# Nearly Lorentz coordinate system

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## Main Question or Discussion Point

Hello! I am reading "A First Course in General Relativity" by Schutz and in chapter 8 (second edition) he introduces Nearly Lorentz coordinate system. He says that we can always find some coordinates such that the metric is: $$g_{\alpha\beta}=\eta_{\alpha\beta}+h_{\alpha\beta}$$ with $$|h_{\alpha\beta}|<<1$$ then he defines $$\bar{h}_{\alpha\beta}=h_{\alpha\beta}-\frac{1}{2}\eta^{\alpha\beta}h^\alpha_\alpha$$ Then using background Lorentz transformations and gauge transformations (which as far as I understand they are coordinate transformations), he obtains $$G^{\alpha\beta}=-\frac{1}{2}\Box \bar{h}^{\alpha\beta}$$ where $G$ is the Einstein tensor and from this he gets: $$\Box\bar{h}^{\alpha\beta}=-16\pi T^{\alpha\beta}$$ Now this seems like a tensorial equation, so it should hold true in any frame. But there are 2 things I am confused about: $\textbf{1.}$ Being a tensor equation, it should hold true in a locally flat frame, too. But there, $h$ is zero because $g=\eta$ which implies that $T=0$. But we can apply this to any point on the manifold, which implies that $T=0$ everywhere, including inside the source (we should be allowed to do this inside the source because, as he does, we have a non-relativistic system so the approximations should hold there, too). But this is clearly not true (right?) so what is wrong with my reasoning? $\textbf{2.}$ In general when we want to use this equation, how do we know what is $T$? For example, in cartesian coordinates, for a perfect fluid, $T=diag(\rho,p,p,p)$. But in order to reach this equation, we did lots of coordinate transformations (which were not explicitly done, but it was only shown that they can be done). Doesn't this means that in this new coordinate system $T$ will have a more complicated form, which we need to know in order to find $h$, which is the purpose of this formalism? How can we get $T$, when we didn't really keep track of all the coordinate transformations we made? Sorry for the long post, but I hope someone can explain to me what I am missing here. Thank you!

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PeterDonis
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this seems like a tensorial equation, so it should hold true in any frame
Yes, but that doesn't mean the condition $h << 1$ will hold in any frame. It will only hold in particular frames that satisfy the "nearly Lorentz coordinate" condition. (Which, btw, can only be satisfied, for any given coordinate chart, in a small neighborhood around a particular point; there is no way, in a general curved spacetime, to find a coordinate chart that is "nearly Lorentz" everywhere.) So you can't apply any conclusions that depend on $h << 1$ being true in an arbitrary coordinate chart. You can only apply them in the particular subset of charts that satisfy $h << 1$.

Yes, but that doesn't mean the condition $h << 1$ will hold in any frame. It will only hold in particular frames that satisfy the "nearly Lorentz coordinate" condition. (Which, btw, can only be satisfied, for any given coordinate chart, in a small neighborhood around a particular point; there is no way, in a general curved spacetime, to find a coordinate chart that is "nearly Lorentz" everywhere.) So you can't apply any conclusions that depend on $h << 1$ being true in an arbitrary coordinate chart. You can only apply them in the particular subset of charts that satisfy $h << 1$.
I see, but taking this into account, my 2 questions (I think) are still not answered. In a locally flat frame $h=0$ so $h<<1$. And for a frame where that holds, how do I find the form of $T$

PeterDonis
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In a locally flat frame $h=0$ so $h<<1$
Only at one particular point. And that, by itself, is not enough to satisfy the "nearly Lorentz" condition. See further comments below.

Which, btw, can only be satisfied, for any given coordinate chart, in a small neighborhood around a particular point; there is no way, in a general curved spacetime, to find a coordinate chart that is "nearly Lorentz" everywhere
Looking more carefully at section 8.3 of Schutz, which is where he introduces "nearly Lorentz" coordinate systems, he only uses them for weak gravitational fields, i.e., spacetimes in which you can find a coordinate chart which is "nearly Lorentz" everywhere. This is a very restricted class of spacetimes; as I noted in the quote just above, a general curved spacetime (not restricted to "weak" fields) will not admit such a chart.

Also, on looking more carefully, I see that Schutz is not claiming that $h$ or $\bar{h}$ is a tensor on the actual curved spacetime; he is only claiming that they are tensors "in SR", which means any equation involving them will only be true approximately, not exactly. (As Schutz later notes, this approximation is known as "linearized theory", since it only keeps terms linear in $h$.) So you can't expect to treat $h$ or $\bar{h}$ as tensors in full generality, the way you can in standard GR tensor equations.

