# Difficulty in learning tensors

1. Aug 23, 2011

### grzz

Let r$_{\mu}$ be a tensor in coordinates x$^{c}$ and R$_{b}$ be a tensor in coordinates X$^{c}$.
Then let r$_{\mu}$ = 0.

Then {$\partial$X$^{\nu}$/$\partial$x$^{\mu}$}R$_{\nu}$ = 0.
I read in a book that one can divide both sides of the last equation by the partial derivative to get R$_{\nu}$ = 0.
I do not understand how this can be done since the partial derivative is summed over together with the R$_{\nu}$.
Can somebody help me!

2. Aug 23, 2011

### grzz

The way I understand to do it is the following:

{$\partial$X$^{\nu}$/$\partial$x$^{\mu}$}R$_{\nu}$ = 0

Multiplying both sides by {$\partial$x$^{\mu}$/$\partial$X$^{\lambda}$} we get

{$\partial$x$^{\mu}$/$\partial$X$^{\lambda}$} {$\partial$X$^{\nu}$/$\partial$x$^{\mu}$}R$_{\nu}$ = 0

i.e. {$\partial$X$^{\nu}$/$\partial$X$^{\lambda}$}R$_{\nu}$ = 0

$\delta$$^{\nu}_{\lambda}$R$_{\nu}$ = 0

R$_{\lambda}$ = 0.

Is the above method too long?

Thanks for any help.

3. Aug 23, 2011

### K^2

It's a theorem in linear algebra. If A is non-singular, and Ab=0, then b=0. If dX/dx is singular, you have other problems.

4. Aug 23, 2011

### grzz

Thanks K^2.

Is the method I used ok?

5. Aug 23, 2011

### Fredrik

Staff Emeritus
Yes it is. Note that what you're doing on the line that starts with "i.e." is to use the chain rule.

6. Aug 23, 2011

### grzz

Is the method above, using Kronecker's delta, considered a long method?

7. Aug 23, 2011

### Fredrik

Staff Emeritus
Nothing that only covers a few short lines is ever considered a long method.

The only way to do it shorter is to note that you started with a matrix equation in component form, and if you multiply it with the inverse...but that might require some explanation, and it would look a lot like what you just said.

8. Aug 23, 2011

### grzz

Thanks a lot, Fredrik