How is Energy Defined for a Massless Particle?

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SUMMARY

The discussion focuses on the energy definition for massless particles, specifically demonstrating that if mass (m) equals zero, then energy (E) is equal to momentum (p) multiplied by the speed of light (c), expressed as E=pc. The equation E^2 = (pc)^2 + (mc^2)^2 simplifies to E=pc when m=0, confirming that massless particles possess energy solely through momentum. The confusion arises from the relationship between momentum (p) and the Lorentz factor (γ), where it is clarified that p = γmu does not apply to massless particles, as γ approaches infinity when mass is zero.

PREREQUISITES
  • Understanding of relativistic energy-momentum relation (E^2 = (pc)^2 + (mc^2)^2)
  • Familiarity with the concept of massless particles in physics
  • Knowledge of the Lorentz factor (γ) and its implications
  • Basic principles of special relativity
NEXT STEPS
  • Study the implications of massless particles in quantum mechanics
  • Learn about the properties of photons and their energy-momentum relationship
  • Explore the concept of the Lorentz transformation in detail
  • Investigate the role of the speed of light (c) in relativistic physics
USEFUL FOR

Students and educators in physics, particularly those studying special relativity, quantum mechanics, or the behavior of massless particles like photons.

khkwang
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Homework Statement



Show that if mass = 0, then E=pc and u=c.

Homework Equations



E^2 = (pc)^2 + (mc^2)^2
B = u/c = pc/E

The Attempt at a Solution



I understand that if m=0, then E^2=(pc)^2 => E=pc.

But isn't p = Ymu? then:
E^2 = (Ymuc)^2 + (mc^2)^2

Plugging in m=0 sets E=0...

It seems strange to me that we can remove (mc^2)^2 from the equation due to m=0, but we can leave (Ymuc)^2 in.

I'm thinking that this definition of p is not the same as Ymu?

EDIT: wait... Y would equal infinity. Which multiplied with 0 makes ? I'm just plain confused now.
 
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p = γmu does not apply to massless particles.
 
Ahhh simple as that lol, thanks.
 

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