Seeking the Equation of Motion for a charged mass attached to a spring

  • #1
snoopies622
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I'm surprised that this question only occurred to me recently. If a have an electrically charged mass attached to a spring and set it oscillating, the resulting production of electromagnetic waves must cause a kind of "friction", a force resisting the motion of the charged mass, so that its oscillation loses energy and eventually stops.

How do I arrive at this force? I know that a changing electrical current (as well as a changing electrical field) will produce a changing magnetic field, which will in turn produce a changing electrical field, and so on, but exactly how this electric field pushes on the charged mass at any given moment seems complicated to me.

I would appreciate a link to any helpful source, thanks.
 

Answers and Replies

  • #2
Ibix
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I would think you could get a force by Googling "power emitted by an oscillating charge" and equating it to ##\vec F\cdot\vec v##. I don't think the result will be particularly tractable, but if you have a numerical integrator to hand it ought to work.
 
  • #3
snoopies622
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This is strange: According to the Larmor formula, the power given off by a charged particle is directly proportional to the square of its acceleration

[tex]

P=\frac{q^2 a^2}{6 \pi \epsilon_0 c^3}

[/tex]

https://en.wikipedia.org/wiki/Larmor_formula

but in a sinusoidal motion case like this, when the particle is farthest away from its equilibrium position its acceleration is at maximum while its velocity is zero, hence [itex]P= \vec F \cdot \vec v =0 [/itex]. So at that moment, the electromagnetic energy coming off the particle is at its highest, while simultaneously no energy is being invested into it.
 
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  • #4
snoopies622
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Wondering if I could get some feedback on this apparent contradiction. How can a charged particle emit energy (a≠0) while not simultaneously absorbing energy (v=0)?
 
  • #5
snoopies622
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A quick calculation shows that setting displacement x= A sin(wt), integrating the power the charged particle (q) emits according to the Larmor formula over one cycle of time, then dividing this energy by the distance it travels in that time gives an [itex] \textit {average} [/itex] force of
[tex]
f= \frac {q^2 A \omega^3}{24 \epsilon_0 c^3}
[/tex]

Of course I see no reason to assume that this would have constant magnitude, but I was curious. In reality I'm guessing that the force is itself sinusoidal, and there must be a lag, a phase difference, between the force the particle moves against and the energy it emits, which is not indicated in the Larmor formula itself.
 
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  • #6
ergospherical
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If you have a charge oscillating as per ##x(t) = Ae(i\omega t)##, then ##\ddot{x} = -\omega^2 x## and the Larmor formula implies that the average power radiated is\begin{align*}
\langle P \rangle = \frac{q^2}{12\pi \epsilon_0 c^3}\mathrm{Re}\left[ \ddot{x}^* \ddot{x} \right] = \frac{q^2 \omega^4 |A|^2 }{12 \pi \epsilon_0 c^3}
\end{align*}The energy of the oscillator is ##E = (1/2) m\omega^2 |A|^2## and we can put\begin{align*}
\dot{E}/E = -\langle P \rangle/E = - \frac{q^2 \omega^2}{6m \pi \epsilon_0 c^3}
\end{align*}which, integrated, is\begin{align*}
E = E_0 e\left(- \frac{q^2 \omega^2}{6m \pi \epsilon_0 c^3}t \right)
\end{align*}i.e. decay with time constant ##\tau = 6m \pi \epsilon_0 c^3/(q^2 \omega^2)##. The amplitude decays as ##|A| \propto \sqrt{E}##.
 
  • #7
snoopies622
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Thanks, ergospherical! That's very helpful.
 
  • #8
snoopies622
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A couple more thoughts here.

1. So if this charged particle oscillates in an exponentially decaying sinusoidal way, as appears to be the case, then that would be consistent with a frictional force that's directly proportional to its speed. But that leaves the problem of the inconsistency of this force with Larmor's formula, which describes energy loss as a function of acceleration only.

2. This problem isn't limited to sinusoidal motion. If for example a particle moves on the trajectory [itex] x=(a/2)t^2 [/itex], then its speed is [itex] at[/itex] , acceleration [itex] a[/itex] , power emission [itex] ka^2[/itex] , and force = power/speed = [itex] ka^2/at = ka/t[/itex] (Here I'm using [itex] k= q^2 / 6 \pi \epsilon_0 c^3 [/itex]) . Thus as it approaches x=0 from the negative direction, the force it would have to overcome would increase without bound and so it could never stop.

I've looked at a couple derivations of Larmor's formula which describe the electromagnetic wave created by an accelerating charged particle and calculate its energy, but so far haven't found one which even mentions this mysterious reactive force which also must accompany the acceleration.
 
  • #9
ergospherical
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then that would be consistent with a frictional force that's directly proportional to its speed.
Not quite; the self-force is known as the Abraham-Lorentz force and is proportional to ##\dddot{x}##.
 
