Relativity Lite

22 | Relativity Lite Please note that this relation is not an “equation,” since it involves an approximation symbol and has γ , on both the left and right sides. It is an iterative relation that is probably best read as a recipe: Take a rough value for γ and multiply it by 1 2 2 2 v c . Add 1 to the result. Stir for 15 seconds. Yield: 1 serving of a slightly better version of γ . We noticed in the last chapter that γ seems to be “1 plus a bit more” for a wide range of velocities. Let us take 1 as our “rough value for γ ,” and call it γ 0 . Our recipe serves up of a slightly better version of γ , which we find to be γ γ 1 2 2 0 2 2 2 2 1 1 2 1 1 2 1 1 1 2 ≡ + = + = + v c v c v c . We could keep on going in this fashion and find a third term that will depend on the fourth power of the velocity. Those of you who are a bit queasy about math can just plug your ears and say “LA LA LA” out loud while I write that one can apply the Pythagorean theorem to figure 1 and discover the exact expression for γ to be γ τ = ≡ − t v c 1 1 2 2 , which involves a square inside a dreaded square root at the bottom of the sea, so it is not as convenient as is our γ 1 , but you can get numbers out of your calculator for high speeds exactly the same shape as the triangle with hypotenuse c t , long side c τ , and short side v t , but rotated so that the miniature triangle’s hypotenuse is facing downward. You may remember the term similar triangles from ninth grade geometry: two such triangles have ratios of sides to each other (and, thus, angles) that are the same. The miniature triangle has hypotenuse v t and short side b so the ratios of these two sides for the miniature and big triangle are equal, b vt vt ct = , or b v t c = 2 . Finally, we note that the f in figures 1a, 1b, and 1c is in each case about 1/2 b , perhaps with a slight correction unnoticeable to the eye. Above, when we said we would “overlay the 9.54 cm c τ line in figure 1b on the 10 cm hypotenuse c t ,” and saw that “the excess is f = 0.46 cm,” we were simply taking the ratio of c t to c τ , which we know is the time-dilation factor γ . Putting these words into ratios, γ τ τ τ τ τ τ ≡ = + = + ≅ + = + = + ct c c f c f c b c v c t v c 1 1 1 2 1 1 2 1 1 2 2 2 2 2 γ . So our guess that f is dependent on the square of the velocity was a good one. If you more or less followed this geometrical method, skip the following. One could instead use the Pythagorean theorem: We square the hypotenuse c t = c τ + f of the original, big triangle (and expand this square). We then equate it to the sum of the squares of the other two sides: f c f fc c vt c + ( ) = + + ( ) = ( ) + ( ) τ τ τ τ 2 2 2 2 2 2 . Since f is small, its square will be much smaller and can be ignored in the above. Canceling the factor of c τ ( ) 2 that appears on both sides and dividing both sides by 2 2 c τ ( ) gives f c v c t τ τ ≅ 1 2 2 2 , where we have kept one factor of t/ τ and set the other one to reproduce the approximation that f is half of b , which gives the fifth version of the equation for γ in the previous paragraph.