Do Photons Have Mass and Momentum?

[juanrga]

Because mass isn't conserved in relativity, only energy and momentum.

In relativity, the mass of an object is given by<br /> m = \sqrt{\frac{E^2}{c^4} - \frac{|\textbf{p}|^2}{c^2}}<br />where E is the total energy of the object and p is the total momentum of the object.

Now consider a photon with momentum p, and therefore energy |p|c (and zero mass), that is absorbed by an object with momentum 0 and energy m₁c². By conservation of momentum the object's momentum becomes p and by conservation of energy its energy becomes |p|c + m₁c². Then its mass becomes<br /> m_2 = \sqrt{\frac{\left(|\textbf{p}|c + m_1c^2 \right)^2}{c^4} - \frac{|\textbf{p}|^2}{c^2}} = \sqrt{\frac{2|\textbf{p}|m_1}{c} + m_1^2} &gt; m_1<br />

It seems that you are taking the absorption of one photon by one electron as an elastic collision.

For the object at rest

m = \sqrt{\frac{E_0^2}{c^4}}

For the object in motion with momentum \textbf{p}

m = \sqrt{\frac{E_1^2}{c^4} - \frac{\textbf{p}^2}{c^2}}

With evidently E_1 &gt; E_0 and m the same because is an invariant. The mass of an electron m=m_e is always the same when it is at rest or when it is moving at 0.99 c.

 

[DrDu]

Consider a highly reflecting cavity where some photons are captured and reflected on and off the walls. The mass of the whole system cavity+photons will depend on the number of photons present, as the photons do increase the rest mass of the whole system.

 

[sophiecentaur]

Consider a highly reflecting cavity where some photons are captured and reflected on and off the walls. The mass of the whole system cavity+photons will depend on the number of photons present, as the photons do increase the rest mass of the whole system.

But does that scenario imply that the photons 'have' mass any more than when you get a measurable Mass Defect after a nuclear reaction? Can one afford to be so literal in these circumstances?

 

[Phrak]

I don't think it's a good idea to use E=mc^2, in general, for the definition of mass as some seem to be doing. This is not really done by most, except within the center of mass frame.

Using m^2 = E^2 - p^2, good in special relativity in cartesian coordinates, then promoting to general relativity, obtains the oriented tensor density consisting of energy and momentum density, invariant in both curved spacetime and curvilinear coordinates, having a norm equal to the mass density.

The four-momentum tensor density, \mathbb{P}_a dx^a is the most fundamental object in this discussion of photon mass. Fortunately, in special relativity, P_\mu dx^\mu \Leftarrow \mathbb{P}_a dx^a, so we can talk about the 4-momentum intensity of a system having a norm equal to the invariant mass as long as we stick to right handed coordinate systems.

 

[DrDu]

But does that scenario imply that the photons 'have' mass any more than when you get a measurable Mass Defect after a nuclear reaction? Can one afford to be so literal in these circumstances?

No, but it shows clearly how photons can contribute to the mass of an object. Furthermore, it should be clear that a standing wave in a cavity is not made up of virtual but rather real photons in contrast e.g. to the photons which are responsible for the Coulomb attraction and add to the mass of an atom.

 

[sophiecentaur]

No, but it shows clearly how photons can contribute to the mass of an object.

I don't think there was ever any question about that. But 'contribution' is not necessarily the same as 'having' or 'being'. A drop of ink is just a drop of ink but it may contribute to a written word.

Furthermore, it should be clear that a standing wave in a cavity is not made up of virtual but rather real photons in contrast e.g. to the photons which are responsible for the Coulomb attraction and add to the mass of an atom.

Except that the photons are bound to a structure. They are interacting with the walls of the cavity - unlike when they are in free space.

 

[Phrak]

No, but it shows clearly how photons can contribute to the mass of an object. Furthermore, it should be clear that a standing wave in a cavity is not made up of virtual but rather real photons in contrast e.g. to the photons which are responsible for the Coulomb attraction and add to the mass of an atom.

