Reconciling 2 expressions for Riemann curvature tensor

In summary: As given in every source I'm aware of. In Misner, Thorne, & Wheeler, for example, it's equations 11.8 and 11.9.
  • #1
Shirish
244
32
I'm reading Carroll's GR notes and I'm having trouble deciphering a particular expression for the Riemann curvature tensor. The coordinate-free definition is (eq. 3.71 in the notes): $$R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z$$ An index-based expression is also given in (eq. 3.66): $$R^{\rho}_{\ \ \sigma\mu\nu}V^{\sigma}=[\nabla_{\mu},\nabla_{\nu}]V^{\rho}+T_{\mu\nu}^{\ \ \ \ \lambda}\nabla_{\lambda}V^{\rho}$$ How do I reconcile these two equations? My attempt so far is as follows: in the first equation (eq. 3.71), I can replace ##X,Y## by fields ##\partial_{\mu},\partial_{\nu}## respectively and ##Z## by ##V##, then I can get the local coordinates for both sides by acting them on ##x^{\rho}##, i.e. (AFAIK to make notation shorter, expressions like ##\nabla_{\partial_{\mu}}## are written as ##\nabla_{\mu}##): $$R(\partial_{\mu},\partial_{\nu})(V)(x^{\rho})=[\nabla_{\mu},\nabla_{\nu}]V(x^{\rho})-\nabla_{[\partial_{\mu},\partial_{\nu}]}V(x^{\rho})$$ So now we can compare this to (eq. 3.66) - the LHS term is the coordinate free version of LHS in eq. 3.66, and the first RHS term is the coordinate free version of the first RHS term in eq. 3.66. Although I have doubts about the 1st RHS term since it's ##([\nabla_{\mu},\nabla_{\nu}]V)(x^{\rho})##, which I guess may not be the same as ##[\nabla_{\mu},\nabla_{\nu}](V(x^{\rho}))=[\nabla_{\mu},\nabla_{\nu}]V^{\rho}##.

The main trouble is with the second term on the RHS. I've tried to reconcile 2nd RHS terms in both the coordinate-free and index-based versions, but no luck. I'd appreciate any help or corrections!
 
Last edited:
Physics news on Phys.org
  • #2
What is your working definition of the torsion tensor?
 
  • #3
dextercioby said:
What is your working definition of the torsion tensor?
A map from 2 vector fields to a third vector field: $$T(X,Y)=\nabla_XY-\nabla_YX-[X,Y]$$ In terms of components, $$T_{\mu\nu}^{\ \ \ \lambda}=\Gamma_{\mu\nu}^{\lambda}-\Gamma_{\nu\mu}^{\lambda}$$
 
  • #4
Shirish said:
A map from 2 vector fields to a third vector field: $$T(X,Y)=\nabla_XY-\nabla_YX-[X,Y]$$ In terms of components, $$T_{\mu\nu}^{\ \ \ \lambda}=\Gamma_{\mu\nu}^{\lambda}-\Gamma_{\nu\mu}^{\lambda}$$
Note that this tensor is zero in standard GR (since the connection ##\Gamma## is symmetric in its two lower indexes). I believe Carroll at this point in his notes is not limiting his analysis to the standard GR case where the metric compatible connection is used.
 
  • Like
Likes PeroK
  • #5
Shirish said:
The main trouble is with the second term on the RHS.
That term is identically zero as you have written it, since ##\partial_\mu## and ##\partial_\nu## commute.
 
  • #6
Shirish said:
The coordinate-free definition is (eq. 3.71 in the notes): $$R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z$$
Note that in this equation, the last term on the RHS is nonzero in standard GR--i.e., it is nonzero even if the torsion is identically zero (because the metric compatible connection is being used).

You appear to be trying to match this equation with equation 3.66 on the assumption that the last term on the RHS of 3.71 only contains a torsion term. That can't possibly be right given the above.
 
  • Like
Likes PeroK
  • #7
PeterDonis said:
I believe Carroll at this point in his notes is not limiting his analysis to the standard GR case where the metric compatible connection is used.
Yes you're right about that.

PeterDonis said:
Note that in this equation, the last term on the RHS is nonzero in standard GR--i.e., it is nonzero even if the torsion is identically zero (because the metric compatible connection is being used).

