sci.physics.relativity

Edward Green wrote:
> Suppose we had two masses in relative motion, so that at least one
> mass must be in motion in any coordinate system (at least in
> approximately locally Lorentz coordinates which encompass both
> masses).
> At least one of the masses therefore has kinetic energy. Must we
> include this kinetic energy in the stress energy tensor, or can we
> simply use the rest mass of each as the energy, so long as we treat
> the masses discretely?

The energy-momentum tensor automatically includes the motion of an
object. One must include both its kinetic energy and its momentum, in
addition to its mass. In relativity, a pointlike mass m contributes to
the energy-momentum tensor:
T^ij(x) = m delta^4(x-X(t))) U^i U^j
Where U is the 4-momentum of the object and X(t) is its trajectory; x
represents the 4 coordinates on spacetime.

When projected onto the locally inertial frame in which it is at rest,
only U^00 is nonzero, and corresponds to its mass density.


> Follow on question: assuming the answer is that we may treat each
> mass as contributing its rest mass alone, [...]

You cannot do so, except in its rest frame.


> Eventually our independent masses will appear to be a swarm of dust
> particles, and we will elect to treat then on average, rather than
> discretely. [...]

Yes. A collection of non-interacting pointlike masses is called "dust",
and the energy-momentum tensor for dust with a mass density \rho is:
T^ij(x) = \rho(x) U^i U^j
Where U is the 4-momentum of the infinitesimal element at x.


> Since both pressure and kinetic energy stem from the same root: that
> the particles are in relative motion, are we over-counting if we
> include both?

Not if you do it correctly.


Tom Roberts