  • #10
snoopies622
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Thanks again, ergospherical, I'm interested in this subject. I'll now look into a derivation of the formula for this force. Already I'm wondering how one resolves the problem of a charge accelerating at a constant rate, since it should both radiate constant energy (according to the Larmor formula) yet experience no reactive force according to the Abraham–Lorentz formula.
 
  • #11
ergospherical
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Already I'm wondering how one resolves the problem of a charge accelerating at a constant rate, since it should both radiate constant energy (according to the Larmor formula) yet experience no reactive force according to the Abraham–Lorentz formula.
I mean, you’ve hit the jackpot. There’s almost an infinitude of literature on this problem (and I’m not even aware whether it is totally resolved). For further reading, try Landau and/or Jackson.
 
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  • #12
snoopies622
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(and I’m not even aware whether it is totally resolved)

I'm getting this impression as well, which bewilders me considering that Maxwell's equations are 160 years old, and since then there have been so many brilliant minds and computing technology, too.
 
  • #13
tech99
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I'm surprised that this question only occurred to me recently. If a have an electrically charged mass attached to a spring and set it oscillating, the resulting production of electromagnetic waves must cause a kind of "friction", a force resisting the motion of the charged mass, so that its oscillation loses energy and eventually stops.
Thank you for a very interesting question. It seems to me that the radiation is caused by acceleration of the charge, so is maximum when the force on the mass is maximum. This is at maximum deflection. The friction of the air (for instance) occurs at max velocity, so is max at zero spring deflection. As radiation represents a loss of energy from the system, it creates an opposing reaction to the accelerating force of the spring. In this respect it is like hysteresis of the spring, maybe like soft elastic. In electrical terms, considering the analogue of an LC circuit, the energy loss has appeared as losses in the dielectric of the capacitor, and so it can be represented as a resistor shunting the capacitor. By contrast, the friction losses in the system, such as air resistance, appear as series resistance between L and C, as they depend on current (velocity). To the outside world, the hysteresis loss of a capacitor can be seen (for AC) as just increased series resistance, and so it is with the radiation resistance arising from acceleration of the charge. These are just my ideas.
An interesting minor point is that the velocity of the charge must create a magnetic field, which is storing energy, so KE will appear to be slightly increased. It must slightly increase the inertia of the mass.
 
  • #14
snoopies622
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Thanks, tech99. It does seem that the energy emitted by the charge would have to be momentarily "stored" somewhere near the charge, implying that Larmor's formula is valid only outside of a certain radius in which this energy is stored. For now the only thing that's clear to me is that without some kind of extra hypothesis like this, [itex] P= \vec F \cdot \vec v [/itex], the Larmor formula and the Abraham-Lorentz formula can't all be true, since one is a function of velocity, one of acceleration and one of jerk.
 
  • #15
tech99
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I think the choice of a mass and spring - a resonant system - makes it hard to explain the action unless someone is very comfortable with AC theory. The difficulty arises from the various phase relations in the system.
An easier explanation is to consider a non resonant system where we have a generator doing work, where there is no stored energy. For instance, consider a circuit consisting of a long piece of resistance wire connected to an AC generator. The applied voltage and the current are in-phase and represent energy dissipated as heat. When the voltage is maximum, the charges have maximum acceleration, so radiation occurs in-phase with voltage and current. This results in an additional resistance, the radiation resistance, against which the generator does work and supplies the radiated energy. If we now use a wire with low resistance, such as copper, the same action regarding radiated energy occurs, and we again see radiation resistance.
 
  • #16
vanhees71
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The problem of this notorious radiation-damping problem is that the interaction between a charged point particle and its own electromagnetic field is not well-defined in classical electrodynamics. The reason is that the point particle concept is an oversimplification. What can be done is to renormalize the mass of the particle, i.e., subtracting the divergence, but the resulting equation (the Abraham-Lorentz equation for the non-relativistic approximation, the Abraham-Lorentz-Dirac equation for the relativistic one) has also very troublesome properties like predicting self-acceleration (runaway solutions) and acausality. A pretty elegant treatment, using a very efficient regularization method can be found in

K. Lechner, Classical Electrodynamics, Springer International Publishing AG, Cham
(2018), https://doi.org/10.1007/978-3-319-91809-9

K. Lechner and P. A. Marchetti, Variational principle and energy-momentum tensor for
relativistic electrodynamics of point charges, Ann. Phys. (NY) 322, 1162 (2007),
https://dx.doi.org/10.1016/j.aop.2006.07.002

C. Nakhleh, The Lorentz-Dirac and Landau-Lifshitz
equations from the perspective of modern renormalization
theory, Am. J. Phys 81, 180 (2013),
https://dx.doi.org/10.1119/1.4773292

I've also a section summarizing the derivation of the LAD equation in my FAQ

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

But I've not yet more examples than the free-particle case and also no discussion of the Landau-Lifshitz approximation, which is considered today the best approximate solution of the problem possible within classical electrodynamics.

