# Stress-energy tensor and pressure

1. Jun 12, 2007

### jostpuur

I learned that stress-energy tensor is defined in first place to be a 16-component object $$T^{\mu\nu}$$, where the first row $$T^{0\nu}$$ tells the energy density current, and the three other rows $$T^{i\nu}$$ tell the momentum density currents.

The Carrol introduces a stress-energy tensor where off diagonal terms are zero, and $$T^{00}$$ and $$T^{ii}$$ (the same value for all i=1,2,3) have some fixed values. He then merely says, that "we can choose to call $$T^{ii}$$ the pressure". Okey, does this have anything to do with earlier pressure concept that we know from elementary physics? If it does, how do you justify that, because it doesn't seem very obvious to me. Those diagonal terms in the first place were supposed to be components momentum currents, and I don't see how they could be interpreted as pressure.

(This question continues were discussion in "photon gas and relativity" lead, but I didn't want to spoil the original topic with my own problems more.)

2. Jun 12, 2007

### pervect

Staff Emeritus
The short answer is that in the rest frame of the body, $T^{ij}$ is equivalent to the classical stress tensor for i,j = 1..3 Or equivalently, the relativistic stress-energy tensor is the relativistic generalization of the classical stress tensor.

Solids can support "shear stresses", simpler ideal fluid models will have a diagonal stress tensor, and hence can be summarized by a single scalar, P, which is the pressure of the fluid.

If you consider a rod under pressure, it does transport momentum across the rod. You can consider T^11 to be the "flow" of x momentum in the x direction, for example (though I find this not very intuitive).

See for instance http://people.hofstra.edu/Stefan_Waner/diff_geom/Sec12.html

The current wikipedia article http://en.wikipedia.org/w/index.php?title=Stress-energy_tensor&oldid=131240365
also covers this and seems fairly sane.

If you have MTW, see pg 131-132, especially box 5.1

3. Jun 12, 2007

### jostpuur

I just realized, that $$p_1$$ momentum travelling in $$x_1$$-direction, and $$-p_1$$ momentum travelling in $$-x_1$$-direction, give a contribution to the momentum current with the same sign. Okey, I'll have to think more about those currents...

4. Jun 12, 2007

### smallphi

As Stefan Waner says (in the link provided by pervect)

$$T^{11}=\frac{\Delta p^1}{\Delta v}$$

where $\Delta p^1$ is the x component of the momentum contained in a 3-volume element with normal vector pointing in x direction i. e. $\Delta v = \Delta t \Delta y \Delta z$. If an observer sits there for time $\Delta t$, he will see the fluid crossing the perpendicular area $\Delta y \Delta z$ carrying x-momentum $\Delta p^1$. Recall that in 4D spacetime, 3-volume elements are parts of hypersurfaces, hence have normal 4-vectors to them. So we get

$$T^{11}=\frac{\Delta p^x}{\Delta t \Delta y \Delta z }=\frac{F^x}{\Delta y \Delta z}= pressure$$

Last edited: Jun 12, 2007