You appear to be trying to match this equation with equation 3.66 on the assumption that the last term on the RHS of 3.71 only contains a torsion term. That can't possibly be right given the above.
So eq. 3.71 is the coordinate-free definition of the Riemann tensor as given in two sources - Carroll's notes and another book I'm studying (Semi-Riemannian Geometry by Newman). "on the assumption that the last term ... a torsion term" - sorry but could you elaborate on that? I didn't understand.
 
  • #8
Shirish said:
So eq. 3.71 is the coordinate-free definition of the Riemann tensor as given in two sources
As given in every source I'm aware of. In Misner, Thorne, & Wheeler, for example, it's equations 11.8 and 11.9.

Shirish said:
"on the assumption that the last term ... a torsion term" - sorry but could you elaborate on that?
My point was that even if the torsion tensor is zero--which it is if the connection ##\Gamma## is metric compatible, as it is in standard GR--the last term on the RHS in equation 3.71, i.e., ##\nabla_[X,Y] Z##, is still nonzero. But in your attempt to match equation 3.71 with equation 3.66 in the OP, you appear to be assuming that the ##\nabla_[X,Y] Z## term in equation 3.71 matches up with the torsion term in equation 3.66. That is not correct; if it were, the ##\nabla_[X,Y] Z## term would have to vanish whenever the torsion vanishes, and it doesn't.
 
  • #9
PeterDonis said:
My point was that even if the torsion tensor is zero--which it is if the connection ##\Gamma## is metric compatible, as it is in standard GR--the last term on the RHS in equation 3.71, i.e., ##\nabla_[X,Y] Z##, is still nonzero. But in your attempt to match equation 3.71 with equation 3.66 in the OP, you appear to be assuming that the ##\nabla_[X,Y] Z## term in equation 3.71 matches up with the torsion term in equation 3.66. That is not correct; if it were, the ##\nabla_[X,Y] Z## term would have to vanish whenever the torsion vanishes, and it doesn't.
Ah I see. I've written two equations. One is $$(R(\partial_{\mu},\partial_{\nu})(V))(x^{\rho})=([\nabla_{\mu},\nabla_{\nu}]V)(x^{\rho})-\nabla_{[\partial_{\mu},\partial_{\nu}]}V(x^{\rho})=([\nabla_{\mu},\nabla_{\nu}]V)(x^{\rho})$$ which follows from the coordinate-free definition for Riemann tensor by replacing ##X,Y## by fields ##\partial_{\mu},\partial_{\nu}## respectively and ##Z## by ##V##, and then acting both sides on ##x^{\rho}##. The LHS of the above equation is ##R^{\rho}_{\ \ \sigma\mu\nu}V^{\sigma}##. Eq. 3.66 is: $$R^{\rho}_{\ \ \sigma\mu\nu}V^{\sigma}=[\nabla_{\mu},\nabla_{\nu}]V^{\rho}+T_{\mu\nu}^{\ \ \ \ \lambda}\nabla_{\lambda}V^{\rho}$$ From what you've said, the term-wise matching that I was referring to earlier is incorrect. So my revised assertion is, the RHS should somehow match. i.e. $$([\nabla_{\mu},\nabla_{\nu}]V)(x^{\rho})=[\nabla_{\mu},\nabla_{\nu}]V^{\rho}+T_{\mu\nu}^{\ \ \ \ \lambda}\nabla_{\lambda}V^{\rho}$$ If this is fine, then the question of my OP is changed to: how do I prove the above equation?
 
  • #10
Shirish said:
which follows from the coordinate-free definition for Riemann tensor by replacing ##X,Y## by fields ##\partial_{\mu},\partial_{\nu}## respectively
Yes, but this is not a good way to compare equations involving the Riemann tensor because, as I have already pointed out, ##\partial_\mu## and ##\partial_\nu## commute. At the very least, you need to pick a pair of vectors ##X## and ##Y## that do not commute. But even that is not really sufficient, since if you want to match the two equations, you need to match them in a way that is valid for any vectors, so you should not pick any particular vectors when doing the matching.
 
