Particle moving in a abelian Maxwell gauge potential can be understood from Lagrangian $$ L=(1/2) m \dot{\vec{r}}^2- q\phi +q \dot{\vec{r}}\cdot \vec{A} $$ or Hamiltonian $$ H=\frac{(\vec{p}-q \vec{A})^2}{2m} +q\phi . $$

We can solve the equation of motion eom and find the Lorentz force law. We can find out the trajectory of particle moving in the magnetic field or electric field. For example,

  1. the particle may move in a circular motion or a helical motion under magnetic field.

  2. the particle may simply accelerate in a straight line under the electric field.

my questions

  • What will be the Lagrangian and Hamiltonian descriptions of particle moving in a non-abelian gauge potential?

  • What are the eom and find the Lorentz force law for non-abelian gauge potential?

We can assume the particle has $N=2$-component coupled to the $SU(N=2)$ gauge field via the fundamental representation.


1 Answer 1


The Electrodynamics Of Moving Bodies
Well, first: let's change this around a little and rewrite it as: $$L = T(𝐯) - V(𝐫,t,𝐯), \hspace 1em V(𝐫,t,𝐯) = e(φ(𝐫,t) - 𝐯·𝐀(𝐫,t)).$$ This has the classic $T = L - V$ form, except that $V$ may now be dependent on the velocity $𝐯 = d𝐫/dt$. So, just as $φ$ serves as a potential energy, pro-rated to unit charge, $𝐀$ serves as potential momentum, likewise pro-rated to unit charge. Correspondingly, the expression for the momentum $$𝐩 = \frac{∂L}{∂𝐯} = 𝐏 + e𝐀, \hspace 1em 𝐏 = \frac{∂T}{∂𝐯}$$ now includes a "kinetic momentum" $𝐏$ and "potential momentum" $e𝐀$. For the Hamiltonian, we find a similar reduction: $$H = 𝐯·𝐩 - L = E + eφ, \hspace 1em E = 𝐯·\frac{∂T}{∂𝐯} - T,$$ into a "kinetic energy" $E$ and a "potential energy" $eφ$. The Euler-Lagrange equations then become: $$\frac{d𝐩}{dt} = -∇V, \hspace 1em \frac{dH}{dt} = \frac{∂V}{dt},$$ or more explicitly, as: $$\frac{d}{dt}(𝐏 + e𝐀) = -∇(e(φ - 𝐯·𝐀)), \hspace 1em \frac{d}{dt}(E + eφ) = \frac{∂}{∂t}(e(φ - 𝐯·𝐀)),$$ the derivatives on the right-hand side being understood to be taken only with respect to the $(𝐫,t)$-dependency in $φ$ and $𝐀$.

Using the equations $$𝐁 = ∇×𝐀, \hspace 1em 𝐄 = -∇φ - \frac{∂𝐀}{∂t},$$ and the chain rule $$\frac{d}{dt} = \frac{∂}{∂t} + 𝐯·∇,$$ on $φ$ and $𝐀$, you may verify that this reconciles with the corresponding equations for the kinetic momentum $𝐏$ and kinetic energy $E$ - i.e. the Lorentz force law and its corresponding power law: $$\frac{d𝐏}{dt} = e(𝐄 + 𝐯×𝐁), \hspace 1em \frac{dE}{dt} = e𝐯·𝐄.$$

