I Diffeomorphism invariance of GR

[Frank Castle]

If you want to stick to a coordinate point of view, isometries are those non-trivial 'coordinate transformations' that don't change the appearance of the gμνg_{\mu \nu}, while more general 'diffeomorphisms' might look like they change these functions, but actually it's just a reformulation of the theory. I used the parantheses to indicate that the statement is not really mathematically correct, but I though it might help you to get on the right track.

In what sense is the statement mathematically incorrect?
To be honest, I think the set of notes that I read (http://web.mit.edu/edbert/GR/gr5.pdf) really threw a spanner in the works for me. It has confused me about the difference between a diffeomorphism and a general coordinate transformation. Are they mathematically and physically equivalent (just philosophically different)? Is the Einstein-Hilbert automatically diffeomorphism invariant by virtue of the fact that it is invariant under general coordinate transformations?

 

[haushofer]

Yes. Numerically, a general coordinate transfo brings you from a tensor T(x) to T'(x'). Given T'(x') you can reset the coordinates' value giving you T'(x). Comparing this with T(x) is exactly what you do if you interpret the gct actively, giving you Lie derivatives.

Note that infinitesimal coordinate transfo's are often used synonymous with the action of Lie derivatives.

 

[Frank Castle]

Yes. Numerically, a general coordinate transfo brings you from a tensor T(x) to T'(x'). Given T'(x') you can reset the coordinates' value giving you T'(x). Comparing this with T(x) is exactly what you do if you interpret the gct actively, giving you Lie derivatives.

Note that infinitesimal coordinate transfo's are often used synonymous with the action of Lie derivatives.

Ah ok, so in the notes that I linked to, is the author simply interpreting a diffeomorphism as a combination of a mapping $p\mapsto\phi(p)=q$ (such that in coordinates $x^{\mu}(p)\mapsto y^{\mu}(q)=(\phi^{\ast}x)^{\mu}(p)$), and a coordinate transformation $y^{\mu}\mapsto x'^{\mu}$, such that the points $p\in M$ on the manifold are moved around, but the coordinates are not, i.e. the coordinates of $q=\phi(p)$ are the same $q$, $x'^{\mu}(q)=x^{\mu}(p)$. Is this what the author is getting at when they claim that under an active coordinate transformation $d^{4}x$ is invariant, but $\sqrt{-g}$ is not?

 

[vanhees71]

Yes. Numerically, a general coordinate transfo brings you from a tensor T(x) to T'(x'). Given T'(x') you can reset the coordinates' value giving you T'(x). Comparing this with T(x) is exactly what you do if you interpret the gct actively, giving you Lie derivatives.

Note that infinitesimal coordinate transfo's are often used synonymous with the action of Lie derivatives.

No, a coordinate transformation doesn't change your tensors at all. That's the key to tensor analysis! It's basis independent!

 

[Frank Castle]

No, a coordinate transformation doesn't change your tensors at all. That's the key to tensor analysis! It's basis independent!

But the components of a tensor will change, right (i.e. $T^{\mu\nu}(x)\rightarrow T'^{\mu\nu}(x')$).

 

[haushofer]

No, a coordinate transformation doesn't change your tensors at all. That's the key to tensor analysis! It's basis independent!

The components do change. That's what I was implying in my shorthand notation and I'm assuming that in this discussion this basic fact about tensors is clear to anyone.

By the way, in the common language of "infinitesimal coordinate transformation" a tensor does change; it is synonymous to the Lie derivative of the tensor. That's why I was also warning the TS for the different usages of the terminology.

 

[haushofer]

But the components of a tensor will change, right (i.e. $T^{\mu\nu}(x)\rightarrow T'^{\mu\nu}(x')$).

Yes.

 

[Frank Castle]

Yes.

Cool, that's what I thought. Is any of what I wrote in my previous post #53 correct at all?

 

[haushofer]

Ah ok, so in the notes that I linked to, is the author simply interpreting a diffeomorphism as a combination of a mapping $p\mapsto\phi(p)=q$ (such that in coordinates $x^{\mu}(p)\mapsto y^{\mu}(q)=(\phi^{\ast}x)^{\mu}(p)$), and a coordinate transformation $y^{\mu}\mapsto x'^{\mu}$, such that the points $p\in M$ on the manifold are moved around, but the coordinates are not, i.e. the coordinates of $q=\phi(p)$ are the same $q$, $x'^{\mu}(q)=x^{\mu}(p)$. Is this what the author is getting at when they claim that under an active coordinate transformation $d^{4}x$ is invariant, but $\sqrt{-g}$ is not?

