Why don't photons possess mass?

[misogynisticfeminist]

Yea, noob question here, if e=mc^2, and mass and energy is the same thing, why is a photon massless when it has energy?

 

[mathman]

Photons always travel at the speed of light and have 0 rest mass. For gravitational purposes, the energy acts like mass.

 

[jcsd]

Cos the full formula is $E^2 = m^2c^4 + p^2c^2$, whic gives you the energy of a particle in an inertial frame. Photons don't have mass but they do have momentum (p), so they have energy.

 

[selfAdjoint]

Furthermore, photons carry EM radiation, and if they had mass EM radiation would have a longitudinal component. But no such component can be detected to ever increasing levels of accuracy. So the photon's mass has to be less than a number starting with a decimal and followed by dozens of zeros before a significant digit, and that's the UPPER LIMIT. So the theoretical requirement that it have zero mass is confirmed in experments up to the limit of experimental accuracy achievable.

 

[Tom McCurdy]

how would you "measure" the mass of a photon? to test it w

 

[misogynisticfeminist]

Furthermore, photons carry EM radiation, and if they had mass EM radiation would have a longitudinal component. But no such component can be detected to ever increasing levels of accuracy. So the photon's mass has to be less than a number starting with a decimal and followed by dozens of zeros before a significant digit, and that's the UPPER LIMIT. So the theoretical requirement that it have zero mass is confirmed in experments up to the limit of experimental accuracy achievable.

so are you saying that photons do possesses mass but it is just extremely miniscule?

 

[aekanshchumber]

E=mC^2 gives us the relation between the mass and energy. according to it mass and eenergy are interchangeble, not that mass always posses energy or anything having energy always have mass.

 

[selfAdjoint]

so are you saying that photons do possesses mass but it is just extremely miniscule?

NO! That's why I put the stuff about an upper bound in caps. Theory says the photon has no mass, and experiments have found no mass, but being experiments they have a margin of error, and that margin is an extremely tiny number. Also whenever they figure out how to do a better measurment, the margin always goes down. All the experiments are consistent with the statement that the mass is zero.

 

[ahrkron]

E=mC^2 gives us the relation between the mass and energy.

As jcsd said, the complete relation has also a term for momentum.

 

[misogynisticfeminist]

Cos the full formula is $E^2 = m^2c^4 + p^2c^2$, whic gives you the energy of a particle in an inertial frame. Photons don't have mass but they do have momentum (p), so they have energy.

so, the energy comes from their momentum instead of their mass? but isn't p=mv, and if mass is 0, momentum is 0. Or is there another formula for it?

NO! That's why I put the stuff about an upper bound in caps. Theory says the photon has no mass, and experiments have found no mass, but being experiments they have a margin of error, and that margin is an extremely tiny number. Also whenever they figure out how to do a better measurment, the margin always goes down. All the experiments are consistent with the statement that the mass is zero.

ok...gottit.

 

[aekanshchumber]

Photons of course have momentum and the momentum is due to the mass, but still the photons do not have any mass this is because,the energy that the photon had, changed in mass. But neither the change in the mass nor in the energy could be noticed, according to the hysenberg's uncertainty principle
dm*de=h/2*pi (transformation of the standerd equation)

 

[aekanshchumber]

Furthermore, photons carry EM radiation, and if they had mass EM radiation would have a longitudinal component. But no such component can be detected to ever increasing levels of accuracy. So the photon's mass has to be less than a number starting with a decimal and followed by dozens of zeros before a significant digit, and that's the UPPER LIMIT. So the theoretical requirement that it have zero mass is confirmed in experments up to the limit of experimental accuracy achievable.

If you are saying that the Photons contains EM radiations then i think you are seriously missconcepted about photon. In fact photon itself is a radiation as photons were introduced to explains the particle character of the EM waves.

 

[arivero]

This is a FAQ in relativity. The m in E=mc^2 was time ago the "relativistic mass". The m in the "complete" formula is the "rest mass". Modernly people prefers to use only the latter, to avoid this kind of misunderstandings.

 

[pmb_phy]

This is a FAQ in relativity. The m in E=mc^2 was time ago the "relativistic mass". The m in the "complete" formula is the "rest mass". Modernly people prefers to use only the latter, to avoid this kind of misunderstandings.

For details on this subject/FAQ see the thread named "Mass of Light" in "General Physics". The same conversation is going on there.

As I've often reminded folks here, many people still refer to "relativistic mass" (aka inertial mass) simply as "mass".

