Time Lapses In Two Different Reference Frames

[Bashyboy]

Hello,

I would like to prove that the time experienced in a moving reference frame is longer than in a stationary frame. Here is my solution:

Suppose that at time t = 0 two reference frames, S and S', origins coincide; similarly, the x,y, and z axes of the S-frame overlap with the corresponding axes of the S'-frame. Now, as time begins to progress, t > 0, S' will begin moving relative to S; and the way in which it will move is such that the origin of the S'-frame will move along the positive x-axis of the S-frame, and the speed at which it will move is \displaystyle \beta = \frac{v}{c}.

Now, suppose two arbitrary events occur, one of which is A, occurring when S and S' coincide, allowing us to infer the coordinates of both frames to be S(x=0, y=0, z=0, t=0) and S'(x'=0, y'=0, z'=0, t'=0); and let the second event be B, which occurs at some later time that is not necessarily the same in both frames.

The invariant interval in S:

\Delta s^2 = (\Delta x^2 + \Delta y^2 + \Delta z^2) - \Delta t^2

\Delta s^2 = - t_B^2

The invariant interval in S':

\Delta s^2 = ((\Delta x^2~' + \Delta y^2~' + \Delta z^2~') - \Delta t^2~'

\Delta s^2 = x_B^2~' - t_B^2~'

Setting the two equal to each other,

- t_B^2 = x_B^2~' - t_B^2~'

Eliminate x_B~' using Lorentz's transformation, x_B~' = \gamma (x_B - \beta t_B) \implies x_B~' = - \gamma \beta t_B

- t_B^2 = \gamma^2 \beta^2 t_B^2 - t_B^2~'

Solving for t-prime

t_B~' = \pm \sqrt{t_B^2(1 + \gamma^2 \beta^2)}

My question is, for what reason can I simply dismiss the negative root, keeping only the positive root?

 

[Chestermiller]

Suppose β were zero, so that the two frames coincided for all times. Would you then lend credence to the negative root? Now, if β were only slightly different from zero, would that really change things significantly? Another answer is that the metrical equations with the squares were derived directly from the Lorentz Transformation. Although these equations are non-linear (leading to your ambiguity with regard to the sign), the Lorentz Transformation is linear and eliminates that ambiguity.