As far as I can tell, that author uses the word coordinate transformation for T(x) --> T'(x'), as is common, and "diffeomorphism" for what I would call T(x) --> T'(x), so a coordinate transformation with the coordinate x' set back to x . See the last few lines on page 9: "the volume element is invariant under a coordinate transformation but not under a diffeomorphism". I read "invariant" as "scalar".

 

[haushofer]

Cool, that's what I thought. Is any of what I wrote in my previous post #53 correct at all?

I think it does; see also Wald's discussion in his appendix on active v.s. passive.

 

[vanhees71]

But the components of a tensor will change, right (i.e. $T^{\mu\nu}(x)\rightarrow T'^{\mu\nu}(x')$).

Of course. It's easy to remember: $\mathrm{d} x^{\mu}$ are vector components. So you get
$$\mathrm{d} x^{\prime \mu}=\frac{\partial x^{\prime \mu}}{\partial x^{\nu}} \mathrm{d} x^{\nu},$$
and each contravariant component of a multilinear indicates transformation like this, i.e.,
$$T^{\prime \mu \nu}= \frac{\partial x^{\prime \mu}}{\partial x^{\rho}}\frac{\partial x^{\prime \nu}}{\partial x^{\sigma}} T^{\rho \sigma},$$
where I use the holonomous basis
$$\boldsymbol{b}_{\mu} = \partial_{\mu},$$
which thus transforms as
$$\boldsymbol{b}_{\mu}'=\boldsymbol{b}_{\nu} \frac{\partial x^{\prime \nu}}{\partial x^{\mu}},$$
i.e., contragrediently to contravariant (upper index) components.

This implies that indeed the tensors themselves are invariant. You can easily check that indeed
$$\boldsymbol{T}=T^{\mu \nu} \boldsymbol{b}_{\mu} \otimes \boldsymbol{b}_{\nu} = T^{\mu \nu} \partial_{\mu} \partial_{\nu}$$
is invariant.

 

[haushofer]

Well, as Wald emphasizes, it's "philosophically differerent" whether you interpret a diffeomorphism as active or passive, but mathematically and physically it's equivalent. That answers the question, why there is so much confusion about "active" vs. "passive". It's only a philosophical confusion!

Well, maybe it's only philosphical, but I've confronted many of my former collegues and even some top-physicists having written textbooks on (quantum) gravity, and I can't say I've ever met a single person who had a clear, conceptual view on all of this and could tell me the relation between all the different interpretations. To me that's a sign that a lot of people can calculate with tensors, but interpreting is a whole different cookie.

To me this business is perhaps the most difficult problem I've ever encountered in my carreer as a PhD-student. In my experience, there are two kinds of reactions to this whole business: people who wave it away as being trivial, and people who admit right away they've never understood it properly in the first place. That explains also why somebody like Einstein was puzzled by the hole-problem for 2 years.

So my message to the Topic Starter would be: don't be discouraged by your confusion. It means you're thinking about. That's a start already.

 

[vanhees71]

Well, ok there's maybe indeed some subtle point in this if it comes to interpretation.

A drastic example is "time-reversal invariance" (let's stick to SR for this example). On the one hand you can easily define it in a passive way by saying, I let run "time in the other direction" by formally setting $t \rightarrow t'=-t, \vec{x} \rightarrow \vec{x}'=\vec{x}$. This implies the time-inversion properties of the usual quantities (if it can be defined at all) like $\vec{p} \rightarrow -\vec{p}$, $E \rightarrow E'=E$, etc. etc. So it's easily defined, and it's easily checked whether your Lagrangian in question is invariant under this transformation (e.g., electromagnetism is invariant, the weak interaction Lagrangian a la Glashow, Salam, Weinberg is not, as it should be).

Now, what's the "active" point of view of this? Of course, I cannot simply let run time backwards. A handwaving idea of making a movie of some process and let it run backwards, hints, however to the solution: What's meant is that if you look at a process under some dynamical law turning an initial state of affairs at time $t=0$ to a final state of affairs at time $t$, then you have time-reversal invariance, if you prepare the system in the exactly time-reversal of the final state at $t=0$ and let it go under the same dynamical laws you should end up at time $t$ with exactly the time-reversal of the initial state of the first setup.

