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cshum00
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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 [tex]\Delta t_p[/tex] to reach a height of [tex]d_h[/tex] or [tex]d_h = v_c \Delta t_p[/tex].
The observer at rest, sees that when the light has reach the height [tex]d_h[/tex], it also has displaced a distance [tex]d_x = v_x \Delta t[/tex] along the x-axis. The observer at rest also sees that the total distance displaced by the light is [tex]d_d = v_c \Delta t[/tex].
One can notice that [tex]d_h[/tex] and [tex]d_d[/tex] has the same velocity [tex]v_c[/tex] but different times [tex]t_p[/tex] and [tex]t[/tex]; 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 [tex]d_d[/tex] and [tex]d_x[/tex] they have the same time [tex]\Delta t[/tex]; while the observer moving at a constant speed experiences a different time [tex]\Delta t_p[/tex];
[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:
[tex](d_x)^2 + (d_h)^2 = (d_d)^2[/tex]
[tex](v_x \Delta t)^2 + (v_c \Delta t_p)^2 = (v_c \Delta t)^2[/tex]
[tex]v_c^2 \Delta t^2 - v_x^2 \Delta t^2 = v_c^2 \Delta t_p^2[/tex]
[tex]\Delta t^2 (v_x^2 - v_c^2) = v_c^2 \Delta t_p^2[/tex]
[tex]\Delta t^2 v_c^2 (1 - \frac{v_x^2}{v_c^2}) = v_c^2 \Delta t_p^2[/tex]
[tex]\Delta t = \frac {1}{\sqrt{(1 - \frac{v_x^2}{v_c^2})}} \Delta t[/tex]
Where [tex]\gamma = \frac {1}{\sqrt{(1 - \frac{v_x^2}{v_c^2})}}[/tex] 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:
[tex]d_x = v_x \Delta t[/tex]
[tex]d_d = v_d \Delta t[/tex] where [tex]v_d[/tex] is no longer [tex]v_c[/tex]
[tex]d_h = v_b \Delta t_p[/tex] where [tex]v_b[/tex] is no longer [tex]v_c[/tex]
And if i apply the Pythagorean theorem i get:
[tex](d_x)^2 + (d_h)^2 = (d_d)^2[/tex]
[tex](v_x \Delta t)^2 + (v_b \Delta t_p)^2 = (v_d \Delta t)^2[/tex]
[tex]v_d^2 \Delta t^2 - v_x^2 \Delta t^2 = v_b^2 \Delta t_p^2[/tex]
[tex]\Delta t^2 (v_x^2 - v_d^2) = v_c^2 \Delta t_p^2[/tex]
[tex]\Delta t^2 v_d^2 (1 - \frac{v_x^2}{v_d^2}) = v_b^2 \Delta t_p^2[/tex]
[tex]\Delta t = \frac {v_b}{v_d \sqrt{1 - \frac{v_x^2}{v_d^2}}} \Delta t[/tex]
Where [tex]\gamma[/tex] becomes [tex]\frac {v_b}{v_d \sqrt{1 - \frac{v_x^2}{v_d^2}}}[/tex]. However, the calculations made on the textbooks they still use [tex]\gamma[/tex] where [tex]v_b = v_d = v_c[/tex] is still at the speed of light when it is no longer the case. Why is that?
The observer at rest, sees that when the light has reach the height [tex]d_h[/tex], it also has displaced a distance [tex]d_x = v_x \Delta t[/tex] along the x-axis. The observer at rest also sees that the total distance displaced by the light is [tex]d_d = v_c \Delta t[/tex].
One can notice that [tex]d_h[/tex] and [tex]d_d[/tex] has the same velocity [tex]v_c[/tex] but different times [tex]t_p[/tex] and [tex]t[/tex]; 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 [tex]d_d[/tex] and [tex]d_x[/tex] they have the same time [tex]\Delta t[/tex]; while the observer moving at a constant speed experiences a different time [tex]\Delta t_p[/tex];
[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:
[tex](d_x)^2 + (d_h)^2 = (d_d)^2[/tex]
[tex](v_x \Delta t)^2 + (v_c \Delta t_p)^2 = (v_c \Delta t)^2[/tex]
[tex]v_c^2 \Delta t^2 - v_x^2 \Delta t^2 = v_c^2 \Delta t_p^2[/tex]
[tex]\Delta t^2 (v_x^2 - v_c^2) = v_c^2 \Delta t_p^2[/tex]
[tex]\Delta t^2 v_c^2 (1 - \frac{v_x^2}{v_c^2}) = v_c^2 \Delta t_p^2[/tex]
[tex]\Delta t = \frac {1}{\sqrt{(1 - \frac{v_x^2}{v_c^2})}} \Delta t[/tex]
Where [tex]\gamma = \frac {1}{\sqrt{(1 - \frac{v_x^2}{v_c^2})}}[/tex] 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:
[tex]d_x = v_x \Delta t[/tex]
[tex]d_d = v_d \Delta t[/tex] where [tex]v_d[/tex] is no longer [tex]v_c[/tex]
[tex]d_h = v_b \Delta t_p[/tex] where [tex]v_b[/tex] is no longer [tex]v_c[/tex]
And if i apply the Pythagorean theorem i get:
[tex](d_x)^2 + (d_h)^2 = (d_d)^2[/tex]
[tex](v_x \Delta t)^2 + (v_b \Delta t_p)^2 = (v_d \Delta t)^2[/tex]
[tex]v_d^2 \Delta t^2 - v_x^2 \Delta t^2 = v_b^2 \Delta t_p^2[/tex]
[tex]\Delta t^2 (v_x^2 - v_d^2) = v_c^2 \Delta t_p^2[/tex]
[tex]\Delta t^2 v_d^2 (1 - \frac{v_x^2}{v_d^2}) = v_b^2 \Delta t_p^2[/tex]
[tex]\Delta t = \frac {v_b}{v_d \sqrt{1 - \frac{v_x^2}{v_d^2}}} \Delta t[/tex]
Where [tex]\gamma[/tex] becomes [tex]\frac {v_b}{v_d \sqrt{1 - \frac{v_x^2}{v_d^2}}}[/tex]. However, the calculations made on the textbooks they still use [tex]\gamma[/tex] where [tex]v_b = v_d = v_c[/tex] is still at the speed of light when it is no longer the case. Why is that?
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