Do Photons Have Mass? Exploring the Particle

[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.

 

[juanrga]

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.
$$m_\Sigma = |\mu_1 + \mu_2 |$$

Beautiful, only to remark that if the particles are not free, neither $m_1$ nor $m_2$ represent the masses of the particles.

 

[juanrga]

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.

Therefore, scattering is only possible if the electron absorb a photon.

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.

See above.

 

[tom.stoer]

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 ...

This is wrong. At each vertex in a Feynman diagram 4-momentum is exactly conserved.

 

[Dale]

... In fact, an isolated electron has never been observed to change momentum due only to the absorption of a real photon.

Therefore, scattering is only possible if the electron absorb a photon.

That is an impressive non-sequitir.

 

[Dale]

This is wrong. At each vertex in a Feynman diagram 4-momentum is exactly conserved.

Oops, sorry, you are right. I meant to say that the absorption of a real photon by an isolated electron cannot conserve momentum.

 

[tom.stoer]

Oops, sorry, you are right. I meant to say that the absorption of a real photon by an isolated electron cannot conserve momentum.

Here I agree :-)

 

[logics]

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

The rationale of this thread is right there: question is not reasonable, it is pointless. If it is so, all answers become pointless too, even though discussion may be interesting; and "light has no mass" may seem more appropriate.
Milk has cheese?, water has ice? can we answer yes or no? The logical problem that makes question meaningless looks like "http://wikipedia.org/wiki/Present_King_of_France" has hair?", but it has a different cause :
there the subject "king" is not, here the subject "EMR" is also the predicate.
Misunderstanding arises because of the inadequacy of term annihilation of a photon or of mass ⁺e ^-e : EMR transforms itself into mass and vice versa.
They say mass is "trapped", "transformed", "a form of" energy ": just like ice of water. water is mobile, ice is not.
does water have ice?

 

[Phrak]

The rationale of this thread is right there: question is not reasonable, it is pointless. If it is so, all answers become pointless too, even though discussion may be interesting; and "light has no mass" may seem more appropriate.
Milk has cheese?, water has ice? can we answer yes or no? The logical problem that makes question meaningless looks like "http://wikipedia.org/wiki/Present_King_of_France" has hair?", but it has a different cause :
there the subject "king" is not, here the subject "EMR" is also the predicate.
Misunderstanding arises because of the inadequacy of term annihilation of a photon or of mass ⁺e ^-e : EMR transforms itself into mass and vice versa.
They say mass is "trapped", "transformed", "a form of" energy ": just like ice of water. water is mobile, ice is not.
does water have ice?

Maybe. But physics does not advance in this way.

 

[Phrak]

Beautiful, only to remark that if the particles are not free, neither $m_1$ nor $m_2$ represent the masses of the particles.

There are as many as two approches to this that I can decern. One is that $\mu_n$ represent the integrals of energy and momentum density over a sufficiently large region of space where a single quanta is to be found and where the values are propperly signed so that they combine as a vector rather than a co-vector. That is,
$$E = \int \rho dx dy dz$$
$$p_i = \int P_{jkt} dx dy dt$$
where $\rho$ is energy density, and $P_{jkt}$ is the momentum density; the infinitessimal flux of mass per time dt through an area $dx^i$ $dx^j$.

The other, is that we presume isolated billiard ball-like particles that have an associated energy-momentum intensity vector. Both representations are elements of a vector space (not the same one) and transform as displacement vectors in Minkowski spacetime.

Unlike the unintegrated density, neither is viable in curved spacetime.

 

[juanrga]

... In fact, an isolated electron has never been observed to change momentum due only to the absorption of a real photon.

Therefore, scattering is only possible if the electron absorb a photon.

That is an impressive non-sequitir.

I was really replying to the part of you message that you have not quoted:

In a Feynman diagram scattering is indeed represented as an absorption and an emission...

Therefore, scattering is only possible if the electron absorb a photon.

In my response I ignored the last part that you quoted, because, as already emphasized in #138, I am not considering an isolated electron but an interacting one. And I did not consider that I may be repeating the same again and again in each new post.

I only want to add that if the electron was isolated, it could not participate in a scattering. And if the electron is being scattered then it is not isolated.

If you want believe that an electron can be scattered (e.g. Compton scattering) without absorbing a photon whereas, at the same time, accepting that in QED that scattering involves a step where a photon is absorbed by an electron, you are welcome.