I don't see any such problems between active and passive interpretations of symmetry principles for the usual continuous symmetries. The mathematics is finally all the same, no matter whether you prefer the active or passive point of view. In my opinion the passive point of view often is more easy to grasp: The physical laws shouldn't depend on our choice of coordinates and other men-made descriptions. On the other hand, the active point of view provides a more intuitive physical picture. E.g., you can state that SR is invariant under Poincare transformations of the coordinates because of the geometrical structure of Minkowski space or you can say (specializing to boost invariance): The physical laws look the same in all inertial reference frames or, even simpler, there is no absolute inertial reference frame since all physical laws are independent of the (constant) velocity of one inertial frame with respect to another.

 

[Frank Castle]

See the last few lines on page 9: "the volume element is invariant under a coordinate transformation but not under a diffeomorphism"

This is what I'm unsure about really. Is the point that the author is making here that under a diffeomorphism we are always free to choose a coordinate transformation, $x\rightarrow x'$, such that the coordinate values of the "new" point $q$ in the "new" coordinate system $x'$ are the same as the coordinate values of the "old" point $p$ in the "old" coordinate system $x$, i.e. $x'^{\mu}(q)=x^{\mu}(p)$. Hence, we have that $d^{4}x\rightarrow d^{4}x'=d^{4}x$ such that $d^{4}x$ is diffeomorphism invariant, however, $\sqrt{-g}$ transforms as $\sqrt{-g}\rightarrow\sqrt{-g}+\mathcal{L}_{X}\sqrt{-g}=\sqrt{-g}\left(1+g^{\mu\nu}\nabla_{\mu}X_{\nu}\right)$ (where $X$ is the vector field generating the diffeomorphism), and so is not invariant under diffeomorphisms?

 

[vanhees71]

If you have a diffeomorphic map of the manifold on itself, the map of a region doesn't need to have the same volume as this region, while a given region's volume of course shouldn't depend on the chosen coordinates.

The volume of a sphere in usual 3D Euclidean space doesn't change, only because you calculate it in terms of spherical coordinates (clever, because simple) or Cartesian ones (a bit lengthy but doable).

 

[Frank Castle]

If you have a diffeomorphic map of the manifold on itself, the map of a region doesn't need to have the same volume as this region, while a given region's volume of course shouldn't depend on the chosen coordinates.

Is this why $d^{4}x\sqrt{-g}$ isn't necessarily invariant under a diffeomorphism?

 

[haushofer]

This is what I'm unsure about really. Is the point that the author is making here that under a diffeomorphism we are always free to choose a coordinate transformation, $x\rightarrow x'$, such that the coordinate values of the "new" point $q$ in the "new" coordinate system $x'$ are the same as the coordinate values of the "old" point $p$ in the "old" coordinate system $x$, i.e. $x'^{\mu}(q)=x^{\mu}(p)$. Hence, we have that $d^{4}x\rightarrow d^{4}x'=d^{4}x$ such that $d^{4}x$ is diffeomorphism invariant, however, $\sqrt{-g}$ transforms as $\sqrt{-g}\rightarrow\sqrt{-g}+\mathcal{L}_{X}\sqrt{-g}=\sqrt{-g}\left(1+g^{\mu\nu}\nabla_{\mu}X_{\nu}\right)$ (where $X$ is the vector field generating the diffeomorphism), and so is not invariant under diffeomorphisms?

Yes. I think that's the point.

 

[Frank Castle]

Yes. I think that's the point.

Ok cool.

So, let me see if I have this right. If we have a diffeomorphism $\phi:M\rightarrow M$ such that $p\mapsto\phi(p)=q$, which in a local coordinate chart is given by, $x^{\mu}(p)\mapsto\left(\phi^{\ast}x\right)^{\mu}(p)=y^{\mu}(q)$. One can then introduce a coordinate transformation, $y^{\mu}(q)\mapsto x'^{\mu}(q)=x^{\mu}(p)$, such that the coordinates of $q=\phi(p)$ in the $x'$ coordinate chart are the same as the coordinates of $p$ in the original coordinate chart $x$.

If the above is correct, then there is still one thing I'm unsure about: say we have some tensor field $T$, then under the diffeomorphism $\phi$ it will transform as $\phi^{\ast}T$. When we make the coordinate transformation (described above) simultaneously with the diffeomorphism , do we not evaluate the transformed tensor $\phi^{\ast}T$ in a basis until we have made the coordinate transformation, such that $\left(\phi^{\ast}T\right)(x')=\left(\phi^{\ast}T\right)(x)$?

 

[haushofer]

I'm not sure what you mean by 'simultaneously', but I'll also be gone for some time and probably not able to respond. If I can with a proper answer, I'll do so.

