Is the Abraham-Lorents force formula in wikipedia correct?

• lukcho
In summary: Not just calculate the radiation reaction. The Abraham-Lorentz equation breaks that down a bit more and it allows you to calculate the reaction without knowing the total energy. It's just a function of the acceleration. This makes the Abraham-Lorentz equation more general, I think.
lukcho
Hello, I am new to the these forums.

I recently read the article in the Wikipedia about this force (Abraham-Lorentz) and I can't understand how it can only depent on the "jerk" (that is the acceleration's derivative).
Maybe i miss something, but this seems quite impossible to me:
What happens if the charge is undergoing a constant acceleratoin?
In this case the "jerk" will be zero and, according to the given there formula, so will be the reaction force.
But still, according to the Larmor formula, the charge will be emitting energy.
So where does that emitted energy come from if there is no reaction force?
This sounds like a way to construct a perpetual motion machine of the first type, doesn't it?

As i inspected the given derivation of that formula, i noticed 2 problems with it:
1) it makes the arbitrary assumption that the motion is periodic and heavily relies on it.
2) at the end the derivation basically says that if the definite integrals of two periodic functions over one period are equal then the functions are one and the same, which is nonsense. So this makes the given "derivation" totally invalid.

What do you people think?

Your points 1 and 2 are correct. The formula, although widely used (often with paradoxical results), is nonsense

There are good derivations of that force (and a lot of bad ones). The end result is that it's pretty much correct. You do get nonsense out if you don't pay attention to when the equation is valid. Otherwise, it's ok.

While it's odd that there's no reaction force for constant acceleration, energy is actually conserved. Note that when you start and stop accelerating, there's a nonzero jerk. Those endpoints exactly compensate for the energy radiated during acceleration. You really have to pay attention to where the field energy is going too. That's how I remember it, at least.

Stingray said:
While it's odd that there's no reaction force for constant acceleration, energy is actually conserved. Note that when you start and stop accelerating, there's a nonzero jerk. Those endpoints exactly compensate for the energy radiated during acceleration. You really have to pay attention to where the field energy is going too. That's how I remember it, at least.

Maybe the endpoints could account for the _total_ energy during some big time period.
But what happens in the mean time?
Note that the period of the constant acceleration could be arbitrary long.
We can't just say that during the constant acceleration we borrow energy from the "nothing" and when we stop with the constant acceleration the "nothing" takes it back from us. This is just not serious.
We should be able to account for the energy all the time. What happens to it, how does it come and go.

In the non-relativistic case, the work done on the particle by the radiation reaction force over some finite time is (integrating by parts)

$$W = \frac{2}{3} q^2 \int_{t_1}^{t_2} \dot{a} \cdot v \mathrm{d}t = \frac{2}{3} q^2 \left[ (a \cdot v)|^{t_2}_{t_1} - \int_{t_1}^{t_2} a^2 \mathrm{d}t \right].$$

The second term agrees with the Larmor formula. The first vanishes if the work is calculated over a sufficiently long time that $a(t_1) = a(t_2) = 0$.

As you said, that's not very satisfying. There shouldn't be transient violations of energy conservation. From what I recall, the solution lies in noticing that the work calculated above finds the change only in a particle's mechanical kinetic energy. It does not take into account changes in the energy of "bound" fields that cannot be interpreted as radiation. This is a little tricky, as some of the bound field is already present as an effective mass.

Stingray said:
From what I recall, the solution lies in noticing that the work calculated above finds the change only in a particle's mechanical kinetic energy. It does not take into account changes in the energy of "bound" fields that cannot be interpreted as radiation. This is a little tricky, as some of the bound field is already present as an effective mass.

Hmm, do you mean that, informally speaking, the two formulae [Larmor and the Abraham-Lorentz] speak in "different languages". For example one of them works with the total energy whereas the other works only with the mechanical energy or something like that?
This could possibly explain it. I can not agree or disagree with this unless you give me exact formulas and conditions
If this is the case then the problem lies in the way the formulae are presented - that the terms in which they operate and the areas of their applicability are not correctly and strictly specified and this leads to confusion.
Is this so?

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Step back and think about this intuitively first. The Larmor equation as normally presented just tells the total energy radiated. If you want the reaction to this, you need to know more than just the total energy in the radiation, you need to know the direction of all the radiation.

So go one step back in the calculation of the total energy radiated. You'll find something like (in spherical coordinates):
$$S = C \frac{q^2 a^2 \sin^2 \theta}{r^2} \ \hat{r}$$
Where S is the Poynting vector, C is just some constant, q is the charge of the particle, a is the acceleration.

