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.

 

[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 $p^{\mu}$.