  • #11
PeterDonis said:
Yes, but this is not a good way to compare equations involving the Riemann tensor because, as I have already pointed out, ##\partial_\mu## and ##\partial_\nu## commute. At the very least, you need to pick a pair of vectors ##X## and ##Y## that do not commute. But even that is not really sufficient, since if you want to match the two equations, you need to match them in a way that is valid for any vectors, so you should not pick any particular vectors when doing the matching.
I agree. The reason I went with that weird choice of ##X,Y,Z## was so that I could reproduce the LHS of Carroll's eq. 3.66 from the LHS of the coordinate-free definition of Riemann tensor (eq. 3.71). And then it would become a matter of reconciling the RHS'es only.

I'm completely new to this as you can probably tell, so any further tips/techniques will be very helpful. For example, how should I typically go about identities like the last equation in my last post?
 
  • #12
Shirish said:
The reason I went with that weird choice of was so that I could reproduce the LHS of Carroll's eq. 3.66 from the LHS of the coordinate-free definition of Riemann tensor (eq. 3.71).
Actually you can't do it that way, because you can't extract components of the Riemann tensor that way. To extract components of the Riemann tensor, or any tensor, you have to plug in basis vectors and 1-forms for all of the indexes of the tensor. So, for example, you could write, somewhat abusing notation,

$$
R^\rho{}_{\sigma \mu \nu} = \left( R(\partial_\mu, \partial_\nu) \partial_\sigma \right) \left( g^{\rho \alpha} \partial_\alpha \right)
$$

where the meaning here is that we evaluate the curvature operator ##R## for the basis vectors ##\partial_\mu## and ##\partial_\nu## on the basis vector ##\partial_\sigma##, and then contract the resulting vector with the basis 1-form ##g^{\rho \alpha} \partial_\alpha##. This gives us a number which is the component ##R^\rho{}_{\sigma \mu \nu}## of the Riemann tensor.
 
  • #13
PeterDonis said:
Actually you can't do it that way, because you can't extract components of the Riemann tensor that way.
My reasoning was like this: in local coordinates, ##R(\partial/\partial x^{\mu},\partial/\partial x^{\nu})\partial/\partial x^{\sigma}## can be expressed as: $$R\bigg(\frac{\partial}{\partial x^{\mu}},\frac{\partial}{\partial x^{\nu}}\bigg)\frac{\partial}{\partial x^{\sigma}}=R_{\ \ \sigma\mu\nu}^{\rho}\frac{\partial}{\partial x^{\rho}}$$ So $$R\bigg(\frac{\partial}{\partial x^{\mu}},\frac{\partial}{\partial x^{\nu}}\bigg)V=R\bigg(\frac{\partial}{\partial x^{\mu}},\frac{\partial}{\partial x^{\nu}}\bigg)V^{\sigma}\frac{\partial}{\partial x^{\sigma}}=V^{\sigma}R\bigg(\frac{\partial}{\partial x^{\mu}},\frac{\partial}{\partial x^{\nu}}\bigg)\frac{\partial}{\partial x^{\sigma}}=V^{\sigma}R_{\ \ \sigma\mu\nu}^{\rho}\frac{\partial}{\partial x^{\rho}}$$

Now if I act this on the ##x^{\rho}## coordinate function, I get $$\bigg(R\bigg(\frac{\partial}{\partial x^{\mu}},\frac{\partial}{\partial x^{\nu}}\bigg)V\bigg)(x^{\rho})=V^{\sigma}R_{\ \ \sigma\mu\nu}^{\rho}\frac{\partial}{\partial x^{\rho}}(x^{\rho})=V^{\sigma}R_{\ \ \sigma\mu\nu}^{\rho}\frac{\partial x^{\rho}}{\partial x^{\rho}}=R_{\ \ \sigma\mu\nu}^{\rho}V^{\sigma}$$

Is any of the above steps wrong?
 
  • #14
Shirish said:
in local coordinates, ##R(\partial/\partial x^{\mu},\partial/\partial x^{\nu})\partial/\partial x^{\sigma}## can be expressed as: $$R\bigg(\frac{\partial}{\partial x^{\mu}},\frac{\partial}{\partial x^{\nu}}\bigg)\frac{\partial}{\partial x^{\sigma}}=R_{\sigma\mu\nu}^{\ \ \ \ \ \ \ \rho}\frac{\partial}{\partial x^{\rho}}$$
No, this is not correct. The curvature operator ##R(X,Y)## applied to a vector ##Z##, which is what the LHS of this describes, gives a vector, but the RHS is not a vector.

