# Tidal tensor

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In Newton's theory of gravitation and in various relativistic classical theories of gravitation, such as general relativity, the tidal tensor represents

General relativity is the geometric theory of gravitation published by Albert Einstein in 1915 and the current description of gravitation in modern physics. General relativity generalizes special relativity and Newton's law of universal gravitation, providing a unified description of gravity as a geometric property of space and time, or spacetime. In particular, the curvature of spacetime is directly related to the energy and momentum of whatever matter and radiation are present. The relation is specified by the Einstein field equations, a system of partial differential equations.

## Contents

1. tidal accelerations of a cloud of (electrically neutral, nonspinning) test particles,
2. tidal stresses in a small object immersed in an ambient gravitational field.

The tidal tensor represents the relative acceleration due to gravity of two test masses separated by an infinitesimal distance. The component ${\displaystyle \Phi _{{a}{b}}}$ represents the relative acceleration in the ${\displaystyle {\hat {a}}}$ direction produced a displacement in the ${\displaystyle {\hat {b}}}$ 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

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

(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 be multiplied by G.)

The gravitational constant, denoted by the letter G, is an empirical physical constant involved in the calculation of gravitational effects in Sir Isaac Newton's law of universal gravitation and in Albert Einstein's general theory of relativity.

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: ${\displaystyle \Phi _{12}=\Phi _{13}=0}$.

In mathematics, multilinear algebra extends the methods of linear algebra. Just as linear algebra is built on the concept of a vector and develops the theory of vector spaces, multilinear algebra builds on the concepts of p-vectors and multivectors with Grassmann algebra.

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.

## Hessian formulation

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:

In classical mechanics, the gravitational potential at a location is equal to the work per unit mass that would be needed to move the object from a fixed reference location to the location of the object. It is analogous to the electric potential with mass playing the role of charge. The reference location, where the potential is zero, is by convention infinitely far away from any mass, resulting in a negative potential at any finite distance.

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

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.

In mathematics, the Laplace operator or Laplacian is a differential operator given by the divergence of the gradient of a function on Euclidean space. It is usually denoted by the symbols ∇·∇, 2, or Δ. The Laplacian Δf(p) of a function f at a point p, is the rate at which the average value of f over spheres centered at p deviates from f(p) as the radius of the sphere grows. In a Cartesian coordinate system, the Laplacian is given by the sum of second partial derivatives of the function with respect to each independent variable. In other coordinate systems such as cylindrical and spherical coordinates, the Laplacian also has a useful form.

A vacuum solution is a solution of a field equation in which the sources of the field are taken to be identically zero. That is, such field equations are written without matter interaction.

In mathematics, mathematical physics and the theory of stochastic processes, a harmonic function is a twice continuously differentiable function f : UR where U is an open subset of Rn that satisfies Laplace's equation, i.e.

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

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

${\displaystyle ds^{2}=d\rho ^{2}+\rho ^{2}\,\left(d\theta ^{2}+\sin(\theta )^{2}\,d\phi ^{2}\right)}$
${\displaystyle 0<\rho <\infty ,\;0<\theta <\pi ,\;-\pi <\phi <\pi }$

### Spherically symmetric field

As an example, we can calculate the tidal tensor for a spherical body using the Hessian. Next, let us plug the gravitational potential ${\displaystyle U=-m/\rho }$ into the Hessian. We can convert the expression above to one valid in polar spherical coordinates, or we can convert the potential to Cartesian coordinates before plugging in. Adopting the second course, we have ${\displaystyle U=-m/{\sqrt {(}}x^{2}+y^{2}+z^{2})}$, which gives

${\displaystyle \Phi _{ab}={\frac {m}{(x^{2}+y^{2}+z^{2})^{5/2}}}\,\left[{\begin{matrix}y^{2}+z^{2}-2x^{2}&-3xy&-3xz\\-3xy&x^{2}+z^{2}-2y^{2}&-3yz\\-3xz&-3yz&x^{2}+y^{2}-2z^{2}\end{matrix}}\right]}$

After a rotation of our frame, which is adapted to the polar spherical coordinates, this expression agrees with our previous result. The easiest way to see this is to set ${\displaystyle y,z}$ to zero so that the off-diagonal terms vanish and ${\displaystyle \rho =x}$, and then invoke the spherical symmetry.

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## References

1. Baldauf, Tobias; Seljak, Uros; Desjacques, Vincent; McDonald, Patrick (13 January 2018). "Evidence for Quadratic Tidal Tensor Bias from the Halo Bispectrum". Physical Review D. 86 (8). arXiv:. Bibcode:2012PhRvD..86h3540B. doi:10.1103/PhysRevD.86.083540.