Notice that the radiation is symmetric around the particle's path, and also invariant to a change in direction ($z \rightarrow -z$). So, despite radiating energy, there is no net reaction force on a charge of constant acceleration.

So, it should make (at least some) intuitive sense that the reaction force can't depend on position, velocity, or acceleration ... but on the change in acceleration (and possibly higher order terms).

clem said:
Your points 1 and 2 are correct. The formula, although widely used (often with paradoxical results), is nonsense
I feel that is an incredibly unfair over-simplification. The equation is not non-sense.

A more appropriate response is probably along the lines of what Stingray said:
Stingray said:
There are good derivations of that force (and a lot of bad ones). The end result is that it's pretty much correct. You do get nonsense out if you don't pay attention to when the equation is valid. Otherwise, it's ok.

Well all this about the radiation direction etc. is very good but i don't see how it is related to my question - where does the emitted energy come from.
Whatever the radiation direction is, the energy preservation law can not be violated.

Let me restate my question again.

Imagine some (macroscopic) ball with net charge.
Imagine that it is attached to some rope and the rope is pulling it so that the ball is in a state of a constant acceleration. Because of the acceleration and the charge, the ball is emitting electromagnethic energy.

Now i will make statements, one by one. Please thell me which one is wrong:

1) Since the emitted energy is proportional to the square of the acceleration and the acceleration is caused by the rope, then the emitted energy should be comming from the rope. Is this correct?

2) Since the derivative of the acceleration is zero, then the reaction force caused by the emitted electromagnethic energy is zero. According to you this is correct.

3) from 2) i conclude that the only reaction force that the ball is applying to the rope is it's inertia, that is its mass myltiplied by the acceleration. Is this right?

4) So from the point of view of the rope the ball's charge is invisible. The ball behaviour is the same regardles of whether it is charged or not - only it's mass matters.

5) from 4) i conclude that a charged ball and an uncharged ball with the same masses should "extract" the same energy (per second) from the rope's pulling. Is this right?

6) The uncharged ball converts all the "extracted" energy to it's own kinetic energy (m*v^2)/2.

7) Since the charged ball has the same mass as the uncharged, the amount of the extracted energy that goes to it's own kinetic energy should be the same.

8) then since the extracted energy in both cases is the same, there is nothing left to be radiated in the charged case.

So where does the radiated energy come from!

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Unfortunately, a thorough answer to your question takes some work. It also requires a fully relativistic formulation to do properly and concisely. Are you familiar stress-energy conservation and other aspects of relativistic mechanics? First I'll give a few brief comments about some of your points:

There is a lot more energy in this system than just radiation and "$mv^2/2$." The charge has non-radiating fields associated with it, and their energy must be taken into account as well.

"Most" of the self-force is really contributing an effective inertia to the particle. There is a force looking like $-m_{\mathrm{eff}} a$ that is usually just absorbed into the "bare" mass to give one observable parameter (only the sum of bare and electromagnetic masses is at least approximately conserved). It is also not true that charged and uncharged bodies have the same mass in any useful sense. Even trying to make an appropriate comparison is very tricky. This at least partially because there is a sense in which stresses contribute to inertia.

Now, I'll give you some flavor for the full relativistic treatment. The energy of a system with stress-energy tensor $T^{ab}$ in a 3-volume $\Sigma$ with respect to a time direction $t^a$ is
$$E(\Sigma) = \int_\Sigma T^{a}{}_{b} t^b \mathrm{d} S_a.$$
The general conservation law is locally $\nabla_a T^{ab} = 0$. Energy conservation specifically means that
$$\oint_H T^{a}{}_{b} t^b \mathrm{d} S_a = 0$$
for any closed 3-surface $H$. Equivalently, the energy on two successive time slices is the same if there is no energy flux through the boundary of your region (which cannot be assumed in the presence of radiation).

Now, the local conservation law may be used together with Maxwell's equations to derive the ALD equation. Energy (and linear and angular momentum) conservation is therefore built in, and can only be violated due to small terms neglected in the various approximations. Consider the energy of the body itself (rather than the entire system). This is calculated as above but with a compactly-supported stress-energy tensor $T^{ab}_{\mathrm{B}}$. With enough work, you could figure out precisely how the transfer balance works between field and mechanical energies.

lukcho said:
Well all this about the radiation direction etc. is very good but i don't see how it is related to my question - where does the emitted energy come from.
Whatever the radiation direction is, the energy preservation law can not be violated.

Let me restate my question again.

Imagine some (macroscopic) ball with net charge.
Imagine that it is attached to some rope and the rope is pulling it so that the ball is in a state of a constant acceleration. Because of the acceleration and the charge, the ball is emitting electromagnethic energy.

