Magnetic and electric fields in relativity

[Frostman]

Homework Statement

In an inertial reference system there is a uniform and constant electric field $\vec E$ parallel to the x-axis. Given a particle of charge $q$ and mass $m$ that at the instant $t = 0$ is at the point $r_0 = (x_0, y_0, 0)$, with relativistic moment $p_0 = (p_{0x}, p_{0y}, 0)$.
1. Solve the relativistic equations of motion and derive the hourly law, that is $x(t)$ and $y(t)$.
2. Using the results of point 1., derive the Newtonian limit of the hourly law. Comment on the results obtained.
3. Using the results of point 2., calculate the hourly law at large $t$. Comment on the results obtained.

Relevant Equations

EOM

I started from:
$$\frac{d}{dt}(m\gamma v_x)=qE\ \ \ \rightarrow \ \ \ m\gamma v_x - p_{0x}=qE(t-0)\ \ \ \rightarrow \ \ \ m\gamma v_x=qEt+p_{0x}
$$$$\frac{d}{dt}(m\gamma v_y)=0\ \ \ \rightarrow \ \ \ m\gamma v_y -p_{0y}=0\ \ \ \rightarrow \ \ \ m\gamma v_y = p_{0y}$$$$\frac{d}{dt}(m\gamma v_z)=0\ \ \ \rightarrow \ \ \ m\gamma v_z = 0\ \ \ \rightarrow \ \ \ v_z=0$$
We observe first of all that the motion takes place only in the xy-plane.
We add and square the two relations:
$$m_vx=\frac{qEt+p_{0x}}{\gamma}\ \ \ \rightarrow \ \ \ m^2v_x^2=\frac{(qEt)^2+p_{0x}+2qEtp_{0x}}{\gamma^2}$$$$mv_y=\frac{p_{0y}}{\gamma}\ \ \ \rightarrow \ \ \ m^2v_y^2=\frac{p_{0y}^2}{\gamma^2}$$
$$m^2(v_x^2+v_y^2)=\frac{(qEt)^2+2qEtp_{0x}+p_{0x}^2+p_{0y}^2}{\gamma^2}\ \ \ \rightarrow \ \ \ \gamma^2=\frac{(qEt)^2+2qEtp_{0x}+p_0^2+m^2}{m^2}$$
$$\gamma=\frac 1m \sqrt{\mathcal{E}_0^2+(qEt)^2+2qEtp_{0x}t}$$

Where:

$v_x^2+v_y^2=v^2=\frac{\gamma^2-1}{\gamma^2}$
$p_{0x}^2+p_{0y}^2=p_0^2$
$\mathcal{E}_0^2=p_0^2+m^2$

Now we can substitute in the initial equations:
$$mv_x=\frac{qEt+p_{0x}}{\gamma}\ \ \ \rightarrow \ \ \ \frac{dx}{dt}=\frac{qEt+p_{0x}}{\sqrt{\mathcal{E}_0^2+(qEt)^2+2qEtp_{0x}t}}$$
$$mv_y=\frac{p_{0y}}{\gamma}\ \ \ \rightarrow \ \ \ \frac{dy}{dt}=\frac{p_{0y}}{\sqrt{\mathcal{E}_0^2+(qEt)^2+2qEtp_{0x}t}}$$

Integrating:
$$x-x_0=\int_0^t dt \frac{qEt+p_{0x}}{\sqrt{\mathcal{E}_0^2+(qEt)^2+2qEtp_{0x}t}}=\frac{\sqrt{\mathcal{E}_0^2+(qEt)^2+2qEtp_{0x}t}}{qE}\Bigg|_0^t=\frac{\sqrt{\mathcal{E}_0^2+(qEt)^2+2qEtp_{0x}t}-\mathcal{E}_0}{qE}$$
$$y-y_0=\int_0^t dt\frac{p_{0y}}{\sqrt{\mathcal{E}_0^2+(qEt)^2+2qEtp_{0x}t}}=\frac{p_{0y} \log({\sqrt{\mathcal{E}_0^2+(qEt)^2+2qEtp_{0x}t}+qEt+p_{0x}})}{qE}\Bigg|_0^t=\frac{p_{0y} \log({\sqrt{\mathcal{E}_0^2+(qEt)^2+2qEtp_{0x}t}+qEt+p_{0x}})-p_{0y}\log (\mathcal{E}_0+p_{0x})}{qE}$$

I find:
$$x(t)=\frac{\sqrt{\mathcal{E}_0^2+(qEt)^2+2qEtp_{0x}t}-\mathcal{E}_0}{qE}+x_0$$
$$y(t)=\frac{p_{0y} \log({\sqrt{\mathcal{E}_0^2+(qEt)^2+2qEtp_{0x}t}+qEt+p_{0x}})-p_{0y}\log (\mathcal{E}_0+p_{0x})}{qE}+y_0$$

But I don't find the error ...

