Geodesic Motion for the Gödel Space-time

[jfy4]

(Hopefully, Part 1 of 2)

This is one of my favorite metrics, and I decided that while tedious, and old-fashioned, I would practice for my GR studies by finding the Christoffell symbols and write out the equations for geodesics using the Gödel metric, then attempting to solve them.

First, the line element:

$$ds^2=a^2[dx^2+dy^2+\frac{1}{2}e^{2x}dz^2-(dt+e^xdz)^2]$$

which I got from Exact Solutions to Einstein's Fields Equations by Stephani et al page 178. Simplifying in order to write down the metric as a matrix

$$ds^2=-a^2dt^2+a^2dx^2+a^2dy^2-\frac{a^2}{2}e^{2x}dz^2-a^22e^{2x}dtdz$$

now as a matrix

$$g_{\alpha\beta}= \begin{pmatrix} -a^2 & 0 & 0 & -a^2e^x \\ 0 & a^2 & 0 & 0 \\ 0 & 0 & a^2 & 0 \\ -a^2e^x & 0 & 0 & -\frac{a^2}{2}e^{2x} \end{pmatrix}$$

After much trial and error and attempting to forecast which Christoffell symbols will be non-zero the following are the result:

$$\Gamma^t_{tt}=2$$

$$\Gamma^t_{tz}= \Gamma^t_{zt}=e^x$$

$$\Gamma^t_{zz}=\frac{e^{2x}}{2}$$

$$\Gamma^z_{tt}=-2e^{-x}$$

$$\Gamma^z_{tz}=\Gamma^z_{zt}=-1$$

$$\Gamma^z_{zz}=-e^{-x}$$

Now inserting these into the geodesic equation we get the following:

$$\frac{du^x}{d\tau}=0$$

$$\frac{du^y}{d\tau}=0$$

$$\frac{du^z}{d\tau}=2e^{-x}(u^t)^2+2u^tu^z+e^x(u^z)^2$$

$$\frac{du^t}{d\tau}=-2(u^t)^2-2e^xu^tu^z-\frac{e^{2x}}{2}(u^z)^2$$

If anyone would take the time to check this, does the procedure seem right and ok?

I am going to attempt to solve these by hand using the symmetries (Killing vectors) and the line element. So I guess more to come.

 

[jfy4]

Geodesic Motion for the Gödel Space-time (part 2)

(Part 2 (now inevitably of 3))

There appear to be Killing vectors associated with time, $$\mathbf{\xi}$$, as well as the z-direction, $$\mathbf{z}$$. There are an additional three Killling vectors, however, we shouldn't need them. These two are

$$\xi^{\alpha}=\langle 1,0,0,0\rangle$$

$$z^{\alpha}=\langle 0,0,0,1\rangle$$

We then have the following equalities:

$$g_{\alpha\beta}\xi^{\alpha}u^{\beta}=g_{tt}u^{t}+g_{tz}u^z\\ =-a^2\frac{dt}{d\tau}-a^2e^x\frac{dz}{d\tau}=\epsilon$$

$$g_{\alpha\beta}z^{\alpha}u^{\beta}=g_{zz}u^{z}+g_{zt}u^{t}\\ =-\frac{a^2}{2}e^{2x}\frac{dz}{d\tau}-a^2e^x\frac{dt}{d\tau}=\varpi$$

where $$\varpi$$ and $$\epsilon$$ are the conserved linear momentum in the z-direction per-unit rest mass, and conserved energy per-unit rest mass. Using these, we have two equations and two unknowns. Using substitution, things clean up to yield

$$\frac{dz}{d\tau}=\frac{2e^{-x}(\varpi e^{-x}-\epsilon)}{a^2}$$

and

$$\frac{dt}{d\tau}=\frac{\epsilon-2\varpi e^{-x}}{a^2}.$$

The other two components of the 4-velocity are constants from the geodesic equation. That is,

$$\frac{dx}{d\tau}=u_{0}^{x}$$

and

$$\frac{dy}{d\tau}=u_{0}^{y}.$$

These are the 4-velocity. Next post I will attempt to solve these for the shape of geodesics. Stay tuned.

EDIT: I decided on a different symbol other than pi.

