# Definition of "tidal forces"

May anyone give me definition of tidal forces? Topic is related to force between two celestial bodies which are tidally locked. I have never heard for such force. Here is quotation which is intriguing me from this comment:

Tidal locking is related to the period of rotation of a body about its own center being synchronized via tidal forces to its rotational period around another body, or their center of mass (either way it's the same). They are not absolutely locked, it's just that the periods are the same on average.

• The Wikipedia article you linked says: "It arises because the gravitational field exerted on one body by another is not constant across its parts: the nearest side is attracted more strongly than the farthest side". Does that make sense to you? May 17, 2019 at 12:07
• en.wikipedia.org/wiki/Libration Also consider libration. The moon is not exactly locked, and can't be because its orbit is not perfectly circular. So it "wobbles" a bit. Brilliant animated image in that link.
– Dan
Feb 28, 2022 at 21:06

Consider two stars (or planets) acting on each other gravitationally, let's call them star $$A$$ and star $$B$$. Now let us assume that star $$A$$ is perfectly spherically symmetric (or point-like) for the time being. It will thus create a gravitational field that acts with a force $$\vec{F}_A = -\frac{G M_A m}{r^2} \vec{n}_r$$ on any particle with mass $$m$$ at a distance $$r$$ from $$A$$, where $$M_A$$ is the mass of star $$A$$ and $$\vec{n}_r$$ a unit radial vector pointing from it.
However, star $$B$$ is not a particle, so its matter elements will feel different forces at different points. The parts of $$B$$ that are closer will feel stronger gravity, while those farther weaker gravity. In the rest frame of the star this is felt as a tide. If $$B$$ is orbiting $$A$$, and if it manages not to get torn apart by the differences of the forces, it will get deformed away from spherical symmetry into a prolate shape. Specifically, the star $$B$$ bulges out towards $$A$$. This means that it will not have a gravitational field corresponding to a spherical mass but rather it will exert a force on $$A$$ that looks approximately like $$\vec{F}_{BA} = \frac{GM_B m}{r_{AB}^2}\left(1 + 3 \frac{M_A}{M_B} k \left(\frac{R_B}{r_{AB}}\right)^5\right) \vec{n}_{AB}$$ where $$M_B$$ is the mass of $$B$$, $$r_{AB}$$ the distance between the stars, $$\vec{n}_{AB}$$ a unit vector pointing from $$A$$ to $$B$$, $$k \sim 0.01$$ some deformability parameter of the star $$B$$, and $$R_B$$ its radius. You surely recognize the first part of the force, which is the usual attraction between two spherical or point-like masses. However, the second term in the force would actually be called a tidal force, since it is there only because of a tidally induced bulge on $$B$$. Generally speaking, a tidal force is a catch-all term for any force that comes about from the fact that the components in a binary have finite size.
I have assumed that star $$B$$ reacts to the tide created by $$A$$ immediately. However, in practice there is a finite response time for the bulges to adjust, and the friction of the stellar or planetary materials with respect to the traveling of theses bulges is the essence of tidal dissipation. One example of that is the friction of Earth's oceans with respect to the surface (and themselves) as they are dragged by the tides created by the Moon. So the tidal force can even be dissipative (but angular-momentum preserving). There is also back-reaction of these dissipative forces on the rotation of the star or the planet, which would be known as a dissipative tidal torque. The total action of these dissipative forces leads to 1) synchronization of rotational and orbital periods in the system, and 2) circularization of the orbits in the binary.