Metric tensor in spherical coordinates

[WannabeNewton]

I can see no reason or justification for this assertion, it seems completely random, like the quantum collapse of some educational wavefunction.

I lol'ed at that haha.

 

[stevendaryl]

I couldn't see it in your derivation. I guess, the exact derivation has to do with covariant derivatives yet, as I said, I cannot grasp it at this stage.

I'm ambivalent about the definition of Riemann in terms of covariant derivatives. It's kind of weird, since to compute $R(U,V,W)$ we have to pretend that $U$, $V$ and $W$ are vector fields, and then do the calculation, and then in the end, nothing matters except $U$, $V$ and $W$ at a single point.

But the definition in terms of covariant derivatives is pretty succinct:

$R(U,V,W) = \nabla_V (\nabla_U W) - \nabla_U (\nabla_V W)$

Then in terms of components:

$(\nabla_U W)^{\mu} = \partial_{\nu} W^{\mu} U^{\nu} + \Gamma^{\mu}_{\nu \lambda} U^{\nu} W^{\lambda}$

$(\nabla_V (\nabla_U W))^{\mu}<br /> = \partial_{\alpha} \partial_{\nu} W^{\mu} U^{\nu} V^{\alpha}<br /> + \partial_{\alpha} (\Gamma^{\mu}_{\nu \lambda} U^{\nu} W^{\lambda}) V^{\alpha}<br /> + \Gamma^{\mu}_{\alpha \beta} (\partial_{\nu} W^{\beta}) U^{\nu} V^{\alpha}<br /> + \Gamma^{\mu}_{\alpha \beta} \Gamma^{\beta}_{\nu \lambda} U^{\nu} W^{\lambda} V^{\alpha}$

$(\nabla_U (\nabla_V W))^{\mu}<br /> = \partial_{\alpha} \partial_{\nu} W^{\mu} V^{\nu} U^{\alpha}<br /> + \partial_{\alpha} (\Gamma^{\mu}_{\nu \lambda} V^{\nu} W^{\lambda}) U^{\alpha}<br /> + \Gamma^{\mu}_{\alpha \beta} (\partial_{\nu} W^{\beta}) V^{\nu} U^{\alpha}<br /> + \Gamma^{\mu}_{\alpha \beta} \Gamma^{\beta}_{\nu \lambda} V^{\nu} W^{\lambda} U^{\alpha}$

Subtract them to get:
$(\nabla_V (\nabla_U W) - \nabla_U (\nabla_V W))^{\mu}<br /> = (\partial_{\alpha} \Gamma^{\mu}_{\nu \lambda}) U^{\nu} W^{\lambda} V^{\alpha}<br /> - (\partial_{\alpha} \Gamma^{\mu}_{\nu \lambda}) V^{\nu} W^{\lambda} U^{\alpha}<br /> + \Gamma^{\mu}_{\alpha \beta} \Gamma^{\beta}_{\nu \lambda} U^{\nu} W^{\lambda} V^{\alpha}<br /> - \Gamma^{\mu}_{\alpha \beta} \Gamma^{\beta}_{\nu \lambda} V^{\nu} W^{\lambda} U^{\alpha}$

Note the miracle that all the derivatives of $W$, $U$ and $V$ cancel out. (I guess it's not a miracle, since the result has to be a tensor, so those cancellations must happen.)

Rename some dummy indices to factor out $U_\nu$, $V_\alpha$ and $W_\lambda$ to get:
$(\nabla_V (\nabla_U W) - \nabla_U (\nabla_V W))^{\mu}<br /> = ((\partial_{\alpha} \Gamma^{\mu}_{\nu \lambda})<br /> - (\partial_{\nu} \Gamma^{\mu}_{\alpha \lambda})<br /> + \Gamma^{\mu}_{\alpha \beta} \Gamma^{\beta}_{\nu \lambda}<br /> - \Gamma^{\mu}_{\nu\beta} \Gamma^{\beta}_{\alpha \lambda})U^{\nu}V^{\alpha} W^{\lambda}$

 

[GRstudent]

It becomes more clear in this way. I appreciate your efforts! Thanks!

