# Homework Help: Calculating the Riemann Tensor

1. May 9, 2012

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

The work I have been following has me very confused... and I am almost sure I am making a mistake somewhere!

After working up to this equation:

$$\delta V = dX^{\mu}\delta X^{\nu} [\nabla_{\mu} \nabla_{\nu}]V$$

I am asked to calculate the curvature tensor. Now the way I did it, turned out different to the way it is shown at the end of the work... it took a bit of time to understand what method I was using was different but I did work it out nonetheless, and what I want to know is which method is correct (most likely mine is wrong but I need some guidance.)

Ignoring $$V$$ and just working out the commutator relationship, I expand:

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

The first part of the calculation, gives

$$\partial_{\nu}\partial_{\mu} - \partial_{\mu} \partial_{\nu}$$

Which is just zero, because the ordinary derivatives commute, so they go to zero. Fine. Now, according to the work I am following, the next set of terms should have been:

$$\Gamma_{\nu} \partial_{\mu} - \partial_{\nu}\Gamma_{\mu}$$

But I ended up with

$$\partial_{\nu}\Gamma_{\mu} - \partial_{\mu} \Gamma_{\nu}$$

and I only arrived at this because it is well known that once you calculate the first set of terms, for instance, using this guide:

$$(a+b)(c+d)$$

ignoring that we are taking this part away from another part, the first term arises because you multiply $$a$$ with $$c$$. Then you multiply $$a$$ with $$d$$.

In the work I am following, it seems that in this:

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

You get

$$\partial_{\nu} \partial{\mu}$$

first of all by following that rule (then the extra's of course $$- \partial_{\mu} \partial_{\nu}$$ but according the second lot, the work has

$$\Gamma_{\nu} \partial_{\mu} - \partial_{\mu} \Gamma_{\nu}$$

My brain agrees with the $$-\partial_{\mu} \Gamma_{\nu}$$ term but I do not understand how it gathers the

$$\Gamma_{\nu} \partial_{\mu}$$

Because for that to be true, it would mean using my expression again for simplicity that

$$(a+b)(c+d) - (a'+b')(c'+d')$$

It seems right to multiply $$(a' \cdot d')$$ but with a steady analysis of the works example, it shows $$b \cdot c$$ which would give the first term $$\Gamma_{\nu}\partial_{\mu}$$... but that isn't right is it? Or am I wrong? Am I doing it wrong?

Thanks

2. May 9, 2012

I've not made a massive amount of headway since I posted this question but I think my confusion exists in some parts I have missed out. There seems to be a sign change for:

$$-[\partial_{\mu} , \Gamma_{\nu}] = -\frac{\partial \Gamma_{\nu}}{\partial X^{\mu}}$$

Where I have a positive solution

$$[\partial_{\nu} , \Gamma_{\mu}] = \frac{\partial \Gamma_{\mu}}{\partial X^{\nu}}$$

Now this commutator would permit the calculation,

$$\partial_{\nu}\Gamma_{\mu}$$ but I am still confused as to why it wasn't calculated straight away, if anyone is following my drift.
......................................................................................

Then upon saying that I think I have worked out my problem, but I need confirmation:

Ah, that was my breaking point - I think I know what is happening now.

Again:

Where I have a positive solution

$$[\partial_{\nu} , \Gamma_{\mu}] = \frac{\partial \Gamma_{\mu}}{\partial X^{\nu}}$$

Now this commutator would permit the calculation,

$$\partial_{\nu}\Gamma_{\mu}$$ ---- so, in the working example, it looked like the person who calculated it was taking the $$\Gamma_{\nu}$$ and multiplying it with $$\partial_{\mu}$$ and that confused me because normally the next multiplication would give $$\partial_{\nu} \Gamma{\mu}$$ but because of this delicate sign change, which exists because

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

The

$$(\partial_{\nu} + \Gamma_{\nu})(\partial_{\mu} + \Gamma_{\mu})$$

Part is positive then it follows that

$$[\partial_{\nu} , \Gamma_{\mu}] = \frac{\partial \Gamma_{\mu}}{\partial X^{\nu}}$$

Which would dictate the multiplication in the form the work gives. If it had been negative, such as this part

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

Then the normal multiplication rule would apply, because as I said, I had no problem agreeing with the product

$$\partial_{\mu} \Gamma_{\nu}$$

Does this seem right now?

