Laws of conservation in special relativity

[LagrangeEuler]

Homework Statement

1. Proton of energy $76,4\mbox{GeV}$ colides with the proton with the proton that staying in the rest. How much proton - antiproton pairs is possible to generate?

Relevant Equations

$$E^2=E_0^2+p^2c^2$$
$E_0=0.94\mbox{GeV}$ - rest energy of proton and antiproton

Momentum $\vec{p}$ before collision is momentum of proton of the energy $E=76.4\mbox{GeV}$. Law of conservation of energy is
$$E+mc^2=E_1+E_2+...+E_n$$
$$mc^2=0.94\mbox{GeV}$$
We could generate only even number of particles after collision because of law of conservation of electric charge. Also there need to be two more protons then antiprotons. So, $n$ is even number and there is two more protons then antiprotons after the collision. Conservation of momentum gives
$$\vec{p}=\vec{p}_1+\vec{p}_2+\vec{p}_3+...+\vec{p}_n$$
I am not sure what to do next?
Also because
$$\frac{E+mc^2}{mc^2}=82,28$$
we now then number of particles is smaller then $82$ because law of conservation of momentum.

 

[PeroK]

I am not sure what to do next?

Do you know about the invariant mass of a system?

 

[LagrangeEuler]

Do you know about the invariant mass of a system?

No. I am not sure what do you think?

 

[PeroK]

No. I am not sure what do you think?

Do you know about the centre of momentum frame?

What should the system look like after the collision if all the excess kinetic energy has been converted to mass? I.e. if we have the threshold energy for $n$ particles.

 

[LagrangeEuler]

Do you know about the centre of momentum frame?

What should the system look like after the collision if all the excess kinetic energy has been converted to mass? I.e. if we have the threshold energy for $n$ particles.

Not really. What that means? Could you just give idea, or some reference?

 

[LagrangeEuler]

So I just google it. https://en.wikipedia.org/wiki/Center-of-momentum_frame

It is the system where some off all momentum is equall to zero. It is here very useful. But how could I go to that system here? I am not sure.

 

[PeroK]

So I just google it. https://en.wikipedia.org/wiki/Center-of-momentum_frame

It is the system where some off all momentum is equall to zero. It is here very useful. But how could I go to that system here? I am not sure.

If you think about the collision in the CoM frame. There is zero total momentum and $n$ particles. It would be good if all the particles were at rest after the collision (in the CoM frame). That would minimise the energy.

That means they are all moving with the same energy-momentum in the lab frame.

Does that help get you started?

 

[LagrangeEuler]

If you think about the collision in the CoM frame. There is zero total momentum and $n$ particles. It would be good if all the particles were at rest after the collision (in the CoM frame). That would minimise the energy.

That means they are all moving with the same energy-momentum in the lab frame.

Does that help get you started?

Yes. Thank you. I will try to finished it now. Could you please just tell me what do you mean by lab frame?

 

[PeroK]

Yes. Thank you. I will try to finished it now. Could you please just tell me what do you mean by lab frame?

The data in the question - one proton at rest and the other with an incident energy - is usually called the lab frame. It assumes that's how the experimenter has set it up.

 

[LagrangeEuler]

Center of mass and center of momentum. Is there any difference?

 

[PAllen]

Another approach it to look up what invariant mass of system means and how it is computed. That should get you to the complete solution very efficiently.

[edit: I see @PeroK already asked this].

 

[LagrangeEuler]

I tried to understand, but still have a problem.
In the laboratory system
$$E+E_0=\sqrt{E_0^2+p_1^2c^2}+\sqrt{E_0^2+p_2^2c^2}+...+\sqrt{E_0^2+p_n^2c^2}$$
$$\vec{p}=\vec{p}_1+\vec{p}_2+...+\vec{p}_n$$
and in COM frame where $$\vec{p}=\vec{0}$$
I get
$$2E_0=\sqrt{E_0^2+p_1'^2c^2}+\sqrt{E_0^2+p_2'^2c^2}+...+\sqrt{E_0^2+p_n'^2c^2}$$
$$\vec{0}=\vec{p}'_1+\vec{p}'_2+...+\vec{p}'_n$$
and it does not have sense. Where is the mistake? It is obvious that the mistake is in the equation
$$2E_0=\sqrt{E_0^2+p_1'^2c^2}+\sqrt{E_0^2+p_2'^2c^2}+...+\sqrt{E_0^2+p_n'^2c^2}$$. Could you please help me to understand where is the error?

 

[PeroK]

I tried to understand, but still have a problem.
In the laboratory system
$$E+E_0=\sqrt{E_0^2+p_1^2c^2}+\sqrt{E_0^2+p_2^2c^2}+...+\sqrt{E_0^2+p_n^2c^2}$$
$$\vec{p}=\vec{p}_1+\vec{p}_2+...+\vec{p}_n$$

What does this equation mean?

 

[LagrangeEuler]

Which one? First one is law of conservation of energy in the lab frame, and the second one is law of conservation of momentum. $E$ is energy of proton in motion, and $E_0$ is energy of proton that staying in rest. So, in the reference frame where momentum is zero vector, $E+E_0$ becoming $2E_0$.

 

[PeroK]

Which one? First one is law of conservation of energy in the lab frame, and the second one is law of conservation of momentum. $E$ is energy of proton in motion, and $E_0$ is energy of proton that staying in rest. So, in the reference frame where momentum is zero vector, $E+E_0$ becoming $2E_0$.

Okay, so $E_0 = mc^2$?

Why don't you use my hint:

If you think about the collision in the CoM frame. There is zero total momentum and $n$ particles. It would be good if all the particles were at rest after the collision (in the CoM frame). That would minimise the energy.

