# If photons have no mass, how can they have momentum?

As an explanation of why a large gravitational field (such as a black hole) can bend light, I have heard that light has momentum. This is given as a solution to the problem of only massive objects being affected by gravity. However, momentum is the product of mass and velocity, so, by this definition, massless photons cannot have momentum.

How can photons have momentum?

How is this momentum defined (equations)?

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Note: one way to resolving this apparent paradox is by noticing that SR doesn't allow massive particle going faster than the speed of light. That means that speed of light is the limit and in certain sense it is infinite (this sense can be made precise by the concept of rapidity). Now, it shouldn't be surprising anymore that the momentum $p = mv$ can be non-zero, because you multiply zero times infinity (again in the above sense) and so in general any result is possible. This is by no means a complete explanation, just something that should give you little intuition. –  Marek Dec 24 '10 at 14:38
This is nine months old, but I feel compelled to post because none of the answers below, including the accepted one, are correct. Here physicsforums.com/showthread.php?t=511175 is a correct explanation. –  Ben Crowell Aug 8 '11 at 15:22
@Marek: A better way to formalize the intuition you've expressed is that $p=m\gamma v$, where $m$ is the invariant rest mass. What approaches infinity is not $v$ but $\gamma$. If you evaluate $\lim_{v\rightarrow c}m\gamma v$, while holding $E=m\gamma c^2$ constant, you do get $E/c$. –  Ben Crowell Aug 8 '11 at 15:55
@Ben: well, it's true that this formalizes the argument but it is still unphysical since now you are letting $m \to 0$ and it's not clear what this means physically. Also, one can increase $\gamma$ simply by moving away from the observed stationary particle (and obviously holding $m$ fixed) which gives $E \to \infty$ but naturally this doesn't tell you anything about the particle itself as all of $E$ comes from your moving away. The bottom line is: the fact can't really be derived as a limit of somenthing else. It can either be understood by intuition or else derived directly from SR. –  Marek Aug 11 '11 at 12:03
OK,, let me get this straight photons have mass, and yet they don't. It seems to me a quandary, but if we consider 1 photon is traveling a C, then any mass in the vector of travel at the speed of light becomes infinite while the mass in any perpendicular vector becomes 0. This basic understanding seems to support both views. and since a photon cannot exist if it isn't moving,, we have our answer. –  user8499 Apr 5 '12 at 23:33

There are two important concepts here that explain the influence of gravity on light (photons).

1. The theory of Special Relativity, proved in 1905 (or rather the 2nd paper of that year on the subject) gives an equation for the relativistic energy of a particle;

$$E^2 = (m_0 c^2)^2 + p^2 c^2$$

where $m_0$ is the rest mass of the particle (0 in the case of a photon). Hence this reduces to $E = pc$. Einstein also introduced the concept of relativistic mass (and the related mass-energy equivalence) in the same paper; we can then write

$$m c^2 = pc$$

where $m$ is the relativistic mass here, hence

$$m = p/c$$

In other words, a photon does have relativistic mass proportional to its momentum.

2. De Broglie's relation, an early result of quantum theory (specifically wave-particle duality), states that

$$\lambda = h / p$$

where $h$ is simply Planck's constant. This gives

$$p = h / \lambda$$

Hence combining the two results, we get

$$m = E / c^2 = h / \lambda c$$

again, paying attention to the fact that $m$ is relativistic mass.

And here we have it: photons have 'mass' inversely proportional to their wavelength! Then simply by Newton's theory of gravity, they have gravitational influence. (To dispel a potential source of confusion, Einstein specifically proved that relativistic mass is an extension/generalisation of Newtonian mass, so we should conceptually be able to treat the two the same.)

There are a few different ways of thinking about this phenomenon in any case, but I hope I've provided a fairly straightforward and apparent one. (One could go into general relativity for a full explanation, but I find this the best overview.)

