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The stress energy tensor is related to the matter Lagrangian:

$$ T_{\mu \nu} = - 2 \frac{\partial (L_M \sqrt{-g})}{\partial g^{\mu \nu}} \frac{1}{\sqrt{-g}}.$$

Now, the stress energy tensor of a point particle tensor is given by:

$$ T^{a b} = mv^a(t)v^b(t) \delta(x-x_p(t)) .$$

So to find the matter Lagrangian:

$$ L_M \sqrt{-g} = -\frac{1}{2}\int mv^a(t)v^b(t) \delta(x-x_p(t)) \sqrt{-g} dg_{a b}. $$

How does one integrate the RHS and proceed to take limits and get the usual Newtonian Lagrangian $$ L = \int\frac{mv^2}{2}dt.$$ (feel free to include what happens to the constant of integration)?

Qmechanic
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More Anonymous
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  • Hint: If $f=f(x_1,x_2,\cdots)$ is some unknown function such that only $\partial f/\partial x_1$ is specified, then $f\neq \int (\partial f/\partial x_1)dx_1$, rather $f=\int\sum_i(\partial f/\partial x_i)dx_i$. Your $f=L_M=L_M(g,x,\dot{x})$ – KP99 Feb 19 '23 at 20:11

1 Answers1

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  1. With the sign convention $(−,+,+,+)$ the Lagrangian for a relativistic point particle is $$\begin{align}L~:=~& -mc\sqrt{-\dot{x}^2}, \cr \dot{x}^2~:=~&g_{\mu\nu}(x)~ \dot{x}^{\mu}\dot{x}^{\nu}~<~0, \end{align} \tag{A}$$ where dot means differentiation wrt. the world-line parameter $\tau$. Here the action is $S=\int \! d\tau~ L $.

  2. In Minkowski spacetime with the static gauge $x^0=c\tau$ it has the non-relativistic limit $$L=\frac{m}{2}\dot{\bf x}^2 -mc^2+{\cal O}(c^{-2}).\tag{B}$$

  3. The Hilbert/metric SEM tensor is defined as $$ T^{\mu\nu}(y)~:=~ \frac{2}{\sqrt{|g(y)|}}\frac{\delta S}{\delta g_{\mu\nu}(y)}~=~\frac{mc}{\sqrt{|g(y)|}} \int \!d\tau \frac{\dot{x}^{\mu}\dot{x}^{\nu}}{\sqrt{-\dot{x}^2}}\delta^4(y\!-\!x(\tau )) .\tag{C} $$ Eq. (C) can be further simplified by taking the world-line parameter $\tau$ to be proper time.

Qmechanic
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