Light Momentum Confusion: Proving with Math/Eqns

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Discussion Overview

The discussion revolves around the mathematical proof of light's momentum, particularly in the context of relativity and quantum mechanics. Participants explore various equations and concepts related to momentum, especially for massless particles like photons.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions how to prove light has momentum using the equation E2=m2c4+p2c2, noting that substituting values for light leads to an undefined result.
  • Another participant introduces the quantum mechanical expression for photon momentum, p=ħk, and points out that the original equation is applicable only to massive particles.
  • A different participant explains that in relativity, momentum is derived from four-momentum rather than the classical mass-velocity product, emphasizing that for photons, the four-momentum must be a null vector.
  • It is noted that the equation p=(mv)/sqrt(1-v^2/c^2) is not valid for light, and that if light carries energy, it must also carry momentum expressed as p=E/c.
  • One participant expresses skepticism about the complexity of the four-momentum explanation for the original poster, suggesting it may be too advanced for the current discussion level.

Areas of Agreement / Disagreement

Participants do not reach a consensus on how to mathematically prove light has momentum, with various competing views and approaches presented. The discussion remains unresolved regarding the best method to demonstrate this concept.

Contextual Notes

Participants highlight limitations in applying certain equations to massless particles and the need for empirical validation in physics, indicating that theoretical predictions must be tested experimentally.

Irfan Nafi
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In E2=m2c4+p2c2
how do you prove light has momentum? I've tried but my answer comes to be undefined. Light has m=0, so you're left with p2c2. p=(mv)/sqrt(1-v^2/c^2) When substituting values of light you get that p=0/0. How do you prove light has momentum using math and equations?
 
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The expression for momentum for photons in QM is ##p=\hbar k##. Where ##k## is the wave number (##k=\frac{2\pi}{\lambda}##) and ##\hbar## is the Planck constant divided by ##2\pi## . Or equivalently, ##p=\frac{h}{\lambda}## . Your equation for ##p## is for massive particles with non zero rest mass.
Edit: Also note that in your expression for p if you put v=c (photon) then you get a division by zero! It is good to check for these stuff when you are doing calculations.
Edit 2: Sorry I thought that this was posted in QM section. Apologies.
 
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In relativity, p is not necessarily obtained by multiplying mass by velocity.

For a photon, instead we work with the four-momentum, the first component ##p^0## of which, in any reference frame, is the energy of the photon in that frame. The other three components ##p^1,p^2,p^3## are momenta in the directions of the three spatial axes of the frame. We want the four-momentum to be a null vector (vector of zero magnitude), because light travels along null geodesics. So that means that not all the other three components of the four momentum can be zero - since the first component (Energy) ##p^0## is nonzero. If we rotate the frame so that the x spatial axis points in the direction of the photon's travel, we know that the other two momentum components ##p^2,p^3## must be zero, and we conclude that ##p^1=p^0## where ##p^0=E=h\nu## is the photon's energy.
 
Irfan Nafi said:
p=(mv)/sqrt(1-v^2/c^2)
This is a relation that only holds fot massive particles. You cannot use it for light.

Irfan Nafi said:
I've tried but my answer comes to be undefined. Light has m=0, so you're left with p2c2
Yes, and so you are done. If light carries energy, then it carries momentum p=E/c.

Irfan Nafi said:
How do you prove light has momentum using math and equations?

You don't. This is not how physics works, it is an empirical science. Your theory makes a prediction and you test it in the lab.
 
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andrewkirk said:
instead we work with the four-momentum
I strongly suspect that this will pass over the head of the OP in a B level thread ...
 

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