Is the Mass of a Photon Affected by its Energy?

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Discussion Overview

The discussion revolves around the concept of whether photons possess mass and how their energy relates to mass, particularly in the context of experiments involving photon sails. Participants explore the implications of energy-mass equations and the nature of momentum transfer from photons to objects like sails.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • Some participants argue that since photons exert force on sails, they must have mass, referencing the equation E=mc^2 and kinetic energy formulas.
  • Others counter that E=mc^2 applies only to objects with non-zero rest mass, asserting that photons have zero rest mass and should be described using the equation E^2=(mc^2)^2 + (pc)^2, which accommodates massless particles.
  • One participant mentions the concept of "relativistic mass" and its historical context, suggesting that while E=mc^2 is often deprecated, it can still be applied under certain interpretations.
  • Another participant highlights the potential confusion surrounding relativistic mass and its implications, noting that misconceptions can lead to erroneous conclusions, such as the idea that relativistic mass could lead to black hole formation.

Areas of Agreement / Disagreement

Participants generally disagree on the interpretation of mass in relation to photons, with competing views on whether photons can be considered to have mass based on their energy and momentum. The discussion remains unresolved, with multiple perspectives presented.

Contextual Notes

Some participants emphasize the limitations of using E=mc^2 for massless particles and the importance of distinguishing between rest mass and relativistic mass. There is also mention of the historical context of these concepts and their relevance in current discussions.

codjohns
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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?
 
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codjohns said:
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 [itex]E^2=(mc^2)^2 + (pc)^2[/itex] 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.
 
codjohns said:
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.
 
Last edited by a moderator:
Nugatory said:
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.
 
Last edited:
Nugatory said:
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 [itex]E^2=(mc^2)^2 + (pc)^2[/itex] 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 [itex]E=mc^2[/itex] where the [itex]m[/itex] refers to the (oft deprecated) "relativistic mass":

[tex]m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex]

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

[tex]m_0 = m \sqrt{1 - \frac{v^2}{c^2}}[/tex]

and the latter factor goes to zero when [itex]v=c[/itex].

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

[tex]E^2=(m_0 c^2)^2 + (p c)^2[/tex]

and

[tex]p = m v[/tex] .

this [itex]E[/itex] is the total energy, the "rest energy" [itex]E_0=m_0 c^2[/itex] (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).

[tex]E = m c^2 = E_0 + T[/tex]

the kinetic energy

[tex]T = E - E_0[/tex]

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

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.
 
Last edited:
"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.
 
Bill_K said:
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

[tex]\frac{M}{r} > \frac{c^2}{2 G}[/tex]

in its own reference frame.
 

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