Photons are traditionally said to be massless. This is a figure of speech that physicists use to describe something about how a photon's particle-like properties are described by the language of special relativity.
No, photons does not have mass in Maxwell theory and this also experimental fact with very big accuracy.
Take into account an equivalence of mass: m c^2=nu h, where m - equivalence of mass, c - velocity of light, frequency of light, h - the Planck constant
it is possible to say No, since a person accept the definition of photons and Mass that we used to explain our research outcomes. However, if a person come across to the opposite (I.e., Yes) he should bring plausible justification and disprove the existing one. for the time being Photon has Zero rest mass.
It depends what you mean by mass. Photons have zero "rest" mass. But in nature, there is no photon at rest. All photons have energy that depends on its frequency. We know energy and mass are the different aspects of the same thing. (Einstein's famous formula) Since photons have energy it also means that photons have mass. But note that, their mass is only coming from their energy. Only, their mass is not intrinsic. In short, yes you can say that, photons have mass! In 1920's Einstein's general relativity is verified based on the fact that photons are affected from gravity.
The electromagnetic force is a long range interaction than the photon must be massless. At quantum level, in 4 dimensions, a mass term for vector fields is inconsistent because it breaks gauge invariance. There is a special case: computing the lowest order vacuum polarization of QED in 2 dimensions (Schwinger 1962), one finds that the photon receives the mass m^2=e^2/pi i.e. the photon of 2-dim QED is a free massive boson.
Photons have zero mass, due to the unbroken U(1) symmetry of the Standard Model, which is preserved by quantum perturbative corrections--and can be confirmed in lattice simulations of (reduced) models. One should distinguish between calibration of experiments and signals. A non-zero mass for the photon would have measurable consequences, that haven't been observed. And photons are *not* ``traditionally said to be massless.'' nor is this ``a figure of speech''. The statement has precise mathematical and experimental consequences that are the topic of standard courses in electromagnetism and field theory at undergradute level.
One such experimental constraint comes from the trilinear and quadrilinear gauge couplings-already at LEP and Fermilab, e.g. photon-W-W vertex and corrections-if gauge invariance is violated there would've been deviations that aren't observed.
Cf. also, http://arxiv.org/pdf/0809.1003.pdf
To describe massive neutrinos we do *not* require Lorentz-violating terms--the seesaw mechanism is sufficient, if there aren't any ``sterile'' neutrinos that could give rise to Dirac-type mass terms. An interesting issue is, whether the neutrinos are Majorana or not-and there's an ongoing experimental effort on ``neutrino less'' double beta decay to test for this.
A non-zero mass for the photon would have measurable consequences, that haven't been observed. (Stam Nicolis).
...yes you can say that, photons have mass! In 1920's Einstein's general relativity is verified based on the fact that photons are affected from gravity. (Alhun Aydin ).
So?
That light is bent by a gravitational field isn't due to the mass of the photon, but to the fact that the graviton (i.e. the metric) couples to energy through the energy-momentum tensor. Massless particles, such as the photon, described by *special* relativity, have non-zero energy and momentum, thus can be affected by spacetime curvature. Standard general relativity.
Off topic (neutrinos): Degenerate in mass neutrinos can't have ``flavor oscillations'' (property of quantum mechanical Hamiltonian evolution). Therefore, *if* they are found to show them (which was measured) this means that, at least, one flavor has different mass from the others and, if the common mass was taken to be zero, this implies that this flavor has mass. No Lorentz violating terms necessary, that would have given rise to many other effects. That's the argument. From there on one can introduce a mass matrix and compute mixing angles-and these have been measured. So the mass *differences* are known, to varying degrees of precision-the masses themselves have much larger error bars, of course.
Once more: experiments can *only* provide limits and confidence intervals, nothing more. However, experiments have a finite resolution and, therefore, provide limits *compatible* with a value, to a given accuracy. This means that it's not possible, with that experimental setup to distinguish the results within that interval, so the meaningful statement is, whether the mass is compatible with zero to a given experimental resolution-the answer is Yes. Claiming that it isn't is then a statement in contradictiton with the experiment.
The issue is that many *different* experiments provide *different* limits and thus test different quantities-and not all of them are uniformly small. These quantities are *not* the photon's mass, they describe relations that are true, when the photon's mass vanishes. This is the topic of quantum field theory and cannot be explained in five sentences, it's what's taught during a year long course.
