Suppose there are two $\gamma$ rays with frequencies $\nu_1$ and $\nu_2$ moving in opposite directions according to a reference frame $S$. I want to find the velocity of the center of mass of this system.

Since photons do not have mass, the center of mass is the frame in which the sums of momenta vanishes.

Let $S'$ be this reference frame. The total momentum in $S'$ is given by:

$$ p'= p_{1}'+p_{2}' = \frac{h}{c}(\nu_{1}'-\nu_{2}')=0 $$

Which implies $\nu_{1}'=\nu_{2}'$.

There frequencies in $S'$ are given by: $$ \nu'= \left(\frac{1+\beta}{1-\beta}\right)^{1/2}\nu $$

Therefore, the condition $\nu_{1}'=\nu_{2}'$ gives $\nu_{1}=\nu_{2}$. But $\nu_{1} \neq \nu_{2}$ because the photons can have different frequencies in $S$.

What has gone wrong in the reasoning?

  • $\begingroup$ You should flip the base of your last expression when calculating the frequency of the other photon. $\endgroup$ – Drake Marquis Nov 20 '14 at 14:17
  • $\begingroup$ @DrakeMarquis, the problem is still there. Because both terms cancel out. $\endgroup$ – Thiago Nov 20 '14 at 14:20
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    $\begingroup$ The center of momentum frame is one with zero total momentum. In Newtonian physics that is synonymous with the center of mass frame, but the former term is preferred in relativistic physics. The center of mass itself is a place. Did you want the location or the frame? $\endgroup$ – dmckee --- ex-moderator kitten Nov 20 '14 at 14:43
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    $\begingroup$ Think about the sign convention for the Doppler shift. The choice of + on top, - on bottom or vice versa depends on the relative motion of the source and observer. So. for boost along the line of the photons' flight one photon should get each convention---they don't change synchronously. $\endgroup$ – dmckee --- ex-moderator kitten Nov 20 '14 at 14:53

This frame is exist. You got wrong result because you ignored that this two photon move in the opposite direction. Set that the first photon move along the z axis and the second photon move against z-axis.$\omega_1$ and $\omega_2$ are the frequency of the first and the second photon correspondingly in the reference frame. In new frame shout be $k'_1=-k'_2$ Let's do the Lorenz transformation for the $k_z$ $$\gamma(\omega_1+\beta\omega_1)=-\gamma(-\omega_2+\beta\omega_2)$$ $$\gamma(\omega_1-\omega_2+\beta(\omega_1+\omega_2))=0$$ Thus we obtain that $\beta=\frac{\omega_2-\omega_1}{\omega_1+\omega_2}$ and $\gamma=\frac{\omega_1+\omega_2}{\sqrt{4\omega_1\omega_2}}$ After that one can check that $\omega'_1=\omega'_2$ $$\omega'_1=\gamma(\omega_1+\beta\omega_1)=\frac{\omega_1+\omega_2}{\sqrt{4\omega_1\omega_2}}(\omega_1+\frac{\omega_2-\omega_1}{\omega_1+\omega_2}\omega_1)=\sqrt{\omega_1\omega_2}$$ $$\omega'_2=\gamma(\omega_2-\beta\omega_2)=\frac{\omega_1+\omega_2}{\sqrt{4\omega_1\omega_2}}(\omega_2-\frac{\omega_2-\omega_1}{\omega_1+\omega_2}\omega_2)=\sqrt{\omega_1\omega_2}$$

  • $\begingroup$ Nice. Another way of looking at this is that the energy-momentum vector of the system as a whole is $(E,p)=(\omega_1+\omega_2,\omega_1-\omega_2)$, in units where $\hbar=1$. The boost needed in order to make this vector purely timelike is $v=p/E$. $\endgroup$ – user4552 Nov 20 '14 at 22:52

"Since photons do not have mass, the center of mass is the frame in which the sums of momenta vanishes" this is incorrect, and only valid in the reference frame where the two photons have the same frequency. To compute the center of mass (if it makes any sense or is useful at all in a relativistic setting), and assuming the two photons are localized, you can use the relativistic mass, $m_{rel}=E/c^2=h\nu/c^2=p/c$, to compute the center of mass in the classical way.


In fact, you do not have to find the frame $S'$ because the center of mass is independent of reference frame. It is $\sqrt{(h\nu_1+h\nu_2)^2-(h\nu_1-h\nu_2)^2}=2h\sqrt{\nu_1\nu_2}$.

  • $\begingroup$ Could you explain how you came up with this equation? $\endgroup$ – Thiago Nov 20 '14 at 14:22
  • $\begingroup$ The total energy is $h\nu_1+h\nu_2$ and the total momentum is $h\nu_1-h\nu_2$. Plug in the formula $m^2=E^2-p^2$ you get this expression. $\endgroup$ – Drake Marquis Nov 20 '14 at 14:26
  • $\begingroup$ The calculates the mass of the system (and demonstrates that a system composed of two massless particles can have a non-zero mass), but not the center of mass. $\endgroup$ – dmckee --- ex-moderator kitten Nov 20 '14 at 14:34
  • $\begingroup$ @dmckee The mass of the system is different from the center of mass? $\endgroup$ – Drake Marquis Nov 20 '14 at 14:38
  • $\begingroup$ Center of mass is a position (for solid system the place where you could push on it in any direction and impart no rotation), mass is a ... well ... masss. $\endgroup$ – dmckee --- ex-moderator kitten Nov 20 '14 at 14:41

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