There is a good reason why I said the equation I wrote in #12 was "somewhat abusing notation": because if you take it too literally, it is very easy to write down wrong things.

Carroll might not be the best source to learn from if you do not already have a good grasp of the coordinate-free notation for vectors and tensors. MTW's treatment of that is much more detailed and might be worth a look for you to see all the subtleties and possible traps that lie beneath the simple-looking notation that Carroll is using.
 
  • #15
Shirish said:
Is any of the above steps wrong?
The very first one is. See post #14.
 
  • #16
PeterDonis said:
The very first one is. See post #14.
I'll need to look into this. I'm reading "Semi-Riemannian Geometry: The Mathematical Language of General Relativity" by Stephen Newman. The first eq. in post #13 is based on eq. (18.7.2) on pg. 489 of the book (note that he uses a different convention of ordering ##R##'s subscripts):
Let ##(M,\nabla)## be a smooth ##m##-manifold with a connection, and let ##(U,(x^i))## be a chart on ##M##. In local coordinates, ##R(\partial/\partial x^i,\partial/\partial x^j)\partial/\partial x^k## can be expressed as $$R\bigg(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\bigg)\frac{\partial}{\partial x^k}=\sum_lR^l_{ijk}\frac{\partial}{\partial x^l}$$ where the ##R^l_{ijk}## are uniquely determined functions in ##C^{\infty}(U)##
Maybe I misinterpreted the above? The way I understand it, the RHS will be a vector field.
 
  • #17
Shirish said:
Maybe I misinterpreted the above?
I would need more context to understand what his notation means, and also what he means by "local coordinates". I don't have that book.
 
  • #18
PeterDonis said:
The curvature operator ##R(X,Y)## applied to a vector ##Z##, which is what the LHS of this describes, gives a vector
Let me expand on this a bit. Suppose we know the components ##R^\rho{}_{\sigma \mu \nu}## of the Riemann tensor in some coordinate chart, and we know the components ##X^\mu##, ##Y^\nu##, and ##Z^\sigma## of three vectors. Then we can express the coordinate-free operation ##R(X,Y)Z## of applying the curvature operator ##R(X,Y)## to the vector ##Z## as follows in index notation:

$$
R(X,Y)Z = R^\rho{}_{\sigma \mu \nu} X^\mu Y^\nu Z^\sigma
$$

Here the LHS and the RHS are both vectors, and it is clear what is being done: we are taking the Riemann tensor and contracting it with three vectors to get a vector.

Consider the above for a while and compare it to what you have been trying to do.
 
  • #19
PeterDonis said:
Let me expand on this a bit.
And to expand a bit more; suppose we leave out the vector ##Z## and just consider the curvature operator ##R(X,Y)##. What is this operator? It is a linear map from vectors to vectors. How do we know? Because if we apply this operator to a vector ##Z##, we get another vector.

Now, in tensor language, a linear map from vectors to vectors is a (1, 1) tensor, i.e., a tensor with one upper and one lower index. How would we obtain the components of this tensor? It should be obvious from what I wrote in post #18, but here it is:

$$
\left[ R(X,Y) \right]^\rho{}_\sigma = R^\rho{}_{\sigma \mu \nu} X^\mu Y^\nu
$$
 
  • #20
I should give more context. The way it's defined in the book so far is
Let ##(M,\nabla)## be a smooth manifold with a connection. The curvature tensor (field) is the map $$R:\mathfrak{X}(M)^3\to\mathfrak{X}(M)$$ defined by $$R(X,Y)Z=\nabla_X(\nabla_Y(Z))-\nabla_Y(\nabla_X(Z))-\nabla_{[X,Y]}(Z)$$ for all smooth vector fields ##X,Y,Z##
The book doesn't really treat ##R(X,Y)## as a separate object (as far as I've read). ##R## is just treated as a map of 3 vector fields and with the weird notation ##R(X,Y)Z## instead of ##R(X,Y,Z)##. Then there is a theorem showing that ##R## is multilinear in all its arguments and anti-symmetric in its first 2 arguments. And that's all that's mentioned in the curvature tensor section before the assertion about its components.

