Kinetic Energy of Colliding Protons

[vela]

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.

 

[Kharrid]

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?

 

[PeroK]

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

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.

 

[PeroK]

@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.

 

[neilparker62]

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.