[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.05,0:00:03.82,Default,,0000,0000,0000,,Pythagoras theorem is one of the\Nmost fundamental theorems in
Dialogue: 0,0:00:03.82,0:00:08.02,Default,,0000,0000,0000,,mathematics. It says that if you\Nhave a right angle triangle like
Dialogue: 0,0:00:08.02,0:00:12.94,Default,,0000,0000,0000,,this. Then the square of the\Nlength of the hypotenuse is
Dialogue: 0,0:00:12.94,0:00:18.10,Default,,0000,0000,0000,,equal to the sum of the squares\Nof the lengths of these other
Dialogue: 0,0:00:18.10,0:00:22.85,Default,,0000,0000,0000,,two sides. And that's something\Nthat has important in
Dialogue: 0,0:00:22.85,0:00:27.50,Default,,0000,0000,0000,,trigonometry as well. For\Nexample, if I have an angle here
Dialogue: 0,0:00:27.50,0:00:32.98,Default,,0000,0000,0000,,theater, and if I choose the\Nunit so that the length of the
Dialogue: 0,0:00:32.98,0:00:34.25,Default,,0000,0000,0000,,hypotenuse is one.
Dialogue: 0,0:00:35.38,0:00:41.32,Default,,0000,0000,0000,,Then This distance is just the\Ncosine of the angle Theta Cause
Dialogue: 0,0:00:41.32,0:00:47.53,Default,,0000,0000,0000,,Theta. And this distance is the\Nsign of the angle theater.
Dialogue: 0,0:00:48.68,0:00:54.27,Default,,0000,0000,0000,,And so Pythagoras Theorem says\Nthat the square of cosine, Cos
Dialogue: 0,0:00:54.27,0:00:58.33,Default,,0000,0000,0000,,squared theater, plus the square\Nof the sign.
Dialogue: 0,0:00:58.94,0:01:00.30,Default,,0000,0000,0000,,Sine squared theater
Dialogue: 0,0:01:00.94,0:01:02.16,Default,,0000,0000,0000,,Equals 1.
Dialogue: 0,0:01:03.44,0:01:08.67,Default,,0000,0000,0000,,Now this has applications in\Nthe real world. If we want to
Dialogue: 0,0:01:08.67,0:01:12.16,Default,,0000,0000,0000,,measure positions and lengths\Nusing systems of coordinates.
Dialogue: 0,0:01:12.16,0:01:15.65,Default,,0000,0000,0000,,So let's suppose that I have\Na pencil.
Dialogue: 0,0:01:17.12,0:01:19.72,Default,,0000,0000,0000,,And I imagine that my pencil is
Dialogue: 0,0:01:19.72,0:01:21.74,Default,,0000,0000,0000,,positioned. Just here.
Dialogue: 0,0:01:22.52,0:01:25.76,Default,,0000,0000,0000,,So it's exactly on the\Nhypotenuse of a triangle.
Dialogue: 0,0:01:26.63,0:01:30.85,Default,,0000,0000,0000,,And it supposed that\NI've got some
Dialogue: 0,0:01:30.85,0:01:35.68,Default,,0000,0000,0000,,coordinate axes. The\Nusual X axis there and
Dialogue: 0,0:01:35.68,0:01:37.48,Default,,0000,0000,0000,,AY axis here.
Dialogue: 0,0:01:38.78,0:01:41.35,Default,,0000,0000,0000,,So now with my pencil back here
Dialogue: 0,0:01:41.35,0:01:44.57,Default,,0000,0000,0000,,this point. Is the
Dialogue: 0,0:01:44.57,0:01:46.28,Default,,0000,0000,0000,,point? XY.
Dialogue: 0,0:01:48.34,0:01:55.46,Default,,0000,0000,0000,,And so. Pythagoras theorem tells\Nme that X squared plus Y
Dialogue: 0,0:01:55.46,0:01:57.94,Default,,0000,0000,0000,,squared must equal 1.
