What is the threshold speed needed for this process to occur?

[bon]

Homework Statement

A particle mass m moves to the right with relativistic speed v. It collides inelastically with another particle mass 3m moving to the left also at speed v. A single body of mass 5m is produced in the collision. What is the threshold speed needed for this process to occur?

Homework Equations

The Attempt at a Solution

So i was just wondering what the quickest way to solve this would be.

I solved using the invariant E^2 - c^2 p^2 looking at the lab frame before the collision and CMF frame after..

i get v to be 3/root21 c..but am really unsure about it..is this right?

Also, what does 'threshold' mean in this context..? Surely the velocity is completely determined by the statement of the problem...i.e. with the given data there aren't a range of possible velocities are there? :S

Thanks

 

[Andrew Mason]

Homework Statement

A particle mass m moves to the right with relativistic speed v. It collides inelastically with another particle mass 3m moving to the left also at speed v. A single body of mass 5m is produced in the collision. What is the threshold speed needed for this process to occur?

Homework Equations

The Attempt at a Solution

So i was just wondering what the quickest way to solve this would be.

I solved using the invariant E^2 - c^2 p^2 looking at the lab frame before the collision and CMF frame after..

i get v to be 3/root21 c..but am really unsure about it..is this right?

Also, what does 'threshold' mean in this context..? Surely the velocity is completely determined by the statement of the problem...i.e. with the given data there aren't a range of possible velocities are there? :S

Thanks

Momentum is conserved in this collision. So, try analysing the collision in the lab frame and applying conservation of momentum:

p = \gamma m_0v

and:

p_1 + p_2 = p_f

AM

 

[bon]

Ok so is my answer wrong?

Also could you answer my question about the "threshold" velocity in this context..thanks

 

[Andrew Mason]

Ok so is my answer wrong?

Also could you answer my question about the "threshold" velocity in this context..thanks

You will have to show your work.

What is the final speed? (hint: what is the rest mass of the combined masses? what is the ratio of relativistic mass to rest mass? => what does that tell you about the speed of the combined masses?)

From that you will be able to determine the initial velocity using conservation of momentum.

That should tell you whether there can be more than one value for v.

AM

 

[bon]

Sorry this isn't helping. Please could you explain what "threshold" means in this context..i think the system is determined..

 

[ehild]

Threshold speed is the minimum speed so as the conditions can be fulfilled. Remember, you will have one m more than before, so the energy of the colliding particles has to be enough to produce that excess mass.


ehild

 

[bon]

thanks ehild - sorry maybe i should have asked my question more carefully - you say that threshold speed in the minimum speed so as the conditions can be fulfilled. My point is this - surely if the two initial masses are m and 3m, and the final mass 5m then there is only ONE possible velocity such that the conditions are fulfilled - i.e. if the velocity was any less the final mass would be less and vica versa. yes?

 

[Andrew Mason]

Sorry this isn't helping.

Can you not post your work?

thanks ehild - sorry maybe i should have asked my question more carefully - you say that threshold speed in the minimum speed so as the conditions can be fulfilled. My point is this - surely if the two initial masses are m and 3m, and the final mass 5m then there is only ONE possible velocity such that the conditions are fulfilled - i.e. if the velocity was any less the final mass would be less and vica versa. yes?

The threshold speed is the minimum speed needed to form a particle of mass 5m. That is the case where no other particles are produced in the collision to carry away momentum. The resulting particle must absorb a great deal of energy (you can work it out) and this usually results in new particles flying off carrying momentum. You could have higher speed collisions producing particles of mass 5m, but not in which all the momentum is carried by the 5m mass.

AM

 

[ehild]

bon,

your v=3/sqrt(21) is correct if both the momentum and the energy is conserved during the collision. It is difficult to read the mind of the person who gave this problem to you. But the collision was said inelastic. The question is what it means. It can mean that although the mechanical energy of the particles is not conserved, the total energy is, but it can mean also that some of the total energy is lost, is transformed to the energy of a photon or vibration of a crystal lattice, something which does not take away momentum, but consumes energy. In that context the threshold energy has sense.

The problem can be even simpler. From conservation of energy,

\frac{4mc^2}{\sqrt{1-v^2/c^2}}=\frac{5mc^2}{\sqrt{1-v'^2/c^2}}

(v' is the velocity of the new particle) you get a relation between (v/c)^2 and (v'/c)^2, and (v'/c)^2 >=0 involves that the original speed is at last 0.6c. Maybe, this was only the question.

ehild

 

[bon]

Ahh so it is possible for the momentum not to be conserved? Since if v'=0 the momentum afterwards is lost!

 

[collinsmark]

Hello bon,

Ahh so it is possible for the momentum not to be conserved? Since if v'=0 the momentum afterwards is lost!

Momentum will be conserved. And that's kind of important to solving this problem.

Using guidance from ehild's last post, you can get an equation (using conservation of energy) to find a relationship between v and v' (see ehild's last post).

But now you still have two unknowns, v and v'. One equation, two unknowns. Wouldn't it be nice to form a second, simultaneous equation, so you'll have two equations and two unknowns? http://www.websmileys.com/sm/love/301.gif. Hmmm, what conservation law can you use to generate the second equation?...

[Edit: btw, v' \ne 0.]

 

[Andrew Mason]

Ahh so it is possible for the momentum not to be conserved? Since if v'=0 the momentum afterwards is lost!

