How to derive equation 27.4 in Dirac's "General Theory of Relativity" book? I've been having trouble following Dirac's logic in deriving equation 27.4 in his general relativity book. If $p^\mu$ is some matter current 4-vector, satisfying $\partial_\mu p^\mu = 0$, Dirac says that if the positions of the matter particles are displaced $x^\mu \to x^\mu + b^\mu$, that
$$
\delta p^\mu = \partial_\nu (p^\nu b^\mu - p^\mu b^\nu).
$$
Does anyone know how to derive this equation? Why isn't it just $\delta p^\mu = b^\nu \partial_\nu p^\mu$?
 A: If we have a single point particle of mass $m$ tracing out a path in spacetime $x(\lambda)$,
\begin{equation}
    p^\mu(x) = m \int d \lambda \; \frac{d x^\mu(\lambda)}{d \lambda} \delta^4( x - x(\lambda) ).
\end{equation}
Let us quickly check that $\partial_\mu p^\mu = 0$. Note that, in 1D
\begin{equation}
    \partial_\lambda \delta\left(x - f(\lambda) \right) = - \frac{df}{d \lambda} \partial_x \delta \left(x - f(\lambda) \right)
\end{equation}
so that
\begin{align}
    \partial_\mu p^\mu(x) &= m \int d \lambda \; \frac{d x^\mu(\lambda)}{d \lambda} \frac{-1}{d x^\mu / d \lambda} \partial_\lambda \delta^4\left( x - x(\lambda) \right) \\
    &= m \int d \lambda (-4) \partial_\lambda \delta^4\left( x - x(\lambda) \right) \\
    &= 0
\end{align}
where in the last line we integrated by parts.
Let's now figure out how the current $p^\mu$ changes under an infinitesimal transformation $\xi^\mu(x)$. If the trajectories of the particles change by
\begin{equation}
    \delta x^\mu(\lambda) = - \xi^\mu(x(\lambda))
\end{equation}
then

\begin{align}
    \delta p^\mu &= \delta \left( m \int d \lambda \frac{dx^\mu(\lambda)}{d \lambda} \delta^4(x - x(\lambda) )\right) \\
    &= -m \int d \lambda \frac{d\xi(x(\lambda))^\mu}{d \lambda} \delta^4(x - x(\lambda) ) + m \int d \lambda \frac{dx^\mu(\lambda)}{d \lambda} \xi^\nu(x(\lambda)) \partial_\nu \delta^4(x - x(\lambda) ) \\
    &= -m \int d \lambda \frac{d x(\lambda)^\nu}{d \lambda} \big(\partial_\nu \xi^\mu (x(\lambda) )\big) \delta^4(x - x(\lambda) ) + m \partial_\nu \; \int d \lambda \frac{dx^\mu(\lambda)}{d \lambda} \xi^\nu(x(\lambda)) \delta^4(x - x(\lambda) ) \\
    &= -m \int d \lambda \frac{d x(\lambda)^\nu}{d \lambda} (\partial_\nu \xi^\mu (x )) \delta^4(x - x(\lambda) ) + m \; \partial_\nu \int d \lambda \frac{dx^\mu(\lambda)}{d \lambda} \xi^\nu(x) \delta^4(x - x(\lambda) ) \\
    &= - m \partial_\nu \xi^\mu (x ) \int d \lambda \frac{d x(\lambda)^\nu}{d \lambda} \delta^4(x - x(\lambda) ) + m \; \partial_\nu \left( \xi^\nu(x) \int d \lambda \frac{dx^\mu(\lambda)}{d \lambda}  \delta^4(x - x(\lambda) ) \right) \\
    &= - (\partial_\nu \xi^\mu) p^\nu + \partial_\nu (\xi^\nu p^\mu) \\
    &= \partial_\nu ( - \xi^\mu p^\nu + \xi^\nu p^\mu ) \label{deltaj}
\end{align}

where in the last line we used $\partial_\nu p^\nu = 0$.
Note that $\partial_\nu ( - \xi^\mu p^\nu + \xi^\nu p^\mu )$ is the only combination of $\xi$ and $p$ that satisfies $\partial_\mu \delta p^\mu = 0$ and contains the term $\xi^\nu \partial_\nu p^\mu$. This observation may help you remember the formula in a pinch.
(By the way, just so we're clear, all of the work above holds in a completely general spacetime metric $g$. In order to have manifestly tensorial equations, we could write
$$
p^\mu = \sqrt{-g} P^\mu
$$
at which point we see
$$
0 = \partial_\mu p^\mu = \partial_\mu (\sqrt{-g} P^\mu) = \frac{1}{\sqrt{-g}} \nabla_\mu P^\mu = 0
$$
for instance.)
