A photon has rest $m=0$, but it is never at rest, so it has no rest mass, then the relativistic mass is according to me non-zero because:

  1. $E = \text{work}=m \times \text{acceleration} \times \text{displacement}$, if $m=0$ then $E=0$, which it is not
  2. Light is trapped in a black hole
  3. In the de-Broglie equation, $\lambda=h / (mc)$, if $m=0$, the expression is undefined, alternatively I can find the mass of a photon applying the wavelenghth of light in lamda
  4. Light has momentum and energy, which is I think impossible with a zero mass.

I dont have a very vast knowledge in physics, please tell me where in each of this 4 cases I am wrong.


1 Answer 1

  1. $E = m \times \text{acceleration} \times \text{displacement}$ is not the whole story even for normal matter once you get into special relativity. The full formula is $$E^2 = p^2 c^2 + m^2 c^4 \tag{1}$$ where $m$ is the rest mass, $p$ is the momentum, and $c$ is the speed of light. Photons have $m=0$ but $p \neq 0$.

  2. I'm not sure why this matters. In general relativity we have bending of space-time so even though light travels on "null paths"$^{[a]}$ when those paths themselves are bent the light goes in a bent trajectory.

  3. The deBroglie equation you've written is a semi-classical equation only applicable to massive particles. It't not even really correct for massive particles though, so don't take it too seriously.

  4. Momentum is not just $p = m v$. One way to think about it is to invert Eq. (1) getting $$pc = \sqrt{E^2 - m^2 c^4}\,. \tag{2}$$ Therefore a photon with energy $E = \hbar \omega$ has momentum $p = \hbar \omega / c$. You can regard the energy of a photon $E = \hbar \omega$ as the fundamental starting point, and then Eq. (2) as another fundamental relation coming from the structure of special relativity. The fact that photons have $m=0$ really means that it is possible to create one with arbitrarily low amounts of energy. With massive particles like electrons creating one with zero momentum costs energy $E = m c^2$ as per Eq. (1). In other words, to create a stationary electron out of the vacuum costs $m c^2$ in energy. For a photon, you can create one with any amount of energy $E=\hbar \omega$ as long as the frequency $\omega$ is low enough. This idea is illustrated in the figure where we show the energy required to create a photon or a particle of mass $m=2$ with specified momentum $p$.

enter image description here

$[a]$: This is just jargon here but it refers to the fact that light moves on special "as fast as possible" paths.

  • $\begingroup$ DanielSank: "[...] the fact that light moves special "as fast as possible" paths." -- This is apparently just jargon, relating to the fact that, in the context of RT, by "light" we're referring to the signal front. Also, to say that a signal front had been "moving along a path" appears to be an approximation or idealization of a large number of signals having been exchanged coherently. $\endgroup$
    – user12262
    Sep 6, 2015 at 8:41
  • $\begingroup$ @user12262 I am very puzzled by your comment. What does "RT" mean? what does coherence have to do with any of this? $\endgroup$
    – DanielSank
    Sep 6, 2015 at 8:43
  • $\begingroup$ DanielSank: "What does "RT" mean?" -- Oh, I mean that as abbreviation of "relativity theory"; i.e. including the special theory of relativity (as the OP question is tagged) as well as general theory of relativity (to which the OP point (2.) was referring, and accordingly point (2.) of your answer, together with its footnote [a].) "what does coherence have to do with any of this?" -- How else do you call distinct elements of one "path", if not "coherent"?? $\endgroup$
    – user12262
    Sep 6, 2015 at 8:56
  • $\begingroup$ @ danielSanki have got the rest of it but h/2pi has units joule-second,and joule itself contains kgnewtonmetre. $\endgroup$ Sep 6, 2015 at 10:47
  • $\begingroup$ @ManasJoyDogra So, what? $\endgroup$
    – DanielSank
    Oct 1, 2015 at 6:19

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