Four-momenta trend as a function of proper time

[Frostman]

Homework Statement

In an inertial reference system, a particle of mass $m$ and charge $q$ is given, with initial speed $v(0) = (v_x (0); v_y (0); v_z (0))$. Furthermore, there is an electric field $E$, parallel to the y-axis, and a magnetic field $B$, parallel to the z-axis, both constant, homogeneous and such that $|E| = |B|$ in natural units.
Calculate the trend of the four-momenta $p^\mu$ as a function of the proper time $\tau$ and of the initial speed.

Relevant Equations

$\frac{dp^\mu}{d\tau}=qF^{\mu\nu}v_\nu$

As a starting point I immediately thought about the equation:

$\frac{dp^\mu}{d\tau}=qF^{\mu\nu}v_\nu$

From this I proceed component by component:

$\frac{dp^0}{d\tau}=qF^{0\nu}v_\nu=q\gamma E_yv_y$
$\frac{dp^1}{d\tau}=qF^{1\nu}v_\nu=q\gamma v_yB_z$
$\frac{dp^2}{d\tau}=qF^{2\nu}v_\nu=q\gamma (E_y-u_xB_z)$
$\frac{dp^3}{d\tau}=qF^{3\nu}v_\nu=0$

Now that I have the values I integrate:

$p^0=\int_{p^0(0)}^{p^0(\tau)}dp^0=\int_{0}^{\tau}d\tau q\gamma E_yv_y$
$p^1=\int_{p^1(0)}^{p^1(\tau)}dp^0=\int_{0}^{\tau}d\tau q\gamma v_yB_z$
$p^2=\int_{p^2(0)}^{p^2(\tau)}dp^0=\int_{0}^{\tau}d\tau q\gamma (E_y-u_xB_z)$
$p^3=\int_{p^3(0)}^{p^3(\tau)}dp^0=\int_{0}^{\tau}d\tau 0=0$

I have a problem solving integrals with respect to $\tau$, maybe I solved the last component, but I'm not sure:

$p^3(\tau) - p^3(0) = 0 \rightarrow p^3(\tau)= \text{cost} = m\gamma v_z(0)$

Can you tell me first of all if everything I have done is correct and if it remains for me to understand how to solve the integrals? If so, how can I proceed in solving these integrals? Is it okay that I analyze component by component or should I write it all in a formula like $\frac{dp^\mu}{d\tau}=qF^{\mu\nu}v_\nu$?

 

[Gaussian97]

My first advice is to write everything in terms of $p$, which is your goal. Notice that $p^\mu = m v^\mu$ so $$\frac{d p^\mu}{d \tau} = \frac{q}{m} F^{\mu\nu}p_{\nu}$$.
Now, using the fact that $|E|=|B|$, look more closely to the equations $\mu = 0$ and $\mu = 1$. Can you spot anything?

 

[Frostman]

My first advice is to write everything in terms of $p$, which is your goal. Notice that $p^\mu = m v^\mu$ so $$\frac{d p^\mu}{d \tau} = \frac{q}{m} F^{\mu\nu}p_{\nu}$$.
Now, using the fact that $|E|=|B|$, look more closely to the equations $\mu = 0$ and $\mu = 1$. Can you spot anything?

$\frac{dp^0}{d\tau}=\frac qmF^{0\nu}p_\nu=\gamma \frac qm E_yp_y$
$\frac{dp^1}{d\tau}=\frac qmF^{1\nu}p_\nu=\gamma \frac qm B_zp_y=\frac{dp^0}{d\tau}$

This from $|E|=|B|$ that is $E_y = B_z$.

 

[Gaussian97]

Ok, perfect, with that you can obtain very useful information that can be used in equation $\mu = 2$. (Specially if you rewrite it in therms of $p$)

 

[Frostman]

Ok, perfect, with that you can obtain very useful information that can be used in equation $\mu = 2$. (Specially if you rewrite it in therms of $p$)

Do you mean that?

$\frac{dp^2}{d\tau}=\frac qm F^{2\nu}p_{\nu}=\gamma \frac qm (E_y-p_xB_z)=\gamma \frac qm E_y(1-p_x)$

 

[Gaussian97]

Do you mean that?

