# To prove the Lorentz invariance of density distribution functions for massless particles in phase space

One defines the density distribution function of a collection of $N$ particles in phase space as follows, $$f(\vec{x},\vec{p},t)=\sum_{i=1}^N\delta^{(3)}(\vec{x}-\vec{x}_i)\delta^{(3)}(\vec{p}-\vec{p}_i).$$ The claim is that the above distribution function is Lorentz invariant. To prove the claim one defines the following function, $$F(\vec{x},\vec{p},t)=\delta(p^2-m^2)\Theta(p^0)f(\vec{x},\vec{p},t).$$ If one is able to prove that $F$ is Lorentz invariant, then $f$ will also be Lorentz invariant. By manipulations one arrives at, $$F(\vec{x},\vec{p},t)=\frac{1}{2E_{\vec{p}}}\sum_{i=1}^N\delta^{(3)}(\vec{x}-\vec{x}_i)\delta^{(4)}(p-p_i).$$ As the delta function restricting the particle momenta on the shell has already converted the 3-dimensional delta function of momenta into a 4-dimensional delta functions, one only worries about the 3-delta function of position. To tackle this problem, integration over coordinate time is introduced, $$F(\vec{x},\vec{p},t)=\frac{1}{2E_{\vec{p}}}\sum_{i=1}^N\int dt_i\delta(t-t_i)\delta^{(3)}(\vec{x}-\vec{x}_i(t_i))\delta^{(4)}(p-p_i(t_i)).$$ Using the relation between proper time and coordinate time $d\tau_i=\frac{m}{E_{\vec{p}}}dt_i$, and suppressing the the index in proper time one arrives at the following equation, $$F(\vec{x},\vec{p},t)=\frac{1}{2m}\sum_{i=1}^N\int d\tau\delta^{(4)}(x-x_i(\tau))\delta^{(4)}(p-p_i(\tau)).$$ The above expression suggests $F$ is Lorentz invariant and therefore, so is $f$. But the above treatment only works for massive particles. My question is how to generalize this method to massless particles for which the mass and proper time both are inevitably zero?

For massless particles, the proper time is inevitably zero and thus cannot be used to parametrize the particle trajectories. One may choose another affine parameter, which by definition keeps the form of the geodesic equation unchanged, say $\lambda$, such that the momentum four-vector is redefined as follows, $$P^\mu\equiv\frac{dx^\mu}{d\lambda}.$$ Now, with the problem one can proceed as follows, $$F(\vec{x},\vec{p},t)=\frac{1}{2E_{\vec{p}}}\sum_{i=1}^N\int dt_i\delta(t-t_i)\delta^{(3)}(\vec{x}-\vec{x}_i(t_i))\delta^{(4)}(p-p_i(t_i))\\ =\frac{1}{2P^0}\sum_{i=1}^N\int d\lambda\left|\frac{dt_i}{d\lambda}\right|\delta(t-t_i(\lambda))\delta^{(3)}(\vec{x}-\vec{x}_i(\lambda))\delta^{(4)}(p-p_i(\lambda))\\ =\frac{1}{2P^0}\sum_{i=1}^N\int d\lambda P^0_i(\lambda)\delta^{(4)}(x-x_i(\lambda))\delta^{(4)}(p-p_i(\lambda)).$$ Here, I have used the following facts, 1) $E_{\vec{p}}=P^0$, 2) $P^0=\frac{dx^0}{d\lambda}=\frac{dt}{d\lambda}$.The four dimensional delta functions are Lorentz invariant; $\lambda$ is a dummy variable, change in which doesn't alter the value of the integral; and now, we have zeroth components of four momentum both in numerator and denominator, the Lorentz transformation factor will precisely cancel for them. Therefore, $F$ is Lorentz invariant.