Dialogue: 0,0:01:59.67,0:02:04.92,Default,,0000,0000,0000,,But of course, in the real world\Nwe don't have a given set of
Dialogue: 0,0:02:04.92,0:02:09.04,Default,,0000,0000,0000,,coordinate axes. I might choose\None set. You might choose a
Dialogue: 0,0:02:09.04,0:02:14.95,Default,,0000,0000,0000,,different set. Let's suppose\Nthat you chose a set UV with the
Dialogue: 0,0:02:14.95,0:02:17.26,Default,,0000,0000,0000,,same origin, but pointing in
Dialogue: 0,0:02:17.26,0:02:22.00,Default,,0000,0000,0000,,different directions. So if I\Ndraw a new set of axes.
Dialogue: 0,0:02:23.05,0:02:29.07,Default,,0000,0000,0000,,I'll have my U access\Ngoing this way.
Dialogue: 0,0:02:29.85,0:02:33.51,Default,,0000,0000,0000,,And my
Dialogue: 0,0:02:33.51,0:02:39.66,Default,,0000,0000,0000,,viaccess\NGoing this
Dialogue: 0,0:02:39.66,0:02:41.56,Default,,0000,0000,0000,,way.
Dialogue: 0,0:02:42.93,0:02:45.97,Default,,0000,0000,0000,,And then with respect to these
Dialogue: 0,0:02:45.97,0:02:53.09,Default,,0000,0000,0000,,new axes. The pencil tip\Nwill be at the point.
Dialogue: 0,0:02:53.09,0:02:54.07,Default,,0000,0000,0000,,UV.
Dialogue: 0,0:02:56.14,0:02:58.89,Default,,0000,0000,0000,,So if I now drop perpendiculars.
Dialogue: 0,0:03:01.01,0:03:04.35,Default,,0000,0000,0000,,Then this distances you
Dialogue: 0,0:03:04.35,0:03:08.06,Default,,0000,0000,0000,,along here. This
Dialogue: 0,0:03:08.06,0:03:14.84,Default,,0000,0000,0000,,distance. His V\Nand this new angle.
Dialogue: 0,0:03:14.84,0:03:15.69,Default,,0000,0000,0000,,Is 5.
Dialogue: 0,0:03:17.16,0:03:20.58,Default,,0000,0000,0000,,But the length of the pencil\Nstill the same, it's still
Dialogue: 0,0:03:20.58,0:03:21.82,Default,,0000,0000,0000,,one that hasn't changed.
Dialogue: 0,0:03:23.26,0:03:30.11,Default,,0000,0000,0000,,And so we must have\Ncaused square 5 plus sign
Dialogue: 0,0:03:30.11,0:03:32.85,Default,,0000,0000,0000,,squared Phi equals 1.
Dialogue: 0,0:03:33.91,0:03:38.64,Default,,0000,0000,0000,,Be cause you squared plus B\Nsquared equals 1.
Dialogue: 0,0:03:39.48,0:03:43.03,Default,,0000,0000,0000,,So called squared Theta plus\Nsign squared theater equals 1
Dialogue: 0,0:03:43.03,0:03:47.64,Default,,0000,0000,0000,,Costco at 5 plus sign. Squared 5\Nequals 1. It doesn't matter what
Dialogue: 0,0:03:47.64,0:03:51.55,Default,,0000,0000,0000,,these angles, Theta or Phi are.\NThis equation must always hold,
Dialogue: 0,0:03:51.55,0:03:55.81,Default,,0000,0000,0000,,so it's not just an equation,\Nit's actually what we call an
Dialogue: 0,0:03:55.81,0:03:58.65,Default,,0000,0000,0000,,identity. Something that's\Nalways true for any angle
Dialogue: 0,0:03:58.65,0:04:00.42,Default,,0000,0000,0000,,theater or any angle fire.
Dialogue: 0,0:04:01.58,0:04:05.65,Default,,0000,0000,0000,,Well, that's two dimensions, but\Nof course we live in a 3
Dialogue: 0,0:04:05.65,0:04:09.66,Default,,0000,0000,0000,,dimensional world. So how would\Npythagoras theorem work there?
