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Two masses

The above diagram represents an isolated system with two masses $M$, at position $X$, and $m$, at position $x$, connected together by an extended spring. Each mass is connected by rigid rods to charges $Q$ and $q$ (of negligible mass) separated by a vertical distance $r$. The masses and charges are constrained to move in a horizontal direction so only horizontal forces affect the dynamics of the system.

The system is released and the two masses are pulled together by the spring.

As there is no external force, $F_{ext}$, acting on the system then we have

$$F_{ext} = M_{tot} \ddot{X}_{com} = 0,$$

where $M_{tot}=M+m$ is the total mass of the system and $X_{com}=(MX+mx)/(M+m)$ is the center of mass of the system.

Thus we have:

$$M \ddot{X} + m \ddot{x} = 0.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$

But let us now consider the masses $M$ and $m$ separately together with the internal forces acting on them.

The force on $M$ consists of the force $F$ from the spring together with a horizontal electromagnetic radiation force on charge $Q$ due to the acceleration of charge $q$.

Therefore we have:

$$M \ddot{X} = F - \frac{qQ}{4\pi\epsilon_0 c^2 r}\ddot{x}.$$

Similarly the force on $m$ consists of the force $-F$ from the spring together with a radiation force on charge $q$ due to the acceleration of charge $Q$.

$$m \ddot{x} = -F - \frac{qQ}{4\pi\epsilon_0 c^2 r}\ddot{X}.$$

Adding these two equations together we obtain:

$$M \ddot{X} + m \ddot{x} = -\frac{qQ}{4\pi\epsilon_0 c^2 r}(\ddot{X} + \ddot{x}).\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$

Equation (2) is not consistent with Equation (1).

There seems to be a paradox here.

One could answer that I have not taken into account the electromagnetic reaction force on the charges due to their emission of electromagnetic radiation. In fact the electromagnetic reaction force on a moving charge $q$ is given by

$$F_{reaction} = -\frac{2}{3}\frac{q^2}{4\pi\epsilon_0 c^3}\dot{a}.$$

Therefore $F_{reaction}$ depends on the rate of change of the acceleration, $\dot{a}$, rather than $a$ itself. As the reaction force term has an extra factor of $c$ in the denominator and the rate of change of acceleration is small (the length of the spring doesn't change very fast if the masses are reasonably large) then we can ignore the possibility of the emission of electromagnetic radiation and the subsequent reaction forces on the charges. Therefore this is not an answer to the paradox.

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Sorry, the Coulomb force isn't proportional to $\ddot X$ in any way. Moreover, the Coulomb force on your picture seems to be vertical, not horizontal. Your calculation seems to be completely wrong. Morever, it's obvious that the total momentum is conserved - action is accompanies by reaction - because both the spring and Coulomb forces obey the third Newton law even separately. A fix is to realize that the factors in eqn (2) aren't $\ddot X+\ddot x$ but they're functions of $x,X$ only and they're manifestly equal in magnitude with opposite signs. –  Luboš Motl Jul 11 '13 at 16:11
    
But I'm disregarding the Coulomb force as it acts vertically and the system can only move horizontally. As well as a static Coulomb field an accelerating charge produces a radiation field whose strength is proportional to the charge's acceleration. –  John Eastmond Jul 11 '13 at 16:21
    
Try working out whether there is more radiation going in one direction than another. If there is a net acceleration then that's where the momentum balance is going. –  Michael Brown Jul 11 '13 at 16:27
    
But because the acceleration of the charges is constant then there is no electromagnetic radiation being emitted by those charges (there is only a radiation field). Thus there is no change of momentum in the electromagnetic field. –  John Eastmond Jul 11 '13 at 16:43
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Where did you find that a constant acceleration does not radiate? cv.nrao.edu/course/astr534/PDFnewfiles/LarmorRad.pdf .It is a many body problem anyway. –  anna v Jul 11 '13 at 19:21
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2 Answers 2

Your expressions for the Coulomb force are wrong. Since the electrostatic repulsion is translation invariant, the force must be a function of the difference $X-x$, and if you do the maths right, using a general force of that form in the system $$M\ddot X=+F-f(x-X)$$ $$m\ddot x=-F+f(x-X)$$ you will find that $M\dot X+m\dot x$ is conserved.

More specifically, your Coulomb force is of the form $$ \pm \frac{Qq}{4\pi\epsilon_0}\frac{L+X-x}{(r^2+(L+X-x)^2)^{3/2}} $$ where $L$ is a length describing the horizontal bits of wire in your diagram.

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True. But I'm saying its not only the Coulomb force that is operating. There is also a radiation force on each charge due to the acceleration of the other one. This is the second term in the expression for the Lienard-Wiechert electric field of an arbitrarily moving charge. –  John Eastmond Aug 12 '13 at 14:14
    
That is only partly correct. If you include the radiation reaction forces on both particles, then total momentum will not be conserved. However, the reason equations (1) and (2) are not consistent is that your Coulomb forces are wrong. –  Emilio Pisanty Aug 12 '13 at 14:36
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There is no paradox here. Apart from the fact that your calculations and your expressions for the electric field are most certainly wrong, Newton's third law is not valid in general for electrodynamics. Hence, your basic premise that $F_{ext} = M_{tot}X_{com}$ itself is not valid.

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In what way are my calculations and my expressions for the electric field wrong? I'm just using the standard expression for the electromagnetic radiation field of an accelerated charge. I also don't see why I can't use Newton's 3rd law if there is no appreciable electromagnetic energy radiated away. –  John Eastmond Jul 12 '13 at 1:20
    
Can you quote your source from where you got that formula? On the one hand you are saying you are using the expression for the 'electromagnetic radiation field', and on the other you are saying 'there is no appreciable electromagnetic energy radiated away'. You appear to be contradicting yourself! –  guru Jul 12 '13 at 10:33
    
Check out the Wikipedia entry on the Liénard–Wiechert potential. There is a difference between electromagnetic radiation that carries energy away in the form of photons and the radiation part of the field produced by an accelerating charge. –  John Eastmond Jul 12 '13 at 18:17
    
In order to radiate actual photons the charge's rate of change of acceleration has to be non-zero. The reaction force back on the charge is then given by the Abraham-Lorentz force. –  John Eastmond Jul 12 '13 at 18:49
    
Well, the expression in the wikipedia entry for the electric field of a point charge derived from the Liénard–Wiechert potential is definitely not the one which you are using! Can you go through the steps of how you derived your expression from the one given in wikipedia? And since we are discussing classical EM here it is better not to bring the quantum aspect, viz photons into the discussion - thats mixing up the two things. –  guru Jul 12 '13 at 21:25
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