About effective (relativistic) mass of photon

Discussion in 'Physics & Math' started by Ultron, Jun 14, 2016.

  1. origin Heading towards oblivion Valued Senior Member

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    Fine, carry on.
     
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  3. Layman Totally Internally Reflected Valued Senior Member

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    It doesn't bother you in any way shape or form that I could derive the same exact equation using the rest mass in E=mc^2? If what you are saying is true then I just discovered that the rest mass is the same as the effective mass or relative mass of the photon... Solving for both gives you the same identical equation.
     
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  5. exchemist Valued Senior Member

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    OK, I've had enough now. We've done it to death.

    Erratum footnote: I realise that I carelessly copied from Layman's post and left out the squares in the formula E²=(mc²)² + (pc)² the first two occasions I quoted it. Sorry about that, readers - a sign of a chemist playing out of position.
     
    Last edited: Jun 15, 2016
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  7. Layman Totally Internally Reflected Valued Senior Member

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    I am just trying to assume that everything you are saying is true and then trying to apply it, like you wanted me too. Feel free to tell me what I am doing wrong...

    Say E = hc/ λ, m = h/ λc, and p = h/ λ, then substitute those varibles into E = mc^2 + pc;

    You get hc/ λ = hc^2/ λc + hc/ λ

    This then reduces to hc/ λ = hc/ λ + hc/ λ

    Then the only way this equation could be true is if hc/ λ = 0; the mass of a photon would then have to be zero and the relative mass would also have to be zero for this equation to be true.

    You can get a rest mass of h/ λc just by substituting hc/ λ for energy in the equation E = mc^2 and solving...
     
  8. exchemist Valued Senior Member

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    Apart from using the wrong formula, as my previous footnote says, I think you are now muddling rest mass with relativistic or effective mass, when you say m=h/λc.

    If you are working with relativistic mass, let's call it "m(r)", you just use E=m(r)c². You only use the full version, E²=(mc²)² + (pc)², if you are working in terms of rest mass "m". That's the point of the equation.

    m(r) =h/λc is correct for a photon, because for that, its rest mass is zero and thus the only contribution to m(r) comes from its momentum due to the EM energy it carries and so p=E/c. For an electron, a proton or any other object with rest mass, this is not what m is. The rest mass of an electron is fixed and thus obviously not a function of its de Broglie wavelength. It is its momentum that is a function of de Broglie wavelength, i.e. the wavelength of the quantum mechanical wave function of the object.

    But back to the photon case, for that it is true that m(r) = h/λc. If you plug that into the correct energy/mass equivalence relation for m(r), namely the simple E=m(r)c², you get E= c².h/λc = hc/λ, which obviously makes sense.
     
    Last edited: Jun 15, 2016
  9. Layman Totally Internally Reflected Valued Senior Member

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    It seems like the equation m = h/λc could only be muddled around with, as you say. If it cannot be used as the mass in E = mc^2, then what mass could it possibly be used for? If you assume that it is only part of the momentum, then it would need to go into the mass of the momentum part of the equation.

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    Then the velocity of the particle would be c, and the bottom of the equation would then be equal to zero. Then you say you are not supposed to use that equation, and you are supposed to use the equation;

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    The problem here is that this equation doesn't have mass in it. So then where, if anywhere, could the relativistic mass of a photon be substituted into anything? I don't know of a momentum equation that considers mass in which it would ever be useful to substitute h/λc for mass to ever get anything useful out of it.
     
  10. exchemist Valued Senior Member

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    Well true, but remember it was you who invented the relation m=h/λc, which you got presumably from p=h/λ. De Broglie does not use this relation and nor did I.

    In fact you don't need relativistic mass for any of this stuff, as I recall saying, somewhere back at the beginning. It seems to be an entirely optional concept. If you read the link I sent you about its history and use that should be clear to you. It does have its uses, in the notion of things appearing to get more massive as they are accelerated and was, so I read, popular in the 1920s, then rather fell from favour, apparently. I'm not an expert on relativity at all, so I can't give you the ins and outs of it. All I do know is that there is nothing wrong with the concept, i.e. it is scientifically respectable, though (as we have seen) you do have to exercise care not to cause confusion with it. Ultron chose to employ it. Well that's fine, he can, so far as I can see.

    Shrugs

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    (to coin a phrase).
     
