Consider two massive charged objects at rest with a large horizontal distance $d$ between them (object $1$: mass $m_1$, charge $q_1$ and object $2$: mass $m_2$, charge $q_2$).

I apply a constant vertical force $\vec{f_1}$ upwards to object $1$ so that it gains an acceleration $\vec{a_1}=\vec{f_1}/m_1$.

The total amount of power $P_1$ that object $1$ radiates is given by the Larmor formula:

$$P_1=\frac{2}{3}\frac{q_1^2 a_1^2}{4\pi\epsilon_0c^3}.\tag{1}$$

Now assume that object $2$ is constrained to move only in the vertical direction. If the horizontal distance $d$ between the objects is large then only the "radiative" part of the Lienard-Wiechert electric field due to object $1$ can do any work on object $2$. The vertical force $\vec{f_2}$ acting on object $2$ is given by:


The power received by object $2$, $P_2$, is given by:


The equation of motion of object $2$ is given by:

$$m_2 \frac{d\vec{v_2}}{dt}=\vec{f_2}.\tag{4}$$

As the vertical force $\vec{f_2}$ is constant and the object $2$ is initially at rest then integrating Eqn.(4) gives:


Substituting Eqn.(2) and Eqn.(5) into Eqn.(3) we find that the power $P_2$ received by object $2$ is given by


Finally, the ratio of the power received by object $2$, $P_2$, to the power emitted by object $1$, $P_1$, is given by


Thus eventually object $2$ receives more power than the total power emitted by object $1$.

What's gone wrong? :)

Additional calculation

I've put some numbers into the equations in such a way that the final velocities $v_1$, $v_2$ remain non-relativistic and only the radiative part of the Lienard-Weichert field from object $1$ to object $2$ is significant: $$ \begin{eqnarray*} a_1&=&1.0\times10^{10}\ \hbox{m/s^2}\\ q_1&=&1.0\ \hbox{C}\\ q_2&=&1.0\times10^{10}\ \hbox{C}\\ m_2&=&1.0\ \hbox{kg}\\ \Delta t&=&2.5\times10^{-5}\ \hbox{s}\\ d&=&7.5\times10^{6}\ \hbox{m}\\ c\Delta t/d&=&1.0\times10^{-3}\\ v_1/c&=&8.3\times10^{-4}\\ v_2/c&=&1.1\times10^{-7}\\ \end{eqnarray*} $$ The energy $E_1$ radiated by object $1$ from $t=0$ to $t=\Delta t$ is: $$E_1=5.5\times10^{-1}\ \hbox{Joules}$$ The energy $E_2$ received by object $2$ from $t=d/c$ to $t=d/c+\Delta t$ is: $$E_2=5.5\times10^{2}\ \hbox{Joules}$$ Therefore object $2$ receives a thousand times more energy than was emitted by object $1$.

Unphysical assumption

The electrostatic energy of body 2 must be at least $q_2^2/4\pi\epsilon_0d=10^{23}$ Joules. This is greater than the rest mass energy of body 2 which equals $m_2c^2=10^{16}$ Joules. This seems to show that my argument is unphysical and therefore not a paradox.

  • $\begingroup$ I think probably this is just a problem of you not using time dependent formulas & propagation delays, but it is also possible that there is a problem with you not including relativistic effects. The Larmor formula was derived from the Lienard-Wiechert potentials and Maxwell's equations conserve energy, so the claim that energy is not conserved here actually explodes into a claim like “vector calculus is inconsistent” which I think is larger than you think you're claiming. So, worth working it out in full detail rather than in sketch form. $\endgroup$ – CR Drost Jan 12 at 16:32

In answer to the last question -- many things, IMHO

  1. Be more careful about forces and accelerations, at high enough speeds constant force will not lead to further constant gains in speed. Also I would suggest not working with some fictitious forces. You are accelerating charges, so let your external force will be supplied by electric field. Less chance to end up with unphysical situation.

  2. You said that distance between objects is large, so why are you using what looks like electrostatic formula to compute the force on charge 2? The logical chain is $q_1$ generates electromagnetic field $\to$ electromagnetic field propagates $\to$ electromagnetic field acts on $q_2$

  3. What about the electromagnetic field generated by $q_2$ and how it acts on $q_1$?

  4. What about recoil due to electromagnetic field? It carries momentum, so it will create a 'drag'?

  5. Careful about 'eventually' statements. You have some unspecified force and you are using non-relativistic formulae. They will only work at low speeds. To get to 'eventually' your treatment has to be relativistic.

A proper way of doing this would be to write Lagrangian for two charges and an external constant electric field, add your constraints as Lagrange multipliers. Then derive coupled differential equations for the motion of two charges. Probably you will also have differential equations for the generated field. Then you solve all of these together. It may be possible to simplify things, but you will have to state carefully under which conditions such simplifications apply.

And use relativistic Lagrangians for the particles.

