# Relativistic hydrostatic

1. Jul 6, 2010

### antoinebret

Hi everybody,

When I have a perfect fluid within a constant gravitational field g, the pressure at depth z simply reads P=\rho g z, where \rho is the fluid density (incompresible).

If now my g is still constant, but my fluid is relativistic. Not because it's moving (everything is still) but because it is hot with k_B T \sim mc^2. What happens to P = \rho g z ?

Thanks!

2. Jul 7, 2010

### Mentz114

Perfect fluids do not contain or conduct heat, so you have some other kind of fluid, perhaps ?

3. Jul 7, 2010

### Ich

They do.
P= k_B T\rho_0/m + \rho g z with \rho=\gamma \rho_0.

You can use [ tex ] and [ /tex ] (click on the formula):
$$P=k_B T\rho_0/m + \rho g z$$

4. Jul 7, 2010

### Mentz114

If you're right, I have been misinformed.

5. Jul 7, 2010

### Ich

Ok, my comment was too general.
What I meant is that it makes sense to ask about a perfect fluid with a certain temperature in the way antoinebret did. From his data you can calculate energy density and pressure and answer the question with the perfect fluid behaviour, the extreme case being a photon gas.
I agree that there's no heat conduction, though.