Photon Rest Mass: How We Know & Mass While Moving

[aleemudasir]

How did we come to the conclusion photon has no rest mass? Does photon have any mass(not talking about rest mass) while moving at usual c?

 

[phinds]

photon has no rest mass because it cannot be at rest so the question is meaningless.

At c, photon has energy equivalent of mass

 

[jtbell]

The "rest" mass, energy and momentum of an object are related by

$$E^2 = (pc)^2 + (mc^2)^2$$

We know to some degree of precision that E = pc for photons. For example, we derive the equations for Compton scattering by assuming that E = pc for both the incoming and outgoing photon, and measurements of Compton scattering agree with those equations.

 

[Naty1]

see the discussion here:

http://en.wikipedia.org/wiki/Photon#Experimental_checks_on_photon_mass

The photon is currently understood to be strictly massless, but this is an experimental question.

 

[jnorman]

is the photon not at rest when it is absorbed by an atom? the mass increase of the atom should represent the rest mass of the photon... (i know this is wrong, but why?)

 

[Matterwave]

The mass increase of the atom corresponds to the energy of the photon (divided by c^2). The photon does not "come to rest" inside the atom, it simply no longer exists.

 

[M Quack]

is the photon not at rest when it is absorbed by an atom? the mass increase of the atom should represent the rest mass of the photon... (i know this is wrong, but why?)

When the photon is absorbed, it ceases to exist.

Similarly, when an atom drop from an excited state to a lower one and emits a photon, that photon is created. It just appears, observing a number of conservation laws (energy, momentum, etc) for the entire (multi)atom-photon system

There is no conservation law for the number of photons. You can just pull them out of your hat, provided that it is not completely black.

 

[danmay]

How did we come to the conclusion photon has no rest mass? Does photon have any mass(not talking about rest mass) while moving at usual c?

It may help to think of a photon is a packet of kinetic energy that's always at c. That's all. When it's absorbed, it's being converted into other forms of energy.

 

[Darwin123]

"Does photon have any mass(not talking about rest mass) while moving at usual c?"
Yes. A photon moving at the usual speed of c has a relativistic mass given by:
m=E/C^2
where "E" is the energy of the photon and "m" is the relativistic mass of the photon.
The relativistic mass, "m", is not zero. So a photon does have a relativistic mass.
Photons are considered "massless" because their rest mass is zero. However,
a photon can't stand still. A photon moves at a speed "c" in all inertial frames. So
the "rest mass" of a photon is not measurable, anyway. The relativistic mass of a
photon is measurable. However, it will vary with inertial frame.

 

[aleemudasir]

If the rest mass of photon is zero then what will happen when we will use the equation
m= m(0)/ sqrt(1-v^2/c^2), which comes upto 0/0. Please explain this situation. And what will happen in this situation when we will go further to use m=E/c^2.
Thanks!

 

[Drakkith]

If the rest mass of photon is zero then what will happen when we will use the equation
m= m(0)/ sqrt(1-v^2/c^2), which comes upto 0/0. Please explain this situation. And what will happen in this situation when we will go further to use m=E/c^2.
Thanks!

You can't use 0 as the mass in those equations, as it comes out to nonsense. (Dividing zero by zero for instance) You must use the equations that apply to a photon, such as e=pc.

 

[Nabeshin]

If the rest mass of photon is zero then what will happen when we will use the equation
m= m(0)/ sqrt(1-v^2/c^2), which comes upto 0/0. Please explain this situation. And what will happen in this situation when we will go further to use m=E/c^2.
Thanks!

You cannot go around willy-nilly applying equations. These equations were derived under certain assumptions, among them being that the object is traveling below c (or equivalently, has mass).

 

[aleemudasir]

You can't use 0 as the mass in those equations, as it comes out to nonsense. (Dividing zero by zero for instance) You must use the equations that apply to a photon, such as e=pc.

Even after using E=pc, the momentum p is defined as, p=mv, so, what now? There still is m!
Correct me if I am wrong.
Thanks.