 

[stevendaryl]

I'm ambivalent about the definition of Riemann in terms of covariant derivatives. It's kind of weird

Just to expand on my complaint; the definition of R in terms of covariant derivatives is very succinct, but it's a little mysterious why it gives the right answer for parallel transport of a vector around a loop.

 

[GRstudent]

Actually, the derivation of the Riemann tensor can be done in much easier way. This was discussed in 8th lecture on GR by Susskind.

It is important to realize that the operator of the covariant derivative is the following:

$\nabla_\mu$$=$$\dfrac{\partial}{\partial x^\mu} + \Gamma_\mu$

Then, the idea of a commutator comes out:

$AB-BA=[A,B]$

$[\dfrac{\partial}{\partial x}, f(x)] = \dfrac{\partial f}{\partial x}$

$[\nabla_\nu , \nabla_\mu]$ is basically the Riemann tensor:

$(\partial_\nu + \Gamma_\nu)(\partial_\mu +\Gamma_\mu)-(\partial_\mu+\Gamma_\mu)(\partial_\nu+\Gamma_\nu)$

When we expand this equation, we are left with

$\partial_\nu \partial_\mu+\partial_\nu \Gamma_\mu + \Gamma_\nu \partial_\mu + \Gamma_\nu \Gamma_\mu - \partial_\mu \partial_\nu - \partial_\mu \Gamma_\nu - \Gamma_\mu \partial_\nu - \Gamma_\mu \Gamma_\nu$

The products of two derivative operators are commute so they cancel each other out.

Then we have

$\partial_\nu \Gamma_\mu + \Gamma_\nu \partial_\mu + \Gamma_\nu \Gamma_\mu - \partial_\mu \Gamma_\nu - \Gamma_\mu \partial_\nu - \Gamma_\mu \Gamma_\nu$

Now, we can notice that there are two commutators written out explicitly in that equation, so we put them together to have:

$-[\partial_\mu, \Gamma_\nu] + [\partial_\nu, \Gamma_\mu]+\Gamma_\nu \Gamma_\mu- \Gamma_\mu \Gamma_\nu$

Using formula for commutator, we have that:

$\dfrac{\partial \Gamma_\mu}{\partial x^\nu}-\dfrac{\partial \Gamma_\nu}{\partial x^\mu} + \Gamma_\nu \Gamma_\mu- \Gamma_\mu \Gamma_\nu$

^
Which is an exact definition of the Riemann tensor.

 

[stevendaryl]

Actually, the derivation of the Riemann tensor can be done in much easier way. This was discussed in 8th lecture on GR by Susskind.

Isn't that basically the same derivation as the one in post #92?

 

[stevendaryl]

Isn't that basically the same derivation as the one in post #92?

I guess not, in the sense that considering $\partial$ and $\Gamma$ as operators and computing the commutator is "cleaner" than sticking in vectors to operate on. Conceptually, it's the same thing, whether you write:

$[\nabla_\mu, \nabla_\nu]$

or

$\nabla_U (\nabla_V W) - \nabla_V (\nabla_U W)$

 

[GRstudent]

^
Correct.

So basically, the covariant derivative shows how the tangent vector varies along the curve?

 

[GRstudent]

The Gammas in polar coordinates are $\dfrac{1}{r}$, what does it really mean?

 

[stevendaryl]

The Gammas in polar coordinates are $\dfrac{1}{r}$, what does it really mean?

