Get Standard Units Flux Multiply c^2: Perfect Fluid/Einstein's Dust

[dsaun777]

TL;DR Summary

Just clarification on components.

The T^0i the components of the energy momentum tensor that correspond to mass flux and/or momentum density. If I wanted to get standard units of flux I would just multiply by c^2 correct? It is the ith component of momentum across the kth surface? I am thinking of an example in particular that relates to either a perfect fluid or Einstein's "dust".

 

[mitochan]

Hi.
Starting from the formalism of T^ij for the case,
$$T^{ij}=\rho u^iu^j$$
, here I omit pressure for simple writing, all the components has dimension same as $\rho$, energy per volume. If you want to give normal dimension to some components as momentum per volume or mass, i.e. energy/c^2, flux, how about dividing them by c?

 

[dsaun777]

Hi.
Starting from the formalism of T^ij for the case,
$$T^{ij}=\rho u^iu^j$$
, here I omit pressure for simple writing, all the components has dimension same as $\rho$, energy per volume. If you want to give normal dimension to some components as momentum per volume or mass, i.e. energy/c^2, flux, how about dividing them by c?

I want to Express flux across a surface on terms of the components of the tensor, how would I do this?

 

[mitochan]

In order to get energy flux whose dimension is energy / area・time from corresponding T^ij components that I have set as energy/volume in my previous post, how about multiplying c ?

 

[mitochan]

Not all the components but T^01,T^02 and T^03 which corresponds to energy flux. Other components correspond to energy density, momentum density and 3D stress tensor.

 

[dsaun777]

Just the T^0i components alone without multiplying by c factor give you momentum flux right

 

[vanhees71]

I'm not sure what you want to achieve since your equation gives the energy-momentum tensor of freely moving particles, where $\rho$ is the density of invariant mass (defined in the usual way as mass per spatial volume in the (local) rest frame of the particles).

It depends a bit on your conventions, concerning the factors of $c^2$. It's not uniquely defined in the literature, what's meant by $u^{\mu}$. In some textbooks one defines the four-velocity as having the dimension of a velocity ("length per time"). Then it's normalized as $u_{\mu} u^{\mu}=c^2$. In other books one uses the (usually more convenient) definition $\tilde{u}^{\mu}=u^{\mu}/c$, because then you can use it to project to local restframes simply without writing all the factors of $c$ since, of course, in this convention you have $\tilde{u}_{\mu} \tilde{u}^{\mu}=1$.

Now let's write
$$T^{\mu \nu} = \rho_0(x) u^{\mu}(x) u^{\nu}(x).$$
In terms of the usual three-velocity $\vec{v}$ you have
$$u^{\mu}=\gamma \begin{pmatrix} c \\ \vec{v} \end{pmatrix}.$$
Now the time-time component is seen to be
$$T^{00}=\rho_0 \gamma^2 c^2$$
This is the energy density component in this reference frame, i.e.,
$$\epsilon=\rho_0 \gamma^2.$$
Then the space-time components are
$$T^{j0}=\rho_0 \gamma^2 c v^j=\epsilon/c v^j.$$
This is the energy current density, and thus
$$g^{j}=T^{j0}/c=\epsilon/c^2 v^j$$
the momentum current density.