# Proof of the transformation equation for electric field

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1. Apr 2, 2015

### Happiness

I want to prove that the transformation equation for electric field holds, namely Ex = Ex', where Ex is the x component of E field in S frame and Ex' is that in S' frame, for the special case in which the charged particle has a vertical velocity Uy' in the S' frame (Ux' = Uz' = 0). It is already proven in the textbook by Robert that the transformation equations (4-5) for Ex, Ey and Ez holds for the special case in which the charged particle is at rest in S' frame (U' = 0).

Equations (4-5) are

$E^\prime_x = E_x$
$E^\prime_y = \gamma(E_y - vB_z)$
$E^\prime_z = \gamma(E_z + vB_y)$

My approach is to have an S'' frame moving with respect to S' frame at V' = Uy' so that the charged particle is at rest in S'' frame. (S' frame is moving in the x direction at V with respect to S frame, while S'' frame is moving in the y direction with respect to S' frame.) Then I could use equations (4-5) to find E'' from E', and then use equations (4-5) again to find E from E'', hence relating E to E'. Since E'' is the frame the charged particle is at rest, equations (4-5) holds.

My result is that Ex and Ex' are not equal, contrary to the claim that equations (4-5) hold in general for any velocity of the charged particle.

Background reading: Introduction to Special Relativity by Robert Resnick (Wiley 1968) Page 163-165

My result:

By Equations (4-5) and considering $S^{\prime\prime}$ and $S^{\prime}$,
$E^{\prime\prime}_x = \gamma^\prime(E^\prime_x+v^\prime B^\prime_z)$
$E^{\prime\prime}_y = E^\prime_y$
$E^{\prime\prime}_z = \gamma^\prime(E^\prime_z-v^\prime B^\prime_x)$

Let $v_0$ be the velocity $S^{\prime\prime}$ moves with respect to S, and $\theta$ be the angle $v_0$ makes from the $x$ axis.
Let $x_0$ be the axis that points in the direction of $v_0$, and $y_0$ be the axis that is $90^\circ$ anticlockwise from $x_0$.

$E^{\prime\prime}_{x_0} = \gamma^\prime\cos\theta\ (E^\prime_x+v^\prime B^\prime_z)+\sin\theta\ E^\prime_y$
$E^{\prime\prime}_{y_0} = -\gamma^\prime\sin\theta\ (E^\prime_x+v^\prime B^\prime_z)+\cos\theta\ E^\prime_y$
$E^{\prime\prime}_{z_0} = \gamma^\prime(E^\prime_z-v^\prime B^\prime_x)$

By Equations (4-5) and considering $S$ and $S^{\prime\prime}$,
$E_{x_0} = E^{\prime\prime}_{x_0} = \gamma^\prime\cos\theta\ (E^\prime_x+v^\prime B^\prime_z)+\sin\theta\ E^\prime_y$
$E_{y_0} = \gamma_0(E^{\prime\prime}_{y_0}+v_0B^{\prime\prime}_{z_0}) = -\gamma_0\gamma^\prime\sin\theta\ (E^\prime_x+v^\prime B^\prime_z)+\gamma_0\cos\theta\ E^\prime_y+\gamma_0v_0B^{\prime\prime}_{z_0}$
$E_{z_0} = \gamma_0(E^{\prime\prime}_{z_0}-v_0B^{\prime\prime}_{y_0}) = \gamma_0\gamma^\prime(E^\prime_z-v^\prime B^\prime_x)-\gamma_0v_0B^{\prime\prime}_{y_0}$

$E_x = E_{x_0}\cos\theta-E_{y_0}\sin\theta = \gamma^\prime(\cos^2\theta+\gamma_0\sin^2\theta)(E^\prime_x+v^\prime B^\prime_z)+(1-\gamma_0)\sin\theta\cos\theta\ E^\prime_y-\gamma_0\sin\theta\ v_0B^{\prime\prime}_{z_0} \neq E^\prime_x$

Handwritten result in more details:

Last edited: Apr 2, 2015
2. Apr 2, 2015

### Orodruin

Staff Emeritus
Please write out the equations you are referring to, there are several people here who would be able to help you, but most will not have a copy of your textbook right next to them.