You seem to have changed the question here, as none of this has to do with the radiation reaction anymore. Where did the energy in the radiation come from? Well, it came from whatever expended energy to accelerate the particle.

lukcho said:
Now i will make statements, one by one. Please thell me which one is wrong:

1) Since the emitted energy is proportional to the square of the acceleration and the acceleration is caused by the rope, then the emitted energy should be comming from the rope. Is this correct?
I don't like the preposition "from the rope" there. But yes, the energy was supplied by whatever accelerated the particle.

lukcho said:
2) Since the derivative of the acceleration is zero, then the reaction force caused by the emitted electromagnethic energy is zero. According to you this is correct.
The derivative of the acceleration is not zero, since the acceleration is constantly changing (since you chose a roughly circular path).

This is like classical descriptions of the electron around a proton ... it will spiral inward.

I'm not sure where you are going with the rest of the questions/statements since this beginning part was wrong, but you seem to be suggesting charge is just 'an extra feature' in addition to mass. Please be careful with this, as Stingray has suggested.

One way to see the difference is to realize that a charge must "carry" its field with it everywhere it goes. So in some sense a charged object is an extended object, while an uncharged mass could just be a "point".

That is just to give you some intuition. If that confuses more than helps, then you'll have to approach it from the math as Stingray is suggesting.

Stingray said:
Unfortunately, a thorough answer to your question takes some work. It also requires a fully relativistic formulation to do properly and concisely. Are you familiar stress-energy conservation and other aspects of relativistic mechanics? First I'll give a few brief comments about some of your points:

There is a lot more energy in this system than just radiation and "$mv^2/2$." The charge has non-radiating fields associated with it, and their energy must be taken into account as well.

"Most" of the self-force is really contributing an effective inertia to the particle. There is a force looking like $-m_{\mathrm{eff}} a$ that is usually just absorbed into the "bare" mass to give one observable parameter (only the sum of bare and electromagnetic masses is at least approximately conserved). It is also not true that charged and uncharged bodies have the same mass in any useful sense. Even trying to make an appropriate comparison is very tricky. This at least partially because there is a sense in which stresses contribute to inertia.

I don't need too extensive and through answer, i just wanted to get some rough (and informal) intuition about where the emitted energy comes from.

If i understand you correctly what you mean, in simpler words, is that the charged ball will appear to the rope as being heavier than the uncharged one (the apparent extra mass comming from the complex interactions between the charged ball and the field it generates around it). Am i right?
Ok, this sounds very satisfactory.

But because the reaction force the rope is experiencing is equal to effective_mass*acceleration, the rope will see this apparent greater mass like a greater reaction force (the acceleration in our setup is always the same). Then we can regard this extra force as being the "radiation reaction force" (it's a matter of interpretation). And it's not zero despite the zero jerk!
Did i miss something here?

JustinLevy said:
The derivative of the acceleration is not zero, since the acceleration is constantly changing (since you chose a roughly circular path).
Hmm, what are you talking about?
In my thought experiment the charged ball was in state of constant acceleration.
This is a matter of experiment setup.
No one was talking about circular paths!

lukcho said:
Hmm, what are you talking about?
In my thought experiment the charged ball was in state of constant acceleration.
This is a matter of experiment setup.
No one was talking about circular paths!
Oops... sorry about that. I thought you had a charge on the end of a rope and you were swinging it around in circles for some reason. Rereading, I'm not sure where I got that specific impression from.

From what you said in the last post, you are starting to get the idea about the charge being a somewhat "extended object" since it needs to 'carry' its field around.

The difficulty with the 'reaction force' is that it is difficult to work with when there isn't some kind of 'pseudo' steady state. Radiation itself is even difficult to define without taking some limit to infinity. (There is even argument in the literature about whether a constant accelerated charge even radiates! Feynman claimed no. In GR it turns out this difficulty leads to radiation being observer dependent.) This is why they usually talk either in terms of conservation laws (so while the details are messy, we can be assured everything works out), or use situations that are constant or cyclical for all time.

As stingray mentioned earlier, you cannot indescriminately apply that reaction force equation. Doing so can lead to paradoxes (run away solutions where a particle 'rides' its own wave, or acausal reactions to an external force).

If things ever get too confusing, definitely fall back on basic intuition and the conservation laws. You are doing work to accelerate the charge. This work must ultimately be what supplies the energy for the kinetic energy in the fields and the particle itself. That being said, if you ever come up with a form of the radiation reaction force that applies in all situations, I and plenty of others would love to see the details.

lukcho said:
Did i miss something here?

