Einstein Summation: Swapping Dummies i & j

[Raptor112]

Homework Statement

My question is regarding a single step in a solution to a given problem. The step begins at:

$\large \frac{\partial \alpha _j}{\partial x ^i} \frac{\partial x^i}{y^p} \frac{\partial x^j}{\partial y^q} - \frac{\partial \alpha _j}{\partial x ^i} \frac{\partial x^i}{\partial y^q} \frac{x^j}{ \partial y^p}$The solution says that the second term has the dummy i and j indices which can be switched in order to factorise which gives us:$\large \big (\frac{\partial \alpha _j}{\partial x ^i} - \frac{\partial \alpha _i}{\partial x ^j} \big ) \frac{\partial x^i}{\partial y^p} \frac{\partial x^j}{\partial y^q}$1. To clarify, the Einstein summation means that changing the index i in the second term has no implication for the index i in the first term?2. If the dummies i and j were switched from the first term then I would result in the negative answer. So how does one know beforehand which term for whcih the indices have to be switched?

 

[George Jones]

In the second term, does replacing index i by index m and replacing index j by index n change anything?

 

[Raptor112]

n the second term, does replacing index i by index m and replacing index j by index n change anything?

Well no because the summation will still be the same. But that makes it diffucult to see the factorization that is needed.

 

[George Jones]

So, now the second term is

$$\frac{\partial \alpha _n}{\partial x ^m} \frac{\partial x^m}{\partial y^q} \frac{\partial x^n}{ \partial y^p}.$$

Does replacing index m by index j and replacing index n by index i change anything?

 

[Raptor112]

So, now the second term is

$$\frac{\partial \alpha _n}{\partial x ^m} \frac{\partial x^m}{\partial y^q} \frac{\partial x^n}{ \partial y^p}.$$

Does replacing index m by index j and replacing index n by index i change anything?

But I still don't see how this answers by second question?

 

[PeroK]

But I still don't see how this answers by second question?

I might suggest a radical approach of putting the sigmas back in so you can see what's going on. The summation convention is only a shorthand, after all.

 

[George Jones]

But I still don't see how this answers by second question?

If you type in the expression that results when this is done, we can discuss the expression.

Also, this might help you to understand things. I sometimes find that my understanding changes when I actually write an expression down. Unfortunately, it can go both ways! Sometime, I don't understand an expression when I visualize it in my mind, but things become clear when I write the expression down. Other times, I think that I understand something, but I when I write it down, my "understanding" fades.

I might suggest a radical approach of putting the sigmas back in so you can see what's going on. The summation convention is only a shorthand, after all.

I only recommend doing this as a short-term aid for learning how to deal with expressions that omit the Sigmas.

 

[Fred Wright]

Your question 1 is true; the Einstein summation convention only applies to terms connected by multiplication . For question 2 there is no way to know before hand and your reasoning must lead to the conclusion that the term in braces equals zero, i.e. ∂α_j/∂xⁱ = ∂α_i/∂x^j