# Kinetic Energy of Colliding Protons

#### Kharrid

Summary: Finding the KE of a two proton collision that creates Kaons. Given the rest KE of protons and kaons, what is the minimum KE of one proton that can create the two kaons.

In high-energy physics, new particles can be created by collisions of fast-moving projectile particles with stationary particles. Some of the kinetic energy of the incident particle is used to create the mass of the new particle. A proton-proton collision can result in the creation of a negative kaon (K−) and a positive kaon (K+):
p+p→p+p+K−+K+

Part A:
Calculate the minimum kinetic energy of the incident proton that will allow this reaction to occur if the second (target) proton is initially at rest. The rest energy of each kaon is 493.7 MeV, and the rest energy of each proton is 938.3 MeV. (Hint: It is useful here to work in the frame in which the total momentum is zero. Note that here the Lorentz transformation must be used to relate the velocities in the laboratory frame to those in the zero-total-momentum frame.)

Part C: Suppose that instead the two protons are both in motion with velocities of equal magnitude and opposite direction. Find the minimum combined kinetic energy of the two protons that will allow the reaction to occur.

So I'm trying to solve part A before doing part C. The answer given for part A is 2494 MeV, but I can't seem to get the right answer. I believe my conceptual understanding of the solution is not correct.

E(p) + E(p) = E(p) + E(p) + E(k) + E(k)

Since, only one proton is moving, I can say:
νmc^2 + 938.3 = 938.3 + 938.3 + 493.7 + 493.7

Solving for lorentz transformation gives me ν = 2.05.
Using ν, I plug it into the KE equation:
K=(ν-1)mc^2 * (1 MeV / 1.6*10^-13 J) = 983.4 MeV

983.4 MeV ≠ 2494 MeV

What am I doing wrong?

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#### vela

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You’re assuming all of the particles after the collision are at rest. That can’t be true if momentum is to be conserved.

#### PeroK

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Part A:
Calculate the minimum kinetic energy of the incident proton that will allow this reaction to occur if the second (target) proton is initially at rest. The rest energy of each kaon is 493.7 MeV, and the rest energy of each proton is 938.3 MeV. (Hint: It is useful here to work in the frame in which the total momentum is zero. Note that here the Lorentz transformation must be used to relate the velocities in the laboratory frame to those in the zero-total-momentum frame.)
The hint/clue was in the question.

#### Kharrid

The hint/clue was in the question.
I admit that I did not understand what the zero total momentum frame was, so I just glossed over it.

I'm trying to think what frame would make the total momentum zero ...

If momentum is p = γmv (where γ is lorentz transformation), then momentum is zero when either v = 0 or γ = 0.

I think the problem wants me to take the frame of the proton and somehow relate a momentum equation like:
p1 (proton1) + p2 (proton2) = p1 (proton 1) + p2 (proton2)
mv1 + 0 = mv2 + mv3

Is this correct or do I have to worry about whether the collision is elastic (first proton bounces back) or inelastic (proton sticks and moves)?

Or am I going in the wrong direction? I don't know where to go from here.

#### PeroK

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I admit that I did not understand what the zero total momentum frame was, so I just glossed over it.

I'm trying to think what frame would make the total momentum zero ...

If momentum is p = γmv (where γ is lorentz transformation), then momentum is zero when either v = 0 or γ = 0.

I think the problem wants me to take the frame of the proton and somehow relate a momentum equation like:
p1 (proton1) + p2 (proton2) = p1 (proton 1) + p2 (proton2)
mv1 + 0 = mv2 + mv3

Is this correct or do I have to worry about whether the collision is elastic (first proton bounces back) or inelastic (proton sticks and moves)?

Or am I going in the wrong direction? I don't know where to go from here.
The "zero-momentum" frame is one in which the total momentum is zero. Also know as the CoM (centre of momentum frame).

In classical physics, this would be easy. If the velocity of the incident proton is $v$, then the CoM frame would be moving at $v/2$ (for equal mass particles).

In SR, it's not so simple. It's based on the Lorentz Transformation.

This problem, therefore, splits into two parts:

1) Analyse the problem in the CoM frame, where the protons have equal and opposite momenta.

2) Transform that result to the lab frame using an "energy-momentum transformation".

Hint: note the total energy-momentum of a system of particles also transforms according to the Lorentz Transformation. Knowing that can be very useful.