In the book by Landau and Lifhitz (vol. 2 classical field theory) you find another very elegant formulation using a clever game with gauge transformations.

The problem is less severe in the quantum-field theoretical context, QED being the paradigmatic example of a Dyson-renormalizable relativistic QFT. There you can evaluate the self-energy to any order of perturbation theory with a finite number of renormalizations (wave-function normalization, electric charge, electron mass).
 
  • #17
tech99
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Sorry I cannot do the mathematics. In simplistic terms, we seem to have a situation resembling Newton's Third Law. The accelerating field applies a force equal and opposite to the reaction due to radiation. In addition, the charge acquires a velocity and then has a magnetic field, which is not radiated but stores energy. The accelerating field and the consequential magnetic field seem to be what we observe as the induction fields.
 
  • #18
snoopies622
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Thanks vanhees71, much for me to take in here.

It occurs to me — aside from all the trouble with the theory, I would think that the relationship between the electrical charge on a light object and its resistance to acceleration could be measured in a lab?
 
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  • #19
vanhees71
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The upshot is: One part of the field is the Coulomb field "bound" to the particle. It's the electrostatic field it has around it when it is observed at rest in an inertial reference frame. If the charge is moving uniformly, i.e., with constant velocity wrt. an inertial frame, you still have only this field, but you observe both electric and magnetic field components, because of the transformation behavior of the electromagnetic field under Lorentz transformations, which describe how to switch from one inertial frame to another, which is moving with constant velocity against the former.

If now the particle is accelerated, e.g., due to some external electromagnetic field, additional to the (boosted) Coulomb field you also have an electromagnetic radiation field, which transports energy, momentum, and angular momentum to infinity, and this has of course an effect on the motion of the particle. Take the simple case of a particle in an external homogeneous magnetic field. Neglecting the radiation you can solve the equation of motion and get a circle as a trajectory the particle moves along with constant angular velocity (the "cyclotron frequenzy ##\omega=\sqrt{1-\beta^2} qB/m##, where ##\beta=v/c=\text{const}##, and kinetic energy is conserved. However, the particle is obviously accelerated, because the direction of the velocity changes (though its speed stays constant). So the particle radiates electromagnetic waves, and this carries away some of the particle's kinetic energy in form of radiation, which in this case is called cyclotron radiation. That means the particle looses the corresponding amount of kinetic energy irradiated by this radiation. Effectively that has the effect of the "radiation damping". It's similar to friction, only that here the energy is lost to radiation rather than to heat of a medium the particle is moving in.

To take the radiation damping into account you have to calculate the force on the particle due to the interaction with its own electromagnetic field. Doing this naively you get a severe divergence. If you try to calculate this force already for a particle at rest, i.e., its Coulomb field, you hit the corresponding singularity of this field at the position of the particle, but this is of just a physically wrong description, because a particle at rest in an inertial frame should stay at rest, no matter whether it is charged and thus has a Coulomb field around it or not. So you have to subtract this diverging force from its own Coulomb field acting on the particle. This you can do after regularizing the Coulomb field somehow to make it finite at the place of the particle. One strategy already followed by Abraham and Lorentz is physically motivated: You just make the point particle an extended particle. Then the Coulomb field is finite everywhere. As Poincare has stressed first, of course you need some stresses holding the particle together against the repulsive Coulomb force between its parts. Then you can work out an equation of motion, which is however very complicated. You get not a nice local equation of motion but a differential-difference equation due to the retardation effect, i.e., the fact that the electromagnetic interaction propagates at the finite speed of light and not instantaneously. So it takes some tiny time that one part of the extended particle can react to changes of other parts. This retardation effect goes of course away in the limit where you make the extension 0 again, i.e., taking the point-particle limit. One part is of course still diverging, but it can be shown that it is effectively just a contribution to the (invariant) mass of the particle (i.e., the inertia due to the energy contained in the particle's Coulomb field, which of course goes to infinity in the point-particle limit). On the other hand, you cannot observe a charged particle without its Coulomb field, i.e., the infinite contribution to the mass must be lumped together with the unobservable "bare mass" of the particle, together leading to the finite physical mass. After this "mass renormalization" you get an equation of motion, the Lorentz-Abraham-Dirac equation (LAD equation), which is formally finite but pretty strange, because it contains a third-order time derivative of the position, and this causes a lot of trouble like "runaway solutions", where a particle at rest in an inertial frame gets rapidly accelerated "out of nothing", which is of course unphysical, because the particle should stay at rest forever. Another bad feature is "pre-acceleration", i.e., when you switch on an external force, starting accelerating the particle, even when having dropped the unphysical runaway solutions, it turns out that in fact the particle starts to accelerate earlier than the force is switched on. This is clearly violating causality. The best one can finally do is to use the socalled Landau-Lifshitz approximation to the LAD equation, which avoids the pre-acceleration as well as the runaway solutions and is of 2nd order in time as a usual mechanical equation of motion should be and it takes into account the radiation reaction.

