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Gruber P. Convex and Discrete Geometry

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The Newtonian Gravitational Field <strong>and</strong> its Potential<br />

12 Special <strong>Convex</strong> Bodies 223<br />

Let C be a compact set in E3 consisting of homogeneous matter. Up to a multiplicative<br />

constant, its (Newtonian) gravitational field g is given by<br />

�<br />

g(y) =<br />

C<br />

x − y<br />

�x − y� 3 dx for y ∈ E3 \C,<br />

where the integral is to be understood componentwise. The corresponding<br />

( Newtonian) gravitational potential P then is given by<br />

�<br />

dx<br />

P(y) =<br />

�x − y� for y ∈ E3 \C,<br />

up to an additive constant. Note that g = grad P.<br />

The Problem <strong>and</strong> One Answer<br />

C<br />

A natural question to ask is whether C is determined uniquely by the gravitational<br />

field g or the gravitational potential P. Pertinent results are due to Novikov [774]<br />

<strong>and</strong> Shahgholian [927]. A special case of the latter’s result is the following (earlier)<br />

result of Aharonov, Schiffer <strong>and</strong> Zalcman [3].<br />

Theorem 12.3. Let C = cl int C ⊆ E 3 be a compact body consisting of homogeneous<br />

matter such that E 3 \C is connected. If the gravitational field of C coincides<br />

in E 3 \C with the gravitational field of a suitable point mass, then C is a ball.<br />

The following proof relies heavily on results from potential <strong>and</strong> measure theory for<br />

which we refer to Wermer [1018] <strong>and</strong> Bauer [82].<br />

Proof. We clearly may suppose that the point mass is located at the origin o. Then,<br />

by assumption, �<br />

C<br />

x − y α y<br />

dx =−<br />

�x − y�3 �y�3 for y ∈ E3 \C,<br />

where α>0 is a suitable constant. Since the fields coincide on the open connected<br />

set E 3 \C, the corresponding potentials coincide on E 3 \C up to an additive<br />

constant, i.e. �<br />

C<br />

dx α<br />

=<br />

�y − x� �y� + β for y ∈ E3 \C,<br />

where β is a suitable constant. Letting �y� →+∞, it follows that β = 0. Thus,<br />

(1)<br />

�<br />

dx α<br />

=<br />

�y − x� �y� for y ∈ E3 \C.<br />

C

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