E.g. the newest modern text that I have is Cosmological Physics, by John A. Peacock. When he speaks of mass in the first chapter he is not referring to rest mass. E.g. he refers to T⁰⁰/c² as mass density where, as you know, T⁰⁰ is energy density.

The misunderstanding that arises on topics like this is due to a lack of knowledge on the subject. It is not due to "poor" or outdated knowledge. Ignoring it can still lead to misunderstandings. Only complete knowledge and understanding will lead to a minimum of misunderstandings.

Pete

 

[humanino]

The most beautiful story about the photon mass can be found in Feynman's lecture on gravitation. I am not going to try to reproduce Feynman's great style. Basically, he is telling the story of a game, where another famous physicist asked him to prove that the photon has no mass. Feynman answered he is willing to play, under the condition : "give me an upper bound, I'll answer. After that, you are not allowed to change the bound !" Of course, the other physicist cheats, and gives Feynman four or five small and smaller bounds.

Anyway, I am not home so I don't have the book with me, I could not even copy the two or three pages if I wanted to. Really, try to find the book. Feynman derives Einstein equations from the assumption "spin 2 massles", as if we did not know about gravitation, but understand QFT already.

 

[kurious]

Photons are massless according to Higgs theory because they do not interact with the Higgs field.Protons and electrons do.The Higgs field has not been proven to exist bu t even if it does not it would seem reasonable that a photon is massless because it does not interact with whatever field does allow protons and electrons to have mass.

 

[misogynisticfeminist]

Photons are massless according to Higgs theory because they do not interact with the Higgs field.Protons and electrons do.The Higgs field has not been proven to exist bu t even if it does not it would seem reasonable that a photon is massless because it does not interact with whatever field does allow protons and electrons to have mass.

interesting, so the answer to this actually lies in the higgs field? any reason as to why the photon don't react with the higgs field?

 

[kurious]

I don't know a physical mechanism but someone on here might have a mathematical reason!

 

[Sariaht]

It's a wave; do waves have mass? You could say they are mass, but they are just particles exchanging velocity (as in all waves?); as written by another scientist here at this forum, mass is caused by particles in movement, and there is really no other solution to the problem. Try to explain it yourself.

By the way: p=mv, so sure, photons are mass

 

[da_willem]

The energy and momentum of a photon are respectively $$hf$$ and $$\frac{h}{\lambda}=\frac{hf}{c}$$. Putting this into the equation $$E^2 = m^2c^4 + p^2c^2$$ you will find m=0.

 

[Sariaht]

But after having reconsidered, hf/c² = m

You dribbled me out there though, the photon is now as said a wave and the mass would be useless to talk about if it wasn't for that the speed of the particles in the wave (and that is the only thing that can cause mass) causes mass which must be true, no?

 

[da_willem]

Again. The equation E=mc^2 is only correct in the rest frame of a particle where p=0 so $$E^2 = m^2c^4 + p^2c^2$$ will yield the famous equation found by Einstein. A photon has no rest frame so you will need the full formula and the results from quantum mechanics which give you the momentum and energy to yield zero 'invariant mass' m.

 

[Sariaht]

Massless things that carry energy and bends in gravityfields, pretty scary, no?

1. Waves are particles exchanging speed with one another
2. Particles moving causes mass

But this does not belong here, it belongs to the theory development forum, so you can forget me ever mentioning it.

 

[da_willem]

See also: https://www.physicsforums.com/showthread.php?t=40279

Massless things that carry energy and bends in gravityfields, pretty scary, no?

By the relation (again...) $m^2c^4 = E^2 - p^2c^2$ mass can be thought of as being part energy and momentum. So it is not weird that a particle can have energy and no mass, as long as it has a equal amount of momentum!

And the general theory of relativity states that not only mass but also energy and momentum generate gravitational fields. And that all particles are influenced by this 'gravitational field' (wich is nothing more than a distortion of the 'normal' flat geometry) by moving in geodesics.

So it's not that scary at all. You just have to get used to the new theories and that might take a while...

 

[pmb_phy]

By the way: p=mv, so sure, photons are mass

That is incorrect. Photons aren't mass, they have mass.

Again. The equation E=mc^2 is only correct in the rest frame of a particle where p=0 ...

You and Sariaht are using "m" to mean different things. Sariaht is using "m" to refer to relativistic mass and you're using it to mean proper mass. Also E refers to energy, not proper energy E₀. As you're using it E does not equal mc². This is rest energy and therefore E₀ = mc².

And the general theory of relativity states that not only mass but also energy and momentum generate gravitational fields.

