Tidal tensor

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

${\displaystyle a=-Gm/r^{2}}$

(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, ${\displaystyle \partial _{r},\partial _{\theta },}$ and ${\displaystyle \partial _{\phi }}$, which are given the subscripts 1, 2, and 3 respectively.

${\displaystyle {\vec {\epsilon }}_{1}=\partial _{r},\;{\vec {\epsilon }}_{2}={\frac {1}{r}}\,\partial _{\theta },\;{\vec {\epsilon }}_{3}={\frac {1}{r\sin \theta }}\,\partial _{\phi }}$

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:

${\displaystyle m/(r+h)^{2}-m/r^{2}=-2mh/r^{3}+3mh^{2}/r^{4}+O(h^{3})}$

Because in discussing tensors we are dealing with multilinear algebra, we retain only first order terms, so ${\displaystyle \Phi _{11}=-2m/r^{3}}$. Since there is no acceleration in the ${\displaystyle \theta }$ or ${\displaystyle \phi }$ direction due to a displacement in the radial direction, the other radial terms are zero: ${\displaystyle \Phi _{12}=\Phi _{13}=0}$.

Similarly, we can compare the gravitational force on two nearby observers lying at the same radius ${\displaystyle r=r_{0}}$ but displaced by an (infinitesimal) distance h in the ${\displaystyle \theta }$ or ${\displaystyle \phi }$ 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

${\displaystyle {\frac {m}{r_{0}^{2}}}\,\sin(\theta )\approx {\frac {m}{r_{0}^{2}}}\,{\frac {h}{r_{0}}}={\frac {m}{r_{0}^{3}}}\,h}$

By using the small angle approximation, we have ignored all terms of order ${\displaystyle O(h^{2})}$, so the tangential components are ${\displaystyle \Phi _{22}=\Phi _{33}=m/r^{3}}$. Again, since there is no acceleration in the radial direction due to displacements in either of the azimuthal directions, the other azimuthal terms are zero: ${\displaystyle \Phi _{21}=\Phi _{31}=0}$.

Combining this information, we find that the tidal tensor is diagonal with frame components ${\displaystyle \Phi _{{\hat {a}}{\hat {b}}}={\frac {m}{r^{3}}}\operatorname {diag} (-2,1,1)}$ This is the Coulomb form characteristic of spherically symmetric central force fields in Newtonian physics.

In the more general case where the mass is not a single spherically symmetric central object, the tidal tensor can be derived from the gravitational potential ${\displaystyle U}$, which obeys the Poisson equation:

${\displaystyle \Delta U=4\pi \,\mu }$

where ${\displaystyle \mu }$ is the mass density of any matter present, and where ${\displaystyle \Delta }$ is the Laplace operator. Note that this equation implies that in a vacuum solution, the potential is simply a harmonic function.

The tidal tensor is given by the traceless part [1]

${\displaystyle \Phi _{ab}=J_{ab}-{\frac {1}{3}}\,{J^{m}}_{m}\,\eta _{ab}}$

of the Hessian

${\displaystyle J_{ab}={\frac {\partial ^{2}U}{\partial x^{a}\,\partial x^{b}}}}$

where we are using the standard Cartesian chart for E³, with the Euclidean metric tensor

${\displaystyle ds^{2}=dx^{2}+dy^{2}+dz^{2},\;-\infty <x,y,z<\infty }$

Using standard results in vector calculus, this is readily converted to expressions valid in other coordinate charts, such as the polar spherical chart

${\displaystyle ds^{2}=d\rho ^{2}+\rho ^{2}\,\left(d\theta ^{2}+\sin(\theta )^{2}\,d\phi ^{2}\right)}$

${\displaystyle 0<\rho <\infty ,\;0<\theta <\pi ,\;-\pi <\phi <\pi }$

In general relativity, the tidal tensor is generalized by the Riemann curvature tensor. In the weak field limit, the tidal tensor is given by the components ${\displaystyle \Phi _{ij}={R}_{i0j0}}$ of the curvature tensor.

• Tidal force
• Stress tensor

• Sperhake, Ulrich. "Part II General Relativity Lecture Notes" (PDF): 19. Retrieved 13 January 2018. {{cite journal}}: Cite journal requires |journal= (help)
• Renaud, F.; Boily, C. M.; Naab, T.; Theis, Ch. (20 November 2009). "Fully Compressive Tides in Galaxy Mergers". The Astrophysical Journal. 706 (1): 68. arXiv:0910.0196. Bibcode:2009ApJ...706...67R. doi:10.1088/0004-637X/706/1/67. S2CID 15831572.
• Duc, Pierre-Alain; Renaud, Florent. "Gravitational potential and tidal tensor". ned.ipac.caltech.edu. Caltech. Retrieved 13 January 2018.