Use of tensor densities in coordinate transformation

[aeson25]

Why use a tensor density transformation when doing a coordinate transformations? What is the advantage? I've always learn that transforming a tensor involves pre and post multiplying by the transformation tensor and it's inverse respectively, but I've come across ones in my research that use the tensor density approach which weights the tensor transformation, and would like to know the justification for using the tensor density. I've looked all over the net and I can only find the pure definition of a tensor density, but not why it's used especially over non weighted transformations. Can anyone add insight?

 

[fresh_42]

Not sure what exactly your question is, but Wikipedia has a long article about it: https://en.wikipedia.org/wiki/Tensor_density

 

[formodular]

It may be the case that when analyzing some physical system and it's behavior under coordinate transformations that one is working not just with scalar/vector/tensor/spinor/... quantities (that transform in the usual way under coordinate transformations) but other non-tensor/spinor quantities, such as the four-volume element $d^4 x$ in GR, which also change when one performs a coordinate transformation, changing the way tensors do but including a Jacobian of the transformation - this is clearly similar but still different to how scalars/vectors/... transform under coordinate transformations and so it's given a new name - a tensor density. In GR you can add a factor (the $\sqrt{-g}$ as mentioned in the limk) into the action to correct for the behavior of the $d^4 x$ tensor density to end up with a scalar, but if for some reason we couldn't do that in a given action and still needed the action to remain invariant, we could define the fields involved (and their derivatives) to transform in such a way that it overall leaves the action invariant, forcing us to work with tensor densities as fields.

 

[Cryo]

Because that's the rule for transforming tensor densities. Perhaps you simply need to work with an object that transforms somewhat like a density and is entirely mathematical, i.e. it is not some charge density, that must be a density by definition.

Consider the determinant of the metric $g=\det\left(g_{\alpha\beta}\right)$. Is it a scalar? Well, it certainly is rank zero, but now look at how it transforms. Consider a different coordinate system, say $\bar{S}$ with metric tensor $\bar{g}_{\alpha\beta}$. So what is the determinant of the metric tensor in that coordinate system?

Well one way to go is to say:

$\bar{g}=\det\left(\bar{g}_{\alpha\beta}\right)$

but also

$\bar{g}=\det\left(\frac{\partial x^\mu}{\partial \bar{x}^\alpha}\frac{\partial x^\nu}{\partial \bar{x}^\beta}g_{\mu\nu}\right)$

After some effor you will find that:

$\bar{g}=\left(\frac{\partial(x)}{\partial (\bar{x})}\right)^2 g$

So $g$ does not transform as a scalar. There are other quantities that will also not transform like scalars, if you want to work with these quantities you need to respect this (or you will get contradictions)