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.

I start with conservation of energy.
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?

[Moderator's note: Moved from a technical forum and thus no template.]

 

[vela]

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]

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]

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]

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]

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?

 

[PeroK]

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]

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]

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]

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.

 

[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]

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]

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]

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]

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.

 

[vela]

You can assume the kaons are at rest because the problem asks you to find the minimum energy of the incident proton. If there were more energy available, part of it would go to giving the kaons kinetic energy. Does that make sense?

What about the kinetic energy of the protons after the collision?

 

[Kharrid]

You can assume the kaons are at rest because the problem asks you to find the minimum energy of the incident proton. If there were more energy available, part of it would go to giving the kaons kinetic energy. Does that make sense?

What about the kinetic energy of the protons after the collision?

Yes that makes sense. We are trying to find enough energy to create the kaons, but not enough to actually pass momentum to them.

Since we only want enough energy to create the kaons, all of the kinetic energy would be transferred to the kaons and their creation. Therefore, the protons after the collision would be at rest since all of the energy was transferred to the kaons.

 

[vela]

Right, so after the collision, the protons and kaons are at rest in the zero-momentum frame. At this point, you should be able to deduce all of the quantities on the right-hand sides of the two equations.

Can you figure out what energy each proton has before the collision?

 

[Kharrid]

Right, so after the collision, the protons and kaons are at rest in the zero-momentum frame. At this point, you should be able to deduce all of the quantities on the right-hand sides of the two equations.

Can you figure out what energy each proton has before the collision?

Rest energy of Kaon is 493.7 MeV, and the rest energy of each proton is 938.3 MeV.

Are these the outcomes of the two equations?
$p_{p1i} + p_{p2i} = p_{p1f} + p_{p2f} + 2*p_{kaon}$
$p_{p1i} + p_{p2i} = 0$
$p_{p1i} = -p_{p2i}$

$E_{p1i} + E_{p2i} = E_{p1f} + E_{p2f} + 2*E_{kaon}$
$E_{p1i} + 938.3 = 938.3 + 938.3 + 2*(493.7)$
$E_{p1i} = 1925.7 MeV$
$E_{p2i} = 938.3 MeV$

This still seems off. I think E_{p2i} should not be the rest value of a proton because in the momentum equation, we found that the initial value of proton 2 is actually some momentum (i.e. not 0). The problem is that I don't know how to relate the energy equation of proton 2 to its momentum in the first equation.

 

[vela]

This still seems off. I think E_{p2i} should not be the rest value of a proton because in the momentum equation, we found that the initial value of proton 2 is actually some momentum (i.e. not 0). The problem is that I don't know how to relate the energy equation of proton 2 to its momentum in the first equation.

You're right about $E_{p2i}$. The momentum equation tells you that the two protons have momenta of the same magnitude, just in opposite directions. What does that imply about their speeds?

 

[Kharrid]

You're right about $E_{p2i}$. The momentum equation tells you that the two protons have momenta of the same magnitude, just in opposite directions. What does that imply about their speeds?

This means that the speed of proton 2 is also the same magnitude but opposite direction of proton 1. Based on that, $E_{p2i} = E_{p1i}$ in the zero magnitude frame. Therefore the kinetic energy of the two protons is
$E_{p1i} + E_{p2i}= 938.3 + 938.3 + 2*(493.7)$
$E_{p1i} + E_{p2i} = 2864$
$E_{p1i} = 1432 MeV$

This isn't the right answer. This is probably because this is the kinetic energy of the proton in the zero momentum frame, but I need the kinetic energy in the lab frame. I think I have to multiply by some ratio. Maybe the lorentz transformation?

 

[vela]

Excellent! Yes, you want to use the Lorentz transformation to calculate the values in the lab frame now. Look at the equations and figure out what else you need to know to do that and see if you can calculate them.

 

[Kharrid]

Given equations are $E=γmc^2$, $p=γmv$, and $K=(γ-1)mc^2$.

To calculate the minimum kinetic energy, the only piece I am missing is the velocity of the proton. I can calculate the velocity of the proton in the zero momentum frame, but I'd have to use a transformation to shift it into the lab frame.