That means they are all moving with the same energy-momentum in the lab frame.

You don't need to use the CoM frame: we only needed that to justify having all the particles moving with the same energy-momentum in the lab frame.

 

[LagrangeEuler]

Yes. I am not sure what to do. I though I was doing what you suggested me.

 

[PeroK]

Yes. I am not sure what to do. I though I was doing what you suggested me.

What does "the particles all have the same energy-momentum" mean?

Let me tell you. It means $E_1 = E_2 = \dots = E_n$ and $p_1 = p_2 = \dots = p_n$.

That simplifies things a lot!

 

[LagrangeEuler]

It means that all particles have same energy.

 

[LagrangeEuler]

It means that energy of all of them is $E'=nE_1$, where $E_1$ is energy of one of them and $n$ is number of them.

 

[PeroK]

It means that energy of all of them is $E'=nE_1$, where $E_1$ is energy of one of them and $n$ is number of them.

That must help!

 

[LagrangeEuler]

Of course. But how I know that?

You said:"If you think about the collision in the CoM frame. There is zero total momentum and n particles. It would be good if all the particles were at rest after the collision (in the CoM frame). That would minimise the energy."
But why that should minimize the energy?

 

[PeroK]

But why that should minimize the energy?

Because there isn't any KE and you can't have less than that!

 

[etotheipi]

The total energy in the lab frame is always greater or equal to the total energy in the CM frame

 

[PeroK]

The total energy in the lab frame is always greater or equal to the total energy in the CM frame

It's fair to ask whether minimising the energy in the CoM frame is equivalent to minimising the energy in the lab frame. With a little bit of work you can see that this is true. However:

@LagrangeEuler I would accept that minimising the energy in the CoM frame is equivalent to minimising the energy in the lab frame. And now try to solve your equations.

 

[LagrangeEuler]

I just want to understand this. Before colision. I have two protons. Its energy is
$$E+mc^2$$
Its momentum is $$\vec{p}$$. What is the energy of this two protons in COM frrame? Thanks.

 

[PeroK]

I just want to understand this. Before colision. I have two protons. Its energy is
$$E+mc^2$$
Its momentum is $$\vec{p}$$. What is the energy of this two protons in COM frrame? Thanks.

You don't need to know that. Stick to the lab frame. It's not too hard to solve this problem in the CoM frame, but it's easier in the lab frame. Working out the transformation to the CoM frame is tricky.

As I said, the reason for thinking about the CoM frame was to convince yourself that the minimum-energy scenario is one where (in the lab frame) all particles move with the same velocity - hence, as they have the same mass, they all have the same energy-momentum.

 

[LagrangeEuler]

And solution is then.
$$E^2=E_0^2+p^2c^2$$
$$p=\frac{\sqrt{E^2-E_0^2}}{c}$$
Momentum of one of the proton after colision is
$$p_1=\frac{p}{n}=\frac{\sqrt{E^2-E_0^2}}{cn}$$
And law of conservation of energy
$$E+E_0=n\sqrt{E_0^2+p_1^2c^2}$$
From that I get
$$n^2=2+\frac{2E}{E_0}$$
and $n=12,8$.
So after colision it is maximum to get 7 protons and 5 antiprotons.

 

[LagrangeEuler]

You don't need to know that. Stick to the lab frame. It's not too hard to solve this problem in the CoM frame, but it's easier in the lab frame. Working out the transformation to the CoM frame is tricky.

As I said, the reason for thinking about the CoM frame was to convince yourself that the minimum-energy scenario is one where (in the lab frame) all particles move with the same velocity - hence, as they have the same mass, they all have the same energy-momentum.

I understand that. But I want to know.

 

[PeroK]

I understand that. But I want to know.

Post another homework problem.

The quickest way is to consider the energy-momentum transformation!

 

[PAllen]

I just want to understand this. Before colision. I have two protons. Its energy is
$$E+mc^2$$
Its momentum is $$\vec{p}$$. What is the energy of this two protons in COM frrame? Thanks.

Compute the invariant mass, as energy (i.e, times c squared if not using c=1) of the system in the lab frame. This equals total energy in the COM frame. Then, the original two protons each simply have half this energy in the COM frame. Get the momentum of each in the COM frame from this energy and their rest energy.

Note, this is why the whole problem becomes trivial using invariant mass or energy. The invariant energy of the system, computed in any frame, divided by the rest mass, is the maximum number of nucleons you can produce. This number, is, of course, frame invariant.

 

[vela]

in COM frame where $$\vec{p}=\vec{0}$$
I get
$$2E_0=\sqrt{E_0^2+p_1'^2c^2}+\sqrt{E_0^2+p_2'^2c^2}+...+\sqrt{E_0^2+p_n'^2c^2}$$
$$\vec{0}=\vec{p}'_1+\vec{p}'_2+...+\vec{p}'_n$$
and it does not have sense. Where is the mistake? It is obvious that the mistake is in the equation
$$2E_0=\sqrt{E_0^2+p_1'^2c^2}+\sqrt{E_0^2+p_2'^2c^2}+...+\sqrt{E_0^2+p_n'^2c^2}$$. Could you please help me to understand where is the error?

Before the collision, the two protons are not at rest, so the total energy is $2E'$, not $2E_0$. After the collision, the particles are at rest, so the energy of each particle is just $E_0$. The equation you wrote down has the initial protons at rest and the final particles all moving.

 

[Vitani1]

What you did in your first post (I think) is okay. Given that you are only required to compute the threshold energy to create an antiproton/proton pair (by threshold I mean you are not concerned with the kinetic energy of the pair after creating - you just need to calculate the energy required for them to exist) then your answer of 82 needs to be divided by 2 since you are dealing with pairs. At least this is what I think.