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Vector Calculus Question in Lagrangian Mechanics

  1. Jul 6, 2014 #1
    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:

    [tex] 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}) [/tex]

    (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:

    [tex]
    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
    [/tex]

    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:

    [tex]
    \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}
    [/tex]

    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.
     
  2. jcsd
  3. Jul 6, 2014 #2

    jedishrfu

    Staff: Mentor

  4. Jul 6, 2014 #3
    Yes I tried using the identity which you mention:

    [tex] \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})[/tex]

    so:
    [tex] \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}})[/tex]

    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.
     
  5. Jul 6, 2014 #4

    samalkhaiat

    User Avatar
    Science Advisor

    This equation is wrong. Instead of the partial time derivative, there should be total time derivative of the vector potential. Then you can use
    [tex]\frac{d \vec{ A }}{ d t } = \frac{ \partial \vec{ A } }{ \partial t } + ( \vec{ v } \cdot \vec{ \nabla } ) \vec{ A }[/tex]
     
  6. Jul 6, 2014 #5
    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.
     
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