#### PeroK

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Is this correct or do I have to worry about whether the collision is elastic (first proton bounces back) or inelastic (proton sticks and moves)?
In SR, the definite of an "elastic" collision is that rest mass is conserved. In this case, as the kaons are produced, it is not an elastic collision.

The assumption in this case is that if you have the threshold (minimum) energy for the kaon production, then all the KE goes to the rest mass of the kaons. (In the CoM frame.) In other words, all four particles are at rest in the final configuration (CoM frame).

#### Kharrid

The "zero-momentum" frame is one in which the total momentum is zero. Also know as the CoM (centre of momentum frame).

In classical physics, this would be easy. If the velocity of the incident proton is $v$, then the CoM frame would be moving at $v/2$ (for equal mass particles).

In SR, it's not so simple. It's based on the Lorentz Transformation.

This problem, therefore, splits into two parts:

1) Analyse the problem in the CoM frame, where the protons have equal and opposite momenta.

2) Transform that result to the lab frame using an "energy-momentum transformation".

Hint: note the total energy-momentum of a system of particles also transforms according to the Lorentz Transformation. Knowing that can be very useful.
So for part 1, the CoM frame gives us $p_{total}=0=p_1+p_2$. From this, the momentum of $p_1$ is equal to the momentum of $p_2$ even though $p_2$ is technically at rest initially.

Before the collision, $p_{total} = p_{i1}+p_{i2} = mv_{i1} + mv_{i2} = mv_{i1} + 0 = mv_{i1}$.
After the collision, $p_{total} = p_{f1}+p_{f2} = mv_{f1} + mv_{f2}$

The "revelation" (I think) is $mv_{i1} = mv_{f1} + mv_{f2}$ --> $v_{i1} = v_{f1} + v_{f2}$

To bring it back to the lab, I think I need to write the momentum equation again:
$p_{i1} + p_{i2} = p_{f1} + p_{f2} + 2E_{kaon}$
$mv_1 = γmv_1 + 2E_{kaon}$

Unfortunately, now I don't even have the KE.

Maybe I should write the energy equation?

$E_{p1} + E_{p2} = E{f1} + E_{f2} + 2E_{kaon}$
$γmv_{i1} + 0 = γmv_{f1} + γmv_{f2} + 2E_{kaon}$
$γmv_{i1} = γmv_{i1} + 2E_{kaon}$

Once again I lose the KE. Something seems wrong.

Also, thanks for the explanation on the collisions.

#### PeroK

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Be careful to use SR expressions in all cases.

The first step is to solve the energy equation in the CoM frame. That should be quite simple, if you look at it the right way.

Note that, for equal mass protons, their momentum is equal and opposite in the CoM frame, hence their energy is equal in that frame.

(Note: I am wondering whether you are a bit out of your depth with this problem, if you don't mind my saying that.)

#### Kharrid

Be careful to use SR expressions in all cases.

The first step is to solve the energy equation in the CoM frame. That should be quite simple, if you look at it the right way.

Note that, for equal mass protons, their momentum is equal and opposite in the CoM frame, hence their energy is equal in that frame.

(Note: I am wondering whether you are a bit out of your depth with this problem, if you don't mind my saying that.)
No problem, I am definitely quite lost. I'm not a physics major, yet circumstances demand I take Phys 3. I will admit I have completely forgotten the CoM concept.

Anyways ... solve the energy equation in the CoM. I think this means:
$KE = 0.5mv^2$
\-> $1/2mv_{i1}^2 + 1/2mv_{i2}^2 = 1/2mv_{f1}^2 + 1/2mv_{f2}^2$

The center of mass velocity is:
$v_{CM} = \frac{mv_1 + mv_2}{2m} = \frac{v_1 + v_2}{2} = \frac{v_1}{2}$

Since energy is conserved in CoM:
$KE = \frac{1}{2}mv_{CM}^2 = \frac{mv_1^2}{8}$

Am I on the right track? Now that I have energy, I have to somehow relate it with an energy-momentum equation ...

#### Kharrid

With the total energy, I have to bring it out of the lab frame. I think:
$E_{p1} + E_{p2} = E_{CoM} + 2E_{kaon}$
$γmc^2 + 938.3 = \frac{γmv_1^2}{8} + 2(493.7)$

Wait, that doesn't make sense, it should be:
$E_{CoM} = E_{p1} + E_{p2} + 2E_{kaon}$
$\frac{γmv_1^2}{8} = γmc^2 + γmc^2 + 2E_{kaon}$

It seems I still have a trailing $v_1$ in my energy equation. Do I still have an error in CoM?

