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aleemudasir
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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?
The photon is currently understood to be strictly massless, but this is an experimental question.
jnorman said: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?)
aleemudasir said: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?
aleemudasir said: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!
aleemudasir said: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 said: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 said: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 said: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 said:Which equation do we use to infer that speed of light is the limit?
aleemudasir said:]
Correct me if I am wrong.
Thanks.
pervect said:People have been trying to correct you all along - you don't give the apperance of actually listening to the corrections, however...
Drakkith said:I believe it is p=h/wavelength, where h is Plancks constant.
aleemudasir said:OK, but it is just the rearrangement of equation λ=h/p, where p=mv.
M Quack said:No, you are wrongly simplifying a general equation to a very special case.
[itex] p = \hbar k[/itex] where k is the wave vector ([itex]\frac{2\pi}/\lambda[/itex]) 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 [itex] E = \frac{1}{2m}p^2[/itex].
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 [itex] E^2 = (m_0 c^2)^2 + (pc)^2[/itex].
Now all experimental observations over a huge huge range of wavelengths/energies/wave numbers match [itex] E=pc = \hbar \omega = \frac{h}{\lambda}[/itex], i.e. they match the above equation with [itex]m_0 = 0[/itex]. 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.
bahamagreen said:Photons have non-zero rest mass inside superconductors...
phinds said:I agree w/ Dead Boss ... you'll need to provide a citation for that. It sounds like nonsense.
M Quack said: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) [itex] E_0=m_0 c^2[/itex] 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).
[itex] E=\frac{h c}{\lambda}[/itex] Sorry for the typo (and I see it is not the only one...)
[itex] k = \frac{2\pi}{\lambda}[/itex]
Stricktly speaking, the non-relativistic limit should also be
[itex]E_{\mathrm{kinetic}} = \frac{1}{2m}p^2 = E - E_0[/itex] where E is the total energy and [itex]E_0=m_0 c^2[/itex] the energy-equivalent of the rest mass.
Paul Bauza said:Rest mass of Photon is 0.
My Q is: 1- why Photon speed is finite if does have no mass?
2- why light is curved passing Black Hole and if is curved because
warped space-time. (not it's mass) how Black Hole can warp Space-Time.
if Space-Time have no mass?
Paul Bauza said:how Black Hole can warp Space-Time.
if Space-Time have no mass?
Photon rest mass refers to the theoretical mass that a photon would have if it were at rest. However, according to the theory of relativity, photons cannot be at rest and therefore do not have a rest mass.
Scientists have conducted numerous experiments and observations that have consistently shown that photons do not have rest mass. Additionally, the theory of relativity, which has been extensively tested and confirmed, also predicts that photons do not have rest mass.
No, according to the theory of relativity, photons cannot have mass while moving. This is because as an object approaches the speed of light, its mass increases infinitely. Since photons travel at the speed of light, they would have infinite mass, which is not possible.
Since photons do not have rest mass, they cannot be measured in the same way as other particles. However, their energy can be measured using various methods, such as the photoelectric effect or Compton scattering. This energy can then be used to calculate the equivalent mass of the photon using Einstein's famous equation, E=mc^2.
No, the concept of photon rest mass is well-established in the scientific community and is not currently a topic of debate. However, scientists continue to study and explore the properties of photons, including their behavior and interactions with matter.