Also generally refrain from attaching photos of handwritten pages. It may be easy for you, but not to the people you are asking for help. Instead use the LaTeX feature to write down your important equations directly into the post.

3. Apr 2, 2015

### Happiness

Equations (4-5) are

$E^\prime_x = E_x$
$E^\prime_y = \gamma(E_y - vB_z)$
$E^\prime_z = \gamma(E_z + vB_y)$

My result:

By Equations (4-5) and considering $S^{\prime\prime}$ and $S^{\prime}$,
$E^{\prime\prime}_x = \gamma^\prime(E^\prime_x+v^\prime B^\prime_z)$
$E^{\prime\prime}_y = E^\prime_y$
$E^{\prime\prime}_z = \gamma^\prime(E^\prime_z-v^\prime B^\prime_x)$

Let $v_0$ be the velocity $S^{\prime\prime}$ moves with respect to S, and $\theta$ be the angle $v_0$ makes from the $x$ axis.
Let $x_0$ be the axis that points in the direction of $v_0$, and $y_0$ be the axis that is $90^\circ$ anticlockwise from $x_0$.

$E^{\prime\prime}_{x_0} = \gamma^\prime\cos\theta\ (E^\prime_x+v^\prime B^\prime_z)+\sin\theta\ E^\prime_y$
$E^{\prime\prime}_{y_0} = -\gamma^\prime\sin\theta\ (E^\prime_x+v^\prime B^\prime_z)+\cos\theta\ E^\prime_y$
$E^{\prime\prime}_{z_0} = \gamma^\prime(E^\prime_z-v^\prime B^\prime_x)$

By Equations (4-5) and considering $S$ and $S^{\prime\prime}$,
$E_{x_0} = E^{\prime\prime}_{x_0} = \gamma^\prime\cos\theta\ (E^\prime_x+v^\prime B^\prime_z)+\sin\theta\ E^\prime_y$
$E_{y_0} = \gamma_0(E^{\prime\prime}_{y_0}+v_0B^{\prime\prime}_{z_0}) = -\gamma_0\gamma^\prime\sin\theta\ (E^\prime_x+v^\prime B^\prime_z)+\gamma_0\cos\theta\ E^\prime_y+\gamma_0v_0B^{\prime\prime}_{z_0}$
$E_{z_0} = \gamma_0(E^{\prime\prime}_{z_0}-v_0B^{\prime\prime}_{y_0}) = \gamma_0\gamma^\prime(E^\prime_z-v^\prime B^\prime_x)-\gamma_0v_0B^{\prime\prime}_{y_0}$

$E_x = E_{x_0}\cos\theta-E_{y_0}\sin\theta = \gamma^\prime(\cos^2\theta+\gamma_0\sin^2\theta)(E^\prime_x+v^\prime B^\prime_z)+(1-\gamma_0)\sin\theta\cos\theta\ E^\prime_y-\gamma_0\sin\theta\ v_0B^{\prime\prime}_{z_0} \neq E^\prime_x$

4. Apr 5, 2015

### pervect

Staff Emeritus
You specified motion in the y-direction. It's not true that Ex = Ex' for motion in the y-direction, it's true for motion in the x direction. See for instance http://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity

5. Apr 5, 2015

### Staff: Mentor

Hi Happiness, this is related to the same problem in that other thread. The composition of two non-colinear boosts is not a boost, it is a boost and a rotation.

6. Apr 5, 2015

### Happiness

Hi DaleSpam, yes, I realised you answered both of my questions together. Thanks a lot! In fact, the problem in that other thread is a spin off from this one.

7. Apr 5, 2015

### Happiness

Hi pervect, thanks for answering! What Dalespam said is the resolution to the contradiction.

8. Apr 6, 2015

### Staff: Mentor

oh, then I am sorry I missed replying to this the first time and I am glad you asked again and I spotted it.