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Action Does NOT Equal Reaction ?

  1. Jul 18, 2003 #1
  2. jcsd
  3. Jul 18, 2003 #2
    No. They assume that the electric field propagates instantaneously, while the magnetic field does not. Which is false .
     
  4. Jul 18, 2003 #3
    a little bit off topic

    The 2 particles are negatively charged, not positively charged, right?
     
  5. Jul 18, 2003 #4

    pmb

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    There are two forms of Newton's third law
    (1) Weak form - forces are equal and opposite but not along a line
    (2) Strong form - forces are equal and opposite and along a line

    The weak form does not hold for all types of forces.

    It'll be simpler if I just quote "Classical Mechanics - Third Edition," Goldstein, Safko and Poole. Page 7-8

    Pete
     
  6. Jul 18, 2003 #5
    Re: Re: Action Does NOT Equal Reaction ?

    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...
     
  7. Jul 18, 2003 #6

    pmb

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    Re: Re: Re: Action Does NOT Equal Reaction ?


    I assume you understood what Goldstein meant. Is that correct? Do you disagree with his conclusion?

    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.

    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.

    If E is the electric field of the moving charge and v is the velocity of the moving charge then the magnetic field of the moving charge is given by

    B = vxE

    For a derivation see Feynman Lecturers Vol II page 26-3. For a picture of the magnetic field see Figure 26-5 in V-II.

    There should be something about this in the Feynman Lectures. Do you have them?

    The nature of relativity prevents an action reaction forces between particles in general - what is equal and opposite in one frame is, in general, not equal and opposite in another.

    French discusses this too. He makes a very important point. From "Special Relativity," A.P. French, page 224

    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.

    Pete
     
  8. Jul 19, 2003 #7
    pmb,
    I admit I didn't put enough thought into this. I'll check - just don't have the time now, hang on please.
     
  9. Jul 19, 2003 #8

    pmb

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    So'kay! Take your time. That's why we're all here! :-)

    Pete
     
  10. Jul 20, 2003 #9
    pmb,
    well I gave it some thought. Here's my (unsatisfactory) results:

    1) There is indeed such a thing as a magnetic field of a single moving particle. The classical approximation is:
    B = q/c * vxr/r3.
    And that was indeed named after Biot/Savart. So you are right, and I was wrong.

    2) Let's now treat the problem completely classical. Let particle 2 be in the path of particle 1, and moving at right angles. Let for simplicity q = c = r = 1. Then, we get indeed
    F1 = r - v1v2,
    F2 = -r.
    In other words, actio=reactio is indeed violated.

    3) Indeed, the missing momentum must go into the field (see e.g. Jackson, chapter 6.8.).

    4) Here my trouble begins. How can any momentum go into the field in such a simple configuration? The only answer I can think of is: Bremsstrahlung. But I don't see how this classical treatment of the problem can produce any e.m. radiation (i.e. sinusoid waves)...!?

    5) My next thought was, we should treat this with full SR. In Jackson, chapter 11.10, I found the following:
    If two particles move at relative speed v, then the field of one particle seen in the rest frame of the other is:
    Elongitudinal = [gamma]qvt/(b2+[gamma]2 v2 t2)3/2
    Etransversal = [gamma]qb/(b2+[gamma]2 v2 t2)3/2
    B = [beta] Etransversal.
    Where b is the shortest connection of both paths, t=0 when they are closest, and B is, of course, at right angles with v and b.

    Now, since b is invariant (because it's at right angles with both paths), this result is indeed symmetric, since all you do is replace v by -v and b by -b when switching from one particle to the other.

    Now this is strange, because in a particle's rest-frame it should not feel any Lorentz force at all, no matter what B is. Plus, I don't know at all how to translate this into forces in the lab (or any other) frame. I still suspect, when doing this properly, actio=reactio will hold. That is, there will be at least one frame of reference in which it will hold. Maybe the center-off-mass one.

    Any help?


    Edit: corrected 1 error, and made some formulae better legible. Couldn't make the gamma small, however. It should be...
     
    Last edited: Jul 20, 2003
  11. Jul 20, 2003 #10
    Re: Re: Re: Re: Action Does NOT Equal Reaction ?

    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?

    Creator
     
    Last edited: Jul 20, 2003
  12. Jul 20, 2003 #11

    pmb

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    Re: Re: Re: Re: Re: Action Does NOT Equal Reaction ?

    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!

    Just made a new web page on the center of mass for a relativistic system

    See -- http://www.geocities.com/physics_world/sr/center_of_mass.htm

    That's in preperation for a new web page on Einstein's "photon in a box" thing. I've defined and described it here

    www.psyclops.com/hawking/forum/printmsg.cgi?period=current&msg=59744

    Pete
     
  13. Jul 20, 2003 #12

    Hurkyl

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    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.

    In particular, this means that there's no reason that the center of mass, as you've defined it, should have a constant velocity (even in the presense of no external forces).
     
  14. Jul 20, 2003 #13

    pmb

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    Please take close note of the very first sentence, i.e.

    "Consider a system of particles, all of which move at constant velocity..."

    Pete
     
  15. Jul 20, 2003 #14

    Hurkyl

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    lol I read the whole thing twice, and never noticed the text up there before the diagram!
     
  16. Jul 20, 2003 #15

    pmb

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    I have it when that happens! :-D

    I've just been going over Einstein's 1906 "photon in a box" paper over the last week. This seems quite compatible with his comments and derivations and conclusions.


    I've just modified that comment. Looking over it I now see that all I needed to do was require that that the speed remain unchanged - not the velocity. So now the particles can move in a cirlce!

    Pete
     
  17. Jul 30, 2003 #16

    jeff

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    Re: Re: Re: Re: Re: Re: Action Does NOT Equal Reaction ?

    Yes yes, photons have non-vanishing relativistic mass but vanishing rest mass.

    The idea you seem to "hate" giving up is that it's wrong to view massless particles as pure energy.

    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.
     
  18. Jul 30, 2003 #17

    pmb

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    Re: Re: Re: Re: Re: Re: Re: Action Does NOT Equal Reaction ?

    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.

    In the second place if you're refering to the notion of a photon having mass = energy/c^2 then you're very much wrong. It's not something I've be reluctant to give up. It's been something I've come to accept - that's quite a big difference. When I first heard of this notion (about 8 years ago as I recall) I assumed people were just misinterpreting Einstein. As I studied Einstein's original work and all the many artilces on it that followed over the years since 1905 I've come to accept that a mass of a photon of m = E/c^2 is exactly what Einstein had in mind and that it makes perfect sense. And this is something which can be found in many relativity texts.

    For example: from "Relativity: Special, General and Cosmological," Rindler, Oxford Univ., Press, (2001), page 120
    re -
    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.


    This notion of an EM field having mass was the view Einstein arrived at in 1906 in what has come to be known as Einstein's photon in a box thought experiment. The paper concluided with Einstein deducing that u = "energy density of EM field" has a mass density of u/c^2.

    In any case I know relativity so I can form my own opinions. I just happen to agree with Rindler and everyone on this point. I gave you another example if you recall

    http://www.geocities.com/physics_world/Guth.jpg

    What are you basing this opinion of that Rindler thinks on?

    Pmb
     
  19. Jul 30, 2003 #18

    Hurkyl

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    Incidentally, just because the sum of the momenta of two or more photons may have mass does not mean each individual photon has mass.
     
  20. Jul 30, 2003 #19

    pmb

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    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.


    Did you get the impression that I thought any different? If so then you were mistaken.


    Pmb
     
  21. Jul 30, 2003 #20

    selfAdjoint

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    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
     
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