Still have questions about photon mass

  • #1
SamRoss
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Just read the FAQ post "Do photons have mass?" and I'm still confused. The post says that all of the photon's energy is in the pc term of the energy-momentum equation. (1) But isn't p equal to mv, implying there is mass? (2) The post also says there is no inconsistency with E=mc^2 but doesn't explain why there is no inconsistency. It still feels pretty darn inconsistent.
 

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  • #2
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Photons are indeed massless, which is why they can travel at the speed of light ; I.e) light itself; it does energy however even if it is massless via E = p•c where p is momentum. According to special relativity, any particle with mass with require an infinite amount of energy to reach the speed of light, however since the photon is massless, it does not invalidate special relativity. Take a Look at the equation with rest mass ( E^2 = (Mo^2)•(c^4)+ (p^2)•(c^2)any particle with zero rest mass still has energy which comes from linear momentum.
 
  • #3
PeterDonis
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But isn't p equal to mv
No. That is a non-relativistic equation that only works as an approximation for things moving very slowly compared to the speed of light. Obviously light itself is not such a thing. :wink:

Even the relativistic generalization of this formula that works for ordinary objects, which is ##p = m v / \sqrt{1 - v^2 / c^2}##, is not valid for light. You have to go back to the more fundamental energy-momentum relation, which is ##E^2 = p^2 c^2 + m^2 c^4##. For light, ##m = 0## so you just have ##E = p c##. (For an object with ##m \neq 0##, you can plug in the relativistic formulas for ##E## and ##p## in terms of ##v## and see how they satisfy the relation.)
 
  • #4
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(1) But isn't p equal to mv, implying there is mass? (2) The post also says there is no inconsistency with E=mc^2 but doesn't explain why there is no inconsistency. It still feels pretty darn inconsistent
For both 1 and 2 the correct formula to use is ##m^2 c^2=E^2/c^2-p^2##. That formula is applicable for all particles, not just photons. For a massive particle at rest, it simplifies to the famous ##E=mc^2## while for a massless particle the same formula simplifies to ##E=pc##
 
  • #5
SamRoss
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No. That is a non-relativistic equation that only works as an approximation for things moving very slowly compared to the speed of light. Obviously light itself is not such a thing. :wink:

Even the relativistic generalization of this formula that works for ordinary objects, which is ##p = m v / \sqrt{1 - v^2 / c^2}##, is not valid for light. You have to go back to the more fundamental energy-momentum relation, which is ##E^2 = p^2 c^2 + m^2 c^4##. For light, ##m = 0## so you just have ##E = p c##. (For an object with ##m \neq 0##, you can plug in the relativistic formulas for ##E## and ##p## in terms of ##v## and see how they satisfy the relation.)
Thank you for your reply but I'm still confused due to the fact that p depends on m even in the case of relativistic momentum. It seems to me that E=pc=γmvc so plugging in m=0 and v=c (including the v in γ) would give E=0/0.
 
  • #6
SamRoss
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For both 1 and 2 the correct formula to use is ##m^2 c^2=E^2/c^2-p^2##. That formula is applicable for all particles, not just photons. For a massive particle at rest, it simplifies to the famous ##E=mc^2## while for a massless particle the same formula simplifies to ##E=pc##
But isn't p still dependent on m?
 
  • #7
SamRoss
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Photons are indeed massless, which is why they can travel at the speed of light ; I.e) light itself; it does energy however even if it is massless via E = p•c where p is momentum. According to special relativity, any particle with mass with require an infinite amount of energy to reach the speed of light, however since the photon is massless, it does not invalidate special relativity. Take a Look at the equation with rest mass ( E^2 = (Mo^2)•(c^4)+ (p^2)•(c^2)any particle with zero rest mass still has energy which comes from linear momentum.
But isn't momentum still dependent on m?
 
  • #8
PeterDonis
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I'm still confused due to the fact that p depends on m even in the case of relativistic momentum.
Only for objects for which ##m \neq 0##. Not for light. Please go back and read my post again, carefully.
 
  • #9
Ibix
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But isn't momentum still dependent on m?
I think the easiest way to understand it is to note that mass isn't really a fundamental concept in relativity. More detail below, but the short version is that the fundamental concept is the energy-momentum four vector. Mass turns out to be a derived quantity. Also, the expression ##p=\gamma mv## can only be derived for things travelling slower than light. The idea of mass being a concept derived from energy and momentum is backwards from the familiar Newtonian thinking - welcome to relativity.

