Solution strategy for linearized gravity problem

[Jonsson]

[Moderator's note: moved to homework forum.]

Hello there,

I am stuck at a problem, and I need some hits/solution strategy to get going. Suppose we consider linearized gravity, and there is some mass, $M$ at the origin at some unknown coordinates the metric is
$$
ds^2 = -(1 + 2 \phi)dt^2 + (1-2\phi)\delta_{ij}dx^i dx^j
$$
for $\phi = -GM/(x^2 + y^2 + z^2) \ll 1$.

Suppose that the mass $M$ now moves in the x direction with velocity $v$.
(1) What is the metric in this case
(2) A photon is falling freely in the $y$ direction. I.e. its undeflected path is $-b\vec e _x + t \vec e_y$. What angle is the actual photon trajectory deflected by?

The first complication is that I don't know what the unknown coordinates are. How do I start this problem? The professor is giving a hint: for (2) Transforming to the rest frame of the moving mass and back is much easier than using the geodesic eqn.

Kind regards,
Marius

 

[Paul Colby]

In linearized gravity I think the Lorentz transformation gives the answer.

 

[Jonsson]

Can you be more specific? There are 3 sets of coordinates: 1 is the coordinates in which we know the placement of mass and the photon. Call these coordiantes $x^\mu$. There is the frame of the stationary mass $\xi^\mu$ and the frame where the mass is moving, in which we somehow obtain the metric from problem (1), call these coordinates $\zeta^\mu$. How do I find the coordinates of the photon in the $\zeta^\mu$ coordinates? It should be possible to boost these to the $\xi^\mu$ coordinates, then somehow work out the trajectory, and finally transform back to $\zeta^\mu$ and eventually to $x^\mu$?

 

[Paul Colby]

Can you be more specific?

The answer to (1) is take your metric and do the transform? Let's pretend I know what the coordinates of a photon are, take them in one frame and transform them to another.

 

[PeterDonis]

In linearized gravity I think the Lorentz transformation gives the answer

No, it doesn't. The Lorentz transformation only works between two coordinate charts both of which are Minkowski, i.e., both of which have the metric $\eta_{\mu \nu}$. That's certainly not true of the chart described in the OP, and I would not expect it to be true of the transformed chart that is being asked for either.

Linearized gravity does not mean spacetime is approximated as flat. It only means that nonlinear effects of spacetime curvature are ignored (assumed to be too small to matter). But linear effects of spacetime curvature are still present.

 

[Paul Colby]

Linearized gravity does not mean spacetime is approximated as flat.

true, what about the metric the OP has given?

 

[PeterDonis]

what about the metric the OP has given?

It's obviously not flat, because of the terms in $\phi$. If you wanted to confirm that it's not flat, you could compute its Riemann tensor; it won't vanish.

 

[Paul Colby]

It's obviously not flat, because of the terms in ϕ.

Isn't $\phi$ specified to be small?

 

[Paul Colby]

Riemann tensor; it won't vanish.

True, are you say linearized gravity doesn't hold ever? The form of $\phi$ given is both small and vanishes at spatial infinity. One may always change coordinates. Why will the Lorentz transform fail here?

 

[PeterDonis]

are you say linearized gravity doesn't hold ever

No, I'm saying that "linearized gravity" is not the same as "spacetime is flat". As I said in post #5.

For a good quick treatment, see the beginning of Chapter 6 in Carroll's online lecture notes:

https://arxiv.org/pdf/gr-qc/9712019.pdf

Note carefully his expressions for the metric, Riemann tensor, and EFE in linearized gravity.

Why will the Lorentz transform fail here?

Mathematically, of course you can apply any coordinate transformation you want to. Whether the transform is telling you anything meaningful physically is a different question. The latter is what I'm referring to when I say the LT does not "work" unless it's between global inertial charts in flat spacetime.

 

[Jonsson]

So what do I do?

 

[Paul Colby]

It's says in the problem statement you provided.

Suppose that the mass $M$ now moves in the x direction with velocity $v$.

My thought was change to the rest frame of $M$. This is where the Lorentz transformation suggestion I was making which PeterDonis seemed to poo poo came from. That still seems reasonable to me given the rest of the problem statement but I wouldn't weight my opinion too much.

Second thing of which I am more certain is to ask what path will the photon take and what equation would this path obey?

 

[Jonsson]

Second thing of which I am more certain is to ask what path will the photon take and what equation would this path obey?

We want to find the path based on the initial conditions given.

Wikipedia says that Lorentz transformation is only correct for inertial coordinates. However, I was flipping though Gravitation by Misner et al.

