Four-momenta trend as a function of proper time

[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.

 

[Frostman]

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.

$\gamma$ in this case is $\frac{1}{\sqrt{1-(v_x(0)^2+v_y(0)^2+v_z(0)^2)}}$

 

[Frostman]

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.

Okay, tomorrow I'll try to put it all together.

What I got for $p^3(\tau)$ is it right?
$p^3(\tau) - p^3(0) = 0 \rightarrow p^3(\tau)= \text{cost} = m\gamma v_z(0)$

 

[Gaussian97]

Yes, but again, you should specify $\gamma(0)$

$\gamma$ in this case is $\frac{1}{\sqrt{1-(v_x(0)^2+v_y(0)^2+v_z(0)^2)}}$

If this is your definition of $\gamma$, then it's okay to write $\gamma$ instead of $\gamma(0)$, but notice that after some time the "relativistic factor" (the one you need to use to compute time dilation, etc...) will no longer be $\gamma$.

 

[italicus]

@Frostman

please check the dimensions, because the mass $m$ is already contained in $\vec p$, so I think there is a dimensional mistake.
Make reference to $\frac{d\vec p}{dt} =q(\vec E +\vec v\times\vec B)$, then introduce the relativistic $\vec p$ and the Faraday tensor. If you take $m$ out of the derivative in the LHS and divide both members by it , the LHS becomes an acceleration, and the RHS contains q/m.
IMO.

 

[Gaussian97]

@Frostman

please check the dimensions, because the mass $m$ is already contained in $\vec p$, so I think there is a dimensional mistake.
Make reference to $\frac{d\vec p}{dt} =q(\vec E +\vec v\times\vec B)$, then introduce the relativistic $\vec p$ and the Faraday tensor. If you take $m$ out of the derivative in the LHS and divide both members by it , the LHS becomes an acceleration, and the RHS contains q/m.
IMO.

Which equation are you referring to?

 

[Frostman]

Good morning to both of you, I try to do a little order since between one answer and the other we had to correct some of my misunderstandings.

Let's start with the relation that describes the motion of a particle in an electromagnetic field:
$$\frac{dp^\mu}{d\tau}=qF^{\mu\nu}p_{\nu}=\frac qm F^{\mu\nu}p_\nu$$
Component by component:

$\frac{dp^0}{d\tau}=\frac qm E_yp_2(\tau)$
$\frac{dp^1}{d\tau}=\frac qm B_zp_2(\tau) \equiv \frac{dp^0}{d\tau}$
$\frac{dp^2}{d\tau}=\frac qm (E_yp_0(\tau)-B_zp_1(\tau))=\frac qmE_y (p_0(\tau)-p_1(\tau))$
$\frac{dp^3}{d\tau}=0$

Now, let's look at the first and second equations, due to the fact that $|E|=|B|$ we have:
$$\frac{dp^1}{d\tau}=\frac{dp^0}{d\tau}$$
Integrating:
$$p^0(\tau)-p^0(0)=p^1(\tau)-p^1(0)$$ $$p^0(\tau)-p^1(\tau)=p^0(0)-p^1(0)$$
This allows us first of all to integrate the following component:

