac{d \tau}{\tau -2 - 2 \sqrt{2}} + \int_0^\tau \frac{d \tau}{\tau -2 + 2 \sqrt{2}}$
$=-\ln{(\tau -2 + 2 \sqrt{2})} + \ln{(2 \sqrt{2} - 2)} + \ln{( \tau -2 - 2 \sqrt{2})} - \ln{(-2-2 \sqrt{2})}$

So $t(4)=\ln{2+ 2 \sqrt{2}} - \ln{2-2 \sqrt{2}} - \ln{ 2 \sqrt{2} - 2} + \ln{-2 - 2 \sqrt{2}}$
$=\ln{ \frac{(2 + \sqrt{2})(-2 - 2 \sqrt{2})}{(2-2 \sqrt{2})( 2 \sqrt{2} - 2 )}$
$=\ln{ \frac{-12 - 8 \sqrt{2}}{-12 + 8 \sqrt{2}}$

How's that?

Thanks!

 

[betel]

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

Now $\frac{1}{( \tau -2 - 2 \sqrt{2})( \tau -2 + 2\sqrt{2})}=\frac{A}{\tau -2 - 2 \sqrt{2}} + \frac{B}{\tau -2 + 2 \sqrt{2}}$

So to get B

$\frac{1}{\tau - 2 - 2 \sqrt{2}} = \frac{A ( \tau - 2 + 2 \sqrt{2})}{ \tau -2 - 2 \sqrt{2}} + B$
set $\tau = 2 - 2 \sqrt{2} \Rightarrow B= -\frac{1}{4 \sqrt{2}}$

and similarly we find $A=\frac{1}{4 \sqrt{2}}$

this means then that

$t (\tau) = -4 \sqrt{2} \frac{1}{4 \sqrt{2}} \int_0^\tau \frac{d \tau}{\tau -2 - 2 \sqrt{2}} - 4 \sqrt{2} (-\frac{1}{4 \sqrt{2}}) \int_0^\tau \frac{d \tau}{\tau -2 + 2 \sqrt{2}}$
$=-\int_0^tau \frac{d \tau}{\tau -2 - 2 \sqrt{2}} + \int_0^\tau \frac{d \tau}{\tau -2 + 2 \sqrt{2}}$
$=-\ln{(\tau -2 + 2 \sqrt{2})} + \ln{(2 \sqrt{2} - 2)} + \ln{( \tau -2 - 2 \sqrt{2})} - \ln{(-2-2 \sqrt{2})}$

So $t(4)=\ln{2+ 2 \sqrt{2}} - \ln{2-2 \sqrt{2}} - \ln{ 2 \sqrt{2} - 2} + \ln{-2 - 2 \sqrt{2}}$
$=\ln{ \frac{(2 + \sqrt{2})(-2 - 2 \sqrt{2})}{(2-2 \sqrt{2})( 2 \sqrt{2} - 2 )}$
$=\ln{ \frac{-12 - 8 \sqrt{2}}{-12 + 8 \sqrt{2}}$

How's that?

Thanks!

I think that is what I got. You should swap the denominator, otherwise you take log of something negative.

 

[latentcorpse]

I think that is what I got. You should swap the denominator, otherwise you take log of something negative.

well I can simplify it to

$\ln{\frac{-3-2 \sqrt{2}}{-3+2 \sqrt{2}}}$

Clearly that argument is negative which causes problems but what do you mean swap the denominator?

Also, why was it convenient (or necessary?) to rewrite the geodesic in that covariant way? Would it not have worked if we had just used the standard geodesic equation and if not, why not?
How did you know you had to rewrite it that way?

Thanks very much for your help on this question!

 

[betel]

well I can simplify it to

$\ln{\frac{-3-2 \sqrt{2}}{-3+2 \sqrt{2}}}$

Clearly that argument is negative which causes problems but what do you mean swap the denominator?

When getting log as the result of an integral you always have to take absolute values. So if you ignore it at the start, you have manually correct later.

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Also, why was it convenient (or necessary?) to rewrite the geodesic in that covariant way? Would it not have worked if we had just used the standard geodesic equation and if not, why not?
How did you know you had to rewrite it that way?

Thanks very much for your help on this question![/QUOTE]
Not neccessary, but this form of the geodesic equation is of more practical use in some cases. Otherwise you have to first calculate all the Christoffel symbols and then sum. Here if your metric does not depend on a couple of coordinates, because of the partial derivative the geodesic equation for such coordinates will be very easy.

And I knew it was a practical way to rewrite it because I gave the question to my students rececently.