On page 439, it says that for linearized gravity

$$
\eta_{\alpha' \beta'} + h_{\alpha' \beta'}= g_{\alpha' \beta'} = \frac{\partial x^\mu}{\partial x^{\alpha'}}\frac{\partial x^\nu}{\partial x^{\beta'}}g_{\mu \nu} = \Lambda^\mu_{\alpha'}\Lambda^\nu_{\beta'}(\eta_{\mu\nu}+ h_{\mu\nu}) = \eta_{\alpha' \beta'} + \Lambda^\mu_{\alpha'}\Lambda^\nu_{\beta'}h_{\mu \nu}
$$
If we can understand this, then it means we can transform the metric coefficients. Problem is, I don't understand that this equality is true:
$$
\frac{\partial x^\mu}{\partial x^{\alpha'}}\frac{\partial x^\nu}{\partial x^{\beta'}}g_{\mu \nu} = \Lambda^\mu_{\alpha'}\Lambda^\nu_{\beta'}(\eta_{\mu\nu}+ h_{\mu\nu})
$$

Can you explain why it is?

 

[Paul Colby]

Can you explain why it is?

Yes. In the linearized gravity where $h_{\mu \nu}$ is about a flat Minkowski metric $\eta_{\mu \nu}$ then $h_{\mu \nu}$ both transform as Lorentz tensors however the $\eta_{\mu \nu}$ is component wise Lorentz invariant.

 

[Paul Colby]

Another question to ask is the bit about $M$ moving part of the problem you were given or did you add that bit as part of your struggle?

 

[Jonsson]

Yes. In the linearized gravity where $h_{\mu \nu}$ is about a flat Minkowski metric $\eta_{\mu \nu}$ then $h_{\mu \nu}$ both transform as Lorentz tensors however the $\eta_{\mu \nu}$ is component wise Lorentz invariant.

I don't understand, how does this quote explain

$$
\frac{\partial x^\mu}{\partial x^{\alpha'}}\frac{\partial x^\nu}{\partial x^{\beta'}}g_{\mu \nu} = \Lambda^\mu_{\alpha'}\Lambda^\nu_{\beta'}(\eta_{\mu\nu}+ h_{\mu\nu})?
$$

We don't know anything about the $\frac{\partial x^\mu}{\partial x^{\alpha'}}$-factors. On the contrary, to me it sounds like your quote is a statement about lorentz invariance. I am confused.

Thanks.

 

[Paul Colby]

The $\frac{\partial x^\mu}{\partial x^{\alpha'}} = \Lambda^\mu_{\alpha'}$ are the Lorentz transformation. Also,

$\Lambda^\mu_\nu\Lambda^\alpha_\beta\eta_{\mu \alpha} = \eta_{\nu \beta}$​

 

[Paul Colby]

Also, ask yourself, is the metric you are given one in which the mass is moving? Looks to me like $M$ hangs out at x=y=z=0.

 

[Jonsson]

The $\frac{\partial x^\mu}{\partial x^{\alpha'}} = \Lambda^\mu_{\alpha'}$ are the Lorentz transformation.

That is remarkable! Can you provide a reference/explain that it is true for the coordinates $x^\mu$ and $x^{\mu'}$?

Also, ask yourself, is the metric you are given one in which the mass is moving? Looks to me like $M$ hangs out at x=y=z=0.

According to the problem description, the metric is for a (stationary mass)/(rest frame of mass)

 

[Paul Colby]

That is remarkable!

Most of this stuff really is. Read the section on linear gravity. MTW is quite good. Also there is a section devoted to Lorentz transformations as I recall.

 

[PeterDonis]

In the linearized gravity where $h_{\mu \nu}$ is about a flat Minkowski metric $\eta_{\mu \nu}$ then $h_{\mu \nu}$ both transform as Lorentz tensors

Having reviewed MTW, yes, I agree this is correct. However, if you look at Chapter 18, you will see an important caveat (which is what I think I was thinking of when I expressed doubt before about the Lorentz transformations being correct for this case): you have to choose the gauge conditions given by equation (18.8a) in MTW. This is a further restriction on the coordinates and the properties of the metric perturbation $h_{\mu \nu}$, over and above the basic assumptions of linearized theory (ignoring quadratic and higher terms). MTW calls this further restriction the "Lorentz gauge" by analogy with electromagnetism; so the fully correct way to express what is said in the above quote is that $h_{\mu \nu}$ transforms like a Lorentz tensor in flat spacetime provided we are working in the Lorentz gauge.

 

[Jonsson]

Having reviewed MTW, yes, I agree this is correct. However, if you look at Chapter 18, you will see an important caveat (which is what I think I was thinking of when I expressed doubt before about the Lorentz transformations being correct for this case): you have to choose the gauge conditions given by equation (18.8a) in MTW. This is a further restriction on the coordinates and the properties of the metric perturbation $h_{\mu \nu}$, over and above the basic assumptions of linearized theory (ignoring quadratic and higher terms). MTW calls this further restriction the "Lorentz gauge" by analogy with electromagnetism; so the fully correct way to express what is said in the above quote is that $h_{\mu \nu}$ transforms like a Lorentz tensor in flat spacetime provided we are working in the Lorentz gauge.

Alright, I take it then that

$$
\frac{\partial x^\mu}{\partial x^{\alpha'}}\frac{\partial x^\nu}{\partial x^{\beta'}}g_{\mu \nu} = \Lambda^\mu_{\alpha'}\Lambda^\nu_{\beta'}(\eta_{\mu\nu}+ h_{\mu\nu}) \tag{1}
$$

follows from using the Lorentz gauge then? Are we able to prove $(1)$?

