No. They assume that the electric field propagates instantaneously, while the magnetic field does not. Which is false .Originally posted by STAii
Is it true ?
There are two forms of Newton's third lawOriginally posted by STAii
From this link http://jnaudin.free.fr/lifters/lorentz/index.htm
Is it true ?
PeteIn a system involving moving charges, the forces between charges predicted by the Biot-Savart law may indeed violate both forms of the action and reaction law.*
*(footnote) If two charges are moving uniformly with parallel velocity vectors that are not perpendicular to the line joining the charges, then the net mutual forces are equal and opposite but do not lie along the vector between the charges. Consider, further, two charges moving (instantaneously) so as to "cross the T." i.e. one charge moving directly at the other, which in turn is moving at right angles to the first. Then the second charge exerts a nonvanishing magnetic force on the first, without experiencing any magnetic reaction force at that instant.
Biot-Savart is for stationary currents - not single moving particles. Because there's nothing 'stationary' when you have a single moving particle. There's no such thing as a 'magnetic field of a single moving particle'. Each particle is affected only by the Lorentz-transformed Coulomb field of the other one, and since Lorentz-transformation is symmetrical, so is the interaction. Don't be fooled...Originally posted by pmb
the forces between charges predicted by the Biot-Savart law ...
Originally posted by arcnets
Biot-Savart is for stationary currents - not single moving particles.
Sure there is. Its a fact of relativity. If a particle has a pure electric field in its rest frame then there will be a magnetic field in a frame in which the particle is moving - that's a basic fact of relativistic electrodynamics. Just look at the the relations for how the fields transform. You'll see that this is so.There's no such thing as a 'magnetic field of a single moving particle'.
That is incorrect. E.g. In frame S there is a moving particle with a charge. In S there is both an electric and a magnetic field due to that charge. If there is another charge moving in S then the force on that charge will be F = q[E + vxB] where E and B are generated by the first charge.Each particle is affected only by the Lorentz-transformed Coulomb field of the other one, and since Lorentz-transformation is symmetrical, so is the interaction. Don't be fooled...
The missing momentum is, of course, in the field. I.e. the EM field has momentum so that the total momentum of Fields + Particles is constant.It is worth pointing out that one of Newton's basic assertions about forces between bodies - the equality of action and reaction - has almost no place in relativistic mechanics. It must essentially be a statement about about force acting on two bodies, as a result of the mutual interaction, *at a given instant.* And, because of the relativity of simultaneity, this phrase has no unique meaning unless the points at which the forces are applied are seperated by a neglegible distance. [...] What the relativistic analysis does do, however, is to compel us to conclude that, according to measurements in a given inertial frame, the forces of action and reaction are in general *not* equal and opposite, and so the total momentum of the interacting particles is no conserved instant by instant.
So'kay! Take your time. That's why we're all here! :-)Originally posted by arcnets
I admit I didn't put enough thought into this. I'll check - just don't have the time now, hang on please.
So you therefore would say that the 'missing' momentum remains in the field? Or would you say the field produces a back reaction on the source particle - thus no real asymmetry?Originally posted by pmb
The missing momentum is, of course, in the field. I.e. the EM field has momentum so that the total momentum of Fields + Particles is constant.
I hear ya! People hate it when you disagree with them and will refuse to give up their old ideas. A person that used to post here is flaming me elsewhere. It's a result of his misunderstanding of Einstein's "photon in a box" gerdanken experiment. He shakes at the notion that light has mass! But alas - it's true!Originally posted by Creator
Thanks for the quotes, Pete. I could have used them last year when on another forum I received a lot of flak for stating that... to adopt relativity one has to be willing to give up the concept of Newton's 3rd law.
J.L. Naquin's site (given in the 1st post) has an excellent graphic which illustrates precisely the point made in your referenece in Goldstein's footnote, which show how when two charges approach each other with orthagonal trajectries, one avoids the magnetic reaction force - at least temporarily.
Which brings us to the next point, which he only briefly mentioned, and you brought up succintly.
So you therefore would say that the 'missing' momentum remains in the field? Or would you say the field produces a back reaction on the source particle - thus no real asymmetry?
Originally posted by Hurkyl
There's a problem with equation 1 on your page. mk is not a constant with respect to t; as time changes, particke k's velocity may change, which will thus change particle k's relativistic mass. Thus, you cannot, in general, perform the last step in equation 1.
I have it when that happens! :-DOriginally posted by Hurkyl
lol I read the whole thing twice, and never noticed the text up there before the diagram!
Yes yes, photons have non-vanishing relativistic mass but vanishing rest mass.Originally posted by pmb
People hate it when you disagree with them and will refuse to give up their old ideas. He shakes at the notion that light has mass
In the first I've never considered a photon as something being "pure energy." The notion of anything being pure energy makes no sense to me.Originally posted by jeff
The idea you seem to "hate" giving up is that it's wrong to view massless particles as pure energy.
According to Einstein, a photon of frequency v has energy hv, and thus (as he came to realize several years later) a finite mass hv/c^2 and a finite momentum hv/c.
Why would you think that? IMHO - The notion of relativistic mass is the closest thing there is to what would consider as being mass. And that definition of "mass," when defined rigorously, demands that light has mass. A box containing a gas of photons of total energy U has inertial properties of that of a mass of energy m = U/c^2. So the inertial mass the box, whose mass is M, is that of an object whose mass is (M+m). According to the equivalence principle itg can be shown that the photon gas has a passive gravitational mass. A beam of light generates a graviational field and therefore has an active gravitational mass.
I'm pretty sure that Rindler would tell you that you're twisting his view about the efficacy of using the term "relativistic mass" into something he never intended or agrees with.
What are you basing this opinion of that Rindler thinks on?I'm pretty sure that Rindler would tell you that you're twisting his view about the efficacy of using the term "relativistic mass" into something he never intended or agrees with.
For a system of two photons moving in opposite directions there is a zero momentum frame. Therefore even though both photons have zero proper mass the system has a non-zero proper mass. In fact I was just explaining this last night to someone. I.e. I was discussing black body radiation - such raidation has an energy-momentum tensor of a perfect fluid.Originally posted by Hurkyl
Incidentally, just because the sum of the momenta of two or more photons may have mass does not mean each individual photon has mass.
Yes. Quite true. And of course there must be another particle around to facilitate the pair production. Otherwise there would be a violation of the conservation of momentum.Originally posted by selfAdjoint
In fact the energy of just one photon, if high enough, can be used to create a new particle and antiparticle. The photon ceases to exist in this pair production
Am I misunderstanding or does that sentence translate into: 'if we ignore some of the forces, the forces are imbalanced'?Note : To be in agreement with Newton's 3rd law, it would be necessary to take into account the moments of the magnetic and electric fields.
Was that something I wrote?Originally posted by russ_watters
From the article: Am I misunderstanding or does that sentence translate into: 'if we ignore some of the forces, the forces are imbalanced'?