Finally, note that a coordinate chart which is "nearly Lorentz" everywhere is not the same as a local inertial chart centered on a particular point. In the latter type of chart, we will have $h = 0$ at the particular chosen point (which can, as you note, be inside the source where $T \neq 0$), but we will not have $h << 1$ everywhere in the spacetime, so this chart will not be a "nearly Lorentz" chart and Schutz's equation will not be valid in it.

Only at one particular point. And that, by itself, is not enough to satisfy the "nearly Lorentz" condition. See further comments below.

Looking more carefully at section 8.3 of Schutz, which is where he introduces "nearly Lorentz" coordinate systems, he only uses them for weak gravitational fields, i.e., spacetimes in which you can find a coordinate chart which is "nearly Lorentz" everywhere. This is a very restricted class of spacetimes; as I noted in the quote just above, a general curved spacetime (not restricted to "weak" fields) will not admit such a chart.

Also, on looking more carefully, I see that Schutz is not claiming that $h$ or $\bar{h}$ is a tensor on the actual curved spacetime; he is only claiming that they are tensors "in SR", which means any equation involving them will only be true approximately, not exactly. (As Schutz later notes, this approximation is known as "linearized theory", since it only keeps terms linear in $h$.) So you can't expect to treat $h$ or $\bar{h}$ as tensors in full generality, the way you can in standard GR tensor equations.

Finally, note that a coordinate chart which is "nearly Lorentz" everywhere is not the same as a local inertial chart centered on a particular point. In the latter type of chart, we will have $h = 0$ at the particular chosen point (which can, as you note, be inside the source where $T \neq 0$), but we will not have $h << 1$ everywhere in the spacetime, so this chart will not be a "nearly Lorentz" chart and Schutz's equation will not be valid in it.
OK, I think I understand what you mean. However, about your statement about the fact that if you manage to have $h=0$ at a particular point, then you will not have $h<<1$ everywhere, could you please explain it a bit? Is this obvious? Why does $h=0$ implies that you can't have $h<<1$, I don't really see it right away. Also, assuming that all the requirements are fulfilled to use the linearized theory, I am still not sure I know how to find $T$. He gives a trivial example for $T$, but I assume this theory applies to more complicated $T$'s, so how do you know what form does $T$ have in that particular coordinate system?

PeterDonis
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Why does $h=0$ implies that you can't have $h<<1$
Having $h = 0$ at one single point means you have to choose coordinates to make $h$ vanish at that point; and that choice of coordinates means you won't be able to have $h << 1$ everywhere in the spacetime; there will be some region of the spacetime (of course not at the point you chose) where $h << 1$ will not hold in those coordinates. Making $h << 1$ everywhere in the spacetime will require a different choice of coordinates, one in which $h = 0$ will not be true anywhere.

assuming that all the requirements are fulfilled to use the linearized theory, I am still not sure I know how to find $T$.
If you know the exact form of $h$ for a particular problem, that tells you $T$. But usually the problem is worked the other way: you start out by making some assumption about $T$ (for example, that it's a perfect fluid), and then deriving the form of $h$. Your assumption about $T$ will also involve making some kind of coordinate choice.

Having $h = 0$ at one single point means you have to choose coordinates to make $h$ vanish at that point; and that choice of coordinates means you won't be able to have $h << 1$ everywhere in the spacetime; there will be some region of the spacetime (of course not at the point you chose) where $h << 1$ will not hold in those coordinates. Making $h << 1$ everywhere in the spacetime will require a different choice of coordinates, one in which $h = 0$ will not be true anywhere.

If you know the exact form of $h$ for a particular problem, that tells you $T$. But usually the problem is worked the other way: you start out by making some assumption about $T$ (for example, that it's a perfect fluid), and then deriving the form of $h$. Your assumption about $T$ will also involve making some kind of coordinate choice.
Thank you for helping me with this, and sorry for asking so many questions... However, it is still not clear to me. For the first issue: $h=0$ at a point and $h<<1$ everywhere don't seem to exclude each other. Having $h=0$ at a point implies automatically that $h<<1$ at that point and tells you nothing about any other point. So why at other points that would not be true, by setting that condition at only one point? (I am aware that it should not be true, otherwise we get a contradiction, but I can't see mathematically why). For the second part what confuses me is the coordinate choice. In order to get $\bar{h}$ to that simple form, you already did some coordinate transformations, transformations which already changed $T$ on the way. But you don't really know these transformations. For example Schutz shows that you can find a gauge to simplify $\bar{h}$, but he doesn't find the transformation explicitly. Also in the beginning, he says that you can always find a coordinate system such that $$g=\eta+h$$ but again, we don't know the form of $T$ in this initial coordinate system. Honestly, I am not sure why he assumes that $T^{00}=\rho$. That is true in a local flat frame, but all these coordinate transformation might lead to a $T$ in which that component is not simply $\rho$