 

[Mentz114]

Is the dtdz term correct in the matrix ? Starting from

$$ds^2=-a^2dt^2+a^2dx^2+a^2dy^2-\frac{a^2}{2}e^{2x}dz^2-a^22e^{2x}dtdz$$

I get

$$g_{\alpha\beta}= \begin{pmatrix} -a^2 & 0 & 0 & -a^2e^{2x} \\ 0 & a^2 & 0 & 0 \\ 0 & 0 & a^2 & 0 \\ -a^2e^{2x} & 0 & 0 & -\frac{a^2}{2}e^{2x} \end{pmatrix}$$

However, using the above or the one you posted I get different Christoffel symbols. For the one you posted

[deleted some wrong stuff]
( I tried hard not mistranscribe these, but there is a possibility I have).

[Edit] I did mistranscribe them. The indexes of the gammas are wrong here]

I hope you aren't doing this by hand ...

 

[jfy4]

Ah, I foolishly mixed up addition with the cross terms as addition with the powers of the exponent...

I am doing this by hand. I'll correct this. My mistake everyone. Please hold.

 

[Mentz114]

My turn to blush - I got the indexes wrong on the gammas, so I deleted them.

Here they are again

$$\begin{align*} {\Gamma^t}_{xt}&= 1\\ {\Gamma^z}_{xt}&= -{e}^{-x}\\ {\Gamma^x}_{zt}&= \frac{{e}^{x}}{2} \\ {\Gamma^t}_{zx}&= \frac{{e}^{x}}{2} \\ {\Gamma^x}_{zz}&= \frac{{e}^{2\,x}}{2}\\ \end{align*}$$

 

[jfy4]

Corrections (part 1)

Corrections

Here is the new Metric with it's pointed out corrections.

$$g_{\alpha\beta}= \begin{pmatrix} -a^2 & 0 & 0 & -a^2e^{2x} \\ 0 & a^2 & 0 & 0 \\ 0 & 0 & a^2 & 0 \\ -a^2e^{2x} & 0 & 0 & -\frac{a^2}{2}e^{2x} \end{pmatrix}$$

I'll post my inverse metric also just to make sure

$$g^{\alpha\beta}= \begin{pmatrix} \frac{1}{a^2(2e^{2x}-1)} & 0 & 0 & \frac{2}{a^2-2a^2e^{2x}} \\ 0 & \frac{1}{a^2} & 0 & 0 \\ 0 & 0 & \frac{1}{a^2} & 0 \\ \frac{2}{a^2-2a^2e^{2x}} & 0 & 0 & \frac{2e^{-2x}}{a^2(2e^{2x}-1)} \end{pmatrix}$$

This will most certainly provide some clarity. Now here are the new Chritoffell symbols. I had some agreement with you Mentz, but not all... I'd be grateful if you looked over this again.

$$\Gamma^{x}_{zz}=\frac{e^{2x}}{2}$$

$$\Gamma^{x}_{zt}=\Gamma^{x}_{tz}=e^{2x}$$

$$\Gamma^{z}_{xz}=\Gamma^{z}_{zx}=1$$

$$\Gamma^{z}_{xt}=\Gamma^{z}_{tx}=\frac{2}{1-2e^{2x}}$$

$$\Gamma^{t}_{xt}=\Gamma^{t}_{tx}=\frac{2e^{2x}}{2e^{2x}-1}$$

With these I recalculated the geodesic equations

$$\frac{du^x}{d\tau}=-\frac{e^{2x}}{2}(u^z)^2-2e^{2x}u^zu^t$$

$$\frac{du^z}{d\tau}=-2u^xu^z-\frac{4}{1-2e^{2x}}u^xu^t$$

$$\frac{du^t}{d\tau}=-\frac{4e^{2x}}{2e^{2x}-1}u^xu^t$$

$$\frac{du^y}{d\tau}=0.$$

I will use the same method as before to get the 4-velocity and then the shape of the geodesic. However, I'll wait a bit before posting (as I should have the first time) to see if I made another mistake. I do apologize.

 

[Mentz114]

This will most certainly provide some clarity. Now here are the new Christoffel symbols. I had some agreement with you Mentz, but not all... I'd be grateful if you looked over this again.

Your Christoffel symbols agree with Maxima. Nice work.
(The CS I posted earlier are for your original metric.)

