How Do Euler-Lagrange Equations Apply in Electromagnetic Theory?

[dingo_d]

Homework Statement

When writing down the Lagrangian and the writing down Euler-Lagrange equation I'm having some difficulties with reasoning something.

Homework Equations

Lagrangian is:

$$\mathcal{L}=\frac{1}{2}mv^2-q\phi+\frac{q}{c}\vec{v}\cdot\vec{A}.$$

Euler-Lagrange eq:

$$\frac{\partial \mathcal{L}}{\partial x_i}=-q\frac{\partial \phi}{\partial x_i}+\frac{q}{c}\frac{\partial}{\partial x_i}(\vec{v}\cdot\vec{A})$$

$$\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial \dot{x}_i}\right)=m\ddot{x}_i+\frac{q}{c}\frac{\partial}{\partial t}A_i+\frac{q}{c}\sum_j\frac{\partial A_i}{\partial x_j}\dot{x}_j$$

Now my teaching assistant wrote that back in form of vectors, rather then component wise, and there was my puzzlement (or huh? moment):

$$\frac{d}{dt}(m\vec{v})+\frac{q}{c}\frac{\partial}{\partial t}\vec{A}+\frac{q}{c}(\vec{v}\cdot\vec{\nabla})\cdot\vec{A}+q\vec{\nabla}\phi-\frac{q}{c}\vec{\nabla}(\vec{v}\cdot\vec{A})=0$$

How is this:

$$\frac{q}{c}\sum_j\frac{\partial A_i}{\partial x_j}\dot{x}_j$$

equal to this:

$$\frac{q}{c}(\vec{v}\cdot\vec{\nabla})\cdot\vec{A}$$

??

Doesn't the derivative in the sum acts on A? And then the whole thing is multiplied with v?

Shouldn't it be:

$$\frac{q}{c}(\vec{\nabla}\cdot\vec{A})\cdot\vec{v}$$?

Because it is not the same if nabla acts on A and v acts on nabla... Or is it? :\

What am I missing?

 

[fzero]

You're just caught in a confusion over notation. Rearrange

$$\sum_j \frac{\partial A_i}{\partial x_j} \dot{x}_j} = \sum_j v_j \frac{\partial A_i}{\partial x_j} = \left( \sum_j v_j \frac{\partial}{\partial x_j}\right) A_i \rightarrow (\vec{v}\cdot \nabla) \vec{A}.$$

Note that $$(\vec{v}\cdot \nabla)$$ is already a scalar operator so there's no 2nd dot product. The derivative in this operator acts on everything to the right, but not on $$\vec{v}$$.

Now consider

$$(\nabla \cdot \vec{A})\cdot \vec{v}$$

There's a few problems with this formula. First of all, $$\nabla \cdot \vec{A}$$ would be a scalar, so the second dot product is incorrect and confusing. Second, if we dot $$\nabla$$ and $$\vec{A}$$ we get the divergence of $$\vec{A}$$ which involves contracting the vector index on $$A_i$$ which is contrary to the term you derived above.

 

[arkajad]

In

$$\frac{q}{c}\sum_j\frac{\partial A_i}{\partial x_j}\dot{x}_j$$

pay attention to the summation index (it is j). It indicates the scalar product.

 

[dingo_d]

Oh! I see now! The second dot product was probably my error in writing :\

Thanks ^^

 

[paranormal]

First of all, it is great that you are questioning and trying to understand the reasoning behind the equations. This shows that you are actively engaging with the material and that is an important aspect of being a scientist.

Now, let's address your confusion. The two expressions are indeed equal and here's why:

In the first expression, you have \frac{q}{c}\sum_j\frac{\partial A_i}{\partial x_j}\dot{x}_j. This can be rewritten as \frac{q}{c}\sum_j\dot{x}_j\frac{\partial A_i}{\partial x_j}. Now, if we define a vector \vec{v}=(\dot{x}_1, \dot{x}_2, \dot{x}_3), then the first expression becomes \frac{q}{c}\vec{v}\cdot\vec{\nabla}\cdot\vec{A}. This is exactly the same as the second expression, \frac{q}{c}(\vec{v}\cdot\vec{\nabla})\cdot\vec{A}. So, both expressions are equal.

The reason we can write the first expression in the form of a vector is because of the chain rule. When we take the derivative of a vector with respect to a scalar, we get a vector. In this case, the scalar is the position, x_i, and the vector is A_i. So, the derivative of the vector A_i with respect to the scalar x_i is a vector, \vec{\nabla}A_i. This is why we can write the first expression in the form of a vector.

In conclusion, both expressions are equivalent and can be used interchangeably. It is just a matter of preference in terms of notation. I hope this explanation helps clear up your confusion. Keep questioning and seeking to understand, it will only make you a better scientist.