Find the photon energy in the center of mass frame and vice versa

[lLehner95]

Homework Statement

In the process $\gamma +p\rightarrow \pi _{0}+p$, in the lab frame the proton is the fixed target, and the photon has energy E_{\gamma }=144,7 MeV. Find the photon energy $E_{\gamma }^{*}$ in the center of mass frame.

Relevant Equations

$E^{*}_{\gamma }=\gamma _{cm}E_{\gamma }-\beta _{cm}\gamma _{cm}p_{\gamma }$
$p_{\gamma }=\frac{E_{\gamma }}{c}$

I tried to use the Lorentz transformation:
$E^{*}_{\gamma }=\gamma _{cm}E_{\gamma }-\beta _{cm}\gamma _{cm}p_{\gamma }$
We have a photon, so it becomes:
$E^{*}_{\gamma }=\gamma _{cm}E_{\gamma }(\beta _{cm}-1)$

Unfortunately, the solutions say that the correct way is to use the inverse transormation:
$E_{\gamma }=\gamma _{cm}E^{*}_{\gamma }(\beta _{cm}+1)$
And the answer becomes:
$E^{*}_{\gamma }=\frac{E_{\gamma}}{\gamma_{cm}(1+\beta_{cm})}$

I already used the procedure to transform quantities (particles energy and momentum) from a reference to another, for example with the process $p+\bar{p}\rightarrow \Lambda +\bar{\Lambda }$. Why it is not possible in this case?

 

[PeroK]

I think the issue is the sign of the momentum. In your solution you have a negative energy. You have assumed that $|p| = p$; whereas in this case $E = |p|c = -pc$.

Alternatively, of course, you could have a negative value for $v$, hence $\beta$. In which case, if you take the modulus in your solution, you get the same as the book answer.

 

[vela]

We have a photon, so it becomes:
$E^{*}_{\gamma }=\gamma _{cm}E_{\gamma }(\beta _{cm}-1)$

It looks like you have a sign error here. It should be $E^{*}_{\gamma }=\gamma _{cm}E_{\gamma }(1-\beta _{cm})$.

You can get this to match the expression from the solutions:
\begin{align*}
E^{*}_{\gamma } &=\gamma _{cm}E_{\gamma }(1-\beta _{cm}) \\
&=\frac{\gamma _{cm}E_{\gamma }(1-\beta _{cm}^2)}{1+\beta_{cm}} \\
&=\frac{E_{\gamma }}{\gamma _{cm}(1+\beta_{cm})}
\end{align*}

 

[lLehner95]

Thank you, what a stupid error. I was so confused. Can you please confirm that, given a particle energy and momentum in the center of mass frame, the transformations to find the same quantities in the lab frame are (in natural units):

$E_{lab}=\gamma _{cm}E_{cm}+\beta _{cm}\gamma _{cm}p_{\parallel,cm}$
$p_{\parallel,lab}=\beta _{cm}\gamma _{cm}E_{cm}+\gamma _{cm}p_{\parallel,cm}$
$p_{\perp,lab}=p_{\perp,cm}$

And viceversa:

$E_{cm}=\gamma _{cm}E_{lab}-\beta _{cm}\gamma _{cm}p_{\parallel,lab}$
$p_{\parallel,cm}=-\beta _{cm}\gamma _{cm}E_{lab}+\gamma _{cm}p_{\parallel,lab}$
$p_{\perp,cm}=p_{\perp,lab}$

The $\beta _{cm}$ and $\gamma _{cm}$ are the same in both reference frames (only the sign of $\beta$ changes):
$\beta_{cm}=\frac{|\sum p_{i,\parallel }|}{\sum E_{i}}$
Where $p_{i,\parallel }$ are the momentum components parallel to the boost, so you have to take into account the sign.

...and if the particle is a photon $E_{\gamma}=p_{\gamma}$

(hope it will be useful for others in the future)

 

[PeroK]

Thank you, what a stupid error. I was so confused. Can you please confirm that, given a particle energy and momentum in the center of mass frame, the transformations to find the same quantities in the lab frame are (in natural units):

$E_{lab}=\gamma _{cm}E_{cm}+\beta _{cm}\gamma _{cm}p_{\parallel,cm}$
$p_{\parallel,lab}=\beta _{cm}\gamma _{cm}E_{cm}+\gamma _{cm}p_{\parallel,cm}$
$p_{\perp,lab}=p_{\perp,cm}$

And viceversa:

$E_{cm}=\gamma _{cm}E_{lab}-\beta _{cm}\gamma _{cm}p_{\parallel,lab}$
$p_{\parallel,cm}=-\beta _{cm}\gamma _{cm}E_{lab}+\gamma _{cm}p_{\parallel,lab}$
$p_{\perp,cm}=p_{\perp,lab}$

The $\beta _{cm}$ and $\gamma _{cm}$ are the same in both reference frames (only the sign of $\beta$ changes):
$\beta_{cm}=\frac{|\sum p_{i,\parallel }|}{\sum E_{i}}$
Where $p_{i,\parallel }$ are the momentum components parallel to the boost, so you have to take into account the sign.

...and if the particle is a photon $E_{\gamma}=p_{\gamma}$

(hope it will be useful for others in the future)

What scenarios are covered by these equations?

 

[lLehner95]

I'm using these equations to solve problems about collisions in particle physics: switching from the laboratory frame to the center of mass frame and viceversa.

 

[PeroK]

I'm using these equations to solve problems about collisions in particle physics: switching from the laboratory frame to the center of mass frame and viceversa.

Okay, I can see that much. So, if I have two particles, A and B, and in the CM frame they have energies $E_{A, cm}, E_{B, cm}$ and momenta $\vec{p}_{A, cm}, \vec{p}_{B, cm}$. Then, if we know the velocity of the CM frame in the lab frame, we can calculate the energy and momentum of the particles in the lab frame using your equations?

How are your equations different from the standard energy-momentum transformations?

Or, are we assuming one particle is at rest in the lab frame? Or, some other assumptions?

 

[lLehner95]

Yes. I wanted to find the most general transformations for energy and momentum of particles between the two reference frames. The energies and momenta are referred to a particle in a specific reference frame. I tried to solve some exercises and i got confused a little with signs. Then, solving more exercises, i understood the logic behind the transformations. The only difference is in the sign of beta. So, now i can say that these transformations are general, and for what concerne particle physics experiments, they hold in every configuration (fixed target or not).

 

[PeroK]

Yes. I wanted to find the most general transformations for energy and momentum of particles between the two reference frames. The energies and momenta are referred to a particle in a specific reference frame. I tried to solve some exercises and i got confused a little with signs. Then, solving more exercises, i understood the logic behind the transformations. The only difference is in the sign of beta. So, now i can say that these transformations are general, and for what concerne particle physics experiments, they hold in every configuration (fixed target or not).

Are you familar with four-vectors? Energy-momentum is a four-vector, hence its components transform according to the Lorentz Transformation.

What you have is one form of this Lorentz Transformation. This is covered here, for example:

https://hepweb.ucsd.edu/ph110b/110b_notes/node54.html
Note that as the sum of four-vectors is itself a four vector, then the total energy-momentum of a system of particles also transforms according to the Lorentz Transformation. This can be very useful. Note, however, that the invariant quantity $E_T^2 - c^2P_T^2$ is not the sum of the rest masses. It is sometimes called the "invariant" mass of the system (because it is the same in all reference frames). This is also very useful