Mathematical Proof that Photons have No Rest Mass

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SUMMARY

This discussion centers on the mathematical proof that photons possess zero rest mass, utilizing key equations such as the Photon Energy Equation (E=hc/λ), the Mass-Energy Equivalence with Momentum Equation (E²=(pc+mc²)²), and the Momentum of a Photon Equation (p=h/λ). The derivation presented leads to the conclusion that m(photon) = 0, reinforcing the established understanding in physics. However, some participants argue that the proof relies on circular reasoning, as the momentum equation assumes mass independence, thus questioning the validity of the derivation.

PREREQUISITES
  • Understanding of the Photon Energy Equation (E=hc/λ)
  • Familiarity with Mass-Energy Equivalence (E²=(pc+mc²)²)
  • Knowledge of the Momentum of a Photon Equation (p=h/λ)
  • Basic principles of mathematical proof and logical reasoning
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  • Study the implications of the Mass-Energy Equivalence in modern physics
  • Explore the derivation of the Momentum of a Photon Equation
  • Investigate the concept of rest mass in quantum mechanics
  • Review critiques of mathematical proofs in physics, focusing on circular reasoning
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Physicists, students of theoretical physics, and anyone interested in the fundamental properties of light and mass in quantum mechanics.

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Using these equations I am about to prove that photons have a rest mass of zero (mathematically)
________________________________________________________________________________________
E=hc/λ Photon Energy Equation

E2=(pc+mc2)2 Mass-Energy Equivalence with Momentum Equation

p=h/λ Momentum of a Photon Equation

________________________________________________________________________________________

Set the First 2 equations equal to each other...
E=hc/λ ------------------------------------------------> E2=(pc+mc2)2(hc/λ)2=(pc+mc2)2

h2c22=p2c2+m2c4

{h2c22}/c2={p2c2+m2c4}/c2

h22= (p2c2/c2)+(m2c4/c2)

h22=(p2)+(m2c2)

(h/λ=p+mc)2

[mc=(h/λ)-p]2

[mc(1/c)=(h/λ)(1/c)-p(1/c)]2

[m=(h/λc)-(p/c)]2

(Enter in the Momentum of a Photon Equation)

{m=(h/λc)-[(h/λ)/c]}2

[m=(h/λc)-(h/λc)]2

(m=0)2

m=0

m(photon) = 0

Is my math correct?
If so, is this legible?
And if so again, has this been proved yet and I am just completely unaware that it has?
I also made a video on YouTube about this if you want to check it out...

 
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(hc/λ)^2=(pc+mc^2)^2

h^2c^2/λ^2=p^2c^2+m^2c^4

Second line does not follow from first, (right side is wrong).
 
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mathman said:
Second line does not follow from first, (right side is wrong).

Thanks. I didn't see that until you pointed it out. So I just did the math again (the right way) and it comes out to be...

0 = m2c2

Could this also prove that the the mass of a photon is zero since we already have a set velocity for the speed of light (2.99x108 m/s)?
 
What you have done is simply proving a circular argument. For example

"p=h/λ Momentum of a Photon Equation"

is ALREADY an assumption that the momentum does NOT depend on ANY mass. In other words, you didn't show that this is true, but rather you adopt this form for the photon.

What you have done is simply doing a mathematical derivation from a known set of criteria, rather than showing that those starting criteria or description is correct. This is not a mathematical proof, but rather a logical consequence.

Zz.
 
ZapperZ said:
What you have done is simply proving a circular argument. For example

"p=h/λ Momentum of a Photon Equation"

is ALREADY an assumption that the momentum does NOT depend on ANY mass. In other words, you didn't show that this is true, but rather you adopt this form for the photon.

What you have done is simply doing a mathematical derivation from a known set of criteria, rather than showing that those starting criteria or description is correct. This is not a mathematical proof, but rather a logical consequence.

Zz.

So in simple terms, are you saying that p=h/λ is an equation DERIVED from the other two equations (and possibly others) ?
 

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