PeterDonis
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For the first issue: $h=0$ at a point and $h<<1$ everywhere don't seem to exclude each other
I agree it's not immediately evident; but your own argument shows why it must nevertheless be true--more precisely, that if you have chosen coordinates such that $h = 0$ anywhere inside the source (where $T \neq 0$), then it cannot be the case that $h << 1$ everywhere in those coordinates. If it were the case, then Schutz's linearized field equation would hold everywhere in those coordinates, which would mean $T = 0$ at the chosen point where $h = 0$. But if that point is inside the source, that can't be the case.

Technically, you could choose coordinates for which $h = 0$ at some point in a vacuum region (where $T = 0$), and that would not preclude having $h << 1$ everywhere in those coordinates. However, I would still expect such a choice to be hard to find in practice. For example, the "natural" coordinates for, say, the solar system in the weak field limit have $h << 1$ everywhere but do not have $h = 0$ anywhere.

I am aware that it should not be true, otherwise we get a contradiction, but I can't see mathematically why
I don't think there is a general mathematical reason why (other than the argument from contradiction that you have already made); I think it's something that will turn out to be true in each case, but the only way to see in detail why it's true in that case would be to work out the details in that particular case.

In order to get $\bar{h}$ to that simple form, you already did some coordinate transformations, transformations which already changed $T$ on the way.
You're looking at it backwards. All of the arguments about coordinate transformations are just to show how you can always find coordinates in which $h$ and $\bar{h}$ and the equations involving them (like the linearized field equation) take particularly simple and easy to use forms. They aren't transformations you would actually do in practice. In practice, as I said, you assume some simple form for $T$--for example, that it's a perfect fluid--and that in itself implies that you have already chosen coordinates in which the simple forms of all the other stuff hold--for example, that in the same coordinates where $T$ takes the simple perfect fluid form, you assume that Schutz's linearized field equation holds, so those coordinates already satisfy all the gauge and other conditions. The only additional check you might make is a sanity check, after you've completed the solution, to verify that all the conditions do in fact hold in the complete solution.

I am not sure why he assumes that $T^{00}=\rho$.
Because, as above, he's assuming that all of the coordinate conditions have already been satisfied in whatever coordinates $T^{00} = \rho$ is true. He's certainly not assuming that $T^{00} = \rho$ will be true in any coordinates whatever--for example, he's not assuming that that will be true in any coordinates in which $g_{\alpha \beta} = \eta_{\alpha \beta} + h_{\alpha \beta}$ with $h << 1$, where none of the other transformation or gauge conditions have been imposed.

I agree it's not immediately evident; but your own argument shows why it must nevertheless be true--more precisely, that if you have chosen coordinates such that $h = 0$ anywhere inside the source (where $T \neq 0$), then it cannot be the case that $h << 1$ everywhere in those coordinates. If it were the case, then Schutz's linearized field equation would hold everywhere in those coordinates, which would mean $T = 0$ at the chosen point where $h = 0$. But if that point is inside the source, that can't be the case.

Technically, you could choose coordinates for which $h = 0$ at some point in a vacuum region (where $T = 0$), and that would not preclude having $h << 1$ everywhere in those coordinates. However, I would still expect such a choice to be hard to find in practice. For example, the "natural" coordinates for, say, the solar system in the weak field limit have $h << 1$ everywhere but do not have $h = 0$ anywhere.

I don't think there is a general mathematical reason why (other than the argument from contradiction that you have already made); I think it's something that will turn out to be true in each case, but the only way to see in detail why it's true in that case would be to work out the details in that particular case.

You're looking at it backwards. All of the arguments about coordinate transformations are just to show how you can always find coordinates in which $h$ and $\bar{h}$ and the equations involving them (like the linearized field equation) take particularly simple and easy to use forms. They aren't transformations you would actually do in practice. In practice, as I said, you assume some simple form for $T$--for example, that it's a perfect fluid--and that in itself implies that you have already chosen coordinates in which the simple forms of all the other stuff hold--for example, that in the same coordinates where $T$ takes the simple perfect fluid form, you assume that Schutz's linearized field equation holds, so those coordinates already satisfy all the gauge and other conditions. The only additional check you might make is a sanity check, after you've completed the solution, to verify that all the conditions do in fact hold in the complete solution.

