4
$\begingroup$

I'm trying to understand the meaning of the Riemann curvature tensor, but I don't seem to be ready yet to understand the detailed rigorous definition.

Anyway, I managed to understand (this gif was really helpful) that it takes two coordinate lines $x^\mu, x^\nu$, parallel transports a vector first on $x^\mu$ and then $x^\nu$ and compares this to the parallel transport along $x^\nu$ followed by $x^\mu$.

I am, however, completely uncertain which index has which meaning in this intuitive example, since every explanation featuring a visual comparison I've seen yet does not explained it detailed enough.

Therefore, my question is: Which Meaning does each index of the Riemann curvature Tensor have?

I am looking for an answer along the lines of „$R^\rho{}_{\sigma\mu\nu}$ tells you how much the $\rho/\sigma/\mu/\nu$-Component of a vector changes when transported […]“.

(Sorry for probably being inaccurate regarding the meaning of “coordinate line”)

$\endgroup$
9
$\begingroup$

I could just give you the answer you want (which I do at the very end), but I think that's no fun. So I will try to give you a flavour of what the Riemann tensor says to show you how the result comes about. I don't know at what level you know the subject, so I apologize if this is too simple or too abstract.

Let $M$ be a (pseudo-)Riemannian manifold. The formal definition of the Riemann tensor is $$R(\mathbf{u},\mathbf v)\mathbf w = \nabla_\mathbf u\nabla_\mathbf v\mathbf w - \nabla_\mathbf v\nabla_\mathbf u\mathbf w - \nabla_{[\mathbf u,\mathbf v]} \mathbf w,$$ where $\mathbf u,\mathbf v,\mathbf w$ are vector fields on $M$ and where $\nabla$ is the Levi-Civita connection. If this looks foreign to you, don't worry, we will soon make it more concrete.

Intuitively, the covariant derivative $\nabla_{\mathbf v} \mathbf w$ tells you the instantaneous rate of change of $\mathbf w$ as it is transported in the direction $\mathbf v$.

Now imagine starting at a point $p$ with a vector $\mathbf w$. Now parallel transport $\mathbf w$ along the direction $\mathbf{v}$ for a small distance, and then continue transporting along the direction $\mathbf{u}$ for another small distance. You will end up with a new vector $\mathbf{w}'$. This vector $\mathbf{w}'-\mathbf{w}$ is given (in the limit) by $$\mathbf{w}'-\mathbf{w}\approx\nabla_\mathbf u \nabla_\mathbf v \mathbf w.$$ Likewise, you could start in the direction $\mathbf{u}$ and continue in the direction $\mathbf{v}$, for which you will end up with another vector $\mathbf{w}''$, given by $$\mathbf{w}''-\mathbf{w}\approx\nabla_\mathbf v \nabla_\mathbf u \mathbf w.$$ This is the picture I imagine you had in mind. Now you can see that, ignoring the last term of the Riemann tensor for now, we have $$R(\mathbf{u},\mathbf{v})\mathbf{w} \approx \nabla_\mathbf u\nabla_\mathbf v\mathbf w - \nabla_\mathbf v\nabla_\mathbf u\mathbf w \approx \mathbf{w}' - \mathbf{w}'',$$ which compares the two differently transported vectors. The difference vector $\mathbf{w}' - \mathbf{w}''$ is roughly the quantity which measures curvature, and this difference vector is roughly what the Riemann tensor gives you.

The last term of the Riemann tensor which we've left out is a correction which complicates the picture, and it has to do with the fact that $\mathbf{u}$ and $\mathbf{v}$ are not just vectors defined at the point $p$, but rather vector fields defined all around $p$, which are themselves changing as you move around $M$. The last term accounts for this change, but luckily we need not talk about it, as you shall see.

Let's move to index notation now. To do so, we pick local coordinates $x^\mu$. Then get a set of coordinate vector fields $\{\partial_\mu\}$, and a set of coordinate covector fields $\{dx^\mu\}$.

From now on, we will take our parallel transport directions $\mathbf u$ and $\mathbf v$ to be coordinate vectors, i.e. $\mathbf{u}=\partial_\mu$ and $\mathbf{v} = \partial_\nu$. It turns out for the special case of coordinate vector fields, the last term of the Riemann tensor vanishes $$\nabla_{[\partial_\mu,\partial_\nu]}\mathbf{w}=0,$$ so we will not have to worry about it, as promised earlier.

We will also abbreviate $\nabla_{\partial_\mu}$ as just $\nabla_\mu$.

Now, by definition we have the components of the Riemann tensor defined as $$R^\rho_{\ \ \sigma\mu\nu}=dx^\rho \left(R(\partial_\mu,\partial_\nu)\partial_\sigma\right) = dx^\rho\left(\nabla_\mu\nabla_\nu \partial_\sigma - \nabla_\nu\nabla_\mu \partial_\sigma\right).$$

Now, we can read off what the indices mean.

The directions $\mu$ and $\nu$ are our two transport directions. The direction $\sigma$ is our initial direction (our initial $\mathbf{w}$). Therefore the Riemann tensor with indices $\mu$,$\nu$, and $\sigma$ tells us the difference of the vectors obtained by transporting $\partial_\sigma$ first along $\nu$ and then along $\mu$ vs. the same vector obtained by first transporting along $\mu$ and then $\nu$. The covector $dx^\rho$ extracts the component of this difference vector in the $\rho$ direction.

In a summary in the way you wanted, $R^\rho_{\ \ \mu\nu\sigma}$ tells us the $\rho$ component of the difference vector obtained by transporting the vector $\partial_\sigma$ along the $\mu$ and $\nu$ directions.

$\endgroup$
4
  • $\begingroup$ This was exactly the kind of answer I was looking for, only better. I love this site! One question though: I've come across this quite some time, but why are we suddenly talking about vector fields instead of vectors? In this manner, I don't understand what things like $\mathbf{X}(\mathbf{Y})$ (I'm applying a vector field to something else…?) would mean, and similarly how vector fields can (or cannot) commute. $\endgroup$ Jun 2 '16 at 7:53
  • $\begingroup$ The definition of a tangent vector in general turns out to be relatively complicated in differential geometry, and the short of it is that the best definition of a tangent vector we can give on a manifold is more or less equivalent to a directional derivative. So all tangent vectors in differential geometry are not only directions, but derivative operators, so they can act on functions defined on the manifold. In your example $Y$ would typically be a scalar function, and $\mathbf{X}$ would be a directional derivative, say $D_\mathbf{X}$, with $\mathbf{X}(Y) = D_\mathbf{X}Y$. $\endgroup$
    – EuYu
    Jun 2 '16 at 8:02
  • $\begingroup$ As for why we consider vector fields, it's because we need some form of dynamical quantity. If you want to look at the curvature, then you must move a vector around, so you don't just have a single vector, but a vector at each point. That's just a vector field. Moving vectors around in a "parallel" manner turns out to be more difficult than you would imagine, and the simplest way to do so is to actually define a vector field which has vanishing covariant derivative in some sense, so a vector which doesn't change direction, as you would intuitively expect of a parallel vector. $\endgroup$
    – EuYu
    Jun 2 '16 at 8:05
  • $\begingroup$ Oh, I forgot that the vectors are derivatives in diff'geo. Thanks for the further explanation! $\endgroup$ Jun 2 '16 at 8:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.