# Is there a momentum for charge?

Since mass and charge behave similarly, so, just like center of mass, I define a point center of charge, that is defined by

$$\vec r_{qm} = \frac {\sum{q_i \vec r_i}} {\sum{q_i}}$$

where $\vec r_i$ is a position vector w.r.t. the origin.

Now suppose just like momentum,there is a quantity we call charge-momentum $q \vec v$ for a system which is changed iff there is an external quantity denoted by $q \vec a$ where $\vec a= \frac{d\vec v}{dt}$ . Let us name this new quantity $q\vec a$ as charge-force.

Now I conjecture, for every charge force, there is an equal and opposite charge force.

And that charge-momentum of an isolated system will be conserved. Just like momentum, can this be a conserved quantity of the universe? Does this make any sense ?

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The center of charge you've defined is something like a normalized dipole moment. The "charge-momentum" is the same as a current. – Muphrid May 7 '13 at 4:20
Muphrid is right. The current caused by a single moving charge is $q\vec{v}$. – Lagerbaer May 7 '13 at 4:51
@Lagerbaer this is even not dimensionally correct – ABC May 7 '13 at 5:26
You're right, but current density $j$ is obtained by multiplying charge density $\rho$ with the charge's drift velocity $\vec{v}$, so they're related in a sense. – Lagerbaer May 7 '13 at 5:31
While "center of charge" is an attractive notion at first, it is not as useful as you'd think: most systems are neutral and the concept is undefined. (See e.g. this question.) – Emilio Pisanty May 7 '13 at 12:47

Of course you can define such a quantity, but the question is: does it mean anything physically?

Contrary to what has been stated in some of the answers/comments, this quantity is not comparable to a "normalized" dipole moment. A dipole is a system of two charges equal in magnitude but opposite in sign. The corresponding dipole moment, which is of great importance for the description of many phenomena in electromagnetism, is defined as the product of the magnitude of one of the charges and their displacement vector. It would be equal to the numerator of your expression if you put the negative charge to the origin, the only term in the sum would be the product of the charge and the position (displacement) vector. So far so good: but what about the statement that your expression would then be a "normalized dipole moment"? Let us look at the denominator: in case of a dipole, the sum of both charges amounts to zero, so we would have a division by zero. This does not give us something normalized, but rather something ill-defined. Hence, this concept does not make much sense.

This problem remains for any system where the sum of charges is equal to zero, i.e. for neutral systems. Thus, it is not defined for many physically important situations and even in cases where it is, it does not tell us anything about the properties of that system.

Your "charge momentum" is related to the current density, which is given by the product of charge density and velocity. Its time derivative would simply be the rate of change of a current, and meaningful for example in a system with time-dependent electrostatic potential: this can be found in electrotechnical application when dealing with alternate currents. But can this be compared to a force?

To answer this question, we have to examine the nature of the comparison of the quantity $q\vec{v}$ to momentum in mechanics. What makes momentum so special is the fact that it is conserved for closed systems, i.e. systems without any external forces. But what is required in order for a certain quantity to be conserved?

One of the key principles of classical mechanics, Noether's theorem, tells us that conserved quantities (also called "conserved charges" or "Noether charges") are related to continuous symmetries of a system (there are various sources on the internet and books which describe this principle in as much detail as one might imagine). The conservation of momentum is a consequence of translational invariance of a physical system. In order to be comparable to momentum, your "charge momentum" would have to correspond to a continuous symmetry of the underlying system, but it turns out that there is none.

However, it is not completely unrelated to that concept either. While your "charge momentum" is no conserved quantity, charge itself is indeed one, corresponding to the $U(1)$ symmetry of electromagnetism. Within the framework of Noether's theorem, there also exists the notion of a so-called conserved four-current $J^\mu$, which has to satisfy

$$\partial_\mu J^\mu=0,$$

i.e. its four-divergence vanishes. In the case of the $U(1)$ symmetry, splitting space and time components and writing out the equation explicitely gives

$$\frac{\partial\rho}{\partial t}+\partial_i j^i=0,$$

which is nothing but the continuity equation where $\rho$ is charge density and $j^i$ is three-dimensional current density, which is related to your "charge momentum", as was already pointed out in the comments. There is no indication that the latter is conserved by itself.

Since there is no conservation of "charge momentum", i.e. no analogue to Newton's first law, applicability of an analogue of Newton's second law, which states that force is defined as the time-derivative of conserved momentum is highly doubtful. Furthermore, there is no reason to assume that the principle of "actio=reactio" (Newton's third law) should hold.

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Not much sense. Your "center of charge" is nothing but the dipole moment divided by the net total charge. "Normalised dipole moment, if you will".

If you take $q|\vec v|$ instead of $q\vec v$, you get something related to current (generally current times a factor). Current is conserved at a junction.

