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If we switch from one inertial frame to a different inertial frame with a relative velocity of $\mathbf{v}$, we could transform the scalar and vector potentials thusly:

$$\varphi' = \gamma \left( \varphi - \mathbf{A}\cdot \mathbf{v} \right) $$

$$\mathbf{A}' = \mathbf{A} - \frac{\gamma \varphi}{c^2}\mathbf{v} + \left(\gamma - 1\right) \left(\mathbf{A}\cdot\mathbf{\hat{v}}\right) \mathbf{\hat{v}}$$

Source: The Cambridge Handbook of Physics Formulas, G. Woan, Cambridge University Press, 2010, ISBN 978-0-521-57507-2.

It seems logical that the expression of the electric field in terms of the potentials would be unchanged before and after the Lorentz transformation:

$\mathbf{E} = -\mathbf{\nabla} \varphi - \frac{\partial \mathbf{A}}{\partial t}$

$\mathbf{E'} = -\mathbf{\nabla} \varphi' - \frac{\partial \mathbf{A'}}{\partial t'}$

Consider a straight line parallel to $\mathbf{\hat{z}}$ along which the scalar potential $\varphi$ and vector potential $\mathbf{A}$ both remain spatially-uniform and changing with time.

Let's have an independent inertial observer that moves with relative velocity $\mathbf{v}$ in a direction parallel to $\mathbf{\hat{z}}$

In this case, we can confirm rather easily on this straight line that minus the spatial-derivative of the electric scalar potential $-\nabla\varphi'$ is zero along $\mathbf{\hat{z}}$ both before and after the transformation (i.e. $\frac{\partial \varphi}{\partial z} = \frac{\partial \varphi'}{\partial z'} = 0$). On the other hand, can we say the same thing about minus the time-derivative of the magnetic vector potential $-\frac{\partial \mathbf{A'}}{\partial t'}$?

As a result of a Lorentz boost of velocity $\mathbf{v}$, the vector potential transforms by the amount:

$$\mathbf{A}' - \mathbf{A} = - \frac{\gamma \varphi}{c^2}\mathbf{v} + \left(\gamma - 1\right) \left(\mathbf{A}\cdot\mathbf{\hat{v}}\right) \mathbf{\hat{v}}$$

On the right-hand side of this equation, our only variables are $\varphi$ and $\mathbf{A}$. In contrast, $\gamma$, $\mathbf{v}$, and $\mathbf{\hat{v}}$ are all properties of the Lorentz boost and are therefore constant, as is $c^2$.

[Edit: Let's not concern ourselves with taking the derivative of this equation with respect to time. Let's consider the derivative of this equation with respect to changes of $\varphi$ or $\mathbf{A}$.]

Expansion of the first term on the R.H.S. reveals a leading term proportional to $\mathbf{v}$ and parallel to $\mathbf{v}$.

Expansion of the second term on the R.H.S. reveals a leading term proportional to $\mathbf{v^2}$ and parallel to $\mathbf{v}$.

This means that the transformation of the magnetic vector potential is such that it differs by a vector equal to $\mathbf{A}' - \mathbf{A}$ that is parallel to the Lorentz boost. Consequently, the derivative of $\mathbf{A}' - \mathbf{A}$ with respect to $\varphi$ or $\mathbf{A}$ is also parallel to the Lorentz boost. Since $\varphi$ and $\mathbf{A}$ may not be constant, this difference implies the existence of a contribution to the component of the electric field parallel to the Lorentz boost depending on our time-varying potentials $\varphi$ and $\mathbf{A}$.

The problem I see here is that if we add a zero-valued contribution to the electric field

$-\mathbf{\nabla_z} \varphi' - (- \mathbf{\nabla_z} \varphi) = 0$

with a non-zero contribution to the electric field

$- \frac{\partial \mathbf{A_z}'}{\partial t'} - (- \frac{\partial \mathbf{A_z}}{\partial t}) \neq 0$

in the direction of the Lorentz boost (parallel to $\mathbf{\hat{z}}$), the sum would be non-zero. How then could these transformations of the electromagnetic potentials [Edit: Clarification - …using the two formulas from "The Cambridge Handbook of Physics Formulas" presented at the top of this post…] be consistent with the fact that $\mathbf{E_\parallel}' = \mathbf{E_\parallel}$?

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  • $\begingroup$ So we cannot take the partial derivative of $\mathbf{A}' - \mathbf{A}$ with respect to $t$. Noted. Thankfully, doing so was not essential to the question I asked. What I really care about is how $-\nabla\varphi'$ and $-\frac{\partial\mathbf{A}'}{\partial t'}$ depend on $\varphi$ and $\mathbf{A}$, which I think should be possible to do without actually doing any derivatives of $t$ of any of the equations I presented. If I understand correctly, it should be possible to create an arbitrary variable for this purpose instead of $t$. $\endgroup$ – Kevin Marinas May 19 at 15:43
  • $\begingroup$ I have fixed the typos related to use of the same time variables (and spatial variables). So I have removed those serious errors. Also, I have now clarified that I am not at all trying to take any time derivative of the equation $\mathbf{A}' - \mathbf{A} = \mathbf{⋯}$. I hope this addresses all the concerns you have pointed out. $\endgroup$ – Kevin Marinas May 19 at 16:37
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It is easier to decompose each vector into parallel and perpendicular components, since the Lorentz transformations leave the perpendicular components unchanged.

