Difficulty in learning tensors

[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 {\partialX^{\nu}/\partialx^{\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!

 

[grzz]

The way I understand to do it is the following:

{\partialX^{\nu}/\partialx^{\mu}}R_{\nu} = 0

Multiplying both sides by {\partialx^{\mu}/\partialX^{\lambda}} we get

{\partialx^{\mu}/\partialX^{\lambda}} {\partialX^{\nu}/\partialx^{\mu}}R_{\nu} = 0

i.e. {\partialX^{\nu}/\partialX^{\lambda}}R_{\nu} = 0

\delta^{\nu}_{\lambda}R_{\nu} = 0

R_{\lambda} = 0.

Is the above method too long?

Thanks for any help.

 

[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.

 

[grzz]

Thanks K^2.

Is the method I used ok?

 

[Fredrik]

Is the method I used ok?

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

 

[grzz]

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

 

[Fredrik]

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.

 

[grzz]

Thanks a lot, Fredrik