This serves the minimize and factor out the distinction between the relativistic and non-relativistic forms. The kinetic part of the Lagrangian, $T$, the kinetic momentum $𝐏$ and kinetic energy $E$ may be, respectively, written as: $$T = m f(𝐯), \hspace 1em 𝐏 = M𝐯, \hspace 1em E = M f(𝐯).$$ In non-relativistic theory, $$f(𝐯) = \frac{|𝐯|^2}2, \hspace 1em M = m,$$ while in relativity, $M$ includes the mass equivalent to $E$, itself: $$M = m + \frac{E}{c^2} ⇒ M = \frac{m}{1 - f(𝐯)/c^2}.$$ If you plug that into the differential equation $$𝐏 = \frac{∂T}{∂𝐯} ⇒ \frac{𝐯}{1 - f(𝐯)/c^2} = \frac{∂f(𝐯)}{∂𝐯},$$ you'll get $$|𝐯|^2 = 2 f(𝐯) - \frac{f(𝐯)^2}{c^2}.$$ Requiring that $f(𝟎) = 0$, and that $f(𝐯)$ be continuous in $𝐯$, selects between the two solutions of this equation for each $𝐯$ and you get: $$f(𝐯) = \frac{|𝐯|^2}{1 + \sqrt{1 - |𝐯|^2/c^2}} = c^2 - c^2 \sqrt{1 - |𝐯|^2/c^2}, \hspace 1em M = \frac{m}{\sqrt{1 - |𝐯|^2/c^2}}.$$ So, relativists like to tack on an extra $-mc^2$ to $L$ and therefore add $mc^2$ to $H$ and $E$, though it's not necessary for anything we do here. Then, the relativists' version of $E$ would be equal to our $E + mc^2$, which works out to $Mc^2$, since $$\frac{f(𝐯)}{\sqrt{1 - (v/c)^2}} + c^2 = \frac{c^2}{\sqrt{1 - (v/c)^2}}.$$

However, we won't do that. Instead, it is tempting to treat $f(𝐯)$, itself, as a velocity component $f(𝐯) = -du/dt$, with respect to an extra fictitious coordinate $u$. Then the quadratic relation for $f(𝐯)$ can be rewritten as: $$\left|\frac{d𝐫}{dt}\right|^2 + 2\left(\frac{dt}{dt}\right)\left(\frac{du}{dt}\right) + \frac{1}{c^2}\left(\frac{du}{dt}\right)^2 = 0.$$

This calls out a geometric invariant of the following form: $$|d𝐫|^2 + 2 dt du + \frac{1}{c^2} du^2 = 0.$$ That is just the relation for proper time, which we'll denote by $s$, in disguise: $$s = t + \frac{u}{c^2}, \hspace 1em ds^2 = dt^2 - \frac{|d𝐫|^2}{c^2}.$$ The coordinate $u = c^2(s - t)$ is time-dilation, ramped up by a factor of $c^2$.

Both it and the other relations continue to have meaning in the non-relativistic case, where they become $$s = t, \hspace 1em |d𝐫|^2 + 2 dt du = 0, \\ \hspace 1em f(𝐯) = -\frac{du}{dt} \hspace 1em⇒\hspace 1em |𝐯|^2 - 2 f(𝐯) = 0 \hspace 1em⇒\hspace 1em f(𝐯) = ½ |𝐯|^2.$$

In both cases, the kinetic part of the Lagrangian then yields the following contribution to the total action: $$\int T(𝐯) dt = \int m f(𝐯) dt = \int \left(-m\frac{du}{dt}\right) dt = -\int m du = -mu,$$ up to a constant offset.

I did this, so that we can write down a Lagrangian dynamics that applies uniformly to both the relativistic and non-relativistic case, when we get to gauge fields.

In the non-relativistic case, the 5-dimensional geometry given by the invariant $|d𝐫|^2 + 2 dt du$ is called Bargmann geometry. The relativistic version of this, which has the additional $(1/c)^2 du^2$ contribution, has no name that I'm aware of. It is a 5-dimensional geometry that contains Minkowski geometry. In both cases, it is the more natural venue for Lagrangian dynamics, owing to the close relation between the $u$ coordinate and total action.

The Gauge Fields Versus Maxwell Fields
Ok, so now we get down to the gauge fields - both Abelian and non-Abelian.

