Lense-Thirring effect - General Relativity

[Pentaquark5]

Let us assume a "toy-metric" of the form
$$ g=-c^2 \mathrm{d}t^2+\mathrm{d}x^2+\mathrm{d}y^2+\mathrm{d}z^2-\frac{4GJ}{c^3 r^3} (c \mathrm{d}t) \left( \frac{x\mathrm{d}y-y\mathrm{d}x}{r} \right)$$
where $J$ is the angular-momentum vector of the source.

Consider the curve
$$ \gamma(\tau)=(x^\mu (\tau))=\bigg(t(\tau),0,0,z(\tau)\bigg) $$

To see wether this is a geodesic, we need to check if the geodesic equation in local coordinates
$$ \frac{\mathrm{d}^2x^\mu}{\mathrm{d}\tau^2}=-\Gamma^\mu_{\:\:\alpha\beta} \frac{\mathrm{d}x^\alpha}{\mathrm{d}\tau}\frac{\mathrm{d}x^\beta}{\mathrm{d}\tau}$$
is fulfilled.

In order to do that, we will need to calculate the Christoffel symbols where $\alpha, \beta$ equals $r$ or $z$:

Along $\gamma$ we have $g_{\mu \nu}=\eta_{\mu \nu}$ and
$$ \Gamma^\mu_{\:\: z\alpha}=\frac{1}{2}\eta^{\mu \nu}\big( \partial_z g_{\mu \alpha}+\partial_\alpha g_{\mu z}-\partial_\nu g_{z \alpha} \big)=0 $$
Indeed,
$$ \partial_z g_{t y}|_\gamma = \partial_z \left( -\frac{4GJx}{c^2r^3} \right)|_\gamma = x\partial_x\left(-\frac{4GJ}{c^2r^3}\right)|_\gamma=0$$There are a few questions I have about this computation, and I would be really thankful for input:
Firstly, can we assume $g_{\mu \nu}=\eta_{\mu \nu}$ along the curve, since spacetime is locally Minkowski along $\gamma$?
Secondly, am I missing something, or should the denominator of $g_{t y}$ be proportional to $r^{-4}$ instead of $r^{-3}$?
And thirdly, since $r=r(x,y,z)=\sqrt{x^2+y^2+z^2}$, shouldn't the derivation $\partial_z g_{t y}|_\gamma$ be nonzero?

 

[PeterDonis]

spacetime is locally Minkowski along $\gamma$?

What makes you think this?

 

[Pentaquark6]

Hey, (I didn't know my PW by heart so I had to create a new account)

See the proof that every Lorentzian Metric is locally Minkowski on p.31 of this script: https://gravity.univie.ac.at/fileadmin/user_upload/i_gravity_physics/material/teaching/rt2/SS_2013/Vienna2013.pdf

 

[PeterDonis]

See the proof that every Lorentzian Metric is locally Minkowski

That doesn't mean that $g_{\mu \nu} = \eta_{\mu \nu}$. For that you have to choose an appropriate coordinate chart. The chart you are using does not do this, even at a point, so you can't use $g_{\mu \nu} = \eta_{\mu \nu}$ in your reasoning if you are using that chart.

 

[PeterDonis]

See the proof that every Lorentzian Metric is locally Minkowski

This proof shows that you can choose coordinates such that $g_{\mu \nu} = \eta_{\mu \nu}$ at a single point. It does not show that you can do so such that $g_{\mu \nu} = \eta_{\mu \nu}$ along an entire curve $\gamma$, which is what you claimed. It is actually possible to show the latter, but doing it requires a different choice of coordinates. And neither of these choices of coordinates is the one you used in the OP, so all this is irrelevant anyway.

 

[Ibix]

Also the metric is singular at x=y=z=0, if I'm not mistaken.

 

[Pentaquark6]

That doesn't mean that $g_{\mu \nu} = \eta_{\mu \nu}$. For that you have to choose an appropriate coordinate chart. The chart you are using does not do this, even at a point, so you can't use $g_{\mu \nu} = \eta_{\mu \nu}$ in your reasoning if you are using that chart.

This proof shows that you can choose coordinates such that $g_{\mu \nu} = \eta_{\mu \nu}$ at a single point. It does not show that you can do so such that $g_{\mu \nu} = \eta_{\mu \nu}$ along an entire curve $\gamma$, which is what you claimed. It is actually possible to show the latter, but doing it requires a different choice of coordinates. And neither of these choices of coordinates is the one you used in the OP, so all this is irrelevant anyway.

Thank you for your answers!

I know that $\eta_{\mu \nu}=g_{\mu \nu}$ only holds point for point, which is why I was asking if this was the justification for setting $\eta_{\mu \nu}=g_{\mu \nu}$: I couldn’t see how this would hold for all of $\gamma$ in those coordinates.

And just to avoid miscommunication: the above is copied from my script, it’s not written by me.

 

[martinbn]

You need to find the Christoffel symbols first, then evaluate them on the curve, when you substitute the in the geodesic equations. And you cannot check that the curve is geodesic, it has unknown functions in it. You can use the geodesic equations to find ODE's for these functions. Their solutions will give you geodesic curves.

 

[Pentaquark6]

You need to find the Christoffel symbols first, then evaluate them on the curve, when you substitute the in the geodesic equations. And you cannot check that the curve is geodesic, it has unknown functions in it. You can use the geodesic equations to find ODE's for these functions. Their solutions will give you geodesic curves.

I’ve already done that. The geodesic is of the form $(x(\tau))=(u^0,0,0,u^3)\tau$ with $u^0=\sqrt{1+(u^z)^2}$. Here $u^0$ and $u^3$ are constant.

However that does not answer the questions formulated in my OP.

 

[martinbn]

However that does not answer the questions formulated in my OP.

Which question! The first was already discussed.

The second.

Secondly, am I missing something, or should the denominator of $g_{t y}$ be proportional to $r^{-4}$ instead of $r^{-3}$?

The denominator of $g_{ty}$ is proportional to $r^{-4}$. So I don't understand your question whether it should be. It is.

The third.

And thirdly, since $r=r(x,y,z)=\sqrt{x^2+y^2+z^2}$, shouldn't the derivation $\partial_z g_{t y}|_\gamma$ be nonzero?

On the curve $x=0$, so the derivative is zero on the curve, although not zero everywhere.

 

[Pentaquark6]

I’m still confused about the reason as to why it is legitimate to assume $\eta_{\mu \nu}=g_{\mu \nu}$ along $\gamma$, as my script does.

 

[martinbn]

It is not an assumption, it is a substitution. The Christoffel symbols and the components of the metric are functions. So to evaluate them at a point you plug in the point's coordinates. If the point is on the curve, the $x$ and the $y$ are zero.

 

[PeterDonis]

I know that $\eta_{\mu \nu}=g_{\mu \nu}$ only holds point for point

It only holds at one point with a particular choice of coordinates. You keep missing this crucial qualifier. With different choices of coordinates, it could be that $g_{\mu \nu} = \eta_{\mu \nu}$ at more than one point. You have to look at the particular coordinates to see.

In the coordinates you gave in the OP, at any point where $x = y = 0$, you will have $g_{\mu \nu} = \eta_{\mu \nu}$. That's obvious just from looking at the line element, since the only term that differs from $\eta_{\mu \nu}$ is zero if $x = y = 0$.