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Is the Abraham-Lorents force formula in wikipedia correct?

  1. Feb 5, 2009 #1
    Hello, im 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?
  2. jcsd
  3. Feb 6, 2009 #2


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    Your points 1 and 2 are correct. The formula, although widely used (often with paradoxical results), is nonsense
  4. Feb 6, 2009 #3


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    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.
  5. Feb 24, 2009 #4
    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.
  6. Feb 24, 2009 #5


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    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 [itex]a(t_1) = a(t_2) = 0[/itex].

    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.
  7. Feb 24, 2009 #6
    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 :smile:
    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?
    Last edited: Feb 24, 2009
  8. Feb 24, 2009 #7
    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):
    [tex] S = C \frac{q^2 a^2 \sin^2 \theta}{r^2} \ \hat{r} [/tex]
    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 ([itex]z \rightarrow -z[/itex]). 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).

    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:
  9. Feb 25, 2009 #8
    Well all this about the radiation direction etc. is very good but i dont 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!
    Last edited: Feb 25, 2009
  10. Feb 25, 2009 #9


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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 "[itex]mv^2/2[/itex]." 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 [itex]-m_{\mathrm{eff}} a[/itex] 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.
  11. Feb 25, 2009 #10


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    Now, I'll give you some flavor for the full relativistic treatment. The energy of a system with stress-energy tensor [itex]T^{ab}[/itex] in a 3-volume [itex]\Sigma[/itex] with respect to a time direction [itex]t^a[/itex] is
    E(\Sigma) = \int_\Sigma T^{a}{}_{b} t^b \mathrm{d} S_a.
    The general conservation law is locally [itex]\nabla_a T^{ab} = 0[/itex]. Energy conservation specifically means that
    \oint_H T^{a}{}_{b} t^b \mathrm{d} S_a = 0
    for any closed 3-surface [itex]H[/itex]. 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 [itex]T^{ab}_{\mathrm{B}}[/itex]. With enough work, you could figure out precisely how the transfer balance works between field and mechanical energies.
  12. Feb 25, 2009 #11
    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.

    I don't like the preposition "from the rope" there. But yes, the energy was supplied by whatever accelerated the particle.

    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.
  13. Feb 25, 2009 #12
    Thanx for your patience! :smile:
    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?
  14. Feb 25, 2009 #13
    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!
  15. Feb 25, 2009 #14
    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.
  16. Feb 26, 2009 #15


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    Nope. It looks right.
  17. Feb 27, 2009 #16
    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.
  18. Feb 27, 2009 #17


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    Last edited: Feb 27, 2009
  19. Feb 27, 2009 #18


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    I'm not sure what you mean. I was agreeing with what you'd said.
  20. Feb 27, 2009 #19
    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.
  21. Feb 27, 2009 #20


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    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).
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