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March 2[edit]

Why aren't we allowed to use the formula of relativistic momentum, for calculating the positive momentum of slow light not moving in a vacuum?[edit]

HOTmag (talk) 16:53, 2 March 2024 (UTC)[reply]

The momenta that light carries are observed to be largely independent of its speed. For example, when light, which are particle-waves, passes from one medium to another and then back again into the original medium such as through a window pane, unless it gets scattered (see Compton scattering), it emerges with the same speed, frequencies, energies and momenta it started with. This is because the photons' momenta and energies are conserved throughout although their wave velocities differ within the different media. Conversely, the momenta of masses are known to depend on their speed. Modocc (talk) 19:55, 2 March 2024 (UTC)[reply]

Yes, I know that the momenta that light carries are observed to be largely independent of its speed. But I was asking about the invalidity of the formula of relativistic momentum: Please notice this formula does not warn: "Don't use me for calculating the positive momentum of slow light not moving in a vacuum", does it? As long as it does not, what makes it void/invalid, for such a calculation, from a formal viewpoint? HOTmag (talk) 20:16, 2 March 2024 (UTC)[reply]

On one hand, individual photons don't have rest mass. So there is nothing really to calculate. On the other hand, two or more counterposing photons do have an invariant mass to which the formula can be applied. Modocc (talk) 20:53, 2 March 2024 (UTC)[reply]

Isn't is common to regard the photon's rest mass as zero? Note that the formula of relativistic momemtum does not warn: "Don't use me when the rest mass is zero", does it? HOTmag (talk) 21:02, 2 March 2024 (UTC)[reply]

The formula + zero mass yields, simply, zero momentum. When we measure lights' momenta, its not zero, unless the system as a whole has a mass. The formula is only used for calculating the momenta of mass. Modocc (talk) 21:18, 2 March 2024 (UTC)[reply]

Do you agree, that the formula should have warned: "Don't use me for any zero mass"? HOTmag (talk) 21:30, 2 March 2024 (UTC)[reply]

An isolated silvered box containing coherent photons will oscillate with respect to transference of the photons' constant momenta even if these are considerably slowed down. Clearly, the formula is not valid for modeling their momenta. Modocc (talk) 21:47, 2 March 2024 (UTC)[reply]

I agree with you that the formula is not valid for modeling those momenta. However, my question is about the formula per se: Why doesn't it warn: "Don't use me for any zero mass"? HOTmag (talk) 21:53, 2 March 2024 (UTC)[reply]

No warning is needed because light carries (or has) momentum without that specific formula. Modocc (talk) 23:17, 2 March 2024 (UTC)[reply]

A warning is needed for the formula to be valid. Just as the well known algebraic formula "If ab=ac then b=c" must warn: "Don't use me if a=o", for this algebraic formula to be valid. Without such a warning, this algebraic formula may be wrong (e.g. in case a=0), whereas a warning makes this algebraic formula valid because it will never be used in cases mentioned in the warning. The same is true for the relativistic formula of momentum: Without the warning "Don't use me if the rest mass is zero", this physical formula may be wrong (e.g. in case the rest mass is zero), whereas a warning makes this physical formula valid because it will never be used in cases mentioned in the warning. HOTmag (talk) 00:47, 3 March 2024 (UTC)[reply]

The fact it's shown to be invalid and said to be invalid, repeatedly by me here, and by others in the literature: suffices. Modocc (talk) 01:25, 3 March 2024 (UTC)[reply]

What? Is it really "said to be invalid...by others in the literature"? I will be glad to see any source ascribing "invalidity" to this formula. HOTmag (talk) 01:46, 3 March 2024 (UTC)[reply]

It is valid for mass only, but it is not valid for individual photons. Correct? Modocc (talk) 02:01, 3 March 2024 (UTC)[reply]

Also, it's because, unlike rest mass, a photon's frequency, energy and momentum are not invariant. Modocc (talk) 02:12, 3 March 2024 (UTC)[reply]

However, physicists usually assume the Energy–momentum relation, which is a work-around that is valid for photons. Modocc (talk) 03:38, 3 March 2024 (UTC)[reply]

"not valid for individual photons"? I've never seen this formula warn: "Don't use me for individual photons" (What about gluons?), but I've only seen this formula warn: "Don't use me for velocities not slower than C (because arithmetically one is not allowed to devide by zero nor to extract square roots of negative numbers)".