Its mass which is the source of gravity where mass and energy are proportional. It is for that reason that energy appears in Einstein's equations. As Einstein phrased it

The special theory has led to the conclusion that inert mass is nothing more or less than energy, which finds its complete mathematical expression in a symmetrical tensor of second rank, the energy tensor.

Mass in one frame is momentum and stress in another frame. Its the stress-energy-momentum tensor which describes the source of gravity. T⁰⁰/c² is (relativistic) mass density. Since mass is proportional to energy it really doesn't matter which one you say is the source since you can replace one with the other in Einstein's equation. In fact there can be a total absence of proper mass and there can still be a gravitational field.

And that all particles are influenced by this 'gravitational field' (wich is nothing more than a distortion of the 'normal' flat geometry) ..

Gravity is not a distortion of spacetime. It can distort it in certain cases. But you can have a gravitational field in flat spacetime ... at least according to Einstein.

Pete

 

[da_willem]

But you can have a gravitational field in flat spacetime ... at least according to Einstein.

How can there be a graviational field in flat spacetime? Can you explain or give me a link to a internet document that explains it?

As I understand it the metric (as determinded by the energy content of space) is the 'replacement' of gravitational potential in general relatvity. In GR particles follow geodesics which are just straight lines with a flat metric (this is $g_{ab} = \eta_{ab}$), and this would be interpreted in the classical theory as the absence of a gravitational field...

 

[pmb_phy]

How can there be a graviational field in flat spacetime?

Note that I said at least according to Einstein. By this I mean that what I said is consistent with the way Einstein interpreted his general theory of relativity, i.e. how he defined quantities in The Foundation of the General Theory of Relativity. According to Einstein the presence of a gravitational field is dictated by the non-vanishing of the affine connection, not the non-vanishing of the Riemann tensor. In fact he stated this explicitly in a letter he wrote to Max Von Laue.

Can you explain or give me a link to a internet document that explains it?

Sure. See Einstein's gravitational field at http://xxx.lanl.gov/abs/physics/0204044

As I understand it the metric (as determinded by the energy content of space) is the 'replacement' of gravitational potential in general relatvity.

I agree 100%.

In GR particles follow geodesics which are just straight lines with a flat metric (this is $g_{ab} = \eta_{ab}$), and this would be interpreted in the classical theory as the absence of a gravitational field...

The relation $g_{ab} = \eta_{ab}$ means only that the frame of reference is inertial. If that holds at all points in spacetime in the given coordinate system then the spacetime is flat. In a non-inertial frame and there is a gravitational field, i.e. according to Einstein that is. And then the metric is not $\eta_{ab}$.

If there is a gravitational field present and the spacetime is flat then the spatial trajectories are not straight lines. The presence of a gravitational field is frame dependant as always. In a uniform gravitational field there are no tidal forces and therefore no spacetime curvature. But there is still a gravitational field unless you transform it away by transforming to a free-fall frame. Any of Einstein's texts is consistent with this.

Pete

 

[misogynisticfeminist]

The energy and momentum of a photon are respectively $$hf$$ and $$\frac{h}{\lambda}=\frac{hf}{c}$$. Putting this into the equation $$E^2 = m^2c^4 + p^2c^2$$ you will find m=0.

the recent talk about relativity in this thread, got me lost somewhat. I don't know **** about relativity lol. Ok, yea, so, is this the eqn which actually explains everything, as in I just sub in vlues of E and P and actually find that the photon has no mass. is this the answer to my question?

 

[pmb_phy]

is this the answer to my question?

Depends on the question. You were not precise about what you meant by "mass." Assuming you meant "proper mass" then there are different ways to answer it. Some may respond by referring to this Higgs thingy and some may respond by mentioning the relationship between the Coulomb force and the photon's proper mass. You were referring to mass and energy so that Higgs point is moot since you didn't ask why a photon has zero proper mass. You asked how it can have no mass when it has energy and E = mc². Your answer lies in the definition of these terms and not having to do with Higgs stuff. That is a different question. After you understand the difference between mass and proper mass and you understand that a photon has zero proper masss and you understand the relationship between Coulomb's law and photon proper mass and you still want to know "why" a photon has zero proper mass then the Higgs thingy may be what you're looking for.

But "why" questions seek reason, not description.

Pete

 

[Haelfix]

Well yes, the photon does not acquire a Higgs mass, but that's somewhat secondary.. The most fundamental reason is that it is a gauge boson, and the U(1) gauge symmetry of electromagnetism is both a global symmetry and crucially a local one. Mathematically, that implies it has zero mass.

