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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)?

Qmechanic
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david4dev
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    In my opinion we should only enquire; why a photon has zero mass, because the photon has momentum is demonstrable via numerous experiments- whereas zero mass is not. – Riad Apr 29 '21 at 16:05
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    You can reverse the logic: momentum is that thing that changes during interactions and is otherwise conserved because space is homogeneous. Given any object (including the photon), you can define its momentum in this way if you can set up a Hamiltonian or Lagrangian description for that object. Yes, for classical non-relativistic point particles momentum is "mass times velocity", but this is just an example of momentum, NOT the definition. – Quillo Jun 11 '22 at 09:54
  • Ref. to comment by Riad: Photons have no rest mass because mass is defined by its acceleration, and photons cannot be accelerated (as it occurs: the latter, as a reason, is not implied in the definition of mass ("kg"). It seems some other empirical finding that there are known no particles, hence no masses, that travel beyond speed of light. No reasons given. "we should enquire" - did you set up a question? – Peter Bernhard Nov 12 '22 at 19:19

10 Answers10

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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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    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. – Haru Fujimura Sep 25 '16 at 23:12
  • "p=E/c" , How do you calculate the energy in this case ? – Hammar Mar 06 '17 at 20:24
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    @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 $E=hf$). – owjburnham Mar 07 '17 at 22:13
  • 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? – Theodore Norvell Oct 19 '18 at 11:56
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    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. – Noldorin Oct 19 '18 at 18:26
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    What does "In units with c=1" mean?? – Ben Wheeler Jan 25 '19 at 18:47
  • 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. – lalala Apr 11 '19 at 13:24
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    @BenWheeler Measuring speed in SI Units like m/s is a common practice. But in scientific notation, it is redundant and carries no extra information. For getting the important information out of the maths, constants like c are written in units that make them dimensionless. So, if c (which was $3 \times 10^8$ m/s) = 1, v = 0.5c (which was $1.5 \times 10^8$ m/s) becomes v = 0.5 – Shailesh Pincha Sep 09 '22 at 06:11
  • @ShaileshPincha So it is just a change of variables? – Tyma Gaidash Oct 17 '22 at 18:39
  • @TymaGaidash Yes. – Shailesh Pincha Oct 21 '22 at 08:41
  • Refering to Hammar, 2nd comment: The implication seems that you need to know the frequency of a photon to tell its momentum (it's not velocity but frequency); and "relativistic mass" of photon is a way to disguise that Newton's "m x a" doesn't help. – Peter Bernhard Nov 12 '22 at 15:22
  • It would be the perfect answer if you indicate your where you get the m² = E² - p² from – Shamal Karunarathne Oct 24 '23 at 21:54
  • I'm confused how p = mv gives an indeterminate form? You get p = 0*c -> 0 = 0, what about that is indeterminate? It's not dividing by 0 or using an infinity anywhere. Oh wait just read up on what the gamma means, how can special relativity just introduce an infinity in there? Whatever – Ethan Dec 02 '23 at 01:46
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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.)

