How can we justify that the boundary integral we get from the following could be ignored, when we want to find the equation of motion?
I consider the energy-momentum of a free particle in special relativity: $$\tag{1} T^{ab} = \int m \, u^a \, u^b \, \delta^4 (x - z) \, d\tau, $$ where $z \equiv z^a (\tau)$ are the spacetime cartesian coordinates of the particle and $\tau$ is its proper time. Of course, $u^a = \frac{d z^a}{d \tau}$ are the components of its four-velocity. I'm using units such that $c \equiv 1$ and metric signature $\eta = (1, -1, -1, -1)$. Then I want to calculate the four-divergence of (1). Integrating by parts give a boundary term : \begin{align} \partial_a \, T^{ab} &= \int m \, u^a \, u^b \, \frac{\partial}{\partial x^a} \, \delta^4 (x - z) \, d\tau \\[1ex] &\equiv - \int m \, u^b \, \frac{d z^a}{d \tau} \, \frac{\partial}{\partial z^a} \, \delta^4 (x - z) \, d\tau \\[1ex] &= - \int m \, \frac{d}{d \tau} \Bigl( u^b \, \delta^4 (x - z) \Bigr) d\tau + \int m \, \frac{d u^b}{d\tau} \, \delta^4 (x - z) \, d\tau \\[1ex] &= - m \, \Bigl( u^b \, \delta^4 (x - z) \Bigr) \Big|_{\tau_1}^{\tau_2} + \int_{\tau_1}^{\tau_2} m \, \frac{d u^b}{d\tau} \, \delta^4 (x - z) \, d\tau. \tag{2} \end{align} The first term isn't zero. From the local conservation of energy-momentum; $\partial_a \, T^{ab} = 0$, we should get the usual free particle equation of motion: $\dot{u}^a = 0$, since the last integral should cancel for any straight path in spacetime. Usually, $\tau_1 = -\, \infty$ and $\tau_2 = +\, \infty$.
My problem is to justify the neglecting of the first part of (2). The Dirac delta $\delta^4$ is 0 everywhere except on the particle's worldline; $z^a(\tau_1)$ and $z^a(\tau_2)$, where the initial and final four-velocity $u^b$ could be anything. This part isn't 0.
How can I find the equation of motion from the local conservation of energy-momentum, i.e. from the equation $\partial_a \, T^{ab} = 0$?