Forces will not be necessary in such treatment

  • $\begingroup$ Regarding point 2: I'm not using the electrostatic component that goes like $q_1/4\pi\epsilon_0d^2$ but rather the radiative component that goes like $q_1a_1/4\pi\epsilon_0c^2d$. $\endgroup$ – John Eastmond Jan 16 at 12:59
  • $\begingroup$ radiative components propagate. where is the retarded time in your expression? $\endgroup$ – Cryo Jan 16 at 20:16
  • $\begingroup$ Ok - I left the propagation time delay out. $\endgroup$ – John Eastmond Jan 17 at 14:11
  • $\begingroup$ Retarted time includes temporal and spatial dependencies. You can't leave it out without justification, what you have for your electric field is not a solution of Maxwell's equations, and statements about 'eventually' are then completely meaningless. $\endgroup$ – Cryo Jan 17 at 18:12
  • $\begingroup$ These observations/objections are superficial and do not address the problem at all. The formulae and assumptions in the question are standard and justified, retardation is of course present in retarded field theory where Larmor's formula is derived but it just isn't relevant here as the problem occurs even if acceleration $a_1$ is constant and the problem occurs sooner (for numbers stated in the question) than any reaction from body 2 can reach the body 1. $\endgroup$ – Ján Lalinský Jan 17 at 21:44

Your calculations are correct, total Larmor's power radiated by body 1 is much lower than power of impressed radiation force due to 1 acquired by 2. That it, is it higher not right from the start of interaction (because impressed power is 0 when body 2 has 0 speed) but after some time. We can derive this time from your formulae to be

$$ t^* = \frac{\frac{2}{3}m_2 c^2}{F_C c} $$ where $F_C$ is Coulomb force of charged body 2 on hypothetical charged body of the same charge separated by distance $d$. Thus $t^*$ is the time needed for the Coulomb force of the body 2 to impress energy $m_2c^2$ to hypothetical body of same charge moving with speed of light away from body 2.

In usual circumstances charge available on single body is very low and energy $m_2c^2$ is very high, so this time is extremely long. Your time $\Delta t$ can be very low because you assume extremely high value of $q_2$ and extremely low value for $m_2$. Maximum net charge we can concentrate on an isolated body of diameter 1m on Earth is probably not bigger than 1e-3 C. Even if this weighted only 1kg the time $t^*$ comes out around 10 billion years. Even if we shorten the distance $d$ to 10m, the time is 26 days. So it is hard to observe this kind of effect. Further decrease of distance will be hard as the repelling force becomes very high.

Larmor's formula gives net EM energy flowing out of accelerated charge body provided there are no other charges anywhere near the region.

In your example, there is such a charged body: the body 2. This makes Larmor's formula derivation assumptions invalid. Total EM field is sum of fields of 1 and 2 and because body 2 is charged, total EM field in its vicinity is much higher than it would be were the body 2 absent. That's why the body 2 is able to acquire much more energy than Larmor's formula would suggest is available. Much more energy is available, because body 2 is charged and has its own strong electric and magnetic field. This fact is entirely ignored by Larmor's formula derivation.

Larmor's formula is pretty much useless when analyzing interaction of charged bodies and energy transfer between them. One must go back to analysis of mutual EM forces as you have done.

  • $\begingroup$ What happens if body 2, with a large charge $q_2$, is enclosed by a very massive charged spherical shell with charge $-q_2$? In that case body 1 would not feel any static field from body 2 and therefore Larmor's formula would be valid for the radiated power from body 1. $\endgroup$ – John Eastmond Jan 17 at 14:33
  • $\begingroup$ I do not follow. Whether body 1 experiences force due to 2 is irrelevant. The time $\Delta t$ given above is so short that radiation field from 2 does not reach 1. $\endgroup$ – Ján Lalinský Jan 17 at 16:25
  • $\begingroup$ In the original setup, before body 1 moves (i.e. $t<0$), there is a static field due to body 2 at the location of body 1. I'm not sure if this causes a problem for the application of Larmor's formula for body 1 when the acceleration is applied to it (i.e. $t>0$). My point was that one could take away this static field by screening body 2 with an equal and opposite charge around it. I don't know if that helps. $\endgroup$ – John Eastmond Jan 18 at 14:49
  • $\begingroup$ The static field of 2 acting on 1 can be made arbitrarily small, just increase the distance $d$. This field does change the Poynting vector near the body 1 but if the field of 2 is very small there, it changes the net Poynting power flowing out of the particle negligibly. However, the field of 2 is not negligible near the body 2 so there the change in Poynting vector is substanial, so the Larmor formula is inapplicable. If body 2 is screened, then we have 3 body problem and the system becomes more complicated, it does not help the analysis. $\endgroup$ – Ján Lalinský Jan 18 at 14:55
  • $\begingroup$ Even with the screening sphere, the field is still strong in the space between the body 2 and the screen, so the same substantial change in the Poynting vector occurs there. $\endgroup$ – Ján Lalinský Jan 18 at 14:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.