Tell me more ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.
up vote 1 down vote favorite

Electromagnetism implies special relativity and then the universal constant "c". And if we set c=1, the coupling constant has units of angular momentum (so in relativistic quantum mechanics we divide by $\hbar$ and we get the adimensional coupling $\alpha$).

Question, loose, is: In which explicit ways does this angular momentum appear in the classical formalism? Has it some obvious, useful meaning?

Edit: some clarifications from the comments.

More explicitly, I have in mind the following. In classical no relativistic mechanics a circular orbit under, ahem, a central force equilibrates $F = K / r^2$ equal to $F= m v^2 / r$, and then we have $K = m r v^2 = L v.$ Thus when introducing relativity we can expect that the angular momentum for a particle orbiting in a central force will have a limit, the minimum possible value being $L_{min} = K/c$. Note that this limit does not imply a minimum radius, we also have classically $K = L^2/ mr $, but m can be argued to be the relativistic mass, so when L goes towards its minimum, m increases and the radius of the orbit goes to zero.

More edit: Given that it seems that my derivation of the Sommerfeld bound $L_{min} = K/c$ risks to be wrong, I feel I should point out that failure, it it is, is completely mine. The original derivation of this relativistic (not quantum!) bound appears in section II.1 of Zur Quantentheorie der Spektrallinien (pages 45-47 here) and also in Kap 5.2 of his book. The usual argument goes about generic ellipses and its stability properties.

share|improve this question
1  
No, electromagnetism does not imply special relativity by itself. The two postulates are independent really. Sure, Maxwell's equations are Lorentz invariant (generally consistent with SR), but that's it. – Noldorin Jan 19 '11 at 18:37
I'm not quite sure what you are looking for here. Surely you must know that angular momentum appears in classical physics as a consequence of Noethers theorem. The rotational invariance of the system under consideration (or more generally flat spacetime) implies the existence of a conserved quantity.. – Columbia Jan 19 '11 at 18:39
1  
Dear Alejandro, electromagnetism doesn't imply special relativity, as Noldorin says. Also, it's true that the coupling constant has units of angular momentum, but one can also say that the coupling constant has units of action or anything else whose units are $\hbar^n$. The fact that two quantities have the same units doesn't imply that they're the same thing. Especially once you set $c=1$, all quantities have units of $\hbar^M G^N$ where $M,N$ are usually integer. So there are only $5^2$ or so different units - but millions of concepts in the world. Your attempt is a meaningless numerology. – Luboš Motl Jan 19 '11 at 18:52
Dear Lubos et al, I had in mind the example of Sommerfeld definition of the fine structure constant. He first tries to use a mix of special relativity and electromagnetism to get closed orbits around the hidrogen atom. He gets some elipses with a bit of rotation themselves, makes a bit of magic, and concludes that there is a natural limit in the angular momentum of an orbit.The he (not me) defines alpha as a quotient between this relativistic angular momentum of electromagnetism and the planck constant. – arivero Jan 19 '11 at 22:19
Good point by Lubos. Dimensional analysis can be useful, but can only tell us so much. It should not "guide" theory, certainly, but should instead be used as a "check". – Noldorin Jan 19 '11 at 23:26
show 4 more comments

2 Answers

up vote 1 down vote
+50

I think R. Ohanian gives the answer to the title question.

share|improve this answer
1  
I guess that you refer to section 2, but could you do a small abstract of the content??? Also, full reference is better: Am J Phys 54, p 500-505 – arivero Jan 19 '11 at 23:27
Yes, my own "abstract" is the following: if you take a multicomponent field (a solution of a wave equation), it is not surprising that it describes a particle with a spin. It follows from the equations, if they are known. And this spin (angular momentum) looks as connected with the energy flux around z-axis. But the spin value is entirely determined with the number of filed components (spinor range) rather than with a particular energy flux value (the latter disappears after integration and using normalizing condition), so I am skeptic about Ohanian's "explanation" about what determines what. – Vladimir Kalitvianski Jan 19 '11 at 23:46
Of course, a spinor describes a particle with a spin; it is just a tautology or definition, not an explanation or a proof. – Vladimir Kalitvianski Jan 19 '11 at 23:47
It seems to me that Ohanian's article does the job. – Carl Brannen Jan 20 '11 at 6:21
It does a lot of the job, and I had accepted the answer if a self-containing explanation had been provided, instead of a pointer. Still, one must be careful about using concept from quantum mechanics, the point was to see where can we go only with maxwell+classical relativistic mechanics. – arivero Jan 20 '11 at 14:22
show 3 more comments
up vote 1 down vote

In SR with a central force (not gravity) you get from Born coordinates, or substitution of classicla experssion $v->gamma*v$
$F=k/r^2=m*(gamma*v)^2/r=L*v*gamma^2/r^2$

With c=1 this gives, $L=k/(v*gamma^2)=k*(1-v^2)/v$

And the limit of that as v->1 gives L=0 as expected.

share|improve this answer
At least a +1 for addressing some part of the question :-) But so your point is that in L= m r v, the velocity and mass are to be taken as no relativistic? – arivero Jan 27 '11 at 3:12
@arivero L=gamma*m*r*v gives same result, which is what is conserved in SR. – Holowitz Jan 27 '11 at 8:19
Well, but which is your take? We have that a version L=F/v allows to recover Sommerfeld bound, and other two versions fail to do it. Probably my "autobahn" to the bound is wrong and one must to do the analysis of trayectory stability as usual, but I would like to know why it is wrong, and which is the right version. – arivero Jan 27 '11 at 10:32
I have put a pointed to the usual statement of the bound F/c, just for reference. It is true that the derivation is not as simple as I take, but I thought it was due to its generality (not circular traj.) – arivero Jan 27 '11 at 10:43
Im not really sure what the question is, but if you introduce parts of SR into an argument, you have to introduce all its implications, meaning the entire theory, SR is not a set of independent statements that you can mix into CM, what I have showed is simply that there is no lower bound on either mrv or mrv*gamma in SR for circular orbits in k/r^2 force, but there is a minimum value of mrv*gamma^2. – Holowitz Jan 27 '11 at 11:09
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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