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The Massless Photon

  1. Nov 3, 2006 #1
    (I hope I've chosen the right sub-forum for this question...)

    Hi folks - I've recently joined here to see if people who are more knowledgeable than me can help me understand some physics issues I have struggled with for a long time.

    My current question is a pretty basic one about how it is possible for a photon to have no mass.

    We have the famous equation, "E = mc squared" My math knowledge is very limited, but from what I know - if I assign the value "0" to m, and multiply 0 by c squared, the answer for E should be zero.

    Yet a photon possesses energy, and is said to have no mass.

    I can see 3 possibilities:

    - "E = mc squared" is not a standard algebra equation, and assigning the value "0" to m doesn't result in E being zero.

    - "E = mc squared" does not apply to photons - something that seems very unlikely to me.

    - photons do, in fact, have mass, or conversely, have no energy.

    Can someone help me understand this? I would be very grateful. Thanks!
  2. jcsd
  3. Nov 3, 2006 #2


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    You need quantum mechanics too... E=h f
  4. Nov 3, 2006 #3
  5. Nov 3, 2006 #4
    Actually I do usually start with Wiki, but it never occurred to me to look up "E = mc squared" there.

    Thanks for the link - it answers my question.

    That was easy, wasn't it? Carry on, carry on... :smile:
  6. Nov 3, 2006 #5
    If you’re just looking for a way to understand; there is a very simple way.
    Don't thing of a photon as "possessing" energy --- thing of a photon AS energy.
    Now when an atom absorbs a photon and moves one its electrons into a higher energy ‘obit’ it gains mass.
    When it drops to a lower energy level the atom loses mass as a photon is emitted in some random direction.

    Where was the mass? in the electron; or in the atom as whole – I’ll let you speculate.
    But that mass cannot just appear and disappear; shouldn’t mass be “conserved”?
    No conservation of mass and conservation of energy is “Old School”
    In modern physics it is the net of Mass and Energy that is conserved.
    So when a bit of mass disappears from an atom we can find it in the energy that escapes from it – we call it a photon.
    When a lot of mass disappears very quickly we can describe it as a lot photons (aka energy) escaping – or more easily described as a nuclear explosion.
  7. Nov 3, 2006 #6

    Doc Al

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    The complete relativistic expression for the energy of a particle is this:
    [tex]E = \sqrt{p^2c^2 + m^2c^4}[/tex]

    Where p is the momentum. Note that for massive particles at rest (momentum = 0) that equation becomes the more familiar [itex]E = mc^2[/itex].

    For a photon: the mass is zero, but the momentum and energy are non-zero. E = pc = hf.
  8. Nov 3, 2006 #7
    Glad I could help. I picked up this firefox extension to make sure I can't avoid searching on wikipedia.

    https://addons.mozilla.org/firefox/2517/ [Broken]
    Last edited by a moderator: May 2, 2017
  9. Nov 4, 2006 #8
    Thanks agains to all. The most fundamental answer to my question is that E=mc squared is not as universally applied as I assumed it was, at least not in its familiar simple form.

    There is something that still confuses me a bit, and that's the concept of "rest mass". I was under the impression that the only time mass could be at rest was at a temperature of absolute zero, which doesn't really exist in nature (like a perfect vacuum). The photon is also never at rest - is the difference that the photon can't conceptually be at rest?

    I've also come across references to the "virtual mass" of a photon. Can anyone shed more light on that?

    The larger question might be - is the massless photon massless by definition? In other words, "mass is a quality of matter, the photon isn't defined as matter despite being a particle, and therefore the photon can't have a quality of matter - mass"?

    Or is it something more concrete - that the photon doesn't exhibit inertia, acceleration and deceleration, and other properties of mass?

    Thanks again.
  10. Nov 4, 2006 #9


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    That's in quantum mechanics. Special Relativity CAN be interfaced to quantum mechanics, but when we talk about SR by itself we mostly mean the classical model, where things can be at rest. You are right about the photon though, it has no rest frame, even in the classical theory, and no mass.

