About effective (relativistic) mass of photon

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

  1. Layman Totally Internally Reflected Valued Senior Member

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    I think the error in this kind of derivation is that you guys are assuming that particles with mass are a wave. By substituting energy with a wave equation (which produces the same equation for mass, h/λc), you are assuming that particles with mass are waves. Particles with mass are not waves, so it then produces the equation for massless particles, E = pc. Then that is wrong, because E = mc^2 + pc, so it is then actually saying that E = pc + pc for particles, which isn't true. It is only E = pc for a photon.
     
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  3. rpenner Fully Wired Valued Senior Member

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    \(m_0 = 0\) means the particle under consideration (photon) has no (invariant or rest-) mass.
    \(m \equiv E/c^2\) means we are defining the term "m" to mean no more and no less than E/c^2 -- this is the so-called relativistic mass which for reasons I posted above does not act like gravitational or inertial mass. It's ridiculously frame-dependent and so can take on any value of \(m_0\) or higher.
    \(\vec{p} \cdot \vec{v} > 0\) means we are ignoring cases where the particle (photon) is not moving or moving anti-parallel to momentum.

    Thus \( E^2 = c^2 \vec{p}^2\) follows from \(E^2 = (m_0 c^2)^2 + (c \vec{p})^2\) and substituting in \(m_0 = 0\). This means either \(E = + c |\vec{p}|\) or \(E = - c | \vec{p} |\).
    \(E = c | \vec{p} | \) follows from \(\vec{p} \cdot \vec{v} > 0\) and \(E \vec{v} = c^2 \vec{p}\)
    \(m = | \vec{p} |/c\) follows by substituting in the definition of relativistic mass: \(m \equiv E/c^2\)
    \(m = h/(\lambda c)\) follows by substituting in the deBroglie relation: \(| \vec{p} | = h/\lambda\)

    But only if \(m_0 = 0\) as for photons.

    Correct.

    \( m_0 = \frac{1}{c^2} \sqrt{ E^2 - c^2 \vec{p}^2 } = \frac{h}{c^2} \sqrt{ f^2 - c^2 / \lambda^2 } \)

    You are confusing invariant mass (rest mass), \(m_0\), with relativistic mass, \(m \equiv E/c^2\).
     
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  5. rpenner Fully Wired Valued Senior Member

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    This framework allows many different ways of expressing itself.
    \(v_{\textrm{phase}} = \frac{\omega}{k} = \frac{E}{p} = \frac{\sqrt{ (m_0 c^2)^2 + (c p)^2 }}{p} \\ v_{\textrm{group}} = \frac{\partial \omega}{\partial k} = \frac{\partial E}{\partial p} = \frac{\partial \quad}{\partial p} \sqrt{ (m_0 c^2)^2 + (c p)^2 } = \frac{ c^2 p}{\sqrt{ (m_0 c^2)^2 + (c p)^2 }} \\ v_{\textrm{phase}} \times v_{\textrm{group}} = \frac{\omega}{k} \times \frac{\partial \omega}{\partial k} = \frac{E}{p} \times \frac{\partial E}{\partial p} = \frac{ 2 E \partial E}{2 p \partial p} = \frac{ \partial E^2 }{ \partial p^2} = c^2 \\ E v = c^2 p \Rightarrow E v / p = v_{\textrm{phase}} v_{\textrm{group}} \Rightarrow v = v_{\textrm{group}} \)

    Since \(\lambda f = v_{\textrm{phase}} = c^2 / v\), the relation between E, v, and λ is \(E = \frac{h c^2}{ v \lambda}\), and we have another expression in v, and λ:
    \(m_0 = \frac{h}{c^2} \sqrt{ \frac{ c^4 }{v^2 \lambda^2} - \frac{c^2}{\lambda^2} } = \frac{h}{c \lambda} \sqrt{ \frac{c^2}{v^2} - 1}\).
     
    Last edited: Jun 21, 2016
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  7. exchemist Valued Senior Member

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    Since you don't believe me, I've left it to others to sort out the issue of relativistic mass with you.

