#### [SOLVED] 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)?

#### @Noldorin 2010-12-24 14:10:23

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

(In the equations below $$p$$ is momentum and $$c$$ is the speed of light, $$299,792,458 \frac{m}{s}$$.)

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

$$E / c^2 = m = \frac{p}{c} = \frac {h} {\lambda c}$$

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.)

#### @b1nary.atr0phy 2016-04-05 22:13:18

Since you aren't defining all your terms, p is momentum and c is the speed of light.

#### @Groo 2017-06-02 23:02:53

I believe relativistic mass is a very confusing term to people, so I would just like to provide some comments: 1) Energy and mass are the same thing, as you wrote. If gravity affects mass, you may as well say it affects energy. 2) Light is moving, therefore it has kinetic energy, therefore it is affected by gravity. 3) Kinetic energy of an object is merely a side effect of changing the inertial frame. Any object "standing still" on Earth at any given moment, is actually drifting away from some distant star at speed c, but this doesn't mean its mass is infinitely large.

#### @Groo 2017-06-02 23:05:45

Which is why I find it really funny when people say that "objects moving near the speed of light would have a near infinite mass", because I don't think they understand what relativistic mass actually is.

#### @Noldorin 2017-06-02 23:20:21

@Groo: Yes, indeed. This is probably why many teachers/authors tend to avoid the concept... I can't say I blame them much, these days.

#### @0tyranny 0poverty 2017-11-08 23:11:49

@Groo I agree that the more distant a galaxy is from us, the light coming from it is proportionately more doppler red shifted since it is moving away from us at a faster speed than a closer galaxy. However, I beg to differ with you that this is anywhere near c.

#### @Groo 2017-11-09 10:11:31

@0tyranny0poverty: due to expansion of universe, each day more and more galaxies are indeed speeding up and leaving the observable universe, but my point wasn't about c really, I just wanted to point that kinetic energy is a matter of reference frame of the observer. So, to leave that issue aside, for galaxies moving at (say) 10% of c relative to Earth, an object "standing still" on Earth will have a large kinetic energy and an increased relativistic mass.

#### @Theodore Norvell 2018-10-19 11:42:37

The last equation says $m = E / c^2 = hc/\lambda$. Shouldn't it be $m = p/c = h/\lambda c$?

#### @Evariste 2019-07-11 14:49:49

@TheodoreNorvell Yep, you're right

#### @Noldorin 2019-07-11 20:19:44

@TheodoreNorvell & Evariste: Yep, unfortunately I had written it correctly before, but when someone tried to improve the appearance of the LaTeX, they accidentally transferred the factor of $c$ from the denominator to the numerator! :-)

This is a fundamental question requiring fundamental thinking. I shall keep away from theories and concentrate on simple facts. From the day we knew of the Brownian motion and realizing that particles of matter are on a continuous motion and not at rest, we should have realized that motion and not rest is the true influential variable of nature. Velocity should therefore be adopted as the prime variable we use to study nature. But velocity has the units of space and time locked in an inseparable format, we should then conclude that space-time is a the variable that need to be considered in our scientific endeavor. But velocity of particles have to involve mass too. This then says that the most fundamental variable of nature is momentum with the units of mass, space and time locked together. As particles possess electrical charges too, we should also add electrical charge unit to get to the fundamental variable of nature.

When we look round we see that E.M radiation has all the above attributes. It has mechanical attributes in the form of energy and momentum flowing along the direction of propagation. This is given by the pointing vector P=E^H. Radiation also have electrical and magnetic attributes in its electric and magnetic fields that are normal to each other and normal to the direction of propagation. These attributes are all verifiable experimentally in the lab by simply directing a beam of radiation onto neutral and charged objects to see them move according to the laws of mechanics and electrodynamics.

Hence, momentum linear or angular is a defining property of our universe, be it in the form of energy or matter. As to why light bends round massive objects, we note that gravity also emerges when radiation condenses into matter. The key idea here is conservation of momentum. This is a fundamental property of our space and an experimental fact. Even elementary particles and radiation can't afford violating this principle. But if momentum is conserved, the forces between any two isolated particles locked in an orbit must be of the inverse square type as given in Bertrand theorem. Actually the theorem allows a spring type force(Hook's spring force) too, but this can be shown to be a limiting case of the inverse square force. Thus Newton's law of gravity and Coulomb's law of static interactions emerge as radiation condenses into matter.

Now, the formula for the bending of a projectile in the vicinity of a massive object in the Newton's theory(the inverse square force theory) have only the speed of the projectile in it, and not its mass. The mass simply cancel's out. According to this fact, Newton proceeded to calculate the deflection of light caused by the sun for example. As it happened, Einstein calculated the same angle and found it to be double that of Newton. People without deeper thinking announced that this meant that Newton's formula is wrong and the whole theory should therefore be discarded- despite the fact that the mass of the sun is not that of a black hole to merit a big modification of the Newton's theory. It turned out that Newton's calculation gives the actual angle of light deflection, whereas what we measure is twice that value due to the symmetry of the problem as shown clearly here; https://file.scirp.org/pdf/JMP_2017102615295175.pdf. The rays that are drawn straight from the source to the sun surface can't cross to the other side- they hit the Sun's surface instead. What we see is rays that come from an angle equal to that after crossing the Sun's surface. The two results support each other in a sense.