    I am one of the ones who do not use "rest mass" because that implies there is some kind of distinguished non-rest mass, which I deny. There are are other posters on this forum who vehemently disagree, and in order to get maximal use out of your experience here you have to be clear in your mind so the disagreement (which is a tempest in a teacup as far as physics is concerned) doesn't confuse you. In my view a particle has one mass, no matter how it is moving in relation to you, and this mass is a "scalar", that is to say the same in every frame.

    I don't know this one. Did you mean virtual photons?

    ST postulates that the speed of light is invariant, the same for all inertial frames. It is a mathematical deduction from that postulate that if light is carried by a particle then that particle must be massless. In fact the two concepts "Massless" and "Moves at the Speed of Light" are synonymous in SR. Gluons in the Standard Model also have this property.

    As far as we can tell, it takes no acceleration to get a massless particle up to the speed c. But the experimental evidence for the masslessness of the photon is mostly in another area. From classical wave theory it follows that if light were carried by a massive particle, it would exhibit longitudinal (compression type) waves. Whereas we only find transverse light waves. Experimentalists do tests with fantastically refined versions of this distinction to examine the possibility that there are some very weak compression waves, that would allow a tiny mass to the photon. They've been doing this, with ever increasing precision, for generations. They've never found anything.

    Of course no experiment can ever proves that the mass is exactly zero; every experiment has error bounds, but the error bounds on these tests have gotten really tiny, off the top of my head I want to say + or - 10-20 electron volts.

    Thanks again.[/QUOTE]
  11. Nov 4, 2006 #10
    You made a good point here. The concept of a "massless particle" is so familiar in present day physics that people forget that it is in fact not such a good concept at all. As you pointed out, the photon is indeed never at rest, and can not be, so why should one talk of rest mass ? It is really nonsense if you think about it and it seems to cause a lot of confusion to many people (judging from the forum questions).So, the photon is massless by definition only with the convenient result that one can insert m=0 into the complete relativistic expression, such that one obtains the correct relation between energy and impulse. Maybe it would be better to use the term "c-particle" to indicate that it is a particle which moves at the speed of light.
  12. Nov 4, 2006 #11


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    There already is a term: "lightlike" particle.
  13. Nov 5, 2006 #12


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    it could be more universally applied if the [itex]m[/itex] in [itex] E = m c^2 [/itex] is always considered the relativistic mass and [itex]E[/itex] is the total energy, kinetic energy plus rest energy. or the energy of the particle in the frame of the observer watching it whiz by. or the momentum of the particle divided by speed. assuming photons travel at the speed of the wavespeed of electromagnetic radiation, [itex]c[/itex], the (relativistic) mass of the photons is

    [tex] m = \frac{E}{c^2} = \frac{h \nu}{c^2} [/tex]

    but here [itex]m[/itex] is not the "rest mass" or "invariant mass". for particles that move more slowly that [itex]c[/itex], special relativity says their momentum is

    [tex] p = m v = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}} [/tex]

    where [itex]m_0[/itex] is the rest mass (and [itex]E_0 = m_0 c^2[/itex] is the rest energy and the total energy [itex]E_0 = m_0 c^2[/itex] is the rest energy plus kinetic energy). so the relationship between rest mass and relativistic mass is

    [tex] m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} [/tex]

    or turned around is

    [tex] m_0 = m \sqrt{1 - \frac{v^2}{c^2}} [/tex]


    [tex] m_0 = \frac{E}{c^2} \sqrt{1 - \frac{v^2}{c^2}} [/tex] .

    now, no matter what finite energy that particle has, if it moves at the speed of light, that equation says that the rest mass of the particle has to be zero.

    the equation

    [tex] E = m c^2 [/tex]

    is just as general as

    [tex] E^2 = m_0^2 c^4 + p^2 c^2 [/tex]

    if the rest mass [itex]m_0[/itex] is related to relativistic mass [itex]m[/itex] as per the equation above and momentum is stiil the same [itex]p = m v[/itex].

    particles moving past an observer at very high speeds "appear" to that observer to have a larger mass than they do if the observer is moving alongside the particle. the mass that the particle has in the same reference frame of the particle is the rest mass.

