This quantum world/Feynman route/Energy and momentum

Energy and momentum

Observe that if $dS$ does not depend on $t$ (that is, $\partial dS/\partial t=0$ ) then

$E=-\frac{\partial dS}{\partial dt}$

is constant along geodesics. (We'll discover the reason for the negative sign in a moment.)

Likewise, if $dS$ does not depend on $𝐫$ (that is, $\partial dS/\partial 𝐫=0$ ) then

$𝐩=\frac{\partial dS}{\partial d𝐫}$

is constant along geodesics.

$E$ tells us how much the projection $dt$ of a segment $d𝒞$ of a path $𝒞$ onto the time axis contributes to the action of $𝒞.$ $𝐩$ tells us how much the projection $d𝐫$ of $d𝒞$ onto space contributes to $S[𝒞].$ If $dS$ has no explicit time dependence, then equal intervals of the time axis make equal contributions to $S[𝒞],$ and if $dS$ has no explicit space dependence, then equal intervals of any spatial axis make equal contributions to $S[𝒞].$ In the former case, equal time intervals are physically equivalent: they represent equal durations. In the latter case, equal space intervals are physically equivalent: they represent equal distances.

If equal intervals of the time coordinate or equal intervals of a space coordinate are not physically equivalent, this is so for either of two reasons. The first is that non-inertial coordinates are used. For if inertial coordinates are used, then every freely moving point mass moves by equal intervals of the space coordinates in equal intervals of the time coordinate, which means that equal coordinate intervals are physically equivalent. The second is that whatever it is that is moving is not moving freely: something, no matter what, influences its motion, no matter how. This is because one way of incorporating effects on the motion of an object into the mathematical formalism of quantum physics, is to make inertial coordinate intervals physically inequivalent, by letting $dS$ depend on $t$ and/or $𝐫.$

Thus for a freely moving classical object, both $E$ and $𝐩$ are constant. Since the constancy of $E$ follows from the physical equivalence of equal intervals of coordinate time (a.k.a. the "homogeneity" of time), and since (classically) energy is defined as the quantity whose constancy is implied by the homogeneity of time, $E$ is the object's energy.

By the same token, since the constancy of $𝐩$ follows from the physical equivalence of equal intervals of any spatial coordinate axis (a.k.a. the "homogeneity" of space), and since (classically) momentum is defined as the quantity whose constancy is implied by the homogeneity of space, $𝐩$ is the object's momentum.

Let us differentiate a former result,

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

with respect to $\lambda .$ The left-hand side becomes

$\frac{d(dS)}{d\lambda }=\frac{\partial dS}{\partial (\lambda dt)}\frac{\partial (\lambda dt)}{\partial \lambda }+\frac{\partial dS}{\partial (\lambda d𝐫)}\cdot \frac{\partial (\lambda d𝐫)}{\partial \lambda }=\frac{\partial dS}{\partial (\lambda dt)}dt+\frac{\partial dS}{\partial (\lambda d𝐫)}\cdot d𝐫,$

while the right-hand side becomes just $dS.$ Setting $\lambda =1$ and using the above definitions of $E$ and $𝐩,$ we obtain

$-Edt+𝐩\cdot d𝐫=dS.$

$dS=-m{c}^{2}ds$ is a 4-scalar. Since $(cdt,d𝐫)$ are the components of a 4-vector, the left-hand side, $-Edt+𝐩\cdot d𝐫,$ is a 4-scalar if and only if $(E/c,𝐩)$ are the components of another 4-vector.

(If we had defined $E$ without the minus, this 4-vector would have the components $(-E/c,𝐩).$)

In the rest frame ${ℱ}^{\prime }$ of a free point mass, $d{t}^{\prime }=ds$ and $dS=-m{c}^{2}d{t}^{\prime }.$ Using the Lorentz transformations, we find that this equals

$dS=-m{c}^2\frac{dt-vdx/c^2}{\sqrt{1-v^2/c^2}}=-\frac{m{c}^2}{\sqrt{1-v^2/c^2}}dt+\frac{m𝐯}{\sqrt{1-v^2/c^2}}\cdot d𝐫,$

where $𝐯=(v,0,0)$ is the velocity of the point mass in $ℱ.$ Compare with the above framed equation to find that for a free point mass,

$E=\frac{m{c}^2}{\sqrt{1-v^2/c^2}}𝐩=\frac{m𝐯}{\sqrt{1-v^2/c^2}}.$

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