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Mass of Photon

  1. May 13, 2014 #1
    We know the equation E=pc or P=mc both of them are avaliable for photon.But photon is a massless particle.How can that be possible ?
    Thanks
     
  2. jcsd
  3. May 13, 2014 #2

    ZapperZ

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    What exactly is the problem here?

    And have you read the FAQ in the Relativity forum?

    Zz.
     
  4. May 13, 2014 #3

    jtbell

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    I guess you got this from p = mv which is non-relativistic. Photons are relativistic objects and therefore follow the relativistic equation ##E^2 = (mc^2)^2 + (pc)^2## where m is the invariant mass a.k.a. "rest mass" (yes, it's a silly name for the mass of a photon which is never at rest), which gives you your E = pc.

    The usual equation for relativistic momentum in terms of velocity is
    $$p = \frac{mv}{\sqrt{1 - v^2/c^2}}$$
    but this doesn't work for photons because it gives p = 0/0 which is undefined mathematically.
     
  5. May 13, 2014 #4
    There happend math error can you wrote it again please ?
     
  6. May 13, 2014 #5

    Dale

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    Please read: https://www.physicsforums.com/showthread.php?t=511175 [Broken]
     
    Last edited by a moderator: May 6, 2017
  7. May 13, 2014 #6

    xox

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    Yes.


    False, in the case of photon $$p=\frac{h}{ \lambda}$$


    It isn't , you are making a mistake, see above.
     
  8. May 13, 2014 #7

    PAllen

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    Is it possible OP means:

    P = E/c = (E/c2)c = mc ?

    Thus a gross abuse of relativistic mass, and another reason it has fallen out of favor.
     
  9. May 13, 2014 #8

    jtbell

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    Are you referring to my equations? I see them OK. Are you using the PF app on a smartphone or other mobile device? I think LaTeX equations are properly visible only in a web browser. I can see them in Safari on my wife's iPad, as well as in Firefox on my main computer.

    However, using the PF app on the iPad, I see only the LaTeX source code.
     
  10. May 14, 2014 #9
    It is not undefined. After substitution with
    [tex]E = \frac{{m \cdot c^2 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}[/tex]
    the result for the removable discontinuity is
    [tex]\left| p \right| = \mathop {\lim }\limits_{\left| v \right| \to c} \frac{{E \cdot \left| v \right|}}{{c^2 }} \cdot \frac{{\sqrt {c^2 - v^2 } }}{{\sqrt {c^2 - v^2 } }} = \frac{E}{c}[/tex]
     
  11. May 14, 2014 #10

    xox

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    You cannot do this types of manipulations since both [tex]E = \frac{{m \cdot c^2 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}[/tex] and [tex]p=\frac{mv}{\sqrt {1 - (v/c)^2}}[/tex] are valid ONLY for [tex]m \ne 0[/tex].

    As such, the formulas cannot be applied to the photon, as you are trying to do in your calculations.
     
    Last edited: May 14, 2014
  12. May 14, 2014 #11
    Did you noticed the limes?
     
  13. May 14, 2014 #12

    xox

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    Yes, I noticed, your STARTING point is wrong, i.e. your error occurs BEFORE you take the limits.
     
  14. May 14, 2014 #13

    pervect

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    Hopefully you've read the FAQ by now.

    When we say that a photon has no mass, we mean that it has no invariant mass or rest mass. When you write P=mc, you are using relativistic mass, which is equal to E/c^2.

    Thus the term "mass" is ambiguous. Modern textbooks and papers tend to use invariant mass, while older textbooks and popularizations still often use relativistic mass.

    Given that we know that the "mass" in your second equation is the old-fashioned "relativistic mass", the second equation is just a restatement of the first

    P = m_relativistic * c is the same as P = (E/c^2)*c = E/c which is the same as E = pc

    As a side note, the general relativistic relationship between energy, momentum, and invariant mass is given by:

    E^2 = (p*c)^2 + (m*c^2)^2
     
  15. May 14, 2014 #14
    What exactly is wrong with my starting point?
     
  16. May 14, 2014 #15

    xox

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    I thought this is quite clear.
     
  17. May 14, 2014 #16
    No, it's not. As I do not set m=0 it does not refer to my calculation. And as you noticed the limes you should be aware of that. So what's your point?
     
  18. May 14, 2014 #17

    xox

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    You CANNOT start with [itex]E = \frac{{m \cdot c^2 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}[/itex] since it DOES NOT apply to photons. One day you'll get it.
     
  19. May 14, 2014 #18
    I demonstrated how to do it. It's quite simple math. One day you'll get it.
     
  20. May 14, 2014 #19

    Dale

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    Closed pending moderation
     
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