# Derivation of Geodetic line by Einstein

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1. Feb 19, 2015

### sujoykroy

I was reading "The foundation of the general theory of relativity" by Albert Einstein. I faced some difficulty to understand his derivation of "geodetic line", stated in "The equation of the geodetic line. The motion of particle" headed section. I will try to describe my problem clearly and in short.

Einstein first stated that along geodetic line $\int ds$ is stationary, which means equation of geodetic should be,
$δ\int_P^{P'}ds=0$​
between two points P and P' of four dimensional continuum, where $ds$ means linear element. Solving this variation equation would give 4 new differential equations describing the geodetic line. This is clear and simple and no difficulty arises here.

However, then he replaced this equation with a parametric version,..

$\int_{λ1}^{λ2}δwdλ=0$
where, $w^2=g_{μν}\frac{dx_μ}{dλ}\frac{dx_ν}{dλ}$​

Einstein, described λ as follows,

Thus, lines, joining P and P', seems to become function of this new "family of surfaces" λ, as mentioned by him in this way,
Upto this point, things are not yet quite that difficult, although some elaboration of λ would be useful. But next he reinterpret δ as follows,
My question is what is this transition? Is it change of line-length? Moreover, how the δ sign will cross the integral ∫ sign, as it happened in the 2nd, replaced, equation, involving λs? although, i guess, knowing the meaning of δ and transition will answer the 2nd question.

I have stressed my brain neurons quite a bit to understand what this transition means and run through some pages of calculus of variations, but still it is hiding at large from me. An enlightenment from you will re-energize my brain cells.

2. Feb 19, 2015

### stevendaryl

Staff Emeritus
Okay, a parametrized path connecting points $P$ and $P'$ gives 4 functions of the parameter, $\lambda$: $x^0(\lambda), x^1(\lambda), x^2(\lambda), x^3(\lambda)$. The meaning of lambda is just that it is a real number indicating how far along the path. So lambda is just a real number that increases continuously from $P$ to $P'$. It doesn't actually matter, but for concreteness, let's assume that at $P$, lambda = 0, and at $P'$, lambda = 1.

Now to vary the path $x^\mu(\lambda)$, you replace $x^\mu(\lambda)$ by a shifted value:$x'^\mu(\lambda) = x^\mu(\lambda) + \delta x^\mu(\lambda)$. (Where $\mu$ runs over the four coordinate indices). $\delta x^\mu(\lambda)$ is just another function of $\lambda$, but since the shifted path has to start at P, and has to end at P', we have to have $\delta x^\mu(0) = \delta x^\mu(1) = 0$. Other than that, $\delta x^\mu$ is an arbitrary function, except that it is assumed to be small. (The shifted path is not much different from the original path).

Then using $x'^\mu$ instead of $x^\mu$, we can compute a shifted value for $w = \sqrt{g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda}}$. Call this shifted value $w'$. Then $\delta w = w' - w$ (only keeping the lowest-order terms in $\delta x^\mu$).

3. Feb 19, 2015

### sujoykroy

As far as my little brain progressed, i figured out as follows.
Since the equation $δ\int_P^{P'}ds=0$ is evaluated between two points of four coordinates continuum, to convert this into a well-known form of variation some parametrization was required and hence λ was introduced. Using λ, all those 4 coordinates are transformed into functions of λ. Much like, λ is acting like the independent variable. So, the former equation, first, transformed into this, -
$δ\int_{λ1}^{λ2}\frac{ds}{dλ}dλ=0$​
As i get to know from ancient book of Dionysius Lardner, (although one could have proved by himself), that, in variation equations interchanging of δ and ∫ does not alter the result. Hence, the last equation can surely be rewritten as
$\int_{λ1}^{λ2}δ(\frac{ds}{dλ})dλ=0$​

Lastly, Einstein did declared that
$ds^2=\sum\limits_{τσ}g_{στ}dx_{σ}dx_{τ}$​
or if following his notation system, it is
$ds^2=g_{στ}dx_{σ}dx_{τ}$​
.
From there we can put dλ into this equation and transform it into
$({\frac{ds}{dλ}})^2=g_{στ}\frac{dx_{σ}}{dλ}\frac{dx_{τ}}{dλ}$​
So, taking $w={\frac{ds}{dλ}}$, the final, previously troublesome, equation gets into the picture,
$\int_{λ1}^{λ2}δwdλ=0$​