I am trying to understand the magnetic vector potential as momentum per unit charge, that is, somehow, the "total" momentum should be $p = mv +qA$.

Here's a part of the paper I was reading - I am facing difficulties understanding the mathematical gymnastics involved, and would appreciate any help! enter image description here

Firstly, let's start with (16). On the RHS, the dot product of velocity with $v$ x $B$ is zero, and only the first term remains. To arrive at (18) from (16), I added the time derivative of $q\phi$ on both sides - the LHS is now what is desired, so manipulation on RHS should yield the RHS of (18); and this is where I'm stuck. How should I take it from here? Please help! A detailed explanation would be great. I'm using Griffiths as the text for my E&M course, and I am only familiar with the level of vector calculus used in the book, and that is probably why I'm stuck where I am.

Also, I need help deriving (19) (trust me, I tried).

Lastly, what is the difference between $p = mv$ and $p = mv + qA$? We call both as momentum, and I'd learnt only the former in a classical mechanics course. So what is the difference in meaning and interpretation of the two different momenta? I am really interested in understanding this concept - vector potential as momentum per unit charge, please help me out! Also, this is a very absorbing concept, one that isn't touched upon in nearly all E&M texts, so this post, when complete with answers, would definitely interest many people on Physics SE!

Here's the link to the entire paper, if the readers would like to know more: Thoughts on the magnetic vector potential

  • $\begingroup$ Equation (14) in your Dokument give you the Lagrangian. So with Euler Lagrangian Methode you get the equations of motion, which give you equation (19). $\endgroup$
    – Eli
    Oct 1, 2019 at 13:08

3 Answers 3


First the derivations:

Equation 18:

The connection between total time derivative $\frac{d}{dt}$ and partial time derivative $\frac{\partial}{\partial t}$ of a scalar function is, using the chain rule,

$$ \frac{d}{dt} = \frac{\partial}{\partial t} + \frac{\partial x}{\partial t} \frac{\partial}{\partial x} + \frac{\partial y}{\partial t} \frac{\partial}{\partial y} + \frac{\partial z}{\partial t} \frac{\partial}{\partial z} $$

Notice that this is equal to

$$ \frac{d}{dt} = \frac{\partial}{\partial t} + \mathbf{v} \cdot \nabla $$

Hence $q \frac{d \phi}{dt} = q \frac{\partial \phi}{\partial t} + q \mathbf{v} \cdot \nabla \phi$, where the last term is the one appearing in Eq. 16, but with a minus sign.

From this we get

$$ \frac{d}{dt} \left( \frac{1}{2} m v^2 \right) = -q \mathbf{v} \cdot \nabla \phi - q \mathbf{v} \cdot \frac{\partial \mathbf{A}}{\partial t}$$

$$ \frac{d}{dt} \left( \frac{1}{2} m v^2 \right) = q \frac{\partial \phi}{\partial t} - q \frac{d \phi}{dt} - q \mathbf{v} \cdot \frac{\partial \mathbf{A}}{\partial t} $$

$$ \frac{d}{dt} \left( \frac{1}{2} m v^2 \right) + q \frac{d \phi}{dt} = q \frac{\partial \phi}{\partial t} - q \mathbf{v} \cdot \frac{\partial \mathbf{A}}{\partial t}$$

$$ \frac{d}{dt} \left(\frac{1}{2} m v^2 + q \phi \right) = q \left(\frac{\partial \phi}{\partial t} - \mathbf{v} \cdot \frac{\partial \mathbf{A}}{\partial t} \right)$$

which is what I think Eq. 18 is supposed to be. Here it is also perfectly clear that if $\phi, \mathbf{A}$ are not explicitly time dependent, then their partial time derivatives are 0, and the whole RHS is zero.

Equation 19:

Here we need the total time derivative of a vector function $\mathbf{A}$. By looking at the components of the vector, we should realize that

$$\frac{d \mathbf{A}}{dt} = \frac{\partial \mathbf{A}}{\partial t} + \left( \mathbf{v} \cdot \nabla \right) \mathbf{A} $$

with the interpretation that e.g.

$$ \frac{d A_x}{dt} = \frac{\partial A_x}{\partial t} + \frac{\partial x}{\partial t} \frac{\partial A_x}{\partial x} + \frac{\partial y}{\partial t} \frac{\partial A_x}{\partial y} + \frac{\partial z}{\partial t} \frac{\partial A_x}{\partial z} $$

We have a useful vector calculus identity link:

$$ \mathbf{v} \times (\nabla \times \mathbf{A}) = \nabla_\mathbf{A} (\mathbf{v} \cdot \mathbf{A}) - (\mathbf{v} \cdot \nabla) \mathbf{A} $$

where the notation $\nabla_\mathbf{A}$ means only the variation due to $\mathbf{A}$ is taken into account in that term.