$\frac{dp^2}{d\tau}=\frac qm F^{2\nu}p_{\nu}=\gamma \frac qm (E_y-p_xB_z)=\gamma \frac qm E_y(1-p_x)$

Well... not quite, the second equality is not correct. Also I'm telling you that the fact that
$$\frac{d p^0}{d \tau} = \frac{d p^1}{d \tau}$$ can be very useful there.

 

[Frostman]

Do you mean this equality?

$\frac qm F^{2\nu}p_\nu=\gamma \frac qm(E_y-p_xB_z)$?

 

[Gaussian97]

Yes, exact

 

[Frostman]

Could it be now?
$\frac qm F^{2\nu}p_\nu=\gamma \frac qm(mE_y-p_xB_z)$?

 

[Gaussian97]

No, still wrong. Try to write here all the intermediate steps, and we'll see where's the error.

 

[Frostman]

I'm using this values:
$F^{\mu\nu}=\begin{pmatrix} 0 & 0 & -E_y & 0\\ 0 & 0 & -B_z & 0\\ E_y & B_z & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}$

$p_\nu = (\gamma m, -\gamma p_x, -\gamma p_y, -\gamma p_z )$

So:

$F^{2\nu}p_\nu=E_y\gamma m - B_z\gamma p_x - 0 - 0=\gamma(E_y m -p_x B_z)$

 

[Gaussian97]

Well, look better than before.
This could be a matter of convention, but I haven't seen anyone writing $p_\nu = \gamma(m, -p_x, -p_y, -p_z)$, anyway since our aim is to compute $p_\nu(\tau)$ it is not a good idea to write them in terms of other things, so I would keep them as simply $p_\nu = (p_0, p_1, p_2, p_3) = (p^0, -p^1, -p^2, -p^3)$

 

[Frostman]

Well, look better than before.
This could be a matter of convention, but I haven't seen anyone writing $p_\nu = \gamma(m, -p_x, -p_y, -p_z)$, anyway since our aim is to compute $p_\nu(\tau)$ it is not a good idea to write them in terms of other things, so I would keep them as simply $p_\nu = (p_0, p_1, p_2, p_3) = (p^0, -p^1, -p^2, -p^3)$

Do you mean the fact of the indices in the four-vector expressed component by component or the minus signs from the second component onwards?

 

[Gaussian97]

I'm saying that I haven't seen anyone writing $p_1 = -\gamma p_x$, but this could be just a matter on how you define things.
In any case, with this definition of $p_\nu$, your expression is correct. But again, since we want to find the functions $p^\mu(\tau)$ I think it is better to keep them in our equations.

So, now you have $\frac{d p^2}{d\tau}=\gamma(E_y m -p_x B_z)$, after expressing them using $p^0$ and $p^1$ and using $|E|=|B|$, what do you notice?

 

[Frostman]

I have:

$\frac{dp^0}{d\tau}=\frac qmF^{0\nu}p_\nu=\gamma \frac qm E_yp_y$
$\frac{dp^1}{d\tau}=\frac qmF^{1\nu}p_\nu=\gamma \frac qm B_zp_y=\frac{dp^0}{d\tau}$
$\frac{dp^2}{d\tau}=\frac qmF^{2\nu}p_\nu=\gamma \frac qm (E_ym-p_xB_z)=\gamma \frac {qE_y}m (m-p_x)=\frac{1}{p_y}\frac{dp^0}{d\tau}(m-p_x)$

I'm not sure if the last step is what you ask me ...

 

[Gaussian97]

No, I'm asking you to write $\frac{d p^2}{d \tau}$ in terms of $p^0$ and $p^1$.

 

[Frostman]

In fact mixing the conventions used confuses me for a moment, it should be like this:

$\frac{dp^2}{d\tau}=\frac qmF^{2\nu}p_\nu=\gamma \frac qm (E_ym-p_xB_z)=\frac {qE_y}m(p^0-p^1)$

 

[Gaussian97]

Ok, that's the equation that will help us with the problem!
Take a moment to look at this equation with $\frac{d p^0}{d \tau} = \frac{d p^1}{d \tau}$ in mind. You should be able to find an interesting conclusion.