$E_{p1i} = γmc^2 = 1432 MeV = 2.294*10^-10 J$
$\frac{1}{√{1-\frac{v^2}{c^2}}} = \frac{2.294*10^-10}{mc^2}$
$1-\frac{v^2}{c^2} = \frac{mc^2}{2.294*10^-10}$
$\frac{v^2}{c^2} = 1-\frac{mc^2}{2.294*10^-10}$
$v_{p1i} = √(c^2(1-\frac{mc^2}{2.294*10^-10}))$
$v_{p1i} =1.76*10^8 \frac{m}{s}$

To convert from one frame of reference to another where the zero momentum frame is "'"
$v_x = \frac{v_x' + u}{1 + \frac{uv_x'}{c^2}}$

The question is, what is u?

 

[vela]

I found a different speed for the proton. Note that you're given $mc^2$ in MeV, so converting to joules is unnecessary.

 

[Kharrid]

Hmm ... this is what I plugged in:
$v_{p1i} = \sqrt{(3*10^8)^2(1-\frac{1.67*10^{-27} * (3*10^8)^2}{2.294*10^{-10}})}$
$v_{p1i} =1.76 * 10^8 \frac{m}{s}$

Does that look right?

 

[vela]

Oh, you made an algebra mistake when solving for the speed. When you flipped the fractions over in the second step, you also squared both sides. You forgot to square the right-hand side.

 

[Kharrid]

$v_{p1i} = \sqrt{(3*10^8)^2(1-(\frac{1.67*10^{-27} * (3*10^8)^2}{2.294*10^{-10}})^2)}$
$v_{p1i} = 2.27 * 10^8 \frac{m}{s}$

Good catch. Note to self: stop skipping steps.

 

[vela]

You have $E_{p1i}$, $p_{p1i}$, and the velocities of the protons in the zero-momentum frame. Now you need to transform back to the lab frame.

Look at what @PeroK wrote back in post 12.

 

[Kharrid]

You have $E_{p1i}$, $p_{p1i}$, and the velocities of the protons in the zero-momentum frame. Now you need to transform back to the lab frame.

Look at what @PeroK wrote back in post 12.

Quick aside, the equations that PeroK listed are not written in my textbook. I assume these are standard equations that students should know? A similar looking equation in my textbook is $E^2 = (mc^2)^2 + (pc)^2$.

$E=γ(E′+vp′)$
$E_{p1i} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}*(E'+vp')$ (note: $p'=mv'$)
$E_{p1i} = \frac{1}{\sqrt{1-\frac{(2.27*10^8)^2}{(3*10^8)^2}}}*(2.294*10^{-10}+(2.27*10^8)(1.67*10^{-27})(2.27*10^8))$
$E_{p1i} = 4.82 * 10^{-10} J = 3013 MeV$

Still not the right answer. Maybe the reason is that the $v$ used in the lorenz transformation should be in the lab frame, but the $v$ I have is in the zero momentum frame. If I need $v$ in lab frame, I need $u$ from the conversion equation in post 33.

If i use the second equation, $E_{tot} = γ(E_{tot}' + vp_{tot})$, $p_{tot}$ is 0 and the equation becomes $E_{tot} = γE_{tot}'$. The question now is if I can reverse the equation so that I don't have to use the speed in the lab frame.

I wonder if this would work?
$E_{tot} = \frac{E_{tot}'}{γ}$ where $γ$ is the lorenz transformation using $v$ of the zero momentum frame.

 

[vela]

You're getting close.

You used the Newtonian expression for momentum. Try using the equation from your textbook to calculate $p_{p1i}$, or use $p = \gamma m v$.

The $v$ that appears in the transformation equation represents the velocity of the lab frame relative to the zero-momentum frame, and the $\gamma$ is the corresponding Lorentz factor. You plugged in the speed of the proton as measured in the zero-momentum frame. Is that the right speed to use there? Remember the lab frame is the reference frame in which the second proton is at rest.

If i use the second equation, $E_{tot} = γ(E_{tot}' + vp_{tot})$, $p_{tot}$ is 0 and the equation becomes $E_{tot} = γE_{tot}'$. The question now is if I can reverse the equation so that I don't have to use the speed in the lab frame.

This approach will work as well, but you still need the speed of the lab frame to calculate $\gamma$.

I wonder if this would work?
$E_{tot} = \frac{E_{tot}'}{γ}$ where $γ$ is the lorenz transformation using $v$ of the zero momentum frame.

No, this won't work. You can't have both $E_\text{tot} = \gamma E'_\text{tot}$, as above, and $E_\text{tot} = E'_\text{tot}/\gamma$.

 

[Kharrid]

You're getting close.

You used the Newtonian expression for momentum. Try using the equation from your textbook to calculate $p_{p1i}$, or use $p = \gamma m v$.