Starting from Newton's second law, $\frac{d}{dt}(m \mathbf{v} ) = q(\mathbf{E} + \mathbf{v} \times \mathbf{B})$ and inserting the potentials, we get

$$ \frac{d}{dt}(m \mathbf{v} ) = q \left(- \nabla \phi - \frac{\partial \mathbf{A}}{\partial t} + \mathbf{v} \times (\nabla \times \mathbf{A}) \right) $$

Using the identity gives

$$ \frac{d}{dt}(m \mathbf{v} ) = q \left(- \nabla \phi - \frac{\partial \mathbf{A}}{\partial t} + \nabla_\mathbf{A} (\mathbf{v} \cdot \mathbf{A}) - (\mathbf{v} \cdot \nabla) \mathbf{A} \right) $$

$$ \frac{d}{dt}(m \mathbf{v} ) + q \left( \frac{\partial \mathbf{A}}{\partial t} + (\mathbf{v} \cdot \nabla) \mathbf{A} \right) = q \left(- \nabla \phi + \nabla_\mathbf{A} (\mathbf{v} \cdot \mathbf{A}) \right) $$

On the LHS we identify the total derivative of the vector potential, so finally

$$ \frac{d}{dt} \left( m \mathbf{v} + q \mathbf{A} \right) = - q \nabla \left(\phi - \mathbf{v} \cdot \mathbf{A} \right) $$

where the gradient on the RHS is understood in accordance to the clarification about the identity we used.

Now the meaning:

The question of the difference between $p=mv$ and $p=mv + qA$ has been asked before on this site, e.g. here. In short, "momentum" is a type of quantity that is often of great importance in physics, just as "energy" is, and there exist several different momenta, just as there exist different important forms of energy.

In the classical theories of Lagrangian and Hamiltonian mechanics, the fundamental variables are so-called generalized coordinates $q_i$ and their time derivatives $\dot{q}_i$. They can be, and are often, real-space coordinates, but they can also be angles or other ways of describing the configurations of a system. For each generalized coordinate, there exists a corresponding quantity called the conjugate momentum, which is given by

$$p_i = \frac{\partial L}{\partial \dot{q}_i} $$

where $L$ is the Lagrangian, usually the kinetic minus the potential energy. In the case of a charged particle in an EM field, the conjugate momentum to position turns out to be exactly $mv + qA$.


a) Eq. 18 does not tell us that $\mathbf{A}$ is momentum per unit charge, Eq. 19 does. Eq. 18 is a conservation law; it tells you that when your charged particle is moving around, the conserved quantity is not just the kinetic energy, but $1/2 m v^2 + q \phi$. Hence we can think of $q \phi$ as potential energy $E_p$, and then

$$ \frac{E_p}{q} = \phi $$

is interpreted as potential energy per charge. Likewise for Eq. 19, it establishes a mathematical form similar to Newton's second law when written on the form

$$ \frac{d p}{dt} = - \nabla U $$

Comparing Eq. 19 to this form, we see that the "total momentum" is now $m \mathbf{v} + q \mathbf{A}$, so kinetic momentum $p_k$ + electromagnetic momentum $p_{EM}$. Hence

$$ \frac{p_{EM}}{q} = \frac{q \mathbf{A}}{q} = \mathbf{A} $$

is electromagnetic momentum per unit charge.

b) I'll explain the chain rule for scalar functions, called fields. Remember that fields $\phi$ depend on position in space, and possibly explicitly on time too:

$$ \phi = \phi(t, x(t), y(t), z(t)) $$

The reason that I'm letting the spatial coordinates depend on $t$, is that the particle we are describing the motion of, will of course move around, and the only value of $\phi$ relevant to the particle is the field value at the particle's position.

