# Weak Field Dynamics

1. Jan 26, 2014

### Tomishiyo

1. The problem statement, all variables and given/known data
The metric for a given particle traveling in the presence of a gravitational field is $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$, where $\eta_{\mu\nu}$ is the Minkowski metric, $h_{00}=-2\phi$ ($\phi$ the Newtonian gravitational potential); $h_{i0}=0$; and $h_{ij}=-2\phi\delta_{ij}$. Units are given in a system such that $c=\hbar=k_B=1$. Find the equations of motion for a massive particle traveling in this field.
(a) Show that $\Gamma^{0}_{\ \ 00}=\partial\phi/\partial t$ and $\Gamma^{i}_{\ \ 00}=\delta^{ij}\partial\phi/\partial x^j$.
(b) Show that the time component of the geodesic equation implies that the energy $p^0+m\phi$ is conserved.
(c) Show that the space part of the geodesic equation lead to $d^2x^i/dt^2=-m\delta^{ij}\partial\phi/\partial x^j$ in agreement with Newtonian theory. Use the fact that the particle is nonrelativistic so $p^0>>p^i$.

2. Relevant equations
Affine Connections (or Christoffer Symbol) Equation:
$$\Gamma^{\mu}_{\ \ \alpha\beta}=\frac{g^{\mu\nu}}{2}\left(\frac{\partial g_{\alpha\nu}}{\partial x^{\beta}}+\frac{\partial g_{\beta\nu}}{\partial x^{\alpha}}-\frac{\partial g_{\alpha\beta}}{\partial x^{\nu}}\right)$$

Geodesic Equation:
$$\frac{d^2x^{\mu}}{d\lambda^2}=-\Gamma^{\mu}_{\ \ \alpha\beta}\frac{dx^{\alpha}}{d\lambda}\frac{dx^{\beta}}{d\lambda}.$$

3. The attempt at a solution
(a) I've managed this part, as long as I make a non-stated (the exercise and the text book didn't say anything about it, I'm not omitting information in the exercise statement) weak field approximation: $\phi<<1$. It seems to be a reasonable assumption, since if it is not the case I can see no other way of the connections yielding such a result, and because I've already seen such approximation in other text books. My trouble starts in the next item.

(b) Here is where trouble starts. First, I write the time component of the geodesic equation as:
$$\frac{d^2x^{0}}{d\lambda^2}=-\Gamma^{0}_{\ \ \alpha\beta}\frac{dx^{\alpha}}{d\lambda}\frac{dx^{\beta}}{d\lambda}=-\Gamma^{0}_{\ \ 00}\left(\frac{dx^{0}}{d\lambda}\right)^2-2\Gamma^{0}_{\ \ 0i}\frac{dx^{0}}{d\lambda}\frac{dx^{i}}{d\lambda}-\Gamma^{0}_{\ \ ij}\frac{dx^{i}}{d\lambda}\frac{dx^{j}}{d\lambda},$$

since Christoffer symbols are symmetric with respect to the bottom indices. Here I am assuming the usual conventions that Greek indices run from 0 to 3 (0 time coordinate; 1,2,3 spatial coordinates), Latin indices run from 1 to 3, and $\lambda$ is a monotonically increasing motion parameter. The first needed Christoffer symbol is given by letter (a). The remaining two are:

$$\Gamma^{0}_{\ \ 0i}=\frac{\partial\phi}{\partial x^i}$$
and
$$\Gamma^{0}_{\ \ ij}=-\delta_{ij}\frac{\partial\phi}{\partial t}.$$

Please, if it is in any way unclear how did I obtained this connections or if you think that I've done it wrong (or my difficulties are related to that), do ask me to show the calculations. I've omitted them not for laziness, but simply because I don't think that is the problem and I think the problem will be better to visualize this way.

So, plugging in the connections the geodesic equation becomes:

$$\frac{d^2x^{0}}{d\lambda^2}=-\frac{\partial \phi}{\partial t}\left(\frac{dx^{0}}{d\lambda}\right)^2-2\frac{\partial \phi}{\partial x^i}\frac{dx^{0}}{d\lambda}\frac{dx^{i}}{d\lambda}-\delta_{ij}\frac{\partial \phi}{\partial t}\frac{dx^{i}}{d\lambda}\frac{dx^{j}}{d\lambda}.$$

We implicitly define the $\lambda$ parameter by:
$$P^{\mu}\equiv(E;p^i)\equiv\frac{dx^{\mu}}{d\lambda},$$
where $E$ is the particle's energy and $p^i$ is the momentum vector. In this way,

$$p^0=E; \ \ \frac{d}{d\lambda}=\frac{dx^0}{d\lambda}\frac{d}{dx^0}=E\frac{\partial}{\partial t}; \ \ p^i=\frac{dx^i}{d\lambda},$$
and thus the geodesic equation yields:

$$E\frac{dE}{dt}=-\frac{\partial \phi}{\partial t}E^2-2E\frac{\partial \phi}{\partial x^i}p^i+\delta_{ij}\frac{\partial \phi}{\partial t}p^i p^j.$$

