Lorentz transformation of electron motion

[PainterGuy]

Homework Statement

I'm trying to understand how different equations are derived for x, y, and z directions of motion of an electron.

Relevant Equations

Please have a look at the posting.

Hi,

It's not homework but I still thought I better post it here.

Please have a look on the attachment. For hi-resolution copy, please use this link: https://imagizer.imageshack.com/img922/7840/CL6Ceq.jpg

I think in equations labelled "12", 'e' is electric charge and Ex is the amplitude of electric field along 'x' direction.

How one could derive 14(i) and 14(ii) from 13(i) and 13(ii) respectively?

Thanks for the help!

 

[anuttarasammyak]

In the Einstein paper referred he said about longitudinal and transvers masses which are not used in current physics. If you are a first learner I do not recommend to go into that way though his physics is right and awesome.

 

[PainterGuy]

Yes, I'm a first learner and thank you for the advice.

I'm still curious about knowing how those expressions are derived.

 

[TSny]

I'm still curious about knowing how those expressions are derived.

Since this is not a homework problem, I will outline one approach.

From $t' = \gamma(t-xv/c^2)$, show $\frac{dt'}{dt} = \gamma(1-\frac{v}{c^2}\frac{dx}{dt})$

So, $\large \frac{dt}{dt'} = \frac{1}{\gamma(1-\frac{v}{c^2} \frac{dx}{dt})}$

From the chain rule, $\large \frac{d}{dt'} = \frac{dt}{dt'} \frac{d}{dt}$
So, $$\frac{d^2x'}{dt'^2} = \frac{dt}{dt'}\frac{d}{dt}\left[ \frac{dt}{dt'} \frac{dx'}{dt}\right]$$ Substitute $x' = \gamma(x-vt)$ and $\large \frac{dt}{dt'} = \frac{1}{\gamma(1-\frac{v}{c^2} \frac{dx}{dt})}$.

When performing the time derivatives $\large \frac{d}{dt}$ , treat $v$ as a constant since $v$ represents the fixed speed between the primed and unprimed inertial frames. Thus, you can also treat $\gamma$ as a constant. After doing all the differentiation, you can let $\frac{dx}{dt} = v$ since the instantaneous x-component of the velocity of the electron in the unprimed frame equals the relative speed of the two frames.

 

[PainterGuy]

Thank you, @TSny .

I tried to do it but getting '0' for the expression below.

\frac{dx^{\prime }}{dt}=\frac{d\left\{ \gamma (x-vt)\right\} }{dt}

I'm sorry if I'm making some obvious mistake.

Shown below is my full attempt.

 

[TSny]

I tried to do it but getting '0' for the expression below.

The mistake is at this point

You can't let $\frac{dx}{dt} = v$ until after you're done taking all of the derivatives.

It's sort of like making the mistake of finding $f''(2)$ for the function $f(x) = x^3$ as follows

$f'(x) = 3x^2$
$\therefore f'(2) = 12$
$\therefore f''(2) = \frac{d}{dx}f'(2) = \frac{d}{dx}(12) = 0$.

 

[PainterGuy]

Thank you!

Do you think I'm on the right path?

 

[TSny]

Your work looks correct to me. There are places where you could have canceled some $\gamma$'s between numerator and denominator. Also, there is a step where you multiply numerator and denominator by $c^2$. I would not have done that, but if it helps you in the simplification, OK.

Good. So, now you have completed the differentiations $\frac {d}{dt}$. You may now let $\large \frac{dx}{dt} = v$. The smoke is about to clear

 

[PainterGuy]

Your work looks correct to me. There are places where you could have canceled some $\gamma$'s between numerator and denominator. Also, there is a step where you multiply numerator and denominator by $c^2$. I would not have done that, but if it helps you in the simplification, OK.

Good. So, now you have completed the differentiations $\frac {d}{dt}$. You may now let $\large \frac{dx}{dt} = v$. The smoke is about to clear

Do you think it's correct? What am I supposed to do next?! Thanks a lot for your help and time.

You can check hi-res image here: https://imagizer.imageshack.com/img922/7309/8cHTMa.jpg

 

[TSny]

Looking good. In your final expression, multiply numerator and denominator by $1/c^4$.

 

[PainterGuy]

Looking good. In your final expression, multiply numerator and denominator by $1/c^4$.

Thank you but I tried to simplify the end expression earlier before I posted it here but multiplying by 1/c^4 doesn't help.

Anyway, you can see it for yourself below. I'm sorry if I'm missing something obvious.

 

[TSny]

$\large \frac 1 {c^4}$ $(c^2-v^2)^2 =$ $\large \frac{c^2-v^2}{c^2} \frac{c^2-v^2}{c^2}$ $= (1-v^2/c^2)(1-v^2/c^2)$

Express $1-v^2/c^2$ in terms of $\gamma$.

 

[PainterGuy]

Thanks a lot for your help and patience!

I think I got it now.

Note to self:
While deriving the expression above, the following relations were used as suggested by you.

Source: https://en.wikipedia.org/wiki/Lorentz_factor#Occurrence

Source: https://en.wikipedia.org/wiki/Lorentz_transformation

To derive 14(ii) m_{0} \gamma^{2} \frac{d^{2} y}{d t^{2}}=e E_{y}^{\prime} from 13(ii) m_{0} \frac{d^{2} y^{\prime}}{d t^{\prime 2}}=e E_{y}^{\prime}, should I proceed as follows.

<br /> \begin{aligned}<br /> &amp;\frac{d^{2} y^{\prime}}{d\left(t^{\prime}\right)^{2}}=\frac{d t}{d t^{\prime}} \frac{d}{d t}\left[\frac{d t}{d t^{\prime}} \frac{d y^{\prime}}{d t}\right]\\<br /> &amp;\text { Subbing the following expressions in the expression above. }\\<br /> &amp;y^{\prime}=y \text { and } \frac{d t}{d t^{\prime}}=\frac{1}{\gamma\left(1-\frac{v}{c^{2}} \frac{d y}{d t}\right)}<br /> \end{aligned}<br />

 

[TSny]

Your result for $\large \frac{d^2x'}{dt'^2}$ looks good.

To derive 14(ii) m_{0} \gamma^{2} \frac{d^{2} y}{d t^{2}}=e E_{y}^{\prime} from 13(ii) m_{0} \frac{d^{2} y^{\prime}}{d t^{\prime 2}}=e E_{y}^{\prime}, should I proceed as follows.

<br /> \begin{aligned}<br /> &amp;\frac{d^{2} y^{\prime}}{d\left(t^{\prime}\right)^{2}}=\frac{d t}{d t^{\prime}} \frac{d}{d t}\left[\frac{d t}{d t^{\prime}} \frac{d y^{\prime}}{d t}\right]\\<br /> &amp;\text { Subbing the following expressions in the expression above. }\\<br /> &amp;y^{\prime}=y \text { and } \frac{d t}{d t^{\prime}}=\frac{1}{\gamma\left(1-\frac{v}{c^{2}} \frac{d y}{d t}\right)}<br /> \end{aligned}<br />

Yes, for $\large \frac{d^2y'}{dt'^2}$ you would let $y' = y$. But $\large \frac{dt}{dt'}$ would still be $\large \frac{1}{\gamma(1-\frac{v}{c^2}\frac{dx}{dt})}$, not $\large \frac{1}{\gamma(1-\frac{v}{c^2}\frac{dy}{dt})}$.