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.
 A: *

*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
$$ m~=~0. \tag{2}$$

*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, 
$$ e~\mathrm{d}\lambda~=~e^{\prime}~ \mathrm{d}\lambda^{\prime}.\tag{5} $$

*We can now address OP's question. The Lagrangian $4$-momentum$^2$ is defined in the standard way:
$$ p_{\mu}~:=~\frac{\partial L}{\partial \dot{x}^{\mu}}~\stackrel{(1)}{=}~ \frac{1}{e}g_{\mu\nu}(x)~\dot{x}^{\nu}.\tag{6}$$

*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.
A: So a massless particle obeys $E = pc$ which is a special case of the relativistic equation $$(m c)^2 = p^\mu p_\mu = (E/c)^2 - |\vec p|^2,$$ when $m=0$. Indeed when we have a massive particle we have $\lambda = \tau/m$, but with a massless particle neither of these are defined: massleses particles do not have mass and do not experience time. Nevertheless you can imagine choosing someone who sees the particle moving at speed $c$ and using their local coordinate of time $t$, describing it as a 4-position path $x^\mu(t) = (ct, \vec x_0 +\vec v~t).$ Not everyone agrees on the $t$-derivative of this because not everyone agrees on $t$, but indeed there is a frame-dependent proportionality constant to take this to the 4-momentum and we can combine that with  $t$ into $\lambda$. So the problem is that there is no $u^\mu$ for massless particles, because there is no $\tau$ for massless particles, but we still have  $p^\mu$ and  $\lambda$ just fine.
A: The tangent vector (field) of a world line is not well-defined since the tangent vector is the derivative with respect to a parameter. Hence the tangent vector is parametrization-dependent. However, whether the tangent is timelike or lightlike (null) is invariant. For massive particles, $ds^2 > 0$, so the space time interval integrated along the curve can be used as a parameter. This is the the proper time, and we get the 4-velocity as $u^\mu = dx^\mu/d\tau$.
For massless particles, $ds^2 = 0$ and there is no notion of proper time. However, it can be shown that any curve can be given an affine parameter $\lambda$ such that the geodesic equation expressed in $\lambda$-derivatives takes a particular form. All such parameters are related affinely $$\lambda_1 = k\lambda_2 + m$$
where $k,m$ are constants. 
This is enough freedom to choose $\lambda$ such that Carroll's condition holds.
(The 4-momentum can be defined in a parametrization-invariant way.)