So how can the field at the position of the particle change with time? Of course it can change explicitly with time in a given fixed point in space, so that is what $\frac{\partial}{\partial t}$ gives you. But if the particle moves a small amount in the $x$ direction over a small time, then the change is

$$ \frac{\partial x}{\partial t} \frac{\partial \phi}{\partial x} $$

by the ordinary chain rule. And since the three spatial directions are not explicitly dependent on each other (instead they all depend on time $t$), we get contributions like this for all three directions.

c) The general rule for the gradient of the dot product of two vector fields is

$$ \nabla(\mathbf{v} \cdot \mathbf{A}) = \mathbf{v} \times (\nabla \times \mathbf{A}) + \mathbf{A} \times (\nabla \times \mathbf{v}) + (\mathbf{v} \cdot \nabla) \mathbf{A} + (\mathbf{A} \cdot \nabla) \mathbf{v} $$

Of course it is symmetric in the two vector fields. The point here is that in our physical situation, it is only $\mathbf{A}$ that is a proper vector field (one vector value for each point in space). The quantity $\mathbf{v}$ is not a field. There is only one value of $\mathbf{v}$ for each point in time, and it is not associated to anything in space, apart from maybe the position of the particle whose motion we are describing. In that sense, we don't consider the "spatial variation" in $\mathbf{v}$, so we drop the two terms where $\nabla$ is acting on $\mathbf{v}$, and keep the two terms where $\nabla$ acts on $\mathbf{A}$. The notation $\nabla_\mathbf{A}$ signifies just that.

  • $\begingroup$ How does equation (18) explicitly tell me that magnetic potential is indeed momentum per unit charge? Is it because the partial time derivative qA yields force, and that dotted with velocity is the rate of work done? Makes sense only dimensionally to me, a little more explanation of (18) would help! $\endgroup$ Oct 2, 2019 at 3:44
  • $\begingroup$ Also, I'm not exactly proficient in multivariable calculus yet (in process of learning for the first time) - so I wonder how you got to the chain rule used in derivation of (18)? $\endgroup$ Oct 2, 2019 at 3:45
  • $\begingroup$ Lastly, I don't understand what grad subscript A means, exactly? Could you explain in a little more detail? $\endgroup$ Oct 2, 2019 at 4:50
  • 1
    $\begingroup$ I've addressed the comments in my answer, please see the edit :) $\endgroup$ Oct 2, 2019 at 6:20
  • $\begingroup$ Great! I understand the chain rule better now! $\endgroup$ Oct 2, 2019 at 6:29

From the linked paper, $$\frac{d}{dt} \left( \frac{1}{2} mv^2 \right) = q \mathbf v \cdot (\mathbf E + \mathbf v \times \mathbf B) \label{16}\tag{16}$$

$$\mathbf E = - \nabla \phi - \frac{\partial}{\partial t} \mathbf A\ \text{ and }\ \mathbf B = \nabla \times \mathbf A \label{17}\tag{17}$$

I assume when taking the total derivative of $\phi$, the following was meant: $$q \frac{d}{dt} \phi(\mathbf r,t) = q \frac{\partial}{\partial t} \phi + q \mathbf v \cdot \nabla \phi \label{n.1}\tag{n.1}$$

If we dot \ref{17}.a with $q \mathbf v$, we get $$q \mathbf v \cdot \mathbf E = - q \mathbf v \cdot \nabla \phi - q \mathbf v \cdot \frac{\partial}{\partial t} \mathbf A$$

Substituting $- q \mathbf v \cdot \nabla \phi$ from \ref{n.1} $$q \mathbf v \cdot \mathbf E = - q \frac{d}{dt} \phi + q \frac{\partial}{\partial t} \phi - q \mathbf v \cdot \frac{\partial}{\partial t} \mathbf A$$

From \ref{16}, we have $$\frac{d}{dt} \left( \frac{1}{2} mv^2 \right) = q \mathbf v \cdot (\mathbf E + \mathbf v \times \mathbf B) = q \mathbf v \cdot \mathbf E$$

since $\mathbf v \cdot (\mathbf v \times \mathbf B) = 0$. We already calculated $q \mathbf v \cdot \mathbf E$: $$\frac{d}{dt} \left( \frac{1}{2} mv^2 \right) = - q \frac{d}{dt} \phi + q \frac{\partial}{\partial t} \phi - q \mathbf v \cdot \frac{\partial}{\partial t} \mathbf A$$

$$\frac{d}{dt} \left( \frac{1}{2} mv^2 + q \phi \right) = q \frac{\partial}{\partial t} \phi - q \mathbf v \cdot \frac{\partial}{\partial t} \mathbf A$$

$$\frac{d}{dt} \left( \frac{1}{2} mv^2 + q \phi \right) = \frac{\partial}{\partial t} q (\phi - \mathbf v \cdot \mathbf A) \tag{18}$$ $$\tag*{$\blacksquare$}$$

$$\frac{d}{dt} (m \mathbf v + q \mathbf A) = - \nabla q (\phi-\mathbf v \cdot \mathbf A) \label{19}\tag{19}$$

