# Difficulty re tensors

1. Apr 22, 2012

### grzz

I am learning about tensors. Can somebody give me some help. Thanks.

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• ###### tensor difficutly.pdf
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2. Apr 22, 2012

### DrGreg

What you did was correct. You used$$\frac{\partial X^\mu}{\partial x^\alpha} \, \frac{\partial X^\nu}{\partial x^\beta} R_{\mu \nu} = r_{\alpha \beta}$$and put $r_{\alpha \beta} = 0$ to prove $R_{\mu \nu} = 0$. You could have done it the other way round, i.e. put $R_{\mu \nu} = 0$ to prove $r_{\alpha \beta} = 0$. As we are talking about two arbitrary coordinate systems, that is equally valid.

3. Apr 23, 2012

### clamtrox

You can kind of see this directly. If all of the tensor components are zero in one coordinate system, they must be zero in all others as well. So to show this, just show that the tensor itself is zero,

$$R = R_{\mu \nu} (dx^{\mu} \otimes dx^{\nu}) = 0$$

and since the tensor itself is a geometric, coordinate independent object, you're done.

4. Apr 23, 2012

### Staff: Mentor

Here's another way of doing the problem. Write the starting equation out in component form. You will realize before you are finished writing that the starting equation represents a set of 16 linear homogeneous algebraic equations in 16 unknowns (the 16 components of R). Since the equations are homogeneous, the only solution to this set of equations is for each and every unknown to be zero.

chet

5. Apr 23, 2012

### grzz

So to remove $\frac{∂X^{μ}}{∂x^{α}}$$\frac{∂X^{σ}}{∂x^{β}}$ from the lhs of $\frac{∂X^{μ}}{∂x^{α}}$$\frac{∂X^{σ}}{∂x^{β}}$ R$_{μσ}$ = 0, do I have to show all the steps as I did before (see the pdf in my origianal post) or can I just say that I divided both sides by $\frac{∂X^{μ}}{∂x^{α}}$$\frac{∂X^{σ}}{∂x^{β}}$?

Thanks everybody for all your help.

6. Apr 23, 2012

### Staff: Mentor

If you have 16 linear algebraic equations in 16 unknowns, and the right hand sides of all the equations are equal to zero, what is the solution for the unknowns?