Relativistic elastic proton collision

[jamesw1]

Homework Statement

A proton moving with speed βc collides elastically with another proton moving in the opposite direction with the same speed. After the collision, the velocity of the first proton (relative to a stationary observer) is perpendicular to the initial trajectory. Find the angle of deflection of this proton from its original trajectory in the frame of reference in which the second proton was initially stationary. In this frame of reference, what is the velocity of the first proton after the collision?

Relevant Equations

p=(E,px,py,pz)
E = γmc^2
E^2 = (pc)^2 + (mc^2)^2

How should I approach this problem? My first thought was to subtract the velocity of the second proton from the velocity of proton going upwards. However, the velocity vectors are perpendicular to each other, therefore I cannot use the SR velocity addition formula.

 

[Orodruin]

You have written down the 4-momentum as one of your relevant equations. What is the 4-momentum of the deflected proton in (a) the lab frame and (b) the other proton’s initial rest frame?

How does (b) relate to the deflection angle?

 

[jamesw1]

$$ P_a=\left(\frac{E}{c},0\right)=\left(\frac{2\gamma _1mc^2}{c},0\right) $$
$$ P_b=\left(\frac{E}{c},0\right)=\left(\frac{mc^2+\gamma _2mc^2}{c},0\right) $$
Where
$$ \gamma _1=\frac{1}{\sqrt{1-\beta ^2}} $$
$$ \gamma _2=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} $$

v is the velocity we are trying to find.
Is this correct?

 

[Orodruin]

Is this correct?

No. The total momentum is zero only in the lab frame.

Either way, the important thing is the momentum of the first proton in the second proton’s rest frame.

 

[jamesw1]

What is the momentum, then? Could anyone walk through a solution, please?

 

[PeroK]

What is the momentum, then? Could anyone walk through a solution, please?

As a first step, you could write down the four-momentum of the protons after the collision in the lab frame.

Then, you could apply the (2D) velocity transformation to the rest frame of one of the protons. Or, apply the energy-momentum transformation.

Does that make sense?

 

[Orodruin]

What is the momentum, then?

4-momentum is given by the formula $p = mV$, where $V$ is the 4-velocity $\gamma(1,\vec v)$. It transforms as a 4-vector under Lorentz transformation and this is all you really need to know.

So what is the 4-momentum of the first proton in the lab frame? What does the Lorentz transformation then say that the 4-momentum is in the rest frame of the second proton before collision?

Could anyone walk through a solution, please?

No, it is against forum rules to provide full solutions before the original poster has solved the problem.

 

[jamesw1]

Is this the correct 4-momentum? Probably not?
$$ P_a=\left(\frac{E}{c},0\right)=\left(\frac{2\gamma _1mc^2}{c},0\right) $$
$$ P_b=\left(\frac{E}{c},\gamma _2mv\right)=\left(\frac{mc^2+\gamma _2mc^2}{c},\gamma _2mv\right) $$
$$ \gamma _1=\frac{1}{\sqrt{1-\beta ^2}} $$
$$ \gamma _2=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} $$
Then I would do something like this
$$ 4\gamma _1^2m^2c^2=m^2c^2\left(1+\gamma _2\right)^2-\gamma _2^2m^2v^2 $$
But that is not correct either, probably

 

[Orodruin]

No. You are trying to write down the total 4-momentum. I am asking specifically for the 4-momentum of the deflected proton after collision. Not the total.

 

[Orodruin]

After the collision, the velocity of the first proton (relative to a stationary observer) is perpendicular to the initial trajectory.

let me just complain a bit about the problem formulation as well …

There is no way of separating the protons as they are identical. It is therefore impossible to conclude that it is the first proton that was reflected in a particular direction. The problem should say ”one of the protons”

 

[PeroK]

I must confess that I don't understand the difficulty with this problem. If the protons are moving at non-relativistic speeds, then the solution ought to be simple. In the lab frame:

Initial velocities: $(v, 0)$ and $(-v, 0)$.

Final velocities: $(0, v)$ and $(0, -v)$.

In the original rest frame of the "second" proton, the velocity of the first proton is:

Initial velocity: $(2v, 0)$

Final velocity: $(v, v)$

What difference does it make if the protons are moving at relativistic speeds?