The answer is that, in principle, the photon cannot have a mass and that principle is unbroken U(1) gauge invariance.
No need to appeal to any authority-U(1) gauge invariance is physics, not religion. It has conceptual consequences that can be tested for theoretical consistency and, also, by experiment. And it is the subject of all electromagnetism courses in physics.
Actually gauge invariance says that the Noether current related to the charge associated with the U(1) group (the EM field) is null, because the Lie algebra is abelian. But this is the electric charge which is null. One cannot deduce that the gravitational charge (the mass) is null.
The photon does not carry electric charge, the metric does carry energy and momentum. However, in both cases, the invariance under gauge transformations for the action of the photon and local translations for the action of the metric preclude the appearance of a mass term for the photon or the graviton.
If you break a symmetry at one scale, then this has consequences at other energy scales, that can be computed-and have been. This is what ``loop corrections'' mean.
So if you imagine that U(1) gauge invariance is broken at some scale, by the appearance of a mass term for the photon, this will have calculable consequences at accessible energies-these are being checked. The statement ``new experiments may reveal that gauge invariance is actually broken'' *means* something very specific: it means the appearance of certain additional terms in the theory. Their effects can be computed and looked for, since they are, of course, signals for new physical effects. This has nothing to do with epistemology and everything to do with real physics, theory and experiment, everyday (and night) work. And they show that such terms do not contribute to the processes that have been measured.
Gauge invariance stems 1) from the principle of relativity according to there the laws of physics are the same for all observers and 2) that the laws of physics can be represented by gauge theories. The first principle is a fundamental, if it was not true it would be difficult to conceive of any physics. The second is still a basic hypothesis, so it could be proven false. As for the gauge group, gravitation is represented by the Lorentz group (more accurately the Spin(3,1) group). The usual U(1)xSU(2)xSU(3) group is related to the other force fields. So we have to distinguish the Noether currents for the gravitational field from the Noether currents related to the other fields.And one cannot say much about the gravitational mass, until we have a great unified theory.
Gauge invariance is an internal symmetry, relativistic invariance is a spacetime symmetry. The two are distinct and, per the Coleman-Mandula theorem, do not mix, for the symmetries of the Standard Model, that are based on Lie algebras.
Both are the result of refining successive approximations: Newtonian mechanics is an approximation to relativistic mechanics and Fermi theory is an approximation to the weak interactions and electrostatics and magnetostatics are approximations to classical electromagnetism and geometric optics is an approximation to wave optics. (Gauge invariance in this context is recognized, already, as expressing charge conservation.)
It is the consistency of the successive approximations that places bounds on what *new* physical effects may look like and could look like. So charge conservation led to the necessity of adding the displacement current term to Ampère's law, that, then implied travelling wave solutions, that were transverse (from Gauss' law and the non-existence of magnetic monopoles)-expressing the fact the photon was massless, since the waves didn't have longitudinal polarization. This result is sensitive to boundary conditions: e.g. in wave guides the waves *do* acquire a longitudinal polarization, and the dispersion relation does seem to give rise to a ``mass term'' of geometric origin-but, precisely, Lorentz invariance has been broken by the boundaries.
Symmetries are related to Noether currents classically, to relations between measurable quantities such as cross sections beyond the classical level. And these deoend on processes that do nor have any classical analog-a pont made by Dirac, already in 1927, that measurements at one energy scale, at the quantum level, are sensitive to physics at energy scales beyond what is ``classically'' accesible.
Dear Stam,
So we agree that we have two kinds of gauge invariance involved : one for gravitation, and the other one for the other fields (including EM). Photon are associated to Noether currents for the EM fields, and the simple implementation of a Yang Mills model lead, because the Lie algebra for U(1) is abelian, to the conclusion that its EM charge is null. But we cannot say anything for its gravitational charge, which is its mass. But I concede that other considerations, as you explain, lead to some upper value of themass of the photon. However I think that it is disturbing to deal with mass seen only its inertial quality (as in the standard model) and forget its other quality, claimed for by GR, that it is also the gravitational charge. So my feeling that the answer would come from a GUT.