Dialogue: 0,0:04:10.61,0:04:11.90,Default,,0000,0000,0000,,Well our pencil.
Dialogue: 0,0:04:12.74,0:04:15.62,Default,,0000,0000,0000,,Would now be sticking\Nup out of the paper.
Dialogue: 0,0:04:17.43,0:04:22.05,Default,,0000,0000,0000,,And we'd have to have not\Njust an X coordinate and AY
Dialogue: 0,0:04:22.05,0:04:25.13,Default,,0000,0000,0000,,coordinate, but also as Ed\Ncoordinate going vertically
Dialogue: 0,0:04:25.13,0:04:25.52,Default,,0000,0000,0000,,upwards.
Dialogue: 0,0:04:26.83,0:04:30.83,Default,,0000,0000,0000,,And then it would no longer be\Ntrue that the X squared plus Y
Dialogue: 0,0:04:30.83,0:04:34.27,Default,,0000,0000,0000,,squared value will be the square\Nof the length of the pencil.
Dialogue: 0,0:04:34.95,0:04:38.29,Default,,0000,0000,0000,,It would be a bit shorter\Nbecause it would be the distance
Dialogue: 0,0:04:38.29,0:04:41.34,Default,,0000,0000,0000,,to where the shadow of the\Npencil falls on the paper.
Dialogue: 0,0:04:42.84,0:04:45.83,Default,,0000,0000,0000,,Instead, we have to add the\Nsquare of the vertical distance
Dialogue: 0,0:04:45.83,0:04:48.28,Default,,0000,0000,0000,,as well to get us to the tip of
Dialogue: 0,0:04:48.28,0:04:52.78,Default,,0000,0000,0000,,the pencil. So in three\Ndimensions we would have not X
Dialogue: 0,0:04:52.78,0:04:56.67,Default,,0000,0000,0000,,squared plus Y squared equals\Nthe length of the pencil squared
Dialogue: 0,0:04:56.67,0:04:59.50,Default,,0000,0000,0000,,plucked, but instead X squared\Nplus Y squared.
Dialogue: 0,0:05:00.04,0:05:03.06,Default,,0000,0000,0000,,Plus, Z squared equals the\Nlength of the pencil squared.
Dialogue: 0,0:05:05.35,0:05:08.83,Default,,0000,0000,0000,,And if the pencil happened to be\Nin three dimensions, but lying
Dialogue: 0,0:05:08.83,0:05:09.99,Default,,0000,0000,0000,,flat in the plane.
Dialogue: 0,0:05:10.80,0:05:14.39,Default,,0000,0000,0000,,That would mean that zed\Nwas zero, and so in our
Dialogue: 0,0:05:14.39,0:05:17.97,Default,,0000,0000,0000,,formula, if said is zero,\Nwe just get the X squared
Dialogue: 0,0:05:17.97,0:05:19.60,Default,,0000,0000,0000,,plus Y squared back again.
Dialogue: 0,0:05:21.70,0:05:24.90,Default,,0000,0000,0000,,Well, that's three dimensions,\Nand what mathematicians like to
Dialogue: 0,0:05:24.90,0:05:28.82,Default,,0000,0000,0000,,do, of course, is to generalize.\NWe've had two dimensions, 3
Dialogue: 0,0:05:28.82,0:05:30.84,Default,,0000,0000,0000,,dimensions. What about four
Dialogue: 0,0:05:30.84,0:05:34.91,Default,,0000,0000,0000,,dimensions? Well, that might\Nseem a bit physically unreal,
Dialogue: 0,0:05:34.91,0:05:39.29,Default,,0000,0000,0000,,but in fact in the middle of the\N19th century physicists did
Dialogue: 0,0:05:39.29,0:05:43.67,Default,,0000,0000,0000,,start to use 4 dimensions to\Ndescribe what was going on in
Dialogue: 0,0:05:43.67,0:05:48.42,Default,,0000,0000,0000,,the world. They have XY and Z as\Nthe three dimensions of space.
