class33 - 1.6.2 Proper time Consider an object moving with...

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1.6.2 Proper time Consider an object moving with a velocity u relative to our rest frame S . Assume two events occur with the object, separated by some time dt ± . In the S ± frame, dt ± = ds ± c (116) since dx ± = 0 in its own rest frame. As discussed before, this is the proper time (i.e. were the clock is moving along with the body), which we now denote τ : so = ds ± c = ds c (117) since ds is invariant and c is a constant. Now let us Fnd the relationship between the proper time and a clock in the frame S . ±rom ds 2 = c 2 2 = c 2 dt 2 - dx 2 - dy 2 - dz 2 (118) giving = dt ± 1 - dx 2 + dy 2 + dz 2 c 2 dt 2 ² 1 / 2 (119) = dt ± 1 - u 2 c 2 ² 1 / 2 (120) whre u is the velocity of the moving clock relative to the S frame; therefore = ds c = dt ± 1 - u 2 c 2 ² 1 / 2 (121) 2 Relativistic Mechanics We now work to modify the structure of classical mechanics brought about by special rel- ativity so that physical laws remain invariant under Lorentz transformation. To begin, we deFne our relevant variables in our four-dimensional world. 2.1 Four-velocity and Four-acceleration The position vector of a point in Minkowski space is a four-vector X =( x 1 ,x 2 3 4 ) (122) where x 4 = ict . This four-vector deFnition leaves the “length” invariant (length) 2 = ³ μ x 2 μ (123) = x 2 + y 2 + z 2 - c 2 t 2 = s 2 (124) 13
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Which is the invariant interval considered previously. The Lorentz transformation for this
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class33 - 1.6.2 Proper time Consider an object moving with...

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