More detail, as promised:

The relevant fundamental concept is the energy-momentum four vector, whose components are the energy (divided by ##c##) and the three components of momentum. The modulus of this four vector (its "length", with scare quotes for the pedantic) is ##L=\sqrt{E^2/c^2-p^2}##, where ##p## is the modulus of the momentum. This is necessarily zero for things travelling at the speed of light, and necessarily real and positive for things travelling slower than light, and is a constant for any given object.

There's another four-vector called the four velocity. It's parallel to the four momentum, but its modulus is defined to be c. Its components are ##\gamma c## and ##\gamma## times the three components of velocity. Since it's parallel to the energy-momentum four vector, though, each of the components must be equal to the corresponding component of the energy-momentum four vector multiplied by ##c/L##, the ratio of their moduli. In particular, that tells you that ##\gamma v=(c/L)p##. A Taylor expansion of ##\gamma## for ##v<<c## will show you that ##L/c## is a sensible generalisation of the Newtonian concept of mass, and hence that ##p=\gamma mv##.

But note that step where I multiplied by ##c/L##. That doesn't work if ##L=mc=0##. In fact, the four velocity can't even be defined in such a case, since its length is defined to be ##c## so it would have to be both this and zero at the same time. The energy-momentum four vector works just fine, but the derivation of ##p=\gamma mv## breaks. So this formula simply doesn't work for things travelling at the speed of light.

Again - this will drive you crazy if you try to think of mass as a concept underlying momentum and energy. It isn't. You have to abandon that way of thinking - it's sooooo 17th century! Mass just tells you about the relationship between momentum and energy.
 
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  • #10
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But isn't p still dependent on m?
No, as the formula shows. Set m=0 and the equation reduces to E=pc. For massless particles momentum depends only on energy, not mass.
 
  • #11
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... plugging in m=0 and v=c (including the v in γ) would give E=0/0.
You're so close to understanding!

Forget relativity for a moment. By the end of the 1800s, we had Newtonian mechanics on the one hand and Maxwell's electromagnetics on the other. Momentum was supposed to be a conserved quantity ##\vec p = m \vec v##, but Maxwell's theory predicted that light carried momentum despite being massless (because otherwise momentum couldn't be conserved). This prediction was verified experimentally just a few years before Einstein's annus mirabilis: https://en.wikipedia.org/wiki/Radiation_pressure#Discovery

So even before special relativity was formulated, it was known that ##\vec p = m \vec v## couldn't be a definition of momentum. Rather, momentum was fundamentally a conserved vector quantity, and for everything except light the formula ##\vec p = m \vec v## held. This was suspicious. Shouldn't there be a way to express momentum that applies to everything? (And wouldn't it be preferable if the formula didn't return a value of zero for light?)

Now, in special relativity we have these two equations:

##E = \gamma m c^2##
##\vec p = \gamma m \vec v##.

But as you note, if you plug in ##m = 0## and ##v = c##, you end up with ##E = 0/0## and ##\vec p = \vec 0 / 0##, which are undefined. What does that tell us? Nothing much. Just that these equations are inapplicable to massless things. Now, this is a big step up from the pre-relativistic situation, when the formula ##\vec p = m \vec v## would give zero momentum for light, and all one could do was say "Well, that's not right, so I guess the formula just doesn't apply to light." With the undefined business, the math is telling us that the formulas don't apply in the massless case.

Fortunately we also have these two equations in SR:

##(mc^2)^2 = E^2 - (pc)^2##
##\vec p = (E/c^2) \vec v##,

both of which DO apply in the massless case.
 
  • #12
SamRoss
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Again - this will drive you crazy if you try to think of mass as a concept underlying momentum and energy. It isn't. You have to abandon that way of thinking - it's sooooo 17th century! Mass just tells you about the relationship between momentum and energy.
Thank you for the post. It was very insightful. It also raised new questions for me, though. If mass is not a concept underlying momentum and energy then how are momentum and energy independently defined? The usual way to derive energy is to take the integral of a force through a distance and the way to define momentum is to simply multiply the velocity four-vector with mass but those strategies don't seem to work.
 
