California Institute of Technology
Question
Asked 12th Jun, 2022
How do we justify replacing total potential with external potential in the Langrangian for electromagnetic fields when the sources are point charges?
For those that have the seventh printing of Goldstein's "Classical Mechanics" so I don't have to write any equations here. The Lagrangian for electromagnetic fields (expressed in terms of scalar and vector potentials) for a given charge density and current density that creates the fields is the spatial volume integral of the Lagrangian density listed in Goldstein's book as Eq. (11-65) (page 366 in my edition of the book). Goldstein then considers the case (page 369 in my edition of the book) in which the charges and currents are carried by point charges. The charge density (for example) is taken to be a Dirac delta function of the spatial coordinates. This is utilized in the evaluation of one of the integrals used to construct the Lagrangian. This integral is the spatial volume integral of charge density multiplied by the scalar potential. What is giving me trouble is as follows.
In the discussion below, a "particle" refers to an object that is small in some sense but has a greater-than-zero size. It becomes a point as a limiting case as the size shrinks to zero. In order for the charge density of a particle, regardless of how small the particle is, to be represented by a delta function in the volume integral of charge density multiplied by potential, it is necessary for the potential to be nearly constant over distances equal to the particle size. This is true (when the particle is sufficiently small) for external potentials evaluated at the location of the particle of interest, where the external potential as seen by the particle of interest is defined to be the potential created by all particles except the particle of interest. However, total potential, which includes the potential created by the particle of interest, is not slowly varying over the dimensions of the particle of interest regardless of how small the particle is. The charge density cannot be represented by a delta function in the integral of charge density times potential, when the potential is total potential, regardless of how small the particle is. If we imagine the particles to be charged marbles (greater than zero size and having finite charge densities) the potential that should be multiplying the charge density in the integral is total potential. As the marble size shrinks to zero the potential is still total potential and the marble charge density cannot be represented by a delta function. Yet textbooks do use this representation, as if the potential is external potential instead of total potential. How do we justify replacing total potential with external potential in this integral?
I won't be surprised if the answers get into the issues of self forces (the forces producing the recoil of a particle from its own emitted electromagnetic radiation). I am happy with using the simple textbook approach and ignoring self forces if some justification can be given for replacing total potential with external potential. But without that justification being given, I don't see how the textbooks reach the conclusions they reach with or without self forces being ignored.
Most recent answer
A revision with a more appropriate title is attached. The Conclusion section is specific about the difference between what is in this report and what is in at least some popular textbooks.
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All Answers (18)
University of Tours
By recognizing that they’re two distinct problems. The only property they have in common is the coupling between the gauge potential and matter, through the current-that's the only thing that really matters, since electrodynamics is,
essentially, uniquely defined by global Lorentz invariance, U(1) gauge invariance and second order equations of motion.
If the gauge potential describes an electromagnetic field, whose source isn't the current, that's the problem of the motion of charged particles in an external electromagnetic field.
If the gauge potential describes an electromagnetic field, whose source is the current, that's another problem. That involves solving Maxwell's equations with given sources and leads to computing the fields from the so-called Liénard-Wiechert potentials. This involves imposing a gauge condition and showing that the fields in fact don't depend on the gauge condition.
Then there's the problem where one solves the coupled equations. This is presented in the article https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.rand.org/content/dam/rand/pubs/research_memoranda/2006/RM2820.pdf&ved=2ahUKEwiJ6vrXoar4AhV6R_EDHRRWCeUQFnoECA8QAQ&usg=AOvVaw3-kIrUgCaLXYX7ig8eMs0N
``Classical electron theory from a modern standpoint", by S. Coleman.
It is in this last case that the self-force becomes relevant; but that doesn't matter, since the quantities that do matter are the particle trajectory and the electromagnetic field. Here, too, a gauge condition must be imposed and it is shown that the trajectory or the fields don't depend on it.
California Institute of Technology
Thank you Stam Nicolis but I haven't found the answer to my question. Let me rephrase the question because I think the answer should be much simpler than the theory in the link you provided.
The particle is a marble with a uniform charge density in its interior. I am happy with approximations that ignore the self force. The problem is that, even without a self force, I still don't see how the marble charge density can be represented by a delta function (even though it is a very small marble) in an integral of the charge density multiplied by total potential (total potential includes that created by the marble and cannot be approximated as constant in the vicinity of the marble). Also, regardless of what is assumed to be the force acting on the marble, the potential used in the Lagrangian for the field must be total potential in order to correctly produce Maxwell's equations. Therefore, even with marbles and ignoring the self force, I still don't see how the charge density can be represented by a delta function in the integral, a part of the field Lagrangian, of charge density times potential.
