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The Geometry of Special Relativity

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The Geometry of Special Relativity THE GEOMETRY OF SPECIAL RELATIVITY Tevian Dray Department of Mathematics Oregon State University The Lorentz transformations at the heart of special relativity are TRIANGLE TRIG LENGTH CONTRACTION COSMIC RAYS just hyperbolic rotations. Special relativity itself can therefore be beautifully described in terms of \hyperbola geometry". ct ct' ct ct' Consider ¹-mesons produced by the collision of cosmic rays with 4 3 gas nuclei in the atmosphere 60 kilometers above the surface of x' x' the earth, which then move vertically downward at nearly the β x x speed of light. The half-life before -mesons decay into other par- HYPERBOLA GEOMETRY 5 ¹ We now recast ordinary triangle trig into hyperbola geometry. ticles is 1.5 microseconds. Assuming it doesn't decay, how long Euclidean distance is based on the unit circle, the set of points which 3 would it take a ¹-meson to reach the surface of the earth? As- Suppose you know tanh ¯ = 5, and you wish to determine cosh ¯. are unit distance from the origin. Hyperbola geometry is obtained One can of course do this algebraically, using the identity suming there were no time dilation, approximately what fraction simply by using a di®erent distance function! Measure the \squared Consider ¯rst a meter stick at rest, as shown in the ¯rst sketch above. of the mesons would reach the earth without decaying? In actual distance" of a point B = (x; y) from the origin using the de¯nition 1 fact, roughly an eighth of the mesons would reach the earth! How cosh2 ¯ = How \wide" is the worldsheet of the stick? The observer at rest of 2 fast are they going? 2 2 2 1 tanh ¯ course measures the length of the stick by locating both ends at the ± = x y ¡ 400 same time, and measuring the distance between them. At t = 0, this ¡ But it is easier to draw any triangle containing an angle whose hy- 3 corresponds to the 2 heavy dots in the sketch, one at the origin and Then, as shown below the unit \circle" becomes the unit hyperbola perbolic tangent is 5. In this case, the obvious choice would be the the other on the unit hyperbola. But all points on the unit hyperbola 400 triangle shown above, with sides of 3 and 5. 9 9 2 2 are at an interval of 1 meter from the origin. The observer at rest x y = 1 What is cosh ¯? Well, we ¯rst need to work out the length ± of the ¡ therefore concludes, unsurprisingly, that the meter stick is 1 meter α hypotenuse. The (hyperbolic) Pythagorean Theorem tells us that α and we further restrict ourselves to the branch with x > 0. If B is a long. point on this hyperbola, then we can de¯ne the hyperbolic angle ¯ 52 32 = ±2 How long does a moving observer think the stick is? This is just Assume the mesons travel at the speed of light. Then it takes them between the line from the origin to B and the (positive) x-axis to be ¡ the \width" of the worldsheet as measured by the moving observer. 60 km the Lorentzian length d of the arc of the unit hyperbola between B so ± is clearly 4. Take a good look at this 3-4-5 triangle of hyperbola This observer follows the same procedure, by locating both ends of the = 200 ¹s 2 2 2 8 m and the point (1; 0), where d = dx dy . We could then de¯ne geometry! Now that we know all the sides of the triangle, it is easy stick at the same time, and measuring the distance between them. 3 10 s ¡ £ the hyperbolic trig functions to be the coordinates (x; y) of B, that to see that cosh ¯ = 5. But time now corresponds to t , not t. At t = 0, this measurement ¯ ¯ 4 0 0 200 400 400 is ¯ ¯ corresponds to the heavy line in the sketch. Since this line fails to to reach the earth. But 200 ¹s is = half-lives, so only 2¡ 3 1:5 3 reach the unit hyperbola, it is clear that the moving observer measures of the mesons would reach the earth! cosh ¯ = x LORENTZ TRANSFORMATIONS the length of a stationary meter stick to be less than 1 meter. Since in fact an eighth reach the earth, which corresponds to 3 half- sinh ¯ = y To determine the exact value measured by the moving observer, use 400=3 lives, time must be dilated by a factor of 3 , so that The Lorentz transformation from a moving frame (x0; ct0) to a frame (hyperbolic) triangle trig. The heavy line in the sketch is the hy- and by symmetry, the point A on this hyperbola has coordinates (x; ct) at rest is given by potenuse of a right triangle, whose horizontal leg measures 1 meter. 