Gravitational Field of a Large Sheet

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Problem: Consider an infinitely large sheet of matter confined to the z = 0 plane. Assume the surface mass density, s₀, as measured in the zero momentum frame S is nearly Newtonian, i.e. the components of the energy-momentum tensor, T_mn, satisfy T₀₀ >> |T_0k| and T₀₀ >> |T_jk|. Assume also that the matter distribution in S does not depend on time. Consider only the region of space above the sheet, i.e. only z > 0.

(a) Find the gravitational field, g_mn, in S.

(b) Find duz/dt at z = 0 for u = 0.

Solution (a): Let the surface proper mass density (mass per unit area) of the sheet, as observed in frame S, be s₀ Since in S the matter distribution is constant then the potentials, f_mn, are given in terms of the stress-energy-momentum (SEM) tensor, T_mn, as

For the matter distribution specified above the volume integral becomes a surface integral, i.e. for f₀₀, we find

Where F is the Newtonian potential given by

Since T₀₀ >> |T_0k | and T₀₀  >> |T_jk|it follows that, f_0k = f_jk = 0. Therefore

Recall

Therefore setting m = n = 0 gives h₀₀

The Newtonian potential F in Eq. (3) may be calculated by constructing a pillbox of height z around a small portion of the sheet and employing Gauss’s law to obtain the gravitational acceleration g. The Newtonian potential may then be obtained by integrating the expression g = -ÑF to obtain F. Gauss’s law yields

Integrating g = -ÑF we obtain

The constant of integration is taken to be zero at z = 0. This means that the coordinate clock is in the z = 0 plane. h_mn then becomes

Substitution into the relation for the metric, i.e.

gives the solution to our problem, i.e. the metric tensor

Solution (b): The acceleration for a particle in the weak field limit is given by

Since we’re looking for the z component set m = 3 and note that U₃ = -u_z. Therefore since z = 0, and u = 0 and therefore U₀ = c₂ and it follows that

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