Noldorin
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    Since you aren't defining all your terms, p is momentum and c is the speed of light. – arkon Apr 05 '16 at 22:13
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    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. – vgru Jun 02 '17 at 23:02
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    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. – vgru Jun 02 '17 at 23:05
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    @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. – Noldorin Jun 02 '17 at 23:20
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    @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. – 0tyranny0poverty Nov 08 '17 at 23:11
  • @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. – vgru Nov 09 '17 at 10:11
  • The last equation says $m = E / c^2 = hc/\lambda $. Shouldn't it be $ m = p/c = h/\lambda c $? – Theodore Norvell Oct 19 '18 at 11:42
  • @TheodoreNorvell Yep, you're right – Evariste Jul 11 '19 at 14:49
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    @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! :-) – Noldorin Jul 11 '19 at 20:19
  • @Groo Till I saw your comment this is how I understood it too. You cannot accelerate an object to the speed of light because as speed increases, relativistic mass increases, and therefore more energy is required to accelerate it and hence infinite energy is required to reach the speed of light. But since you say mass doesn't increase can you shed a little more light on this? – Lost Apr 19 '21 at 12:06
  • @Lost: you can look at it this way: since motion is relative, when a spaceship is travelling at 0.7c relative to Earth, Earth is also travelling a -0.7c relative to the spaceship. From the spaceship's perspective, it's Earth that has a huge relativistic mass, even if the person on the ship is trying to increase ship's speed relative to Earth. Also, if there is another ship next to this one, travelling at the same constant speed relative to Earth, both of them are stationary from their reference frames, and have their "relativistic masses" equal to... – vgru Apr 19 '21 at 19:18
  • ...rest masses. Meaning that they can accelerate easily relative to each other, none of them is thinking about "infinite masses", they can have a plain old fashioned space race and none of them will notice anything weird, unless their relative speed becomes very large. So you can see that "mass approaching infinity" cannot be the reason why you cannot accelerate an object past the speed of light, because there isn't an absolute frame of reference towards which you could say an object has infinite mass. – vgru Apr 19 '21 at 19:21
  • The other way to reason it is through kinetic energy. If you are sitting, your kinetic energy is zero, relative to your stationary frame. Even if you are sitting inside a train travelling at 300mph relative to Earth. Even if you are sitting on Earth, which is travelling at 150000mph relative to some other galaxy, v = 0, E = 0. In a way, relativistic mass is the idea that you can take this kinetic energy (which can be zero or millions of joules, depending on the frame) and add it to rest energy (in that first expression, E2 = (mo2c2)2 + p2c2), and then pretend that the p2c2 part disappeared. :) – vgru Apr 19 '21 at 19:32
  • @Groo Okay. So, then I guess the argument that 'to accelerate it further, more enrgy is required' still holds (?) since if other thing is accelerating its 'relative vel' only changes and what I said would still apply even when the "infifnite mass" is an ambiguious thing. Ryt? – Lost Apr 20 '21 at 06:40
  • @Groo If I see an object's vel increasing, what that means is its relative velocity is increasing and then its relativistic mass increases wrt my frame and hence wrt my frame it can never reach velocity of c. Same argument can exactly be applied to other frames and hence speed of light can't be reached from any ither frame. Is this line of thought correct? – Lost Apr 20 '21 at 06:42
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    @Lost: your reasoning is correct, yes. The "further" word in "to accelerate it further, more energy is required" is perhaps not the best way to reason about it. There will always be some part of the universe which is currently moving at close to c relative to you, and no matter how much energy you put into accelerating yourself, this relative speed won't pass c. But at the same time, if you are not accelerating, there will be a part of the universe which is almost stationary relative to you. Relative to this reference frame, nothing will stop you to get "further" from zero to 0.1c. – vgru Apr 20 '21 at 09:33
  • "... where m0 is the rest mass of the particle (0 in the case of a photon). Hence this reduces to E=pc." - Why had rest mass been introduced in the first place, in that additive formula, if it is zero? I'd say: To reconcile with Newtonian law (m x v2) seen anew: Mass the rest mass of which is more than zero never completely (conversely seen) puts all its energy into impulse. A photon may be absorbed by an electron - it puts in all all of its momentum. It's a paradox that regular mass is some "lack" of momentum and photons may be seen as mass turned into momentum only. – Peter Bernhard Nov 12 '22 at 09:56
  • @PeterBernhard Because that is the general formula (the energy-momentum relation), the one that is usually learnt/quoted. Simple as that. – Noldorin Nov 12 '22 at 16:34
  • @Noldorin. I feel corroborated. I just uploaded some other answer focusing on that aspect. – Peter Bernhard Nov 12 '22 at 17:13
  • Refering to "they have gravitational influence. " This doesn't seem to imply that photons are not accelerated by gravitation. Unlike mass on both plates of a scale photons do not gain velocities. – Peter Bernhard Nov 12 '22 at 17:21
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"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 $p=E/c$).

Johannes
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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.

Sklivvz
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  • Please note that Newton himself calculated the deflection of light and it is his results that Einstein used to compare with. See also my answer where a reference is given that the Newoton and Einstein results with a factor of 2 are both correct if interpretted correctly. regards. – Riad Apr 29 '21 at 16:00
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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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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.

AccidentalFourierTransform
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Alberto
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  • Let's you keep things simple when relativity doesn't come into play. – Alex Jasmin Aug 21 '18 at 03:11
  • "light can collide with matter; a beam of light can set a small wheel (in vacuum) rotating" - this only works if it's not a complete vaccume because the light is heating the air which pushes on the wheel. It's not light pushing the wheel directly. – Systemsplanet May 17 '23 at 23:20
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"How can photons have momentum?" (if they do not have mass)

How is this momentum defined (equations)?"

Wikipedia on momentum says: "In Newtonian mechanics, momentum (...) is the

product of the mass and velocity

of an object."

p=mv

Interestingly, Wikipedia does not mention,in that context, the De-Broglie wavelength equation which defines momentum alternatively as:

p=h/λ

Both equations put in one give mv = h/λ, which shows that p (momentum) is not derived from mass alone. One and the same momentum may come from masses of different velocities or, different from that, from different waves of different lengths. Momentum (and energy) unifies both causes in one and the same effect.

It's an empirical finding that photons have momentum which is being made use of by the solar sail, for instance, and it is a definition of terms to say that photons have "mass", if this is some other way to express the equivalency of both formulas above. To be precise, photons are not said to have mass, they are said to have "rest mass" zero and they are said to have "relativistic mass".