Regarding your equal-and-opposite situation, the closest I can come up with is this: If you have a body emitting charge in free space (no magnetic field), there will be an equal-and-opposite displacement current (and thus an equal and opposite "displacement charge force"). Not that displacement current is not a real current, it's more of a mathematical convenience.

Firstly, the parallel between charge and mass comes from gravitational mass. Basically, Coulomb's law and the law of gravity are similar: $$\frac{kq_1q_2\hat r}{r^2}\sim-\frac{Gm_1m_2\hat r}{r^2}$$

These similarities are quite useful, for example, the shell theorem and Gauss' law work for both systems. There even are things like gravitomagnetism, though that deals with the similarities in the relativistic formulation. The similarity is only limited by the absence of "negative" gravitational mass.

However, these similarities don't extend to the inertial property of mass. Newton's laws deal with the inertial properties. The fact that attraction due to gravity is proportional to the inertial nature of a body can be said to be "chance" in the classical formulism. (More modern models explore and explain this link)

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The center of charge is the "Monopole" moment, not the dipole moment! – Martin J.H. Dec 2 '13 at 9:59
@MartinJ.H. True, but the monopole moment is a scalar. And I'm talking about the center of charge proposed by the OP above; $\sum q\vec r$ is the dipole moment – Manishearth Dec 2 '13 at 10:36

Mass and charge are not so similar for the charge having "center of charge".

The notion of "center of mass" appears in many applications when number of bodies move. In this situation, the movement can be splitted into movement of center of mass and individual movements of bodies relative to the center of mass.

This occurs because of dual role of mass: it both (1) produces a gravity and reacts to it, and (2) guides the reaction of body to ANY force.

This principle is not true for charge. So, your point of center of charge plays no role in movement.

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I rather like this answer. It gets to the point. – David Z Nov 30 '13 at 21:20

The "center-of-charge" is part of a more general concept that is used quite often in physics: Multipole expansion.

The general idea of multipole expansion is the following: If you view a charge (or mass) distribution from a large distance, then most of its internal structure is irrelevant to you. Instead, it suffices to do all calculations based on a few, low order moments.

• To 0th order in the expansion, you only consider the net charge of the distribution. Take the nucleus of an atom: To 0th order you treat it as a point charge with charge +Ze and to ignore the internal structure. The location of the point charge is at the center-of-charge. (In atoms, the net charge is 0 so you can safely ignore this part.)
• To 1st order in the expansion, you additionally look at field caused by the dipole moment of the charge distribution. As far as I know, permanent electric dipole moments have not been observed in nuclii yet, but many nuclii have a magnetic dipole moment. Many molecules have an electric dipole moment.
• To 2nd order, you additionally consider the quadrupolar moments of the distribution.
• To 3rd order, the octupolar, ...

As for charge force and charge momentum: You are absolutely free to define such quantites! However, when conjecturing that the charge momentum is conserved, you should have a proof up your sleeve. Rigorous proofs can be difficult. Often, it is much easier to disprove something, because you only need to find one example where the theory is wrong.

Let me try to give such an example: Take a system of two particles moving towards each other. Let them have equal charges, but very unequal masses, i.e. let the first particle be very, very light (an electron), and the other particle be very heavy (a proton).

Let them fly towards each other at non-relativistic speeds. Because the second particle is much, much heavier, it simply pushes the lighter particle out of its way, and it does not change its velocity much. The sketch below shows this for the initial conditions (1), for when the lighter particle has speed 0 (2), and for the "final" state where the light particle flies away at twice the speed.

Now, let's look at the charge moment at the three different times:

1. The total "charge momentum" of the system is 0
2. Since the lighter particle is at rest, and the hevier particle barely slowed down, the total charge momentum is -q*v_initial.
3. The light particle flies away with twice the intial speed, and the heavy particle still barely changed course, so the charge momentum in -3*q*v_initial.

So in this case, the total charge momentum changes over time, so it cannot be conserved.

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In regards to the current question, for a single charge going in a loop, $i = qf$. $f = \frac {1}{(2\pi \sqrt{LC})}$, $\omega = \frac{1}{\sqrt{ LC}} =\frac V R$. After the corresponding substitutions,$i = \frac{qV}{2\pi R}$.

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I've edited your post to improve its format, please check if the formulas are correct. – Ali Nov 29 '13 at 14:53

Well I say charge indeed has momentum...

Given $F = qE$ and realising that $F = \frac{dP}{dt}$ It follows $\Delta P = qE \Delta t$. We define $\Delta P$ to be the charge momentum and $E\Delta t = (iota)$$From this it is clear that the charge momentum (which equal mass momentum) are two different entities, in fact it can be showed that the two exist in two different field spaces. The reader must note also that iota is very large, and$$iota = \frac{m}{q} \frac{dr}{dt}$$....the$m/q$ratio implies a relation between two different field space properties,pretty much like$E/B = v\$....it is a way of relating two different fields(field spaces).

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This is nonsense. – user23503 Feb 11 '14 at 20:19

## protected by Qmechanic♦May 11 '13 at 22:06

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