WEBVTT 00:00:07.100 --> 00:00:09.938 MARKUS KLUTE: Welcome back to 8.20, special relativity. 00:00:09.938 --> 00:00:11.480 In the previous section, we have seen 00:00:11.480 --> 00:00:15.170 how we can look at energy and momentum of particles 00:00:15.170 --> 00:00:16.219 in a decay. 00:00:16.219 --> 00:00:19.220 Here we now want to, in collisions of particles, 00:00:19.220 --> 00:00:21.230 create new particles. 00:00:21.230 --> 00:00:23.360 The example, the first example here, 00:00:23.360 --> 00:00:27.830 is the collision of two protons to create a proton, a neutron, 00:00:27.830 --> 00:00:29.510 and a charged pion. 00:00:29.510 --> 00:00:31.080 The masses are given there. 00:00:31.080 --> 00:00:34.040 So the question now is, what is the minimal energy 00:00:34.040 --> 00:00:37.790 needed in order for this process to occur 00:00:37.790 --> 00:00:39.950 in a fixed-target experiment? 00:00:39.950 --> 00:00:43.550 Fixed-target experiment is we have an accelerated proton 00:00:43.550 --> 00:00:44.990 and another proton at rest. 00:00:44.990 --> 00:00:48.570 This might just be a hydrogen target just sitting there. 00:00:48.570 --> 00:00:50.420 So the question is, how much energy-- 00:00:50.420 --> 00:00:52.460 how much do we have to accelerate 00:00:52.460 --> 00:00:56.930 the proton for this process to be possible? 00:00:56.930 --> 00:01:01.700 Now, again, stop the video, and try to work this out. 00:01:01.700 --> 00:01:06.350 The important part here is to realize 00:01:06.350 --> 00:01:10.100 that minimal energy here means that, after the decay 00:01:10.100 --> 00:01:12.890 or the decay after the process occurred, 00:01:12.890 --> 00:01:15.770 all the new particles need to be addressed. 00:01:15.770 --> 00:01:20.610 That is when the process requires minimal energy. 00:01:20.610 --> 00:01:23.300 So, instead of analyzing this in the laboratory frame, 00:01:23.300 --> 00:01:25.825 we want to analyze this in the center-of-mass frame. 00:01:25.825 --> 00:01:27.200 All right, the momentum has to be 00:01:27.200 --> 00:01:29.220 conserved in this discussion. 00:01:29.220 --> 00:01:31.440 So there needs to be some sort of momentum. 00:01:31.440 --> 00:01:33.720 But, in the center-of-mass frame, that's not required. 00:01:33.720 --> 00:01:36.890 So, in that frame, the momentum of all outgoing particles 00:01:36.890 --> 00:01:38.720 can be 0. 00:01:38.720 --> 00:01:40.710 And that's how we start the discussion here. 00:01:40.710 --> 00:01:42.170 So, in this S prime frame-- 00:01:42.170 --> 00:01:44.840 here S prime is the center-of-mass frame-- 00:01:44.840 --> 00:01:49.800 the energy, the minimal energy required, is 2 times 00:01:49.800 --> 00:01:52.890 the mass of the proton times gamma. 00:01:52.890 --> 00:01:54.990 So here, two protons are colliding 00:01:54.990 --> 00:01:57.690 with the same velocity. 00:01:57.690 --> 00:01:59.250 And that's then equal to the energy 00:01:59.250 --> 00:02:03.675 after this process, c squared times the sum of the masses, 00:02:03.675 --> 00:02:05.880 the sum of the mass of the proton, the neutron, 00:02:05.880 --> 00:02:08.190 and the charged pion. 00:02:08.190 --> 00:02:10.710 And then you just have to solve this for gamma 00:02:10.710 --> 00:02:18.360 to find gamma equal to 1.08 or beta in this frame of 0.37. 00:02:18.360 --> 00:02:22.410 Note, this is the gamma, relativistic gamma, 00:02:22.410 --> 00:02:26.220 or the velocity beta of the protons, two protons 00:02:26.220 --> 00:02:27.630 in the center-of-mass frame. 00:02:27.630 --> 00:02:30.100 So we're not quite there yet with our answer. 00:02:30.100 --> 00:02:32.430 The answer then needs to be boosted back 00:02:32.430 --> 00:02:34.140 into the laboratory frame. 00:02:34.140 --> 00:02:37.470 And we have seen how we can do this for beta or velocities 00:02:37.470 --> 00:02:38.050 in general. 00:02:38.050 --> 00:02:42.510 We find beta in the laboratory frame is 2 times-- 00:02:42.510 --> 00:02:46.890 or just result, 0.37, over 1 plus 0.37 00:02:46.890 --> 00:02:49.680 squared, which is 0.65. 