Photon Momentum
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Courtney McHargue 1I
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- Joined: Fri Sep 28, 2018 12:17 am
Photon Momentum
In today's lecture we talked about De Broglie's Wave Equation, I don't understand how a photon can have momentum when it doesn't have a mass?
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MaanasO 1A
- Posts: 72
- Joined: Fri Sep 28, 2018 12:26 am
Re: Photon Momentum
Hi Courtney!
This is a question with a rather weird answer. Photons have mass... technically. If you remember from lecture, Professor Lavelle used a term: "rest mass." This is mass in the traditional sense, the way we're used to thinking about it. Photons do not have a rest mass, but they do have a nifty property called relativistic mass. This is just mass that's relative to its momentum (m = p/c). Yes, it's really weird. But that doesn't necessarily explain how photons have momentum.
This idea I just mentioned comes from the full version of E = mc^2. Turns out the full version of this equation is
E^2=(m0*c^2)^2 + (p*c)^2 where E = energy, m0 = rest mass, c = speed of light, and p = momentum.
For a photon, the rest mass is 0, meaning E^2 = (p*c)^2, or Energy = momentum * speed of light (just take the square root on both sides). So one way momentum can be calculated for a photon is, p = E/c. This particular equation demonstrates that if a photon has energy (which it does), then it will have momentum as a by-product.
Since we're dealing with the energy of a photon, we can use a formula we've seen before (E = hv). Additionally, we known that λv = c, which means v = c/λ.
Pulling a substitution, we get that E = hc/λ. This is all stuff we've done in class already. So using the fact that E = pc, we can create an equation as follows:
pc = hc/λ
If you eliminate c from both sides of the equation, we get a familiar formula:
p = h/λ (De Broglie's Wavelength!)
So this is how a photon can have momentum without having a rest mass, and incidentally how De Broglie's equation can be derived.
If you'd like, I read through a couple articles that explain this. If what I said didn't make much sense, I hope these posts might make things a little clearer.
https://physics.stackexchange.com/quest ... e-momentum
https://www.quora.com/Do-photons-have-momentum
Hope that helps!
This is a question with a rather weird answer. Photons have mass... technically. If you remember from lecture, Professor Lavelle used a term: "rest mass." This is mass in the traditional sense, the way we're used to thinking about it. Photons do not have a rest mass, but they do have a nifty property called relativistic mass. This is just mass that's relative to its momentum (m = p/c). Yes, it's really weird. But that doesn't necessarily explain how photons have momentum.
This idea I just mentioned comes from the full version of E = mc^2. Turns out the full version of this equation is
E^2=(m0*c^2)^2 + (p*c)^2 where E = energy, m0 = rest mass, c = speed of light, and p = momentum.
For a photon, the rest mass is 0, meaning E^2 = (p*c)^2, or Energy = momentum * speed of light (just take the square root on both sides). So one way momentum can be calculated for a photon is, p = E/c. This particular equation demonstrates that if a photon has energy (which it does), then it will have momentum as a by-product.
Since we're dealing with the energy of a photon, we can use a formula we've seen before (E = hv). Additionally, we known that λv = c, which means v = c/λ.
Pulling a substitution, we get that E = hc/λ. This is all stuff we've done in class already. So using the fact that E = pc, we can create an equation as follows:
pc = hc/λ
If you eliminate c from both sides of the equation, we get a familiar formula:
p = h/λ (De Broglie's Wavelength!)
So this is how a photon can have momentum without having a rest mass, and incidentally how De Broglie's equation can be derived.
If you'd like, I read through a couple articles that explain this. If what I said didn't make much sense, I hope these posts might make things a little clearer.
https://physics.stackexchange.com/quest ... e-momentum
https://www.quora.com/Do-photons-have-momentum
Hope that helps!
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