Lienard's Formula from the Larmor Formula and Relativity

Question

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 (times 1/c) 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 $k^\mu$ or 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.

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.

https://media.physicsisbeautiful.com/resources/2019/02/18/solutions_manual.pdf $$\\$$ Also, see my related post: https://www.physicsforums.com/threads/use-relativity-and-the-larmor-formula-to-calculate-lienards-formula.1063692/

Answer 1

It looks like you had a fruitful discussion in the chat. Here, I talk about stuff unrelated to that.

I think you should be familiar that any 3-vector quantity in NR physics has to be upgraded to 4-vectors in SR. For example, the momentum $\vec p$ turns into $$\tag1p^\mu= \begin{pmatrix}E/c\\\vec p \end {pmatrix}=m_0\eta^\mu\qquad\implies\qquad\eta^\mu= \begin{pmatrix}c/\sqrt{1-v^2/c^2}\\\vec v/\sqrt{1-v^2/c^2} \end {pmatrix}$$ You also know that the appropriate way to upgrade power is $P=\frac{\mathrm dE}{\mathrm d\tau}=c\frac{\mathrm dp^0}{\mathrm d\tau}$, which motivates us trying to find a $K^\mu=\frac{\mathrm dp^\mu}{\mathrm d\tau}$ in the first place, allowing us to read off the time-component for the power.

Now, you seem to be confused as to where and what supplies the $a^2$. You are correct that, in Problem 12.38, there is a derivation that the 4-acceleration's time component is $$\tag2\alpha^t=\frac{\vec u\cdot\vec a}{c(1-u^2/c^2)^2}$$ and thus would vanish in the $\vec u=\vec0$ limit. But this is only just one kind of misunderstanding that you are having. The next part of Problem 12.38, it derived $\vec\alpha$ so that in part (b) it can compute $$\tag3\alpha^\nu\alpha_\nu=\alpha_\mu\alpha^\mu=\gamma^4\left[a^2+\frac{(\vec u\cdot\vec a)^2}{c^2-u^2}\right]$$

The rest of the argument is purely assembling some stuff. It is actually a guess, not a proper derivation. In the end, an argument like what Griffiths's is asking you to do, is only a generation of a possible 4-vector relation, and it is not at all proven that such a relation is what Nature would actually use. Instead, it just so happens that it is often going to give you the correct answer, because there are stringent criteria for alternative 4-vector relations to hold. But, because it is not a proper derivation, one has to either rederive the same relations properly in some other way, or compare that equation with experiment, and let experiment decide if it is correct or not.

One thing that you are likely confused about, is that, earlier, we said we want $K^\mu=\frac{\mathrm dp^\mu}{\mathrm d\tau}$, and we had $p^\mu=m_0\eta^\mu$. Then it stands to reason that we should have $K^\mu\propto\frac{\mathrm d\eta^\mu}{\mathrm d\tau}\propto\alpha^\mu$; that is, you are led to think that $K^t\propto\alpha^t$ and thus arose your confusion, that $\alpha^t$ vanishes as $\vec u=\vec0$ as in Equation (2)

However, the solution to Problem 12.71 is saying that $K^\mu\propto\eta^\mu$ contrary to your expectations. Instead, the requisite dependence upon acceleration is given by $\alpha^\nu\alpha_\nu$; That is, the wanted formula is actually a multiplication of Equation (3) with Equation (1) until something tolerable appears. There is no good rhyme or reason, except that we are interested in any ridiculous upgrade of the Larmor formula to 4-vector form.