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This heuristic argument may help convince the questioner that there is some notion of momentum (if not rest mass) attached to photons, but it is incorrect to invoke Newton's theory of gravity to say they have gravitational influence. They do not affect Newton's theory of gravity. The curved trajectories of photons can only be described with special relativity. There is no mechanism within Newton's theory to assert that a shiny flashlight will attract a marble. (In GR, the flashlight beam has a non-zero stress-energy tensor, which will affect nearby trajectories. Hard to say exactly how.) –  Eric Zaslow Dec 24 '10 at 15:39
@space_cadet: To give an accurate quantitative response, that is true. The concept can however be explained without resorting to all the fancy shit. –  Noldorin Dec 26 '10 at 14:54
E=pc has nothing to do with quantum mechanics. It's a purely classical feature of SR. To derive E=pc, just set $m_o=0$ in your first equation, and you're done. The discussion about gravity is also irrelevant, since E=pc is about SR, not GR. It's also incorrect to imagine that the energy $E$ of a photon can be used to find an equivalent gravitational mass via $E=mc^2$; the source of gravitational fields in GR is the stress-energy tensor, not the scalar mass-energy. –  Ben Crowell Aug 8 '11 at 15:18
@Ben: the "stress energy tensor" has the energy density as it's 00 component, and the integral of that over space is the relativistic mass. A light wave has a complicated gravitational field, because it is moving fast, but two light-waves moving in opposite directions which are at a box gravitate exactly like a point mass with their "relativistic mass" GM/r^2. "Relativistic mass" is just another name for the energy for a person who has not internalize their equivalence yet. –  Ron Maimon Aug 11 '11 at 7:50
@Ron Maimon: No, the integral of $T_{00}$ over space is not the relativistic mass in GR (and you can't get away with SR here because Noldorin is talking about gravitational fields). There are various ways of defining a scalar, conserved mass-energy in GR (ADM mass, Komar mass, ...), and none of them are simply a volume integral of $T_{00}$. –  Ben Crowell Aug 14 '11 at 23:26

The answer to this question is simple and requires only SR, not GR or quantum mechanics.

In units with $c=1$, we have $m^2=E^2-p^2$, where $m$ is the invariant mass, $E$ is the mass-energy, and $p$ is the momentum. In terms of logical foundations, there is a variety of ways to demonstrate this. One route starts with Einstein's 1905 paper "Does the inertia of a body depend upon its energy-content?" Another method is to start from the fact that a valid conservation law has to use a tensor, and show that the energy-momentum four-vector is the only tensor that goes over to Newtonian mechanics in the appropriate limit.

Once $m^2=E^2-p^2$ is established, it follows trivially that for a photon, with $m=0$, $E=|p|$, i.e., $p=E/c$ in units with $c \ne 1$.

A lot of the confusion on this topic seems to arise from people assuming that $p=m\gamma v$ should be the definition of momentum. It really isn't an appropriate definition of momentum, because in the case of $m=0$ and $v=c$, it gives an indeterminate form. The indeterminate form can, however, be evaluated as a limit in which $m$ approaches 0 and $E=m\gamma c^2$ is held fixed. The result is again $p=E/c$.

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+1 for sensibility. Most answers seem to think we need to justify light having some kind of positive mass for some reason. Though that can be done, relativistic mass isn't very useful as a concept. –  Stan Liou Aug 12 '11 at 1:25
I agree with Stan Liou, this is the best answer to the question. There is absolutely no need to introduce relativistic mass, which can be a misleading concept even to students of physics. I guess it's mostly an attempt to stay within our classical way of thinking about movement, but it provides more misconceptions than insights (e.g. people thinking a higher velocity implies a stronger interaction with the Higgsfield). –  Wouter Jun 15 '13 at 19:24

If Newton's gravitation could define the bending of light by gravity, then the general relativity wouldn't have come up. Photons don't have mass and it's clear from the fact that it travels at the speed of light. Gravity is an illusion that seems to attract things but in fact it bends spacetime; which is why a straight path seems curved. Newton's law of gravitation is still used because it's simple and we seldom encounter such massive objects like black holes in practical life, for which it does not hold.