And I'm stuck in the above equation. I can't express the second term in terms of $E$, $m$ and $\phi$. I can't think of any possible relation to that. The third term is another problem. At first, I thought to use the energy-momentum relation $E^2=m^2+p^2$, but that relation holds only for free-particles, and so I can't state that the 4-momentum magnitude is $g_{\mu\nu}P^{\mu}P^{\nu}=-m^2$. My text books does it in a similar exercise: in the solved exercise, it was showing that the time-component of the geodesic equation entails that the energy of a massless particle decays as the scale-factor in a Friedman-Robertson-Walker expanding universe. In this exercise, it used the fact that the four-momentum magnitude for a massless particle was zero (I got a little bit confused with the argument, because I thought you needed to correct this result with the different metric, but in the text book it seems the author just assumed the magnitude to be $-m^2$ and then stated it was zero since the mass is zero). So, I don't know what $g_{\mu\nu}P^{\mu}P^{\nu}$ is supposed to be in this modified metric and thus I can't express the last term in terms of the quantities of interest. A possible solution that has occurred me if to redefine $\lambda$ in a more convenient way, but I couldn't think in any good definition to this case.

If you read me up to now, I deeply appreciate your patience, and I'll be very grateful if you can help me.

Thank you!

2. Jan 31, 2014

### Tomishiyo

3. Jan 31, 2014

### strangerep

Well, it's been a while since I did any GR calculations, so I'm quite rusty.
But,... since no one else has replied... :uhh:

Firstly, where did the question come from? And which textbook(s) are you using?

Are you sure about that sign?

Here you seem to start going wrong. If $\lambda$ is the proper time along the particle's trajectory, then it's 4-velocity is $u^\mu := dx^\mu/d\lambda$. The magnitude of 4-velocity is $u^\mu u_\mu = 1$. If $m$ is the scalar mass of the particle, then its 4-momentum is $P^\mu := m u^\mu$, from which we get: $g_{\mu\nu}P^\mu P^\nu = m^2 u^\mu u_\mu = m^2$.

To perform part (b), you first need to state precisely what it means (mathematically) for something to be "conserved" in this context. Without that, you'll get nowhere. So... what does it mean when I say that a quantity "X" associated with the particle is "conserved"? Also, what do you know about the energy-momentum (tensor) in general?

You should be able to find this information in your textbook(s) or note(s) somewhere.

Last edited: Jan 31, 2014
4. Feb 4, 2014

### Tomishiyo

The question comes from Dodelson's "Introduction to Modern Cosmology". This is the only textbook I'm using at the moment.

Positive. Christoffer symbol are given by:

$$\Gamma^{\mu}_{\ \ \alpha\beta}=\frac{g^{\mu\nu}}{2}\left[\frac{\partial g_{\alpha\nu}}{\partial x^{\beta}}+\frac{\partial g_{\beta\nu}}{\partial x^{\alpha}}-\frac{\partial g_{\alpha\beta}}{\partial x^{\nu}}\right]$$

If we set $\mu=0$, then the only non-vanishing term is the term with $\nu=0$, so:

$$\Gamma^{0}_{\ \ \alpha\beta}=\frac{g^{00}}{2}\left[\frac{\partial g_{\alpha 0}}{\partial x^{\beta}}+\frac{\partial g_{\beta 0}}{\partial x^{\alpha}}-\frac{\partial g_{\alpha\beta}}{\partial x^{0}}\right].$$

Setting now $\alpha=i$, $\beta=j$ leads to:

$$\Gamma^{0}_{\ \ ij}=\frac{g^{00}}{2}\left[\frac{\partial g_{i0}}{\partial x^{j}}+\frac{\partial g_{j0}}{\partial x^{i}}-\frac{\partial g_{ij}}{\partial x^{0}}\right].$$

First and second term vanishes due to the metric being diagonal. Thus:
$$\Gamma^{0}_{\ \ ij}=-\frac{g^{00}}{2}\frac{\partial g_{ij}}{\partial x^{0}}=-\frac{1}{2}\left(-\frac{1}{1+2\phi}\right)\frac{\partial}{\partial t}\left(\delta_{ij}(1-2\phi)\right)\approx +\frac{1}{2}\delta_{ij}\frac{\partial}{\partial t}(1-2\phi)=-\delta_{ij}\frac{\partial\phi}{\partial t},$$
where we used the weak field approximation.