Dialogue: 0,0:05:48.42,0:05:50.60,Default,,0000,0000,0000,,They also have an extra fourth
Dialogue: 0,0:05:50.60,0:05:56.09,Default,,0000,0000,0000,,dimension. T for time and in\Nfact this was very important in
Dialogue: 0,0:05:56.09,0:05:59.86,Default,,0000,0000,0000,,Einstein's theory of Relativity,\Nwhich he published in 19105.
Dialogue: 0,0:06:00.93,0:06:02.09,Default,,0000,0000,0000,,So how does that work?
Dialogue: 0,0:06:02.66,0:06:09.46,Default,,0000,0000,0000,,Well, if we look\Nat what a four
Dialogue: 0,0:06:09.46,0:06:12.86,Default,,0000,0000,0000,,dimensional length might be.
Dialogue: 0,0:06:13.71,0:06:18.59,Default,,0000,0000,0000,,Well, it's Square would be X\Nsquared plus Y squared plus set
Dialogue: 0,0:06:18.59,0:06:23.07,Default,,0000,0000,0000,,squared. And then you might\Nthink it would be plus T
Dialogue: 0,0:06:23.07,0:06:25.11,Default,,0000,0000,0000,,squared, but actually there's a
Dialogue: 0,0:06:25.11,0:06:29.63,Default,,0000,0000,0000,,little twist. Because we don't\Nhave plus T squared.
Dialogue: 0,0:06:30.29,0:06:35.67,Default,,0000,0000,0000,,In fact we have minus T squared\Nbecause time is physically
Dialogue: 0,0:06:35.67,0:06:37.62,Default,,0000,0000,0000,,different from the spatial
Dialogue: 0,0:06:37.62,0:06:42.45,Default,,0000,0000,0000,,coordinates. And really, in\Nthere there's a hidden factor
Dialogue: 0,0:06:42.45,0:06:47.34,Default,,0000,0000,0000,,of C squared. Really, it's X\Nsquared plus Y squared +6
Dialogue: 0,0:06:47.34,0:06:51.35,Default,,0000,0000,0000,,squared minus C squared T\Nsquared. To get the
Dialogue: 0,0:06:51.35,0:06:55.36,Default,,0000,0000,0000,,dimensions right and see is\Nthe speed of light.
Dialogue: 0,0:06:56.42,0:07:01.88,Default,,0000,0000,0000,,But I suppose that we've chosen\Nour units so that the speed of
Dialogue: 0,0:07:01.88,0:07:07.34,Default,,0000,0000,0000,,light is actually equal to 1,\Nand so X squared plus Y squared
Dialogue: 0,0:07:07.34,0:07:12.80,Default,,0000,0000,0000,,plus Z squared minus T squared\Nis the formula for the square of
Dialogue: 0,0:07:12.80,0:07:14.48,Default,,0000,0000,0000,,the four dimensional length.
Dialogue: 0,0:07:15.83,0:07:18.49,Default,,0000,0000,0000,,And now, as in the three\Ndimensional case, we could ask
Dialogue: 0,0:07:18.49,0:07:22.68,Default,,0000,0000,0000,,what happens if we. Just look at\Nlength line one particular
Dialogue: 0,0:07:22.68,0:07:29.15,Default,,0000,0000,0000,,plane. Well, let's take the XY\Nplane then. That would mean that
Dialogue: 0,0:07:29.15,0:07:32.16,Default,,0000,0000,0000,,zed and T were both 0.
Dialogue: 0,0:07:32.90,0:07:37.60,Default,,0000,0000,0000,,And so we would get for\Nour formula X squared plus
Dialogue: 0,0:07:37.60,0:07:38.45,Default,,0000,0000,0000,,Y squared.
Dialogue: 0,0:07:40.80,0:07:47.56,Default,,0000,0000,0000,,But what if we look instead not\Nat the XY plane, but at the XT
Dialogue: 0,0:07:47.56,0:07:49.77,Default,,0000,0000,0000,,plane. And then.