3. May 10, 2012

Can no one help?

4. May 10, 2012

### clamtrox

I don't understand your question. Remember that derivatives don't commute, and they only operate on things on their right. So
$$(\partial + \Gamma)(\partial + \Gamma) = \partial \partial + \partial \Gamma + \Gamma \partial + \Gamma \Gamma$$

5. May 14, 2012

If I have:

$$(a+b)(c+d) -(a'+b')(c'+d')$$

multiplying it out, I begin with

$$ac - a'c'$$

Just as in the OP equation, for the GR curvature tensor, we have

$$\partial_{\nu} \partial_\mu - \partial_{\mu} \partial_{\nu}$$

I'm fine with that. It's what comes next is muddling my brain up. According to my simple equation above,

$$(a+b)(c+d) -(a'+b')(c'+d')$$

The next thing I compute is

$$ad - a'd'$$

but we don't seem to be doing that when calculating the curvature tensor.

What I am told I should have to calculate next is

$$\Gamma_{\nu}\partial_{\mu} - \partial_{\mu}\Gamma_{\nu}$$

Instead that looks like

$$bc - a'd'$$

are you following now?

6. May 14, 2012

Compare it with

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

In other words. The first lot of computations I can agree with. I don't understand when it goes to compute the next set of terms, it seems to take

$$\Gamma_{\nu}$$

with

$$\partial_{\mu}$$

but then on the other side it would correctly take

$$\partial_{\mu}\Gamma_{\nu}$$

This is what has caused the confusion.

7. May 14, 2012

This I agree with. But my work seems to be skipping the second part in your equation, the

$$\partial \Gamma$$ part. Look at the way I have presented the equations from the work. It does not follow your pattern.

8. May 14, 2012

Let me show you the whole thing now, naturally we compute

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

It says when multiplying out the terms, we should have

$$\partial_{\nu} \partial_{\mu} - \partial_{\mu} \partial_{\nu} \Gamma_{\nu \beta}^{\alpha} \partial_{\mu} - \partial_{\nu}\Gamma_{\nu \beta}^{\alpha} + \Gamma_{\nu \delta}^{\alpha}\Gamma_{\mu \beta} - \Gamma_{\nu \delta}^{\alpha}\Gamma_{\mu \beta}$$

The important bit begins at 38:30. Is he calculating this right? And if he is, what is it I am failing to understand?

Last edited: May 14, 2012
9. May 14, 2012

It took me a while to find the video. But I found it.

10. May 14, 2012

### clamtrox

So what happens when you try to calculate it properly? Remember also that you get two terms from the second covariant derivative (because $\nabla V$ is a 1-1-tensor).

$$\nabla_{[\mu} \nabla_{\nu]} V^{\alpha} = \nabla_{\mu} \nabla_{\nu} V^{\alpha} - [\mu \leftrightarrow \nu] = \partial_{\mu} (\nabla_\nu V^\alpha) - \Gamma^{\lambda}_{\mu \nu} \nabla_{\lambda} V^{\alpha} + \Gamma^\alpha_{\mu \delta} \nabla_\nu V^\delta - [\mu \leftrightarrow \nu]$$

Oh, and that notation that Susskind is really confusing. I think you will be better off working it out without any shorthands

Last edited: May 14, 2012
11. May 14, 2012

Well take a look at your example:

$$(\partial + \Gamma)(\partial + \Gamma) = \partial \partial + \partial \Gamma + \Gamma \partial + \Gamma \Gamma$$

By the time he's cancelled out the first terms, he should be writing what you have here:

$$\partial \Gamma$$

Compare it with

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

In other words. The first lot of computations I can agree with. I don't understand when it goes to compute the next set of terms, he seems to take

$$\Gamma_{\nu}$$

with

$$\partial_{\mu}$$

but then on the other side it would correctly take

$$\partial_{\mu}\Gamma_{\nu}$$

This is what has caused the confusion. Shouldn't it be

$$\partial_{\nu}\Gamma_{\mu} - \partial_{\mu}\Gamma_{\nu}$$

12. May 14, 2012

### clamtrox

You are still explaining very poorly. Can you show your work using proper indices step by step?

13. May 14, 2012

I'd rather not. I think that would complicate it.