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#### PeroK

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No problem, I am definitely quite lost. I'm not a physics major, yet circumstances demand I take Phys 3. I will admit I have completely forgotten the CoM concept.

Anyways ... solve the energy equation in the CoM. I think this means:
$KE = 0.5mv^2$
\-> $1/2mv_{i1}^2 + 1/2mv_{i2}^2 = 1/2mv_{f1}^2 + 1/2mv_{f2}^2$

The center of mass velocity is:
$v_{CM} = \frac{mv_1 + mv_2}{2m} = \frac{v_1 + v_2}{2} = \frac{v_1}{2}$

Since energy is conserved in CoM:
$KE = \frac{1}{2}mv_{CM}^2 = \frac{mv_1^2}{8}$

Am I on the right track? Now that I have energy, I have to somehow relate it with an energy-momentum equation ...
This is all, I'm sorry to say, classical physics. You'll need the SR expressions for energy and momentum.

$E = \gamma mc^2$ and $\vec p = \gamma m \vec{v}$

Let me do the easy bit for you:

In the CoM frame. I'll use $'$ to denote quantities in this frame. Ready for the inverse Lorentz Transformation.

By conservation of energy we have:

$E'_i = 2E'_p$ (initial energy is twice the energy of each proton).

$E'_f = 2m_p c^2 + 2m_k^2$ (final energy is the rest energy of the four particles, as in the minimum energy case they are all at rest after the collision)

Therefore:

$E'_p = (m_p + m_k)c^2$

The total energy of each proton is the rest energy of a proton plus the rest energy of a kaon. In other words, the KE of each proton is the rest energy of a kaon. And that's all very logical. All of the proton's KE is transformed into the rest energy of the kaon. That's where the kaon comes from.

That was the easy bit! The tricky bit is to transform this back to the lab frame.

But, I'd say you need to revise the Lorentz Transformation and SR energy-momentum. Try to get a grip on the concepts with some simpler problems.

#### PeroK

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With the total energy, I have to bring it out of the lab frame. I think:
$E_{p1} + E_{p2} = E_{CoM} + 2E_{kaon}$
$γmc^2 + 938.3 = \frac{γmv_1^2}{8} + 2(493.7)$

Wait, that doesn't make sense, it should be:
$E_{CoM} = E_{p1} + E_{p2} + 2E_{kaon}$
$\frac{γmv_1^2}{8} = γmc^2 + γmc^2 + 2E_{kaon}$

It seems I still have a trailing $v_1$ in my energy equation. Do I still have an error in CoM?
That's not the idea. The particles have energy-momentum in the lab frame $E_p, E_k$ etc. And, they have energy-momentum in the CoM frame $E'_p, E'_k$ etc.

These energy-momenta are related by a Lorentz Transformation (between the lab and CoM frames).

$E = \gamma(E' + vp')$

For each particle; and for the total:

$E_{tot} = \gamma(E'_{tot} + vp_{tot})$

Note that it's the inverse transformation to go from CoM to lab frame.

You need to know this stuff.

#### Kharrid

That's not the idea. The particles have energy-momentum in the lab frame $E_p, E_k$ etc. And, they have energy-momentum in the CoM frame $E'_p, E'_k$ etc.

These energy-momenta are related by a Lorentz Transformation (between the lab and CoM frames).

$E = \gamma(E' + vp')$

For each particle; and for the total:

$E_{tot} = \gamma(E'_{tot} + vp_{tot})$

Note that it's the inverse transformation to go from CoM to lab frame.

You need to know this stuff.
Unfortunately, the textbook doesn't go over many examples and sticks to the formulas with very little explanations. In class, we are so wrapped up with time dilation and length dilation that this isn't covered in enough depth. In any case, I'll read through the chapter one more time to hopefully get a better understanding and then come back here to solve the problem. Thanks for all the help so far!

#### Orodruin

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The first step is to solve the energy equation in the CoM frame.
It can be done in a pretty straightforward fashion without ever referring to the CoM frame. Unfortunately, that solution involves using Lorentz invariants, which is likely beyond the OP’s current level. It is a mystery to me why SR is generally taught in a convoluted way with focus on arbitrary coordinate dependent statements prone to extended misunderstandings instead of a concise geometric approach.