 

[Paul Colby]

Are we able to prove (1)

This is the standard general coordinate transformation specialized to the Lorentz case. It's more of a definition than a proof? Have you tried the second part of the problem. This would be finding the photon trajectory when the mass is stationary. You sound unsure about coordinate transforms and what they mean. You may need to back up a bit and work it through. Have a look at MTW section 2.9 page 66 and following. I bought this book back in the 70's and found it a steep learning curve. I should probably reread much of it.

 

[Paul Colby]

MTW calls this further restriction the "Lorentz gauge" by analogy with electromagnetism; so the fully correct way to express what is said in the above quote is that hμνhμνh_{\mu \nu} transforms like a Lorentz tensor in flat spacetime provided we are working in the Lorentz gauge.

This is a really important deal and I'm aware of it. Mr Jonsson need not be concerned because he's been given a geodesic problem which is solvable given a metric. I'm just trying to get him to say "geodesic" so he can get on track with his homework.

 

[Jonsson]

Ok, I have been working away at this. I found the metric in the non moving frame. But how do I go from there. Suppose I write the up the geodesic equation or equations for conservation of momentum along a geodesic
$$
0 = g_{\mu \nu} \frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda},
$$
We don't have an equation for $\frac{dx^\mu}{d\lambda}$? How can we find this quantity in the unknown coordinates when they are unknown? Here is a drawing of the situation

 

[Paul Colby]

I'm confused. The coordinates your metric is written in are, $x$, $y$, $z$, and ever popular, $t$. Again the $g_{\mu \nu}(x,y,z,t)$ you've given is that of a mass at the origin for all time, $t$. This is about as "stationary" as it gets for mass $M$. So, you are looking for 4 functions, $x(\lambda)$, $y(\lambda)$, etc. where $\lambda$ is some handy parameter you are free to choose. Now, as an added confidence builder, I've never solved this problem myself (that I recall). There have to be an quite a few of these solutions because every light ray possible obeys the equation you've given. You need to pick the one that of these meets the initial conditions you've also been given. Oh, it's time to drop the general index sums and write out the equation to be solved using your particular metric.

[edit] You might try a simpler metric $g_{\mu \nu} = \eta_{\mu \nu}$ first. What do the geodesics look like for this case?

 

[Paul Colby]

Hey, is the equation you've given the "geodesic" equation? I thought it is something like,

$\frac{d^2x^\mu}{d\lambda^2} = \Gamma^\mu_{\alpha\beta}\frac{dx^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda} = 0$

what gives?​

 

[Jonsson]

I added a drawing to my previous post. There are three sets of coordinates.

(1) $(t,x,y,z)$ as I have drawn in the picture above
(2) $(t,x^1,x^2,x^3)$ where the mass is stationary
(3) $(t,\zeta^1,\zeta^2,\zeta^3)$ where the mass is moving.

It may well be that $(t,x,y,z) = (t,x^1,x^2,x^3)$, but it is really difficult to tell from the problem.

 

[Paul Colby]

Okay, what I don't like about, (*) $g_{\mu\nu}\dot{x}^\mu\dot{x}^\nu = 0$, where, $\dot{f}=\frac{df}{d\lambda}$, is; while it's true for null paths, all null paths are not a solution to the problem you're trying to solve. Make life simple and choose a flat metric, $M=0$. Then the path $x(\lambda) = R \cos(\lambda)$, $y(\lambda)=R \sin(\lambda)$, and $t(\lambda) = R \lambda$, obey the null condition (*). The photon going in circles in flat space isn't what ever happens.

 

[Jonsson]

Okay, what I don't like about, (*) $g_{\mu\nu}\dot{x}^\mu\dot{x}^\nu = 0$, where, $\dot{f}=\frac{df}{d\lambda}$, is; while it's true for null paths, all null paths are not a solution to the problem you're trying to solve.

That is a good thing, because we are not interested in all null paths.

Then the path $x(\lambda) = R \cos(\lambda)$, $y(\lambda)=R \sin(\lambda)$, and $t(\lambda) = R \lambda$, obey the null condition (*). The photon going in circles in flat space isn't what ever happens.

Yes, I agree these solve (*), but why is it important that you've found exactly one solution that don't interest us?

Can you propose some strategy that will work?

 

[Paul Colby]

That is a good thing, because we are not interested in all null paths.

Isn't a photon trajectory a null path?

Yes, I agree these solve (*), but why is it important that you've found exactly one solution that don't interest us?

Because what I found isn't a geodesic but is a solution to (*). So (*) isn't a complete specification of the geodesic problem statement.

Can you propose some strategy that will work?

yes, solve the geodesic problem for the metric you've been given then transform the answer to the $M$ is in motion frame?

 

[Jonsson]

yes, solve the geodesic problem for the metric you've been given then transform the answer to the $M$ is in motion frame?

Thanks. I found the deflection angle is $\frac{4GM}{b}$. How do you say I transform this quantity back?