Lorentz factor for slow speeds

[cshum00]

One way to derive Lorentz factor is imagining the experiment of the light clock. This experiment is about two observers. One observer is moving at a constant speed on the x-axis and the other observer standing at rest. The observer moving along the x-axis carries a light clock which shoots a light beam vertically with respect to him. The light beam takes \Delta t_p to reach a height of d_h or d_h = v_c \Delta t_p.

The observer at rest, sees that when the light has reach the height d_h, it also has displaced a distance d_x = v_x \Delta t along the x-axis. The observer at rest also sees that the total distance displaced by the light is d_d = v_c \Delta t.

One can notice that d_h and d_d has the same velocity v_c but different times t_p and t; that is because according to Einstein's second postulate, light travels at the same speed for all inertial frames. As for time, the observer at rest is watching d_d and d_x they have the same time \Delta t; while the observer moving at a constant speed experiences a different time \Delta t_p;

[PLAIN]http://img263.imageshack.us/img263/1949/lightclock.png

So, since i have a triangle i can do the sum of the distances using the Pythagorean theorem:
(d_x)^2 + (d_h)^2 = (d_d)^2

(v_x \Delta t)^2 + (v_c \Delta t_p)^2 = (v_c \Delta t)^2

v_c^2 \Delta t^2 - v_x^2 \Delta t^2 = v_c^2 \Delta t_p^2

\Delta t^2 (v_x^2 - v_c^2) = v_c^2 \Delta t_p^2

\Delta t^2 v_c^2 (1 - \frac{v_x^2}{v_c^2}) = v_c^2 \Delta t_p^2

\Delta t = \frac {1}{\sqrt{(1 - \frac{v_x^2}{v_c^2})}} \Delta t

Where \gamma = \frac {1}{\sqrt{(1 - \frac{v_x^2}{v_c^2})}} is Lorentz factor.

Now, let's try the same experiment with a clock that is not light but something else traveling at a really slow speed like for example a ball with no force acting onto it so that it moves at a constant speed upward. So the equations becomes as follows:
d_x = v_x \Delta t

d_d = v_d \Delta t where v_d is no longer v_c

d_h = v_b \Delta t_p where v_b is no longer v_c

And if i apply the Pythagorean theorem i get:
(d_x)^2 + (d_h)^2 = (d_d)^2

(v_x \Delta t)^2 + (v_b \Delta t_p)^2 = (v_d \Delta t)^2

v_d^2 \Delta t^2 - v_x^2 \Delta t^2 = v_b^2 \Delta t_p^2

\Delta t^2 (v_x^2 - v_d^2) = v_c^2 \Delta t_p^2

\Delta t^2 v_d^2 (1 - \frac{v_x^2}{v_d^2}) = v_b^2 \Delta t_p^2

\Delta t = \frac {v_b}{v_d \sqrt{1 - \frac{v_x^2}{v_d^2}}} \Delta t

Where \gamma becomes \frac {v_b}{v_d \sqrt{1 - \frac{v_x^2}{v_d^2}}}. However, the calculations made on the textbooks they still use \gamma where v_b = v_d = v_c is still at the speed of light when it is no longer the case. Why is that?

 

[JesseM]

Now, let's try the same experiment with a clock that is not light but something else traveling at a really slow speed like for example a ball with no force acting onto it so that it moves at a constant speed upward. So the equations becomes as follows:
d_x = v_x \Delta t

d_d = v_d \Delta t where v_d is no longer v_c

d_h = v_b \Delta t_p where v_b is no longer v_c

And if i apply the Pythagorean theorem i get:
(d_x)^2 + (d_h)^2 = (d_d)^2

(v_x \Delta t)^2 + (v_b \Delta t_p)^2 = (v_d \Delta t)^2

v_d^2 \Delta t^2 - v_x^2 \Delta t^2 = v_b^2 \Delta t_p^2

\Delta t^2 (v_x^2 - v_d^2) = v_c^2 \Delta t_p^2

\Delta t^2 v_d^2 (1 - \frac{v_x^2}{v_d^2}) = v_b^2 \Delta t_p^2

\Delta t = \frac {v_b}{v_d \sqrt{1 - \frac{v_x^2}{v_d^2}}} \Delta t

Where \gamma becomes \frac {v_b}{v_d \sqrt{1 - \frac{v_x^2}{v_d^2}}}. However, the calculations made on the textbooks they still use \gamma where v_b = v_d = v_c is still at the speed of light when it is no longer the case. Why is that?

I think it's just because the textbooks aren't usually trying to calculate what you're calculating, namely the time dilation as a function of velocities measured both in the clock rest frame (the velocity v_b) and velocities in the frame where the clock is moving (v_x and v_d). Normally the idea is that the equation expresses time dilation in terms of the the horizontal velocity of the clock in the frame where it's in motion, which would be equal to v_x for the ball. In other words:

\Delta t = \frac{1}{\sqrt{1 - v_x^2/c^2}} \Delta t_p

If the ball has a vertical velocity of v_b in the clock rest frame, then the standard version of the time dilation equation says it must have a slower vertical velocity of v_h = (\sqrt{1 - v_x^2/c^2}) v_b in the frame where the clock is moving (you can derive this from the Lorentz transformation, but more simply you can derive it from the fact that the first postulate of relativity requires that if a ball-clock is keeping pace with a light-clock when it's at rest in one frame, then an identically-constructed ball-clock and light-clock in a different frame must still keep pace when at rest in a different frame, and we know that a light clock slows down by a factor of \sqrt{1 - v_x^2/c^2} when it has a horizontal velocity of v_x in our frame so a ball-clock must do the same). And since v_d represented the diagonal velocity, in this frame we have v_d^2 = v_h^2 + v_x^2, so substituting in the above gives v_d^2 = (1 - v_x^2/c^2)*v_b^2 + v_x^2 which can be reduced to (v_d^2 - v_x^2)/v_b^2 = (1 - v_x^2/c^2) or \frac{1}{\sqrt{1 - v_x^2/c^2}} = \frac{v_b}{\sqrt{v_d^2 - v_x^2}} = \frac{v_b}{v_d \sqrt{1 - v_x^2/v_d^2}}. Substituting this into the standard time dilation equation I gave above gives:

\Delta t = \frac{v_b}{v_d \sqrt{1 - v_x^2/v_d^2}} \Delta t_p

Which is the same as what you got.