This quantum world/Feynman route/Geodesic equations

Geodesic equations

Consider a spacetime path $𝒞$ from $A$ to $B.$ Let's change ("vary") it in such a way that every point $(t,𝐫)$ of $𝒞$ gets shifted by an infinitesimal amount to a corresponding point $(t+\delta t,𝐫+\delta 𝐫),$ except the end points, which are held fixed: $\delta t=0$ and $\delta 𝐫=0$ at both $A$ and $B.$

If $t\to t+\delta t,$ then $dt=t_2-t_1⟶t_2+\delta t_2-(t_1+\delta t_1)=(t_2-t_1)+(\delta t_2-\delta t_1)=dt+d\delta t.$

By the same token, $d𝐫\to d𝐫+d\delta 𝐫.$

In general, the change $𝒞\to {𝒞}^{\prime }$ will cause a corresponding change in the action: $S[𝒞]\to S[{𝒞}^{\prime }]\ne S[𝒞].$ If the action does not change (that is, if it is stationary at $𝒞$ ),

$\delta S=\underset{{𝒞}^{\prime }}{\int }dS-\underset{𝒞}{\int }dS=0,$

then $𝒞$ is a geodesic of the geometry defined by $dS.$ (A function $f(x)$ is stationary at those values of $x$ at which its value does not change if $x$ changes infinitesimally. By the same token we call a functional $S[𝒞]$ stationary if its value does not change if $𝒞$ changes infinitesimally.)

To obtain a handier way to characterize geodesics, we begin by expanding

$dS({𝒞}^{\prime })=dS(t+\delta t,𝐫+\delta 𝐫,dt+d\delta t,d𝐫+d\delta 𝐫)$

$=dS(t,𝐫,dt,d𝐫)+\frac{\partial dS}{\partial t}\delta t+\frac{\partial dS}{\partial 𝐫}\cdot \delta 𝐫+\frac{\partial dS}{\partial dt}d\delta t+\frac{\partial dS}{\partial d𝐫}\cdot d\delta 𝐫.$

This gives us

${(}^{*})\underset{{𝒞}^{\prime }}{\int }dS-\underset{𝒞}{\int }dS=\underset{𝒞}{\int }[\frac{\partial dS}{\partial t}\delta t+\frac{\partial dS}{\partial 𝐫}\cdot \delta 𝐫+\frac{\partial dS}{\partial dt}d\delta t+\frac{\partial dS}{\partial d𝐫}\cdot d\delta 𝐫].$

Next we use the product rule for derivatives,

$d\left(\frac{\partial dS}{\partial dt}\delta t\right)=\left(d\frac{\partial dS}{\partial dt}\right)\delta t+\frac{\partial dS}{\partial dt}d\delta t,$

$d(\frac{\partial dS}{\partial d𝐫}\cdot \delta 𝐫)=\left(d\frac{\partial dS}{\partial d𝐫}\right)\cdot \delta 𝐫+\frac{\partial dS}{\partial d𝐫}\cdot d\delta 𝐫,$

to replace the last two terms of (*), which takes us to

$\delta S=\int [(\frac{\partial dS}{\partial t}-d\frac{\partial dS}{\partial dt})\delta t+(\frac{\partial dS}{\partial 𝐫}-d\frac{\partial dS}{\partial d𝐫})\cdot \delta 𝐫]+\int d(\frac{\partial dS}{\partial dt}\delta t+\frac{\partial dS}{\partial d𝐫}\cdot \delta 𝐫).$

The second integral vanishes because it is equal to the difference between the values of the expression in brackets at the end points $A$ and $B,$ where $\delta t=0$ and $\delta 𝐫=0.$ If $𝒞$ is a geodesic, then the first integral vanishes, too. In fact, in this case $\delta S=0$ must hold for all possible (infinitesimal) variations $\delta t$ and $\delta 𝐫,$ whence it follows that the integrand of the first integral vanishes. The bottom line is that the geodesics defined by $dS$ satisfy the geodesic equations

$\frac{\partial dS}{\partial t}=d\frac{\partial dS}{\partial dt},\frac{\partial dS}{\partial 𝐫}=d\frac{\partial dS}{\partial d𝐫}.$

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