  • #13
Orodruin
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Thank you for the post. It was very insightful. It also raised new questions for me, though. If mass is not a concept underlying momentum and energy then how are momentum and energy independently defined? The usual way to derive energy is to take the integral of a force through a distance and the way to define momentum is to simply multiply the velocity four-vector with mass but those strategies don't seem to work.
You have to understand that the definition of mass in relativity is not the same as the definition of mass in classical mechanics. Whereas the (inertial) mass in classical mechanics is the inertia of an object, mass in relativity is just a relationship between an object's energy and its momentum. Energy and momentum are intimately tied together in relativity as they transform into each other under Lorentz transformations. They are conserved quantities as long as your theory is translationally invariant in time and space. (If you have studied Lagrangian mechanics, they are the corresponding conserved quantities from Noether's theorem.)

Now, the quantity ##E^2 - p^2 c^2## for an object is a Lorentz invariant and it would therefore make very much sense to refer to it simply as the squared rest energy ##E_0## of an object, as it is equal to ##E_0^2## in the frame where ##p = 0##. However, when you do the non-relativistic (small speeds) limit of relativistic kinematics, something absolutely amazing happens - you recover classical mechanics with the inertia of an object being given by ##m = E_0/c^2##, which is nothing but a restructuring of the mass-energy equivalence ##E_0 = mc^2##. This is the reason we write ##E^2 - p^2c^2 = m^2 c^4## instead of using ##E_0^2##, the rest energy is just ##c^2## multiplied by the classical inertia in the object's rest frame. I find that this is very seldom appreciated enough and you will often see statements such as "##E = mc^2## is just a special case of ##E^2 - p^2 c^2 = m^2 c^4##" without the appreciation that it really did not have to be, but that it is a conclusion from the classical limit of the theory that the inertia is indeed equal to the rest energy (divided by ##c^2##). Because of this fact, we refer to ##\sqrt{E^2/c^4 - p^2/c^2}##, not an object's inertia, as the definition of an object's mass in relativity and it is equal to the object's inertia in its rest frame.
 
  • #14
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The usual way to derive energy is to take the integral of a force through a distance.
That's how you derive kinetic energy (or rather the change in kinetic energy). This is still true in special relativity, actually. See my post here if you're interested in seeing it done: Change of constants of integration for relativistic energy (when you get to ##mc^2 \Delta \gamma## at the end there, you can do a binomial expansion of ##\gamma = (1 - \beta^2)^{-1/2}##, and you'll find that the lowest-order term is just the change in the classical kinetic energy, ##.5 mc^2 \Delta (\beta^2) = .5 m \Delta (v^2)##).

and the way to define momentum is to simply multiply the velocity four-vector with mass but those strategies don't seem to work.
That's how you can define the four-momentum of something with mass. Massless things have four-momenta too, though: ##\vec P \equiv (E, \vec p c)##.
 
  • #15
PeterDonis
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how are momentum and energy independently defined?
They aren't. There is only one "thing" here in relativity, which is called the 4-momentum or the energy-momentum 4-vector in most textbooks. If you choose a particular inertial frame, then "energy" is the "time" component of this 4-vector in that frame and "momentum" is the 3-vector "space" part of this 4-vector in that frame. They are not independent of each other since the whole thing has to transform as a 4-vector under Lorentz transformations, which means that "energy" and "momentum" transform into each other when you change frames the same way that "time" and "space" transform into each other when you change frames.

@Orodruin gave an excellent explanation of how the concept of "mass" fits into all this in post #13.
 
  • #16
SamRoss
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Now, in special relativity we have these two equations:

##E = \gamma m c^2##
##\vec p = \gamma m \vec v##.

But as you note, if you plug in ##m = 0## and ##v = c##, you end up with ##E = 0/0## and ##\vec p = \vec 0 / 0##, which are undefined. What does that tell us? Nothing much. Just that these equations are inapplicable to massless things. Now, this is a big step up from the pre-relativistic situation, when the formula ##\vec p = m \vec v## would give zero momentum for light, and all one could do was say "Well, that's not right, so I guess the formula just doesn't apply to light." With the undefined business, the math is telling us that the formulas don't apply in the massless case.

