[Special relativity - Mathematical background] Tensor and pseudo-tensor

[wizard85]

Hi to all,

Let $$A^{ik}$$ be an antisymmetric tensor of rank 2; Why is $$A^{*ik}=1/2e ^{iklm}A_{lm}$$ defined its dual? $$e^{iklm}$$ is the completely antisymmetric unit tensor.

Furthermore, $$e^{iklm}$$ is a pseutotensor, what does it mean? Conversely, why the product $$e^{iklm} e_{prst}$$ is a true tensor?

Thanks in advance... ;)

 

[haushofer]

I believe it has to do with the fact that in n dimensions the spaces of p forms and of (n-p) forms are diffeomorphic, which you can easily show by yourself by counting how many independent elements a p form and a (n-p) form have in n dimensions.

A hint: a p form is described in terms of a wedged product of p differentials,

$$\omega = \omega_{\mu_{1}\ldots\mu_{p}}dx^{\mu_{1}}\wedge\ldots\wedge dx^{\mu_{p}}$$

The order of the p differentials doesn't matter; you just pick up a minus sign by rearranging them, so changing order doesn't give you another independent basis. So you can choose p differentials out of n differentials in which the order doesn't matter. Now also apply this reasoning for (n-p) forms. (Maybe it helps to do it explicitly for 3 forms in 4 dimensions first, and then for 1 forms in 4 dimensions)

So the dual is defined as such to go between these two spaces of forms, and this is done via the Hodge dual, which depends on the metric in general. This hodge dual uses the epsilon "tensor" symbol. It's kindoff nasty, because in general it's not a tensor! The symbol is defined as being +1,0, or -1, depending on the indices. If you write down a coordinate transformation, you'll see that the Jacobian comes into play (just look at how determinants of matrices are defined in terms of epsilon symbols!). This means the symbol is a tensor DENSITY, which is indicated by "pseudo". However, you canv take the tensor product between two tensor densities with opposite weights to obtain an honest tensor. In special relativity texts this is not always emphasized.

With the Hodge dual you use the determinant of the metric to obtain an honest tensor out of the epsilon symbol.

You product of two epsilon symbols however IS a true tensor, because if you write down the tensor transformation law for this product, the two Jacobian terms cancel. This is again because these are tensor densities of opposite weights.

So, just write down the tensor transformation laws for your objects! :) You'll have to use

$$A = [A]_{ij}, \ \ \ \ \ \ \ det(A) = \epsilon_{i_1\ldots i_n}[A]_{1i_1}\ldots [A]_{ni_n}$$

 

[pervect]

I think Clifford algebra helps a lot to understand the Hodges dual. There's a pretty good and short introduction to Clifford algebra's at http://www.mrao.cam.ac.uk/~clifford/introduction/intro/intro.html