Well, here I have three questions:
1) The four-velocity magnitude is ALWAYS 1, regardless of the fact that we aren't operating in a Minkowski metric? Shouldn't it somehow be different due to the metric?
2) In the same sense, is it always true that the magnitude of the four-momentum is $m^2$, regardless of the metric?
3) I'm not sure the $\lambda$ I used is supposed to be the proper time. I followed the textbook example when he used the same definition of $\lambda$ to show that the energy of a photon, in a FRW Universe, falls with the inverse of the scale factor $a$.

I don't recall of a precise definition from the textbook, but again, using the example it provided when discussing the dependence of a photon energy on the scale factor of a FRW Universe, it should mean that the quantity is equal to a constant in time, so the time-derivative of the quantity in question should vanish. The textbook does say something about the energy-momentum tensor and conservation laws, namely, it says that conservation implies that the covariant derivative of the energy-momentum tensor must vanish, but since it states in the exercise that it wants me to show that the conservation of energy derives from the time component of the geodesic equation, I don't think this approach will solve the problem.

Last edited: Feb 4, 2014
5. Feb 4, 2014

### strangerep

The only Dodelson cosmology book I know of is "Modern Cosmology", i.e., this one. Is that the correct book? (If so, it would enable faster turnaround if you had given a more specific reference, e.g., page number, exercise number, or whatever.)

Oh, I see. You're using the (-,+,+,+) metric convention. OK.

Since this is for a massive particle, we can transform (locally) to a frame of reference in which the particle is at rest. (Think of free-fall.) In that frame, the equations of special relativity apply, and the 4-velocity vector is (1,0,0,0), hence magnitude 1. We want such scalar quantities to remain invariant under arbitrary coordinate transformations (diffeomorphisms, actually), so that leads to the generalizations I gave in my previous post. The same idea goes for for $m^2$.

I don't have a copy of Dodelson, but Amazon allowed me to read a little of it. On p31, eq(2.24) I now see how you got
But there, Dodelson is treating a massless particle. That's distinctly different from massive particles since, e.g., $ds^2 = 0$ in the massless case, so we can't use $s$ as a proper time parameter, and one must use something else to parameterize a lightlike curve.

But you're working with a massive particle, so you can't just transcribe stuff that was intended for the massless case. Unfortunately, Amazon doesn't let me see enough pages to know whether he treats the massive case somewhere else. If not, then... maybe you need an "Intro to GR" book as well.

Well, since I can see very much of Dodelson right now, I can't make useful suggestions -- except that you need to adapt the method properly to the massive case instead of blindly trying to use massless formulae directly.

Edit: I got a copy of Dodelson. I guess your exercise is his "Exercise 3" on p54? I'll have to read a bit more of that chapter to get the context properly.

Last edited: Feb 4, 2014
6. Feb 5, 2014

### Tomishiyo

Yeah, it is Modern Cosmology. I got it confused with Ryden's "Introduction to Cosmology". Also, I'm sorry for forgetting to mention that page number and exercise, those are 54 and 3, respectively.

I see. Thanks by the explanation.

But the parameter $\lambda$ isn't supposed to be arbitrary? I mean, from what I've read in Dodelson, the only "requirement" for $\lambda$ was that it should increase monotonically along the particle's path. That is why I thought I could use the definition from the quote. Well, I'll try to use proper time then.

As for the massive case, Dodelson does not treat it, for he leaves it as an exercise (I think it is the following exercise). I got "General Relativity" from M.P. Hobson, G.Efstathiou and A.N.Lasenby as a GR book.

Last edited: Feb 5, 2014
7. Feb 5, 2014

### strangerep

That's only because we can't use proper time as the parameter along a lightlike path. Normally, one uses proper time for massive particles, although one could of course rescale that parameter if one wished. But here, there's not much point.

Edit: I see that sections 5.7 and 5.8 of your HEL GR book discuss this more extensively.

Have you looked at section 7.6 of HEL yet?

In Dodelson, I'm now wondering if maybe you're meant to use the nonrelativistic condition [that seems to be for part(c) of the exercise] also in part(b). Or so I think, since I've just spent many hours trying to make it work out without such an assumption, but failed. :grumpy:

So maybe you should do part(c) first, and then try again to do part(b).

BTW, are you aware of the Dodelson errata list? It might come in handy later.

Last edited: Feb 6, 2014
8. Feb 6, 2014

### Tomishiyo

Reading HEL I realized why you ask me for a formal definition for energy conservation. It is indeed not well defined in GR context, forgot about that. I'll try to read textbook again and search for some clue about that. All I could do for the moment was to show that, in a non-relativistic case:

$$\frac{dt}{d\tau}=cte$$
and proper time is related to coordinate time by
$$d\tau=\sqrt{(1+2\phi)}dt$$
But since I don't know the exact definition I don't see how this will help me. One thing to mention, and to worry about, is that $\frac{dt}{d\tau}=cte$ is not even close to something like $p^0+m\phi=cte$.

