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Lorentz Force Law - Tensorial Form

  1. Sep 27, 2016 #1
    1. The problem statement, all variables and given/known data
    Show that the Lorentz Force Law, [itex]\frac{dp^{\nu}}{d \tau} = -q U_{\mu} F^{\mu \nu}[/itex], is consistent with [itex]P^\mu P_\mu= -m^2.[/itex] Here U is the 4-velocity, F is the Electromagnetic Tensor, and p is the 4-momentum. (Minkowski Space)

    2. Relevant equations
    As stated above.

    3. The attempt at a solution
    I initially tried differentiating [itex]p^\nu p_\nu = -m^2[/itex] w.r.t. tau. This gave me, through the product rule,
    [tex]p_\nu \frac{d p^\nu}{d \tau} + p^\nu \frac{d p_\nu}{d \tau} = 0 [/tex]. Since the only difference between the covariant and contravariant forms of the Lorentz Force Law is whether the [itex]\nu[/itex] on the F is raised of lowered. Substituting the Lorentz Force Law into the above, we get:
    [tex] -q ( U_\mu F^{\mu \nu} p_\nu + U_\mu F^\mu_\nu p^\nu) = 0 [/tex]
    -q cancels, leaving
    [tex] ( U_\mu F^{\mu \nu} p_\nu + U_\mu F^\mu_\nu p^\nu) = 0 [/tex]

    This is where I'm stuck. Both the left hand term and the right hand term are constants, and if I showed that one was the negative of the other I'd probably be set, but I'm not sure where to go from here.

    Thanks!
     
  2. jcsd
  3. Sep 28, 2016 #2

    Orodruin

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    First off, the metric is constant. What does this tell you about the derivative ##dp^2/d\tau##?

    Hint: use the fact that you can raise and lower indices using the metric.

    Second, what is the 4-momentum expressed using the mass and 4-velocity?
     
  4. Sep 28, 2016 #3
    Well, first off, considering that [itex] p^2 = p^\mu p_\mu = \eta_{\mu \nu} p^\mu p^\nu = m^2[/itex], the constancy of the metric just tells me that the derivative is zero.

    I suppose what I could do is take advantage of the fact that [itex] p_\nu = \eta_{\nu \alpha} p^{\alpha} = m \eta_{\nu \alpha} U^\alpha [/itex], and also of the fact that [tex] U^{\mu}U_{\mu} = -1[/tex], but I'm not sure of exactly how that would work with the indices.
     
  5. Sep 28, 2016 #4
    Well, first off, considering that [itex] p^2 = p^\mu p_\mu = \eta_{\mu \nu} p^\mu p^\nu = m^2[/itex], the constancy of the metric just tells me that the derivative is zero.

    I suppose what I could do is take advantage of the fact that [itex] p_\nu = \eta_{\nu \alpha} p^{\alpha} = m \eta_{\nu \alpha} U^\alpha [/itex], and also of the fact that [tex] U^{\mu}U_{\mu} = -1[/tex], but I'm not sure of exactly that would work with the indices.

    EDIT: oops! Accidental double post. I can't seem to delete it, so if someone with that ability could do that, that would be nice. Thanks.
     
  6. Sep 28, 2016 #5

    Orodruin

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    No it doesnt. It also tells you something about the form of the derivative regardless of the fact that p^2 = m^2.
     
  7. Sep 28, 2016 #6
    I've got to admit, I'm very confused. If I add the metric in, the form of [itex]\frac{\partial p^2}{\partial \tau}[/itex] is [tex]\frac{\partial p^2}{\partial \tau} = \frac{\partial \eta_{\mu \nu} p^\mu p^\nu}{\partial \tau} = \eta_{\mu \nu} p^\mu \frac{\partial p^\nu}{\partial \tau} + \eta_{\mu \nu} p^\nu \frac{\partial p^\mu}{\partial \tau}. [/tex]
    This can be re-written as [tex]\eta_{\mu \nu} p^\mu f^\nu + \eta_{\mu \nu} p^\nu f^\mu...[/tex]

    Or as [itex] 2 p_\mu f^\mu = -q U_\nu F^{ \nu \mu} p_{\mu} = 0. [/itex]

    I sort of just figured it out while typing. I feel like I'm probably (big if) on the right path now. Thank you!
     
  8. Sep 28, 2016 #7

    Orodruin

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    Right (apart from a factor of two in the middle step). Now you can use the relation ##p = mU##. What does the result tell you?

    Edit: Note that what you want to show is that left-hand side expression is equal to zero given the Lorentz force (otherwise it is incompatible with a fixed mass). Do not go around assuming that this is the case for the Lorentz force, you have to show it!
     
  9. Sep 28, 2016 #8
    I get [itex] m U_\mu f^\mu = -q m U_\nu F^{\nu \mu} U_\mu [/itex]. Looking just at the left hand side, we can see that [tex] m U_\mu f^\mu = m (f^0 U_0 + f^i U_i) = m(\frac{\partial E}{\partial \tau} \gamma + \frac{\partial E}{\partial \tau} \gamma^{-1}) = m \frac{\partial E}{\partial \tau} (\frac{\gamma^2+1}{\gamma})[/tex]

    However, in an object's rest frame, [itex]\frac{\partial E}{\partial \tau} = 0[/itex] I believe, due to conservation of energy. Therefore, the left hand side of the equation is equal to zero, showing that the law is consistent.
     
  10. Sep 28, 2016 #9

    Orodruin

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    You are seriously overthinking this problem. Just look at the symmetry properties of the right-hand side.

    Also, what you want to show is that the right-hand side is equal to zero. That the left-hand side is follows from ##p^2## being a constant.
     
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