$$\int_{p^2(0)}^{p^2(\tau)}dp^2=\int_0^\tau d\tau\frac qmE_y (p_0(\tau)-p_1(\tau))$$
Using the result just obtained:
$$\int_{p^2(0)}^{p^2(\tau)}dp^2=\int_0^\tau d\tau\frac qmE_y (p_0(0)-p_1(0))$$$$p^2(\tau)=\frac qmE_y (p_0(0)-p_1(0))\tau+p^2(0)$$$$p^2(\tau)=\gamma(0)\bigg[\frac qmE_y (m-mv_x(0))\tau+mv_y(0)\bigg]$$
Now we can go to evaluate the first two components of $p_\nu$:
$$\int_{p^0(0)}^{p^0(\tau)}dp^0=\int_0^\tau d\tau\frac qm E_yp_2(\tau)$$$$\int_{p^0(0)}^{p^0(\tau)}dp^0=\int_0^\tau d\tau\frac qm E_y\gamma(0)\bigg[\frac qmE_y (m-mv_x(0))\tau+mv_y(0)\bigg]$$$$p^0(\tau)=\frac qm E_y\gamma(0)\bigg[\frac qmE_y (m-mv_x(0))\frac{\tau^2}2+mv_y(0)\tau\bigg]+p^0(0)$$$$p^0(\tau)=\frac {q^2}{m^2} E_y^2\gamma(0)(m-mv_x(0))\frac{\tau^2}2+\frac qm E_y\gamma(0)mv_y(0)\tau+\gamma(0)m$$$$p^0(\tau)=\gamma(0)\bigg[\frac {q^2}{m} E_y^2(1-v_x(0))\frac{\tau^2}2+q E_yv_y(0)\tau+m\bigg]$$
Now for $p^1(\tau)$ we can use this relation that we find:
$$p^0(\tau)-p^1(\tau)=p^0(0)-p^1(0)$$ $$p^1(\tau)=p^0(\tau)-p^0(0)+p^1(0)$$$$p^1(\tau)=\gamma(0)\bigg[\frac {q^2}{m} E_y^2(1-v_x(0))\frac{\tau^2}2+q E_yv_y(0)\tau+mv_x(0)\bigg]$$
Dulcis in fundo:
$$\frac{dp^3}{d\tau}=0$$$$\int_{p^3(0)}^{p^3(\tau)}dp^3=0$$$$p^3(\tau)-p^3(0)=0$$$$p^3(\tau)=\gamma(0)mv_z(0)$$
So summing up the various components we have:
$$p^0(\tau)=\gamma(0)\bigg[\frac {q^2}{m} E_y^2(1-v_x(0))\frac{\tau^2}2+q E_yv_y(0)\tau+m\bigg]$$$$p^1(\tau)=\gamma(0)\bigg[\frac {q^2}{m} E_y^2(1-v_x(0))\frac{\tau^2}2+q E_yv_y(0)\tau+mv_x(0)\bigg]$$$$p^2(\tau)=\gamma(0)\bigg[qE_y (1-v_x(0))\tau+mv_y(0)\bigg]$$$$p^3(\tau)=\gamma(0)mv_z(0)$$
I hope everything is correct and did not make any miscalculations, let me know!

 

[Gaussian97]

Yes! This is almost perfect.
Fist a little mistake that is not important:

Let's start with the relation that describes the motion of a particle in an electromagnetic field:
$$\frac{dp^\mu}{d\tau}=qF^{\mu\nu}p_{\nu}=\frac qm F^{\mu\nu}p_\nu$$

It should be
$$\frac{dp^\mu}{d\tau}=qF^{\mu\nu}u_{\nu}=\frac qm F^{\mu\nu}p_\nu$$
But fortunately, this doesn't affect the rest of the computation.
Unfortunately, there is another mistake that does affect the final result

$$p^2(\tau)=\gamma(0)\bigg[\frac qmE_y (m-mv_x(0))\tau+mv_y(0)\bigg]$$
Now we can go to evaluate the first two components of $p_\nu$:
$$\int_{p^0(0)}^{p^0(\tau)}dp^0=\int_0^\tau d\tau\frac qm E_yp_2(\tau)$$$$\int_{p^0(0)}^{p^0(\tau)}dp^0=\int_0^\tau d\tau\frac qm E_y\gamma(0)\bigg[\frac qmE_y (m-mv_x(0))\tau+mv_y(0)\bigg]$$

Notice that you have computed $p^2(\tau)$, but inside the integral you have $p_2(\tau)$. What's the relation between these two?

 

[Frostman]

Yes! This is almost perfect.
Fist a little mistake that is not important:

It should be
dpμdτ=qFμνuν=qmFμνpν
But fortunately, this doesn't affect the rest of the computation.

Yep, I just forgot to change the name!

Unfortunately, there is another mistake that does affect the final result
Notice that you have computed p2(τ), but inside the integral you have p2(τ). What's the relation between these two?

A sign! I forgot to keep in mind that $p^2(\tau)=-p_2(\tau)$, right?

 

[Gaussian97]

Yes, and I think, after that everything is ok, unless @italicus wants to add something else that I have missed.

 

[Frostman]

Yes, and I think, after that everything is ok, unless @italicus wants to add something else that I have missed.

Okay, perfect. I await @italicus! In the meantime, I thank you and sorry for the stupid mistakes that I made yesterday, I was practicing all day and I was destroyed.

 

[Gaussian97]

No worries, we are here to help!

 

[italicus]

No worries, we are here to help!

I just wanted to evidence what @Gaussian97 told in #42. The mass m is part of $p^{\mu}$.