Dialogue: 0,0:07:50.57,0:07:55.73,Default,,0000,0000,0000,,Y and said will be 0 and\Nwe would have.
Dialogue: 0,0:07:55.73,0:07:59.06,Default,,0000,0000,0000,,X squared not plus T squared.
Dialogue: 0,0:07:59.79,0:08:03.04,Default,,0000,0000,0000,,But minus T squared.
Dialogue: 0,0:08:04.69,0:08:07.19,Default,,0000,0000,0000,,And now that gives us a problem.
Dialogue: 0,0:08:07.88,0:08:13.35,Default,,0000,0000,0000,,Give us a problem if we try\Nto express the angle between
Dialogue: 0,0:08:13.35,0:08:17.00,Default,,0000,0000,0000,,our 4 dimensional length and\Na fixed direction.
Dialogue: 0,0:08:18.55,0:08:21.36,Default,,0000,0000,0000,,Because if we say that the angle
Dialogue: 0,0:08:21.36,0:08:26.58,Default,,0000,0000,0000,,was theater. Then we would\Nbe looking at Cos squared
Dialogue: 0,0:08:26.58,0:08:27.46,Default,,0000,0000,0000,,Theta minus.
Dialogue: 0,0:08:28.84,0:08:30.80,Default,,0000,0000,0000,,Sine squared Theta.
Dialogue: 0,0:08:32.35,0:08:35.43,Default,,0000,0000,0000,,And the problem is that this is\Nnot a constant for different
Dialogue: 0,0:08:35.43,0:08:40.39,Default,,0000,0000,0000,,values of Theta. If there were a\Nplus there, it would be constant
Dialogue: 0,0:08:40.39,0:08:44.22,Default,,0000,0000,0000,,with a minus. It's not a\Nconstant, so that doesn't work.
Dialogue: 0,0:08:44.22,0:08:47.35,Default,,0000,0000,0000,,We can't use trigonometric\Nfunctions to talk about these
Dialogue: 0,0:08:47.35,0:08:52.64,Default,,0000,0000,0000,,angles. Instead, we have to use\Nnew kinds of functions called
Dialogue: 0,0:08:52.64,0:08:57.75,Default,,0000,0000,0000,,hyperbolic functions. And these\Nare similar to the trigonometric
Dialogue: 0,0:08:57.75,0:09:03.43,Default,,0000,0000,0000,,functions, calls them sign, but\Nwe write them with the letter H
Dialogue: 0,0:09:03.43,0:09:06.26,Default,,0000,0000,0000,,on the end. So this one.
Dialogue: 0,0:09:06.87,0:09:12.00,Default,,0000,0000,0000,,Is caution and this one is a bit\Nharder to pronounce, so I'll
Dialogue: 0,0:09:12.00,0:09:13.58,Default,,0000,0000,0000,,pronounce it as shine.
Dialogue: 0,0:09:14.66,0:09:18.76,Default,,0000,0000,0000,,And for the angle I'll put you\Nrather than theater, because
Dialogue: 0,0:09:18.76,0:09:23.24,Default,,0000,0000,0000,,this isn't a real angle. It's a\Nhyperbolic angle, and it works
Dialogue: 0,0:09:23.24,0:09:25.48,Default,,0000,0000,0000,,in a different sort of way.
Dialogue: 0,0:09:26.54,0:09:29.16,Default,,0000,0000,0000,,And now.
Dialogue: 0,0:09:30.29,0:09:34.30,Default,,0000,0000,0000,,We can use these to obtain a\Nformula like this one that
Dialogue: 0,0:09:34.30,0:09:36.97,Default,,0000,0000,0000,,didn't work for the\Ntrigonometric functions, but one
Dialogue: 0,0:09:36.97,0:09:38.64,Default,,0000,0000,0000,,which does work with the
Dialogue: 0,0:09:38.64,0:09:40.88,Default,,0000,0000,0000,,hyperbolic functions. The reason
Dialogue: 0,0:09:40.88,0:09:47.32,Default,,0000,0000,0000,,it works? Is to do with where\Nthese functions come from. The
Dialogue: 0,0:09:47.32,0:09:49.39,Default,,0000,0000,0000,,trick functions come from.
