How to simplify \nabla (A.v) in the derivation of Lorentz force?

[jag]

Homework Statement

The problem I'm working on is on deriving the Lorentz force from a given relativistic Lagrangian. I have figured out most of it. The specific part that I'm unable to figure out is how to simplify $\nabla A \cdot v$ where A is the magnetic vector potential and v is the velocity.

Relevant Equations

Given in the attempted solution section

I know that $∇(A⋅v)=(A⋅∇)⋅v+(v⋅∇)⋅A+v×(∇×A)+A×(∇×v)$

The third term $v×(∇×A)$ simplifies to $v×B$. I'm just now sure how to "get rid" of the other terms. I tried checking for some vector identities but couldn't make any headways. Any guidance?

 

[Steve4Physics]

I know that $∇(A⋅v)=(A⋅∇)⋅v+(v⋅∇)⋅A+v×(∇×A)+A×(∇×v)$

The third term $v×(∇×A)$ simplifies to $v×B$. I'm just now sure how to "get rid" of the other terms. I tried checking for some vector identities but couldn't make any headways. Any guidance?

My (probably naive and definitely unrigorous) view is that $A$ is a vector-field with a vector-value at each point in space - but $v$ is not.

In fact $v$ is a simple vector - the instantaneous velocity of the particle; $v$ is not the local velocity of an element in a fluid say; there is no velocity-field.

Consequently, spatial derivatives of $v$ must be taken to be zero, so any terms which include div, grad or curl $v$ simply vanish.

Hopefully someone more knowledgable than I will confirm that - or correct it if it's wrong!

 

[jag]

@Steve4Physics Thank you for the reply. I had some thoughts similar to you but I am not sure whether it is correct. Another thought I was having is whether the terms are just "canceled" because of some vector identities but I'm not sure on this one too. Looking forward to some clarification from someone.

 

[TSny]

In general, the Lagrangian is a function of generalized coordinates $q_i(t)$, their time derivatives $\dot q_i(t)$, and the time $t$: $L \left[ q_i(t), \dot q_i(t), t \right]$.

For the particle moving in the EM field, it is natural to take the coordinates to be the Cartesian coordinates $x_i(t)$ of the position of the particle $\mathbf{x}(t)$.

So we have $L \left[x_i(t), \dot {x}_i(t), t \right]$; or, $L [ \mathbf{x}(t), \dot{\mathbf x}(t), t ]$.

If the term $\mathbf A \cdot \mathbf v$ occurs in the Lagrangian, then this should be interpreted as $\mathbf {A} [ \mathbf{x}(t), t] \cdot \dot{\mathbf x} (t)$ where the argument of $\mathbf{A}$ is shown explicitly. Here, $\mathbf{x}(t)$ and $\dot{\mathbf x} (t)$ are the position and velocity vectors of the particle at time $t$.

For less strain on the eyes, we can write $\mathbf {A} [ \mathbf{x}(t), t] \cdot \dot{\mathbf x} (t)$ as $\mathbf {A} ( \mathbf{x}, t) \cdot \dot{\mathbf x}$.
But it is very important to keep in mind that $\mathbf{x}$ and $\dot{\mathbf x}$ are functions of $t$.

For example, when setting up the Euler-Lagrange equation of motion for the coordinate $x_i\,$, you will need $$\frac {d}{dt} \left[\frac {\partial}{\partial \dot{x}_i} \left[ \mathbf {A} ( \mathbf{x}, t) \cdot \dot{\mathbf x} \right] \right] = \frac {d}{dt} A_i\left(\mathbf{x}, t\right) $$ In evaluating the right side, note that $t$ occurs in the argument of $A_i$ explicitly and also implicitly in $\mathbf{x}$.

You will also need $$\frac{\partial}{\partial x_i} \left[ \mathbf {A} (\mathbf{x}, t) \cdot \dot{\mathbf x} \right] = \left[ \frac{\partial \mathbf{A}( \mathbf{x}, t)}{\partial x_i} \right] \cdot \dot{\mathbf x} $$ In the Euler-Lagrange equations, the positions $x_i$ and the velocities $\dot{x}_i$ are treated as independent variables. So, the derivative $\large \frac{\partial}{\partial x_i}$ does not act on $\dot{\mathbf x}$, as already pointed out by @Steve4Physics.

 

[jag]

@TSny Thank you for the answer. It is clear to me why the spatial derivatives doesn't act on $\dot x$. One question, for the term (v⋅∇)⋅A, does the spatial derivative also doesn't act on $A$?

 

[PeroK]

@TSny Thank you for the answer. It is clear to me why the spatial derivatives doesn't act on $\dot x$. One question, for the term (v⋅∇)⋅A, does the spatial derivative also doesn't act on $A$?

$A$ is a function of position.

 

[TSny]

One question, for the term (v⋅∇)⋅A, does the spatial derivative also doesn't act on $A$?