Because, as above, he's assuming that all of the coordinate conditions have already been satisfied in whatever coordinates $T^{00} = \rho$ is true. He's certainly not assuming that $T^{00} = \rho$ will be true in any coordinates whatever--for example, he's not assuming that that will be true in any coordinates in which $g_{\alpha \beta} = \eta_{\alpha \beta} + h_{\alpha \beta}$ with $h << 1$, where none of the other transformation or gauge conditions have been imposed.
So the first part now makes sense, thank you so much for that. For the second part, the fact that I look at it backwards actually made me understand it more, now, and based on your first part I do indeed understand why Schutz did that. However, one last thing I am confused about is how can you just take the simplest form for T and assume that the linearize would work. I mean it is desirable, but how can you know that for a given (randomly picked) $T$, the $h$ will be just by itself and it will not contain any of the other derivatives? That gauge that gets rid of the derivatives is one of the many possible (which don't get rid of the derivatives), so statistically you have a lot more chances to not be in the right gauge that to be in it.

PeterDonis
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how can you just take the simplest form for T and assume that the linearize would work.
Remember this whole scheme assumes weak fields; that means $T$ can't be very large anywhere either, because a strong source would create strong fields. So assuming a simple form for $T$ amounts to assuming, for weak fields, that you can find coordinates in which the fluid is everywhere at rest (since that's what the simple diagonal form for $T$ requires) and for which $h << 1$ everywhere.

how can you know that for a given (randomly picked) $T$, the $h$ will be just by itself and it will not contain any of the other derivatives?
First, you can't just "randomly pick" a $T$. It has to satisfy the weak field requirement.

Second, you're not just picking $T$; you're picking $T$ plus a coordinate chart in which Schutz's linearized field equation holds. In other words, you're assuming that you've picked coordinates in which all of the gauge and other transformations have already been applied so that the field equation only contains $\bar{h}$, as it does in Schutz's linearized field equation. That's a much stronger requirement than just picking a weak field $T$.

statistically you have a lot more chances to not be in the right gauge
Statistics has nothing to do with it. You're not randomly picking coordinates any more than you're randomly picking $T$. See above.

Remember this whole scheme assumes weak fields; that means $T$ can't be very large anywhere either, because a strong source would create strong fields. So assuming a simple form for $T$ amounts to assuming, for weak fields, that you can find coordinates in which the fluid is everywhere at rest (since that's what the simple diagonal form for $T$ requires) and for which $h << 1$ everywhere.

First, you can't just "randomly pick" a $T$. It has to satisfy the weak field requirement.

Second, you're not just picking $T$; you're picking $T$ plus a coordinate chart in which Schutz's linearized field equation holds. In other words, you're assuming that you've picked coordinates in which all of the gauge and other transformations have already been applied so that the field equation only contains $\bar{h}$, as it does in Schutz's linearized field equation. That's a much stronger requirement than just picking a weak field $T$.

Statistics has nothing to do with it. You're not randomly picking coordinates any more than you're randomly picking $T$. See above.
Sorry I didn't formulate that right. I meant randomly picking a coordinate system, not a $T$ (as you said $T$ is given by the physics and it must be small in order to fulfill the linearization requirements). By random, I just meant, from the physics point of view, a coordinate system in which the linearization works is not privileged over one where it doesn't (i.e. the physics is the same, only the maths is easier). So why would a coordinate system in which $T$ is simple (or as you said and also Schutz used, the fluid is everywhere at rest) necessary allows linearization. My point is that you want a simple $T$ and linearization. But they don't necessary imply one another so why when solving problems you assume you have both?

PeterDonis
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I meant randomly picking a coordinate system
You can't randomly pick the coordinate system either.

a coordinate system in which the linearization works is not privileged over one where it doesn't
Not in the sense that you have to use a nearly Lorentz coordinate system to make physical predictions; but if fields are in fact weak, it's a lot easier to make those predictions in a nearly Lorentz coordinate system in which the linearized equations hold. That's why they are used. The fact that no coordinate system is "privileged" does not mean physics is just as easy to do in any coordinate system. The main reason for picking particular kinds of coordinates is to make it easier to do the physics.

why would a coordinate system in which $T$ is simple (or as you said and also Schutz used, the fluid is everywhere at rest) necessary allows linearization
Because you can always find a nearly Lorentz coordinate system if the fields are weak. The proof of that, which Schutz describes, makes no assumptions whatever about $T$. So you can assume $T$ has whatever form you want (consistent with the general requirement that fields have to be weak), and you will be able to find a nearly Lorentz coordinate system in which $T$ takes that form.