I hope the above is helpful in giving context
 
  • #21
PeterDonis said:
Carroll might not be the best source to learn from if you do not already have a good grasp of the coordinate-free notation for vectors and tensors.
Another reason for this, in addition to what I pointed out before, is that Carroll leaves out useful information that, for a person not familiar with the field, will not be obvious. For example, consider these two expressions:

$$
[\nabla_\mu, \nabla_\nu] V^\rho
$$

$$
[\nabla_X, \nabla_Y]Z
$$

These two expressions look the same, so you might be misled into thinking that they are the same, or at least that they express the same kind of relationship. But they don't!

The first expression is an expression for a (1, 2) tensor--a tensor with one upper and two lower indexes.

The second expression is an expression for a vector! It must be, because it is a piece of the curvature operator ##R(X,Y)## applied to the vector ##Z##, which, as we have already seen, gives a vector. But how can this be?

The answer is actually given by Carroll, but he tosses it off so quickly that if you're not already familiar with all this you will most likely miss the implications. Right after equation 3.71 Carroll says that the expression ##\nabla_X## means the covariant derivative along vector field ##X##, and gives the equation ##\nabla_X = X^\mu \nabla_\mu##.

Consider what that means. First, it makes it obvious that ##\nabla_X## and ##\nabla_\mu## are not the same thing, since the former is obtained by contracting the latter with a vector. Second, it shows that the operator ##\nabla_X## is actually a scalar operator--i.e., it has no index, so applying it to a vector just gives another vector. That is why the second expression above is an expression for a vector.

Btw, that same paragraph in Carroll also contains another tossed off statement that is highly relevant to this thread topic. He says:

The last term in (3.71), involving the commutator ##[X,Y]##, vanishes when ##X## and ##Y## are taken to be the coordinate basis vector fields...which is why this term did not arise when we originally took the commutator of two covariant derivatives.

In that last part, "when we originally took the commutator of two covariant derivatives", Carroll is referring to...equation 3.66! In other words, he is saying that you should not expect to be able to match equation 3.66 with equation 3.71, because the latter contains a term that the former does not.
 
  • #22
Shirish said:
The way it's defined in the book so far is
Ok, so, as you say, that book is treating ##R(X,Y)Z## as a map from sets of 3 vectors, to vectors. It gives the same coordinate-free equation as Carroll (and MTW) do, but note the statement after it: "for all smooth vector fields". That means it doesn't actually mean quite the same thing by the notation ##R(X,Y)Z## as Carroll (and MTW) do: it means the map from sets of 3 vectors to vectors, without choosing any particular vectors ##X##, ##Y##, ##Z##, whereas Carroll (and MTW) by that same notation ##R(X,Y)Z## mean the vector that results from applying the map to a particular set of 3 vectors ##X##, ##Y##, and ##Z##.

Unfortunately, notational differences like this are common in the literature, so one has to be very careful when interpreting different sources and comparing them. In this particular case, since the book you referenced doesn't mean the same thing as Carroll by the notation ##R(X,Y)Z##, you can't use the book's version to help you interpret Carroll's version.
 
  • #23
PeterDonis said:
Ok, so, as you say, that book is treating ##R(X,Y)Z## as a map from sets of 3 vectors, to vectors. It gives the same coordinate-free equation as Carroll (and MTW) do, but note the statement after it: "for all smooth vector fields". That means it doesn't actually mean quite the same thing by the notation ##R(X,Y)Z## as Carroll (and MTW) do: it means the map from sets of 3 vectors to vectors, without choosing any particular vectors ##X##, ##Y##, ##Z##, whereas Carroll (and MTW) by that same notation ##R(X,Y)Z## mean the vector that results from applying the map to a particular set of 3 vectors ##X##, ##Y##, and ##Z##.

Unfortunately, notational differences like this are common in the literature, so one has to be very careful when interpreting different sources and comparing them. In this particular case, since the book you referenced doesn't mean the same thing as Carroll by the notation ##R(X,Y)Z##, you can't use the book's version to help you interpret Carroll's version.
So if ##X,Y,Z## are vector fields, Carroll's (and MTW's) versions of ##R(X,Y)Z## actually mean ##R(X_p,Y_p)Z_p## for some ##p\in M##? I'll also look at your post #21 in a bit more detail when I get back home - seems like it addresses my main point of confusion.
 
