Vector Calculus Question in Lagrangian Mechanics

[Travis091]

Hi guys. I hope this isn't a bad place to post my question, which is:

I'm reading some lecture notes on Lagrangian mechanics, and we've just derived the Euler-Lagrange equations of motion for a particle in an electromagnetic field. It reads:

$$m \ddot{\vec{r}} = -\frac{e}{c} \frac{\partial \vec{A}}{\partial t}-e\nabla \phi(\vec{r})+ \frac{e}{c} \nabla (\dot{\vec{r}}\cdot \vec{A})$$

(A and phi are the vector and scalar potentials, respectively). Now the author switches to index notation, and I get lost in the process. The author gives:

$$m \ddot{r}^a = - \frac{e}{c} \frac{\partial A^a}{\partial t} - e \frac{\partial \phi(\vec{r})}{\partial r^a} + \frac{e}{c} \left( \frac{\partial A^b}{\partial r^a} - \frac{\partial A^a}{\partial r^b}\right) \dot{r}^b$$

My problem is with the third term above. Is the summation over b implied? I thought that it is bad form to repeat a summation index three or more times. Also, if we are to look at the a-th component of the gradient of the inner product, it should be:

$$\left(\nabla (\dot{\vec{r}}.\vec{A})\right)_a = \left(\nabla (\dot{r}_i A^i)\right)_a=\frac{\partial}{\partial r^a} \left ( \dot{r}_i A^i \right) = \dot{r_i}\frac{\partial A^i}{\partial r^a}$$

which is clearly not equal to the text. I have a feeling I'm doing something stupid, and it would be great if someone can point out my mistake(s).

Thanks.

 

[jedishrfu]

There a vector identity for that:

DEL (A.B) = (A.DEL) B + (B.DEL) A + A x (CURL B) + B x (CURL A)

see http://en.wikipedia.org/wiki/Vector_calculus_identities

 

[Travis091]

Yes I tried using the identity which you mention:

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

so:
$$\nabla(\mathbf{\dot{r}} \cdot \mathbf{A}) = (\mathbf{\dot{r}} \cdot \nabla)\mathbf{A} + (\mathbf{A} \cdot \nabla)\mathbf{\dot{r}} + \mathbf{\dot{r}} \times (\nabla \times \mathbf{A}) + \mathbf{A} \times (\nabla \times \mathbf{\dot{r}})$$

but I didn't get anywhere. I could have made a mistake while using this identity, I will try again later, but more importantly, why is my expression for the third term incorrect? That is what I would really like to understand. I want to clear out my misconception.

 

[samalkhaiat]

Hi guys. I hope this isn't a bad place to post my question, which is:

I'm reading some lecture notes on Lagrangian mechanics, and we've just derived the Euler-Lagrange equations of motion for a particle in an electromagnetic field. It reads:

$$m \ddot{\vec{r}} = -\frac{e}{c} \frac{\partial \vec{A}}{\partial t}-e\nabla \phi(\vec{r})+ \frac{e}{c} \nabla (\dot{\vec{r}}\cdot \vec{A})$$

Thanks.

This equation is wrong. Instead of the partial time derivative, there should be total time derivative of the vector potential. Then you can use
$$\frac{d \vec{ A }}{ d t } = \frac{ \partial \vec{ A } }{ \partial t } + ( \vec{ v } \cdot \vec{ \nabla } ) \vec{ A }$$

 

[Travis091]

That's it! Many thanks for your help. It was actually my mistake, the author skipped a step and jumped from the Lagrangian to the second equation above. I filled the missing step incorrectly, mixing up the total and partial derivatives.