The $v$ that appears in the transformation equation represents the velocity of the lab frame relative to the zero-momentum frame, and the $\gamma$ is the corresponding Lorentz factor. You plugged in the speed of the proton as measured in the zero-momentum frame. Is that the right speed to use there? Remember the lab frame is the reference frame in which the second proton is at rest.

$E_{p1i} = \frac{1}{\sqrt{1-\frac{(v)^2}{(3*10^8)^2}}}*(2.294*10^{-10}+(2.27*10^8)(1.67*10^{-27})(2.27*10^8)(\frac{1}{\sqrt{1-\frac{(2.27*10^8)^2}{(3*10^8)^2}}}))$

The mystery is what $v$ is. If the lab frame is stationary, then the CoM frame is moving at some velocity $u$. I'm not sure what this $u$ is. Can I say that $u=v$ for the second proton since the second proton is at rest in the lab frame but moving in the CoM frame? This sounds off ...

Wait a sec, I know the energy of the second proton. I also know the velocity of the second proton is $-v$ in the zero momentum frame. Therefore, I could use it to find the velocity of the proton in the lab frame. But, I already know that the velocity of the second proton in the lab frame is 0 so this doesn't help.

It seems I need to find $u$, but I don't know how to find this value.

Or maybe I do. I could use $v_x' = \frac{v_x - u}{1 - \frac{uv_x}{c^2}}$ where "'" is the zero momentum frame.
$v_x' = \frac{v_x - u}{1 - \frac{uv_x}{c^2}}$
$-2.27*10^8 = \frac{0- u}{1 - \frac{u(0)}{c^2}}$
$u=2.27*10^8$

Plug that into the top equation, but that would still give me a wrong answer ...

 

[PeroK]

$E_{p1i} = \frac{1}{\sqrt{1-\frac{(v)^2}{(3*10^8)^2}}}*(2.294*10^{-10}+(2.27*10^8)(1.67*10^{-27})(2.27*10^8)(\frac{1}{\sqrt{1-\frac{(2.27*10^8)^2}{(3*10^8)^2}}}))$

The mystery is what $v$ is. If the lab frame is stationary, then the CoM frame is moving at some velocity $u$. I'm not sure what this $u$ is. Can I say that $u=v$ for the second proton since the second proton is at rest in the lab frame but moving in the CoM frame? This sounds off ...

Wait a sec, I know the energy of the second proton. I also know the velocity of the second proton is $-v$ in the zero momentum frame. Therefore, I could use it to find the velocity of the proton in the lab frame. But, I already know that the velocity of the second proton in the lab frame is 0 so this doesn't help.

It seems I need to find $u$, but I don't know how to find this value.

Or maybe I do. I could use $v_x' = \frac{v_x - u}{1 - \frac{uv_x}{c^2}}$ where "'" is the zero momentum frame.
$v_x' = \frac{v_x - u}{1 - \frac{uv_x}{c^2}}$
$-2.27*10^8 = \frac{0- u}{1 - \frac{u(0)}{c^2}}$
$u=2.27*10^8$

Plug that into the top equation, but that would still give me a wrong answer ...

A couple of points:

1) It's rarely a good idea to go looking for velocities in these energy-momentum transformations. It rarely leads anywhere.

Tip: stick with energy-momentum: $E, p$ for each particle. Resist the temptation to use $v$.

2) You've got yourself bogged down in complicated notation and plugging in the numbers has just made things worth.

For example, many posts ago I actually gave you the answer for the calculation in the CoM frame. I'll repeat it here:

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 equals 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's it. Try to undestand why that's the answer. The whole concept of particle collisons is that if they have enough energy they don't just bounce elastically off each other; they create new particles. Some or all of their KE is lost and becomes the rest energy of new particles.

Before you go any further you need to understand this post.

 

[vela]

The mystery is what $v$ is. If the lab frame is stationary, then the CoM frame is moving at some velocity $u$. I'm not sure what this $u$ is. Can I say that $u=v$ for the second proton since the second proton is at rest in the lab frame but moving in the CoM frame? This sounds off ...

Wait a sec, I know the energy of the second proton. I also know the velocity of the second proton is $-v$ in the zero momentum frame. Therefore, I could use it to find the velocity of the proton in the lab frame. But, I already know that the velocity of the second proton in the lab frame is 0 so this doesn't help.

It seems I need to find $u$, but I don't know how to find this value.