Can people who give negative ratings also give reasons for doing so, so the reader will be better informed. regards.

#### @anna v 2019-08-20 06:00:50

I have not voted, but the reason you get downvotes is that your answer is not within mainstream physics, which has quantum mechanics as the underlying level of nature from which the classical frameworks emerge. Classical mechanics and classical electrodynamics emerge from the quantum. The photon is a quantum particle and a quantum mechanical superposition of zillions of photons give the classical electromagnetic wave.

Thank you anna. I read for you and you seemed a reasonable person. Do you think this is reasonable. Are we back to the actions of the dark ages. I have peer reviewed articles that support any of the many ideas in my answer if anyone cares to ask for. regards.

#### @anna v 2019-08-20 06:43:20

it is just the policy of this site, to keep on mainstream physics for the answers, and we obey it.

The peer reviewed articles I referred to, are mainstream physics and some by esteemed authors like David Hestene and others. I think the correctness of the logic of a statement should be our guide to accept or reject anything said. regards.

#### @anna v 2019-08-20 18:41:15

It is possible also to get negative votes if people do think that the points in the question are not answered adequately.

Anna, I don't think I can agree here. The question asks how can photons have momentum when they have no mass. I showed that momentum is an essential attribute for anything to exist and felt. You have linear momentum if you are energy, and circular momentum or spin if you are a mass.

If a mass is moving, then it has angular momentum(spin) plus linear momentum(energy). This is clear from Einstein equation; E^2=(pc)^2 + (m0c^2)^2 =(pc)^2 + (p0c)^2=(p^2+p0^2)c^2, where m0 is rest mass. The p0 is the momentum in circulation and the p is that of motion. The square addition to generate the total momentum is due to the two being normal to each other. One tangential and the second is radial.

#### @anna v 2019-08-21 02:51:59

Well, you assume that the questioner can reason further and answer him/her/self the two questions in the body of the question, including the way photons, quantum mechanical entities, make up the classical light beam .

I promise this is my last comment; That things should have momentum to exist is one nature's facts and secrets. It is fundamental and can't be explained using other simpler facts. What is more intriguing is that the sum of momentum of the whole universe is zero. If not, another universe need to exist to balance it. So momentum seemed to be self generated, which can give the answer to how everything has come out of nothing.

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)

#### @Alchimista 2017-10-22 14:03:20

The point is that the mirror coated box will be heavier, not the photons

#### @Alberto 2012-12-17 20:50:55

In my opinion it is not necessary to evoke the theory of relativity or quantum physics to explain how light can have momentum but not mass. In the 19th century, it was already known that light can collide with matter; a beam of light can set a small wheel (in vacuum) rotating.

The key parameter for the study of collisions under classical mechanics is the momentum :

$$q= mv$$

(Momentum always being conserved in an isolated system)

The natural question is: Can the principle of conservation of the momentum be extended to electromagnetic radiations also?

From experience you know that the answer is positive, provided you define the momentum of light as

$$q = \frac{L}{c}$$

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

Can you extend the analogy assuming that light has mass too?

The assumption is reasonable. In case of positive answer, you get the Einstein equation

$$m = \frac{L}{c^2}$$

However you are not allowed to make such extensions 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 this paradox?

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

#### @Alex Jasmin 2018-08-21 03:11:32

Let's you keep things simple when relativity doesn't come into play.

#### @Johannes 2014-06-02 16:31:01

"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 $p=E/c$).

#### @AYAN BHUYAN 2012-08-12 19:04:27

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.

#### @Tom 2012-06-20 10:38:48

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.

#### @dmckee --- ex-moderator kitten 2012-07-06 13:17:33

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.

#### @user4552 2011-08-08 16:59:16

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 $p=E/c$.

#### @Haru Fujimura 2016-09-25 23:12:37

This is the best answer, other answers that try to insist that photons have mass, (of any form, relativistic or otherwise) should be voted down in my opinion, because it obscures the fact that energy bends space-time and thus changes the direction of the lightwave.

#### @Hammar 2017-03-06 20:24:35

"p=E/c" , How do you calculate the energy in this case ?

#### @owjburnham 2017-03-07 22:13:47

@Hammar The Planck Relation will give you the energy: $E=h\nu$, where $h$ is Planck's constant, and $\nu$ is the frequency of the light (so you'll sometimes see this written as $E=hf$).

#### @Theodore Norvell 2018-10-19 11:56:32

The second part of the question asks about light being bent by gravity. Does it follow from the fact that light has momentum that it is affected by gravity, as the question suggests?

#### @Noldorin 2018-10-19 18:26:39

Quantum mechanics cannot be ignored when talking about elementary particles like photons; that is a ludicrous statement. From a conceptual standpoint it’s necessary, even if “quantum mechanical” formulae aren’t explicitly used.

#### @Ben Wheeler 2019-01-25 18:47:24

What does "In units with c=1" mean??

#### @lalala 2019-04-11 13:24:47

As far as I remember the bending of light according to SR only gives half the angle predicted by GR. So for bending of light to be evidence of GR a (for 1915) high precision measurement was needed to distinguish the effects.

#### @Sklivvz 2010-12-24 13:30:09

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.