# Why is it that light has momentum?

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1. Aug 14, 2015

### Cole

The way that I understand it is this. But correct me if I'm wrong. Because P=m*v, sense infinity * 0 in calculus is 1, and because the speed of light is the limit of velocity, we can express it as infinity, so p = 0 kg * infinity m/s. Is that at all right?

2. Aug 14, 2015

### Staff: Mentor

I wouldn't say it that way. If you have zero times infinity then you simply cannot use that way to find the quantity of interest. It could be finite or it could be 0 or infinite. You just cannot tell.

So the physical reason that we say light has momentum is because it can change the momentum of matter. We can then work through the math and find a quantity with units of momentum that is conserved.

3. Aug 14, 2015

### DEvens

No. The speed of light is finite, being just a tad under 3E8 m/s. Also, as DaleSpam referred to, the idea of multiplying a zero by an infinity and getting a non-zero thing only makes sense in terms of a limit. You don't have any such limit.

Asking "why" in science is a troublesome thing. What sort of answer will you find satisfying? Eventually in science we come back to the postulates of a theory, and those postulates are "because I said so, and because they make a theory that works."

The important subject of special relativity relevant to the momentum of light is the idea of four-momentum and conservation of energy and momentum. One finds that, in order to conserve momentum and energy and keep relativistic forms of everything, one needs light to carry momentum E/c.

4. Aug 14, 2015

### Drakkith

Staff Emeritus
A photon has momentum because it has energy. The equation p = mv simply doesn't apply to a massless object. Instead, you have to use one of the following relationships:

Consider the full form of Einstein's equation: e2 = mc4 + p2c2
If m = 0 then the equation reduces to: e2 = p2c2
Taking the square root of both sides gives us: e = pc
Rearranging: p = e/c
Since the energy of a photon is given by e = hv, we can replace e in the momentum equation to get p = hv/c, just as shown above. (remember that the v here is the greek letter nu, which is used for frequency, not velocity) The other relationships are different ways of calculating the momentum.

5. Aug 14, 2015

### Mentz114

There is overwhelming experimental evidence for the momentum of the EM field. The Nobel prize winning experiment to create a Bose-Einstein condensate used emission and absorption of light-quanta to slow ions to almost zero velocity (in the lab frame) reaching temperatures very close to 0oK.

6. Aug 14, 2015

### PAllen

Light having both energy momentum was deduced from classical EM theory before either special relativity or quantum mechanics. Arguments like Dalespam suggested helped in the development of how the EM field could carry energy and momentum. This led to partial anticipations of a few features of SR one or two decades before 1905.

7. Aug 15, 2015

### HallsofIvy

And "0 times infinity" is NOT 1 in Calculus or any other standard mathematics!

8. Aug 15, 2015

### vanhees71

Photons are among the most abstract notions and cannot be understood properly without quantum field theory! So one should not argue with photons concerning phenomena which can be explained within classical electrodynamics. The classical answer is also not simple, but it's way more correct than to use an oversimplified photon picture, talking about photons as if they were just massless classical particles, which is as wrong a picture as you can have.

So here's the answer within classical field theory. As I said it's also not so simple either. First of all we must ask ourselves, why we need fields at all, and the answer is that in nature space and time are to be described in terms of Einstein's theory of relativity rather than Newton's space-time model. The idea of fields goes back to Faraday, who grasped the idea very intuitively from his careful experiments concerning electromagnetic phenomena.

Any lecture about electricity and magnetism starts with the interaction of charged particles and Coulomb's Law, according to which two pointlike charges, which are at rest (for a sufficiently long time!) attract or repell each other with a force acting along the line connecting the charges and a magnitude that is inversely proportional to its distance squared, but the particular shape of this force law is not so important here.

The important point now is what happens if one of the charges starts to move (e.g., because some electromagnetic field is switched on and accelerates this charge). What happens to the other charge concerning the mutual interaction? In Newtonian physics there's no problem to assume that instantaneously the force to this charge changes to the Coulomb force law. That's the "action-at-a-distance" picture of interactions, but this contracts the fact that within relativistic spacetime there can be no signal faster than the speed of light, and thus the charge interaction force cannot adapt corresponding to the other charge's change of position instantaneously.

