# Einstien notation question

1. Oct 12, 2007

### ehrenfest

1. The problem statement, all variables and given/known data
Is $$\partial ^ {\nu} F_{\mu \nu} + m^2 A_{\mu}$$ the same expression as $$\partial _{\nu} F^{\mu \nu} + m^2 A^{\mu}$$?

What form of the metric do I need to hit them with to show that?

Also is

$$\partial ^{\nu} \partial_{\mu} A_{\nu} = \partial_{\mu} \partial ^{\nu} A_{\nu}$$

?

2. Relevant equations

3. The attempt at a solution

EDIT: I changed the second expression so it makes sense.

Last edited: Oct 12, 2007
2. Oct 12, 2007

### nrqed

No. The second expression does not make sense to start with since in the F term the two mu are downstairs. Did you mean F with with indices upstairs? If so, the tewo expressions could still not be equal since one would be a covariant vector and the other one a contravariant vector.

Two tensors may only be equal if they have the same number of covariant and contravariant indices.

3. Oct 12, 2007

### ehrenfest

Yes. That's the danger of copying and pasting. I changed it so it makes sense.

4. Oct 13, 2007

### CompuChip

Ad 1) You wrote down the covariant and contravariant form of the same expression. You can convert between them by just writing metric tensors in between, e.g.
$$m^2 A^{\mu} = m^2 g^{\mu\nu} A_\nu$$
and similar for the first term. But the expressions are not the same, so do not expect the metric to drop out in the end.

$$g^{\nu\lambda} \partial_{\lambda} \partial_{\mu} = \partial_{\mu} g^{\nu\lambda} \partial_{\lambda} A_{\nu},$$
that is, do partial derivatives commute?
AFAIK they do, and the identity at least holds when the metric is constant (that is, $g_{\mu\nu}$ does not explicitly depend on the coordinate w.r.t which you are differentiating).

Last edited: Oct 13, 2007
5. Oct 13, 2007

### ehrenfest

But if they are both set equal to 0 then the metric will drop out and they are indeed precisely the same equations, right?

6. Oct 13, 2007

### nrqed

Ah, if the expression is equal to zero, then it's a special case. In general, one cannot say that a covariant expression is equal to its contravariant equivalent. But if the covariant expression is equal to zero then its contravariant counterpart will be equal to zero, indeed.

Last edited: Oct 13, 2007
7. Oct 13, 2007

OK. Thanks.