A passenger is at rest on a train. If the train is going 40 mph relative to the ground, how fast is the passenger moving relative to the ground?

Say a person is moving 60 mph relative to the ground. He’s at rest relative to the car he’s in. How fast is the car moving relative to the ground?

 

[Kharrid]

That's it. Try to undestand why that's the answer. The whole concept of particle collisons is that if they have enough energy they don't just bounce elastically off each other; they create new particles. Some or all of their KE is lost and becomes the rest energy of new particles.

Before you go any further you need to understand this post.

In the CoM frame, one can separate the total energies before and after the collision. Before the collision, there are only two energies of the protons. These energies are the exact same because the momentum of the protons are equal and opposite and velocity is squared in the energy equation. So, $E_i' = 2E_p'$.

After the collision, the energies of the final product are the energies of the protons and the kaons. NOTE: the rest energy of a proton and kaon DOES NOT MEAN their energies are 0. Therefore, $E_f' = 2E_p' + 2E_k'$. In the expanded form, the lorentz transformation is present, but it is 1 because v=0.

A passenger is at rest on a train. If the train is going 40 mph relative to the ground, how fast is the passenger moving relative to the ground?

Say a person is moving 60 mph relative to the ground. He’s at rest relative to the car he’s in. How fast is the car moving relative to the ground?

1) The passenger is going 40mph relative to the ground.

2) The car is moving 60 mph relative to the ground.

 

[PeroK]

In the CoM frame, one can separate the total energies before and after the collision. Before the collision, there are only two energies of the protons. These energies are the exact same because the momentum of the protons are equal and opposite and velocity is squared in the energy equation. So, $E_i' = 2E_p'$.

After the collision, the energies of the final product are the energies of the protons and the kaons. NOTE: the rest energy of a proton and kaon DOES NOT MEAN their energies are 0. Therefore, $E_f' = 2E_p' + 2E_k'$. In the expanded form, the lorentz transformation is present, but it is 1 because v=0.

Okay, so you know the energy of each proton in the CoM frame. There are a number of ways of getting the energy in the lab frame. By far the most difficult is to try to find the velocity of the CoM frame. That would work in classical mechanics, but in SR it is MUCH simpler to use energy-momentum.

The simplest way is to use the "invariant" quantity associated with energy-momentum. The quantity $E^2 - p^2c^2$ is invariant. That means it is the same in all inertial reference frames. In this case it is the same in the lab and CoM frames.

This applies to each individual particle (where the invariant quantity is its rest energy). But, it also applies to the whole system of particles. Therefore, you have:

$(E_{tot})^2 - (p_{tot})^2c^2 = (E'_{tot})^2 - (p'_{tot})^2c^2$

Where this is the total energy-momentum of ALL particles in each frame.

Can you work with that equation?

Note that $p'_{tot} = 0$, so that makes it even simpler.

 

[Kharrid]

So from the discussion, I assume that I should stay away from velocities? I'm not sure what next step I should use if I go down this path.

If I do need the velocity of the particle, I need the speed of the zero momentum frame relative to the lab frame. Since the velocity of the proton 2 in the lab frame is 0 and in the zero momentum frame is the equal and opposite velocity of proton 1, then the velocity of the zero momentum frame is the velocity of proton 1 in the zero momentum frame. Therefore, can I use the velocity of proton 1 in the zero momentum frame that I know as my $u$ and solve using the $v'$ equation?

Also, PeroK, that equation is not in the textbook but I don't mind learning it. If I take your word that $E^2-(pc)^2$ is invariant, then the $p_{tot}' = 0$, $E_{tot} = E_{p1} + E_{p2} + 2E_{kaon}$, and $E_{tot}' = 2E_p'+2E_k'$. The only question in this formula is that I need to look up the mass of a kaon for $E_k'$, which I wouldn't have on a formula sheet. Doesn't this mean I need to go the velocity route?

 

[PeroK]

You can't do anything without the mass of the kaon. The final expression, logically, must be a function of the mass of the proton and the mass of the kaon.

You need a new textbook.

 

[PeroK]

Also, PeroK, that equation is not in the textbook but I don't mind learning it. If I take your word that $E^2-(pc)^2$ is invariant, then the $p_{tot}' = 0$, $E_{tot} = E_{p1} + E_{p2} + 2E_{kaon}$, and $E_{tot}' = 2E_p'+2E_k'$. The only question in this formula is that I need to look up the mass of a kaon for $E_k'$, which I wouldn't have on a formula sheet. Doesn't this mean I need to go the velocity route?