Now this has another serious consequence: In Newton's mechanics the interaction between two particles must be such that the force on particle 1 excerted on particle 2 $\vec{F}_{12}$ must be just the oposite to the force on particle 2 excerted by particle 1, i.e., $\vec{F}_{12}=-\vec{F}_{21}$. The consequence is that the total momentum of the particles is conserved in Newtonian mechanics as long as only the interaction forces are at work. Indeed the equations of motion for the two particles read
$$m_1 \frac{\mathrm{d}^2}{\mathrm{d} t^2} \vec{x}_1=\vec{F}_{12}, \quad m_2 \frac{\mathrm{d}^2}{\mathrm{d} t^2} \vec{x}_2=\vec{F}_{21}=-\vec{F}_{12}.$$
Now adding the two equations and defining the total momentum to be
$$\vec{P}=m_1 \frac{\mathrm{d}}{\mathrm{d} t} \vec{x}_1 + m_2 \frac{\mathrm{d}}{\mathrm{d} t} \vec{x}_2$$
one finds
$$\frac{\mathrm{d}}{\mathrm{d} t} \vec{P}=0, \quad \vec{P}=\text{const}.$$
But for a relativistic interaction-force law, this cannot work, because when changing the position of particle 2 the force on a distant particle 1 cannot adapt to the new situation instantaneously, and vice versa, so one cannot have Newton's 3rd law for non-instantaneous interactions between the particles and thus there is a problem with momentum conservation.

On the other hand it is known, that momentum conservation is a consequence of the homogeneity of space, i.e., the assumption underlying both the Newtonian and the special-relativistic space-time model that the physical laws are not dependent on position. So there should be somehow a momentum-conservation law also in relativistic physics, and indeed the way out is the field-theoretical picture of interactions. So, according to Faraday, there is no action at a distance but some entity called the electromagnetic field that is around any charged body, and the electromagnetic force acting between two charges is not instantaneous at a distance but due to the field around the particles, i.e., particle 2 feels a force, because of the presence of the field due to the charge of particle 1 and vice versa. Now one can solve the problem with momentum conservation by assuming that the field itself is a dynamical quantity as the particles and thus can carry energy and momentum (the same argument holds true for angular momentum and center-of-momentum velocity due to the invariance under rotations and Lorentz boosts). Indeed as it turns out, Maxwell's equations are precisely describing this: accelerated charges are the source of electromagnetic waves that carry energy, momentum, and angular momentum with them, and the total amount of all these quantities is conserved.

There is only one problem left, and this is the socalled "radiation-reaction problem". If you have a single charge and accelerate it somehow (e.g., by switching on some electromagnetic fields created by other far-distant charges), also the field of the particle itself cannot always be a Coulomb field everywhere, because for the field to change at some distance because of the acceleration of the charge it is created from, needs some time, namely at least the time the light takes to travel from the position of the charge to the distant point, where we observe the field (btw. also somehow using another "test charge" to measure the forces due to this field!). Again this means that the field change is in terms of an electromagnetic wave which propagates with a finite speed, carrying away energy, momentum, and angular momentum; but this must influence the motion of the charged particle itself, i.e., it is not simply accelerated by the external field but it gives away some energy, momentum, and angular momentum to its own radiation field! So there must be forces (and torques) acting on the charge due its own radiation field. Using point particles, this leads to a desaster as is known since the late 19th century when Abraham and Lorentz investigated precisely this question in more detail in terms of the just discovered "electron theory", because at this time the electron was discovered in experiments with gas discharge tubes (Thomson 1897), and at this time of course the electrons where considered to be point particles in the sense of classical mechanics. Today, we still do not have a complete solution for this problem within classical mechanics, but this is another story.

9. Aug 15, 2015

### Staff: Mentor

@vanhees71, we have recently added a feature to PF for posters to be able to identify the level of response that they would like. The little green letter "B" next to the thread title indicates that the OP would like answers at a "Basic" level. I think that your answer is likely to be more "Intermediate" or "Advanced". Do you think that you can simplify your response a bit for the OP?

10. Aug 15, 2015

### Cole

That's very nice of you! Yes please that would help.
From what I've got so far is that, because light is a changed particle, it can still push against other charged particles because of its energy despite it having mass. I'm trying here.

11. Aug 15, 2015

### PAllen

Light is not a particle in any classical sense, and it is certainly not charged. I was trying to motivate the basics of the idea of light as propagating changes of electric and magnetic fields carrying energy and momentum had been worked out by the late 1800s. There is no need to bring in photons, massless particles, or relativity into this understanding.

12. Aug 16, 2015

### Markus Hanke

Look at it this way - suppose you have a ( classic ) test particle somewhere that is initially completely at rest ( no momentum ). Now you fire a photon at that test particle. What will happen ? The test particle will start to move very slightly - after the collision, it hence has momentum. We know already that momentum ( like energy ) is a conserved quantity, it cannot be created or destroyed; therefore, we must conclude that the photon has transferred momentum to our test particle during the collision, so it must have carried momentum.

Does this make sense ?

13. Aug 17, 2015

### Cole

Yes that makes a lot of sense! Thanks you so much!