  • #24
Shirish said:
So if ##X,Y,Z## are vector fields, Carroll's (and MTW's) versions of ##R(X,Y)Z## actually mean ##R(X_p,Y_p)Z_p## for some ##p\in M##?
Not just for some ##p##, but for some specific ##X_p##, ##Y_p##, ##Z_p## at that ##p##.

In some cases, the notation might also refer to picking specific vector fields ##X##, ##Y##, and ##Z##, but not picking a specific event ##p##. Sometimes one has to look at the specific context to determine which interpretation is meant.

Generally speaking, a tensor equation is an equation that relates tensor quantities at a particular event. It should be valid at any event you choose, but once you've picked an event, the equation should only involve tensor quantities at that event. (The only wrinkle here is that any equation involving derivatives assumes that you know the values of the tensor fields whose derivatives are being taken, not just at the chosen event, but in some open neighborhood of that event. But you are using the values in the open neighborhood to define/compute values of derivatives at the chosen event only, not everywhere in the open neighborhood.)
 
  • #25
PeterDonis said:
Yes, but this is not a good way to compare equations involving the Riemann tensor because, as I have already pointed out, ##\partial_\mu## and ##\partial_\nu## commute. At the very least, you need to pick a pair of vectors ##X## and ##Y## that do not commute. But even that is not really sufficient, since if you want to match the two equations, you need to match them in a way that is valid for any vectors, so you should not pick any particular vectors when doing the matching.
Yes. To piggyback on this, the last term drops out if ##X## and ##Y## are commuting vector fields (and doesn't if they aren't). Sometimes you see people nake this assumption that the two are vector fields associated with a coordinate frame hence commuting. I believe Wald even does this in his definition of the curvature tensor.
 
  • Like
Likes dextercioby
  • #26
@jbergman , @PeterDonis : Alright then following your advice, I've done the calculation for Riemann tensor acting on general vector fields instead of basis vector fields. To recap, Carroll's equation 3.66 is $$R^{\rho}_{\ \ \sigma\mu\nu}Z^{\sigma}=[\nabla_{\mu},\nabla_{\nu}]Z^{\rho}+T_{\mu\nu}^{\ \ \ \ \lambda}\nabla_{\lambda}Z^{\rho}$$ I'm starting with the Riemann curvature tensor definition: $$R(X,Y)Z=\nabla_X(\nabla_YZ)-\nabla_Y(\nabla_XZ)-\nabla_{[X,Y]}Z$$ Now I can use the identity ##\nabla_X(\nabla_YZ)=\nabla_{\nabla_XY}Z+\nabla^2_{X,Y}Z## and the definition ##\nabla^2_{X,Y}Z(\ldots)=\nabla^2Z(\ldots,Y,X)## to get
$$\begin{align*}
X^{\mu}Y^{\nu}Z^{\sigma}R^{\rho}_{\ \ \sigma\mu\nu}\partial_{\rho} &= \nabla_{\nabla_XY}Z+\nabla^2_{X,Y}Z-\nabla_{\nabla_YX}Z-\nabla^2_{Y,X}Z-\nabla_{[X,Y]}Z
\\ &=\nabla_{\nabla_XY}Z-\nabla_{\nabla_YX}Z-\nabla_{[X,Y]}Z+\nabla^2_{X,Y}Z-\nabla^2_{Y,X}Z
\\ &=\nabla_{T(X,Y)}Z+\nabla^2_{X,Y}Z-\nabla^2_{Y,X}Z
\\ &=X^{\mu}Y^{\nu}T_{\mu\nu}^{\ \ \ \ \lambda}\nabla_{\partial_{\lambda}}Z+\nabla^2_{X,Y}Z-\nabla^2_{Y,X}Z
\\ &=X^{\mu}Y^{\nu}T_{\mu\nu}^{\ \ \ \ \lambda}\nabla_{\partial_{\lambda}}Z+\nabla^2Z(\text{d}x^{\rho},Y,X)\partial_{\rho}-\nabla^2Z(\text{d}x^{\rho},X,Y)\partial_{\rho}
\\ &=X^{\mu}Y^{\nu}T_{\mu\nu}^{\ \ \ \ \lambda}(\nabla_{\partial_{\lambda}}Z)^{\rho}\partial_{\rho}+X^{\mu}Y^{\nu}\nabla^2Z(\text{d}x^{\rho},\partial_{\nu},\partial_{\mu})\partial_{\rho}-X^{\mu}Y^{\nu}\nabla^2Z(\text{d}x^{\rho},\partial_{\mu},\partial_{\nu})\partial_{\rho}
\\ &=X^{\mu}Y^{\nu}\big(T_{\mu\nu}^{\ \ \ \ \lambda}(\nabla_{\partial_{\lambda}}Z)^{\rho}+\nabla^2Z(\text{d}x^{\rho},\partial_{\nu},\partial_{\mu})-\nabla^2Z(\text{d}x^{\rho},\partial_{\mu},\partial_{\nu})\big)\partial_{\rho}
\end{align*}$$ From the above calculation, I can only conclude that
  1. In Carroll's convention, ##T_{\mu\nu}^{\ \ \ \ \lambda}\nabla_{\lambda}Z^{\rho}## actually means ##T_{\mu\nu}^{\ \ \ \ \lambda}(\nabla_{\partial_{\lambda}}Z)^{\rho}## and NOT ##T_{\mu\nu}^{\ \ \ \ \lambda}\nabla_{\partial_{\lambda}}(Z^{\rho})##
  2. What Carroll is calling ##\nabla_{\mu}\nabla_{\nu}Z^{\rho}## is actually index notation version of ##\nabla^2Z(\text{d}x^{\rho},\partial_{\nu},\partial_{\mu})##
Do the above calculation and conclusions seem fine to you?
 