The gauge fields have the same form as the electromagnetic field, except there are multiple copies of the potentials and field strengths: one for each dimension of the underlying Lie algebra. Therefore, they would be written as: $$φ = \sum_{a=0,⋯,N-1} φ^a Y_a, \hspace 1em 𝐀 = \sum_{a=0,⋯,N-1} 𝐀^a Y_a,$$ the decomposition taking place with respect to a Lie algebra basis $\left(Y_a: a = 0,⋯,N-1\right)$, where $N$ is the dimension of the Lie algebra. In here and the following, we will use the Einstein summation convention, that a matching upstairs and downstairs index in any monomial is to be summed over the range of the index, so that this may be written more briefly as: $$φ = φ^a Y_a, \hspace 1em 𝐀 = 𝐀^a Y_a.$$ The corresponding $𝐁$ and $𝐄$ fields would be: $$𝐁 = 𝐁^a Y_a, \hspace 1em 𝐄 = 𝐄^a Y_a.$$

The Lie algebra is endowed with a (bi-linear) Lie bracket $\left[\_,\_\right]$ that satisfies the properties: $$[u,v] = -[v,u], \hspace 1em [u,[v,w]] + [v,[w,u]] + [w,[u,v]] = 0.$$ This both gives rise to and can be defined by a set of structure coefficients $f_{ab}^c$: $$\left[Y_a, Y_b\right] = f_{ab}^c Y_c,$$ that satisfy properties reflecting the two defining properties of the Lie algebra: $$f_{ab}^c = -f_{ba}^c, \hspace 1em f_{ae}^d f_{bc}^d + f_{be}^d f_{ca}^d + f_{ce}^d f_{ab}^d = 0.$$ The Lie algebra is Abelian, if the Lie bracket is 0, and non-Abelian if it is not. The cardinal example of the latter is given by the basis $\left(Y_0, Y_1, Y_2\right) = (X,Y,Z)$, with $N = 3$ is $$[Y,Z] = X, \hspace 1em [Z,X] = Y, \hspace 1em [X,Y] = Z,$$ which is $su(2)$, the Lie algebra of the Lie group $SU(2)$.

The equations connecting the field strengths and potentials are different from those of electromagnetism in that they are now non-linear and contain a contribution arising from the structure coefficients: $$𝐁^c = ∇×𝐀^c + ½ f_{ab}^c 𝐀^a×𝐀^b, \hspace 1em 𝐄^c = -∇φ^c - \frac{∂𝐀^c}{∂t} + f_{ab}^c φ^a 𝐀^b,$$ the last term can be rewritten as $$f_{ab}^c φ^a 𝐀^b = ½f_{ab}^c (φ^a 𝐀^b - 𝐀^a φ^b),$$ owing to the anti-symmetry $f_{ab}^c = -f_{ba}^c$ of $f$ in $a$ and $b$.

Assume that the Lie algebra has been expanded to include an ordinary product operation that makes the Lie bracket a commutator: $$[u,v] = uv - vu \hspace 1em⇒\hspace 1em Y_aY_b - Y_bY_a = \left[Y_a,Y_b\right] = f_{ab}^c Y_c.$$ Such an expansion is, for instance, accomplished with a matrix representation of the $Y$'s. More generally, the expansion is called an "enveloping algebra" of the Lie algebra; and the one that stands out, which corresponds to "faithful" matrix representations, is the "universal enveloping algebra".

