Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

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)?

share|cite|improve this question

9 Answers 9

up vote 47 down vote accepted

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.)

share|cite|improve this answer

"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 $p=E/c$).

share|cite|improve this answer

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)

share|cite|improve this answer

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

share|cite|improve this answer

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.

share|cite|improve this answer

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.

share|cite|improve this answer
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

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 $p=E/c$.

share|cite|improve this answer

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.

share|cite|improve this answer
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

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


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.

share|cite|improve this answer

protected by dmckee Mar 20 '13 at 19:15

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.