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 $\partial_a \, T^{ab} = 0$?


1 Answer 1


To avoid taking delicate limits, let's keep the interval $[\tau_1,\tau_2]$ finite. Then the initial and final boundary terms (2) are not zero and should not be ignored: They correspond to energy-momentum source/sink terms [in the energy-momentum continuity equation] because of the incoming/outgoing particle along the world-line.

Concerning the equivalence between energy-momentum conservation and the EOM of the particle, see e.g. this related Phys.SE post.

  • $\begingroup$ Then, how do we find the equation of motion from the equation $\partial_a \, T^{ab} = 0$ ? The local conservation of energy-momentum should give $0$! $\endgroup$
    – Cham
    Commented Feb 8, 2022 at 23:32
  • $\begingroup$ $\partial_a \, T^{ab}$ is not zero, but instead given by the source/sink terms. $\endgroup$
    – Qmechanic
    Commented Feb 8, 2022 at 23:39
  • $\begingroup$ Well, it should be! Local conservation of energy-momentum ask for it. $\endgroup$
    – Cham
    Commented Feb 8, 2022 at 23:40
  • $\begingroup$ To find the equation of motion, how can we justify that the boundary terms can be ignored? Most authors that are using this method doesn't say anything about why the boundary terms could be ignored! $\endgroup$
    – Cham
    Commented Feb 8, 2022 at 23:53
  • $\begingroup$ For a free particle, the start and end positions are at infinity: $z^a(\pm\, \, \infty)$. The Dirac deltas should give 0 there, or aren't well determined, I think. $\endgroup$
    – Cham
    Commented Feb 9, 2022 at 0:09

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