Assume, also, that the $Y$'s freely intermix with the scalars and vectors; e.g. $φ^a Y_b = Y_b 𝐀^a$, $𝐀^a Y_b = Y_b 𝐀^a$. Then this makes it possible to write everything in simpler and more direct form: $$\begin{align} 𝐀×𝐀 &= 𝐀^a×𝐀^bY_aY_b \\ &= ½\left(𝐀^a×𝐀^b - 𝐀^b×𝐀^a\right)Y_aY_b \\ &= ½𝐀^a×𝐀^b\left(Y_aY_b - Y_bY_a\right) \\ &= ½f_{ab}^c𝐀^a×𝐀^bY_c, \\ φ𝐀 - 𝐀φ &= φ^a 𝐀^b Y_a Y_b - 𝐀^b φ^a Y_b Y_a \\ &= φ^a 𝐀^b \left(Y_a Y_b - Y_b Y_a\right) \\ &= f_{ab}^c φ^a 𝐀^b Y_c \\ &= ½f_{ab}^c \left(φ^a 𝐀^b - 𝐀^a φ^b\right) Y_c, \end{align}$$ using the anti-symmetry of the vector product $𝐀^a×𝐀^b = -𝐀^b×𝐀^a$.

Then, the equations for the field versus potential can be rewritten as: $$𝐁 = ∇×𝐀 + 𝐀×𝐀, \hspace 1em 𝐄 = -∇φ - \frac{∂𝐀}{∂t} + φ𝐀 - 𝐀φ.$$

The Gauge Dynamics Of Moving Bodies
So, in answer to the second part of your question: the same equations hold, by analogy - but now with the charge $e$ also replicated into $N$ components $\left(e_a: a=0⋯,N-1\right)$: one for each set of fields $\left(𝐀^a,φ^a,𝐁^a,𝐄^a\right)$. The total momentum and total energy now include the sum of all the contributions from the respective potentials: $$H = E + e_a φ^a, \hspace 1em 𝐩 = 𝐏 + e_a 𝐀^a;$$ and now $V$ includes a sum from all the contributions: $$V = e_a \left(φ^a - 𝐯·𝐀^a\right).$$

The charge $e = e_a y^a$ is now referenced to a dual basis $\left(y^a: a=0⋯,N-1\right)$, and we assume that the $y$'s can freely intermix with the vectors and scalars. Associated with it, is a contraction operator: $$Y_a·y^b = y^b·Y_a = δ_a^b = \left(\begin{align}1 & \hspace 1em (a = b)\\0 & \hspace 1em (a ≠ b)\end{align}\right),$$ so, we can rewrite these equations as: $$H = E + e·φ, \hspace 1em 𝐩 = 𝐏 + e·𝐀;$$ and $V$ as: $$V = e·\left(φ - 𝐯·𝐀\right).$$

The analogous equations of motion hold both in terms of the total energy and total momentum: $$ \frac{d𝐩}{dt} = -∇V \hspace 1em⇒\hspace 1em \frac{d\left(𝐏 + e·𝐀\right)}{dt} = -∇\left(e·\left(φ - 𝐯·𝐀\right)\right), \\ \frac{dH}{dt} = \frac{∂V}{∂t} \hspace 1em⇒\hspace 1em \frac{d\left(E + e·φ\right)}{dt} = \frac{∂}{∂t}\left(e·\left(φ - 𝐯·𝐀\right)\right), $$ and in terms of the kinetic energy and kinetic momentum: $$ \frac{d𝐏}{dt} = e·\left(𝐄 + 𝐯×𝐁\right), \hspace 1em \frac{dE}{dt} = e·\left(𝐯·𝐄\right). $$

However, they can't be reconciled this time, because the $𝐁$ and $𝐄$ now have the extra non-linear contributions to them ... unless you also assume that the charge $e$ precesses and has a non-zero time derivative $de/dt$ that matches up with the non-Abelian parts of the field: $$ \frac{de}{dt}·𝐀 = e·(𝐀φ - φ𝐀 - 𝐯×(𝐀×𝐀)) = e·(𝐀(φ - 𝐯·𝐀) - (φ - 𝐯·𝐀)𝐀), \\ \frac{de}{dt}·φ = e·(𝐯·(𝐀φ - φ𝐀)) = e·(φ(φ - 𝐯·𝐀) - (φ - 𝐯·𝐀)φ). $$