  11. Layman Totally Internally Reflected Valued Senior Member

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    No, I got the relation m = h/λc from the original post, which was the whole thing I was arguing against... That was the final answer that Ultron got from his proof. I just took the E/c^2 out of the middle of it. I just provided an alternate proof that doesn't assume that the mass is zero, without removing mc^2 in my post #38. So, you have been arguing about this the whole time with me? And, you didn't even know that I was arguing against this equation? Funny thing is that you almost convinced me that the equation was actually true. Now you are telling me that I was the one that made it up, lol.
     
  12. Layman Totally Internally Reflected Valued Senior Member

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    I couldn't help but notice that the relativistic mass that you called \(m_r\) is equal to p/c, so \(m_r = p/c\). Then if you assume that \(m_r = p/c\), then \(p = m_r c\) for a photon which would be in the same classical form as p = m v where the velocity is the speed of light. But, if you assume that \(m_r\) is equivalent to the rest mass, then plugging p/c into \(E = mc^2\) gives back the equation E = pc. Although, you couldn't have a rest mass and relativistic mass be the same here because then you wouldn't get the correct amount of energy by using both. You would end up getting twice the amount of energy (the situation I got in my post #44). So it got me scratching my head thinking that it could be possible that the energy is E = pc, because the mass of the photon is \(m_r = h/{λc}\). Then you blew it by saying that I was the one that invented the equation, lol.
     
    Last edited: Jun 15, 2016
  13. Nacho Registered Senior Member

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    LOL! At first I thought this was an interesting topic .... but I'm not so sure now?? I'm so very CONFUSED!!! Somebody PLEASE tell me the anwer!
     
  14. Layman Totally Internally Reflected Valued Senior Member

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    I think you are confused, because this whole thread is just a bunch of complete nonsense. I think exchemist just realized that the OP derived a new equation for relativistic mass of the photon that has never been accepted by the scientific community. It might be possible that the equation is correct, but society doesn't have the technology to experimentally verify if it is correct or not. I think it would be necessary to check to see if the value of mass here would fall under the experimental range. It seems like it would be a really small value. It may be worth considering if we do end up getting the technology that could detect it.

    Personally, I don't think it would be necessary for the photon to have a relativistic mass, because a photon wouldn't even need to have it's measuring rods changed in order to measure another photon traveling the speed of light. Then again, my alternate derivation of it could show that it is actually the rest mass, and it could be used as the relativistic mass or the rest mass, not both... Then there wouldn't ever be able to be an experiment that could prove that either way, since it would just be E = pc on both accounts. The world may never know...
     
  15. rpenner Fully Wired Valued Senior Member

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    You can't argue with definitions since those are the ground rules for making sure that we are all talking about the same thing. If you wish to invent new definitions, you carry all the burden of demonstrating your new terms are part of a precise, communicable framework for describing the behavior of a large class of related phenomena or you are not doing physics.

    So what physics tells us is that distinct particles have distinct quantum numbers, including electric charge, spin and rest mass, \(m_0\).

    For any plane wave: \( \omega = 2 \pi f , \quad |\vec{k}| = 2 \pi / \lambda, \quad v_{\textrm{phase}} = \lambda f = \frac{\omega}{ |\vec{k}|}, \quad v_{\textrm{group}} = \frac{\partial \omega}{\partial |\vec{k}|} \)
    For any free particle in special relativity: \(E^2 = (m_0 c^2)^2 + (c \vec{p})^2 , \quad E \vec{v} = c^2 \vec{p} \)
    For any free particle in quantum mechanics: \(E = h f = \hbar \omega , \quad | \vec{p} | = h / \lambda , \quad \vec{p} = \hbar \vec{k}\)

    These relations are mutually compatible, all hold in quantum field theory of free particles, and work equally well for particles with zero or finite rest mass. The discovery of the Higgs boson did not change this in the least, because the mass that the Higgs adds to fundamental particles operates in a way indistinguishable from any other form of rest mass like the extra dynamical mass of nucleons and the missing mass of nuclei. Rest mass is rest mass is rest mass, so empirical observation and thus theory tells us.

    But if you wish to introduce relativistic mass \(m \equiv c^{-2} E\), to what end?
    1) It doesn't act like a source of gravitation in GR since both \(E\) and \(\vec{p}\) contribute to space-time curvature in ways that \(m \equiv c^{-2} E\) has no room to supplement.
    2) It doesn't act like inertia since the relation between force and acceleration is \(\gamma^3 m_0\) if the force is parallel to the momentum and \(\gamma m_0\) if they are perpendicular, so for arbitrary angles, force isn't even parallel to acceleration.
    3) Therefore \(c^2 m\) and \(E\) are just two ways of writing the same quantity without the special significance of \(m_0\) to gravitation and inertia and since physicists often use units where \(c=1\) in many contexts they have no convenience even in terms of units.
    4) Since \(m \equiv c^{-2} E\) seems to be sterile for improving our understanding of the behavior of reality, its a definition without justification, so it would be simpler to ignore antique references which use it and allow us to drop the subscript from \(m_0\) when we wish to discuss the property of rest mass common to all particles of a distinguishable "type".
     
