How Do You Correctly Apply Indices in Tensor Calculus for Curvature?

In summary, the video discusses a problem with a parallel transport issue and how to solve it. The first derivative is easy to solve, but the second is more difficult. The running index on the second derivative is confusing and needs to be replaced. The final solution uses a different substitution for the running index on the second derivative.
  • #1
BiGyElLoWhAt
Gold Member
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Here's what I'm watching:

At about 1:35:00 he leaves it to us to look at a parallel transport issue. Explicitly to caclculate
##D_s D_r T_m - D_r D_s T_m##
And on the last term I'm having some difficulties, the second christoffel symbol.
So we have
##D_s [ \partial_r T_m - \Gamma_{rm}^t T_t]## after taking the first derivative. The first part of the second derivative is easy, but the second, I think I may have figured it out, but I'm not 100%, and would like someone to look at this to see if I'm doing this correctly, and if not, to correct me.
Second derivative:
##\partial_s [\partial_r T_m - \Gamma_{rm}^t T_t ] - \Gamma_{sm}^q [\partial_r T_q - \Gamma_{rq}^t T_t]##
My question is the running index (I think that's what it's called) on the second term, and how to replace the indices on, particularly, the very last christoffel symbol. I believe I need a different running index on the second derivative than I do for the first, so hence the q on ##\Gamma_{sm}^q## . However, my concern is in my ability to change the christoffel symbol ##\Gamma_{rm}^t## in the first derivative to ##\Gamma_{rq}^t## in the second. Part of me wants to do this:
##\Gamma_{sm}^q [\partial_r T_m - \Gamma_{rm}^t T_t]_q## **Edit** I realized after looking that I messed up with the m's here. Too many m's in the lower indices.
But I'm not sure if that's applicable. Also not sure what that would mean. I don't like changing anything within the [...] brackets, but I'm not sure how to introduce a new running index, as I'm relatively positive my result should be of the form ##S_{srm}## with S some tensor. Perhaps I should use T, but it's a different tensor of different rank, so I used S.

Hellp, someone learn me some knowledge.
 
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  • #2
Ok, so after watching a little bit further, it appears that my answer should be:
##R_{sr\ m}^{\ \ \ \ t} V_t ##
(In the video, he uses n, I used m).
Should I just define new tensors at each step? That's almost what it seems like he's doing/did.
Also, slightly off topic, his t is between the r and m, but latex wouldn't let me do that. Is that important? Also, can you do that in latex? (I botched it with spaces...)
 
  • #3
Perhaps I figured it out. I think redefining the tensor, then substituting it back in helped. Lemme try again:
##D_s [\partial_r T_m - \Gamma_{rm}^t T_t] \to D_s V_{rm}##
##D_s V_{rm} = \partial_s V_{rm} - \Gamma_{sr}^q V_{qm} - \Gamma_{sm}^q V_{rq}##
##= \partial_s [\partial_r T_m - \Gamma_{rm}^t T_t ] - \Gamma_{sr}^q [ \partial_q T_m - \Gamma_{qm}^t T_t ] - \Gamma_{sm}^q [ \partial_r T_q - \Gamma_{rq}^t T_t ] ##
I'm relatively confident that this is accurate, however, unnecessary. Perhaps, upon doing the second term (the second half of the commutator) I will get some cancellation, but I'm wondering if I can simplify this. I'm thinking that I should be able to use similar indices on some of the terms, since I should be able to "run" an index separately for each different term separately. So my t's in the first term, I should be able to run them separately from my t's in the second term and the 3rd term, fill the matrices, and then sum them all together. With that in mind, do I actually need all of this explicitly?
 
  • #4
Eh, I made a mistake.
This is my final result:
##-\partial_s \Gamma_{rm}^t T_t - \Gamma _{sm}^q [\partial_r T_q - \Gamma _{rq}^t T_t ] + \partial_r \Gamma_{sm}^t T_t + \Gamma_{rm}^q [\partial_s T_q - \Gamma_{sq}^t T_t ] ##

My first derivative is above, in post #3 D_s D_r T_m
Here's my second, D_r D_s T_m:
##\partial_r [\partial_s T_m - \Gamma_{sm}^t T_t ] - \Gamma_{rs}^q [\partial_q T_m - \Gamma_{qm}^t T_t ] - \Gamma_{rm}^q [ \partial_s T_q - \Gamma_{qs}^t T_t] ##
The final result is the difference between the two.
So I'm left with 4 partial derivatives instead of 2, and 2 dual christoffel terms. I have 2 partial terms too many. If I can replace my q's with t's on the running indices, then all the partial terms cancel out, but I can only do that assuming that multiplication by a christoffel commutes with partial differentiation, which I'm not so sure, since christoffel symbols are functions of the metric, which can depend on the position...
The second derivative I solved using the same exact method as the first, except instead of ##V_{rm}## as the substitution, it's ##V_{sm}##, for obvious reasons.

I'm really lost here, and can't seem to find what I did wrong. Any advice, people of PF?
 

1. What is a second derivative?

A second derivative is the rate of change of the slope of a function. It is calculated by taking the derivative of the first derivative of the function.

2. How is the second derivative useful in exploring curvature?

The second derivative helps us understand the concavity of a function, which is a measure of the curvature of the graph. It tells us whether the function is convex or concave, and helps us identify points of inflection.

3. Can the second derivative be negative?

Yes, the second derivative can be negative. This indicates that the function is concave down and has a negative curvature at that point.

4. How do I interpret the second derivative graphically?

The second derivative can be interpreted as the rate of change of the slope of the original function. If the second derivative is positive, the slope of the function is increasing. If it is negative, the slope is decreasing. The concavity of the function can also be determined from the second derivative graph.

5. Are there any real-world applications of the second derivative?

Yes, the second derivative has many real-world applications, including predicting the behavior of stock prices, analyzing the acceleration of objects in motion, and understanding the rate of change in chemical reactions.

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