Why do you say photons rather than electromagnetic field?

Pervect had something interesting to say about the mass of a confined electomagnetic field. Apparently, the calculation of the field energy will yield a value exactly twice the energy put into the system. The field pressure put on the cavity walls puts the material elements in tension, reducing their mass by the correct amount.

 

[dchris]

Because mass isn't conserved in relativity, only energy and momentum.

In relativity, the mass of an object is given by<br /> m = \sqrt{\frac{E^2}{c^4} - \frac{|\textbf{p}|^2}{c^2}}<br />where E is the total energy of the object and p is the total momentum of the object.

Now consider a photon with momentum p, and therefore energy |p|c (and zero mass), that is absorbed by an object with momentum 0 and energy m₁c². By conservation of momentum the object's momentum becomes p and by conservation of energy its energy becomes |p|c + m₁c². Then its mass becomes<br /> m_2 = \sqrt{\frac{\left(|\textbf{p}|c + m_1c^2 \right)^2}{c^4} - \frac{|\textbf{p}|^2}{c^2}} = \sqrt{\frac{2|\textbf{p}|m_1}{c} + m_1^2} &gt; m_1<br />

thanks Greg, now i get it. Can you just tell me why do we take |p|c in the place of E?

 

[DrGreg]

Now consider a photon with momentum p, and therefore energy |p|c (and zero mass), that is absorbed by an object with momentum 0 and energy m₁c²...

It seems that you are taking the absorption of one photon by one electron as an elastic collision.

You have misread my mind. I never said the second object was an electron. It could be an atom, or, for that matter, an apple.

 

[DrGreg]

In relativity, the mass of an object is given by<br /> m = \sqrt{\frac{E^2}{c^4} - \frac{|\textbf{p}|^2}{c^2}}<br />where E is the total energy of the object and p is the total momentum of the object.

Now consider a photon with momentum p, and therefore energy |p|c (and zero mass)...

thanks Greg, now i get it. Can you just tell me why do we take |p|c in the place of E?

For a photon, E = pc. If you didn't already know that, you can get it from the equation above with m = 0. See photon.

 

[juanrga]

You have misread my mind. I never said the second object was an electron. It could be an atom, or, for that matter, an apple.

I read what your wrote. You wrote general claims for an «object», in general, and I showed that your claims are not valid for a kind of objects, including electrons.

If your object is an atom or apple then the total energy E is not the kinetic energy {*} but includes U and V (if any) and your equation is not valid.

{*} Including the rest term.

 

[DrGreg]

I read what your wrote. You wrote general claims for an «object», in general, and I showed that your claims are not valid for a kind of objects, including electrons.

What I said was:

Now consider a photon with momentum p, and therefore energy |p|c (and zero mass), that is absorbed by an object with momentum 0 and energy m₁c².

It is a matter of simple verbal logic that whatever follows applies only to those objects that have the capability to absorb a photon (and therefore excludes electrons). (If I'd said "consider a red object", then obviously I'd be excluding blue objects.)

If your object is an atom or apple then the total energy E is not the kinetic energy {*} but includes U and V (if any) and your equation is not valid.

{*} Including the rest term.

I stand by what I said. You haven't defined what you mean by "U" and "V", but I'm guessing you are referring to the internal potential energies and kinetic energies of the object's constituent particles, but I don't care about that. All I am interested in is the total momentum 0 and total energy m₁c² of the entire composite object, in which case I can apply the equation of post #117 and conclude that the total "system mass" of the object is (by definition) m₁.

In case there is still confusion, the total system mass is not the sum of the individual particle masses, it's defined as the total energy (as measured in the zero-momentum frame) divided by c².