For \ref{19}, you need to use (again, I am not sure about this) $$q \frac{d}{dt} \mathbf A = q \frac{\partial}{\partial t} \mathbf A + q (\mathbf v \cdot \nabla) \mathbf A \label{n.2}\tag{n.2}$$

From $\mathbf F = m \mathbf a = m \frac{d}{dt} \mathbf v = q(\mathbf E + \mathbf v \times \mathbf B)$, $$m \frac{d}{dt} \mathbf v = q(\mathbf E + \mathbf v \times \mathbf B)$$

Adding both sides $\frac{d}{dt} q \mathbf A$, $$\frac{d}{dt} (m \mathbf v + q \mathbf A) = q(\mathbf E + \mathbf v \times \mathbf B + \frac{\partial}{\partial t} \mathbf A + (\mathbf v \cdot \nabla) \mathbf A) \label{n.3}\tag{n.3}$$

Using $\mathbf A \times (\mathbf B \times \mathbf C) = (\mathbf A \times \mathbf C) \mathbf B - (\mathbf A \times \mathbf B) \mathbf C$ and \ref{17}.b, $$\mathbf v \times \mathbf B = \mathbf v \times (\nabla \times \mathbf A) = \nabla (\mathbf v \cdot \mathbf A) - (\mathbf v \cdot \nabla) \mathbf A$$

Inserting this into \ref{n.3}, $$\frac{d}{dt} (m \mathbf v + q \mathbf A) = q(\mathbf E + \frac{\partial}{\partial t} \mathbf A + \nabla (\mathbf v \cdot \mathbf A))$$

Using \ref{17}.a, we have $\mathbf E + \frac{\partial}{\partial t} \mathbf A = - \nabla \phi$, so $$\frac{d}{dt} (m \mathbf v + q \mathbf A) = - q(\nabla \phi - \nabla (\mathbf v \cdot \mathbf A))$$

$$\frac{d}{dt} (m \mathbf v + q \mathbf A) = - \nabla q (\phi-\mathbf v \cdot \mathbf A)$$ $$\tag*{$\blacksquare$}$$

I will not be able to address your interpretation question at all since I am not accustomed to it either. Also sorry for the backwards proof. Any and all improvements are welcome.

  • $\begingroup$ Please explain (n.1) and (n.2) $\endgroup$ Oct 2, 2019 at 5:17
  • $\begingroup$ @arya_stark Meyer explained it more clearly than I could, what I advise you is to try to get the intuition of derivatives of a (vector/scalar) field as well. I kind of made these up while I was answering the question. $\endgroup$
    – acarturk
    Oct 2, 2019 at 7:41

The potential energy deu to the Lorenz force is (Goldstein book):


and the kinetic energy is:


so the total energy $E=T+U$ is explicit independent from t,so:

$$\frac{d}{dt}\,E=0=\frac{d}{dt}\left(\frac{1}{2}\,m\,v^2+q\,\phi(x)-q\,A(x,t)\cdot\,v(t)\right)$$ or: $$\frac{d}{dt}\left(\frac{1}{2}\,m\,v^2+q\,\phi(x)-\underbrace{q\,\phi(x)}_{\ne f(t)}-q\,A(x,t)\cdot\,v(t)\right)=0$$

$\Rightarrow\quad$equation (18)



the equations of motion

$$\frac{d}{dt}\left(\frac{\partial L}{\partial v}\right)+\frac{\partial L}{\partial x}=0$$


$$\frac{\partial L}{\partial v}=m\,v+q\,A$$

$$\frac{\partial L}{\partial x} =\frac{\partial }{\partial x} \left[q\,(\phi-A\cdot v)\right]$$

$\Rightarrow\quad$equation (19)


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