Proof of Transformation Equation for Electric Field Ex

[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:

 

[Orodruin]

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.

 

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

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

 

[pervect]

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

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

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

 

[Dale]

Let v0 v_0 be the velocity S′′ S^{\prime\prime} moves with respect to S, and θ \theta be the angle v0 v_0 makes from the xx axis.
Let x0 x_0 be the axis that points in the direction of v0 v_0 , and y0 y_0 be the axis that is 90∘ 90^\circ anticlockwise from x0 x_0 .

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.

 

[Happiness]

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.

Hi DaleSpam, yes, I realized 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.

 

[Happiness]

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

My result:

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

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

 

[Dale]

Hi DaleSpam, yes, I realized 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.

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