- #1

Frostman

- 115

- 17

- 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 ...

$$\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 ...

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