In the lab frame, the final velocities apply also at relativistic speeds. There's no difference there. Conservation of energy and momentum demand it.

The only difference is that we must use the Lorentz Transformation/Relativistic Velocity Transformation when we transform to the rest frame of the second proton.

Why do anything more complicated than that?

 

[Orodruin]

The only difference is that we must use the Lorentz Transformation/Relativistic Velocity Transformation when we transform to the rest frame of the second proton.

I mean, this is what I am trying to get the OP to do. I would certainly opt for the Lorentz transformation route though. I'm not really a fan of the velocity addition formulas for the simple reason that they obscure the physics more.

 

[jamesw1]

I have zero intuition for the second equation (y transformation) but I did something

$$ u'_x=\frac{u_x-v_x}{1-\frac{v_xu_x}{c^2}}=\frac{0+\beta c}{1}=\beta c $$

$$ u'_y=\frac{u_y\sqrt{1-\frac{v_x^2}{c^2}}}{1-\frac{v_xu_x}{c^2}}=\frac{\beta c\sqrt{1-\frac{\beta ^2c^2}{c^2}}}{1}=\beta c\sqrt{1-\beta ^2} $$

$$ \tan \theta =\frac{u'_y}{u'_x}=\frac{\beta c\sqrt{1-\beta ^2}}{\beta c}=\sqrt{1-\beta ^2} $$

$$ u'=\sqrt{u'^2_x+u'^2_y}=\sqrt{\beta ^2c^2\left(1-\beta ^2\right)+\beta ^2c^2}=\sqrt{\beta ^2c^2\left(2-\beta ^2\right)}=\beta c\sqrt{2-\beta ^2} $$

 

[PeroK]

I have zero intuition for the second equation (y transformation) but I did something

$$ u'_x=\frac{u_x-v_x}{1-\frac{v_xu_x}{c^2}}=\frac{0+\beta c}{1}=\beta c $$

That ought to make sense. If the x-component of the velocity of a particle is zero in the lab frame and you transform to a frame "moving at $v$" in the x-direction, then the x-component of the velocity of the particle in that frame must be $-v$.

That said, it does no harm to check using the velocity transformation.

$$ u'_y=\frac{u_y\sqrt{1-\frac{v_x^2}{c^2}}}{1-\frac{v_xu_x}{c^2}}=\frac{\beta c\sqrt{1-\frac{\beta ^2c^2}{c^2}}}{1}=\beta c\sqrt{1-\beta ^2} $$

You might recognise $\sqrt{1 - \beta^2} = \dfrac 1 \gamma$. Where $\gamma$ is associated with the transformation between frames. In the non-relativistic limit, we have $\gamma = 1$ and $u'_y = v$.

$$ \tan \theta =\frac{u'_y}{u'_x}=\frac{\beta c\sqrt{1-\beta ^2}}{\beta c}=\sqrt{1-\beta ^2} $$

And, again, the gamma factor appears and the non-relativistic limit is $\tan \theta = 1$. As it should be.

$$ u'=\sqrt{u'^2_x+u'^2_y}=\sqrt{\beta ^2c^2\left(1-\beta ^2\right)+\beta ^2c^2}=\sqrt{\beta ^2c^2\left(2-\beta ^2\right)}=\beta c\sqrt{2-\beta ^2} $$

Again, this agrees in the non-relativistic limit with $u' = v\sqrt 2$.

It's just a question of getting familiar with these equations. It might be an interesting exercise to check that energy and momentum are conserved in the particle frame using these velocities. Although, if you trust the velocity transformation formulas, then it must. But, it might be worth getting as much practice as possible.

 

[Orodruin]

I have zero intuition for the second equation (y transformation) but I did something

This is why I strongly suggest going for the 4-vector approach. Projecting out 3-vector relations generally obscures the geometric structure of the underlying theory for little gain.

 

[Orodruin]

It's just a question of getting familiar with these equations.

I’d actually advice against that. Instead working to understand the 4-vector formalism will generally be more productive in understanding the theory. Particular cases projected onto 3-vector quantities such as velocity addition formulas drop out from there directly, but generally do not offer any deeper insight.