Momentum cannot be lost. If the momenta of the colliding masses before and after are not equal, you are missing some other particle carrying momentum. This is one way to discover new particles - eg. the neutrino.

AM

 

[ehild]

Ahh so it is possible for the momentum not to be conserved? Since if v'=0 the momentum afterwards is lost!

The momentum is always conserved if all the participants in a collision are taken into account. I answered the question about the "threshold speed". One equation, conservation of energy, requires that the original speed of the colliding particles is at least 0.6c. The second equation is conservation of momentum, and the two conservation laws together result in a single value for the initial speed, the same you have given in your very first post:
v=3/sqrt(21) c = 0.65c.

Since your first post, the discussion is about "threshold".

3. The Attempt at a Solution

So i was just wondering what the quickest way to solve this would be.

I solved using the invariant E^2 - c^2 p^2 looking at the lab frame before the collision and CMF frame after..

i get v to be 3/root21 c..but am really unsure about it..is this right?

Also, what does 'threshold' mean in this context..? Surely the velocity is completely determined by the statement of the problem...i.e. with the given data there aren't a range of possible velocities are there? :S

As there is a unique solution of the problem, I still say that I can not read the mind of other people, although I have tried. Your solution was correct, you were right, but next time please write out your solutions in detail.

ehild

 

[bon]

Ok I see - very clear now. Thanks for your help :)

I'll try to post full working new time.

Thanks

 

[Andrew Mason]

The momentum is always conserved if all the participants in a collision are taken into account. I answered the question about the "threshold speed". One equation, conservation of energy, requires that the original speed of the colliding particles is at least 0.6c. The second equation is conservation of momentum, and the two conservation laws together result in a single value for the initial speed, the same you have given in your very first post:
v=3/sqrt(21) c = 0.65c.

Since your first post, the discussion is about "threshold".

I get a different answer.
p_1 + p_2 = p_f

(1) \gamma 3mv - \gamma mv = 5mv' = \gamma' 4mv'

Since:

5mv' = \gamma' 4mv' then,

\gamma' = 5/4,

1 - v'^2/c^2 = (4/5)^2

v' = \sqrt{1 - (4/5)^2}c = \sqrt{.36} = .6c

Substituting into (1):

\gamma 3mv - \gamma mv = \gamma 2mv = 5mv' = 5m(.6c) = 3mc

\gamma v/c = 3/2

Squaring both sides and multiplying by 1/\gamma^2 = (1 - v^2/c^2)[/tex]<br /> <br /> (3/2)^2(1 - v^2/c^2) = (v/c)^2<br /> <br /> (1 + (3/2)^2)(v/c)^2 = (3/2)^2<br /> <br /> v^2/c^2 = (3/2)^2/(1 + (3/2)^2) = (9/4)/(1 + 9/4) = 9/(4 + 9) = 9/13<br /> <br /> v = 3c/\sqrt{13} = .832c<br /> <br /> AM

 

[bon]

No - I think you are wrong because the particle that is produced has rest mass 5m but is still moving, right?

 

[Andrew Mason]

No - I think you are wrong because the particle that is produced has rest mass 5m but is still moving, right?

The resulting particle must have a rest mass of 3m + 1m. How can the rest mass be more than the rest masses of the original particles?

AM

 

[bon]

It says a single body of mass 5m is produced in the collision..it doesn't say that this is the total energy of the body..

 

[ehild]

Well, the problem also does not say if the given masses are the rest masses of the particles or not. Bon assumed that all of them are rest masses.

ehild

 

[bon]

so here is rest mass not conserved?

 

[ehild]

The resulting particle must have a rest mass of 3m + 1m. How can the rest mass be more than the rest masses of the original particles?

AM

What about electron-positron pair production from a photon?
Rest mass is not a quantity that is conserved in relativity.

By the way, this is only a problem, and a very undefined one. The creator should have been more specific about those masses and about what is meant on inelastic scattering.

ehild

 

[bon]

What about electron-positron pair production from a photon?
Rest mass is not a quantity that is conserved in relativity.

By the way, this is only a problem, and a very undefined one. The creator should have been more specific about those masses and about what is meant on inelastic scattering.

ehild

Ok thanks for all your help.

I just wanted to check my understanding by attempting another problem:

A proton of total energy 3 GeV makes a head-on collision with a 5 GeV electron.
Calculate the available energy in the centre-of-mass system to create any new additional
particles in the collision.

My solution:

Can't we just use the invariant E^2 - c^2p^2

This is the total energy available after the collision in CMF, since there p=0, E^2-c^p^2 = Etot^2

so the energy available is: 1189 MeV?

 

[Andrew Mason]

What about electron-positron pair production from a photon?
Rest mass is not a quantity that is conserved in relativity.

By the way, this is only a problem, and a very undefined one. The creator should have been more specific about those masses and about what is meant on inelastic scattering.

ehild

One does not generally refer to a photon as having mass. So I had assumed we were talking about particles having a rest mass - ie both before and after particles. Since the problem refers to a mass of 5m after the collision, I assumed it was referring to the relativistic mass of a particle of non-zero rest mass.

I was ignoring the inertia of the 'lost' energy - ie the energy 'lost' in the inelastic collision. This was a mistake because if this 'lost' energy is momentarily contained within the combined incident masses, the resulting mass would have a rest mass greater than 4m.

If 5m is the rest mass of the resulting particle, the energy lost in the collision would have to be equal to the increase in rest mass. I have a feeling that would be rather complicated to work out.

AM