 

[anuttarasammyak]

From the top three lines I observe
\frac{v_x}{\sqrt{1-v^2}}=\frac{qEt}{m}+\frac{p_{0x}}{m}
\frac{v_y}{\sqrt{1-v^2}}=\frac{p_{0y}}{m}
where
v^2=v_x^2+v_y^2
Adding squared formula side by side we get
v^2=\frac{Q}{1+Q}
where
Q:=(\frac{qEt}{m}+\frac{p_{0x}}{m})^2+(\frac{p_{0y}}{m})^2
So
v_x=[\frac{qEt}{m}+\frac{p_{0x}}{m}]\frac{1}{\sqrt{1+Q}}
v_y=\frac{p_{0y}}{m}\frac{1}{\sqrt{1+Q}}
The second equation says $v_y$ decreases with time. Conservation of y-momentum requires reducing $v_y$.

x(t)=\int_0^t [\frac{qEt}{m}+\frac{p_{0x}}{m}]\frac{1}{\sqrt{1+Q}}dt+x_0
y(t)=\int_0^t \frac{p_{0y}}{m}\frac{1}{\sqrt{1+Q}}dt +y_0

Introducing T
\frac{qE}{m}t+\frac{p_{0x}}{m}=\frac{qE}{m}T

x(t)=\int_{\frac{p_{0x}}{qE}}^{t+\frac{p_{0x}}{qE}} \frac{T}{\sqrt{T^2+\frac{1+(\ \frac{p_{0y}}{m}\ )^2}{(\frac{qE}{m})^2}}}dT+x_0
=[(t+\frac{p_{0x}}{qE})^2+\frac{1+(\frac{p_{0y}}{m})^2}{(\frac{qE}{m})^2} ]^{1/2}-[(\frac{p_{0x}}{qE})^2+\frac{1+(\frac{p_{0y}}{m})^2}{(\frac{qE}{m})^2} ]^{1/2} +x_0

y(t)=\int_{\frac{p_{0x}}{qE}}^{t+\frac{p_{0x}}{qE}} \frac{\frac{p_{0y}}{qE}}{\sqrt{T^2+\frac{1+(\frac{p_{0y}}{m})^2}{(\frac{qE}{m})^2}}}dT+y_0
=\frac{p_{0y}}{qE}[sinh^{-1} (\frac{|qE|T}{\sqrt{m^2+p_{0y}^2}})]_{\frac{p_{0x}}{qE}}^{t+\frac{p_{0x}}{qE}}+y_0

At the last integral calculation, you use log function form and I use arcsinh function form which are equivalent. Is this consistent with your result?

 

[Frostman]

From the top three lines I observe
\frac{v_x}{\sqrt{1-v^2}}=\frac{qEt}{m}+\frac{p_{0x}}{m}
\frac{v_y}{\sqrt{1-v^2}}=\frac{p_{0y}}{m}
where
v^2=v_x^2+v_y^2
Adding squared formula side by side we get
v^2=\frac{Q}{1+Q}
where
Q:=(\frac{qEt}{m}+\frac{p_{0x}}{m})^2+(\frac{p_{0y}}{m})^2
So
v_x=[\frac{qEt}{m}+\frac{p_{0x}}{m}]\frac{1}{\sqrt{1+Q}}
v_y=\frac{p_{0y}}{m}\frac{1}{\sqrt{1+Q}}
The second equation says $v_y$ decreases with time. Conservation of y-momentum requires reducing $v_y$.

x(t)=\int_0^t [\frac{qEt}{m}+\frac{p_{0x}}{m}]\frac{1}{\sqrt{1+Q}}dt+x_0
y(t)=\int_0^t \frac{p_{0y}}{m}\frac{1}{\sqrt{1+Q}}dt +y_0

Introducing T
\frac{qE}{m}t+\frac{p_{0x}}{m}=\frac{qE}{m}T

x(t)=\int_{\frac{p_{0x}}{qE}}^{t+\frac{p_{0x}}{qE}} \frac{T}{\sqrt{T^2+\frac{1+(\ \frac{p_{0y}}{m}\ )^2}{(\frac{qE}{m})^2}}}dT+x_0
=[(t+\frac{p_{0x}}{qE})^2+\frac{1+(\frac{p_{0y}}{m})^2}{(\frac{qE}{m})^2} ]^{1/2}-[(\frac{p_{0x}}{qE})^2+\frac{1+(\frac{p_{0y}}{m})^2}{(\frac{qE}{m})^2} ]^{1/2} +x_0

y(t)=\int_{\frac{p_{0x}}{qE}}^{t+\frac{p_{0x}}{qE}} \frac{\frac{p_{0y}}{qE}}{\sqrt{T^2+\frac{1+(\frac{p_{0y}}{m})^2}{(\frac{qE}{m})^2}}}dT+y_0
=\frac{p_{0y}}{qE}[sinh^{-1} (\frac{|qE|T}{\sqrt{m^2+p_{0y}^2}})]_{\frac{p_{0x}}{qE}}^{t+\frac{p_{0x}}{qE}}+y_0

At the last integral calculation, you use log function form and I use arcsinh function form which are equivalent. Is this consistent with your result?

I followed your steps and it all comes back to me, but I can't get from my result to yours.
In any case, I evaluated both in mine and in your result also the case in which $t = 0$. So both $x$ and $y$ are equal to their initial components.

Now, if I want to solve the II and III point, I should evaluate in the first case the fact that $v\ll 1$ and $t\rightarrow +\infty$ in the second case.

But I don't know how to proceed other than replacing $\mathcal{E}_0$ with $m$. How can I develop my $x(t)$ and $y(t)$?