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

AI Thread Summary
The discussion centers on the momentum of photons, which have no mass, leading to the equation p=mv=0. Instead, momentum for photons is expressed as p=h/lambda, derived from the energy-momentum relation p=E/c. The de Broglie relation connects energy and momentum, showing consistency with experimental evidence from diffraction patterns of x-rays and electrons. The conversation emphasizes the need for a deeper understanding of electromagnetic momentum through Maxwell's equations and the Poynting vector. Overall, the thread highlights the complexities of proving photon momentum experimentally while encouraging further exploration of the topic.
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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