Lorentz Force Law - Tensorial Form

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Homework Statement


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)

Homework Equations


As stated above.

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!
 

Answers and Replies

  • #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?
 
  • #3
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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.
 
  • #4
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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.
 
  • #5
Orodruin
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constancy of the metric just tells me that the derivative is zero
No it doesnt. It also tells you something about the form of the derivative regardless of the fact that p^2 = m^2.
 
  • #6
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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.
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!
 
  • #7
Orodruin
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Or as [itex] 2 p_\mu f^\mu = -q U_\nu F^{ \nu \mu} p_{\mu} = 0. [/itex]
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!
 
  • #8
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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!
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.
 
  • #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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