How will u justify rest mass of photon is 0?
Answers
Answered by
0
The concept of rest mass of a particle originates from Einstein's Special Theory of Relativity.
The rest mass is the mass of a particle as measured by an observer who observes the particle as stationary or moving with zero velocity.
The relativistic expressions for Eand→p obey the relativistic energy–momentum relation as below
E2−(→pc)2=(m0c2)2 ......(1)
where m0 is the rest mass of the particle, cis velocity of light in vacuum and E is its total energy.
The expression becomes
E=m0c2√1−v2c2=γm0c2 .....(2)
We also get the following
→p=m0→v√1−v2c2=γm0→v .....(3)
From (2) and (3) we see that any particle which has non-zero rest mass can not have velocity =c as the denominator reduces to zero and the value of momentum or total energy becomes infinite or indeterminate.
One of the postulates of special theory of relativity states that
The speed of light is constant in all frames and in vacuum and =c. As photons can not be brought to rest therefore, above equations are applicable to particles having speed less than conly and NOT to photons.
(We also that the equations (2) and (3) take the form 00 for a particle having rest mass 0 and moving with velocity c.)
Consider two cases of equation (1)
For the particle be at rest ⇒→p=0,andE=m0c2.Setting rest mass equal to zero we get E=∣∣→p∣∣c.
From classical electromagnetic theory we have the expression which relates energy E and momentum →p of light as E=∣∣→p∣∣c. Thus we see that even though (2) and (3) are not applicable to photons but (1) can be made applicable to photons by considering photons to have no mass.
It needs to be remembered that momentum of a photon is a function of its energy. It is not proportional to its velocity, which is always c.
The rest mass is the mass of a particle as measured by an observer who observes the particle as stationary or moving with zero velocity.
The relativistic expressions for Eand→p obey the relativistic energy–momentum relation as below
E2−(→pc)2=(m0c2)2 ......(1)
where m0 is the rest mass of the particle, cis velocity of light in vacuum and E is its total energy.
The expression becomes
E=m0c2√1−v2c2=γm0c2 .....(2)
We also get the following
→p=m0→v√1−v2c2=γm0→v .....(3)
From (2) and (3) we see that any particle which has non-zero rest mass can not have velocity =c as the denominator reduces to zero and the value of momentum or total energy becomes infinite or indeterminate.
One of the postulates of special theory of relativity states that
The speed of light is constant in all frames and in vacuum and =c. As photons can not be brought to rest therefore, above equations are applicable to particles having speed less than conly and NOT to photons.
(We also that the equations (2) and (3) take the form 00 for a particle having rest mass 0 and moving with velocity c.)
Consider two cases of equation (1)
For the particle be at rest ⇒→p=0,andE=m0c2.Setting rest mass equal to zero we get E=∣∣→p∣∣c.
From classical electromagnetic theory we have the expression which relates energy E and momentum →p of light as E=∣∣→p∣∣c. Thus we see that even though (2) and (3) are not applicable to photons but (1) can be made applicable to photons by considering photons to have no mass.
It needs to be remembered that momentum of a photon is a function of its energy. It is not proportional to its velocity, which is always c.
Similar questions