It can be written more succinctly, if we assume that dual basis and basis can be combined into a single algebra satisfying: $$y^a·[u,v] = y^a·(uv - vu) = y^a·uv - y^a·vu = y^au·v - uy^a·v = (y^au - uy^a)·v.$$ In particular, we assume that the product operation is expanded to include the following: $$y^c Y_a - Y_a y^c = f_{ab}^c y^b.$$

For instance, in a matrix representation, we may take $y^a·u = \text{Tr}\left(y^au\right)$, with a "trace operator" $\text{Tr}\left(\_\right)$ that satisfies $\text{Tr}\left(xy\right) = \text{Tr}\left(yx\right)$. Then $y^a·vu = \text{Tr}\left(y^avu\right) = \text{Tr}\left(uy^av\right) = uy^a·v$.

Then, we may write the precession equations as: $$ \frac{de}{dt}·𝐀 = \left((φ - 𝐯·𝐀)e - e(φ - 𝐯·𝐀)\right)·𝐀, \\ \frac{de}{dt}·φ = \left((φ - 𝐯·𝐀)e - e(φ - 𝐯·𝐀)\right)·φ. $$ These equations are satisfied, with a charge precession given by: $$\frac{de}{dt} = (φ - 𝐯·𝐀)e - e(φ - 𝐯·𝐀).$$ That's Wong's Equation. Since this is to hold for arbitrary potentials, then the equation is both necessary and sufficient, to compensate for the contribution of the non-linear part of the field strength to the force and power laws.

Thus, for charges moving in a gauge field, you have the usual Lorentz force (and power) laws, extended to include multi-component charges, both in terms of the total {energy, momentum} and kinetic {energy, momentum}, plus an extra equation for charge precession that makes the total versus kinetic versions compatible and accounts for the non-linear part of the equations governing field strength.

Lagrangian And Hamiltonian Formulations
As to the first part of your question: Hamiltonian formulations are much easier to arrive at than Lagrangian formulations. The latter problem is still considered open, since the only formulations for it that have been posed are rather arcane and alien to physicists' eyes. However, there is one exception of note - by Duviryak - that can be repurposed (and generalized) to a form more friendly to physicists.

Classical Mechanics Of Relativistic Particle With Color

I reworked it into a form using ordinary index and tensor notation and generalized it to work in curved space-times. It applies to the relativistic case, but I think I can also refurbish it further here to make it work in non-relativistic form, too ... at least for the case of flat space-times.

The Hamiltonian case requires setting up Poisson brackets: $$\left\{e_a, e_b\right\} = f_{ab}^c e_c, \hspace 1em \left\{e_a, p_ν\right\} = 0 = \left\{p_μ, e_b\right\}, \hspace 1em \left\{p_μ, p_ν\right\} = 0, $$ with the component $-p_0$ being the relativists' $E$ (our $E + mc^2 = Mc^2$) and $\left(p_1, p_2, p_3\right) = 𝐏$ being the kinetic momentum. The charge components $e_a$ are treated as momentum components $p_a$ with the $N$ dimensions of the Lie group tacked onto the 4-dimensions of space-time.

For this to work, you need the conjugate coordinates: $x^0 = t$, $\left(x^1, x^2, x^3\right) = 𝐫$. But then there are constraints, such as $M^2 - 𝐏^2/c^2 = m^2$ - the "mass shell constraint". So, that means you're working with Poisson brackets with constraints.

