photons - General relativity: For massless particles, is the momentum and velocity 4-vectors equal?

I am following Carroll's GR book. He explain that it is convention to parameterize geodesics of photons by a parameter $\lambda$ such that

$$p^\mu ~=~ \frac{d x^{\mu}}{d \lambda}.\tag{3.62}$$ But this is the definition of 4-velocity for a massive particle in the case of $\lambda=\tau$ equal to proper time. My question then, is $$u^\mu = p^\mu$$ for all massless particles? I ask because it's well-known that proper time $\tau$ freezes for massless particles.

Answer

1. 
   
   

   How should we define momentum of a massless point particle? It seems most systematic to use the Lagrangian formalism. The Lagrangian of a massive or massless point particle is$^1$

   $$L~=~\frac{\dot{x}^2}{2e}-\frac{m^2e}{2}, \qquad \dot{x}^2~:=~g_{\mu\nu}(x)~ \dot{x}^{\mu}\dot{x}^{\nu}~<~0, \qquad \dot{x}^{\mu}~:=\frac{dx^{\mu}}{d\lambda} ,\tag{1}$$ cf. e.g. this, this and this Phys.SE posts. We now restrict to the massless case $$m~=~0. \tag{2}$$
   
   


2. 
   

   In eq. (1) $\lambda$ is a worldline (WL) parameter, and $e>0$ is a WL einbein field introduced to make the action

   $$S[x,e]~=~ \int\! \mathrm{d}\lambda ~L\tag{3}$$ gauge invariant under WL reparamerizations $$\lambda\longrightarrow \lambda^{\prime}~=~f(\lambda).\tag{4}$$ In more detail the WL einbein field $e$ transforms as a WL co-vector/one-form, $$e~\mathrm{d}\lambda~=~e^{\prime}~ \mathrm{d}\lambda^{\prime}.\tag{5}$$
   
   


3. 
   

   We can now address OP's question. The Lagrangian $4$-momentum$^2$ is defined in the standard way:

   $$p_{\mu}~:=~\frac{\partial L}{\partial \dot{x}^{\mu}}~\stackrel{(1)}{=}~ \frac{1}{e}g_{\mu\nu}(x)~\dot{x}^{\nu}.\tag{6}$$
   
   


4. 
   

   Carroll's eq. (3.62), which in our notation reads$^3$

   $$p^{\mu}~=~\dot{x}^{\mu}, \tag{7}$$ is eq. (6) in the gauge $e=1$, cf. e.g. this Phys.SE post.
   
   
   

--

$^1$ In this answer we put the speed of light $c=1$ to one and use the sign convention $(−,+,+,+)$.

$^2$ It is fun to check that in the massive case the Lagrangian momentum (6) becomes the standard $4$-momentum

$$p^{\mu}~\approx~\frac{m\dot{x}^{\mu}}{\sqrt{-\dot{x}^2}}\tag{8}$$ in the static gauge $\lambda=x^0$ in Minkowski space. [Here the $\approx$ symbol means equality modulo eom.]

$^3$ Note that the notion of $4$-velocity

$$u^{\mu}~:=~\frac{dx^{\mu}}{d\tau} \tag{9}$$ is not defined for massless particles. Here $\tau$ denotes proper time, which doesn't change for a massless particle. So OP's equation $p^{\mu}=u^{\mu}$ does not make sense if $u^{\mu}$ is supposed to be the conventional 4-velocity.