 

[juanrga]

Beautiful, only to remark that if the particles are not free, neither $m_1$ nor $m_2$ represent the masses of the particles.

There are as many as two approches to this that I can decern. One is that $\mu_n$ represent the integrals of energy and momentum density over a sufficiently large region of space where a single quanta is to be found and where the values are propperly signed so that they combine as a vector rather than a co-vector. That is,
$$E = \int \rho dx dy dz$$
$$p_i = \int P_{jkt} dx dy dt$$
where $\rho$ is energy density, and $P_{jkt}$ is the momentum density; the infinitessimal flux of mass per time dt through an area $dx^i$ $dx^j$.

The other, is that we presume isolated billiard ball-like particles that have an associated energy-momentum intensity vector. Both representations are elements of a vector space (not the same one) and transform as displacement vectors in Minkowski spacetime.

Unlike the unintegrated density, neither is viable in curved spacetime.

Valid only if the particles are free.

In previous posts I have considered the mass of generic objects including non-isolated apples.

 

[Dale]

If you want believe that an electron can be scattered (e.g. Compton scattering) without absorbing a photon whereas, at the same time, accepting that in QED that scattering involves a step where a photon is absorbed by an electron, you are welcome.

Yes, this is exactly the point. QED models the scattering that way, also involving a virtual particle which by definition can never be detected. So it is an unfalsifiable part of QED. However QED is not the only model which accurately describes scattering, and other models do not involve an abbsorption. It is an unfalsifiable, model dependent claim. Therefore, scattering does not imply absorption.

Btw I am not objecting to your overall claim, just the logic that the observed acceleration of free electrons implies absorption. It is simply not true of real photons and free electrons, to make it true requires one specific model in which it is unfalsifiably true.

 

[Phrak]

Valid only if the particles are free.

I don't know what you mean, unless you missed my use of the conditional "isolated".

 

[juanrga]

Valid only if the particles are free.

I don't know what you mean, unless you missed my use of the conditional "isolated".

It was only a redundant emphasis of what I wrote just after of the paragraph that you quoted above:

In previous posts I have considered the mass of generic objects including non-isolated apples.

 

[juanrga]

However QED is not the only model which accurately describes scattering, and other models do not involve an abbsorption.

I am curious, could you say what models please?

 

[Dale]

I am curious, could you say what models please?

Rutherford scattering, Mott scattering, Compton scattering, for example.

 

[juanrga]

However QED is not the only model which accurately describes scattering, and other models do not involve an abbsorption.

I am curious, could you say what models please?

Rutherford scattering, Mott scattering, Compton scattering, for example.

In #148 I cited «Compton scattering», «Møller scattering», «Bhabha scattering» and others. And remarked as they «are all based in the existence of a photon absorption step».

When you said me that scattering could be explained with the same accuracy by other models different than QED, I asked you waiting the name of that alternative to QED that also explains scattering... but your response has been «scattering». Fine .

 

[Phrak]

Try other formulations of quantum field theory.

 

[logics]

See - https://www.physicsforums.com/showthread.php?t=511175

The quick answer: NO

question is ...pointless.
does water have ice?

Maybe. But physics does not advance in this way.

Physics, Science in general, advances looking for, finding, and accepting the "truth".
what is your quick "yes/no...because" answer?,
what is your argument?

 

[juanrga]

Try other formulations of quantum field theory.

Well I was said above:

However QED is not the only model which accurately describes scattering, and other models

It seems that he was talking about alternatives to QED model, not about another formulation of the same QED, but in any case after reading your reply:

what other formulation of QED(QFT) you claim that does not use photons to describe electron electron scattering for instance?

 

[logics]

I am curious, could you say what models please?

... Compton scattering, for example.

and what is your argument?
Compton scattering is an elastic collision between light and a "free" electron", with a transfer of Energy from γ to ^-e: can we say this is "absorption" in any model?
If we can accept that E [=]→ M, a fortiori we must accept that E [=]→ KE, that "EMR" can transform itself not only into mass but also into kynetic energy.
The "absorption" of extra energy causes an increase of KE and then this causes an increase of velocity and then this causes an increase of momentum.
[EMR]E → E_k → v → p is not an identity but a chain of causal relationships
Is there any problem? do photons have mass? do photons have momentum?