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

Question

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.

Qmechanic · Accepted Answer · 2019-02-09T11:14:10.477

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.

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$^1$ 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.

score 0 · Answer 2 · answered Apr 13 '17 at 19:57

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.

But in special relativity we have $u^\mu$ for massless particles. In this case, is $p^\mu = u^\mu$? — Mikkel Rev, Apr 13 '17 at 20:19
No. There is no $u^\mu$ for massless particles in special relativity. There just isn't. If someone told you that there is, they were telling you a half-truth at best. There is a path, ${\forall s: x^\mu(s)},$ which everyone agrees on (subject to reparameterizations of $s$ and Poincaré transforms of the 4-vectors in the set, of course), but the tangent vectors to this path are all null vectors. $u^\mu(s)$ is defined as the tangent vector corresponding to $x^\mu(s)$ normalized so that $u^\mu~u_\mu = c^2$ (in the $(+ - - -)$ metric) and you simply cannot normalize null vectors this way. — CR Drost, Apr 13 '17 at 20:46

score 0 · Answer 3 · answered Apr 13 '17 at 20:11

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.)

So is $u^\mu = p^\mu$ or not? – Mikkel Rev Apr 13 '17 at 20:17 — Mikkel Rev, Apr 13 '17 at 20:17
$u^\mu$ cannot be defined for light-like worldlines – Christoph Apr 13 '17 at 20:48 — Christoph, Apr 13 '17 at 20:48

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

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