Then you have the issue of what the coordinates conjugate to $e_a$ should be. For that, you need the entire group manifold. Thus, for instance, of the Lie algebra is $su(2)$, then the gauge group will be $SU(2)$, which can be represented geometrically as the unit hypersphere $𝔴^2 + 𝔵^2 + 𝔶^2 + 𝔷^2 = 1$. Then the basis vectors $X$, $Y$, $Z$ in $su(2)$ would be represented by differential operators $$\left(X, Y, Z\right) = ½ \left(𝔴 \frac{∂}{∂𝖗} - 𝖗\frac{∂}{∂𝔴} - 𝖗×\frac{∂}{∂𝖗}\right), \hspace 1em 𝖗 = (𝔵, 𝔶, 𝔷).$$ Denoting the Lie group coordinates $\left(g^a:a=0,1,⋯,N-1\right)$, then I think, the extra brackets would be: $$ \left\{p_μ, x^ν\right\} = δ_μ^ν, \hspace 1em \left\{p_μ, g^b\right\} = 0 = \left\{e_a, x^ν\right\}, \hspace 1em \left\{e_a, g^b\right\} = Y_a g^b, \\ \left\{x^μ, x^ν\right\} = 0, \hspace 1em \left\{x^μ, g^b\right\} = 0 = \left\{g^a, x^ν\right\}, \hspace 1em \left\{g^a, g^b\right\} = 0, $$ where $δ_μ^ν = 1$ if $μ = ν$ and $δ_μ^ν = 0$ if $μ ≠ ν$. The equation of motion would then come off the operator: $$\frac{d}{dt} = \left\{\_, H\right\} + \frac{∂}{∂t},$$ where the Hamiltonian $H = Mf(𝐯) + e_a φ^a$ has to be rewritten with the equation $𝐏 = M𝐯$ solved in terms of $𝐯 = 𝐏/M$; $$H = \frac{|𝐏|^2}{m + M} + e·φ, \hspace 1em M = m \sqrt{1 + \frac{|𝐏|^2}{m^2c^2}}, \hspace 1em 𝐏 = 𝐩 - e·𝐀,$$ and the Poisson brackets, themselves, have to be reduced to a form that eliminates the mass-shell constraint. I don't think the substitution, above, of $M$ in terms of $𝐏$, alone, will do the trick.

For the Lagrangian case, Duviryak essentially adopts an action of the form: $$S = \int F(U) ds, \hspace 1em U = \left(U^a:a=0,1,⋯,N-1\right), \hspace 1em U^a = 𝐀^a·\frac{d𝐫}{ds} - φ^a \frac{dt}{ds},$$ using the Minkowski proper time $s$. The function $F$ is left unspecified. The actual form of $F$ does not enter the picture at all ... except to generate a relation between $e$ and $m$.

The charge and mass are defined by $$e_a = \frac{∂F}{∂U^a}, \hspace 1em -mc^2 = F - e_a U^a.$$ This reduces the action to: $$S = \int \left(-mc^2 + e_a \left(𝐀^a·\frac{d𝐫}{ds} - φ^a \frac{dt}{ds}\right)\right) ds = \int \left(-mc^2 ds + e_a \left(𝐀^a·d𝐫 - φ^a dt\right)\right).$$ In terms of our $u$ coordinate, using $ds = dt + du/c^2$, and rewriting $d𝐫 = 𝐯 dt$ and $du = -f(𝐯) dt$, this would become: $$S = \int \left(-mc^2 + mf(𝐯) + e_a \left(𝐀^a·𝐯 - φ^a\right)\right) dt.$$

It has the extra $-mc^2$ in it, so it won't work for the non-relativistic case, as is.

To make this work uniformly for both Relativity and non-relativistic theory, we'll have to adapt the action to the following form: $$S = \int F(U) μ dt, \hspace 1em U = \left(U^a:a=0,1,⋯,N-1\right), \hspace 1em U^a = \frac{𝐀^a·𝐯 - φ^a}μ.$$ Previously, the choice was $μ = 1 - f(𝐯)/c^2$. Here, we choose $μ = f(𝐯)$.

