Can we experimentally prove the momentum of a photon without mass?

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SUMMARY

The discussion centers on the momentum of photons, which are massless particles. The momentum of a photon is defined by the equation p = E/c, where E represents the energy of the photon. The de Broglie relation, p = h/λ, is derived from the Planck relation and is consistent with the energy-momentum relationship for photons. Experimental evidence for these principles can be observed in the diffraction patterns of x-rays and electrons.

PREREQUISITES
  • Understanding of the energy-momentum relation in physics
  • Familiarity with Maxwell's equations and the Poynting vector
  • Knowledge of the de Broglie relation and Planck's equation
  • Basic concepts of electromagnetic radiation and its properties
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  • Study the derivation and applications of the Poynting vector in electromagnetic field theory
  • Explore experimental evidence for the de Broglie relation through diffraction patterns
  • Learn about the implications of massless particles in quantum mechanics
  • Investigate advanced topics in energy-momentum relations for various particle types
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Kehsibashok
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1.photon has no mass . so m=0.hence, p=mv=0.by doing some calculations , we can get that
p=h/lambda.we can prove p=mv experimentally.but how can we prove the second one experimentally?
 
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Kehsibashok said:
1.photon has no mass . so m=0.hence, p=mv=0.by doing some calculations , we can get that
p=h/lambda.we can prove p=mv experimentally.but how can we prove the second one experimentally?
p = mv is only useful for massive particles at relatively low speeds. For photons, you'll need the relativistic energy-momentum relation: Energy–momentum relation
 
There is more than one type of momentum - p=mv is the mechanical form.

Momentum for electromagnetic radiation is determined by the momentum that it can impart to a mechanical system. The derivation follows from energy transport properties of radiation derived from Maxwell's equations - look up "Poynting Vector". The electromagnetic radiation momentum is found to be p=E/c; this relation also holds for the photon.

The equation p=h/lambda is the de Broglie relation; it is "derived" from the Planck relation (E=h*f), then divide by c to get E/c=p=h*f/c=h/lambda. Of course this is not a derivation - it merely shows that the two are algebraicly consistent. The experimental proof of the de Broglie relation can be seen experimentally: x-rays and electrons both give diffraction patterns in accordance with the above.

For more detail see http://hyperphysics.phy-astr.gsu.edu/hbase/debrog.html
 
only rest mass of photon is zero

momentum of photon can be calculated by


p = E/c

where E is energy of photon
don't ask me any proof please because I'm an tenthee! studying for iit and was told this on a chemistry lecture and would study more on it guess in my research later
further proofs are encouraged and needed by me as my teacher said it would come later on:-p

and a kind request to UltrafastPED pls do no\umerically in separate lines or it's feels scrambled
 
"and a kind request to UltrafastPED pls do no\umerically in separate lines or it's feels scrambled"

I prefer inline formulas, especially when there are chains of implication. This makes for more concise, "unscrambled" chains of logic.

BTW, for the proof of p=E/c see any text on electromagnetic field theory (upper level undergraduate physics); you will arrive at the Poynting vector sometime in the second semester!
 

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