Let us use the convention that bold-faced symbols are standard three component vectors. Denoting $\boldsymbol{\beta} = \mathbf v/c$ and a general four-vector by $f$ we shall use the following four-vector transformations:

\begin{align} \tag{1} \nabla &=\left(\frac{\partial}{\partial(ict)},\boldsymbol{\nabla}\right)\\[5pt] \tag{2} A & =\left(\frac{i \phi}{c},\mathbf{A} \right) \\[5pt] \tag{3} \mathbf{f'_{\parallel}} &= \gamma(\mathbf{f_{\parallel}}+i \boldsymbol{\beta} f_0)\\[5pt] \tag{4} f'_0 &= \gamma(f_0 - i \boldsymbol{\beta} \cdot \mathbf{f_{\parallel}}) \end{align}

Now we shall use the definition $$\mathbf{E'_{\parallel}} = - \nabla'_{\parallel} \phi' - \dfrac{\partial \mathbf{A'_{\parallel}}}{\partial t'}$$

and substitute eqn. $(1) - (4)$ in the above,

\begin{align} \tag{5} \mathbf{E'_{\parallel}} = -\gamma\left(\nabla_{\parallel}+\frac{\mathbf v}{c^2}\frac{\partial}{\partial t}\right)\gamma(\phi - \mathbf{v \cdot A}) - \gamma \left(\frac{\partial}{\partial t}+\mathbf v \cdot \nabla_{\parallel}\right)\gamma\left(\mathbf{A_{\parallel}}-\frac{\mathbf v}{c^2}\phi\right) \end{align}

The right side of eqn. $(5)$ when simplified generates eight terms, out of which two cancel each other because of $\pm (\mathbf{v}/c^2) \partial{\phi}/\partial{t}$. Also, the term with $(\mathbf{v}\cdot\nabla)\mathbf{A}_{\parallel}$ cancels with $-\nabla_{\parallel}(\mathbf{v}\cdot\mathbf{A})$ since $\mathbf{v}$ is a constant. In the end only four terms remain.

I will let you figure out that those terms can be reduced to $$ \tag{6} \mathbf{E'_{\parallel}} = -\gamma^2 \left(\nabla_{\parallel} \phi - \dfrac{\partial \mathbf{A_{\parallel}}}{\partial t} \right)(1 - v^2/c^2). $$

Note that we can use $\gamma^2 (1 - \beta^2) = 1$ in $(6)$ yielding us the desired result, $$ \tag{7} \mathbf{E'_{\parallel}} = -\nabla_{\parallel} \phi - \dfrac{\partial \mathbf{A_{\parallel}}}{\partial t} = \mathbf{E_{\parallel}}. $$

EDIT: In fact the transformations used here are same equations OP cited from The Cambridge Handbook of Physics Formulas.

Proof:

From eqns. $(2)$ and $(4)$, \begin{align} \frac{i\phi'}{c} &= \gamma\left(\frac{i\phi}{c} - i \frac{\mathbf{v}}{c}\cdot \mathbf{A}\right) \\[5pt] \phi' &= \gamma\left(\phi - \mathbf{v}\cdot\mathbf{A}\right) \end{align}

Also, from eqns. $(2), (3)$ and $(5)$, \begin{align} \mathbf{A'} &= \mathbf{A'_{\perp} + \mathbf{A'_{\parallel}}}\\[5pt] &= \mathbf{A_{\perp}} + \gamma \left(\mathbf{A_{\parallel}} + i \frac{\mathbf{v}}{c} \frac{i \phi}{c}\right)\\[5pt] &= \left(\mathbf{A_{\perp}} + \mathbf{A_{\parallel}}\right) + (\gamma-1)\mathbf{A_{\parallel}} - \frac{\gamma \phi}{c^2}\mathbf{v}\\[5pt] &= \mathbf{A} - \frac{\gamma \phi}{c^2} \mathbf{v} + (\gamma-1)(\mathbf{A}\cdot\mathbf{\hat{v}})\mathbf{\hat{v}} \end{align} where the last equality comes from resolving $\mathbf{A}$ into parallel component along $\mathbf{\hat v}$.

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  • $\begingroup$ @Frobenius, I have modified the wrong question and hopefully it is now at least "less wrong". That being said, I see that exp-ikx gave a through attempt answer, and while I appreciate it a lot, it still bothers me that the two equations from The Cambridge Handbook of Physics Formulas at the very beginning of my post could not somehow be used in the answer as these equations themselves are at the core of my inquiry, although admittedly I could have made that point clearer. $\endgroup$ – Kevin Marinas May 19 at 16:21
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    $\begingroup$ @KevinMarinas, see the latest edit. I have used the same as you cited. $\endgroup$ – Abhay Hegde May 19 at 19:27

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