I didn't understand your second remark about the invariance: Note this formula does not demand, of the frequency or of the energy or of the momentum, to be invariant. It only demands of the rest mass (needed for calculating the relativistic momentum) to be invariant. Apparently, every photon's rest mass is invariant, so apparently this formula doesn't warn against using it for slow photons, because no division by zero (nor any extraction of square root of any negative number) is made for them. HOTmag (talk) 09:52, 3 March 2024 (UTC)[reply]

For m₀=0, the Energy–momentum relation

E^{2}=(p{\textrm {c}})^{2}+\left(m_{0}{\textrm {c}}^{2}\right)^{2}\,

is reduced to

E=pc

and light's momentum is thus given by E/c in accordance with its energy which is related to its frequency per the Planck relation,

E=h\nu .

Furthermore, don't use

\mathbf {p} =\gamma m_{0}\mathbf {v} \,

for light because its momenta are not zero. From what I've read and recall at this point, slow light interaction models typically invoke either group waves, their photons' coupling to quasi-particles and/or lattice calculations involving Quantum electrodynamics. Stuff that is far too complex to expound on. Modocc (talk) 14:17, 3 March 2024 (UTC)[reply]

I already know I shouldn't use the formula of relativistic momentum, for any slow photon, because (as you explain well): "its momenta are not zero". That's obvious, and I've known all of that. I remind of you, that my question was only about why this formula doesn't warn: "Don't use me for slow photons", while all other formulas I know - do warn againt using them in forbidden cases. HOTmag (talk) 14:41, 3 March 2024 (UTC)[reply]

It is not clear at all how to define the photon momentum in medium. See this. Ruslik_Zero 20:51, 2 March 2024 (UTC)[reply]

Thank you for the article, which actually makes a distinction between the kinetic momentum and the canonical momentum.

On the other hand, a given slow photon's momentum could be easily determined (hence defined) - as zero, if we were allowed to use the formula of relativistic momemta, which does not warn: "Don't use me when the rest mass is zero". The question is, why aren't we allowed to use this formula, when it comes to a given slow photon? HOTmag (talk) 21:02, 2 March 2024 (UTC)[reply]

(ec) What makes you think that a photon in a medium has mass zero? Please don't forget that a photon is not a simple classical particle but an excitation of a quantum field which interacts with the medium in an a priori rather complicated way. I assume that we can effectively continue to treat it as a particle, certainly with non-zero momentum, possibly with an effective mass due to the interaction with the medium, but I'm out of my depth here. --Wrongfilter (talk) 21:35, 2 March 2024 (UTC)[reply]

Don't forget, that by "mass" we actually mean "invariant mass", so if it's zero in a vacuum, it's apparently zero in any medium. Just as my own invariant mass does not change when I pass from a vacuum to any medium. HOTmag (talk) 21:50, 2 March 2024 (UTC)[reply]

I wouldn't bet on that. --Wrongfilter (talk) 22:05, 2 March 2024 (UTC)[reply]

That's why I was careful when I wrote about the photon's invariant mass: "it's apparently zero in any medium". HOTmag (talk) 22:11, 2 March 2024 (UTC)[reply]

That seems like a misinterpretation of "invariant". --Wrongfilter (talk) 22:16, 2 March 2024 (UTC)[reply]

Besides photons, about which we can argue whether their invariant mass may change when passing from a given medium to another, do you have in mind any other example of a particle whose invariant mass may change when passing from a given medium to another? HOTmag (talk) 22:28, 2 March 2024 (UTC)[reply]

To be clear, both Minkowski and Abraham arguments seem to assume some sort of complicated transference of the photons' momenta to and from the propagation media. Personally, FWIW, I've come to reject both arguments per Occam's razor in favor of any models in which the photons retain their momenta... But that's on me. Modocc (talk) 21:30, 2 March 2024 (UTC)[reply]

The photon, with it's rest mass of zero, always travels at the speed of light. Note that even inside a medium most of the space is empty. If "light" travels more slowly in a medium, that is not because of a slower speed of individual photons, but either because the interaction with the medium causes photons to travel a more complicated path ("bouncing between atoms"), or because the photon is absorbed and re-emitted. --Stephan Schulz (talk) 14:30, 3 March 2024 (UTC)[reply]

Thanks. This was actually my suspicion.

By the way, do you have any new answer to a close question in one of my previous threads? HOTmag (talk) 14:42, 3 March 2024 (UTC)[reply]

~~Perhaps individual photons do slow down, or if not at least couple with polaritons, and gain effective mass and attract each other due to QED interactions.~~ Upon reading their article they used an effective field theory and I quote: "...where vg is the group velocity inside the medium,..." and a function of vg "...is the effective photon mass...". It is somewhat interesting though. Modocc (talk) 16:17, 3 March 2024 (UTC)[reply]