  • #27
Shirish said:
Carroll's equation 3.66
Is an equation for the Riemann tensor acting on basis vector fields. So if you want to understand the Riemann tensor acting on general vector fields (in particular, vector fields that don't commute--coordinate basis vector fields always commute), you cannot use Carroll's Equation 3.66.

I suggest re-reading the last part of my post #21.

Shirish said:
Do the above calculation and conclusions seem fine to you?
No. I think you're going off in a wrong direction that will only confuse you further. See above.
 
  • #28
Thanks! I'll re-read your post more carefully.
PeterDonis said:
No. I think you're going off in a wrong direction that will only confuse you further. See above.
Exactly which step of the calculation is wrong? And in case the calculation is correct, then which of the two conclusions is wrong? That feedback will help improve my understanding
 
  • #29
Shirish said:
Exactly which step of the calculation is wrong?
The whole calculation is wrong because it's based on a wrong assumption: that you can obtain Carroll's equation 3.66 by applying the Riemann curvature tensor to general (non-commuting) vector fields. You can't. You obtain Carroll's Equation 3.66 by applying the Riemann curvature tensor to coordinate basis vector fields only.
 
  • #30
Shirish said:
the identity ##\nabla_X(\nabla_YZ)=\nabla_{\nabla_XY}Z+\nabla^2_{X,Y}Z## and the definition ##\nabla^2_{X,Y}Z(\ldots)=\nabla^2Z(\ldots,Y,X)##
Where do this identity and definition come from?
 
  • #31
PeterDonis said:
The whole calculation is wrong because it's based on a wrong assumption: that you can obtain Carroll's equation 3.66 by applying the Riemann curvature tensor to general (non-commuting) vector fields. You can't. You obtain Carroll's Equation 3.66 by applying the Riemann curvature tensor to coordinate basis vector fields only.
(Hope I don't sound argumentative, it's just that I always try to get full clarity) So a couple of points:
  1. I was asking about calculation on a standalone basis (the calculation itself doesn't refer to Carroll's equation) - look at it as a generic differential geometry calculation. I'm not sure if (and where) I've gone wrong in that
  2. The identity is a result on second order covariant derivatives and the definition is that of a total covariant derivative. Both can be found in Lee's Introduction to Riemannian Manifolds (resp. Proposition 4.21 and Proposition 4.17)
 
  • #32
Shirish said:
I was asking about calculation on a standalone basis
Then I don't understand what you are trying to show with the calculation. I also don't understand what in Carroll you are trying to compare it to, if it isn't Equation 3.66.

(I note, btw, that while you use general vectors ##X##, ##Y##, and ##Z##, you use the coordinate basis vector ##\partial_\rho## instead of a general covector. You should use a general covector.)
 
