Graduate What does Dirac mean by "the suffixes would not balance"?

[Kostik]

TL;DR

Dirac is saying that $A_{\mu;\nu}-A_{\nu;\mu} = A_{\mu,\nu} - A_{\nu,\mu}$ only works for covariant vectors. Why?

In Dirac ("GTR") p. 39 he says, "For a covariant vector $A_\mu$, we have
$$A_{\mu;\nu}-A_{\nu;\mu} = A_{\mu,\nu} - \Gamma^\rho_{\mu\nu} A_\rho - \left( A_{\nu,\mu} - \Gamma^\rho_{\nu\mu}A_\rho \right) = A_{\mu,\nu} - A_{\nu,\mu}.$$ This result may be stated: covariant curl equals ordinary curl. It holds only for a covariant vector. For a contravariant vector we could not form the curl because the suffixes would not balance."

Yet a similar calculation shows that $A^\mu_{\,\,;\nu}-A^\nu_{\,\,;\mu} = A^\nu_{\,\,,\mu}-A^\mu_{\,\,,\nu}$, which is the parallel result. So, what is he trying to say?

EDIT: Sloppy error - my mistake!

 

[Orodruin]

Yet a similar calculation shows that $A^\mu_\nu-A^\nu_\mu = A^\nu_\mu-A^\mu_\nu$. So, what is he trying to say?

No it doesn’t. At least not if done correctly. You cannot have $\mu$ covariant in one term and contravariant in the other. It simply does not make sense. Same for $\nu$.

 

[Kostik]

I am simply computing: $$A^{\mu}_{\,\,;\nu}-A^{\nu}_{\,\,;\mu} = A^{\mu}_{\,\,,\nu} + \Gamma^\mu_{\alpha\nu} A^\alpha - \left( A^{\nu}_{\,\,,\mu} + \Gamma^\mu_{\alpha\nu} A^\alpha \right) = A^{\mu}_{\,\,,\nu} - A^{\nu}_{\,\,,\mu}.$$ Can you please explain why this calculation cannot be made?

EDIT: Sloppy error -- the second derivative is wrong.

There are countless examples of an index covariant in one term and contravariant in another. $A_\mu B^\nu - A^\mu B_\nu=0$.

 

[Orodruin]

I am simply computing: $$A^{\mu}_{\,\,;\nu}-A^{\nu}_{\,\,;\mu} = A^{\mu}_{\,\,,\nu} + \Gamma^\mu_{\alpha\nu} A^\alpha - \left( A^{\nu}_{\,\,,\mu} + \Gamma^\mu_{\alpha\nu} A^\alpha \right) = A^{\mu}_{\,\,,\nu} - A^{\nu}_{\,\,,\mu}.$$ Can you please explain why this calculation cannot be made?

Because (i) you did not get the second covariant derivative correct and (ii) the expression does not make sense to begin with.

There are countless examples of an index covariant in one term and contravariant in another. $A_\mu B^\nu - A^\mu B_\nu=0$.

No, there aren’t. Not if the source is correct. Name one.

 

[Kostik]

Oops, yes, I got the second derivative wrong. I see exactly what Dirac means by "the suffixes would not balance". Sloppy ... apologies.

 

[Orodruin]

Oops, yes, I got the second derivative wrong. I see exactly what Dirac means by "the suffixes would not balance". Sloppy ... apologies.

No, he means that the expression doesn’t make sense to start with because the indices are not balanced between the terms. $\mu$ is contravariant in one term and covariant in the other. That makes the expression coordinate system dependent and therefore inconsistent.

 

[Vanadium 50]

There are countless examples

By which you mean zero?

 

[Orodruin]

By which you mean zero?

To be honest, the OP’s belief that this is possible is far more worrying than faulty algebra …

 

[DrGreg]

I am simply computing: $$A^{\mu}_{\,\,;\nu}-A^{\nu}_{\,\,;\mu} = ...$$ Can you please explain why this calculation cannot be made?

The subtraction makes no sense. It's as if you are trying to subtract 3 cats from 5 dogs and don't understand why an answer of 2 is wrong.

It makes no sense to subtract $Y_\mu$ from $X^\mu$ no matter what $\mathbf{X}$ and $\mathbf{Y}$ are.

Oops, yes, I got the second derivative wrong.

You did, but that's irrelevant to why the expression makes no sense.

 

[PeterDonis]

I am simply computing: $$A^{\mu}_{\,\,;\nu}-A^{\nu}_{\,\,;\mu} = A^{\mu}_{\,\,,\nu} + \Gamma^\mu_{\alpha\nu} A^\alpha - \left( A^{\nu}_{\,\,,\mu} + \Gamma^\mu_{\alpha\nu} A^\alpha \right) = A^{\mu}_{\,\,,\nu} - A^{\nu}_{\,\,,\mu}.$$ Can you please explain why this calculation cannot be made?

EDIT: Sloppy error -- the second derivative is wrong.

Even with that fixed, as has already been said, your equation makes no sense because the indexes don't match.

Here's a suggestion: take Dirac's expression with all indexes covariant (lower), and try to obtain your expression by raising indexes properly, using the metric. You won't be able to do it. (Which, btw, is another way of stating Dirac's point in the passage you quoted.)

There are countless examples of an index covariant in one term and contravariant in another. $A_\mu B^\nu - A^\mu B_\nu=0$.

What makes you think that expression is any more valid than your other one?

 

[Kostik]

To be honest, the OP’s belief that this is possible is far more worrying than faulty algebra …

I understand the issue, no need for worry. Just mentioning in passing the equation in Weinberg "Gravitation and Cosmology", on p. 100, $$\varepsilon_{\rho\sigma\eta\zeta}=-g\varepsilon^{\rho\sigma\eta\zeta}.$$

 

[Orodruin]

I understand the issue, no need for worry.

Are you sure? Your posts in this thread really suggest that you do not.

Just mentioning in passing the equation in Weinberg "Gravitation and Cosmology", on p. 100, $$\varepsilon_{\rho\sigma\eta\zeta}=-g\varepsilon^{\rho\sigma\eta\zeta}.$$

This is a very particular case. First of all, $\epsilon$ is fully anti-symmetric of order 4. The space of such tensors is one-dimensional so it is necessary that the two relate by some multiplicative factor. Second, neither of those is actually a tensor.

 

[Orodruin]

It should also be pointed out that both $\epsilon$ (as defined by Weinberg) are tensor densities of weight -1. The metric determinant is not a scalar, but a scalar density, meaning that the sides of the equation have different weights. As such, multiplying by powers of $g$ will never turn it into a tensor equation.

 

[Kostik]

Yep I noticed that, sorry being very sloppy in this thread. Appreciate all the replies.