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Computing a term in worldlines

  1. Sep 5, 2012 #1
    1. The problem statement, all variables and given/known data

    L is the langrangian

    [tex]\dot{q}[/tex] is the velocity with time derivative


    2. Relevant equations

    [tex]\frac{\partial L}{\partial \dot{q}} = -M \frac{1}{2} \frac{1}{\sqrt{1- \dot{q} \cdot \dot{q}}} \frac{\partial}{\partial q_i} (-\dot{q} \cdot \dot{q})[/tex]

    3. The attempt at a solution

    A few questions here. My work is telling me to calculate the right hand term

    [tex]\frac{\partial}{\partial q_i} (-\dot{q} \cdot \dot{q})[/tex]

    First of all, what is the dot in [tex]\dot{q} \cdot \dot{q}[/tex], is this just the dot product which would be the velocity squared?

    Secondly, the work demonstrates how to calculate the right hand side as

    [tex]\frac{\partial}{\partial q_i} (-\dot{q} \cdot \dot{q}) = -\frac{\partial}{\partial q_i}(\dot{q}_j \cdot \dot{q}_j) = -\delta_ij \dot{q}_j - \dot{q}_j \delta_ij[/tex]

    This part has me totally confused. What exactly is happening here... ? How do you get, mathematically from

    [tex]\frac{\partial}{\partial q_i} (-\dot{q} \cdot \dot{q})[/tex]

    then subscripts of j appear and a negative sign shows up..

    [tex]-\frac{\partial}{\partial q_i}(\dot{q}_j \cdot \dot{q}_j)[/tex]

    Then what exactly is this

    [tex]-\delta_ij \dot{q}_j - \dot{q}_j \delta_ij[/tex]

    Can some one help me please, it has me stuck in my tracks... :(
     
  2. jcsd
  3. Sep 5, 2012 #2

    vela

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    There's something wrong with your equation. You can't have a free index on one side of the equation and not the other.
    You're also missing dots all over the place in this post.

    All of the derivatives in your post should be with respect to ##\dot{q}_i##, not ##q## or ##q_i##.

    Yes.

    I'm not sure why you're confused about the negative sign. It was there right from the beginning, no?

    You're using implied-summation notation. Repeated indices indicate they should be summed over. In "regular" notation, you'd write
    $$\dot{\vec{q}}\cdot\dot{\vec{q}} = \sum_{j=1}^{n} \dot{q}_j \dot{q}_j.$$ Now what do you get if you differentiate this sum with respect to ##\dot{q}_i##? Keep in mind that the ##\dot{q}_i##'s are independent. Try it and compare what you get to the expression below.
    If you don't understand what ##\delta_{ij}## means, look up "Kronecker delta."
     
    Last edited: Sep 5, 2012
  4. Sep 7, 2012 #3
    Yes, I now realize that negative sign was there all along my fault.

    No, I am not asking about the Kronacker delta, I am asking why are we taking it away in this form? How do we end up with this

    [tex]\delta_{ij}\dot{q} - \dot{q}\delta_{ij}[/tex]

    How does one end up with that mathematically? Is there a significant name for this form? How does one get from [tex]\dot{q}\cdot \dot{q}[tex] to that above?
     
  5. Sep 7, 2012 #4

    vela

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    You calculate the derivative. There's nothing tricky going on here really other than the notation. Just differentiate ##\dot{\vec{q}}\cdot\dot{\vec{q}}## with respect to ##\dot{q}_i##.
     
  6. Sep 11, 2012 #5
    Just differentiation? I take it I will use the product rule?
     
  7. Sep 11, 2012 #6

    vela

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    Yes, since you're differentiating a sum of products, you need to use the product rule.
     
  8. Sep 14, 2012 #7
    See when I read this and thought, ''ok product rule'' I thought everything was sorted, but I when to calculate it but I am a bit lost.

    I got as far as [tex]-\dot{q} \cdot \dot{q} = -\dot{q}' \cdot \dot{q} - \dot{q} \cdot \dot{q}'[/tex]

    But what am I doing now exactly? As I said before, I know what the Kronacker Delta is, but why has it been attached to both sides of this?

    [tex]-\delta_{ij}\dot{q}_j - \dot{q}_j \delta_{ij}[/tex]

    What is the meaning behind this? I understand why the coefficient of 2 shows up, this is because the derivative of [tex]\dot{q}^2[/tex] is [tex]2\dot{q}[/tex], but why the delta sign?
     
  9. Sep 17, 2012 #8
    Hi Vela, just in case you forgot about my thread, above I had some more questions!

    Thanks!
     
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