Nope. It looks right.

Stingray said:
Nope. It looks right.
Then am i right to conclude that the Abraham-Lorentz formula uses different definition of "reaction force" than the one i mentioned (the force caused by the extra apparent mass' inertia)?

If this is so, then i wonder what that different definition of "reaction force" might be and what is its meaning in terms of practical use.

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lukcho said:
Then am i right to conclude that the Abraham-Lorentz formula uses different definition of "reaction force" than the one i mentioned (the force caused by the extra apparent mass' inertia)?

If this is so, then i wonder what that different definition of "reaction force" might be and what is its meaning in terms of practical use.

I'm not sure what you mean. I was agreeing with what you'd said.

atyy said:
There's a related discussion in Appendix 1 of Stephen Parrott's http://arxiv.org/abs/gr-qc/9303025.

Hmm, when i started this thread i thought that the subject is known and clear to the people from ages as it is classical physics! I just meant to clear the things for myself.
I never imagined there can still be such gray areas in the clasical physics with still standing unanswered questions, lack of consensus and not universally accepted theory, and even paradoxes!
I'm amazed.

Classical physics is certainly not completely understood. Still, there is a lot more that has been worked out than most people appreciate. Many theoretically difficult problems in classical physics were worked out so long ago that most people today are not aware of what has been accomplished (or even that an interesting problem exists).

Stingray said:
I'm not sure what you mean. I was agreeing with what you'd said.

I mean this:
The charged ball looks heavier to the rope than the uncharged ball (because of the extra apparent mass caused by interractions between the charge and it's field - this is according to you).

By "looks heavier" here i mean that the rope must apply greater force in order to maintain the same acceleration.

The ball applies the same force to the rope but in the reverse direction - this is what we call the "reaction force"

We determined that the reaction force of the charged ball is greater than the uncharged.
Consider the difference between the two.
Aren't we right to name it "radiation reaction force" and if not, why.

Now this definition of "radiation reaction force" seems to be the "right" one to me as it sounds very natural, and i can't really think of another meaningful way to define it.
But note that it does not vanish when the acceleration is constant.

So i asked, since the Abraham-Lorentz gives zero result when the acceleration is constant, does it use different definition for the term "radiation reaction force"?

--
In fewer words:
If the apparent (or the effective) mass of the charged ball is not the same as the uncharged one, then it follows that the Abraham-Lorentz formula is wrong unless it uses some very odd definition of the term "radiation reaction"

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ops i think i know where i may be wrong here.
I may have misunderstood what you mean when you say the charged ball has greater apparent mass

Did you mean that this apparent mass actually behaves as a real mass in the sense that it converts the incomming (from the rope) energy to a kinetic energy, just like a real mass?

What i understood was that this apparent mass looks like such only from the POV of the rope, but it could actually convert the energy to radiation.

Ok, in the case that the extra mass caused by the charge actually behaves just like a real mass, then it is irrelevant to my thought experiment - we can just increase the mass of the uncharged ball until the effective masses of the two become equal. And then the other considerations in the experiment will be the same as before.

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I have a basic question about the Larmor power equation which may be appropriate to throw in here. I hope it isn't redundant. Isn't the power going into an accelerating particle equal to the dot product of the force acting on it and its velocity? The Larmor equation has no velocity term. Does this mean that there is a time delay between when power is invested in a charged particle and when it radiates energy? If so, where is the energy during the delay?

Why there should be a delay if the Larmor formula does not depend on the velocity? I don't quite follow you.

Well, the formula predicts that a charged particle with non-zero acceleration should emit radiation even when its velocity is zero. If the velocity is zero than the dot product of the applied force and velocity is zero, so the instantaneous power being put into the particle is zero. So for it to emit energy at a moment when energy is not being invested into it, I thought that perhaps it was emitting energy that had been put into it a moment earlier, when its velocity was not zero.

That was my reasoning, anyway.

Hmm, i see what you mean. This sure is another interesting question.

The energy for radiation is taken from the external force that (linearly) accelerates your charged particle. If you apply simply a uniform electric field, the charge acceleration will not be constant, so the Lorentz friction term is not zero. To have a strictly constant acceleration, you have to apply a tricky external force that "compensates" the friction influence.

The problem with the energy-momentum conservation is not here. The real problem is that the third derivative leads to self-accelerating solutions, that is why it is wrong.

