Kinetic Energy of Colliding Protons

In summary: 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 minimum kinetic energy is found to be 983.4 MeV. In summary, the minimum
  • #36
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.
 
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  • #37
##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.
 
  • #38
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.
 
  • #39
vela said:
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.
 
  • #40
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.

Kharrid said:
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##.
 
  • #41
vela said:
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 ...
 
  • #42
Kharrid said:
##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:

PeroK said:
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.
 
  • #43
Kharrid said:
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?
 
  • #44
PeroK said:
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.

vela said:
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.
 
  • #45
Kharrid said:
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.
 
  • #46
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?
 
  • #47
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.
 
  • #48
Kharrid said:
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!
 
  • #49
Kharrid said:
vela said:
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?
 
  • #50
PeroK said:
You need a new textbook.

At this point, I really do.

PeroK said:
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.

PeroK said:
##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.

PeroK said:
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.

vela said:
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).
 
  • #51
Kharrid said:
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.
The fact that proton 2 is at rest in the lab frame is critical here.

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}##
That's not right. Check your calculation. Note the speed you got is greater than ##c##. That's a dead giveaway it can't be right.

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##
You must have gotten a negative number inside the radical, which is another sign you made a mistake, so I'm not sure how you got an answer at all here.
 
  • #52
vela said:
The fact that proton 2 is at rest in the lab frame is critical here.That's not right. Check your calculation. Note the speed you got is greater than ##c##. That's a dead giveaway it can't be right.You must have gotten a negative number inside the radical, which is another sign you made a mistake, so I'm not sure how you got an answer at all here.

Haha yea, I focused on the problem and forgot the basics. Also, I "forced" it to work, which should have been a big sign that it wasn't right. The error was that I multiplied the bottom by 2 instead of squaring.

##v = 2.89 * 10^8 \frac{m}{s}##

Substituting the new value in:
##E_{p1} = ((\frac{1}{\sqrt{1 - \frac{(2.89*10^8)^2}{(3*10^8)^2}}})(1.67*10^{-27})(2.89*10^8))^2(3*10^8)^2 + (1.67*10^{-27})^2(3*10^8)^4##
##E_{p1} = 3.14*10^-19 J = 1.96*10^{-6} MeV##

Hmm, maybe I wrote the equation wrong?
 
  • #53
Kharrid said:
##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.

Note that energy and magnitide of momentum of a particle are related. If you know one, you know the other. Assuming you know the mass of the particle.

In SR, we have ##E^2 = p^2 c^2 + m^2c^4##

You must remember this. If you have the energy of a particle, then you have the magnitude of its momentum. And vice versa.

In this case, using the relationship between energy and momentum of the incident proton, we have:

##(E_{tot})^2 - (p_{tot})^2c^2 = (E_1 + m_pc^2)^2 - (p_1)^2c^2 = E_1^2 + 2E_1m_pc^2 + m_p^2c^4 - (E_1^2 - m^2c^4) = 2(E_1m_pc^2 + m_p^2c^4)##

Setting this equal to the invariant quantity from the CoM frame, we have:

##2(E_1m_pc^2 + m_p^2c^4) = 4(m_p+m_k)^2c^4##

And that gives you the energy of the incident proton, ##E_1##, in terms of the mass of the proton and the mass of the kaon only. You just need to rearrange that formula.

Note that this is a general result. The proton and the kaon could be replaced by any other particle. There is nothing special about these particles. Also, if you look at the way this was derived, there is no limitation on the number of particles that result. In other words, the RHS of that equation is, more generally:

##(\sum m_i)^2c^4##

Where this is the sum of the masses of the final set of particles produced.
 
  • #54
@Kharrid I'll show you a final trick. If you have the particle masses in MeV/c^2, then using the approach I suggested, everything simplifies. The final equation:

##E_1 = \frac{2(m_p+m_k)^2c^4 - m_pc^4}{m_p^2c^2}##

Then, replacing the masses with the values in MeV, we have:

##E_1 = \frac{2(m_p+m_k)^2 - m_p^2}{m_p}##

If you plug the masses (in MeV) into that equation you should get the answer.
 
  • #55
Here's another trick (if you're still watching this thread). The equation being used is an adaptation of equation 34 in this reference. $${\left(\frac{m_f}{m_i}\right)}^2=1+\frac{2(\gamma-1)m_1m_2}{{\left(m_1+m_2\right)}^2}$$ where ##m_1## and ##m_2## are the masses of high energy particles involved in a 'head-on' collision - protons in this instance. $$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$v is the relative velocity of the colliding particles and the inelastic collision results in the generation of mass from energy - ie kaons in this particular problem. If the masses of the colliding particles are the same, the formula simplifies further: $${\left(\frac{m_f}{m_i}\right)}^2=1+\frac{\gamma-1}{2}=\frac{\gamma+1}{2}$$Since this is a ratio formula it doesn't matter what units you are using for 'mass' be it kg, Mev or amu. For part A) , you can easily solve for ##\gamma-1## and then multiply by rest energy (proton) to obtain the minimum KE required.
 
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