Is the Mass of a Photon Affected by its Energy?

[codjohns]

I am wondering about photons having mass. The new thing on all the space shows is the sailing through space with a sail that harnesses photons as a ship harnesses wind. They did an experiment and the photons exerted a force on the sail. Am I wrong to think that E=mc^2 (E can not be 0 so M can not be 0) and since the photons exert kinetic energy on the sail I can assume that KE=1/2MV^2 therefor KE (how its exerting force on the sail) is = 1/2Mc^2. Since the energy is not 0 then the mass can not be 0, calculate the number of photons and the amount of force and you will have the mass of a photon.

Am I wrong to think this?

 

[Nugatory]

I am wondering about photons having mass. The new thing on all the space shows is the sailing through space with a sail that harnesses photons as a ship harnesses wind. They did an experiment and the photons exerted a force on the sail. Am I wrong to think that E=mc^2 (E can not be 0 so M can not be 0) and since the photons exert kinetic energy on the sail I can assume that KE=1/2MV^2 therefor KE (how its exerting force on the sail) is = 1/2Mc^2. Since the energy is not 0 then the mass can not be 0, calculate the number of photons and the amount of force and you will have the mass of a photon.

Am I wrong to think this?

You are wrong. E=mc^2 only applies to objects with non-zero rest mass and photons have a zero rest mass. The equation that you're looking for is E^2=(mc^2)^2 + (pc)^2 where p is the momentum and m is the rest mass; this one works even if the rest mass is zero, as it is for a photon.

This equation allows the photon to carry momentum, which is transferred to the sail when the photon hits it.

 

[ZapperZ]

I am wondering about photons having mass. The new thing on all the space shows is the sailing through space with a sail that harnesses photons as a ship harnesses wind. They did an experiment and the photons exerted a force on the sail. Am I wrong to think that E=mc^2 (E can not be 0 so M can not be 0) and since the photons exert kinetic energy on the sail I can assume that KE=1/2MV^2 therefor KE (how its exerting force on the sail) is = 1/2Mc^2. Since the energy is not 0 then the mass can not be 0, calculate the number of photons and the amount of force and you will have the mass of a photon.

Am I wrong to think this?

Please read our FAQ subforum.

https://www.physicsforums.com/showthread.php?t=511175

Zz.

 

[lightarrow]

You are wrong. E=mc^2 only applies to objects with non-zero rest mass ...

... and which are stationary in the frame of reference considered.

 

[rbj]

You are wrong. E=mc^2 only applies to objects with non-zero rest mass and photons have a zero rest mass. The equation that you're looking for is E^2=(mc^2)^2 + (pc)^2 where p is the momentum and m is the rest mass; this one works even if the rest mass is zero, as it is for a photon.

even though it is deprecated in most, nearly all, current texts, there is most certainly a usage to the famous E=mc^2 where the m refers to the (oft deprecated) "relativistic mass":

m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}

and m_0 is the "rest mass" or "invariant mass" of the body. a photon is assumed to move at a speed of c in anyone's frame of reference, so it cannot have a non-zero rest mass because

m_0 = m \sqrt{1 - \frac{v^2}{c^2}}

and the latter factor goes to zero when v=c.

if m used in E=mc^2 is the expression above, then E=mc^2 is perfectly consistent with

E^2=(m_0 c^2)^2 + (p c)^2

and

p = m v .

this E is the total energy, the "rest energy" E_0=m_0 c^2 (which is the "E=mc^2" you're referring to) plus the kinetic energy (from the POV of the observer in the frame of reference).

E = m c^2 = E_0 + T

the kinetic energy

T = E - E_0

goes to the classical approximation T = \frac{1}{2}m_0 v^2 when v<<c.

This equation allows the photon to carry momentum, which is transferred to the sail when the photon hits it.

a quantity of the dimension "mass" can be derived from dividing the momentum of the photon by its speed (which is c).

"wrong" and "currently deprecated" are not the same thing.

 

[Bill_K]

"wrong" and "currently deprecated" are not the same thing.

A better characterization would be "highly misleading". One need only count up the queries here on PF that were produced as a result of this confusion. Many posters have imagined, for example, that the relativistic mass increase must eventually turn the particle into a black hole.

 

[rbj]

Many posters have imagined, for example, that the relativistic mass increase must eventually turn the particle into a black hole.

point taken. a simple black hole needs to satisfy

\frac{M}{r} > \frac{c^2}{2 G}

in its own reference frame.