Both terms are intricate, and the following remarks might not explain:

Rest mass as a term literally refers to weights put on both sides of a scale: masses are weighted "at rest" to each other. However, the unit of mass is "kg", and mass is measured by acceleration, by gravitational acceleration (of the earth). Thus, photons' rest mass is considered zero as particles that live at speed of light cannot be accelerated: there do not own "mass" because they cannot be gauged one against another. A particle of wavelength x cannot be put on scales against particle of wavelength y. The term "rest" in fact, in my opinin, may be considered misleading, as it is earth's acceleration, thus "speed not rest" that is being measured. If this is understood it becomes clear that photons which have speed of light cannot be accelerated. The speed of light separates the realm of mass (in strict sens) from the realm of waves. Waves differ in frequency, not velocity, Masses differ in velocity, not frequency. As photons cannot be gauged by different velocities they gain by gravitational (and inertial) acceleartion, they are said to have "rest" mass zero.

On the other hand, there is the term "relativistic" mass. This term is opposed to "rest mass" which is no trivial knowledge, it is just the opposite. Relativistic mass is a term to acknowledge that restmassless photons do have mass in the form of "relativistic mass" as they own momentum (solar sail impulse) just like Newtonian masses have. It is a mode of putting it.

How can photons have momentum if they do not have mass?

The de Broglie formula shows that mass is not a prerequisite for momentum. This formula is based on empirical findings; there is no "reason why". To reconcile these findings with Newtonian formula photons are said to have "rest mass" zero (no accelaration to variation in speed) on one hand, and on the other hand, they are said to have "relativistic" mass (in that sense they do have mass(!)) as they exert mechanical momentum (do not only transfer energy).

  • No, $p=\frac{h}{\lambda}$, as mentioned in several other answers. Also, modern treatments of SR tend to avoid the concept of relativistic mass because it can be misleading and confusing. There are a few demonstrations of that confusion on this page, but also see https://physics.stackexchange.com/q/3436/123208 There's more on the deprecation of relativistic mass on https://physics.stackexchange.com/a/133395/123208 and the links on that page. – PM 2Ring Nov 12 '22 at 17:43
  • Immediately I corrected the error you stated, thank you. I will read through the links you gave. It is very helpful to advise as you do on the term relativistic mass. However, is "rest mass" any better termed? Admittedly, the question above assumes that photons have "no mass", hence, answering on relativistic in contrast to rest mass is in contradiction to the "premise" of the question. By the way "rest" mass denotes "acceleratable" mass, which is even more confusing, in my opinion (that picture is two plates of one scale, in rest, they are being accelerated though: "rest") – Peter Bernhard Nov 12 '22 at 19:06
  • Modern physics uses the term "mass" or occasionally "invariant mass" in preference to "rest mass", but sometimes the term "rest energy" is used. – PM 2Ring Nov 12 '22 at 19:40
  • All new words you teach us are intersting. "Invariant" is better termed then "rest". In Einstein's language the photon "rast" - runs crazily fast, at speed of light, it's a verb: rasen -, thus it sounds as just the opposite of "rest". Again, invariant mass is invariant thus at rest to other mass related to when measuring, both are being accelerated by the floor of the scales. Then, there is "rest energy", even more interesting, as energy not moved seems mass., that's another aspect that appears.
    • Many related questions at StE sortable now by their premise mass or no mass of photon.
    – Peter Bernhard Nov 12 '22 at 19:51
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Something that hasn't been mentioned is the concept of electromagnetic momentum and the Poynting vector.

The Poynting vector is defined as $${\bf S} = \frac{1}{\mu_0}\,{\bf E}\times{\bf B} $$ and "depicts the direction and rate of transfer of energy, that is power, due to electromagnetic fields in a region of space which may or may not be empty" and if energy is being transferred this suggests a momentum flow.

A useful video for you that goes into greater detail is due to Fermilab.

jim
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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.

That radiation is the fundamental ingredient of nature is supported by astronomical observations- which showed that radiation is all that is there at the start of the life of our universe. It is also supported by experiments in the lab wherein pure radiation(gamma rays) can produce matter, and pure matter(anti matter included) can produce pure radiation. The process is fully reversible and indefinitely. We can thus say that radiation could be considered as evaporated matter and matter as condensed radiation. If radiation condenses by going round (at the same speed) in closed loops, we get trapped radiaton, or rest energy or rest mass as a result. The circulation of the momentum produces the intrinsic spin, and that of the electric field(radial direction) creates the electric charge(Gauss theorem). The magnetic field vector is normal to the other two- producing the magnetic dipole moment along the spin direction. This completes all the required matter attributes- emerging as a result of radiation condensing into matter- thus supporting the radiation origin of matter.

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.

Riad
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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)

AccidentalFourierTransform
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