# Why isn't there a centre of charge?

When determining the gravitational attraction between 2 solid bodies, we can simplify computations by taking their masses to be concentrated at their respective centres of mass. However, had they been electrically charged bodies and we needed to compute electrostatic attraction, there is no "centre of charge" notion available to us. ( We would evaluate a 2-dimensional integral, or equivalently, apply Gauss's law. )

Is there any simple reason for the absence of such a centre of charge? Frequently, I've got the answer: "charge just doesn't work that way".

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There is a simple reason why we talk about centers of mass and never centers of charge. The analogy would be sensible if the center of mass were the location where the gravitational force - acting on masses - acts. The gravitational force is given by $Gm_1m_2/r^2$, and similarly, the electrostatic one could be $Q_1Q_2/4\pi\epsilon r^2$. However, that's not why we use center of mass. We use the center of mass to simplify the right hand side of $F=ma$ - when we calculate the acceleration $a$ - and it appears in the combination $ma$ even for electric forces. –  Luboš Motl Apr 7 '11 at 20:42
So we may always imagine that the forces - including gravitational ones, electrostatic ones, magnetic ones, and others - are acting "at the center of mass". The mass is special because it is what determines the inertia of objects. ... You may also calculate the average location of a change, e.g. by $\int \rho\, \vec r\, dV / \int \rho \, dV$. However, this quantity won't enter any important equations or their simplifications - and moreover, it may be ill-defined because the denominator, the total charge, may vanish. The mass denominator, the total mass, is always positive for objects. –  Luboš Motl Apr 7 '11 at 20:44
"When determining the gravitational attraction between 2 solid bodies, we can simplify computations by taking their masses to be concentrated at their respective centres of mass." This is wrong. This only works for a sphere. For a nonspherical object, you don't get the right answer this way. Also, the center of mass is not in general the same thing as the center of gravity (in cases where the field is nonuniform). –  Ben Crowell Aug 19 '11 at 2:54

There is no "center of charge" that simplifies calculations completely because there is no center gravity that does this, either.

Two rigid bodies feel gravity that is quite similar to that of point masses, but not exactly the same. Unless the objects are perfect spheres, they feel tidal forces, which depend in second-order and higher derivatives of the potential.

We can only calculate gravitational interactions as points at the center of mass if we want to ignore these second-order effects. On a day-to-day basis, using gravity for things like playing baseball with your nephew, that's fine. The tidal forces are usually very small in applications that we encounter because the phenomena are small compared to the characteristic length scales involved (i.e. a baseball's trajectory is small compared to the radius of Earth, Earth's diameter is fairly small compared to the Earth-Sun distance, etc.)

We can indeed make the same first-order approximation in electrostatics that we do in Newtonian gravity. In practice, a slightly different procedure is normally used, called the multipole expansion. First, a reference point is fixed. The total charge, called the monopole moment, is independent of this reference point, so if you want your first-order approximation to be good, it is wise to choose a reference point near the center of your charge distribution. Higher-order effects are then calculated relative to the reference point by computing dipole, quadrupole, and higher moments of the charge distribution. These higher orders are more likely to come into play in everyday life in electrostatics than in gravity because the size of the charge distributions is similar to the separation between charged bodies (i.e. two balloons you rubbed on your head and are playing with are not very far separated from each other, compared to their diameters).

You might also be interested in the hard sci-fi novel "Incandescence", in which a species of beings living inside a large asteroid orbiting a black hole observe tidal effects to discover general relativity without ever seeing the outside world. It's an interesting demonstrations of what higher-order gravity effects look like beyond the center-of-mass approximation.

The author, Greg Egan, has a web page explaining the tidal effects described in his novel here.

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-1 This answer is totally wrong. The acceleration of a rigid sphere E in response to point/sphere source S is exactly the same as the response of a point mass to the same sphere/point source. The reason is that the point and the sphere make the same outside potential, so a point E and a sphere E have exactly the same reaction force on S, therefore they must feel the same action force from S. There is no way around it. Tidal forces come from shape distortion, like oceans rising and falling, not from the rigid spherical shape. –  Ron Maimon Aug 17 '11 at 2:17
@Ron, Yes, you are right. Thanks for pointing it out. Maybe you could be more circumspect with your criticism, though, because the answer is not "totally wrong", as you claim. It was wrong in its specific example, but correct in its general idea. I'm updating the answer now. –  Mark Eichenlaub Aug 19 '11 at 0:23
I could be wrong, but I'd say that the analogue of center of mass is the dipole moment, not the monopole moment. The monopole moment is the analog of the total mass, but total mass is not the same as center of mass. Just compare the formulas for the center of mass and the dipole moment: they are the same, but with charges instead of masses. –  becko Aug 19 '11 at 0:45
@becko The analog of the monopole moment is the total mass located at the center of mass. Both are involved; I considered it implicit the the amount of mass to be located at the center of mass was the total mass. Center of mass itself is simply a location and could not have a well-defined potential. –  Mark Eichenlaub Aug 19 '11 at 0:48
@becko Counterexample: the dipole moment is well-defined and usually non-zero even if the total charge is zero. –  Mark Eichenlaub Aug 19 '11 at 1:17

In the same way that bodies that are far away from you can be approximated as point masses, charge distributions that are sufficiently far away can be approximated as point charges, with total charge $Q$, and then you would, indeed, compute something like the center of charge.

This is the first part of the so-called multipole expansion: You begin by approximating a charge distribution $\varrho(x)$ by a point charge of charge $Q = \int dx \varrho(x)$. The next step would be to look at the dipole moment, $p = \int dx x\varrho(x)$, followed by the Quadrupole moment etc

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Right. I was about 90% done when you posted. +1 –  Mark Eichenlaub Apr 7 '11 at 19:33

Charges for electrostatic interaction determine the interaction force. In this respect they are similar to masses that determine the gravitational force.

However when we look at the particle dynamics, it is the particle mass that determines the particle inertia whatever the force nature is. In this respect the masses are different from charges. According to the Newton equations, the center of mass makes sense as a quasi-particle position (the system as a whole).

Instead, charge distribution can be described and is often described as system charge form-factors.

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You can define something like the center of mass for charge, but it can be undefined, when the total charge is equal to zero.

Take a look at the definition of the dipole moment as well.

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Defining a centre of mass relies on the symmetry of particles with a mass charge. Namely that they are all of the same sign. This often allows the problem to be reduced to a geometrically simple description. For electric charge there are two signs and therefore this simple symmetry doesnt exist at short distances, though as mentioned above it can be approximated to at large distance.

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