University of Tours
The charge density of a point is a delta function; the charge density of a marble, of finite radius isn’t. This doesn’t change the fact that the coupling of the 4-current with the vector potential is given by J_μ A_ν η^{μν}.
In the rest frame of the marble, the 4-current J_μ=(ρ,0) and ρ=ρ0, if 0<=r<Rmarble and 0 if r>Rmarble. If the marble is a point (or can be taken to be a point) Rmarble=0, then ρ=qδ(x). Here q = integral over the volume of the marble.
In any event, for a point charge, q, Jμ(x)=q(dxμ/dλ), where λ parametrizes the worldline of the point charge. This is the only term one can write that can combine with Aμ to give a term consistent with (a) Lorentz invariance and (b) gauge invariance.
The worldline remains invariant under reparametrizations, so, in fact, one does need to impose an additional gauge fixing condition. One such condition is to take as parameter λ the coordinate time, so dxμ/dλ=(1,v).
So, if the marble is at rest, one solves Maxwell's equations for the fields, produced by this source. If the marble moves with uniform velocity, it suffices to perform a Lorentz transformation on the fields, since the statement that the marble moves with uniform velocity means that it's related to the marble at rest by a Lorentz transformation.
If the marble moves with nonuniform velocity, then there doesn't exist a globally defined Lorentz transformation to the marble at rest. But the solution is known and presented in all the textbooks. The delta functions just express the fact that the motion of the source, be it point or marble, is known, once and for all and isn't affected by the fields the source produces.
California Institute of Technology
I figured out the answer. The marble charge density can be represented by a delta function in an integral of charge density times external potential (the potential created by all marbles except the one under discussion) if the marble dimensions are much smaller than the spacing between marbles, because the external potential is nearly constant in the marble interior. I couldn't figure out why it is okay to replace total potential in the integral with external potential in order to use the delta function representation. The answer is as follows. The total integral is the integral of marble charge density times external potential plus the integral of marble charge density times marble potential. If the total integral is used to construct a Lagrangian for particles, translational invariance makes the integral of marble charge density with marble potential a constant. If the total integral is used to construct a Lagrangian for fields, the integral of marble charge density with marble potential is constant when varying total fields with the marble field held fixed in this variation. So whether the integral is used to construct a Lagrangian for particles or a Lagrangian for fields, replacing total potential with external potential simply changes the Lagrangian by an additive constant which has no effect on the implied equations for either fields or particle motion.
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University of Tours
No, the potential is the output of the calculation, not the input and no assumption that it is constant needs to be made.
University of Tours
The only assumptions required are Lorentz invariance and gauge invariance. The delta function just imposes constraints, nothing more or less.
California Institute of Technology
Sorry Stam Nicolis but I don't think that you ever understood the question. Your replies have been factually correct but not answers to the question. My answer is an answer to the question. You would see that if you understood the question.
University of Tours
Once more: The Lagrangian, that describes the interaction of ANY system with the electromagnetic field is given by the expression JμAνημν.
Here Aν is the 4-vector potential of the electromagnetic field-that is defined by Fμν=dμAν-dνΑμ (here d=partial derivative...) and Jμ is the current of the matter fields, whether these are point charges, or marbles.
If the potential is ``external'', this, just, means that it's a known, fixed, function of the coordinates, as is the corresponding electromagnetic field. It, obviously, doesn't mean that the potential is constant!
For point charges Jμ=qdxμ/dt=q(1,v) (in ``proper time'' gauge) and the dynamics of the point charge is defined by its Lagrangian, -m sqrt(1-v.v).This reduces to the expression (m/2)v.v in the non-relativistic limit, ||v||<<1.
(The velocity is expressed in units where the speed of light c=1.)
So the equation of motion is md2xμ/dt2=qFμν dxν/dt and that's the equation to be solved. The mass is defined by the constraint pμpνημν=m2 .
In the non-relativistic limit this reduces to solving
mdv/dt = qE+q v x B
For more than one particles some care is necessary in order to avoid trivial errors.
For marbles the equation of motion is a bit more complicated, obviously, but not by much.