    Last edited: Jun 16, 2016
  16. exchemist Valued Senior Member

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    Not at all. Your determinedly adversarial approach seems to blind you to what I have been actually saying. In my view there is nothing wrong with writing relativistic mass as equal to h/λc , for a photon. I thought you had introduced it but evidently I had not registered that that Ultron had already used it in his OP. Either way it does not matter, as it is an obvious relation - for a photon. What I then said was that it is not required in showing the connection between light, matter and energy embodied in de Broglie's and Planck's relations. I am in fact agreeing with you, in part, that its utility is limited.

    rpenner's contribution above seems to me to express the modern view, as described in the link I gave you, that relativistic mass is a concept that, after initially being used quite a bit when relativity was young, has progressively fallen from favour. That does not make it wrong, though, so, as far as I'm concerned, Ultron is quite at liberty to use it if he or she wants to. It's Ultron's thread. As the link I gave points out, the idea of relativistic mass is still used today in some popularisations of relativity, however much the physics community may frown on this practice.

    What I do find extraordinary is that, from your latest post, you still seem to think that relativistic mass is something that can be determined to exist or otherwise by experiment. As both I and rpenner have explained, it is a mere ratio of two experimentally verifiable properties (for a photon, either E/c² or p/c - the two are equivalent).
     
    Last edited: Jun 16, 2016
  17. exchemist Valued Senior Member

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    Yes this is an admittedly terribly muddled debate about whether or not relativistic mass is a legitimate concept.

    There is what seems to me a good summary of the issue in the link I gave Layman earlier in the thread, i.e. this one here: http://sasuke.econ.hc.keio.ac.jp/~ken/physics-faq/mass.html

    And rpenner expresses the modern view towards the end of the thread.
     
  18. Nacho Registered Senior Member

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    Thanks for that link exchemist. I actually understood that pretty good ... at least I think I did.
     
  19. James R Just this guy, you know? Staff Member

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    As was pointed out above, relativistic mass is not a concept that is used very often these days by professional physicists. The problem with it is that its value varies with the frame of reference, unlike rest mass. Applying the concept to photons is even more problematic, because it is impossible to move into the reference frame of a photon. Photons have no rest frame and no rest mass.
     
  20. Layman Totally Internally Reflected Valued Senior Member

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    I don't see how your opinion could mean anything really when you don't even know what we are even talking about, other than you seem to trust Ultron's opinion over mine...
    I don't have a problem with relativistic mass. A theory on how to apply relativistic mass increases to massless particles that travel the speed of light does not exist. If you haven't understood what I have been arguing against this whole time, then you probably have misunderstood this as well.
    No, I know that the mass of the photon, relativistic or otherwise, has never been shown to exist by experiment. A scientific theory cannot be proven unless it has been shown to exist through experimentation. Therefore, it could only be a theoretical possibility at best. I was only saying that it would then take an advancement in technology to further narrow down the possible mass of a photon (which would break quantum physics on so many levels, if it was discovered).
     
    Last edited: Jun 19, 2016
  21. Layman Totally Internally Reflected Valued Senior Member

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    To help clear things up with exchemist, do you know that the photon has a relativistic mass which has been defined as m = h/λc, like the OP's proof claims?
     
  22. rpenner Fully Wired Valued Senior Member

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    \(m_0 = 0 \wedge m \equiv E / c^2 \wedge \vec{p} \cdot \vec{v} > 0 \Rightarrow E^2 = c^2 \vec{p}^2 \Rightarrow E = c p \Rightarrow m = p / c \Rightarrow m = h / ( \lambda c )\)
     
  23. Layman Totally Internally Reflected Valued Senior Member

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    So all that time I said a photon could have little or no mass, you believed that the photon has a mass of \({h}/{λc}\)? \({h}/{λc}\) is not equal to zero. So, you are saying that light does have mass! That is not a known scientific fact.

    I am starting to get the feeling that you guys just like to argue about everything. If you assume that is true, then \( mc^2 = (h/{λc}){c^2}\), and then that would reduce to, pc, for all particles, but the statement that E = pc for all particles is not true.​
     
    Last edited: Jun 19, 2016

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