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The reason why the path of photons is bent is that the space in which they travel is distorted. The photons follow the shortest possible path (called a geodesic) in bent space. When the space is not bent, or flat, then the shortest possible path is a straight line. When the space is bent with some spherical curvature, the shortest possible path lies actually on an equatorial circumference.

Note, this is in General Relativity. In Newtonian gravitation, photons travel in straight lines.

We can associate a momentum of a photon with the De Broglie's relation

$$p=\frac{h}{\lambda}$$

where $h$ is Planck's constant and $\lambda$ is the wavelength of the photon.

This also allows us to associate a mass:

$$m=p/c=h/(\lambda c)$$

If we plug in this mass into the Newtonian gravitational formula, however, the result is not compatible with what is actually measured by experimentation.

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This is a somewhat confused answer. The most obvious explanation for photon momentum is due to special relativity. –  Noldorin Dec 24 '10 at 13:34
Actually, many people thought light IS bent in Newtonian gravity, just like any massive particle with finite velocity. The finite velocity of light had been known since Galileo. –  Jeremy Dec 24 '10 at 14:34
@Jeremy: except that Newtonian gravity can't explain GR at all. There are some accidental equalities, like the equity of Schwarzschild radius in both theories. But in other cases values differ. E.g. Newtonian picture predicts twice smaller bending of light than GR (this is again related to the above-mentioned fact that light behaves differently in GR than massive particles) and this shows that any similarity is purely accidental and trying to explain GR effects with Newtonian picture serves only to hide the true concepts that happen in the curved space-time. –  Marek Dec 24 '10 at 14:47
@Bruce: I have to disagree purely on the grounds that this answer is very sketchy. It gives next to no justification or depth unfortunately. GR definitely has something to say here, but it's perhaps not the primary point. –  Noldorin Dec 24 '10 at 15:43
@Noldorin: We'll just have to agree to disagree than. I think that an explanation that gets the numbers clearly wrong is a wrong explanation (wrong as in: it coincidentally predicted the existence of this event, but trying to do the same for other stuff can lead the person to predict the existence of nonexistent phenomena). And I think that using this explanation without disclaimers in bold to warn the reader is, indeed, harmful. –  Malabarba Dec 24 '10 at 16:14

Until a good specialist comes and clean up this mess, I am going to touch in point that was not raised yet.

Momentum is the Noether charge of translational invariance. If you consider only massive entities in system Lagrangian, it will be noticed that the Lagrangian is not invariant over shift translation unless you add the electromagnetic field Lagrangian.

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This answer is way too confused, so -1. Both massive and massless particles can have translation invariant dynamics (indeed, all of our state-of-the-art theories respect even much stronger Poincaré symmetry) and so there will be conserved charges called momentum in every case. But what precisely is the momentum depends on the precise dynamics of the theory. This is why these notions of momentum differ between classical and special relativistic theories and between massive and massless theories. –  Marek Dec 25 '10 at 0:18

In my opinion it is not necessary to evoke the theory of relativity or quantum physiscs to explain how light can have momentum but no mass

In the 19th century it was already known that light can collide with matter A beam of light can make a small wheel, in vacuum, rotating

the classic mechanics key parameter for the study of collisions is the momentum : q = mv, Momentum being alwayes conservative in an insolated system The natural queation is: Can the principle of conservation of the momentum be extended also to electromagnetic radiations ? From the experience you know that the answer is positive provide you define the momentum of light as

               q = L/c


Where L is the energy of light and c the light speed

Can you extend the analogy assuming that light has also mass ? The assumption is reasonable, in case of positive answer you get the Einstein's equation

           m = L/c^2


However you are not allowed to make such extension since in Physics you must stick to the experimental evidences There is no evidence that light has also mass If so , how do you solve the paradox ?