Let's try this again. Look at what you have

$$(\partial + \Gamma)(\partial + \Gamma) = \partial \partial + \partial \Gamma + \Gamma \partial + \Gamma \Gamma$$

I agree with this. Now, take a look at the RHS of this equation, the second term

$$\partial \Gamma$$

This comes into play when you have calculated the first terms

$$\partial_{\nu}\partial_{\mu} - \partial_{\mu}\partial_{\nu}$$

That cancels out anyway, so it is of no consequence. I'm interested on the second term on the RHS of your equation, again, it is

$$\partial \Gamma$$

You get that from multiplying, I presume, the first partial in the first set of paranthesis with the last term in the next set of paranthesis, the Gamma. But Susskind is clearly doing something different, because after the first lot of terms vanish, he continues his multiplication but begins with the Gamma term in the first set of paranthesis and then he mutliplies that with the first partial in the next lot.

So effective what susskind is doing is

$$\Gamma_{\mu} \partial_{\nu}$$

you do see this yes?

My question is why. I thought it would have been $$\partial_{\mu} \Gamma_{\nu}$$

I can't explain this any more. Please read carefully, follow what I am saying and compare it with what susskind is doing.

Last edited: May 14, 2012
14. May 14, 2012

Writing out my simple example,

$$(a+b)(c+d) - (a'+b')(c'+d')$$

Following what susskind is doing, he seems to be doing this:

$$ac - a'c' \cdot bc - a'd' \cdot bd - b'd'$$

15. May 14, 2012

My problem lies therein with the bc-term straight after the

$$ac - a'c'$$

(which cancels anyway. Not missing a part in case I am unclear)

16. May 14, 2012

### clamtrox

As you know, summation is commutative, so it doesn't matter what order you write your sums in. A+B = B+A. Multiplication however, isn't. So AB != BA

17. May 14, 2012

### clamtrox

Finally, I should say that I don't understand Susskinds notation, and I think you would be doing yourself a favour by doing the calculation properly

18. May 14, 2012

Oh I know that, I repeated some mu-nu indices which has caused a little confusion:

I'd rather not. I think that would complicate it.

Let's try this again. Look at what you have

$$(\partial + \Gamma)(\partial + \Gamma) = \partial \partial + \partial \Gamma + \Gamma \partial + \Gamma \Gamma$$

I agree with this. Now, take a look at the RHS of this equation, the second term

$$\partial \Gamma$$

This comes into play when you have calculated the first terms

$$\partial_{\nu}\partial_{\mu} - \partial_{\mu}\partial_{\nu}$$

That cancels out anyway, so it is of no consequence. I'm interested on the second term on the RHS of your equation, again, it is

$$\partial \Gamma$$

You get that from multiplying, I presume, the first partial in the first set of paranthesis with the last term in the next set of paranthesis, the Gamma. But Susskind is clearly doing something different, because after the first lot of terms vanish, he continues his multiplication but begins with the Gamma term in the first set of paranthesis and then he mutliplies that with the first partial in the next lot.

For a note:

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

So effective what susskind is doing is

$$\Gamma_{\nu} \partial_{\mu}$$

you do see this yes?

My question is why. I thought it would have been $$\partial_{\nu} \Gamma_{\mu}$$

I can't explain this any more. Please read carefully, follow what I am saying and compare it with what susskind is doing.

19. May 14, 2012

You should be able to follow that now surely? Susskind isn't using a strange notation - he's just leaving out some indices for simplicity, whereas he'd have to include them if the had multiplied it by $$V^{\beta}$$.

20. May 14, 2012

I'll tell you what, Ill write it even more straightforward than that - I'll show you what I think should be there. The following is concerned only when the first set of terms are made to disappear as we have seen before:

For a note:

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

So effective what the work is doing is

$$\Gamma_{\nu} \partial_{\mu}$$

you do see this yes?

My question is why. I thought it would have been $$\partial_{\nu} \Gamma_{\mu}$$ - This quantity is taken away from $$\partial_{\mu}\Gamma_{\nu}$$ and I agree with that... so what it has is

$$\Gamma_{\nu} \partial_{\mu} -\partial_{\mu}\Gamma_{\nu}$$

Where I thought the next lot of terms should be

$$\partial_{\nu} \Gamma_{\mu}-\partial_{\mu}\Gamma_{\nu}$$

Now which one is right?