 

[Frostman]

Since they are infinitesimal quantities, I think it means that even if integrated for a "proper time" these quantities remain the same.

 

[Gaussian97]

Mmm... I am not sure what quantity you refer to as infinitesimal...
In any case if you still haven't spotted, try to compute $$\frac{d^2 (p^2)}{d \tau ^2}$$

 

[Frostman]

Mmm... I am not sure what quantity you refer to as infinitesimal...
In any case if you still haven't spotted, try to compute $$\frac{d^2 (p^2)}{d \tau ^2}$$

Yeah
$$\frac{d^2 (p^2)}{d \tau ^2}= \frac {qE_y}m\bigg(\frac{dp^0}{d\tau}-\frac{dp^1}{d\tau}\bigg)=0$$

 

[Gaussian97]

Ok, now should be straightforward to compute $p^2(\tau)$, which can the be used to compute $p^0$ and $p^1$.

 

[Frostman]

Yes, I can find:
$$p^0(\tau)= p^1(\tau) =\gamma \frac qm E_yp_y \tau=\gamma \frac qm B_zp_y \tau$$
$$p^2(\tau)= 0$$
$$p^3(\tau)= 0$$
Right?
I think we need to find an another way to write them as function not only of proper time but also of initial speed.

 

[Gaussian97]

No, that's not correct, how do you reach such conclusions?

 

[Frostman]

I just integrated this two times $\frac{d^2 (p^2)}{d \tau ^2}= 0$ but yes, it makes non-sense since $\frac{d (p^2)}{d \tau }$ must be a constant and $p^2$ a linear term in $\tau$...
Which quantity I need to integrate so? I understand the fact that $\frac{d^2 (p^2)}{d \tau ^2}= 0$ because of $\frac{d p^0}{d \tau}= \frac{d p^1}{d \tau}$

 

[Gaussian97]

I just integrated this two times $\frac{d^2 (p^2)}{d \tau ^2}= 0$ but yes, it makes non-sense since $\frac{d (p^2)}{d \tau }$ must be a constant and $p^2$ a linear term in $\tau$...
Which quantity I need to integrate so? I understand the fact that $\frac{d^2 (p^2)}{d \tau ^2}= 0$ because of $\frac{d p^0}{d \tau}= \frac{d p^1}{d \tau}$

Well, I don't know how you integrated, but as you say. The integral should give you $p^2$ as a linear function of $\tau$. In the same way the equation for $p^3$ doesn't imply that $p^3=0$.

 

[Frostman]

Well, I don't know how you integrated, but as you say. The integral should give you $p^2$ as a linear function of $\tau$. In the same way the equation for $p^3$ doesn't imply that $p^3=0$.

I have these relation:

$\frac{dp^0}{d\tau}=\gamma \frac qm E_yp^2$
$\frac{dp^1}{d\tau}=\gamma \frac qm B_zp^2$
$\frac{dp^2}{d\tau}=\frac {qE_y}m(p^0-p^1)$

$\frac{dp^0}{d\tau}=\frac{dp^0}{d\tau}$
$\frac{dp^2}{d\tau}=\frac {qE_y}m(p^0-p^1)$

I can integrate this:
$$\frac{dp^2}{d\tau}=\frac {qE_y}m(p^0-p^1)$$
$$\int_{p^2(0)}^{p^2(\tau)}dp^2=\int_{0}^{\tau}\frac {qE_y}m(p^0-p^1)$$
$$p^2(\tau)-p^2(0)=\frac {qE_y}m(p^0-p^1)\tau$$

 

[Gaussian97]

Ok, now you should relate $p^2(0)$ and $p^0-p^1$ with $\vec{v}(0)$.
Also, now you have $p^2(\tau)$ you can substitute in the equations for $\frac{d p^0}{d \tau}$ and $\frac{d p^1}{d \tau}$.

 

[Frostman]

Ok, now you should relate $p^2(0)$ and $p^0-p^1$ with $\vec{v}(0)$.
Also, now you have $p^2(\tau)$ you can substitute in the equations for $\frac{d p^0}{d \tau}$ and $\frac{d p^1}{d \tau}$.

