Derivation of Geodesics Eq from EM Tensor of Point Particle

[sergiokapone]

The energy-momentum tensor of a free particle with mass $m$ moving along its worldline $x^\mu (\tau )$ is
\begin{equation}
T^{\mu\nu}(y^\sigma)=m\int d \tau \frac{\delta^{(4) }(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}.\tag{2}
\end{equation}
The covariant derivative of tensor gives
\begin{equation}
\nabla_{\mu} T^{\mu\nu} = \frac{1}{\sqrt{-g}} \frac{\partial \left( \sqrt{-g} T^{\mu\nu}\right) }{\partial y^{\mu}} + \Gamma^{\nu}_{\mu\lambda}T^{\mu\lambda}
\end{equation}

And a derivation of geodesics equation from energy-momentum tensor of point particle in GR $\nabla_{\mu} T^{\mu\nu} = 0$ looks like:

\begin{align}

0 &~~~~~~=~ \frac{1}{\sqrt{-g(y)}}\partial^{(y)}_{\mu} \left(\sqrt{-g(y)}T^{\mu\nu}(y)\right)
+\Gamma^{\nu}_{\mu\lambda}(y) T^{\mu\lambda}(y) \cr
&~~~~~~=~\frac{m}{\sqrt{-g(y)}} \int \!\mathrm{d}\tau ~\dot{x}^{\nu}\dot{x}^{\mu}\partial^{(y)}_{\mu}\delta^4(y\!-\!x(\tau )) ~+~\Gamma^{\nu}_{\mu\lambda}(y) T^{\mu\lambda}(y) \cr
&~~~~~~=~\underbrace{-\frac{m}{\sqrt{-g(y)}} \int \!\mathrm{d}\tau ~\dot{x}^{\nu} \frac{d}{d\tau}\delta^4(y\!-\!x(\tau ))}_\text{this term we integrating by parts}
~+~\Gamma^{\nu}_{\mu\lambda}(y) T^{\mu\lambda}(y) \cr
&\stackrel{\text{int. by parts}}{=}~\frac{m}{\sqrt{-g(y)}}\underbrace{ \dot{x}^{\nu}\delta^{(4)}(y−x(τ))}_\text{shoul be =0! Why?} + \cr
&~~~~~\frac{m}{\sqrt{-g(y)}} \int \!\mathrm{d}\tau\underbrace{\left[\ddot{x}^{\nu}+ \Gamma^{\nu}_{\mu\lambda}(x(\tau))\dot{x}^{\mu}\dot{x}^{\lambda} \right]}_\text{=0 geodesics equation}\delta^4(y\!-\!x(\tau )) .
\end{align}Can anyone explain where did the term $uv=\dot{x}^{\nu}\delta^{(4)}(y−x(τ))$ go after integration by parts $\int udv=uv−\int vdu$ in the last term (5)?

 

[Gaussian97]

Well, you can think that when evaluating the limits of the integral $\delta^{(4)}(y-x(\tau))$ will become 0 because it's basically 0 everywhere. For more rigorous ways you should study generalized functions and it's derivatives. You can look at the Gel'Fand and Shilov book "Generalized Functions Volume I".

 

[sergiokapone]

Well, you can think that when evaluating the limits of the integral $\delta^{(4)}(y-x(\tau))$ will become 0 because it's basically 0 everywhere. For more rigorous ways you should study generalized functions and it's derivatives. You can look at the Gel'Fand and Shilov book "Generalized Functions Volume I".

Can you write appropriate formulae from Gel'Fand and Shilov book's here?

 

[Gaussian97]

Well, is not a formula, is an entire book dedicated to generalized functions, but maybe you need the equation of Example 7 of Chapter 2:

$$\int \delta'(x-h)\varphi(x)\text{d}x=-\varphi'(h)$$

 

[sergiokapone]

Yes, this formula explain the transition from (4) to (6), but why then mentioned integration by parts?

 

[Gaussian97]

Maybe because the author knows that the lector may not be familiarized with derivatives of generalized functions and then uses the argument I said before of "integrating by parts and supposing that $\delta$ outside an integral is 0."

 

[sergiokapone]

Well, you can think that when evaluating the limits of the integral δ(4)(y−x(τ))δ(4)(y−x(τ))\delta^{(4)}(y-x(\tau)) will become 0 because it's basically 0 everywhere.

"integrating by parts and supposing that δδ\delta outside an integral is 0."

Then I conclude,
$\dot{x}\delta (x) = 0$ everywrehe, even in $x=0$. Does it right?

 

[Gaussian97]

Well, you have to be very careful because $\delta$ is not a usual function, but most of the times you can think of it as a function that is 0 everywhere and is not defined at $x=0$. Usually having a $\delta$ outside an integral is not a good idea.