ehild said:
Try to derive the relation between the final KE and the generator energy instead of calculating with numbers from the beginning. You get the momentum of the accelerated proton from the formula above, using E=m0c2+E(generator). No need to calculate the speed.
And let me know, what you learned from the Science article.
I got the formula from Giancoli: Physics for Scientist and Engineers.
ehild
Ahh! Much quicker technique. And it gets the same answer as I was getting with my more laborious technique (after correcting the boo-boo you found, and thanks for that).
The
Science article doesn't provide much illumination beyond explaining why one would be interested in this sort of problem. It seems that at the time the article was written, particle accelerators typically accelerated a particle toward a stationary target particle, thereby wasting a considerable amount of kinetic energy in conserving momentum. At nonrelativistic speeds, a perfectly inelastic collision wastes half the initial kinetic energy this way. At relativistic speeds, the waste is considerably greater, and that is the point of the exercise. But the article doesn't explain how to calculate the amount of relativistic kinetic energy that is taken up conserving momentum; it only gives a table showing the amounts for various accelerators that were in operation at the time. Calculating these amounts is left as an exercise for the student! Literally.
So here's how I think it must go.
E
proton = T
i + mc
2
E
proton = 3 + .938
E
proton = 3.938 GeV
E
proton2 = p
i2c
2 + (mc
2)
2
3.938
2 = p
i2c
2 + .938
2
p
i2c
2 = 14.628
p
i = 3.825/c
The total initial energy is:
E
i = T
i + (2)mc
2
E
i = 3 + (2)(.938)
E
i = 4.876 GeV
Now E
i = E
f and p
i = p
f. So the "Remainder Energy" (which could be rest energy of whatever particles exist after collision, radiation energy, etc.) after the collision that is not taken up conserving momentum is:
E
f2 = (p
f/c)
2c
2 + RE
2
4.876
2 = 3.825
2 + RE
2
RE = 3.024 GeV
Therefore, the kinetic energy after the collision is:
T
f = E
f - RE = 4.876 - 3.024 = 1.85 GeV
As a proportion of the initial kinetic energy, this is:
1.85/3 = .617
The chart in the article (reproduced for this exercise in my special relativity textbook) gives a value of .60. It bothers me to think there could be this much rounding error, but I don't know what else to think.
The values provided versus what I compute are:
Accelerator energy / Loss due to COM (published) / Loss due to COM (computed by me)
3 / .60 / .62
7 / .72 / .69
25 / .80 / .79
200 / .92 / .91
1000 / .96 / .96
There is a follow-up exercise. The article suggests that the lost kinetic energy can be recovered by building an accelerator that shoots opposing beams of protons at each other instead of shooting one beam of protons at a stationary mass. Thus, for example, if each beam is accelerated at 3 GeV, the total initial kinetic energy will be 6 GeV instead of 3 GeV, and since the total p
i of the opposed protons is 0, the total p
f must also be 0 and there should be no energy lost conserving momentum; it can all go into inelastic interactions.
The task we are given is to find the energy an old model, stationary target accelerator would require to produce the same amount of energy available for inelastic interactions as a new model opposing beam accelerator where each beam has an energy of (a) 3 GeV (b) 25 GeV (c) 31 GeV.
So suppose that the energy in each beam of a colliding beam accelerator is 3 GeV. Then the total initial energy is:
E
i = 2(T
i + mc
2)
E
i = 2(3.938) = 7.876 GeV
And since p
f = p
i = 0, kinetic energy after collision is 0 and all of E
f = E
i is available for inelastic interactions. This is what we called RE in the above, stationary target case. Using the above formulas, we need to solve for the initial T
i of the single, moving proton that will produce the same E
f as we just found in the colliding beam case.
We have:
E
f2 = (p
f/c)
2c
2 + RE
2
E
proton2 = p
i2c
2 + (mc
2)
2
E
proton = T
i + mc
2
E
i = T
i + (2)mc
2
Bearing in mind that E
f = E
i and p
f = p
i, we can rewrite the first equation as:
E
i2 = (p
i/c)
2c
2 + RE
2
We know RE and mc
2, so the four equations become:
E
i2 = (p
i/c)
2c
2 + 7.876
2
E
proton2 = p
i2c
2 + .938
2
E
proton = T
i + .938
E
i = T
i + (2).938
Thus, we have four equations and four unknowns. We solve for T
i:
T
i(2).938 + [(2).938]
2 = 7.876
2
T
i = 31.19 GeV
The value published in the
Science article is 31 GeV. The table published in the article versus the values I computed are:
Colliding beam energy / Single beam energy (published) / Single beam energy (computed by me)
3 / 31 / 31.19
25 / 1360 / 1432.62
31 / 2040 / 2173.04
Again, the discrepancy between the published values and what I'm getting bothers me, but I don't know what to do about it. Anyway, the energy savings achieved by the colliding beam technique are very impressive!