It should be:
$$p^2(\tau)=\gamma (qE_y(1-v_x(0))\tau+mv_y(0))$$

One question: $p^0-p^1$ are $p^0(0)-p^1(0)$ or $p^0(\tau)-p^1(\tau)$?

 

[Gaussian97]

Yes, but just to be 100% clear, you should write $\gamma(0)$.

 

[Frostman]

I don't know if you see my last question in the previous post.

$$p^2(\tau)=\gamma(0) (qE_y(1-v_x(0))\tau+mv_y(0))$$

If now I integrate this:
$$\frac{dp^0}{d\tau}=\gamma(0) \frac qm E_yp^2$$
$$p^0(\tau)=\gamma(0) \bigg(\frac qm E_yp^2\tau + m\bigg)$$

$p^2$ in this case is $p^2(0)$ or $p^2(\tau)$? If is $p^2(\tau)$, why for $p^0-p^1$ we used $p^0(0)-p^1(0)=\gamma m (1-v_x(0))$?

 

[Gaussian97]

One question: $p^0-p^1$ are $p^0(0)-p^1(0)$ or $p^0(\tau)-p^1(\tau)$?

$p^0(\tau)-p^1(\tau)$, of course.

If now I integrate this:
$$\frac{dp^0}{d\tau}=\gamma(0) \frac qm E_yp^2$$
$$p^0(\tau)=\gamma(0) \bigg(\frac qm E_yp^2\tau + m\bigg)$$

Where does the $\gamma$ come from? And also how do you do that integral?

$p^2$ in this case is $p^2(0)$ or $p^2(\tau)$? If is $p^2(\tau)$, why for $p^0-p^1$ we used $p^0(0)-p^1(0)=\gamma m (1-v_x(0))$?

Again, $p^2 = p^2(\tau)$. The second question can be answered in different ways, but you should be able to answer it by looking at
$$\frac{d p^0}{d \tau} = \frac{d p^1}{d \tau}$$

 

[italicus]

Hi Frostman

have a look at this, mainly chapter 6 from page 89 on:

http://www.dfm.uninsubria.it/fh/FHpages/Teaching_files/appSR2.pdf

Furthermore, I’ like to carry your attention to the fact that the velocity $\vec v$ has three components , so what is $\gamma$ ? To which component does it refer ? No, you can’t do that way, the motion of the charged particle isn’t in the $x$ axis direction only.
Just use the law that you have written for $\frac { dp^{\mu}}{d\tau}$ using the Faraday tensor, the last one that takes into account that $\vec E$ is parallel to $y$ axis and $\vec B$ is parallel to the $z$ axis , and they have the same magnitude.

 

[Frostman]

Where does the $\gamma$ come from? And also how do you do that integral?

$\frac{dp^0}{d\tau}=\gamma \frac qm E_yp^2$

$p^0(\tau)-p^0(0)=\int_{0}^{\tau}d\tau\gamma \frac qm E_yp^2(\tau)$

$p^0(\tau)-p^0(0)=\int_{0}^{\tau}d\tau\gamma \frac qm E_y\gamma(0) (qE_y(1-v_x(0))\tau+mv_y(0))$

I need to solve this right?

Again, $p^2 = p^2(\tau)$. The second question can be answered in different ways, but you should be able to answer it by looking at
$$\frac{d p^0}{d \tau} = \frac{d p^1}{d \tau}$$

Yes. I understand this step.

 

[Gaussian97]

$\frac{dp^0}{d\tau}=\gamma \frac qm E_yp^2$

$p^0(\tau)-p^0(0)=\int_{0}^{\tau}d\tau\gamma \frac qm E_yp^2(\tau)$

$p^0(\tau)-p^0(0)=\int_{0}^{\tau}d\tau\gamma \frac qm E_y\gamma(0) (qE_y(1-v_x(0))\tau+mv_y(0))$

I need to solve this right?

Yes, that's the idea, but you have an extra $\gamma$ factor in $\frac{dp^0}{d\tau}=\gamma \frac qm E_yp^2$ which will complicate the integral. Try to compute $\frac{dp^0}{d\tau}$ again. Once you've done that, the integral becomes quite trivial and the problem is almost done.