Yes, the spatial derivative does act on $\mathbf{A}$. But I would not write the second dot in the expression. There is only one scalar product in the expression, which is the $\mathbf{v} \cdot \mathbf{\nabla}$ part. I would write the term as $(\mathbf{v} \cdot \mathbf{\nabla}) \mathbf{A}$.

You can show that this term gets canceled by other terms in the Euler-Lagrange equations. For example, the i^th component of $(\mathbf{v} \cdot \mathbf{\nabla}) \mathbf{A}$ will get cancelled by part of $\large \frac {d}{dt} \left[\frac {\partial}{\partial \dot{x}_i} \left[ \mathbf {A} ( \mathbf{x}, t) \cdot \dot{\mathbf x} \right] \right]$, which occurs in the Euler-Lagrange equation for $x_i(t)$.

Note that $\frac {d}{dt} \left[\frac {\partial}{\partial \dot{x}_i} \left[ \mathbf {A} ( \mathbf{x}, t) \cdot \dot{\mathbf x} \right] \right] =\frac {d}{dt} A_i\left(\mathbf{x}, t\right)$. The time derivative $\large \frac{d}{dt}$ is a total derivative, not a partial derivative. So, you need to take into account that $\mathbf{x}$ in the argument of $A_i$ depends on time.

 

[Orodruin]

But I would not write the second dot in the expression.

I would go so far as calling it wrong to write the second dot. The correct notation with respect to the vector structure is important in vector analysis.

 

[vanhees71]

@TSny Thank you for the answer. It is clear to me why the spatial derivatives doesn't act on $\dot x$. One question, for the term (v⋅∇)⋅A, does the spatial derivative also doesn't act on $A$?

I think you are a victim of the confusion with the partial derivatives wrt. to $q$ and $\dot{q}$ in the Lagrange formalism, which is partially due to the physicists' sloppy notation, which however is very convenient when you've gotten used to it.

The Lagrangian is a function of the generalized coordinates $q=(q_1,\ldots,q_f)$ and velocities $\dot{q}=(\dot{q}_1,\ldots,\dot{q}_f)$ (where $f$ is the number of degrees of freedoms), i.e., $L=L(q,\dot{q})$. Now when taking partial derivatives $\partial/\partial q_j$ and $\partial/\partial \dot{q}_j$ you have to consider the $q$ and $\dot{q}$ simply as independent variables (by definition of the physicists' sloppy notation).

On the other hand $q(t)$ are trajectories parametrized with time as the parameter, and then $\dot{q}(t)=\mathrm{d} q(t)/\mathrm{d} t$ are time derivatives of these trajectories. This time derivative is written as a total derivative, and then you get an expression like the Euler-Lagrange equations
$$\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{q}_j}=\frac{\partial L}{\partial q_j}.$$
On the left-hand side you first have to interpret the $\dot{q}_j$ as the one independent variable of the Lagrangian wrt. which you have to differentiate the Lagrangian, which gives again a function of the $q$ and $\dot{q}$ of course. Then the total time derivative tells you to now plug in the trajectory $q(t)$ for the $q$ and the time-drivatives $\mathrm{d} q(t)/\mathrm{d} t$ for the $\dot{q}$ and differentiate this expression with respect to $t$.

Now in your example you have the piece
$$L_{\text{mag}}=q \dot{\vec{x}} \cdot \vec{A}(\vec{x},t),$$
where now we use the cartesian coordinates $\vec{x}$ for the $q$'s, and we write $\vec{\nabla}=\partial/\partial \vec{x}$ for the partial derivatives with respect to these $q$'s. The $\dot{\vec{x}}$ have to be considered as indepenent variables when such partial derivatives are to be calculated. In this case it gets more clear in the Ricci notation, i.e., you write (with Einstein summation convention applied)
$$L_{\text{mag}}=q \dot{x}_j A_j (\vec{x},t)$$
Then
$$\frac{\partial L_{\text{mag}}}{\partial x_k}=q \dot{x}_j \frac{\partial}{\partial x_k} A(\vec{x},t).$$
Also the contribution of this term to the left-hand side of the Euler-Lagrange equations. You first need
$$\frac{\partial L_{\text{mag}}}{\partial \dot{x}_k} = q A_k(\vec{x},t).$$
Then you have to take the total time derivative, now putting $\vec{x}(t)$ into $A_k$ and use the chain rule,
$$\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L_{\text{mag}}}{\partial \dot{x}_k}=q \frac{\mathrm{d}}{\mathrm{d} t} A_k [\vec{x}(t),t] = q \dot{x}_l \frac{\partial}{\partial x_l} A_k[\vec{x}(t),t] + q \frac{\partial}{\partial t} A_k[\vec{x}(t),t].$$
In the final expression the partial time derivative only refers to the explicit time dependence of $A_k$!

 

[jag]

Hey All, thank you very much for the explanation. I took some time to digest all the information you gave me but it is super clear now. I successfully derived the Lorentz force!