14. Aug 17, 2015

### Cole

15. Aug 17, 2015

### bcrowell

Staff Emeritus
This is a "why" question, and a "why" question in science is always tricky. To answer it, you have to spell out what fundamental facts you are willing to take as your initial assumptions. Different people may have different preferences about these foundational choices. As a matter of taste, I would say that an answer to this question should not invoke photons or quantum mechanics, since relativity can stand on its own as a classical theory, and quantum field theory is hard. Also, my preference is for an answer that works for other massless phenomena, not just electromagnetic waves, because we (unlike Einstein in 1905) now know that light waves are only one of several different equally fundamental massless building blocks of the universe.

So as starting assumptions, let's take just the following: (1) we've already established all the well-known facts about kinematics in special relativity; (2) we're talking about an object O that is massless; (3) a massless object behaves the same as a massive object with undetectably small mass; and (4) when a constant force is applied to an object, the change in its momentum equals the force multiplied by the time interval, $\Delta p=F\Delta t$. Given assumption 4, one can also show that the equation for work $W=Fd$ applies without any need for relativistic corrections.

We do not assume Newtonian relations such as $p=mv$ or $p=\sqrt{2mE}$, which as we'll see below cannot hold relativistically.

So now suppose that we take an object of mass $m>0$, start it from rest, and apply a constant force to it for time $t$, stopping when we have done some desired amount of work $W$. During this time, it moves a distance $d=W/F$, which is finite. By assumption 1, we know that the object will always go slower than $c$, so $t>d/c=W/cF$. Assumption 4 then tells us that the object's momentum is $p>p_{\text{min}}=W/c$. As we make the mass smaller and smaller, $t$ get smaller, and so does $p$, but because we always have $p>p_{\text{min}}$, assumption 3 tells us that the result for a massless object must still be a momentum greater than or equal to $p_{\text{min}}$. Therefore a massless object with energy $E$ has a momentum greater than or equal to $E/c$. (It turns out that it exactly equals $E/c$, although I won't show that here.) Therefore any massless object with nonzero energy has a nonzero momentum.

Our argument proves that the Newtonian relations $p=mv$ and$p=\sqrt{2mE}$ need to be modified relativistically, since clinging to either one would give $p=0$ when $m=0$, leading to a contradiction. Although it's outside the scope of this answer to show it, it turns out that relativistically, the correct definition of mass is $m^2=E^2-p^2$ (in units such that $c=1$), where $E$ is the mass-energy. In the special case of $m=0$, we find $p=E$, or $p=E/c$ if we revert to units with $c\ne1$.

Last edited: Aug 17, 2015
16. Aug 18, 2015

### vanhees71

I wonder which other massless building blocks you have in mind. The only one which might exist is one of the three neutrinos, but that's not clear yet. We don't know the neutrino masses very well yet. We'll see what comes out in the near future in experiments like KATRIN, measuring the endpoint of the tritium-decay spectrum with unprecedented precision.

Electromagnetic waves are described very probably by an exactly massless vector field. At least the upper bound for the photon mass is very small.

Nevertheless, let me stress again, that particle pictures of electromagnetic waves are very tricky at best. One should not argue about them in terms of particles at all. The short answer is that electromagnetic waves in the classical picture are dynamical entities that carry energy, momentum, and angular momentum. The reasons, why such entities are natural in a relativistic world, I tried to give previously. I agree that this is not a simple explanation, and I'm sure that somebody else can express it in simpler terms then I. In any case, they are not like pointlike objects with a defined position and momentum but fields and as such present everywhere.

You can have wave packets of the free electromagnetic field which behave in some sense like a particle, when you take the (energy-density weighted) mean position of the packet and its total momentum, but one must not think about this as a particle in a literal sense.

17. Aug 18, 2015

### Markus Hanke

I can't speak for bcrowell, but gluons should also carry momentum, though I must admit that I have never seen an explicit expression for this.

18. Aug 18, 2015

### vanhees71

Sure, gluons are massless and carry also momentum. However, it's a bit tricky, because the gluons are confined inside hadrons, and you cannot observe them as free massless quanta like the photons, which are massless quanta of the electromagnetic field (note that intentionally I explicitly avoid to call them "particles"; "massless quanta" seems to be save to alert the readers to be careful when thinking about them). For the same reason (confinement) there's also no analogon to classical electromagnetic fields for the gluon field.

19. Aug 18, 2015

### bcrowell

Staff Emeritus
I had in mind gluons and gravitational waves.

20. Aug 18, 2015

### vanhees71

Sure, gravity is also a massless field. Gravitational waves are not directly observed yet, but as far as I heard from my colleagues involved in GR the hopes are high that they will soon be detected.