The equivalence principle implies that the gravitational mass is equal to the inertial mass -which applies to ``slowly'' moving (with respect to the velocity of light) objects in the same reference frame-and implies that any physical system ``interacts'' with spacetime, i.e. the metric, through the energy--momentum tensor only. That's all. The current is the energy-momentum tensor and the charge is the 00 component thereof. However, since the metric, as a field, does carry energy and momentum, the energy momentum tensor is ``covariantly'' conserved and an ``energy'' can only be defined at infinity-what's known as the Arnowitt-Deser-Misner (ADM) mass. It is gravitational and inertial (satisfying the equivalence prnciple).
And this can be tested (and has been tested) once more, theoretically and experimentally-which implies constraints for new physics.
So, as long as a field theory couples to the spacetime metric through the energy-momentum tensor only, it automatically realizes the equivalence principle.
Dear Manuel. To go back to your question... Photons have no REST mass, because they only exist travelling at the light velocity c. However, they do have a mass m (not a rest one !) given by the equation mc2 = h.nu, and there momentum is mc = h.nu/c as it should be.
No, that statement is false. Photons have *energy*, but not *mass*. The difference is that mass is Lorentz invariant, so doesn't depend on the reference frame, whereas energy is the fourth component of a 4-vector and does depend on the reference frame. So the statement that photons are massless is valid in all inertial reference frames. Their energy depends on the reference frame and satisfies E^2-p^2c^2=0, where p^2 is the square of the modulus of the 3-momentum.
You can deny the status of "mass" to the photon if you wish, or you can define an "effective" mass through the momentum. What is correct is that the rest mass of photons is zero.
This ``effective mass through the momentum'' does not have an invariant meaning, since it depends on the reference frame. Furthermore, if it *did* have the properties of mass, this would imply that the speed of light would not be constant, since light-like trajectories in one frame wouldn't be light-like in another. Therefore such a ``mass'' doesn't have the physical properties of mass, defined through inertial reference frames-it's not the mass of any physical object.
The mass can be defined as the Lorentz invariant E^2-p^2c^2=m^2c^4. Obviously all terms in this equation have the appropriate dimensions, but only the right hand side is, by construction, invariant under Lorentz transformations. E and p, individually, are not. (And, in general relativity, where Lorentz invariance is *local*, mass can only be defined on the boundary of the spacetime.)
It's like arguing over the ``correct'' value of the x-component of the 3-momentum: there isn't any correct value for this quantity. That's why it's useful to ask whether there do exist quantities that do *not* depend on such choices. When Lorentz invariance is global, such a quantity does exist and has, not only, the right dimensions, but, also, the right physical properties-the invariant length of the energy-momentum 4-vector. For a photon this quantity vanishes-in all inertial frames.
According to GTR any body with non zero rest mass, if travels with speed =c (speed of light), its mass tends to infinite. (M=Mo/sqrt(1-(v*v/c*c))).Mo-rest mass,M apparent mass, v- velocity of moving object.
So if photon has a mass even just greater than zero, its mass appears to be infinite when it is travelling in vacuum. So, we can say that there is no rest mass for photon. Same logic holds for Neutrino also. But some theories says Neutrino should have some rest mass. So, there is some conflicts running on it.
But If we observe GTR, mathematical framing of any equation allows some precision and theoretical limitations. So, I feel we should not interpret this logic here.
But for Photon, definition says photon is a particle which exists only at 'C'. So, we cant afford to discuss experimental results of rest mass of photon. So, I feel mass of photon is just situational dependent. Can be taken as Zero in experimental cases and can consider some finite but small value for theoretical usage.
To answer on the question - Do photons have mass? is necessary to define – what is the mass before. Now there are seems 4 different views of this [physical] notion.
(1) – the mass is a measure of the inertia (“inertial mass”): if no forces impact on a body, the body is at rest or at rectilinear uniform motion. After the body is impacted with a transmission to it of a momentum, the state above changes; at that the more the mass the lesser the change [of the speed, momentum, and energy of the body];
(2) – the mass is the source of the gravity attractive force (“gravitational mass”);
(3) – the mass is an equivalent of the energy – since they are proportional each other through standard universal constant seed of light; for moving body the mass, as something that counteracts to impacts, depends on the body’s speed also;
(4) – the mass is a “relativistic invariant” in the SR, m=1/c*sqrt[E2-(pc)2], which is equal to the inertial mass of a body at rest in given reference frame.
