This quantum world/Feynman route/Whence the classical story

Whence the classical story?

Imagine a small rectangle in spacetime with corners

$A=(0,0,0,0),B=(dt,0,0,0),C=(0,dx,0,0),D=(dt,dx,0,0).$

Let's calculate the electromagnetic contribution to the action of the path from $A$ to $D$ via $B$ for a unit charge ($q=1$) in natural units ( $c=1$ ):

$S_{ABD}=-V(dt/2,0,0,0)dt+A_x(dt,dx/2,0,0)dx$

$=-V(dt/2,0,0,0)dt+[A_x(0,dx/2,0,0)+\frac{\partial A_x}{\partial t}dt]dx.$

Next, the contribution to the action of the path from $A$ to $D$ via $C$:

$S_{ACD}=A_x(0,dx/2,0,0)dx-V(dt/2,dx,0,0)dt$

$=A_x(0,dx/2,0,0)dx-[V(dt/2,0,0,0)+\frac{\partial V}{\partial x}dx]dt.$

Look at the difference:

$\mathrm{\Delta }S=S_{ACD}-S_{ABD}=(-\frac{\partial V}{\partial x}-\frac{\partial A_x}{\partial t})dtdx=E_{x}dtdx.$

Alternatively, you may think of $\mathrm{\Delta }S$ as the electromagnetic contribution to the action of the loop $A\to B\to D\to C\to A.$

Let's repeat the calculation for a small rectangle with corners

$A=(0,0,0,0),B=(0,0,dy,0),C=(0,0,0,dz),D=(0,0,dy,dz).$

$S_{ABD}=A_z(0,0,0,dz/2)dz+A_y(0,0,dy/2,dz)dy$

$=A_z(0,0,0,dz/2)dz+[A_y(0,0,dy/2,0)+\frac{\partial A_y}{\partial z}dz]dy,$

$S_{ACD}=A_y(0,0,dy/2,0)dy+A_z(0,0,dy,dz/2)dz$

$=A_y(0,0,dy/2,0)dy+[A_z(0,0,0,dz/2)+\frac{\partial A_z}{\partial y}dy]dz,$

$\mathrm{\Delta }S=S_{ACD}-S_{ABD}=(\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z})dydz=B_{x}dydz.$

Thus the electromagnetic contribution to the action of this loop equals the flux of $𝐁$ through the loop.

Remembering (i) Stokes' theorem and (ii) the definition of $𝐁$ in terms of $𝐀,$ we find that

${\oint }_{\partial \mathrm{\Sigma }}𝐀\cdot d𝐫=\underset{\mathrm{\Sigma }}{\int }\text{curl}𝐀\cdot d\mathrm{\Sigma }=\underset{\mathrm{\Sigma }}{\int }𝐁\cdot d\mathrm{\Sigma }.$

In (other) words, the magnetic flux through a loop $\partial \mathrm{\Sigma }$ (or through any surface $\mathrm{\Sigma }$ bounded by $\partial \mathrm{\Sigma }$ ) equals the circulation of $𝐀$ around the loop (or around any surface bounded by the loop).

The effect of a circulation ${\oint }_{\partial \mathrm{\Sigma }}𝐀\cdot d𝐫$ around the finite rectangle $A\to B\to D\to C\to A$ is to increase (or decrease) the action associated with the segment $A\to B\to D$ relative to the action associated with the segment $A\to C\to D.$ If the actions of the two segments are equal, then we can expect the path of least action from $A$ to $D$ to be a straight line. If one segment has a greater action than the other, then we can expect the path of least action from $A$ to $D$ to curve away from the segment with the larger action.

Compare this with the classical story, which explains the curvature of the path of a charged particle in a magnetic field by invoking a force that acts at right angles to both the magnetic field and the particle's direction of motion. The quantum-mechanical treatment of the same effect offers no such explanation. Quantum mechanics invokes no mechanism of any kind. It simply tells us that for a sufficiently massive charge traveling from $A$ to $D,$ the probability of finding that it has done so within any bundle of paths not containing the action-geodesic connecting $A$ with $D,$ is virtually 0.

Much the same goes for the classical story according to which the curvature of the path of a charged particle in a spacetime plane is due to a force that acts in the direction of the electric field. (Observe that curvature in a spacetime plane is equivalent to acceleration or deceleration. In particular, curvature in a spacetime plane containing the $x$ axis is equivalent to acceleration in a direction parallel to the $x$ axis.) In this case the corresponding circulation is that of the 4-vector potential $(cV,𝐀)$ around a spacetime loop.

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