 

[sophiecentaur]

What I said was:

It is a matter of simple verbal logic that whatever follows applies only to those objects that have the capability to absorb a photon (and therefore excludes electrons). (If I'd said "consider a red object", then obviously I'd be excluding blue objects.)I stand by what I said. You haven't defined what you mean by "U" and "V", but I'm guessing you are referring to the internal potential energies and kinetic energies of the object's constituent particles, but I don't care about that. All I am interested in is the total momentum 0 and total energy m₁c² of the entire composite object, in which case I can apply the equation of post #117 and conclude that the total "system mass" of the object is (by definition) m₁.

In case there is still confusion, the total system mass is not the sum of the individual particle masses, it's defined as the total energy (as measured in the zero-momentum frame) divided by c².

People seem to think that electrons absorb photons because they are seem unaware that it is the electron as part of a charge system that is doing the 'absorbing'. This may account for the confusion.

 

[juanrga]

What I said was:

It is a matter of simple verbal logic that whatever follows applies only to those objects that have the capability to absorb a photon (and therefore excludes electrons). (If I'd said "consider a red object", then obviously I'd be excluding blue objects.)

Electrons can absorb photons, otherwise they could not be accelerated.

Moreover, what you wrote in previous posts do not apply to other objects as atoms, nuclei, apples...

I stand by what I said. You haven't defined what you mean by "U" and "V", but I'm guessing you are referring to the internal potential energies and kinetic energies of the object's constituent particles, but I don't care about that. All I am interested in is the total momentum 0 and total energy m₁c² of the entire composite object, in which case I can apply the equation of post #117 and conclude that the total "system mass" of the object is (by definition) m₁.

In case there is still confusion, the total system mass is not the sum of the individual particle masses, it's defined as the total energy (as measured in the zero-momentum frame) divided by c².

The case of an elementary particle was considered before and I will not repeat.

For a composite object being a collection of free particles the expression for its rest mass is

m = \sqrt{\frac{E_0^2}{c^4}} = \sqrt{\frac{E^2}{c^4} - \frac{\textbf{p}^2}{c^2}}

Where E and p are total energy and momentum, respectively, of the object.

Of course, the object mass m is not just «the sum of the individual particle masses» as you correctly state, but m does not vary with the total momentum p of the object, as you affirm. The mass of the object at rest, p=0, is the same than the mass of the object with a nonzero p.

For a free microscopic object being a collection of bound particles, the expression for its rest mass is the same

m = \sqrt{\frac{E_0^2}{c^4}} = \sqrt{\frac{E^2}{c^4} - \frac{\textbf{p}^2}{c^2}}

where again m is not just the sum of the individual particle masses and, again, m is not a function of the total momentum p, as you affirm. The mass of a nucleus is independent of its p. The mass defect, i.e. the difference between nuclei m and the sum of the mass of the nucleons, is a function of the binding energy, neither of E nor p of the nuclei.

For objects as apples, the above expression does not apply, because the total energy of an apple is not just kinetic, it is lacking U and V contributions.

For an apple falling with non-relativistic speeds within Earth gravitational potential, the mass is

m = \frac{p^2}{2(E-E_0-V)}

Once again this m is not just the sum of the individual particle masses and, again, m is not a function of the total momentum p, as you affirm. The mass of the apple is independent of p. In group theory it is often said that m is a central charge, i.e. one of the basic invariants.

It seems that you are confounding the rest mass m as defined above with the mass M

M = \frac{E}{c^2}

In some (old?) texts you can find this definition of mass, which of course varies with the total momentum p of the object.

M = M(p)

 

[Phrak]

Because mass isn't conserved in relativity, only energy and momentum.

I don't know what you mean by this. The mass of a system is conserved in special relativity. We can form a mass continuity equation in special relativity, and promote it to general relativity with the use of densities.

 

[Phrak]

Electrons can absorb photons, otherwise they could not be accelerated.

