1 00:00:03,660 --> 00:00:06,390 When we analyzed a one-dimensional elastic 2 00:00:06,390 --> 00:00:09,360 collision in any frame and we had 3 00:00:09,360 --> 00:00:15,060 a velocity V1 initial and V2 initial, 4 00:00:15,060 --> 00:00:16,650 that we saw that we could reduce-- 5 00:00:16,650 --> 00:00:20,490 this is a one-dimensional elastic collision-- 6 00:00:20,490 --> 00:00:22,950 we always have the energy condition and the momentum 7 00:00:22,950 --> 00:00:23,950 condition. 8 00:00:23,950 --> 00:00:26,970 But when we combine the energy movement together, 9 00:00:26,970 --> 00:00:31,770 we found the following idea that the relative velocity 10 00:00:31,770 --> 00:00:36,240 V1, 2, which we called the relative velocity initial, 11 00:00:36,240 --> 00:00:40,860 was just equal to the final relative velocity. 12 00:00:40,860 --> 00:00:45,480 The statement there was that V1 initial minus V2 13 00:00:45,480 --> 00:00:53,220 initial is equal to minus V1 final minus V2 final. 14 00:00:53,220 --> 00:01:01,480 And we call this the energy momentum relation 15 00:01:01,480 --> 00:01:04,480 for our classical mechanics. 16 00:01:04,480 --> 00:01:09,160 And then we could combine that with our conservation of energy 17 00:01:09,160 --> 00:01:12,910 law, conservation of momentum. 18 00:01:12,910 --> 00:01:16,150 And we've got a linear system of equations 19 00:01:16,150 --> 00:01:20,020 that is much easier to solve than the quadratic system. 20 00:01:20,020 --> 00:01:22,960 Now what I'd like to show is that this concept 21 00:01:22,960 --> 00:01:28,690 of relative velocity, that V1, 2 relative velocity 22 00:01:28,690 --> 00:01:34,870 is independent of the choice of reference frame. 23 00:01:37,780 --> 00:01:41,240 And so if we want to analyze a collision 24 00:01:41,240 --> 00:01:44,229 in any other reference frame, then we always 25 00:01:44,229 --> 00:01:49,030 can keep this result. So now let's look at that. 26 00:01:49,030 --> 00:01:52,479 So again, let's imagine that we have two particles, where we 27 00:01:52,479 --> 00:01:55,930 have particle 1 and particle 2. 28 00:01:55,930 --> 00:02:01,360 We have some origin, r1 and r2. 29 00:02:01,360 --> 00:02:04,450 And now we want to choose another reference frame. 30 00:02:04,450 --> 00:02:09,490 So suppose that we pick a second reference frame, which has 31 00:02:09,490 --> 00:02:13,570 maybe some origin over here. 32 00:02:13,570 --> 00:02:20,230 And we have the vectors r1 prime and r2 prime. 33 00:02:20,230 --> 00:02:24,520 And the relative vector from the center of one reference 34 00:02:24,520 --> 00:02:28,010 frame to the center of the other reference frame. 35 00:02:28,010 --> 00:02:29,920 So what we have are the two conditions 36 00:02:29,920 --> 00:02:37,050 that r1 is equal to capital R plus r1 prime. 37 00:02:37,050 --> 00:02:40,240 r1 is capital R plus r1 prime. 38 00:02:40,240 --> 00:02:45,070 r2 is capital R plus r2 prime. 39 00:02:45,070 --> 00:02:48,050 And now if we subtract these two equations, 40 00:02:48,050 --> 00:02:54,490 we have r1 minus r2 equals r1 prime minus r2 prime. 41 00:02:54,490 --> 00:02:59,440 And this shows us that the relative position 42 00:02:59,440 --> 00:03:06,910 vectors-- and let's draw that this one from 2 to 1. 43 00:03:06,910 --> 00:03:12,850 And even in our diagram, we can see that the vector from 2 to 1 44 00:03:12,850 --> 00:03:18,400 does not depend on the choice of reference frame. 45 00:03:18,400 --> 00:03:21,579 And even more importantly, when we differentiate, 46 00:03:21,579 --> 00:03:28,720 we get that V1 minus V2 is equal to V1 prime minus V2 prime. 47 00:03:28,720 --> 00:03:33,130 And that becomes our statement that the relative velocity 48 00:03:33,130 --> 00:03:38,380 vector is independent of the choice of reference frame.