 

[Frank Castle]

I'm not sure what you mean by 'simultaneously'

I was meaning that we carry out the diffeomorphism $x^{\mu}(p)\mapsto\left(\phi^{\ast}x\right)^{\mu}(p)=y^{\mu}(q)$ and the coordinate transformation $y^{\mu}(q)\mapsto x'^{\mu}(q)=x^{\mu}(p)$ at the same time.

If I can with a proper answer, I'll do so.

Ok. Thanks for all your help so far, it's much appreciated!

 

[haushofer]

I'm not sure about that. A diffeomorphism is a transformation on the manifold $M$. A coordinate transformation is a transformation in $R^n$. But a diffeomorphism does induce a coordinate transformation, so they are not independent of each other (that's why there is a 1-to-1 correspondence between diffeormorphisms and coordinate transformations, "active and passive transformations", in the first place! See again Wald). So the only thing you can do imo is to first perform a diffeomorphism (which induces a coordinate transformation), and only after that perform a coordinate transformation (or the other way around).

Maybe I'm pedantic here, but from experience I know that there is no overload of being pedantic in these discussions. ;) Maybe it helps for you to do all these steps explicitly (and in both orders, to see the hang of it).

 

[Frank Castle]

I'm not sure about that. A diffeomorphism is a transformation on the manifold MM. A coordinate transformation is a transformation in RnR^n. But a diffeomorphism does induce a coordinate transformation, so they are not independent of each other (that's why there is a 1-to-1 correspondence between diffeormorphisms and coordinate transformations, "active and passive transformations", in the first place! See again Wald). So the only thing you can do imo is to first perform a diffeomorphism (which induces a coordinate transformation), and only after that perform a coordinate transformation (or the other way around).

Maybe I'm pedantic here, but from experience I know that there is no overload of being pedantic in these discussions. ;) Maybe it helps for you to do all these steps explicitly (and in both orders, to see the hang of it).

Ok. So if one starts with a diffeomorphism $\phi:M\rightarrow M$ such that $p\mapsto\phi(p)=q$, then in a local coordinate chart this induces a coordinate transformation $x^{\mu}(p)\mapsto\left(\phi^{\ast}x\right)^{\mu}(p)=y^{\mu}(q)$. We can then choose a coordinate transformation such that $y^{\mu}(q)\mapsto x'^{\mu}(q)=x^{\mu}(p)$. Would this be correct at all?

 

[haushofer]

Yes, I think that 's exactly what is done in your notes. First the diffeo, and then the coordinatetranso (or, first the active and then the passive transformation).

 

[Frank Castle]

Yes, I think that 's exactly what is done in your notes. First the diffeo, and then the coordinatetranso (or, first the active and then the passive transformation).

Ok cool. So does one then transform a tensor $T\mapsto\phi^{\ast}T$ and subsequently evaluate it in the final coordinate chart such that $\left(\phi^{\ast}T\right)(x')=\left(\phi^{\ast}T\right)(x)$?

With a diffeomorphism is the whole philosophy that one can either view it in an active sense as actually moving points around on the manifold, such that for an infinitesimal diffeomorphism, points are mapped to different positions on the manifold, with the coordinates $x^{\mu}$ and $x'^{\mu}$ labelling two different points in the same coordinate chart. Or, one can view it passively as a simple coordinate transformation between two coordinate charts such that the coordinates $x^{\mu}$ and $x'^{\mu}$ label the same point in different coordinate charts?

In the former case, i.e. the active sense, the points on the manifold are actually moved around, with $x^{\mu}$ and $x'^{\mu}$ labelling two different points $p$ and $q$ in the same coordinate chart, however, we can always carry out a coordinate transformation, $x'\rightarrow\tilde{x}$, such that in this new coordinate chart the coordinate values of the point $q$ are the same as the coordinates of the point in the old coordinate chart, i.e. $\tilde{x}^{\mu}=x^{\mu}$? For an infinitesimal diffeomorphism, does this happen as follows: An infinitesimal diffeomorphism is generated by a vector field $X$ such that the orbit of a point $p$ under this diffeomorphism is along the integral curves of $X$. This induces a coordinate transformation $$x^{\mu}\rightarrow y^{\mu}=x^{\mu}+\epsilon X^{\mu}$$ where $\epsilon<<1$. Thus if we carry out the coordinate transformation $$y^{\mu}\rightarrow x'^{\mu}=y^{\mu}-\epsilon X^{\mu}$$ then we find that $x'^{\mu}=x^{\mu}$. Is this then why the author claims that $d^{4}x$ does not transform under a diffeomorphism?