0 = 0?

Reply Mon 20 May, 2019 05:50 pm
I used to hear some people said that photon is massless while has momentum. I was confused and surprising. I didn’t know why they said that.
Recently, I got their idea. It’s based on the equation below:
E² = p²C² + (m0C²)²
They assume m0 = 0 (photon is massless), then E = pC, or p = E/C (photon has momentum)
And according to Quantum, p = h/λ, then E = hC/λ = hγ
People can detect the energy of photon, so they can calculate the momentum of photon p.
It seems nice. But I think it carefully and found that it’s only a tricky mathematical game. My analysis is below:
Because the component p²C²and the component (m0C²)²are in equal position. If people can assume m0 = 0 while p ≠ 0, they can also assume p = 0 while m0 ≠ 0. That will be fair.
But a photon can not be both m0 = 0 and m0 ≠ 0 (or be both p ≠ 0 and p = 0 ), so the solution is m0 = 0, p = 0. It will be fair.
Then the equation E² = p²C² + (m0C²)² is actually talking 0 = 0. Or say, the frequency of the “photon” is zero (the wavelength is infinite). It does not exist.
Thank you.
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Reply Mon 20 May, 2019 06:59 pm
I think you are confusing m0 (i.e. multiplication) with m0 (i.e. subscript). The rest of what you wrote is nonsense.

Reply Mon 20 May, 2019 08:07 pm
Oh,,,m0 in the topic is subscript. I wrote fast.
That kind of reply such as just saying "you are nonsense","you are wrong",etc,without analysis are nonsense in discussion.
I think that nobody in a forum want such a reply.
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Reply Tue 21 May, 2019 05:01 pm
There are algebraic "proofs" that 0 = 1 that stump high school math students. They usually involve the algebraic equivalent of dividing by zero, or occasionally ignoring the constant term of an integral (in the calculus version).

I thought this was going to be one of those.
Reply Tue 21 May, 2019 07:27 pm
Dear,,,...math should not try to trick physics.
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