Help with derivation of electric field of a moving charge

[GilSE]

Homework Statement

Derive electric field of moving charge formulas from Lienard-Wiechert potentials.

Relevant Equations

Formulas of electromagnetic field of moving charge to be derived:
$$\vec E=\frac q {\mu^3}\left((1-\beta^2)\vec R-R(1-\beta^2)\vec\beta\right)+\frac q{c\mu^3}\left(\vec R\times\left((\vec R-R\vec\beta)\times\vec\beta\right)\right)$$
$$\vec B =\frac{\vec R}{R}\times\vec E$$
$$\mu=R-\vec\beta\cdot\vec R$$
$\vec R$ is a difference between point, where the fields are calculated, and the point, where moving charge are.
Here and further, all values are considered taken in $t'$ moment (except those in general formulas for calculating EMF from potentials)
Values in the right side of these equations should be calculated at the moment $t'$ given by equation: $R(t')=c(t-t')$.
One can derive field equations from Lienard-Wiechert potentials:
$$\varphi = \frac q {\mu}$$
$$\vec A=\frac{q\vec\beta}{\mu}$$
By general formulas:
$$\vec E = -\nabla\varphi-\frac 1 c \frac{\partial\vec A}{\partial t}$$
$$\vec B=\nabla\times\vec A$$
Also I have these relations from my textbook:
$$\frac{\partial R}{\partial t'}=-c\frac{\vec\beta\cdot\vec R}R$$
$$\frac {\partial t'}{\partial t}=\frac R{\mu}$$

At first, I derived that:
$$\nabla \frac 1{\mu}=-\frac 1{{\mu}^3}\left((1-\beta^2)+\frac{\dot{\vec\beta}\cdot\vec R}c\right)\vec R$$
(dot means differentiation with respect to $t'$).
I assume this result is true because it gives valid result for magnetic field.

To find electric field one should also derive partial derivative of $\vec A$ with respect to $t$. I've used chain rule, substituted $\vec A$ and used derivative of product formula.
$$\frac {\partial \vec A}{\partial t}=\frac {\partial \vec A}{\partial t'}\frac {\partial t'}{\partial t}=q\frac {\partial t'}{\partial t}\left(\frac1\mu\frac{\partial\vec\beta}{\partial t'}-\frac 1{\mu^2}\frac{\partial\mu}{\partial t'}\vec\beta\right)=q\frac {\partial t'}{\partial t}\left(\frac1\mu\dot{\vec\beta}-\frac 1{\mu^2}\frac{\partial\mu}{\partial t'}\vec\beta\right)$$
Then calculated separately:
$$\frac{\partial\mu}{\partial t'}=\frac{\partial R}{\partial t'}-\frac{\partial(\vec\beta\cdot\vec R)}{\partial t'}=\frac{\partial R}{\partial t'}-\vec\beta\cdot\frac{\partial\vec R}{\partial t'}-\vec R\cdot\frac{\partial\vec\beta}{\partial t'}$$
Now use the fact that:
$$\frac{\partial R}{\partial t'}=-c\frac{\vec\beta\cdot\vec R}R$$
Also, obviously:
$$\frac{\partial \vec R}{\partial t'}=-c\vec\beta$$
Substituting these two formulas in the above one:
$$\frac{\partial\mu}{\partial t'}=-c\frac{\vec\beta\cdot\vec R}R+c\beta^2-\vec R\cdot\dot{\vec\beta}$$
And finally:
$$\frac {\partial \vec A}{\partial t}=q\frac {\partial t'}{\partial t}\left(\frac1\mu\dot{\vec\beta}-\frac 1{\mu^2}\frac{\partial\mu}{\partial t'}\vec\beta\right)=$$
$$=q\frac R\mu\left(\frac 1 \mu\dot{\vec\beta}+\frac 1 {\mu^2}\left(c\frac{\vec\beta\cdot\vec R}R-c\beta^2+\vec R\cdot\dot{\vec\beta}\right)\vec\beta\right)=$$
$$=q\frac R{\mu^3}\left(\mu\dot{\vec\beta}+\left(c\frac{\vec\beta\cdot\vec R}R-c\beta^2+\vec R\cdot\dot{\vec\beta}\right)\vec\beta\right)$$
Using expression for $\nabla\frac 1{\mu}$, electric field is:
$$\vec E=\frac q{{\mu}^3}\left((1-\beta^2)+\frac{\dot{\vec\beta}\cdot\vec R}c\right)\vec R-q\frac R{\mu^3}\left(\frac{\mu\dot{\vec\beta}}c+\left(\frac{\vec\beta\cdot\vec R}R-\beta^2+\frac{\vec R\cdot\dot{\vec\beta}}c\right)\vec\beta\right)=$$
$$=\frac q {\mu^3}\left((1-\beta^2)\vec R-R\left(\frac{\vec R\cdot\dot{\vec\beta}}R-\beta^2\right)\vec\beta\right)+\frac q {c\mu^3}\left((\dot{\vec\beta}\cdot R)\vec R-\mu R\dot{\vec\beta}-R(\dot{\vec\beta}\cdot R)\vec\beta\right)$$
The second part after simplification matches the second part of Lienard-Wiechert E-field equation, but the first slightly differs, and I can't find an error causing it. I'll appreciate any help with finding it.

 

[anuttarasammyak]

The first term has dimension of T^-1 though it should be non dimensional. You should investigate its derivation.

 

[TSny]

Besides the error pointed out by @anuttarasammyak, check your work for the following:

At first, I derived that:

I get an additional term proportional to $\vec{\beta}$. $$\nabla \frac 1{\mu}=-\frac 1{{\mu}^3}\left((1-\beta^2)+\frac{\dot{\vec\beta}\cdot\vec R}c\right)\vec R + \frac 1 {\mu^2} \vec{\beta}$$

I assume this result is true because it gives valid result for magnetic field.

Verify that the additional term does not alter the result for the magnetic field.

 

[GilSE]

The first term has dimension of T^-1 though it should be non dimensional. You should investigate its derivation.

Unfortunately, must be a typo. In my handwritten notes there is no dot above beta in this term. But still thank you for your answer!

 

[GilSE]

Besides the error pointed out by @anuttarasammyak, check your work for the following:

I get an additional term proportional to $\vec{\beta}$. $$\nabla \frac 1{\mu}=-\frac 1{{\mu}^3}\left((1-\beta^2)+\frac{\dot{\vec\beta}\cdot\vec R}c\right)\vec R + \frac 1 {\mu^2} \vec{\beta}$$

Verify that the additional term does not alter the result for the magnetic field.

Thank you very much, with that additional term the result looks correct at last! I will be searching why I haven't obtained it when was calculating gradient.

 

[TSny]

Thank you very much, with that additional term the result looks correct at last! I will be searching why I haven't obtained it when was calculating gradient.

Good. If you wish, you can post your work for $\nabla \mu$ and we can review it.