This quantum world/Feynman route/Action

Action

Let's go back to the propagator

$⟨B|A⟩=\int 𝒟𝒞Z[𝒞:A\to B].$

For a free and stable particle we found that

$Z[𝒞]=e^{-(i/\hslash )m{c}^{2}s[𝒞]},s[𝒞]=\underset{𝒞}{\int }ds,$

where $ds=\sqrt{d{t}^2-(d{x}^2+d{y}^2+d{z}^2)/c^2}$ is the proper-time interval associated with the path element $d𝒞$. For the general case we found that the amplitude $Z(d𝒞)$ is a function of $t,x,y,z$ and $dt,dx,dy,dz$ or, equivalently, of the coordinates $t,x,y,z$, the components $cdt/ds,dx/ds,dy/ds,dz/ds$ of the 4-velocity, as well as $ds$. For a particle that is stable but not free, we obtain, by the same argument that led to the above amplitude,

$Z[𝒞]=e^{(i/\hslash )S[𝒞]},$

where we have introduced the functional $S[𝒞]=\underset{𝒞}{\int }dS$, which goes by the name action.

For a free and stable particle, $S[𝒞]$ is the proper time (or proper duration) $s[𝒞]=\underset{𝒞}{\int }ds$ multiplied by $-m{c}^2$, and the infinitesimal action $dS[d𝒞]$ is proportional to $ds$:

$S[𝒞]=-m{c}^{2}s[𝒞],dS[d𝒞]=-m{c}^{2}ds.$

Let's recap. We know all about the motion of a stable particle if we know how to calculate the probability $p(B|A)$ (in all circumstances). We know this if we know the amplitude $⟨B|A⟩$. We know the latter if we know the functional $Z[𝒞]$. And we know this functional if we know the infinitesimal action $dS(t,x,y,z,dt,dx,dy,dz)$ or $dS(t,𝐫,dt,d𝐫)$ (in all circumstances).

What do we know about $dS$?

The multiplicativity of successive propagators implies the additivity of actions associated with neighboring infinitesimal path segments $d{𝒞}_1$ and $d{𝒞}_2$. In other words,

$e^{(i/\hslash )dS(d{𝒞}_1+d{𝒞}_2)}=e^{(i/\hslash )dS(d{𝒞}_2)}{e}^{(i/\hslash )dS(d{𝒞}_1)}$

implies

$dS(d{𝒞}_1+d{𝒞}_2)=dS(d{𝒞}_1)+dS(d{𝒞}_2).$

It follows that the differential $dS$ is homogeneous (of degree 1) in the differentials $dt,d𝐫$:

$dS(t,𝐫,\lambda dt,\lambda d𝐫)=\lambda dS(t,𝐫,dt,d𝐫).$

This property of $dS$ makes it possible to think of the action $S[𝒞]$ as a (particle-specific) length associated with $𝒞$, and of $dS$ as defining a (particle-specific) spacetime geometry. By substituting $1/dt$ for $\lambda$ we get:

$dS(t,𝐫,𝐯)=\frac{dS}{dt}.$

Something is wrong, isn't it? Since the right-hand side is now a finite quantity, we shouldn't use the symbol $dS$ for the left-hand side. What we have actually found is that there is a function $L(t,𝐫,𝐯)$, which goes by the name Lagrange function, such that $dS=Ldt$.

Template:Noprint