The choice for the factor $μ$ will reduce action integral to the form $$S = \int -F(U) du = \int L dt, \hspace 1em L = F(U)μ = F(U) f(𝐯).$$ This time, define the charge and mass by: $$e_a = \frac{∂F}{∂U^a}, \hspace 1em m = F - e_a U^a = F - e·U.$$ Then, the Lagrangian reduces to $$L = e·(𝐯·𝐀 - φ) + mμ = T - V, \hspace 1em T = m f(𝐯), \hspace 1em V = e_a \left(φ^a - 𝐀^a·𝐯\right) = e·(φ - 𝐀·𝐯).$$

The total variational in $L$ is: $$ΔL = e·ΔU μ + F Δμ = e·Δ\left(𝐯·𝐀 - φ\right) - e·U Δμ + F Δμ = e·Δ(𝐯·𝐀 - φ) + m Δμ.$$ Since $L = e·(𝐯·𝐀 - φ) + mμ$, then it follows from this that $$0 = Δe·(𝐯·𝐀 - φ) + Δm μ,$$ which will be used to show that $m$ is a constant of motion.

From the variational, we may read off the derivative of $L$ with respect to $𝐯$: $$𝐩 = \frac{∂L}{∂𝐯} = e·𝐀 + m \frac{∂μ}{∂𝐯} = e·𝐀 + M𝐯.$$ So, we recover the decomposition of the momentum into the kinetic momentum and potential momentum. For its Euler-Lagrange equation, we have: $$\frac{d𝐩}{dt} = ∇L = e·∇(𝐯·𝐀 - φ) - m∇μ = -∇(e·∇(φ - 𝐯·𝐀)) = -∇V.$$

The remaining degree of freedom is covered by the gauge transformation, which for the gauge potentials can be written for non-Abelian gauge fields as: $$Δ𝐀 = -∇χ + χ𝐀 - 𝐀χ, \hspace 1em Δφ = \frac{∂χ}{∂t} + χφ - φχ.$$ Upon substitution, this yields $$\begin{align} ΔL &= e·\left(𝐯·Δ𝐀 - Δφ\right) + mΔμ \\ &= e·\left(𝐯·(-∇χ + χ𝐀 - 𝐀χ) - \left(\frac{∂χ}{∂t} + χφ - φχ\right)\right) \\ &= e·\left(-\left(𝐯·∇ + \left(\frac{∂}{∂t}\right)\right)χ - χ(φ - 𝐯·𝐀) + (φ - 𝐯·𝐀)χ \right) \\ &= -e·\frac{dχ}{dt} + e·((φ - 𝐯·𝐀)χ - χ(φ - 𝐯·𝐀)) \\ &= -\frac{d}{dt}(e·χ) + \frac{de}{dt}·χ - ((φ - 𝐯·𝐀)e - e(φ - 𝐯·𝐀))·χ. \end{align}$$

Under the action integral variational $ΔS = \int ΔL$, the total differential $d(e·χ)/dt$ drops out from the analysis, leaving behind $$ΔS = \left(\frac{de}{dt} - ((φ - 𝐯·𝐀)e - e(φ - 𝐯·𝐀))\right)·χ.$$ Since $χ$ can be arbitrarily chosen, then the factor in the integrand is zero, and we get: $$\frac{de}{dt} = (φ - 𝐯·𝐀)e - e(φ - 𝐯·𝐀).$$

Finally, substituting this into the identity $Δe·(𝐯·𝐀 - φ) + Δm μ = 0$, we get: $$\begin{align} \frac{dm}{dt} μ &= \frac{de}{dt}·(φ - 𝐯·𝐀) \\ &= ((φ - 𝐯·𝐀)e - e(φ - 𝐯·𝐀))·(φ - 𝐯·𝐀) \\ &= e·((φ - 𝐯·𝐀)(φ - 𝐯·𝐀) - (φ - 𝐯·𝐀)(φ - 𝐯·𝐀)) \\ &= 0. \end{align}$$

From this, follows $dm/dt = 0$ ... at least for $𝐯 ≠ 𝟎$. We'll probably need something more to plug up the hole for $𝐯 = 𝟎$, since $μ = f(𝐯) = 0$, when $𝐯 = 𝟎$. So, that part might not be water-tight.


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