I wrote a quite understandable article on this subject and on another way of taking the radiative losses into account. Yes, it is possible to respect the energy-momentum conservation in the system (charge+radiated field) in another way. I invite you to read my publications on it at arxiv:0811.4416, "Reformulation instead of Renormalizations" and at arxiv:0806.2635, "Atom as a "Dressed" Nucleus" by Vladimir Kalitvianski (available also in the Central European Journal of Physics, Volume 7, N 1, pp. 1-11, 2009).

Bob.

If I accelerate an electrically charged particle from point a to point b with constant acceleration, then according to the Larmor formula, the energy invested in the particle and emitted as radiation will be the square of this acceleration times the time it took the particle to go from a to b, times a constant. If neither the acceleration nor the time for the trip from a to b was zero, this energy must also be non-zero.
JustinLevy said:
Where did the energy in the radiation come from? Well, it came from whatever expended energy to accelerate the particle.
I agree. Since

$$W = \int \bold {F} \cdot d \bold {s}$$

the average force applied to the particle during this trip is the total energy supplied to it divided by the distance from a to b. Since the energy is non-zero and the distance from a to b is finite, this average applied force is also non-zero. Yet
Stingray said:
there's no reaction force for constant acceleration
Using the same charge and path, a higher rate of (constant) acceleration would -- again according to the Larmor formula -- produce a higher radiation power and a larger total energy. And dividing this larger energy by the same distance between points a and b would yield a larger average applied force, which would also not be zero.

What is my error?

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snoopies622 said:
$$W = \int \bold {F} \cdot d \bold {s}$$

w=f.ds
dw/dt=d(f.ds)/dt=f.(ds/dt)+(df/dt).ds?

At the start, the force ramps from 0 to something, so df/dt is not zero.

Another thing to consider is that a charged particle in a conservative field doesn't move the same way as an uncharged particle. If you have an uncharged particle circling the Earth due to gravity, the magnitude of the force on the particle is constant. If you have a charged particle, it will radiate and spiral in as it loses energy, so the magnitude of the force on the particle is not constant.

What is baffling me is not the idea that to move a charged particle with constant acceleration one needs to apply a force that is not constant, but rather the claim - and apparent implication of the Abraham-Lorentz formula -that it requires no force at all (besides f = ma where m is the particle's rest mass).

If I am misunderstanding things and a force is required, I would like to know what it is. Dimensional analysis suggests

$$\vec F = k \mu _0 \big {(} \frac { q a }{c} \big {)} ^2 \big {(} \frac {\vec a }{a} \big {)}$$

as a possibility where k is some number and $$a = | \vec a |$$, but this is purely a guess and not based on the Larmor formula or on anything else. Maybe if there is such a formula, velocity would have to be a part of it, too.

snoopies622 said:
Abraham-Lorentz formula

Maybe the formula is problematic? Parrott suggests there's a problem with the relativistic form of the equation (Appendix 1, http://arxiv.org/abs/gr-qc/9303025). Maybe also try a related discussion by Rohrlich (http://arxiv.org/abs/0804.4614). I've read Parrott's discussion, and it seems reasonable to me, I'm still trying to figure out Rohrlich's.

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Thanks for the references, atyy. I take it that classical EM has no answer to this question?

snoopies622 said:
Thanks for the references, atyy. I take it that classical EM has no answer to this question?

No, at least what is written in textbooks. The AL formula is bad. If at some moment the acceleration is constant, the third derivative makes such a solution unstable. You are bound to obtain a self-accelerating solution.

Mathematically you need three initial constants.

Try to make numerical 1D calculations. You will be obliged to specify the initial acceleration.
If you consider the third derivative by perturbation theory, it means singularly perturbed equation. There is a whole science of it.

Bob.

So if I ask, "I have a particle of rest mass m0 and charge q and I want to accelerate it at rate a, what force must I apply to it?", classical electromagnetism cannot answer this question? To me this seems a little like asking about the self-inductance of a long straight wire.

If

$$V = L \frac {di}{dt}$$

and

$$\frac {di}{dt} = \rho \frac {dv}{dt}$$

where $$\rho$$ is charge per unit length, then there is an electric potential and therefore a reactive force even if the acceleration is constant, though I don't know if this approach is valid when dealing with a single charged particle.

The problem with one charge is different. We have mechanical equations and wave equations. We know from experience how external fields act on the electron. Then we make two suppositions:

1) The radiated field should be incerted into the particle equation,

2) The electron is pointlike.

The latter supposition (electron model) gives the exact "friction" force expressed only via particle variables. Now the question is if this modelling is good physically and mathematically.

The model without friction is even better since the losses are small, so most people use just external forces.

The ideas of self-action and pointlikenes of the electron are bad. Please read my articles on these subjects (arxiv:0811.4416 and arxiv:0806.2635).

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