no, it has zero rest mass m' so its mass is undefined at any speed.
As you know of course it does have finite energy and momentum.
according to einstein's mass energy relation
M= m'/root[1-(v^2/c^2)]
Where M is mass of particle when it moves with velocity 'v'
And m' is its initial mass.
When particle moves with velocity of light i.e. When 'v=c' then
M= m'/0 =infinity i.e. Infinite mass
But a photon of light can have velocity of light.
So does this possess infinite mass?!!
2 answers
see Wikipedia:
" .... In empty space, the photon moves at c (the speed of light) and its energy and momentum are related by E = pc, where p is the magnitude of the momentum vector p. This derives from the following relativistic relation, with m = 0:[15]
E^{2}=p^{2} c^{2} + m^{2} c^{4}.
The energy and momentum of a photon depend only on its frequency (ν) or inversely, its wavelength (λ):
E=\hbar\omega=h\nu=\frac{hc}{\lambda}
\boldsymbol{p}=\hbar\boldsymbol{k},
where k is the wave vector (where the wave number k = |k| = 2π/λ), ω = 2πν is the angular frequency, and ħ = h/2π is the reduced Planck constant.[16]
Since p points in the direction of the photon's propagation, the magnitude of the momentum is
p=\hbar k=\frac{h\nu}{c}=\frac{h}{\lambda}. ......"
" .... In empty space, the photon moves at c (the speed of light) and its energy and momentum are related by E = pc, where p is the magnitude of the momentum vector p. This derives from the following relativistic relation, with m = 0:[15]
E^{2}=p^{2} c^{2} + m^{2} c^{4}.
The energy and momentum of a photon depend only on its frequency (ν) or inversely, its wavelength (λ):
E=\hbar\omega=h\nu=\frac{hc}{\lambda}
\boldsymbol{p}=\hbar\boldsymbol{k},
where k is the wave vector (where the wave number k = |k| = 2π/λ), ω = 2πν is the angular frequency, and ħ = h/2π is the reduced Planck constant.[16]
Since p points in the direction of the photon's propagation, the magnitude of the momentum is
p=\hbar k=\frac{h\nu}{c}=\frac{h}{\lambda}. ......"
0 answers