Photon: momentum without mass?

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Discussion Overview

The discussion revolves around the concept of photon momentum and energy, particularly addressing the implications of a massless particle possessing these properties. It explores theoretical frameworks, definitions, and the relationship between mass, momentum, and energy in the context of special relativity.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • Some participants assert that a photon, having zero mass, can still possess momentum and energy, referencing the relationship between these properties.
  • One participant emphasizes that the formula E=mc^2 is applicable only to objects at rest, suggesting that a more general formula, E^2 = m^2 c^4 + p^2 c^2, is relevant for massless particles like photons.
  • Another participant argues that the equation for momentum for photons is p=hv/c, indicating that traditional definitions of momentum (p=mv) do not apply to massless particles.
  • Several participants challenge the application of classical mechanics to photons, noting that the concept of "relativistic mass" is often misunderstood and is not commonly used in modern physics discussions.
  • There is a suggestion that formulas from nonrelativistic mechanics are inadequate for describing the behavior of light and fast-moving objects.

Areas of Agreement / Disagreement

Participants do not reach a consensus, as there are competing views regarding the application of classical mechanics to photons and the definitions of mass in this context.

Contextual Notes

The discussion highlights ambiguities in the term "mass" and the limitations of classical mechanics when applied to relativistic scenarios, particularly concerning light and massless particles.

Gabriele Pinna
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The mass of a photon is zero but why does it have a momentum and an energy (E=mc^2=hv) ?
 
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Gabriele Pinna said:
E=mc^2
This is only true for objects at rest, and photons are not at rest.
The more general formula is ##E^2 = m^2 c^4 + p^2 c^2##. There is nothing wrong with a massless particle having momentum and energy.
 
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Gabriele Pinna said:
The mass of a photon is zero but why does it have a momentum and an energy (E=mc^2=hv) ?

There's nothing in the laws of physics that prevent a particle from having zero mass and nonzero momentum and energy. Mass, momentum, and energy are properties, so they are related. That relationship allows for zero mass along with nonzero energy and momentum. In fact, the same relationship asserts that if the mass is zero the energy and momentum have to equal each other.
 
But momentum is equal to p=mv so if m=0→p=0
 
Gabriele Pinna said:
But momentum is equal to p=mv so if m=0→p=0

No. See mfb's post #2.

I suspect you are getting confused by the so-called "relativistic mass" which almost no physicists use nowadays, but nevertheless survives in introductory (especially popular-level) treatments of relativity, versus the "invariant mass / rest mass" which Orodruin is referring to.
 
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Gabriele Pinna said:
But momentum is equal to p=mv so if m=0→p=0

That only applies to objects with mass. For photons, the equation for momentum is p=hv/c, where h is Planck's constant, v is frequency, and c is the speed of light.
 
jtbell said:
the "invariant mass / rest mass" which Orodruin is referring to.
It sounds like something I would have said, but I have been silent in this thread so far. :wink:

There are Insights FAQs on both relativistic mass and on photons which are relevant. I suggest OP reads them.
 
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Drakkith said:
But momentum is equal to p=mv so if m=0→p=0
That only applies to objects with mass.
And even for those, it is just an approximation for slow speeds.

@Gabriele Pinna: Formulas from nonrelativistic mechanics are just an approximation, they are good for slow objects, but they do not work for fast objects (a large fraction of the speed of light) and they are completely meaningless for light.
 
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Gabriele Pinna said:
But momentum is equal to p=mv so if m=0→p=0

That used to be true in Newtonian mechanics, but it's not true in special relativity. See also the various comments about the ambiguities in the word "mass".
 
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Thanks
 

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