Fortunately we also have these two equations in SR:

##(mc^2)^2 = E^2 - (pc)^2##
##\vec p = (E/c^2) \vec v##,

both of which DO apply in the massless case.
Thanks very much for the post. I was not imagining light as having momentum if it does not have mass and after reading your post I searched the internet for a little while. Would it be correct for me to say that for massive objects, energy and momentum are defined as E=γmc2 and p=γmv while for light they are defined as E=hc/λ and p=h/λ and that both of these definitions satisfy the equation m2c4=E2-p2c2 which is why that formula is seen as more fundamental?
 
  • #17
Orodruin
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Would it be correct for me to say that for massive objects, energy and momentum are defined as E=γmc2 and p=γmv
It would be more appropriate to state that the mass is defined as ##m = \sqrt{E^2/c^4 - p^2/c^2}## and that the velocity is given by ##pc^2/E##. This holds for both massive and massless particles. You can easily derive the special cases of ##m = 0## and ##m \neq 0## from there.
 
  • #18
SamRoss
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It would be more appropriate to state that the mass is defined as ##m = \sqrt{E^2/c^4 - p^2/c^2}## and that the velocity is given by ##pc^2/E##. This holds for both massive and massless particles. You can easily derive the special cases of ##m = 0## and ##m \neq 0## from there.
I am used to thinking of a world in which, in any given frame, length, time, and mass are fundamental units and other quantities (such as speed, momentum, and energy) are built from them. I can measure length with rods, time with clocks, and mass through collisions. Should I rather think of energy and momentum as fundamental units? If so, how are they measured?
 
  • #19
Ibix
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Thank you for the post. It was very insightful. It also raised new questions for me, though. If mass is not a concept underlying momentum and energy then how are momentum and energy independently defined? The usual way to derive energy is to take the integral of a force through a distance and the way to define momentum is to simply multiply the velocity four-vector with mass but those strategies don't seem to work.
As others have noted, energy and momentum aren't independently defined. All four vectors have one component that is different from the other three - the one associated with the time-like direction in spacetime (I've got my fingers crossed behind my back here, but it's a true enough statement for this level of discussion). As far as I'm aware, the original chain of thought was that relativity must look like Newtonian physics in the low speed limit. The quantity mass times velocity is interesting in Newtonian physics. I wonder if mass times four velocity will be interesting? Oh, hey, the three spatial components of the resulting vector look like momentum and the time-like one looks like kinetic energy plus some constant term. Ooh! And it's conserved too! That four vectors are the fundamental concept in relativity was a realisation that came later. It also leads you in more complicated directions, like massless objects that are somewhat dubious in Newtonian physics, but work fine in relativity.

Starting with a blank sheet, though, I suppose you'd just have to notice that there's some four vector that is conserved in collisions (a kinetic theory of gases should get you there pretty sharpish). Some systematic experimentation would get you to the right thing - a four vector normalised to some quantity m that remains constant (modulo nuclear stuff) for a given object.
 
  • #20
PeterDonis
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All four vectors have one component that is different from the other three - the one associated with the time-like direction in spacetime (I've got my fingers crossed behind my back here, but it's a true enough statement for this level of discussion).
Probably, yes, but as your intuition is apparently telling you, there are plenty of complexities lurking just beneath the surface. :wink:

If you prefix your statement with In an inertial frame, that restricts things enough to prevent most of the complexities from surfacing.
 
  • #21
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If mass is not a concept underlying momentum and energy then how are momentum and energy independently defined?
Usually energy is defined as the conserved quantity associated (through Noether’s theorem) with time translation symmetry, and momentum is the conserved quantity associated with space translation symmetry.
 
  • #22
SamRoss
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Usually energy is defined as the conserved quantity associated (through Noether’s theorem) with time translation symmetry, and momentum is the conserved quantity associated with space translation symmetry.
Yes, I think that's a better way of looking at those concepts for this problem. Thanks.
 
  • #23
Ibix
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Probably, yes, but as your intuition is apparently telling you, there are plenty of complexities lurking just beneath the surface. :wink:
Nothing to do with intuition, everything to do with knowing of light cone coordinates, ##t\pm x##. Although that would be an effective rebuttal of the idea that energy and momentum have independent existences, it's probably a bit abstract for the present discussion... Good point about covering myself by specifying inertial frames - thanks.
 
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