Special Rel Colliding Particles Problem

1. The problem statement, all variables and given/known data
This should be quite a simple problem, i'm tying myself in knots with it though regardless. Anyway,

An electron of energy 9.0 GeV and a positron of energy E collide head on to produce a B meson and an anti B meson (B nought mesons), each with a mass of 5.3 GeV/c^2 . What is the minimum positron energy required to produce the B Meson pair? (You may neglect the rest mass energies of the electron and the positron).

2. Relevant equations
Invarience of the interval? Lorentx transforms for energy and momentum?

3. The attempt at a solution

Obviously not a linear subtraction (I wish). In the CM (ZM/COM) frame, it seems to me that the electron and the positron have equal energies, E, where E= 5.3GeV

Their momenta are equal and opposite, and the value for the invarient of the whole system is 4*(5.3 GeV)^2

gamme = g
If I then use E' = g(E - vp) and take p to be zero as the unprimed frame is the cm frame, I can work out the velocity - but then I get stuck, and I'm a bit dubious about this wole last step. (The idea would then be to transform the total energy by the same amount and subtract the 9 from it)

Any help would be greatly appreciated, as would any quicker (non 4 vector based please because this is first year undergrad stuff), methods.

Thanks
Cpfoxhunt

Reply 1

Morbo

17

So, the first thing you did right was transform to the centre of mass/centre of momentum frame, and stated that the energy in the centre of momentum frame,

E_{CM}

, is the minimum energy required for the creation of our two mesons, which will be at rest in the CM frame i.e. the rest mass energy of the two mesons.

Remember this important rule that the centre of momentum energy is the energy available for particle creation.

So we know

E_{CM} = 2 \times 5.3\mathrm{Gev} = 10.6\mathrm{Gev}

Now, the invariance of the interval is the easiest way. So I presume you know that

E^2 - p^2c^2 = m^2c^4

is invariant between inertial frame of reference. Know also that this invariance applies both to single particles AND to systems of particles, where the interval becomes:

(\sum E_i)^2 - (\sum p_i)^2c^2

So let's find an expression for the interval for both the frames, and then we can equate them.

Centre of Momentum Frame
This is easy, since the total momentum is zero,

\sum p_i=0

in the CM frame! So the interval is simply:

E_{CM}^2 - (\sum p_i)^2c^2 = E_{CM}^2

Lab Frame
So again, we write down the interval for the system of particles. Let:

E_{e}

and

p_{e}

be the energy and momentum of the electron respectively.

E

and

-p

be the energy and momentum of the positron respectively (since the positron has momentum in the opposite direction to the electron).

So:

\\(\sum E_i)^2 - (\sum p_i)^2c^2[br]\\= (E_{e} + E)^2 - (p_{e} - p)^2c^2[br]\\= E_{e}^2 + 2E_{e}E + E^2 - (p_{e}^2 - 2p_{e}p + p^2)c^2[br]\\= [E_{e}^2 - p_{e}^2c^2] + [E^2 - p^2c^2] + 2E_{e}E + 2p_{e}pc^2[br]

Now, look at the square brackets. They are just the intervals for the electron and the positron, which we know are just equal to the square of their rest mass energies. We are told that their rest mass energies are negligible, so we are allowed to take their intervals to be zero:

E_{e}^2 - p_{e}^2c^2 \approx 0

E^2 - p^2c^2 \approx 0

This simplifies the expression for the lab frame interval to:

2E_{e}E + 2p_{e}pc^2

And by using their intervals again, we can substitute for

p_{e} = \frac{E_{e}}{c}

and

p = \frac{E}{c}

, yielding the lab frame interval to be:

2E_{e}E + 2\frac{E_e}{c}\frac{E}{c}c^2 = 4E_{e}E

And finally...
We use the invariance of the interval between these two frames in order to equate the two expressions we have, and get:

E_{CM}^2 = 4E_{e}E

We can now rearrange this and find

\\E = \frac{E_{CM}^2}{4E_{e}}[br]\\= \frac{(10.6\mathrm{GeV})^2}{4 \times 9.0\mathrm{GeV}}[br]\\= 3.12\mathrm{GeV} \text{ to 3 s.f.}

Hope this helps!

Reply 2

Cpfoxhunt

OP

Helps abolutely hugely thank you very much!
Just one thing - is the answer 2.83 or 3.12 GeV?
Cheers
Chris

Reply 3

Morbo

17

EDIT: Yeah, it's

3.12\mathrm{GeV}

. I have no idea how I got 2.83 before!

Reply 4

Wangers

16

Sorry to hijack the thread, but:

Considering special relativity, as an object approachs C, the energy needed to propel the object to greater velocities approaches infinity - hence no object can exceed V=c in space.

As space time distorts, is the energy physiclly manifesting in an increase in mass - hence causing the difficulty in increasing speed, or is it distorting the intangible gravimetric fields of space? ie is the impossibility to do with a change in the object, or in the fabric of space?

Reply 5

Morbo

17

Wangers

As space time distorts, is the energy physiclly manifesting in an increase in mass - hence causing the difficulty in increasing speed, or is it distorting the intangible gravimetric fields of space? ie is the impossibility to do with a change in the object, or in the fabric of space?

Firstly, people tend to have trouble with relativity because they seek to modify Newtonian concepts in order to understand relativistic ones. Phrases that give this away here are:

"is the energy phyiscally...an increase in mass"
"difficulty in increasing speed"

Newtonian mechanics - the mechanics we perceive to be "logical" because of our experience - is, in fact, a modification of relativity, and once you look at the subject that way round, things become a lot easier. Accept the relativistic mechanics and then see how Newtonian mechanics emerges.

So, if I may address your questions in this light, I will try to explain. The key concepts to understand this are energy density and spacetime. Think of mass as simply an area of high energy density, and know that when there are non-uniform energy densities, spacetime is not flat. (I hope you understand what I mean by flat spacetime, in terms of a the path of a light beam being that of a straight spatial line).

When an object is travelling at relativistic speeds relative to us, the increased energy we add in order to "accelerate" it does not go into what we would normally see as an "increase in speed". This is simply a fact we must accept! In other words,

E \neq \frac{1}{2}mv^2

.

If we look at it like this, then the energy we add is both an "increase in mass" (if we seek to modify Newtonian mechanics - although this is not an idea physicists like to use for the reasons I outlined initially), and a distortion of spacetime, because the increased energy density distorts spacetime as a mass would, and hence the two effects are indistinguishable.

This is related to the indistinguishability of a linearly accelerating frame and a gravitational field - the very concepts of mass and of acceleration are inextricably linked.