# How to deal with the index in Einstein summation?

1. Oct 6, 2005

### yukcream

Given U^k_i, the components of U is a delta function i.e for i=k U^i_k =1,
to prove it is invariant under Lorentz transformation~~

I don't know how to express it in Einstein summation notation, I am very confused with the upper-lower index, is it right to write the transformation in this?

U'^k_i = T^i_m T^n_k U^k_i ? where T is the Lorentz transformation~~

yukyuk

2. Oct 6, 2005

### Jimmy Snyder

Not exactly. What you have is:

$$U'^k{}_i = T^i{}_m T^n{}_k U^k{}_i$$

better, but still not right would be:

$$U'^n{}_m = T^i{}_m T^n{}_k U^k{}_i$$

This balances the indices. Note that after summation, the i's and k's will disappear. In that case, the n's and m's will be the same on both sides of the equation. This balance of indices is what you are aiming for.

The problem with this notation is that you can't tell the difference between the two T's. I recommend the following notation which brings out that difference:

$$U^{n'}{}_{m'} = T^i{}_{m'} T^{n'}{}_k U^k{}_i$$

Notice that I took the prime off of the tensor U and put it on the indices. There is a lot of justification for this. This notation indicates what you are doing, you are expressing the same tensor in different coordinates.

I also think that this notation lays bare the solution to the problem.

By the way, your description of the delta 'function' is incomplete. You have:

for $i = k \ U^i{}_k = 1$

but what you should have is:

for $i = k \ U^i{}_k = 1$ and for $i \ne k \ U^i{}_k = 0$

Last edited: Oct 6, 2005
3. Oct 6, 2005

### robphy

$$U'^k_i = T^i_m T^n_k U^k_i$$ is not correct because the "free indices" aren't the same on both side... on the left it's $$\blacksquare^k_i$$ whereas it is $$\blacksquare^n_m$$ on the right.

The correct form is
$$U'^n{}_m= T^i{}_m T^n{}_k U^k{}_i$$ (where each side has $$\blacksquare^n{}_m$$ )
As jimmysnyder suggests, for these coordinate transformations, it may be better to use these primed- and unprimed-indices.

This may help (and I'll use the standard $$\delta^a{}_b$$ as the Kronecker delta). The Kronecker delta acts like an "index-substitution operator":
$$Q^a=\delta^a{}_bQ^b$$, which "substitutes a for b".
$$Q^a{}_b=\delta^a{}_m\delta^n{}_bQ^m{}_n$$, which "substitutes a for m, and b for n".

Try googling "index gymnastics" tensor
http://mathworld.wolfram.com/IndexGymnastics.html

Working with tensors made more sense to me after I was introduced to the "abstract index notation"
See http://en.wikipedia.org/wiki/Abstract_index_notation http://www.ima.umn.edu/nr/abstracts/arnold/einstein-intro.pdf [Broken]

Last edited by a moderator: May 2, 2017
4. Oct 6, 2005

### yukcream

I want to prove that the given martix
$$U^k{}_i$$
is invariant under Lorentz transformation~ am I correct to prove in following way?
Express U in delta as its really a delta function~

$$\delta'^k{}_i=T^i{}_mT^n{}_k\delta^m{}_n$$
$$\delta'^k{}_i=T^i{}_mT^m{}_k$$ so
$$=\delta^k{}_i$$

You both help me a lot and the article linked is very useful thx

yukyuk

Last edited: Oct 6, 2005
5. Oct 6, 2005

### Jimmy Snyder

This equation is not balanced in the sense that I described to you. Can you try this a second time? To determine if an equation is balanced, perform the following steps:

$$\delta'^k{}_i=T^i{}_mT^n{}_k\delta^m{}_n$$

2. From each side of the equation, eliminate summation indices:

$$\delta'^k{}_i=T^iT_k\delta$$

3. Look at the list of superscripted indices on the left and right of the equal sign:

k on the left, i on the right

4. If they are not the same, the equation is not balanced. In that case, stop right here, redo the equation and start again with step 1. If you pass this test, go on to step 5.

5. Look at the list of subscripted indices on the left and right of the equal sign:

i on the left, k on the right

6. If they are not the same, the equation is not balanced. In that case, stop right here, redo the equation and start again with step 1. If you pass this test, then the equation is balanced.

6. Oct 6, 2005

### yukcream

O~~~~I get it
The correct answer is this, right?
$$\delta^{i'}_{k'}=T^{i'}_{m}T^{n}_{k'}\delta^{m}_{n}$$
$$\delta^{i'}_{k'}=T^{i'}_{m}T^{m}_{k'}=\delta^{i'}_{k'}$$

This time the indices on both side is balance?!
thank you very much~~~

yukyuk

Last edited: Oct 6, 2005
7. Oct 7, 2005

### Jimmy Snyder

Right!

Extra text added to satisfy an unnecessary criterion.