#### vela

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With the total energy, I have to bring it out of the lab frame. I think:
$E_{p1} + E_{p2} = E_{CoM} + 2E_{kaon}$
$γmc^2 + 938.3 = \frac{γmv_1^2}{8} + 2(493.7)$

Wait, that doesn't make sense, it should be:
$E_{CoM} = E_{p1} + E_{p2} + 2E_{kaon}$
$\frac{γmv_1^2}{8} = γmc^2 + γmc^2 + 2E_{kaon}$

It seems I still have a trailing $v_1$ in my energy equation. Do I still have an error in CoM?
Step back a bit here. It's easier if you write everything in terms of energies and momentums. Let's define the various energies that appear in the problem.
\begin{align*}
E_{p1i} &= \text{energy of proton 1 before the collision} \\
E_{p2i} &= \text{energy of proton 2 before the collision} \\
E_{p1f} &= \text{energy of proton 1 after the collision} \\
E_{p2f} &= \text{energy of proton 2 after the collision} \\
E_{k1f} &= \text{energy of kaon 1 after the collision} \\
E_{k2f} &= \text{energy of kaon 2 after the collision}
\end{align*} Conservation of energy then says
$$E_{p1i} + E_{p2i} = E_{p1f} + E_{p2f} + E_{k1f} + E_{k2f},$$ which is the idea you expressed in your original post, albeit with incorrect values for the energy.

This equation holds in all inertial frames of reference, but the actual values that you'd plug into the equation depends on the specific frame of reference you're working in. For example, in the lab frame, S, proton 2 is at rest, so $E_{p2i}$ would just be the rest energy, $m_p c^2$. In the zero-total-momentum frame, S', however, the two protons would have equal but opposite momenta before the collision, so the second proton would have to be moving so $E_{p2i} > m_p c^2$.

We can similarly define the various momenta that appear in the problem:
\begin{align*}
p_{p1i} &= \text{momentum of proton 1 before the collision} \\
p_{p2i} &= \text{momentum of proton 2 before the collision} \\
p_{p1f} &= \text{momentum of proton 1 after the collision} \\
p_{p2f} &= \text{momentum of proton 2 after the collision} \\
p_{k1f} &= \text{momentum of kaon 1 after the collision} \\
p_{k2f} &= \text{momentum of kaon 2 after the collision}
\end{align*} and conservation of momentum requires
$$p_{p1i} + p_{p2i} = p_{p1f} + p_{p2f} + p_{k1f} + p_{k2f}.$$

For simplicity, forget about the lab frame for now and concentrate on the zero-total-momentum frame. Let's say proton 1 moves initially with speed $v$ in the $+x$ direction. What's the initial speed and direction of proton 2? Tell us all of the values of energy and momenta you can deduce given that the protons are moving at the minimum speed to produce the two kaons.

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#### Kharrid

For simplicity, forget about the lab frame for now and concentrate on the zero-total-momentum frame. Let's say proton 1 moves initially with speed $v$ in the $+x$ direction. What's the initial speed and direction of proton 2? Tell us all of the values of energy and momenta you can deduce given that the protons are moving at the minimum speed to produce the two kaons.
This will probably be wrong, but hopefully you can understand my thought process.

\begin{align*}
E_{p1i} &= \text{γmc^2} \\
E_{p2i} &= \text{938.3 MeV} \\
E_{p1f} &= \text{moving with velocity v in +x direction?} \\
E_{p2f} &= \text{moving with velocity v in +x direction?} \\
E_{k1f} &= \text{493.7 MeV} \\
E_{k2f} &= \text{493.7 MeV}\\
p_{p1i} &= \text{momentum is v in +x, so } mv_{i1} \\
p_{p2i} &= \text{at rest, so v = 0} \\
p_{p1f} &= \text{shared momentum, so } mv_{f1} \\
p_{p2f} &= \text{shared momentum, so } mv_{f2} \\
p_{k1f} &= \text{at rest, so 0} \\
p_{k2f} &= \text{at rest, so 0}
\end{align*}

#### Kharrid

Part of the problem is that I don't know when to use γ lorentz transformation and when not to. For example, now that I reread some of the posts, I'm not sure if $p_{p1i}$ should be $γmv_{i1}$ or $mv_{i1}$. I think it SHOULD include the γ because the momenta are happening at some very high speeds where SR rules affect the outcome. This means all my momenta should some form of $γmv$.