    dunno if it means

    [tex] m = \frac{h \nu}{c^2} [/tex]

    of if it means that the jury might be out as to whether photons, the particle manifestation of light, travels as fast as the wavespeed of light. some have posted is upper bound for the rest masses of photons (if i recall this upper bound was somewhere around 10-52 kg which is virtually nothing. so the difference in speeds are not measureable if there is such a difference at all. if it turns out that photons are known to travel at precisely the speed of light (waves), then the rest mass of the photons would have to be zero.

    it's because they (are believed to) move at the speed of light that their rest mass has to be zero.

    it exhibits inertia, but no acceleration or deceleration if it always flies by at a speed of [itex]c[/itex], no matter who the observer is (this is a postulate of special relativity). they have momentum of

    [tex] p = m v = \frac{h \nu}{c^2} v [/tex]

    which, if they move at speed [itex]v = c[/itex], then the momentum is

    [tex] p = \frac{h \nu}{c} [/tex]

    they have non-zero momentum if that is what you mean by "inertia".

    i know that i am presenting this from a POV that is discouraged in modern physic pedagogy (i don't think that Doc Al will like it), but it's correct given the definition of relativistic mass.
  14. Nov 6, 2006 #13
    Thank you, folks.

    I wish I could tell you where I saw the "virtual mass" for a photon mentioned. Most of my physics reading is books for laymen, but once I did try to wade through articles by physicsts themselves in the library - I think it might have be in an article by Bohr where I saw it.

    Here's a related question - if a photon has no mass, what pushes a solar sail? Is it other forms of radiation, as opposed to photons?
  15. Nov 6, 2006 #14


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    Photons have momentum, as well as energy.
  16. Nov 6, 2006 #15
    The answer to your question regards your confusion as to the definition of the term "mass." Sometimes the term is used to refer to a particles proper mass while sometimes its used to refer to inertial mass (aka relativistic mass). The photon has zero proper mass and an inertial mass m = E/c2. For details please see


    Best wishes

  17. Nov 6, 2006 #16


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    Note that Pete's paper, with the URL given above, while mostly correct, has not been peer reviewed, and that some of us (like me) have disagreements with him on certain technical points and usages. Most of these are rather "fine" points, though.
  18. Nov 6, 2006 #17
    Pervect - given what you said above, my question would be - do you agree with Pete that photons have inertial mass?
  19. Nov 6, 2006 #18


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    Photons have momentum and carry energy.

  20. Nov 6, 2006 #19


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    I would say that photons have a zero invariant mass. If pressed, I would admit that they have a non-zero "relativistic mass", though I would be quick to point out that I personally did not like relativistic mass.

    But you asked about "inertial mass". I'm not quite sure what you mean by "inertial mass". At a guess, you are trying to divide the momentum of a photon by its speed (which is always 'c') and come up with a number. This number will depend on the frame of reference - it will not be a property of the photon alone.

    My general remark would be this: photons carry momentum and energy. This should be clearly understood. The idea of "mass" is actually somewhat of an "umbrella concept" - the name "mass" is an "umbrella" which covers a large number of closely related, but different, concepts. You'll really need to learn about mass in Newtonian mechanics, mass in special relativity (invariant mass and perhaps relativistic mass) and mass in general relativity (ADM mass, Bondi mass, Komar mass) separately. To quote Max Jammer, "Mass is a mess".

    For some online reading, samples of Max Jammer's two books on mass are available on Google. They appear to be some of the better non-technical references out there.


    You can find the remark about "mass is a mess" at http://books.google.com/books?vid=I...x+jammer+mass&sig=rvmUlL1YxTTmzI_MxSEwoJB7fzo

    The umbrella analogy is AFAIK mine, however.
  21. Nov 6, 2006 #20
    That question cannot be answered until you have a definition of inertial mass. A.P. French in his SR text defines "inertial mass" as the ratio of the particle's momentum to its speed. Many others use this definition as well. So according to Frenchl anything that has momentum has speed. Therefore mass, just like momentum, will be dependant on the observer.

    Best wishes

    Last edited: Nov 6, 2006
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