I'll check the EOMs much later tonight or tomorrow but I doubt if they're wrong.

$$\frac{du^x}{d\tau}$$

I find this notation strange. This looks clearer,

$$\frac{dx}{d\tau}$$

In fact it is a bit naughty to use coordinate labels as tensor indexes, but ${\Gamma^t}_{xz}$ is clear and unambiguous.

 

[jfy4]

The first expression you wrote is for the proper acceleration in the x direction. The second expression you wrote is for the proper velocity. At least That is how I meant them.

 

[martinbn]

Just a small remark, you can get the equations of the geodesics (and from them you can read off the Christoffel symbols) as the Euler-Lagrange equations for the functional

$$L=\frac12 g_{\alpha\beta}\dot{x}^\alpha\dot{x}^{\beta},$$

which saves a lot of computations.

 

[Mentz114]

The first expression you wrote is for the proper acceleration in the x direction. The second expression you wrote is for the proper velocity. At least That is how I meant them.

Yes, my mistake. $du^x/d\tau$ is $d^2x/d\tau^2$. I came back to change my post but you'd already seen it. I'm also doing things in a rush.

 

[jfy4]

Just a small remark, you can get the equations of the geodesics (and from them you can read off the Christoffel symbols) as the Euler-Lagrange equations for the functional

$$L=\frac12 g_{\alpha\beta}\dot{x}^\alpha\dot{x}^{\beta},$$

which saves a lot of computations.

Could you make a thread where you do that? I would really appreciate seeing that in detail.

 

[jfy4]

Yes, my mistake. $du^x/d\tau$ is $d^2x/d\tau^2$. I came back to change my post but you'd already seen it. I'm also doing things in a rush.

No problem.

 

[jfy4]

As I said earlier, I am using the Killing vectors to solve for the 4-velocity components. The first two somewhat easy ones are the time and z components. The corresponding Killing vectors are:

$$\xi^{\alpha}=\langle 1,0,0,0\rangle$$

$$\eta^{\alpha}=\langle 0,0,0,1\rangle$$

Using these we can get two expressions to solve for the t and z components of velocity.

$$g_{\alpha\beta}\xi^{\alpha}u^{\beta}=g_{tt}u^{t}+g _{tz}u^z\\ =-a^2u^t-a^2e^{2x}u^z=\epsilon$$

$$g_{\alpha\beta}\eta^{\alpha}u^{\beta}=g_{zz}u^{z}+g_{zt}u^{t}\\ =-\frac{a^2}{2}e^{2x}u^z-a^2e^{2x}u^t=\varpi$$

Here $$\epsilon$$ and $$\varpi$$ are (like before) the conserved energy per-unit rest mass and the conserved linear momentum in the z direction per-unit rest mass (I decided on this by inspection...). With these two equations and the two unknown velocities, they can be solved to get

$$u^t=\frac{\varpi-\frac{\epsilon}{2}}{a^2(\frac{1}{2}-e^{2x})}$$

and

$$u^z=\frac{\epsilon-e^{-2x}\varpi}{a^2(\frac{1}{2}-e^{2x})}$$

The y component is easy

$$u^y=u_{y_0}$$

However the x component appears more difficult. And, in fact, ended up extremely messy, and so I am a little nervous. Here was my procedure. I started with the line element. I then multiplied through by a $$-1$$ to get it in terms of $$d\tau^2$$. I then divided through by $$d\tau^2$$ and $$a^2$$ to get

$$\frac{1}{a^2}=\left(\frac{dt}{d\tau}\right)^2-\left(\frac{dx}{d\tau}\right)^2-\left(\frac{dy}{d\tau}\right)^2+\frac{1}{2}e^{2x}\left(\frac{dz}{d\tau}\right)^2+2e^{2x}\left(\frac{dt}{d\tau}\right)\left(\frac{dz}{d\tau}\right)$$

With this expression I substituted back in my solutions for the other three components of velocity and simplified as much as I could imagine. The x component of the 4-velocity turned out to be

$$u^x=\sqrt{\frac{2e^{-2x}\varpi^2}{a^4(1-2e^{2x})}-\frac{1}{a^2}-u_{y_0}^2-\frac{\epsilon(\epsilon-4\varpi)}{a^2(2e^{2x}-1)}}.$$

Quite messy I know... For the shape of the geodesics I was going to divide the non-time 4-velocity components by $$dt/d\tau$$ in order to get all the geodesic shapes in terms of t. And while this seems possible for y and z, x seems a little intense. But I will attempt something and see how it goes from there. I may need to take a different strategy for the x component. Stay tuned.

