Is power Lorentz invariant

  • Thread starter Heirot
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  • #1
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Power, defined as P = dE/dt is Lorentz invariant according to

http://farside.ph.utexas.edu/teaching/em/lectures/node130.html, Eq. 1645

But, considering another equation for the power, P = q E v, where E and v are electric field and velocity vectors, respectively; this is obviously not the case since in the momentarily rest frame of the charge P = 0?

What is wrong with the above reasoning?

Thanks!
 

Answers and Replies

  • #2
Don't think so, dE/dt is basically the time derivative of the 0th component of the four momentum integrated over a 3-volume. Components change from frame to frame.
Furthermore, the concept of "lorentz invariance" basically means that the geometric object stays the same in frames, only it's basis' and components transform to fit this new frame.
I could be wrong, anyone bother reading the entire article?
 
  • #3
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I would agree with you but I can't find an error in Richard Fitzpatrick's reasoning...
 
  • #4
PeterDonis
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But, considering another equation for the power, P = q E v, where E and v are electric field and velocity vectors, respectively; this is obviously not the case since in the momentarily rest frame of the charge P = 0?

Where are you getting this formula from?
 
  • #5
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The Lorentz force is given by F = q (E + v x B), and the power is P = F.v. This gives P = q E.v which is nowhere near being Lorentz invariant.
 
  • #6
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In general, power is certainly not invariant. It is the 0'th component of four-force, [itex]dp^\mu /d\tau[/itex].
 
  • #7
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Actually, the power is defined as a derivative wrt the coordinate, and not proper time.
 
  • #8
Everything needs to be put into covariant form before you can use them in SR, E and B fields are combined into the faraday tensor which is invariant. Classical equations should not be referred to in relavistic calculations.
 
  • #9
PAllen
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That website is not arguing that power, in general, is Lorentz invariant (that is absurd, as elfmotat has pointed out). It is arguing that:

Power radiated by a charged dipole (which has, among other features, that no momentum is carried away) is invariant.

The particular features of this problem are used throughout.

Radiated power for a situation where the radiation carries away net momentum would not meet the conditions of the derivation.
 
  • #10
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Everything needs to be put into covariant form before you can use them in SR, E and B fields are combined into the faraday tensor which is invariant. Classical equations should not be referred to in relavistic calculations.

I disagree. It is true that covariant Lorentz scalars are (by definition) Lorentz invariant. But that doesn't mean that there aren't other noncovariant Lorentz invariant quantities.
 
  • #11
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That website is not arguing that power, in general, is Lorentz invariant (that is absurd, as elfmotat has pointed out). It is arguing that:

Power radiated by a charged dipole (which has, among other features, that no momentum is carried away) is invariant.

The particular features of this problem are used throughout.

Radiated power for a situation where the radiation carries away net momentum would not meet the conditions of the derivation.

I fail to see where the argument that the power is Lorentz invariant is used only in connection with the radiating charge. As far as I can see the argument is related to the transformation properties of energy and time and not to the radiating charge.
 
  • #12
PAllen
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I fail to see where the argument that the power is Lorentz invariant is used only in connection with the radiating charge. As far as I can see the argument is related to the transformation properties of energy and time and not to the radiating charge.

Right here, the logic is only true for radiation carrying away no net momentum:

It follows from Sect. 10.22 that we can write $P^\mu = (d{\bf p}, dE/c)$, where $d{\bf p}$ and $dE$ are the total momentum and energy carried off by the radiation emitted between times $t=0$ and $t=dt$, respectively. As we have already mentioned, $d{\bf p} = 0$ in the instantaneous rest frame $S$. Transforming to an arbitrary inertial frame $S'$, in which the instantaneous velocity of the charge is $u$, we obtain
\begin{displaymath} dE^{'} = \gamma(u) \left(dE + u dp^1\right) = \gamma dE. \end{displaymath} (1644)

Note, this refers back to the earlier observation:

This is known as Larmor's formula. Note that zero net momentum is carried off by the fields (1628) and (1629).

In the general case, there is no frame where radiation carries no momentum (you can't talk about the 'rest frame' of a laser beam).
 
  • #13
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Right you are! The mystery is solved!

Thank you, PAllen :)
 

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