    But I have to correct a misunderstanding of yours about particles and waves here. Particles with mass most certainly are also waves. That is the whole basis of quantum theory, of which de Broglie's relation was a fundamental part.
     
  8. James R Just this guy, you know? Staff Member

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    Nobody has yet convinced me of the usefulness of talking about the photon's relativistic mass in that way. If somebody can show me how that equation leads to useful results or predictions, I'd be happy to see it.
     
  9. James R Just this guy, you know? Staff Member

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    That's not an error. That's standard quantum physics. Actually, things are usually a bit more complicated than de Broglie's simple explanation of a particle as a wave with a single wavelength. A more realistic description is one of particles as wave packets made of an appropriate combination (superposition) of waves of many wavelengths.

    First, notice that there's an error in the equation you've cited there. The correct one is:
    \(E^2 =(pc)^2 + (m_0 c^2)^2\)
    Two special cases of this:

    1. Photons have \(m_0=0\), so for them we have \(E=pc\).
    2. Very fast massive particles have values of \(pc\) much greater than \(m_0 c^2\), so it is approximately true for them that \(E=pc\).
     
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  10. Layman Totally Internally Reflected Valued Senior Member

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    That is what the wiki said, and that is partly why I was saying that the mass of the photon was E = pc and not m = h/λc. That is the whole argument here, and it sounds like you agree with me. Then the value of the momentum can be undefined for the photon as well, since the equation for momentum doesn't allow calculations for anything traveling the speed of light. For me, that is even questionable, because you wouldn't be able to calculate the momentum in the first place, since that calculation involves speed and mass with the Lorentz factor in the denominator. Even though Einstein predicted momentum for a photon, I don't think it is something that has been widely accepted for those reasons. It is generally not understood why or how a photon could have any type of mass or momentum or how it should be dealt with. You even said yourself that you don't know how to find any useful results or predictions thinking about the mass of the photon in the way the OP described.
     
  11. rpenner Fully Wired Valued Senior Member

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    E is not mass. pc is not mass. Presumably you mean \(m_0 = 0 \; \Rightarrow \; m_{\textrm{rel}} \equiv E/c^2 = p/c \) in which case you have no logical reason to reject \(m_0 = 0 \; \Rightarrow \; m_{\textrm{rel}} = \frac{h}{\lambda c} \) because \( p = h/\lambda \).

    No one is agreeing with you. Everyone is saying \(m_{\textrm{rel}} \equiv E/c^2\) is a lousy definition because while \(m_{\textrm{rel}}\) has units of mass, it is not a mass in either the inertial or gravitational sense.

    Obvious nonsense. The equations for momentum have been given in post #52, both how momentum relates to rest mass, velocity, and energy and how momentum relates to wavelength (or equivalently, wavenumber, \(\vec{k}\)).

    Then you seem ignorant that the momentum of photons has been confirmed to be as calculated in Compton scattering.

    Then you would be speculating on the opinions of professionals in complete opposition to every paper that has considered the subject.

    Why do you say such things when you, as a layman, can't give an opinion on what is "generally understood" by professionals?

    James R was saying he had no use for \(m_{\textrm{rel}} \equiv E/c^2\) when \(m_0 = 0\) is one of the crucial defining properties of photons. It the latter that is nearly universally meant by scientists when they use the phrase "the mass of the photon".
     
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  12. James R Just this guy, you know? Staff Member

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    Layman:

    The value of the photon momentum is \(p=E/c\). According to quantum mechanics, a photon has energy \(E=hf\), where \(f\) is its frequency. Also \(c=f\lambda\), where \(\lambda\) is the wavelength. Thus, we can write the momentum of the photon as \(p=hf/c=h/\lambda\).

    The expression \(p=\gamma mv\) doesn't apply to photons, which have zero mass.

    As I stated previously, most physicists these days don't use the concept of relativistic mass, because it is a frame-dependent quantity. Instead, they just tend to use rest mass. Photons have zero rest mass.
     
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