If the electromagnetic field isn't external, its Lagrangian is -(1/4)F.F-(1/2)(d.A)2 in Lorenz-Feynman gauge and its dynamics is defined by Βοχ Αν=Jν, where Box is the d'Alembertian operator. So the equations are now coupled.
The way to solve the coupled equations is presented in Coleman's article.
University of Tours
In the rest frame of the point charge, ρ(t,x)=qδ(x) and J(t,x)=0. What's the problem with that?
If you put that charge in an external electromagnetic field, defined by the fields E(t,x) and B(t,x), it's going to move-its motion will be described by the solution to the equation of motion
mdv/dt=qE+qvxB
with v =dx/dt and initial conditions x(0)=0 and v(0)=0.
On the other hand, if you take the point charge, in its rest frame, as a source of an electromagnetic field, you'll find the Coulomb potential, the radial electric field and zero magnetic field.
California Institute of Technology
Most of what you say Stam Nicolis is bloviation, again indicating that you do not understand the question. You also seem to be confused as to what is constant.
Suppose the integral being discussed is used to construct a Lagrangian for particle motion in a given fixed external field. The Lagrangian that I construct per the above discussions in this thread produces the correct equations of motion for particles in a given external field when self forces can be neglected, and that is the bottom-line requirement for the Lagrangian for particles when self forces can be neglected.
Now suppose the integral being discussed is used to construct a Lagrangian for the fields. The appropriate variational principle, when total fields seen by the marble are replaced by external fields, gives independent variations to the total scalar potential and each component of the total vector potential while holding the self fields (fields created by the marble of interest) fixed. In this variational method, the external potentials are taken to be the total potentials (the unknowns to be solved) minus the fixed known self potentials. Contrary to your interpretation, external potentials are not fixed known functions. Total potentials are the quantities to be solved with only the self potentials fixed known functions.
If I may be permitted to bloviate, it is interesting that the scalar potential and each component of the vector potential are given independent variations even if a gauge condition was selected. A gauge condition can actually be imposed after deriving Lagrange's equations that give all vector components independent variations. This works because a requirement of any allowed gauge condition is that it does not disturb the four inhomogeneous equations for the four components of the potential.
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California Institute of Technology
I am writing a report to explain the answer in detail and will post it when done. As a heads-up for what is coming, Stam Nicolis wrote
"In the rest frame of the point charge, ρ(t,x)=qδ(x) and J(t,x)=0. What's the problem with that? "
The problem with that is when the charge density is in an integral where it multiplies the total scalar potential that has a singularity at the same location as the charge density. Also, if we avoid infinities by making the particle size greater than zero, there are other reasons already explained as to why the charge density cannot be represented by a delta function in that integral regardless of how small the particle is. This is what the entire discussion was suppose to be about. I'm sure this issue was resolved more than a hundred years ago but it was easier to figure out the answer than to search the literature for the answer. Details of the answer will be posted soon.
University of Tours
There’w nothing in the paper that’s new and it doesn’t address the self-field problem, that’s solved in Coleman’s paper.
University of Tours
Either the potentials-and the fields deduced from them-are given, and the equations to be solved are those for the particle’s trajectory; or the the particle’s motion is given and the fields due to this motion are to be found-these are solutions to Maxwell’s equations; or the coupled system is to be solved. The first two problems are in any textbook (that includes the variational principle, that relies only on gauge invariance, Lorentz invariance and the requirement that the equations are second order) and the last is the subject of Coleman’s article.
California Institute of Technology
A reply to Stam Nicolis who said
"There’w nothing in the paper that’s new ..."
There is nothing new except an explanation of how to remove an infinity. This might not be new to some people but it was new to me.
The explanation replaces the field Lagrangian given by Eq.(4.2) with the one given by Eq.(4.3). You claim that the paper has nothing new but I never saw this replacement in anything that I studied. I can't be the first person to have invented this but it was new to me. This replacement also produces better conformity with the Lagrangian for particle mechanics.
In spite of your criticisms and complaints, I am happy with this answer to the question of how to justify replacing potential with external potential. It works for me. The only purpose of the paper was to answer that question. It wasn't intended to have anything else that is new. You can continue to look for things to complain about but my reply will continue to be "It works for me."
California Institute of Technology
However, the title of the paper gives the misleading impression that it covers more theory than just an answer to the question asked in this thread. The title should have been something more descriptive of the fact that the only purpose of the paper is to answer that question. I apologize for not selecting a more appropriate title.
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