Is a Photon Massless or Does it Have Relativistic Mass?

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photon mass??

if a photon is a particle and the wavelength of a particle is h/mass*velocity ie
h/momentum . what is the mass to be substituted in the formula??aren't photons pointlike and maseless??
 
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Don't know how energy and photon wavelength are related?
 


i do know that. but since photon is a particle shouldn't the other thing work too??
 


Yes a photon is massless, that is, it's believed to have zero rest mass. The equation you are considering is supposed to show the wave nature of a particle with mass and isn't applicable to a photon.

You can't just arbitrarily use formulas as you propose...formulas, for example, applicable to an electron or photon may not be applicable to each other, nor for example, to a basketball. A photon appears as a "particle" only under certain measurement conditions...other times it appears as a wave with field like characteristics. And the electromagnetic force applicable to photons in quantum theory of course can't be used with, say, neutrons and protons in a nucleus which are bound by the strong force.
 


The general "formula" for wavelenght is:

Lambda = h/momentum

Momentum of a massive, non relativistic, particle is m*v

Momentum of a massive, Relativistic, particle is: ?

Momentum of MASSLESS particle is: ?

Try to fill in those question marks :-)
 


Glenn.tanQ!.and ll try to c if i can come up with formulas for those:wink:.. and Naty1...ok the same formula can't be used for a lot other things ..:smile:and ll look upto that link..
 


Nphysicist said:
if a photon is a particle and the wavelength of a particle is h/mass*velocity ie
h/momentum . what is the mass to be substituted in the formula??aren't photons pointlike and maseless??


Some people like to say that the photon has mass because the photon has energy [math]E=hf[/math], where [math](h)[/math] is 'Planck’s constant' and [math](f)[/math] is the frequency of the photon. Thus, they tend to assume that because it has energy [math](E)[/math] it must have mass [math](M)[/math] because of Einstien’s mass-energy equivalence equation [math]E=Mc^2[/math]...

They also say that the photon has momentum, and momentum is related to mass [math]p = Mv[/math] where [math](v)[/math] is velocity and [math](p)[/math] is for momentum. Yet, you cannot justify it having mass using this argument. This is actually 'relativistic mass' - which is nothing but the measure of energy which will change with velocity. It isn't actually mass, even though mass and energy are related. In physics jargon, the mass of an object is called its 'invariant mass,' and the photon has no invariant mass. Now, a massless particle can have energy and it can have momentum, simply because mass is related to these through the equation [math]E^2 = M^2c^4 + p^2c^2[/math], which is subsequently zero-mass for a photon because [math]E = pc[/math] for massless radiation (remember, c means the speed of light). So yes, the photon has momenta and energy, and can deliver a punch out of it when it hits a surface, but it doesn't have mass.
 


Nphysicist said:
if a photon is a particle and the wavelength of a particle is h/mass*velocity ie
h/momentum . what is the mass to be substituted in the formula??aren't photons pointlike and maseless??
For a photon momentum IS NOT m*v but E/c where E is its energy. So wavelenght = h/(E/c) = hc/E.
 
  • #10


Can the photon transfer gravity from one point in space-time to another?
 
  • #11


Primordial said:
Can the photon transfer gravity from one point in space-time to another?

Define "transfer".

But anyway, no, the photon is not the force carrier of Gravity.
 
  • #12


Primordial said:
Can the photon transfer gravity from one point in space-time to another?
As malawi_glenn said, you have to define what you mean with "transfer gravity". It could also be intended in the sense that an object at rest (e.g. an atom) loses a little of its mass trough the emission of a photon, which is then absorbed by another, far object at rest (another atom) which then increase its mass a little (this because an atom's mass is much greater than E/c^2, E = photon's energy, for photons emitted or absorbed by atoms).
 
  • #13


Light Arrow : Thank you for your response, I understand a photon to have relativistic mass due to the initial source relativistic momentum and due to the relativistic momentum of the inertial reference system that absorbs that energy, does this sound logical to you?
 
  • #14


Primordial said:
Light Arrow : Thank you for your response, I understand a photon to have relativistic mass due to the initial source relativistic momentum and due to the relativistic momentum of the inertial reference system that absorbs that energy, does this sound logical to you?
I sincerely haven't understood very much what you say; however the fact is very simple: every system at rest before and after absorption (emission) of an amount of energy E (any kind of energy), increases (decreases) its mass of the amount E/c^2. This is the correct use of the famous equation E = mc^2, and comes from the equation E^2 = (mc^2)^2 + (cp)^2.
 
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