Also, PeroK wrote earlier that the excess KE goes into the creation of the kaons. Does this mean that the final velocity of the protons is 0 and that final momentum is also 0? This doesn't make sense, of course, in the equations because the momenta would disappear on the right side of the equations, but I'm having trouble understanding what happens after the collision.

Now that I think more about it, I guess the protons would lose their kinetic energy, but would still have some potential energy since they are same charges and would want to repel each other. So the momentum of the protons is transferred either into the creation of the kaons (not sure how to represent this in the conservation of momentum equation) or the protons retain some momentum to repel each other.

#### vela

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You’re thinking from the perspective of the lab frame.

Consider the zero total momentum frame. If you set the total momentum to be zero, you get
$$p_{p1i}+p_{p2i} = 0.$$ If the momentum of the first proton isn’t 0, the momentum of the second cannot be zero.

#### Kharrid

You’re thinking from the perspective of the lab frame.

Consider the zero total momentum frame. If you set the total momentum to be zero, you get
$$p_{p1i}+p_{p2i} = 0.$$ If the momentum of the first proton isn’t 0, the momentum of the second cannot be zero.
Ok, let me rewrite the momentum equations then:

\begin{align*}
p_{p1i} &= -p_{p2f} \\
p_{p2i} &= -p_{p1i} \\
p_{p1f} &= -p_{p2f} \\
p_{p2f} &= -p_{p1f} \\
p_{k1f} &= \text{at rest, so 0} \\
p_{k2f} &= \text{at rest, so 0}
\end{align*}

#### vela

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Part of the problem is that I don't know when to use γ lorentz transformation and when not to. For example, now that I reread some of the posts, I'm not sure if $p_{p1i}$ should be $γmv_{i1}$ or $mv_{i1}$. I think it SHOULD include the γ because the momenta are happening at some very high speeds where SR rules affect the outcome. This means all my momenta should some form of $γmv$.
The Lorentz transformations are a set of equations that let you relate quantities in one frame to the same quantities in another. The quantity $\gamma$ is the Lorentz factor. You’re correct that the factor should always be there in the expression for momentum in this problem. You have to be careful, though, because there are multiple $\gamma$’s in this problem.

Also, PeroK wrote earlier that the excess KE goes into the creation of the kaons. Does this mean that the final velocity of the protons is 0 and that final momentum is also 0? This doesn't make sense, of course, in the equations because the momenta would disappear on the right side of the equations, but I'm having trouble understanding what happens after the collision.
Remember if you’re working in the zero-momentum frame, you need all the momenta to sum to zero after the collision.

Now that I think more about it, I guess the protons would lose their kinetic energy, but would still have some potential energy since they are same charges and would want to repel each other. So the momentum of the protons is transferred either into the creation of the kaons (not sure how to represent this in the conservation of momentum equation) or the protons retain some momentum to repel each other.

#### Kharrid

Remember if you’re working in the zero-momentum frame, you need all the momenta to sum to zero after the collision.
I am guessing that the momentum values are correct.

So, that leaves me with figuring out how to bring the momentum into the lab frame.

Conservation of Energy:
$E_{p1i} + E_{p2i} = E_{p1f} + E_{p2f} + 2E_{kaon}$
$γmc^2 + (938.3) = E_{p1f} + E_{p2f} + 2(493.7)$

My question is where to go from here because I don't know $v_{p1i}$ so I can't find $E_{p1i}$, but that's the unknown so that's ok. On the right side though, I don't know what to put for $E_{p1f}$ and $E_{p2f}$. What do I do next?

#### vela

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I am guessing that the momentum values are correct.
Kind of. You said the momentum of the kaons is 0. That's correct, but why?

So, that leaves me with figuring out how to bring the momentum into the lab frame.
Not yet. First you want to finish the analysis in the zero-momentum frame.

#### Kharrid

Kind of. You said the momentum of the kaons is 0. That's correct, but why?

Not yet. First you want to finish the analysis in the zero-momentum frame.
The momentum of the kaons is 0 because, when they are created, they are at rest. Since the equation for $p = γmv$ and v = 0, the momentum for the kaons is 0.

#### vela

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How do you know they are created at rest?

#### Kharrid

How do you know they are created at rest?
If you put it that way, then I guess I don't. I just reasoned that since they are created by the collision, their initial state is at rest. One of Newton's Laws states that an object in motion will stay in motion unless acted on by an equal and opposite force. Conversely, an object at rest will stay at rest unless acted on by an outside force. The Kaons are created, but are not affected by any other force.