 

[aleemudasir]

You cannot go around willy-nilly applying equations. These equations were derived under certain assumptions, among them being that the object is traveling below c (or equivalently, has mass).

Which equation do we use to infer that speed of light is the limit?

 

[Drakkith]

Even after using E=pc, the momentum p is defined as, p=mv, so, what now? There still is m!
Correct me if I am wrong.
Thanks.

Which equation do we use to infer that speed of light is the limit?

I believe it is p=h/wavelength, where h is Plancks constant.

 

[pervect]

]
Correct me if I am wrong.
Thanks.

People have been trying to correct you all along - you don't give the apperance of actually listening to the corrections, however...

 

[aleemudasir]

People have been trying to correct you all along - you don't give the apperance of actually listening to the corrections, however...

I listen, I just want to clear my doubts. I hope that is the purpose of this forum!

 

[aleemudasir]

I believe it is p=h/wavelength, where h is Plancks constant.

OK, but it is just the rearrangement of equation λ=h/p, where p=mv.

 

[M Quack]

OK, but it is just the rearrangement of equation λ=h/p, where p=mv.

No, you are wrongly simplifying a general equation to a very special case.

p = \hbar k where k is the wave vector (\frac{2\pi}/\lambda) is used pretty much everywhere in QM, including massive particles such as electrons. Note that momentum is conserved, but not velocity, so momentum is a much more fundamental property than velocity.

In the non-relativistic limit for massive particles you get E = \frac{1}{2m}p^2.

For a photon, the non-relativisitc limit obviously does not make any sense. The most reasonalbe thing you can do is (as jtbell pointed out) use E^2 = (m_0 c^2)^2 + (pc)^2.

Now all experimental observations over a huge huge range of wavelengths/energies/wave numbers match E=pc = \hbar \omega = \frac{h}{\lambda}, i.e. they match the above equation with m_0 = 0. In other words, all available experimental data supports the hypothesis that the photon is massless.

But then again that is just a theory, just as evolution is just a theory.

 

[vin300]

The magnitude of photon four momentum is zero, and the magnitude equals -m²c², that implies the photon rest mass, m, is zero.

 

[aleemudasir]

No, you are wrongly simplifying a general equation to a very special case.

p = \hbar k where k is the wave vector (\frac{2\pi}/\lambda) is used pretty much everywhere in QM, including massive particles such as electrons. Note that momentum is conserved, but not velocity, so momentum is a much more fundamental property than velocity.

In the non-relativistic limit for massive particles you get E = \frac{1}{2m}p^2.

For a photon, the non-relativisitc limit obviously does not make any sense. The most reasonalbe thing you can do is (as jtbell pointed out) use E^2 = (m_0 c^2)^2 + (pc)^2.

Now all experimental observations over a huge huge range of wavelengths/energies/wave numbers match E=pc = \hbar \omega = \frac{h}{\lambda}, i.e. they match the above equation with m_0 = 0. In other words, all available experimental data supports the hypothesis that the photon is massless.

But then again that is just a theory, just as evolution is just a theory.

OK.
But we know p=h/λ and you have proved E=h/λ, so are E and P equal?

 

[M Quack]

We know that E=pc, so they are proportional to each other. And that implies that the rest mass is zero. That is the whole point. There are several different ways at arriving at this conclusion, as several people have pointed out.

If there is a rest mass, then there has to be a rest energy (or equivalent energy) E_0=m_0 c^2 at rest, i.e. at p=0 or k=0. For the photon, all datapoints fall onto the line E=pc which give E=0 at p=0. Hence no energy at p=0, which in turn means no mass at p=0 (at rest).

E=\frac{h c}{\lambda} Sorry for the typo (and I see it is not the only one...)

k = \frac{2\pi}{\lambda}

Stricktly speaking, the non-relativistic limit should also be

E_{\mathrm{kinetic}} = \frac{1}{2m}p^2 = E - E_0 where E is the total energy and E_0=m_0 c^2 the energy-equivalent of the rest mass.

