momentum of photons
2011-11-20 9:47 pm
圖片參考：http://imgcld.yimg.com/8/n/HA06578099/o/701111200039413873403450.jpg
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Let's just focus on option two. According to the suggested answer, option 2 is correct, but I totally don't understand how the equation (P=E/c) come out?
Beside, I think it is wrong according to my own explanation. And, I hope you guys can point out my mistakes. When f doubles, energy of photon doubles, kinetic energy doubles, then v becomes (sqrt2)v. (because K.E.=(1/2)Mv^2).
Then, new momentum should be sqrt2 times the original one, but not double. What's wrong with this explanation?
回答 (2)
2011-11-20 11:39 pm
✔ 最佳答案
The equation p = E/c, where E is the energy of a photon and c is the speed of light, is the definition of photon momentum.If we assume the classical formulation of momentum, i.e. mass x velocity, applies to a photon, then accordng to Einstein's famous mass-energy equation E = mc^2, we would have,
momentum of photon p = mc = mc^2/c = E/c
You can't apply the clssical equation of kinetic energy (1/2)mv^2 to a photon, simply because a photon has no rest mass. A photon only have energy and momentum.
2011-11-25 6:39 pm
光子動量=E/c係定義。
但唔係Out C既咩?
但唔係Out C既咩?
收錄日期: 2021-04-29 17:45:38
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20111120000051KK00394