The same result you can get much easier with the regularization used by Lechner et al used in the papers quoted above and my SRT FAQ article.

This is of course not a full self-consistent description of the dynamics of a point particle and the electromagnetic field, and so far nobody has come up with a solution for that problem, and it is pretty likely that there is no consistent description, because the notion of a point particle is unphysical. Quantum theory teaches us that you cannot localize a particle definitely, and thus the concept of a point particle doesn't make sense. If you turn to quantum theory, i.e., quantize both the particle and the electromagnetic field, an alternative strategy to derive a classical equation of motion for a "point particle" is to analyze the problem from the point of view of open quantum systems, i.e., you consider the particle and the quantized em. field and consider only the particle. This leads to a non-Markovian quantum Langevin equation for the motion of the particle's average position and momentum, which can be further approximated again by the Landau-Lifshitz approximation of the LAD equation. Of course also in the quantum case you have to renormalize the self-energy of the charged particle as in the classical treatment. For this treatment (though only in the non-relativistic limit), see

G. W. Ford, J. T. Lewis and R. F. O’Connell, Quantum
Langevin equation, Phys. Rev. A 37, 4419 (1988),
https://doi.org/10.1103/PhysRevA.37.4419
 
  • #20
tech99
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I can see that the mathematics are really difficult, so how about the experiment? It is very difficult to obtain sufficient acceleration over a significant distance using a purely mechanical system. But we can do it by using electrons alone, and this was the basis of Hertz's discovery of radio waves. If we take a copper rod which is 1 metre long, it contains a dense cloud of charges in the form of free electrons. The charge is very large, thousands of Coulombs. If we cut the rod in the centre and apply a very high frequency alternating potential, we can cause intense acceleration of this very large charge (although the displacement is very small). An alternating current flows into the rods, made possible by the capacitance between them. We observe an opposition to the acceleration - inertia - in the form of electrical resistance. For example, if we use a frequency of 100 MHz, the resistance amounts to approx 30 Ohms, and is called the Radiation Resistance.
 
  • #21
snoopies622
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Kind of an epilogue - I just noticed an odd fact:

If i let the position of the oscillating charged particle be [itex] x=Ae^{i \omega t} [/itex] and take the time derivative three times, then use [itex] \ddot{x} [/itex] for the acceleration term in the Larmor power formula and [itex] \dddot{x} [/itex] and [itex] \dot{x} [/itex] for [itex] P = \vec F \cdot \vec v [/itex] (using this jerk for the Abraham-Lorentz force term), i get exactly the same function for the power emitted [tex] P = \frac {q^2}{6 \pi \epsilon_0 c^3} \omega^4 A^2 e^{2i \omega t} [/tex] which means that they're not only the same magnitude (not the surprising part) but also in phase with one another. This part is strange to me since the particle's velocity, acceleration, and jerk are all [itex] \textbf {out of} [/itex] phase with one another. I see how mathematically this arises from the use of complex numbers to represent wave functions - for instance, if [itex]\theta = 0 [/itex] then the product of [itex] \dddot{x} [/itex] and [itex] \dot{x} [/itex] is a vector that's also in the [itex]\theta = 0 [/itex] position even though at that moment velocity is at [itex]\theta = (1/2) \pi [/itex] and jerk is at [itex]\theta = (3/2) \pi [/itex]. And yet physically it doesn't make sense for [itex] P = \vec F \cdot \vec v [/itex] to be non-zero at a moment when both jerk and velocity are zero physically - that is, when the real component of their complex representations are both zero.

I'm not sure what to make of this.
 
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  • #22
vanhees71
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Be careful! You can stick with the complex quantities only as long as you consider the linear equations of motion. To evaluate the power, where you mutliply quantities you have to take the physical real parts first!
 
  • #23
snoopies622
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Thanks vanhees71! I was thinking of this question in my sleep and it seemed to me that the only sane solution is indeed to leave out the complex numbers altogether.

So if instead i let x=Acos(wt) and take the time derivatives of that, i get [itex] \ddot{x}^2 =A^2 \omega^4 cos^2(\omega t) [/itex] and [itex] \dddot{x} (\dot{x}) = -A^2 \omega^4 sin^2(\omega t) [/itex], and so the only difference between using the Larmor power formula and the Abraham Lorentz approach is a phase difference of pi/2.
 

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