400 (x; y) = (sinh ¯; cosh ¯); see the ¯gure below. A little work shows v cosh ® = If the observer is moving with speed c = tanh ¯, then the length of 9 that this de¯nition is exactly the same as the standard one, namely v x = (x0 + ct0) the hypotenuse in meters must be 1= cosh ¯. c But, as shown in the ¯rst drawing above, ¯ ¯ v What if the stick is moving and the observer is at rest? This situation e + e¡ ct = ct0 + x0 cosh ¯ = c is shown in the second sketch above. The worldsheet now corresponds 2 2 2 v p400 9 ³ ´ to a \rotated rectangle", indicated by the parallelograms in the sketch. = tanh ® = ¡ :99974684 e¯ e ¯ We can simplify things still further. Introduce the rapidity ¯ via c 400 ¼ sinh ¯ = ¡ ¡ The fact that the meter stick is 1 meter long in the moving frame is 2 v shown by the distance between the 2 heavy dots (along t0 = 0), and A more accurate argument would use the fact that the mesons travel = tanh ¯ the measurement by the observer at rest is indicated by the heavy line To see this, use x2 y2 = 1 to compute c 60 km in 4:5 ¹s (of proper time). Thus, ¡ (along t = 0). Again, it is clear that the stick appears to have shrunk, Inserting this into the expression for we obtain m dx2 dy2 since the heavy line fails to reach the unit hyperbola. Furthermore, (60 km)(1000 ) 400 2 2 2 2 sinh ® = km = d¯ d = dy dx = 2 = 2 the heavy line is again the hypotenuse of a right triangle (!), whose 6 8 m ´ ¡ x 1 y + 1 2 (4:5 10¡ s)(3 10 s ) 9 ¯ ¯ ¡ 1 cosh ¯ spacelike leg again measures 1 meter. £ £ ¯ ¯ = = = cosh ¯ then take the square root¯ of either ¯expression and integrate. (The 1 tanh2 ¯ scosh2 ¯ sinh2 ¯ Thus, a moving object appears shorter by a factor 1= cosh ¯. so that, as shown in the second drawing above, integrals are hard.) Finally, solve for x or y in terms of ¯. ¡ ¡ p v 400 and = tanh ® = :99974697 y y' v c p4002 + 92 ¼ = tanh ¯ cosh ¯ = sinh ¯ DOPPLER EFFECT A c β B x' Inserting these identities into the Lorentz transformations above β TIME DILATION x brings them to the remarkably simple form x 4 x ct ct' x cosh ¯ sinh ¯ x0 4-x = 2 ct sinh ¯ cosh ¯ ct 2 µ ¶ µ ¶ µ 0 ¶ 4 α α Lorentz transformations are just hyperbolic rotations! 2 x' ACKNOWLEDGMENTS A rocket sends out ashes of light every 2 seconds in its own rest REFERENCES frame, which you receive every 4 seconds. How fast is it going? x This approach to special relativity grew out of class notes for a Draw a horizontal line as shown in the enlarged second drawing, so Using hyperbolic triangle trig, it is straightforward to show that mov- course on Reference Frames, which in turn forms part of a major [1] Corinne A. Manogue and Kenneth S. Krane, The Oregon State University that ing clocks run slow by the same factor as before, namely cosh ¯. Can Paradigms Project: Re-envisioning the Upper Level, you ¯nd the necessary right triangles for both cases in the ¯gure? upper-division curriculum reform e®ort, the Paradigms in Physics Physics Today 56, 53{58 (2003). x project [1], begun in the Department of Physics at Oregon State Uni- tanh ® = The Geometry of Special Relativity 4 x [2] Tevian Dray, , in preparation ¡ versity in 1997, and supported in part by NSF grants DUE{965320 (http://www.math.oregonstate.edu/~tevian/geometry). (4 x)2 x2 = 22 ¡ ¡ and DUE{0231194. This presentation is largely excerpted from a [3] Tevian Dray, The Geometry of Special Relativity, 3 v 3 book in preparation [2], where further details can be found, and has Physics Teacher 46, 144{150 (2004). which is easily solved for x = 2, so that c = tanh ® = 5. also appeared in [3].
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