The coefficient $\Gamma^{i}_{jk}$ is defined implicitly via the equation:

$\nabla_{j} e_{k} = \Gamma^{i}_{jk} e_{i}$

where $e_{i}$ is the ith basis vector. To illustrate, let's make it simpler, and use 2D polar coordinates. In rectangular coordinates, the coordinates are $x$ and $y$, with basis vectors $e_x$ and $e_y$. Now we can do a coordinate change to coordinates $r$ and $\theta$ defined by:
$x = r cos(\theta), y = r sin(\theta)$ with corresponding basis vectors $e_r = cos(\theta) e_x + sin(\theta) e_y$ and $e_{\theta} = - r sin(\theta) e_x + r cos(\theta) e_y$. We compute derivatives:

$\nabla_{r} e_{r} = 0$

So, $\Gamma^{r}_{rr} = \Gamma^{\theta}_{rr} = 0$

$\nabla_{\theta} e_{r} = (\partial_{\theta} cos(\theta)) e_x + (\partial_{\theta} sin(\theta)) e_y$
= $- sin(\theta) e_x + cos(\theta) e_y$
= $e_{\theta}/r$

So, $\Gamma^{r}_{\theta r} = 0$, and $\Gamma^{\theta}_{\theta r} = 1/r$

$\nabla_{r} e_{\theta} = (\partial_{r} (-r sin(\theta))) e_x + (\partial_{r} (r cos(\theta))) e_y$
= $-r sin(\theta) e_x + cos(\theta) e_y$
= $e_\theta/r$

So, $\Gamma^{r}_{r \theta} = 0$, and $\Gamma^{\theta}_{r \theta} = 1/r$

$\nabla_{\theta} e_{\theta} = (\partial_{\theta} (-r sin(\theta))) e_x + (\partial_{\theta} (r cos(\theta))) e_y$
= $- r cos(\theta) e_x - r sin(\theta) e_y$
= $-r e_{r}$

So, $\Gamma^{r}_{\theta \theta} = -r$, and $\Gamma^{\theta}_{\theta \theta} = 0$

 

[GRstudent]

stevendaryl,

One of the components of the Riemann tensor in Schwarzschild solution is

$R^{r}_{\theta \theta r}{} = \dfrac{M}{R}$

What does this component mean in real sense? Susskind told that, here, two $\theta \theta$ indexes (which are downstairs) have something to do with two vectors which define a plane at some point in space. What other two r-r represent? Basically, Riemann tensor takes as input 4 vectors and outputs the curvature value, right?

Thank you.

Is this a true picture of basis vectors? http://www.springerimages.com/Images/RSS/1-10.1007_978-1-4614-0706-5_6-0

 

[Dale]

$R^{r}_{\theta \theta r}{} = \dfrac{M}{R}$

What does this component mean in real sense?

I am not completely certain that I have this right, but I believe that it means that if you move a $dr$ vector around a closed differential $d\theta$, $d\theta$ loop then it will change by a differential amount M/R in the $dr$ direction.

 

[GRstudent]

^
This makes some sense.

 

[stevendaryl]

I am not completely certain that I have this right, but I believe that it means that if you move a $dr$ vector around a closed differential $d\theta$, $d\theta$ loop then it will change by a differential amount M/R in the $dr$ direction.

I'm not sure about the convention for the order of the indices, but if you have a loop (parallelogram) in which the two sides are parallel (both $d\theta$), then the loop has area 0, and so the parallel transport gives 0.

I think that the correct interpretation is this: Make a parallelogram with one side equal to $dr\ e_r$ and another side equal to $d\theta \ e_\theta$. March the vector $V = e_\theta$ around the parallelogram to get a new vector $V'$ that will be slightly different from $e_\theta$. Then letting $\delta V = V' - V$,

$\delta V^r = R^r_{\theta \theta r} dr d\theta$

 

[GRstudent]

This is an excellent representation of basis vectors in Spherical coordinates:

http://en.citizendium.org/wiki/File:Spherical_unit_vectors.png

I tried to reproduce the Riemann tensor's plane according to stevendaryl's suggestions here

http://img826.imageshack.us/img826/5614/605pxsphericalunitvecto.png