In Newton's theory of gravitation and in various relativistic classical theories of gravitation, such as general relativity, the tidal tensor represents:
tidal accelerations of a cloud of (electrically neutral, nonspinning) test particles tidal stresses in a small object immersed in an ambient gravitational field The tidal tensor describes the relative acceleration due to gravity between two test masses separated by an infinitesimal distance. The component Φ a b {\displaystyle \Phi _{ab}} represents the relative acceleration in the a ^ {\displaystyle {\hat {a}}} direction produced by a displacement in the b ^ {\displaystyle {\hat {b}}} direction.
Tidal tensor for a spherical body The most common example of tides is the tidal force around a spherical body (e.g., a planet or a moon). Here we compute the tidal tensor for the gravitational field outside an isolated spherically symmetric massive object. According to Newton's gravitational law, the acceleration a at a distance r from a central mass m is
a = − G m / r 2 {\displaystyle a=-Gm/r^{2}}
(to simplify the math, in the following derivations we use the convention of setting the gravitational constant G to one. To calculate the differential accelerations, the results are to be multiplied by G.) Let us adopt the frame in polar coordinates for our three-dimensional Euclidean space, and consider infinitesimal displacements in the radial and azimuthal directions, ∂ r , ∂ θ , {\displaystyle \partial _{r},\partial _{\theta },} and ∂ ϕ {\displaystyle \partial _{\phi }} , which are given the subscripts 1, 2, and 3 respectively.
ϵ → 1 = ∂ r , ϵ → 2 = 1 r ∂ θ , ϵ → 3 = 1 r sin θ ∂ ϕ {\displaystyle {\vec {\epsilon }}_{1}=\partial _{r},\;{\vec {\epsilon }}_{2}={\frac {1}{r}}\,\partial _{\theta },\;{\vec {\epsilon }}_{3}={\frac {1}{r\sin \theta }}\,\partial _{\phi }}
We will directly compute each component of the tidal tensor, expressed in this frame. First, compare the gravitational forces on two nearby objects lying on the same radial line at distances from the central body differing by a distance h:
m / ( r + h ) 2 − m / r 2 = − 2 m h / r 3 + 3 m h 2 / r 4 + O ( h 3 ) {\displaystyle m/(r+h)^{2}-m/r^{2}=-2mh/r^{3}+3mh^{2}/r^{4}+O(h^{3})}
Because in discussing tensors we are dealing with multilinear algebra, we retain only first order terms, so Φ 11 = − 2 m / r 3 {\displaystyle \Phi _{11}=-2m/r^{3}} . Since there is no acceleration in the θ {\displaystyle \theta } or ϕ {\displaystyle \phi } direction due to a displacement in the radial direction, the other radial terms are zero: Φ 12 = Φ 13 = 0 {\displaystyle \Phi _{12}=\Phi _{13}=0} . Similarly, we can compare the gravitational force on two nearby observers lying at the same radius r = r 0 {\displaystyle r=r_{0}} but displaced by an (infinitesimal) distance h in the θ {\displaystyle \theta } or ϕ {\displaystyle \phi } direction. Using some elementary trigonometry and the small angle approximation, we find that the force vectors differ by a vector tangent to the sphere which has magnitude
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