A photon: (1) - if no forces impact on a photon, the photon is at rectilinear uniform motion. . After the photon is impacted with a transmission of momentum, the sate above changes; the more the photon’s momentum (energy) the lesser the change [of momentum, and energy of the photon]; and (4) since for the photon E.=pc=hr*omega (“r” is “reduced”), the photon’s mass is a “relativistic invariant”, which is equal to zero, the photon is “(4)-massless”.
Or, in the SR, – “since the relativistic mass of a photon – if it would have a “rest mass” - must be infinite, so this rest mass is equal to zero”.
The masses (1) and (2) are quite different – because of the inertia and the gravity are evidently different phenomena. But they are, for static conditions, seems totally equivalent for “massive” bodies; for that at first sight follows that photons should have zero gravitational mass – as that is posited in the GR.
In reality the Matter’s spacetime is 4D Cartesian manifold where every particle moves with constant speed (“speed f light”) and for every particle the equation E=pc is true; for the massive bodies the body’s momentum p is proportional to the speed and so p=mc=E/c, where the coefficient “m” is equal to the “relativistic mass”.
For the photon the relation p=E/c exists only.
So photons don’t differ from massive particles practically. Moreover, it seems as rather reasonable to suggest that, since any particle is some constantly running close-chain algorithm, that any particle is some “gyroscope”, which rotates with a frequency (omega), so, that the particle’s (so – a body’s) energy is proportional to this frequency and is E=hr*omega; so the main difference between particle and photon is that particles at rest move along t-axis, when photons move in the space only.
And the inertia is a realization of the fact that gyroscopes resist to attempts to change their frequency – both the frequency and its axis’s direction.
From the similarity of all particles in Nature, including photons, follows, that photons – in contrast to the GR - should have gravitational mass.(As well as inertial and gravitational masses are equivalent indeed).
And it seems, that it is so – more see
“The Informational Model – Gravity”: http://vixra.org/abs/1409.0031
There are 2 versions:Engl. –pages 1-12, and Russian – pages 13-24”
Cheers
I'm sorry but Photons differ deeply from massive particle. The quanta absorption emission doesn't occur with other particles, and there is the key to most of the difference.
I seriously doubt that photons can cause a curvature of space-time while they travel, some paradoxes would be originated. GRT doesn't abolish gravitation or gravitational field at all, the gravitational field is only a side effect of the curvature. Since GRT is strictly based on "mass-energy" this doesn't mean that forces are no more present. In general, forces are only consequences of the energies at stake and how they evolve according to the least action principle.
I think that from a theoretical point of view, the photon is strictly massless. In fact, existing particles can be classified using the representations of the Poincare group (group of symmetries of Minkowski metric) and the mass is one of the invariants which characterize the particle. Representations of Poincare group are caharacterized by the two so-called Casssimir invariants, the mass and the spin. The Maxwell equations (i.e. equations for the photon, in quantum field theoretical sense) are pretty uniquely determined by these two invariants: mass zero and spin 1. Since the mass is zero, it is actually rather helicity than the spin, because photon cannot have the spin projection equal to zero. This can be understood as the consequence of the fact that you cannot associate a frame with the photon. so there is no other relevant direction than the wave vector of the photon. All these facts are consequences of the zero mass of the photon.
Formally you can use Einstein's relation E = m c^2 to calculate the "mass" of the photon, but to me this is completely incorrect. The relation to be used is
E^2 - p^2 c^2 = m^2 c^4,
where the photon has m = 0. Thus, there is momentum p = E/c associated with the photon, but not the mass.
It seems that the pages in this thread are too short…
- When, in fact, on the last posts in this thread there are some comments in my post above (yesterday) already. Though some additional comments possibly would be necessary.
So, when answering on the question “Do photons have mass? ”it is necessary to remember that there are at least four (mainstream) conceptually different views of this [physical] notion. And, e.g., the declaration as “The relation to be used is E^2 - p^2 c^2 = m^2 c^4” is only one of those views; it is only definition (as that Alexei Bykov says above) of some relativistic invariant in the SR. Which, as other invariants (e.g., the interval s [s2], proper time, speed of light, etc.) is nothing more then some sequence from some mathematical construction – Minkowski formalism in the SR. And this invariant for the photon is equal to zero – that’s all. But that doesn’t relate to other views practically – that isn’t a measure of inertia (or is applicable only for bodies at rest), and doesn’t relate to gravity.
As well as it doesn’t relate to “5-th” view – in the informational model, where total momentum of every particle in the 4D Cartesian spacetime is always p=mc and for all particles E=pc – as that is true for photons (the “SR momentum” in the Martin’s equation above is equal p=mv, where v is the body’s speed in the 3D space).
