bernhard.rothenstein said:
I read in “N. David Mermin, It’s about time, (Princeton University Press, 2005) p.144:
”Furthermore, even when you put together identical bricks, it turns out (in the relativistic case) that the mass of the object you construct depends on how you put the bricks together.”
In order to test the way in which the students understand the statement I proposed them the following problem:
Consider a high sensitivity balance with equal arms. Put on its left pan a cube of volume a3. Put on its right pan a sphere of radius r made from the same material (density ). The balance is located in a uniform vertical gravitational field (g). The balance being in a state of mechanical equilibrium and in thermal equilibrium with the surrounding, determine the relationship between a and r.
what is the solution you propose?
The solution I propose is probably too advanced for the class :-(.
First off, I would propose replacing the sphere and the cube with a cube (1x1x1) and a non-cubical box - (4x.5x.5) for example, with the first number being the height.
Then we ask: given the metric of a "uniform gravitational field", what are the equations of continuity for the stress-energy tensor?
(modern notation: \nabla_a T^{ab} - semicolon notation T^{ab}{}_{;a})
i.e.
given the metric dx^2 = -(1+gz)dt^2 + dx^2 + dy^2 + dz^2
let us assume the stress-energy tensor is
T^{tt} = rho(z)
T^{zz} = P(z)
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Physically, we want to make (1+gz)^2*rho(z) constant to represent a density that appears constant in the orthonormal basis of any observer regardless of height, z, i.e. we want
rho(z) = rho_0 / (1 + gz)^2
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[note that this doesn't include the stress terms due to the self-gravity of the box, only the stress terms due to the "uniform gravitational field"]
We compute \nabla_a T^{ab}
There is only one non-trivial equation in the result. Non-relativistically, we expect that z component of pressure would be proportional to depth.
i.e. non-relativistiaclly
g*rho(z) + dP/dz = 0
So non-realtivistically, the pressure * area of the tall box (equal to the force exerted on the scale) would be the same as the pressure*area of the cube, because the solution for pressure vs depth would be linear.
But relativistically, we get instead from the above
g (1+gz) * rho(z) + g*P(z)/(1+gz) + dP(z)/dz = 0
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in terms of rho_0 this becomes
g * (rho_0 + P(z) ) /(1+gz) + dP(z)/dz = 0
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Thus, we get an exponential solution for pressure vs depth and not a linear solution. Therfore pressure * area is not equal for the tall box and the cube - their weights are different.
The way I would describe this is that "pressure causes gravity"
Note that I've used geoemtric units. In standard units, pressure gets divided by c^2, so the contribution of pressure to gravity is very small.
This is only a toy version of the problem, but I don't think there's any simple way to find the stress-energy tensor (and the associated metric!) if the self-gravity of the box is included.
It probably would be more to the spirit of the original problem if we assumed that most of the pressure was due to the self-gravity of the system, and that the acceleration was very small, rather than my assumption that most of the pressure is due to the "uniform gravitational field". But I'm not sure how to set up that case easily.
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This also illustrates how the pressure changes associated with the problem setup (the "uniform gravitational field" affect the system. The case that would probably best illustrate Mermin's original point is one in which the acceleration is very very low, minimizing the effects of the pressure changes and disturbing the system as little ias possible. In this case, the balance should essentially measure the Komar mass - which won't be equal for the different shapes.
The pressure terms though, are one good way of explaining why the intergal of the energy density / c^2 does not give the mass. Note that the discrepancy is very small.
