# Magnetic and electric fields in relativity

• Frostman
In summary, the conversation discusses the motion of an object in the xy-plane with given initial conditions and a constant electric field. Using equations for momentum and energy, the expert summarizer presents a derivation for the expressions of x(t) and y(t) in terms of the initial conditions and time. The expert also notes that there may be a discrepancy in the final result when compared to a solution using arcsinh function. Further evaluation is needed for the cases of small velocity and large time.
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 ...

Last edited:
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?

Last edited:
anuttarasammyak said:
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)##?

## 1. What is the difference between magnetic and electric fields in relativity?

In relativity, magnetic and electric fields are actually two aspects of the same phenomenon known as the electromagnetic field. The main difference between them is the frame of reference in which they are observed. In one frame of reference, the field may appear purely electric, while in another it may appear purely magnetic. This is due to the effects of relativity on the perception of these fields.

## 2. How do magnetic and electric fields interact in relativity?

In relativity, magnetic and electric fields are not independent of each other. They are intertwined and can transform into one another depending on the observer's frame of reference. This is known as electromagnetic duality and is a fundamental concept in relativity.

## 3. Can magnetic and electric fields affect the passage of time in relativity?

Yes, magnetic and electric fields can affect the passage of time in relativity. This is because these fields are part of the fabric of spacetime and can warp it, causing time dilation. This effect is observed in high-speed particles moving through magnetic and electric fields.

## 4. How do magnetic and electric fields behave in the presence of massive objects in relativity?

In relativity, massive objects can also generate their own magnetic and electric fields. These fields can interact with the fields of other objects, causing complex interactions and effects. This is why understanding the behavior of magnetic and electric fields in the presence of massive objects is crucial in many areas of physics, such as astrophysics and particle physics.

## 5. Are magnetic and electric fields affected by the curvature of spacetime in relativity?

Yes, magnetic and electric fields are affected by the curvature of spacetime in relativity. This is because the electromagnetic field is a fundamental part of the fabric of spacetime, and any curvature or warping of spacetime will also affect the behavior of these fields. This is why Einstein's theory of general relativity is essential for understanding the behavior of magnetic and electric fields in the presence of massive objects.

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