 

[WannabeNewton]

Could you make a thread where you do that? I would really appreciate seeing that in detail.

Have you worked with the euler lagrange equations before?

 

[jfy4]

Have you worked with the euler lagrange equations before?

Yes...

EDIT: I said that in a voice that would resemble a skeptic, not in a sassy offensive way FYI.

 

[WannabeNewton]

Yes...

EDIT: I said that in a voice that would resemble a skeptic, not in a sassy offensive way FYI.

Just asking because the variational method for geodesics posted before is just taking half the line element and using the euler lagrange equations with that to read off the christoffel symbols.

 

[jfy4]

Just asking because the variational method for geodesics posted before is just taking half the line element and using the euler lagrange equations with that to read off the christoffel symbols.

Solid, I've have not done that with with a specific metric. I have derived the geodesic equation using the Lagrangian, however that was general, and not for a specific metric.

I will try it. It sounds great.

 

[jfy4]

Hopefully Final

Well, besides all of the mistakes and retakes, not that bad...

Here is everything together after my attempts to correct everything.

Here is the metric we are using: $$g_{\alpha\beta}= \begin{pmatrix} -a^2 & 0 & 0 & -a^2e^x \\ 0 & a^2 & 0 & 0 \\ 0 & 0 & a^2 & 0 \\ -a^2e^x & 0 & 0 & -\frac{a^2}{2}e^{2x} \end{pmatrix}$$

This is the Gödel space-time.

From this metric one can calculate the geodesic equations using either the Christoffell symbols and inverse metric, or, as I recently tried out in full swing with a couple metrics for fun, Langrange's equations on the Lagrangian (really cool!). The geodesic equations are

$$\ddot{t}=-2\dot{x}\dot{t}-e^{x}\dot{x}\dot{z}$$

$$\ddot{z}=2e^{-x}\dot{x}\dot{t}$$

$$\ddot{x}=-e^{x}\dot{t}\dot{z}-\frac{e^{2x}}{2}(\dot{z})^2$$

$$\ddot{y}=0.$$

Then using the Killing vectors $$\xi^{\alpha}=\langle 1,0,0,0\rangle$$

$$\eta^{\alpha}=\langle 0,0,0,1\rangle$$

$$\Psi^{\alpha}=\langle 0,0,1,0\rangle$$

one can put the 4-velocity components in terms of conserved quantities. Specifically in terms of the conserved energy per-unit rest mass, $$\epsilon$$, conserved linear momentum in the z-direction per-unit rest mass, $$\varpi$$, and the conserved linear momentum in the y-direction per-unit rest mass,$$\psi$$.

Then using the line element it is possible to solve for the x component of velocity. They all are:

$$\dot{t}=\frac{e^x\epsilon-2\varpi}{a^2e^x}$$

$$\dot{x}=\frac{\sqrt{4\epsilon\varpi e^x-a^2e^{2x}-\epsilon^2 e^{2x}-2\varpi^2-e^{2x}\psi^2}}{a^2e^x}$$

$$\dot{y}=\frac{\psi}{a^2}$$

$$\dot{z}=\frac{2(\varpi e^{-x}-\epsilon)}{a^2e^x}$$

With these I attempted to solve for the 3-velocity in terms of time, and was unsuccessful. I ran the x differential equation through Mathematica, and there was a closed symbolic form! however, to put it lightly, it was horrific... I will not post it up because of it's appearance. Thanks for waiting and helping out a lot. I learned a lot during this process. Cheers.

