Solving for Extrema of Proper Time Integral

[Kashmir]

The question is to extremize the proper time : $\begin{aligned}\int d\tau=\int ( dt^{2}-dx^{2}\\ -dy^{2} \\ -dz^2)^{1/2} \end{aligned}$

I've studied calculus of variations somewhat and a can solve a similar problem which I found in the mathematical methods book the author Mary Boas. The way she solves a similar problem is shown below, however I can't use that method for my problem of extremising the proper time above.

I would like to get some help to solve the above integral similar to the way shown below :

Given a problem to find $y$ that makes the integral stationary $I=\int_{x_1}^{x_2} F\left(x, y, y^{\prime}\right) d x$
where $F$ is a given function. The $y(x)$ which makes $I$ stationary is called an extremal whether $I$ is a maximum or minimum or neither. We consider a set of varied curves $Y(x)=y(x)+\epsilon \eta(x)$
just as before. Then we have
$I(\epsilon)=\int_{x_1}^{x_2} F\left(x, Y, Y^{\prime}\right) d x,$
and we want $(d / d \epsilon) I(\epsilon)=0$ when $\epsilon=0$. Remembering that $Y$and $Y^{\prime}$ are functions of $\epsilon$, and differentiating under the integral sign with respect to $\epsilon$, we get
$\frac{d I}{d \epsilon}=\int_{x_1}^{x_2}\left(\frac{\partial F}{\partial Y} \frac{d Y}{d \epsilon}+\frac{\partial F}{\partial Y^{\prime}} \frac{d Y^{\prime}}{d \epsilon}\right) d x .$

Substituting (2.1) and (2.5) into (2.11), we have
$\frac{d I}{d \epsilon}=\int_{x_1}^{x_2}\left[\frac{\partial F}{\partial Y} \eta(x)+\frac{\partial F}{\partial Y^{\prime}} \eta^{\prime}(x)\right] d x$
We want $d I / d \epsilon=0$ at $\epsilon=0$; recall that $\epsilon=0$ means $Y=y$ Then (2.12) gives
$\left(\frac{d I}{d \epsilon}\right)_{\epsilon=0}=\int_{x_1}^{x_2}\left[\frac{\partial F}{\partial y} \eta(x)+\frac{\partial F}{\partial y^{\prime}} \eta^{\prime}(x)\right] d x=0$

Assuming that $y^{\prime \prime}$ is continuous, we can integrate the second term by parts just as in the straight-line problem:
$\int_{x_1}^{x_2} \frac{\partial F}{\partial y^{\prime}} \eta^{\prime}(x) d x=\left.\frac{\partial F}{\partial y^{\prime}} \eta(x)\right|_{x_1} ^{x_2}-\int_{x_1}^{x_2} \frac{d}{d x}\left(\frac{\partial F}{\partial y^{\prime}}\right) \eta(x) d x .$
The integrated term is zero as before because $\eta(x)$ is zero at $x_1$ and $x_2$$\left(\frac{d I}{d \epsilon}\right)_{\epsilon=0}=\int_{x_1}^{x_2}\left[\frac{\partial F}{\partial y}-\frac{d}{d x} \frac{\partial F}{\partial y^{\prime}}\right] \eta(x) d x=0 .$
since $\eta(x)$ is arbitrary, we must have
$\frac{d}{d x} \frac{\partial F}{\partial y^{\prime}}-\frac{\partial F}{\partial y}=0 . \quad \text { Euler equation }$

 

[vanhees71]

You have to parametrize the worldline with an arbitrary parameter $\lambda$. The Lagrangian then reads
$$L=\sqrt{\dot{x}^{\mu} \dot{x}^{\nu} \eta_{\mu \nu}},$$
where $\dot{x}^{\mu}=\mathrm{d}_{\lambda} x^{\mu}$. The "canonical momenta" are
$$p_{\mu} = \partial_{\dot{x}^{\mu}} L = \frac{\dot{x}^{\mu}}{\sqrt{\dot{x}^{\rho} \dot{x}^{\sigma} \eta_{\rho \sigma}}}=\mathrm{d}_{\tau} x^{\mu}.$$
Then the Euler-Lagrange equations say
$$\dot{p}_{\mu} = 0.$$
Now
$$\dot{p}_{\mu} = (\mathrm{d}_{\lambda} \tau) \mathrm{d}_{\tau} p_{\mu}=0 \; \Rightarrow \; \mathrm{d}_{\tau} p_{\mu}=0 \; \Rightarrow \; \mathrm{d}_{\tau}^2 x^{\mu}=0 \; \Rightarrow \; x^{\mu}=p_{0}^{\mu} \tau + x_0^{\mu}$$
with $p_0^{\mu}=\text{const}$ and $x_0^{\mu}=\text{const}$.

 

[Kashmir]

Can you please simplify it ? I can't understand the terminology.

 

[PeroK]

Can you please simplify it ? I can't understand the terminology.

There's a proof of the Euler-Lagrange equations in the more general case in Neuenschwander's book Emmy Noether's Wonderful Theorem. We have a functional:
$$J = \int_a^b L(t, x^{\mu}, \dot x^{\mu}) \ dt$$The $\{x^{\mu}(t)\}$ that make $J$ extremal satisfy:
$$\frac{\partial L}{\partial x^{\mu}} = \frac{d}{dt}\bigg (\frac{\partial L}{\partial \dot x^{\mu}} \bigg )$$Hartle doesn't give a proof of this but obviously uses the result, e.g. in equations (5.56) and later.

Additionally for GR, of course, you have to take $t = x^0$ and use some dummy parameter $\sigma$. So, you have something like:
$$\tau_{ab} = \int_a^b d\tau = \int_a^b\big [-g_{\alpha \beta}dx^{\alpha}dx^{\beta} \big]^{\frac 1 2}$$$$= \int_0^1\big [-g_{\alpha \beta}\frac{dx^{\alpha}}{d\sigma}\frac{dx^{\beta}}{d\sigma} \big]^{\frac 1 2} \ d\sigma$$Where, without loss of generality, I've assumed $\sigma \in [0,1]$.

Note that we have $$L = \frac{d\tau}{d\sigma} = \big [-g_{\alpha \beta}\frac{dx^{\alpha}}{d\sigma}\frac{dx^{\beta}}{d\sigma} \big]^{\frac 1 2}$$And, if you have $g_{\alpha \beta} = \eta_{\alpha \beta}$, then the Euler-Lagrange equations yield:$$\frac{d^2x^{\mu}}{d\tau^2} = 0$$which is equation (5.62) in my edition of Hartle.

 

[PeroK]

PS The first few chapters of Neuenschwander's book (or equivalent) are almost essential prerequisites for GR (and, in fact, most advanced modern physics). The basic calculus of variations in Boas may leave too much of a gap.