00:02:49.680 --> 00:02:52.440 That velocity, we can then take and calculate 00:02:52.440 --> 00:02:58.210 the gamma factor of the proton in the fixed-target experiment. 00:02:58.210 --> 00:03:00.210 All right, so we analyzed this situation 00:03:00.210 --> 00:03:02.310 in the center-of-mass frame and then 00:03:02.310 --> 00:03:06.060 did a Lorentz transformation by just looking at the velocity 00:03:06.060 --> 00:03:09.890 into the fixed-target frame. 00:03:09.890 --> 00:03:13.750 So this means now, numerically, that the proton colliding 00:03:13.750 --> 00:03:18.100 with the proton at rest has a total energy of this one 00:03:18.100 --> 00:03:27.210 proton of gamma m0 c squared, which is 0.32 times 938 MeV 00:03:27.210 --> 00:03:29.230 over c-- 00:03:29.230 --> 00:03:30.970 MeV. 00:03:30.970 --> 00:03:36.370 And so that results in 1.238 GeV. 00:03:36.370 --> 00:03:39.110 But we're interested in the kinetic energy. 00:03:39.110 --> 00:03:40.660 So the kinetic energy here is given 00:03:40.660 --> 00:03:44.770 by gamma minus 1 m0 c squared, which is 300 MeV. 00:03:44.770 --> 00:03:48.730 So we have to accelerate a proton to 300 MeV 00:03:48.730 --> 00:03:54.430 in order to be able to have this process to occur. 00:03:54.430 --> 00:03:57.250 All right, very similar problem now, but here 00:03:57.250 --> 00:03:58.750 we want to produce anti-matter. 00:03:58.750 --> 00:04:04.870 So we have a process of proton plus proton into three protons 00:04:04.870 --> 00:04:06.450 and an antiproton. 00:04:06.450 --> 00:04:07.930 Charge is conserved. 00:04:07.930 --> 00:04:09.760 In the initial state, the charge was 2. 00:04:09.760 --> 00:04:12.730 In the final stage, the charge was plus 2 as well. 00:04:12.730 --> 00:04:17.089 OK, this works very similar as in the previous problem. 00:04:17.089 --> 00:04:20.360 But what we want to do here is compare the fixed target 00:04:20.360 --> 00:04:22.640 with symmetric collisions. 00:04:22.640 --> 00:04:25.370 OK, so, again, the question is, what is the minimal energy 00:04:25.370 --> 00:04:27.980 needed in order to produce antiprotons 00:04:27.980 --> 00:04:30.270 in proton-proton collisions? 00:04:30.270 --> 00:04:35.730 OK, so, exactly following the same procedure as before, 00:04:35.730 --> 00:04:38.670 in the center-of-mass energy, the energy is 2 times 00:04:38.670 --> 00:04:41.400 the mass of the protons times gamma times c squared. 00:04:41.400 --> 00:04:45.430 And that's 4 times the mass of the proton. 00:04:45.430 --> 00:04:48.640 OK, gamma prime, so the gamma factor 00:04:48.640 --> 00:04:50.380 in the center-of-mass frame is 2. 00:04:50.380 --> 00:04:52.180 Beta is 0.75. 00:04:52.180 --> 00:04:54.370 And then we just do the very same thing again. 00:04:54.370 --> 00:04:57.190 We calculate the velocity in the fixed-target frame. 00:04:57.190 --> 00:05:03.330 And we find the velocity of beta of 0.96 and gamma of 3.57. 00:05:03.330 --> 00:05:07.880 So, if we compare this now, we need a pair of 1 GeV-- 00:05:07.880 --> 00:05:10.140 remember, gamma minus 1 is the kinetic energy-- 00:05:10.140 --> 00:05:15.630 protons in a collider experiment or 2.57 GeV 00:05:15.630 --> 00:05:18.940 protons in a fixed-target experiment. 00:05:18.940 --> 00:05:22.080 OK, so you see that, in fixed-target experiment, 00:05:22.080 --> 00:05:23.910 in order to produce new particles, 00:05:23.910 --> 00:05:27.556 the energy has to be much larger, a factor of 2.5 00:05:27.556 --> 00:05:31.110 here in this example, than a colliding experiment. 00:05:31.110 --> 00:05:36.360 And that explains why we use collider experiments in order 00:05:36.360 --> 00:05:40.170 to test the energy frontier, in order to produce the largest 00:05:40.170 --> 00:05:41.190 possible energies. 00:05:41.190 --> 00:05:43.770 And the LHC is one example where we 00:05:43.770 --> 00:05:47.430 have proton-proton collisions in a circular ring where 00:05:47.430 --> 00:05:48.960 those protons are brought together 00:05:48.960 --> 00:05:51.890 in symmetrical collisions.