  • #33
PeterDonis said:
Then I don't understand what you are trying to show with the calculation.
So the calculation shows that $$X^{\mu}Y^{\nu}Z^{\sigma}R^{\rho}_{\ \ \sigma\mu\nu}\partial_{\rho}=X^{\mu}Y^{\nu}\big(T_{\mu\nu}^{\ \ \ \ \lambda}(\nabla_{\partial_{\lambda}}Z)^{\rho}+\nabla^2Z(\text{d}x^{\rho},\partial_{\nu},\partial_{\mu})-\nabla^2Z(\text{d}x^{\rho},\partial_{\mu},\partial_{\nu})\big)\partial_{\rho}$$
$$\implies X^{\mu}Y^{\nu}\big(Z^{\sigma}R^{\rho}_{\ \ \sigma\mu\nu}\big)\partial_{\rho}=X^{\mu}Y^{\nu}\big(T_{\mu\nu}^{\ \ \ \ \lambda}(\nabla_{\partial_{\lambda}}Z)^{\rho}+\nabla^2Z(\text{d}x^{\rho},\partial_{\nu},\partial_{\mu})-\nabla^2Z(\text{d}x^{\rho},\partial_{\mu},\partial_{\nu})\big)\partial_{\rho}$$
$$\implies Z^{\sigma}R^{\rho}_{\ \ \sigma\mu\nu}=T_{\mu\nu}^{\ \ \ \ \lambda}(\nabla_{\partial_{\lambda}}Z)^{\rho}+\nabla^2Z(\text{d}x^{\rho},\partial_{\nu},\partial_{\mu})-\nabla^2Z(\text{d}x^{\rho},\partial_{\mu},\partial_{\nu})$$ and this is what I was trying to compare to Carroll's eq. 3.66
 
  • #34
Shirish said:
So the calculation shows that $$X^{\mu}Y^{\nu}Z^{\sigma}R^{\rho}_{\ \ \sigma\mu\nu}\partial_{\rho}=X^{\mu}Y^{\nu}\big(T_{\mu\nu}^{\ \ \ \ \lambda}(\nabla_{\partial_{\lambda}}Z)^{\rho}+\nabla^2Z(\text{d}x^{\rho},\partial_{\nu},\partial_{\mu})-\nabla^2Z(\text{d}x^{\rho},\partial_{\mu},\partial_{\nu})\big)\partial_{\rho}$$
$$\implies X^{\mu}Y^{\nu}Z^{\sigma}\big(R^{\rho}_{\ \ \sigma\mu\nu}\big)\partial_{\rho}=X^{\mu}Y^{\nu}\big(T_{\mu\nu}^{\ \ \ \ \lambda}(\nabla_{\partial_{\lambda}}Z)^{\rho}+\nabla^2Z(\text{d}x^{\rho},\partial_{\nu},\partial_{\mu})-\nabla^2Z(\text{d}x^{\rho},\partial_{\mu},\partial_{\nu})\big)\partial_{\rho}$$
$$\implies R^{\rho}_{\ \ \sigma\mu\nu}=T_{\mu\nu}^{\ \ \ \ \lambda}(\nabla_{\partial_{\lambda}}Z)^{\rho}+\nabla^2Z(\text{d}x^{\rho},\partial_{\nu},\partial_{\mu})-\nabla^2Z(\text{d}x^{\rho},\partial_{\mu},\partial_{\nu})$$ and this is what I was trying to compare to Carroll's eq. 3.66
You still do not get it. It makes no sense to compare what you calculated with Carroll's Equation 3.66. I have already explained why several times. There is no point in explaining it again. You are evidently on a fool's errand and you are not listening when I try to explain that.
 
  • #35
The OP question is based on a misunderstanding, which has been repeatedly explained. Enough is enough. Thread closed.
 

Similar threads

  • Differential Geometry
Replies
3
Views
1K
Replies
2
Views
2K
  • Differential Geometry
Replies
1
Views
2K
  • Differential Geometry
Replies
3
Views
3K
  • Special and General Relativity
Replies
2
Views
1K
  • Special and General Relativity
Replies
10
Views
779
  • Special and General Relativity
4
Replies
124
Views
7K
  • Special and General Relativity
Replies
7
Views
2K
  • Differential Geometry
Replies
7
Views
2K
  • Special and General Relativity
2
Replies
62
Views
4K
Back
Top