Dialogue: 0,0:09:50.14,0:09:53.27,Default,,0000,0000,0000,,A circle. Where is the
Dialogue: 0,0:09:53.27,0:09:56.27,Default,,0000,0000,0000,,hyperbolic functions? Come
Dialogue: 0,0:09:56.27,0:09:59.37,Default,,0000,0000,0000,,from. A hyperbola.
Dialogue: 0,0:10:00.98,0:10:05.46,Default,,0000,0000,0000,,And if you know about the\Nequations of these curves, the
Dialogue: 0,0:10:05.46,0:10:09.93,Default,,0000,0000,0000,,equation of a circle is X\Nsquared plus Y squared equals
Dialogue: 0,0:10:09.93,0:10:11.16,Default,,0000,0000,0000,,the radius squared.
Dialogue: 0,0:10:13.50,0:10:16.06,Default,,0000,0000,0000,,But for I hyperbola like this.
Dialogue: 0,0:10:16.59,0:10:20.21,Default,,0000,0000,0000,,The equation is similar, but\Nwithout a plus sign. Instead it
Dialogue: 0,0:10:20.21,0:10:21.52,Default,,0000,0000,0000,,has a minus sign.
Dialogue: 0,0:10:22.16,0:10:24.16,Default,,0000,0000,0000,,Uncaught a minus sign\Nis just what we want.
Dialogue: 0,0:10:25.37,0:10:30.47,Default,,0000,0000,0000,,And we find that rather than\Nhaving this formula, which
Dialogue: 0,0:10:30.47,0:10:35.06,Default,,0000,0000,0000,,doesn't work for trig functions,\Nthe corresponding one for
Dialogue: 0,0:10:35.06,0:10:38.12,Default,,0000,0000,0000,,hyperbolic functions, cosh\Nsquared U minus.
Dialogue: 0,0:10:39.10,0:10:40.71,Default,,0000,0000,0000,,Shine squared you.
Dialogue: 0,0:10:41.21,0:10:43.84,Default,,0000,0000,0000,,Does work and that this is
Dialogue: 0,0:10:43.84,0:10:49.45,Default,,0000,0000,0000,,actually equal? To one, no\Nmatter what the hyperbolic angle
Dialogue: 0,0:10:49.45,0:10:51.01,Default,,0000,0000,0000,,you might be.
Dialogue: 0,0:10:52.44,0:10:55.44,Default,,0000,0000,0000,,So the way that scientists think\Nabout the four dimensions of
Dialogue: 0,0:10:55.44,0:10:58.72,Default,,0000,0000,0000,,space time is very similar to\Nthe way in which they think
Dialogue: 0,0:10:58.72,0:11:00.08,Default,,0000,0000,0000,,about 3 dimensions of ordinary
Dialogue: 0,0:11:00.08,0:11:05.17,Default,,0000,0000,0000,,space. But there is this little\Ntwist that in the formula for
Dialogue: 0,0:11:05.17,0:11:09.76,Default,,0000,0000,0000,,the square of the length,\Nthere's a minus sign in front of
Dialogue: 0,0:11:09.76,0:11:13.96,Default,,0000,0000,0000,,tea, and that means that when\Nthey look at rotations involving
Dialogue: 0,0:11:13.96,0:11:17.01,Default,,0000,0000,0000,,the time coordinate, they're not\Nordinary rotations using
Dialogue: 0,0:11:17.01,0:11:19.69,Default,,0000,0000,0000,,trigonometric functions, they\Nare hyperbolic rotations using
Dialogue: 0,0:11:19.69,0:11:21.22,Default,,0000,0000,0000,,these new hyperbolic functions.