 

[bahamagreen]

Photons have non-zero rest mass inside superconductors...

 

[Dead Boss]

Got any reference for that?

 

[phinds]

Photons have non-zero rest mass inside superconductors...

I agree w/ Dead Boss ... you'll need to provide a citation for that. It sounds like nonsense.

 

[Darwin123]

It is irrelevant whether "photons" have a nonzero rest mass in a superconductor.
Electromagnetic radiation takes on all sorts of properties in condensed matter.
However, photons in a canonical vacuum have a zero rest mass.
This may be semantics. However, I am not sure whether it is precisely
correct to call anything a photon inside any sort of conductor. Inside a
conductor, the electromagnetic waves form plasmons and polaritons. What
you may be describing may be a plasmon-polariton. A plasmon-polariton may
have a nonzero rest mass. However, it couldn't exist outside the conductor.
I have also read of "photons with nonzero rest mass" in quantum wells. Again,
this is not a "true" photon. It is a "dressed" photon. The photon is coupled with
some other excitation, which greatly changes its properties.
A photon in a vacuum has a zero rest mass. The photon inside condensed matter
can have any number of properties.

 

[bahamagreen]

I agree w/ Dead Boss ... you'll need to provide a citation for that. It sounds like nonsense.

The end of the "Experimental checks on photon mass" section of the Wikipedia Photon page at http://en.wikipedia.org/wiki/Photon

"Photons inside superconductors do develop a nonzero effective rest mass; as a result, electromagnetic forces become short-range inside superconductors."

 

[M Quack]

Inside a plasma you can also get electromagnetic waves where the energy at k=0 does not vanish. But these are collective oscillations of the electrons relative to the ion background, so hardly just photons. Identifying this energy as a "rest mass"/c^2 is completely misleading.

http://en.wikipedia.org/wiki/Waves_in_plasmas

 

[aleemudasir]

We know that E=pc, so they are proportional to each other. And that implies that the rest mass is zero. That is the whole point. There are several different ways at arriving at this conclusion, as several people have pointed out.

If there is a rest mass, then there has to be a rest energy (or equivalent energy) E_0=m_0 c^2 at rest, i.e. at p=0 or k=0. For the photon, all datapoints fall onto the line E=pc which give E=0 at p=0. Hence no energy at p=0, which in turn means no mass at p=0 (at rest).

E=\frac{h c}{\lambda} Sorry for the typo (and I see it is not the only one...)

k = \frac{2\pi}{\lambda}

Stricktly speaking, the non-relativistic limit should also be

E_{\mathrm{kinetic}} = \frac{1}{2m}p^2 = E - E_0 where E is the total energy and E_0=m_0 c^2 the energy-equivalent of the rest mass.

p=0 (at rest), but at rest velocity is also zero and only by virtue of velocity being zero we can get the momentum as zero, so how could you be strict in saying in concluding the rest mass as zero.

 

[vanhees71]

First of all, free photons have with overwhelming accuracy indeed zero rest mass. The particle-date booklet value for the upper bound is m_{\gamma}<1 \cdot 10^{-18} \; \mathrm{eV}..

Second, in superconductors the photon indeed acquires a mass through the Andersen-Higgs mechanism. Nowadays better known in the context of the Higgs-Kibble mechanism in the standard model of elementary particles which gives mass to the electrons, quarks, and the W and Z bosons such as not to violate the basic underlying non-abelian (chiral) gauge symmetry, in fact this mechanism that a spontaneously broken local gauge symmetry implies massive gauge bosons has been first formulated by Andersen in the context of superconductivity. You can understand superconductivity on a phenomenological level by just assuming that the electromagnetic U(1) gauge symmetry is spontaneously broken (see Weinberg, QT of Fields, vol 2). Here, of course we know the microscopic mechanism behind it, thanks to the famous BCS theory of superconductivity.