 

[da_willem]

is this the answer to my question?

I think it is the answer to your question, but maybe not the answer to why a photon has no mass! Your question came from a misunderstanding of the formula relating mass and energy. As said before E=mc^2 anly aplies to an object having zero momentum and otherwise you would need the full equation with all these squares and c's...

So I think it gives a satisfactory answer to your original question. But if you would want to know why a photon has no mass you would have to read some of the other posts more carefully.

 

[pmb_phy]

As said before E=mc^2 anly aplies to an object having zero momentum ..

That is incorrect. If one is using m to mean proper mass then E₀ = mc², not E = mc².

Pete

 

[humanino]

pmb_phy : I don't see your point here. I think it has been posted several times that $$E^2 = \vec{p}\,^2 + m^2$$ and that the expressions you use are valid only in the rest frame where $$p=0$$. Unfortunately in the case of photon, there is no rest frame (I am aware that you know it...).

One can always formally define $$m=\sqrt{E^2 - \vec{p}\,^2} = \sqrt{p^2}$$ but no physical interpretation in a rest frame can be obtained from it for a real photon (for which $$m=0$$). Then, for a virtual photon the mass defined in this way is perfectly legitimate.

I am sorry to be confused here As far as I undestand, Healfix's argument is sound (gauge symmetry). Now could you elaborate on several things :

you say :

Gravity is not a distortion of spacetime. It can distort it in certain cases. But you can have a gravitational field in flat spacetime ... at least according to Einstein.

In a flat spacetime there is no gravitational field. Usually one defines the graviton field as a perturbation from the flat metric (even though it might not be the best convention : see Rovelli which defines the graviton field as the vierbein instead).

you also said :

The relation $$g_{ab}=\eta_{ab}$$ means only that the frame of reference is inertial. If that holds at all points in spacetime in the given coordinate system then the spacetime is flat. In a non-inertial frame and there is a gravitational field, i.e. according to Einstein that is. And then the metric is not $$\eta_{ab}$$.

The discussion might not be clear to me. This seems correct to me : consider a non-empty paracompact region A of spacetime : if the metric is globally flat in A then their can't be any gravitational field in it. I believe this is the meaning of your post. But, maybe due to foreign language, I feel I might be wrong. Is this the meaning of your posts ?

If you are saying : one can have a locally flat metric in a gravitational field, then OK, that's true, but that is the essence of GR.

 

[humanino]

Once again about the photon mass question : it is not the first thread about that. Really, Feynman's discussion in his "lecture on gravitation" is the best.

___________________________
EDIT : the discussion one can find there is entirely experimental. No theoretical argument there to motivate a vanishing mass of the photon.

 

[zefram_c]

Allow me to throw in my two cents:

SR allows for clear distinctions between massless and massive objects and describes the behavior of both. It doesn't, however, predict what has mass and what doesn't. Nor does it provide a fundamental explanation for mass; it is simply taken as an intrinsic parameter of the object under study. I'm not going into the various interpretations of E=mc²; the complete formula posted by jcsd leaves much less room for error - it's just not the one that's been popularized.

GR (someone correct me if I'm wrong) doesn't explain mass either. It establishes the relation between spacetime curvature and the stress-energy tensor, and the source's mass goes somewhere in the latter.

Maxwell's equations predict that the speed of light is always c. This, combined with SR, predicts photons are massless.

QFT doesn't explain mass either. Mass is a free parameter in the field equations for free fields as well as QED and QCD.

The weak interaction theory is somewhat different. Since the local gauge transformation rules are different for two components that make up the free particle wave function, the mass term breaks the invariance. Hence traditional mass terms cannot be allowed. To allow mass terms to appear in a gauge invariant way, another field - the Higgs field - must be introduced that is nonzero and asymmetric in the vacuum state. Now when the interaction with this field is accounted for, new terms appear in the equations that have the form of mass terms. In fact, they originate from the coupling of the original massless fields with the non-zero vacuum expectation of the Higgs field. The vacuum expectation is chosen in such a way as to be asymmetric with respect to the gauge mediators, but symmetric wrt a linear combination of them. This remaining symmetry of both the interaction and the vacuum is the U(1) symmetry of EM and accounts for a massless photon. (I note in passing that the local gauge symmetry is still present for the interaction). More importantly, the whole phenomenon of mass can be given a new interpretation as arising from the interaction of massless particles with a very strange vacuum.

All this theory doesn't offer any profound reason for masslessness of photons - the theory is built to be consistent with it. The current experimental upper bound on photon mass is on the order of 10^-16eV for lab experiments.