DrGreg was speaking of an electron in isolation. In isolation, an electron is in uniform motion. It will not radiate. The converse is equally true; it will not absorb a quanta of light and not accelerate [speaking in the lingo where elementary particles are presumed little bee bee balls of stuff].

The rest of your post is incoherent.

 

[sophiecentaur]

Not light, of course, because the energy change would have to be too high. There would be a quantum of very low energy electromagnetic energy, though, in the case of interaction with a near-static electric field - which is something I had forgotten in my last post. It is a good idea not to restrict one's ideas about photons to those of light.
You can often shed more light on the nature of photons by looking beyond light.

 

[juanrga]

DrGreg was speaking of an electron in isolation. In isolation, an electron is in uniform motion. It will not radiate. The converse is equally true; it will not absorb a quanta of light and not accelerate [speaking in the lingo where elementary particles are presumed little bee bee balls of stuff].

The rest of your post is incoherent.

If you read the posts and check the context of the quote that you extract, you will see that was that of an electron initially at rest, absorbing a photon so that finally it is moving with momentum p.

 

[Dale]

Electrons can absorb photons, otherwise they could not be accelerated.

An electron can be accelerated by scattering. Absorption is not the only process that can result in acceleration.

 

[sophiecentaur]

An electron can be accelerated by scattering. Absorption is not the only process that can result in acceleration.

Does that mean that the scattered photon has a different frequency? (If it's transferred some KE to the electron) It could be a very small difference, I suppose.

 

[Dale]

Does that mean that the scattered photon has a different frequency? (If it's transferred some KE to the electron) It could be a very small difference, I suppose.

In the COM frame the scattered photons frequency is the same. In all other frames it changes.

 

[sophiecentaur]

COM of the electron?

 

[Dale]

No, COM of the electron+photon.

 

[sophiecentaur]

But, if the photon has no mass . . . .?
(Reasonable question?)

 

[jtbell]

COM = center of momentum; the frame in which the total momentum is zero.

 

[Dale]

But, if the photon has no mass . . . .?
(Reasonable question?)

Oops, sorry, when you asked your follow-up question I should have realized that the acronym was unclear. As jtbell mentioned COM means center of momentum, not center of mass.

 

[sophiecentaur]

All clear now. Thanks.

 

[juanrga]

An electron can be accelerated by scattering. Absorption is not the only process that can result in acceleration.

Compton scattering, Møller scattering, Bhabha scattering... are all based in the existence of a photon absorption step.

But my point was really other. I was said that electrons cannot absorb photons, which is not true.

 

[Phrak]

I find it cleaner and more obvious to avoid square roots where possible and use the vector equations that are good in Minkowski coordinates for adding masses.

define\mu for a particle.

\mu = (E/c^2, \textbf{p}/c)

The bases vectors are dropped for convenience.

For a two particle system.
\mu_1 = (E_1/c^2, \textbf{p}_1/c)
\mu_2 = (E_2/c^2, \textbf{p}_2/c)
Vector addition.
\mu_1+\mu_2 = ([E_1 + E_2 ]/c^2, [\textbf{p}_1 + \textbf{p}_2 ]/c)
The particle masses.
m_1 = |\mu_1|
m_2 = |\mu_2|
The combined mass.
m_\Sigma = |\mu_1 + \mu_2 |

 

[Dale]

Compton scattering, Møller scattering, Bhabha scattering... are all based in the existence of a photon absorption step.

In a Feynman diagram scattering is indeed represented as an absorption and an emission, but there is a key difference between absorption and scattering. The absorption step, by itself, cannot conserve momentum, so by itself it is a "virtual" event which can never be observed directly, much like virtual photons. On the other hand scattering can be directly observed. In fact, an isolated electron has never been observed to change momentum due only to the absorption of a real photon.

But my point was really other. I was said that electrons cannot absorb photons, which is not true.

I wasn't responding to your other point. Merely correcting the assertion that the only way for an electron to change momentum is by absorption of a photon.