Just an observation that when you use algebra you don't seem to resolve things. For example, twice I've pointed out that the energy in the CoM frame is $2(m_p + m_k)c^2$. Yet, you don't use this. You retain the notation $E_{tot}' = 2E_p'+2E_k'$. That seems not to acknowledge what we've already calculated.

Also, we want the initial energy of the first proton, so we want to equate the initial energy-momentum in the lab frame with the final energy-momentum in the CoM frame.

To give you a bit more help:

$E_{tot} = E_{p1} + m_pc^2$

$p_{tot} = p_{p1}$

$E'_{tot} = 2(m_p + m_k)c^2$

And, it's $E_{p1}$ that your after.

And, for any particle we have $E^2 = p^2c^2 + m^2c^4$, just in case this is another formula your textbook thinks you can do without!

 

[vela]

Say a person is moving 60 mph relative to the ground. He’s at rest relative to the car he’s in. How fast is the car moving relative to the ground?

2) The car is moving 60 mph relative to the ground.

The proton is moving with speed $v$ relative to the zero-momentum frame. It's at rest relative to the lab frame. How fast is the lab frame moving relative to the zero-momentum frame? This is essentially the same question as above. You guessed this answer earlier, but you weren't sure if you were right. How about now?

 

[Kharrid]

You need a new textbook.

At this point, I really do.

Just an observation that when you use algebra you don't seem to resolve things. For example, twice I've pointed out that the energy in the CoM frame is $2(m_p + m_k)c^2$. Yet, you don't use this. You retain the notation $E_{tot}' = 2E_p'+2E_k'$. That seems not to acknowledge what we've already calculated.

That's on me. I should be more accurate and expand the equations fully.

$E_{tot} = E_{p1} + m_pc^2$

$p_{tot} = p_{p1}$

$E'_{tot} = 2(m_p + m_k)c^2$

And, it's $E_{p1}$ that your after.

I understand how each equation is derived, so now moving to solve. If I am looking for $E_{p1}$, then the top equation becomes $E_{p1} = E_{tot} - m_pc^2$. From the equation below, $E_{tot}^2 = p_{tot}^2c^2 + m_p^2c^4$. Therefore, $E_{p1} = p_{tot}^2c^2 + m_p^2c^4$.

$E_{p1} = p_{p1i}^2c^2 + m_p^2c^4$.
$E_{p1} = (γmv_{p1i})^2c^2 + m_p^2c^4$

Looks like I need the velocity of the proton in the lab frame. I will get back to this at the end. For now, let's try equating the invariant equation for particles $E^2-(pc)^2$.

$E_{p1}^2-(p_{p1}c)^2=E_{p1}^2-(p_{p1}c)^2$

It seems that either way, I need the $p_{p1}$ that I do not have because I don't know the speed of proton 1 in the lab frame.

And, for any particle we have $E^2 = p^2c^2 + m^2c^4$, just in case this is another formula your textbook thinks you can do without!

Fortunately, the authors did include that equation.

The proton is moving with speed $v$ relative to the zero-momentum frame. It's at rest relative to the lab frame. How fast is the lab frame moving relative to the zero-momentum frame? This is essentially the same question as above. You guessed this answer earlier, but you weren't sure if you were right. How about now?

The lab frame is moving at speed $v$ relative to the zero momentum frame because the two frames differ by a speed of $v$. This can be determined by comparing the speeds of proton 2 in both frames. Let's try solving for the velocity of proton 1 using this equation:

$v = \frac{v_x' + u}{1+\frac{uv_x'}{c^2}}$
$v = \frac{2.27*10^8 + 2.27*10^8}{1+\frac{(2.27*10^8)(2.27*10^8)}{(3*10^8)^2}}$
$v = 4.54 * 10^8 \frac{m}{s}$

Plugging v into the previous equation after quote 2:
$E_{p1} = (γmv_{p1i})^2c^2 + m_p^2c^4$
$E_{p1} = ((\frac{1}{\sqrt{1 - \frac{(4.54*10^8)^2}{(3*10^8)^2}}})(1.67*10^{-27})(4.54*10^8))^2(3*10^8)^2 + (1.67*10^{-27})^2(3*10^8)^4$
$E_{p1} = 6.27 * 10^{-20} J = 0.00000039MeV$

What a long equation. Too bad it's wrong ... At this point, either I calculated the velocity of the proton 1 wrong or it was a calculation error. (I think).