The light momentum and the momentum of a material particle are not the same thing

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In the 19th century it was already known that light can collide with matter A beam of light can make a small wheel, in vacuum, rotating No, this is a historical misconception. See the WP article "Crookes radiometer." In my opinion it is not necessary to evoke the theory of relativity or quantum physiscs to explain how light can have momentum but no mass You invoke the relation $q=L/c$, which is a relativistic relation. –  Ben Crowell Sep 1 '13 at 14:15
In the 19th century the TeX typesetting was certainly unknown. But IMHO in the 19th century momentum was already denoted by $p$, whereas $q$ denoted “position”. –  Incnis Mrsi Aug 20 at 19:15

"momentum is the product of mass and velocity, so, by this definition, massless photons cannot have momentum"

This reasoning does not hold. Momentum is the product of energy and velocity.

"How is this momentum defined (equations)?"

Inserting factors of $c$, the relativistically correct relation between momentum $p$ and velocity $v$ is $$c^2 p = E v$$ This holds for non-relativistic massive particles (total energy dominated by rest-energy: $E = m c^2$, and therefore $p=mv$) as well as for massless particles like photons ($v = c$ and hence $p=E/c$).

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Light doesn't have momentum in the normal sense that matter has. Frequency and Wavelength soak up momentum. The more energetic the higher the frequency. Wavelength can change even though light stays at C .. Light doesn't bend, but space can be deformed. A better question is how can space be deformed when space has even less energy, mass, or wavelength than light does?? Space has no mass, momentum, yet changes (grows) between galaxies in the voids between them. Space takes a tremendous amount of energy to deform it, but what is causing space to be deformed? What is gravity exactly? There is something about space and time that is linked together. Time slows down as you encounter a gravitational field. How can time have energy to distort space? Where does the energy that time has come from and where does it go? What is time exactly? Time and gravity are unknowns yet have a constant predictable nature in physics, except quantum physics. A singular particle can phase into different multiverses popping back into existence by probability. How can a particle exist in two different locations at the same time? Where are multi-verses in relation to this universe located? Is the Weak Nuclear Force weak because it exists in all multi-verses simultaneously in comparison to the other forces in physics that can only exist in one universe at time? I have about a half dozen more questions but I will just stop at this point. The more we know about universe the bigger the unknown grows.

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Even pure classical (i.e. Maxwell's equations) light has well defined momentum, a fact which has been known since the end of the 19th century. –  dmckee Jul 6 '12 at 13:17

Of course they have mass. When saying "photons have no mass" in LHC rap, they were referring to the rest mass, it just didn't rhyme.

(If you pack a bunch of photons into your mirror-coated box, it will be heavier, by E/c^2 as usual)

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Note that particle physicists, cosmologist and other specialist in relativity do not use the term "relativistic mass" at all finding the term to be definable but not useful. There is just $\text{mass} = (E,\vec{p})^2$ and energy and momentum and kinetic energy. –  dmckee May 29 '11 at 1:44
"If you pack a bunch of photons into your mirror-coated box, it will be heavier, by E/mc^2 as usual." This is sort of true, but not really relevant. What is different between a photon and a photon-filled box is that the former has zero invariant mass, while the latter has a nonzero invariant mass. (Invariant mass isn't additive.) "Heavier" is also somewhat irrelevant and misleading. The question relates to inertial mass, not gravitational mass. –  Ben Crowell Aug 8 '11 at 17:36
In the general case, it won't even be more massive "by $E/c^2$ "as usual." Mass is just the magnitude of the four-momentum vector, and of course the sum of two null (lightlike) four-vectors have a positive magnitude. This explains the 'heavier' box, but the mass of the box will increase by $E/c^2$, E total energy of the photons, only in the frame in which the momenta of the photons balance each other. –  Stan Liou Aug 9 '11 at 6:09

## protected by dmckee♦Mar 20 '13 at 19:15

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