 

[Mentz114]

Things get a bit simpler in the frame field of the comoving observer. Using this coframe basis

$$\vec{e_0}= a\ dt -a\,{e}^{x}dz,\ \ \ \vec{e_1}= a\ dx,\ \ \ \vec{e_2}= a\ dy,\ \ \ \vec{e_3}= -\frac{a\,{e}^{x}}{\sqrt{2}}dz$$

for the stationary frame [tex]V^\mu=\langle -1,0,0,0 \rangle[/itex]( with covector $U$) the acceleration vector $U_{mu;\nu}V^\nu$ is zero. This means comoving observers are on geodesics. There is an interesting twist ( no pun intended ) because the rotation tensor $\omega_{mn} &= u_{[m;n]} + \dot{u}_{[m}u_{n]}$ ( writing $u$ for $U$ ) has non-zero components

[tex]
\omega_{13}=-\omega_{31}=\frac{1}{\sqrt{2}\,a}
[/itex]

this gives a vorticity vector
$$\omega^m=\frac{1}{2} \epsilon^{mbnk}\ U_b\ \omega_{nk} = \left(0 , 0 , -\frac{1}{\sqrt{2}\,a} , 0} \right)$$

which means this observer is experiencing some kind of rotation around the y-axis, with angular velocity $-\frac{1}{\sqrt{2}\,a}$.

 

[jfy4]

Things get a bit simpler in the frame field of the comoving observer. Using this coframe basis...

Wow, that is really cool.

I have a question. How did you know what the co-frame basis was explicitly? Was it from a reference, or is there a general procedure to build the co-frame basis?

 

[Mentz114]

I used guesswork and a procedure. Start by positing a coframe field with some unknowns, then work out the metric from these, then solve algebraic equations to get the unknowns.

The key is that

$$g_{\mu\nu}= -e_0\otimes e_0 +e_1\otimes e_1 +e_2\otimes e_2 +e_3\otimes e_3$$

I can post the whole calculation if you want to see it.

People with a lot of experience can read the frame and coframe basies from the metric but I still need to work them out if the metric has off-diagonal terms.

 

[jfy4]

If you wouldn't mind that would be great. It seems extremely useful and cool.

 

[Mentz114]

OK. Having done this for boosted frames, I guessed that the coframe would be of the form

$$\vec{e_0}= a\ dt\ -\ e03\ dz ,\ \ \ \vec{e_1}= a\ dx,\ \ \ \vec{e_2}= a\ dy,\ \ \ \vec{e_3}= e33\ dz$$

where e03 and e33 are unknown. Working out the sum of the tensor products shown in my previous post gives

$$\left[ \begin{array}{cccc} -{a}^{2} & 0 & 0 & -a\ e03 \\ 0 & {a}^{2} & 0 & 0 \\ 0 & 0 & {a}^{2} & 0 \\ -a\ e03 & 0 & 0 & {e33}^{2}-{e03}^{2} \end{array} \right]$$

and it's trivial to solve for e03 and e33 so this matrix is the required metric.

I've also done this with a frame basis where the target is the inverse metric.

 

[WannabeNewton]

Just some quick questions (hopefully). Did you get the killing vectors by solving killing's equation or did you use known properties of the space - time described by the godel metric? Also, what property of the space - time results in the metric admitting a killing vector for the z - direction (why is the linear momentum conserved in the z direction)? Thanks.

EDIT: Never mind I was too stupid to look at the actual metric and see that it doesn't depend on y or z. But out of curiosity how come you didn't use conservation of the y component of linear momentum when you took the isometries into account to simplify the geodesic equations?

 

[jfy4]

...out of curiosity how come you didn't use conservation of the y component of linear momentum when you took the isometries into account to simplify the geodesic equations?

I'm not sure I see what your getting at. Could you be more specific. Do you mean you want me to substitute the 4-velocity components into the geodesic equations to put them in terms of the conserved quantities?

 

[jfy4]

for the stationary frame [tex]V^\mu=\langle -1,0,0,0 \rangle[/itex]( with covector $U$) the acceleration vector $U_{mu;\nu}V^\nu$ is zero...

What kind of acceleration vector is $U_{mu;\nu}V^{\nu}$. What are the components of a rank 2 acceleration vector (tensor)? That seems a little odd to me.

 

[Mentz114]

What kind of acceleration vector is $U_{\mu;\nu}V^{\nu}$. What are the components of a rank 2 acceleration vector (tensor)? That seems a little odd to me.

I mis-typed it. It's a covariant vector ( 1-form).

$$\dot{U}_\mu = U_{\mu;\nu}V^{\nu}$$

also written

$$\nabla_\nu\ U_\mu\ V^\nu$$