So, again, there are no principal differences between photon and “massive” particles practically, including – relating to all views on the mass notion. Nature is built by some rather universal scheme.
Including relating to the “gravitational mass”, the attempts to declare that “photons don’t interact gravitationally (don’t change the frequency/ energy in the field and don’t create gravitational field since photon’s mass is equal to zero”) – with a great probability aren’t correct, since don’t consistent with well-based suggestions. For example (see other thread here, “What is the gravity field of a photon”) – it seems as rather correct (evident) to suggest that if in some “massless” closed container there are two equal numbers of electrons and positrons, with total mass 2m, to after annihilation the container will weight 2m also, at that in both cases the gravitational forces by which the container will act on Earth will be equal also.
Besides, there is other and rather serious ground, that the GR isn’t correct in the case of photons (more – see again http://vixra.org/abs/1409.0031 ): the GR states that “gravitational time dilation”, what reveals as, e.g., in the shifts of energy levels on (Earth surface, Pound et. al. experiments) distance h, is equal to GMh/(rc2). But the shift is a result of the gravitational mass defect, which is equal to both – Earth and body in this system. Thus, if the GR is correct, the mass defect – or the gravitational (negative) energy must be twice more then it is in reality.
So – there exist a number of reasons to understand that photons have practically the same masses as “usual” or “massive” particles. The absence of “the spin projection equal to zero” isn’t a problem – that is simply because of that photons are born after purely spatial impacts (transforming spatial momentum) and so move in the 4D spacetime only in the 3D space. “Massive” particles are born if the impact has temporal component, but they can also move in the space also; so their spins have all possible spatial projections. However we don’t see till now the “massive” particles’ spins’ temporal projections…
Cheers
" it seems as rather correct (evident) to suggest that if in some “massless” closed container there are two equal numbers of electrons and positrons, with total mass 2m, to after annihilation the container will weight 2m also, at that in both cases the gravitational forces by which the container will act on Earth will be equal also. "
after the annihilation the beams will be reflected by the container. I doubt that the weight of the container would be the same. This is the question Eddington posed in his book but he felt he could not give the answer on the weight of light.
The interaction with the gravitational field is with the momentum only, in the bending of light for example.
“Yes, mass can be defined in several ways.” – that is correct.
But “It is not about "point of views", it's just about convenience.” – that isn’t correct principally; the “mass” notion has a number of different definitions not because of “convenience”, but because of every definition relates to different physical phenomenon; and, for example, the problem of the equivalence of “inertial” and “gravitational” masses is one of the utmost important problems in physics; as well as the problems “what is the inertia” and “what is the gravity”, though.
“About container - if you will calculate mass defined as above of system of photons created after annihilation it will be exactly 2m, when mass of each photon is zero.” – if there were, let, N electrons and N positrons, having the mass 2m kg and further appeared 2N photons having mass 0 kg, then the mass of photons (and, of course of “container”) is 2N*0=0…
“General relativity does not predict doubled redshift” – indeed, that is so, the GR predicts doubled blue shift of photon detector (as that is in Pound et. al. experiment), from what immediately follows doubled value of the gravitational potential energy of a bodies system.
And –so the attempt to establish unique definition for the mass as “relativistic invariant” is wrong, since, first of all, that it remains out consideration of fundamental physical problems. Moreover, again, - the using of the “relativistic invariant” version, which is convenient in the SR and the GR, is non- convenient in the rest physics, because of the SR is evidently self-non-consistent and so has rather limited in this science. And the GR with rather great probability has the problems with the adequacy to the reality also – see the examples above in the SS posts and papers; besides – the base of the GR is pseudo Riemannian 4D spacetime (which followed from pseudo Euclidian 4D spacetime in the SR), where some of dimensions and some spacetime intervals are imaginary. Since the real 4D spacetime is real manifold, from that follow next questions to the applicability of the GR…
Cheers
@ Aleksey >“General relativity does not predict doubled redshift” – indeed, that is so, the GR predicts doubled blue shift of photon detector (as that is in Pound et. al. experiment), from what immediately follows doubled value of the gravitational potential energy of a bodies system." You were kidding maybe...
I think this question should be reconfigured to; “if we assume that light has a mass by what constraints would we observe it?”
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