Tensor Calculus Problem: Simplifying Terms with Index Exchange

Mentz114
Messages
5,429
Reaction score
292
If you don't like indexes, look away now. I got these terms from a tensor calculus program as part of a the transformed F-P Lagrangian.
<br /> \begin{align}<br /> {g}^{b a}\,{g}^{d e}\,{g}^{f c}\,{X}_{a,b c}\,{X}_{d,e f}\\<br /> +{g}^{b a}\,{g}^{c f}\,{g}^{e d}\,{X}_{a,b c}\,{X}_{d,e f}\\<br /> +{g}^{b a}\,{g}^{c e}\,{g}^{d f}\,{X}_{a,b c}\,{X}_{d,e f}\\<br /> +{g}^{a b}\,{g}^{c e}\,{g}^{d f}\,{X}_{a,b c}\,{X}_{d,e f}<br /> \end{align}<br />

I think I can substitute ##g^{pq}## with ##g^{qp}## without harm. Also ##,{X}_{p,q r}={X}_{p,r q}## so I can exchange ##q## and ##r##. But can I do this if ##q## and ##r## are in different ##g##'s (like swapping ##e## and ##f## in the fourth term) ?

If these gymnastics are allowed then the terms are equal and there is a good simplification.
 
Physics news on Phys.org
The metric tensor is symmetric, so there's never any harm with changing ##g^{ab}\leftrightarrow g^{ba}##. And the same goes for partial derivatives (important to note though that this is not true for covariant derivatives, so if it was ##X_{p;qr}## instead, you can't just arbitrarily make this swap). In addition, every index seems to be summed over, so they are all dummy indices anyways, so within each term you are allowed to make arbitrary index substitutions (as long as you replace both instances of said index simultaneously).
 
  • Like
Likes Mentz114
Matterwave said:
The metric tensor is symmetric, so there's never any harm with changing ##g^{ab}\leftrightarrow g^{ba}##. And the same goes for partial derivatives (important to note though that this is not true for covariant derivatives, so if it was ##X_{p;qr}## instead, you can't just arbitrarily make this swap). In addition, every index seems to be summed over, so they are all dummy indices anyways, so within each term you are allowed to make arbitrary index substitutions (as long as you replace both instances of said index simultaneously).

Thank you. I thought it would be OK but not certain. I should have mentioned that ##g## is ##\eta##, the Minkowski metric.

It's a pity there isn't a change of sign so some of these pesky things could cancel ...

(you wouldn't like to look over the other 80 terms, by any chance ? :-)
 
Nope, no change in sign for either the metric or the derivatives term. Both are symmetric. Although, since they are all added together...and they are all summed over every index...my suspicion is that all 4 terms are the same term...

Certainly the first two terms are identical, and the bottom two are identical. I'm not sure if the top and bottom are identical though.

I hope someone can check this result. It's been a while since I've done much index gymnastics.
 
Matterwave said:
Nope, no change in sign for either the metric or the derivatives term. Both are symmetric. Although, since they are all added together...and they are all summed over every index...my suspicion is that all 4 terms are the same term...

Certainly the first two terms are identical, and the bottom two are identical. I'm not sure if the top and bottom are identical though.

I hope someone can check this result. It's been a while since I've done much index gymnastics.

The iTensor program agrees with you. Those terms got amalgamated into 2 after I tidied the symmetry declarations. In fact The four terms in the massless Lagrangian only have 28 terms after canonicalising (?). I can make about 8 cancel, but the program disagrees.

The problems are cause by the programs inabilty to handle a contravariant derivative index. So I have to write ##\partial^\lambda \phi^{\mu\nu}## as ##g^{k\lambda}\partial_k \phi^{\mu\nu}##. When the gauge transformation done the humber of dummy indexes rises to 10. In the canonical form though it drops to 6 which is the same as the untransformed Lagrangian.

This is what I get for the four terms in the first post

##2{g}^{\%1 \%2}\,{g}^{\%3 \%5}\,{g}^{\%4 \%6}\,{X}_{\%1,\%2 \%3}\,{X}_{\%4,\%5 \%6}+2{g}^{\%1 \%2}\,{g}^{\%3 \%6}\,{g}^{\%4 \%5}\,{X}_{\%1,\%2 \%3}\,{X}_{\%4,\%5 \%6}##

(yes, it looks horrible). If we swap ##\%4## and ##\%5## in the first term it is the same as the second. This is the same procedure used to amalgamate the 4 into 2, isn't it ?

I don't know why the program can't see this. If it was legal the first time, why not now ?

All good fun.
 
In this video I can see a person walking around lines of curvature on a sphere with an arrow strapped to his waist. His task is to keep the arrow pointed in the same direction How does he do this ? Does he use a reference point like the stars? (that only move very slowly) If that is how he keeps the arrow pointing in the same direction, is that equivalent to saying that he orients the arrow wrt the 3d space that the sphere is embedded in? So ,although one refers to intrinsic curvature...
ASSUMPTIONS 1. Two identical clocks A and B in the same inertial frame are stationary relative to each other a fixed distance L apart. Time passes at the same rate for both. 2. Both clocks are able to send/receive light signals and to write/read the send/receive times into signals. 3. The speed of light is anisotropic. METHOD 1. At time t[A1] and time t[B1], clock A sends a light signal to clock B. The clock B time is unknown to A. 2. Clock B receives the signal from A at time t[B2] and...
So, to calculate a proper time of a worldline in SR using an inertial frame is quite easy. But I struggled a bit using a "rotating frame metric" and now I'm not sure whether I'll do it right. Couls someone point me in the right direction? "What have you tried?" Well, trying to help truly absolute layppl with some variation of a "Circular Twin Paradox" not using an inertial frame of reference for whatevere reason. I thought it would be a bit of a challenge so I made a derivation or...

Similar threads

Replies
4
Views
957
Replies
1
Views
800
Replies
1
Views
2K
Replies
1
Views
900
Replies
3
Views
1K
Back
Top