Yes, I've looked at it. There's a mistake in part (c), where that $m$ isn't supposed to be there. But everything is ok in part (b).

Last edited: Feb 6, 2014
9. Feb 6, 2014

### strangerep

I was over-thinking it. I see now that Dodelson is not a particularly difficult book, and a simpler, sloppier, interpretation of those exercises is more likely to match his intent. I must remember that if stuff from Dodelson comes up again in the future.

Yeah. Here's what I think you should do...

1. Restore factors of $c$ in all the equations you're starting from, including the equation that relates mass to 4-momentum. (Dodelson uses $c=1$, but that can get confusing and/or sloppy when one tries to take a non-rel limit carefully.) You can probably get some clues about where the $c$'s should be from HEL. E.g., $P^0 := E/c$, and so on. Then...

2. Start again from the geodesic equation, with $\tau$ (proper time) as the parameter.

3. Use the definition $P^\mu := m dx^\mu/d\tau$ to convert the geodesic equation to be an equation in $P^\mu$ rather than $x^\mu$.

4. You should get a version of the geodesic equation that's very close to what you got in your original post (but your lhs there is not quite right and will be different).

5. Use the mass definition to simplify part of the rhs. Remember that only 1st order terms in $\phi$ need be retained.

6. Take the non-relativistic limit carefully, and try to get rid of your middle term on the lhs.

7. That should get you very close to the finish. (I've already checked all of the above for myself, except for the part about carefully restoring factors of $c$ and hence performing the non-rel limit properly.)

10. Feb 7, 2014

### Tomishiyo

What do you mean by "use the mass definition"? Up to point 6, what I got was:

$$\frac{1}{m}\frac{dP^0}{d\tau}=-\frac{1}{m^2 c^2}\frac{\partial \phi}{\partial t}(P^0)^2$$

The equation agrees with HEL, his expression on p.154 reads:

$$\frac{d^2x^{\mu}}{d\tau^2}+\Gamma^{\mu}_{\ \ 00}c^2\left(\frac{dt}{d\tau}\right)^2=0$$

Now, trouble is converting that derivative in proper time in a derivative with respect to coordinate time. They are related by the interval, thus $d\tau^2=(-c^2-2\phi)dt^2$, where I used the fact that $\frac{dx^i}{dt}<<c$. But here, if I try to use this relation, my final equation turns out to be:

$$\left(\frac{1}{P^0}\right)^2\frac{dP^0}{dt}+\frac{\sqrt{-c^2-2\phi}}{c^2m}\frac{\partial \phi}{\partial t}=0$$

Which can be integrated, but yields the wrong answer. I've also tried to pick only first order term in $\phi$ in this relation, but the result isn't better:

$$\left(\frac{1}{P^0}\right)^2\frac{dP^0}{dt}+\frac{c}{m(2\phi-c^2)}\frac{\partial\phi}{\partial t}=0.$$

I couldn't integrate this one, but I think the only possible outcome is a log on $\phi$, which is clearly wrong. So, I'm stuck again :(.

Also, could you elaborated that part of the magnitude of the four-momentum being always $-m^2$ again? I'm asking it because Dodelson explicitly states that the magnitude of a four-vector is metric dependent (and to me that makes sense, once "length" is something that is defined by the metric). He says that in page 24, in case you wanna check.

11. Feb 7, 2014

### strangerep

Hmm. You've taken the non-relativistic approximation sooner than I intended. But maybe that doesn't matter.

I don't think you need that complication, at least not yet. Get the "mass relation" stuff sorted out first (see below), and then other things might fall into place.

He's talking about general displacements in spacetime. But here, we're restricting to the intrinsic properties of a particle moving along its worldline. Remember that here, $m$ is the rest mass of the particle (hence HEL denote it as $m_0$). I prefer the term "invariant mass" since that also makes sense for the photon. The point is that a physical particle in free-fall doesn't change its rest mass (since there's a local coordinate system at every point on its trajectory in which it is at rest). But $m$ is invariant, so if you change the coordinate system you can only change the metric components, and the components of $u^\mu$. But an expression like $m^2 = g_{\mu\nu} P^\mu P^\nu$ is a scalar under such transformations (provided you use the definition of $P^\mu$ as a scalar multiple of $u^\mu$, the proper velocity).

Edit: Don't rely on Dodelson too much for GR stuff. He only gives a very brief and incomplete summary, with emphasis on light-like worldlines -- since the latter are more important for relating Cosmology models to what we can actually observe. Other GR textbooks have wider emphasis, as you'd expect.

Last edited: Feb 7, 2014