# Use relativity and the Larmor formula to calculate Lienard's formula

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Homework Statement
Use relativity and the Larmor formula to calculate Lienard's formula.
Relevant Equations
Larmor Formula, Lienard Formula
I am trying to understand the solution to exercise 12.71 in the document linked below which accompanies Griffith's book on electrodynamics. The problem states that we are to use the Larmor formula and relativity to derive the Lienard formula. $$Larmor \ {} formula: P = \frac{\mu_0q^2a^2}{6\pi c} \ {} when \ {} v = 0$$ $$Lienard's \ {} formula: P = \frac{\mu_0q^2\gamma^6}{6\pi c}\left (a^2 - \left|\frac{\mathbf v \ {} \times \ {} \mathbf a}{c}\right|^2\right)$$ In the book it mentions that ##k^\mu = \frac {dp^\mu}{d\tau}## is a four-vector and ##k^0 = mc\frac {d^2t}{d\tau^2} = \frac {1}{c}\frac{dE}{d\tau}##. As you can see in the solution, from this Griffiths begins searching for any four-vector he can think of whose time component is equal to the Larmor formula when v = 0, concluding that the time component of such a four-vector should be ##\frac{1}{c}\frac{dE}{d\tau}##. The four-vector he gives is $$k^\mu = \frac{1}{4 \pi \epsilon_0}\frac{2}{3}{q^2}{c^5}\alpha^v\alpha_v\eta^\mu$$ The time component equals the Larmor formula (times 1/c) when v = 0, and he calculates P in terms the time component and finds that P is equal to Lienard's formula.
$$\\$$ It's hard for me to see what motivated Griffiths to start searching for any four-vector he could think of whose time component is the Larmor formula (times 1/c) for v = 0, and then reach the conclusion that the time component must be ##\frac{1}{c}\frac{dE}{d\tau}##. Moreover, ##\frac{d^2t}{d\tau^2}## = 0 if v = 0, so isn't the proper power and therefore the ordinary power zero when v = 0, in contradiction to the Larmor formula? The four-vector he's constructed is not ##\frac {dp^\mu}{d\tau}##, and the time component is not ##mc\frac{d^2t}{d\tau^2}##, although his notation suggests otherwise, writing ##k^0## at the beginning of his solution, and later ##k^\mu## for his four-vector. So, I can't understand how he came up with this idea and conclusion, and how these formulas are consistent with ##\frac{d^2t}{d\tau^2} = 0## when v = 0 which should imply that the proper power and therefore the ordinary power are zero when v = 0.

https://media.physicsisbeautiful.com/resources/2019/02/18/solutions_manual.pdf

Last edited:
To clarify, in the book by Griffiths, we have ##p^\mu \equiv m\frac{d}{d\tau}\eta^\mu## and ##E \equiv cp^0##, so by definition, ##\frac{dE}{d\tau} = c\frac{dp^0}{d\tau} = m\frac{\mathbf u \cdot \mathbf a}{(1 - u^2/c^2)^2}##. Evaluating the proper power in a coordinate system where the particle is instantaneously at rest, ##\mathbf u = 0## we have ##\frac{dE}{d\tau} = 0##, which seems to disagree with the Larmor formula. I do not question the Larmor formula, but I'm lost on how we're defining energy and the power of the accelerating charged particle. Obviously, it couldn't be the presentation given above. Yet, that was simply how the energy is defined in the book and shouldn't be wrong by definition.

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