# Time Dilation Back to Original Frame is Inconsistent?

I am watching someone jog at nearly the speed of light, and they snap their fingers. In my frame (A), it takes $\Delta t$ seconds.

Using $\Delta \bar{t}=\frac{\Delta t}{\gamma}$ to bring you from frame A to B, and then dividing by gamma again to get back to frame A (because the gammas are the same) doesn't give you the same number. What's wrong? (same with length contraction)

I am aware that Lorentz transforms can do this successfully. Why can't the "ordinary" time dilation formula?

The time dilation equation can only be used when you are comparing the time $\Delta \bar{t}$ between two events that occur at the same spatial location in one frame (like two ticks of a clock at rest in that frame, in which case $\Delta \bar{t}$ can be interpreted as the clock's own proper time between ticks), and the time $\Delta t$ between these same to events in a different frame where they occur at different locations, and in this frame the (difference in position)/(difference in time) between the two events is v--in this case, $\Delta \bar{t} = \Delta t / \gamma = \Delta t \sqrt{1 - v^2/c^2}$ as you said. You can rearrange this equation to $\Delta t = \Delta \bar{t} \times \gamma$ if you know the time in the frame where the events happened at the same location (the runner's own time between snaps in your example) and want to solve for the time in the frame where they did not (your frame), but what you can't do is just use the same equation but reverse the roles of which time-coordinate is supposed to be the one in the frame where the events happened at the same location.

More generally, if you want to translate time intervals for arbitrary pairs of events that aren't necessarily located at the same position in either frame, then you can use the full Lorentz transformation--if one frame is labeled with unprimed coordinates and the other is labeled with primed coordinates, and the spacetime origins of the two frames coincide and their spatial axes are aligned, and the primed frame is moving at speed v in the positive x-direction of the unprimed frame, the Lorentz transform equations for intervals between events would be:

$\Delta x^{\prime} = \gamma(\Delta x - v \Delta t)$

$\Delta t^{\prime} = \gamma(\Delta t - v \Delta x/c^2)$

$\Delta y^{\prime} = \Delta y$

$\Delta z^{\prime} = \Delta z$

and for transforming from primed back to unprimed:

$\Delta x = \gamma(\Delta x^{\prime} - v \Delta t^{\prime})$

$\Delta t = \gamma(\Delta t^{\prime} - v \Delta x^{\prime} /c^2)$

$\Delta y = \Delta y^{\prime}$

$\Delta z = \Delta z^{\prime}$

It's easy to check that if you pick a pair of events that occur at the same spatial location in the unprimed frame, so $\Delta x = \Delta y = \Delta z = 0$, then you get $\Delta t = \Delta t^{\prime} / \gamma$, the time dilation equation. Likewise, for two events that occur at the same spatial location in the primed frame, $\Delta t^{\prime} = \Delta t / \gamma$.

In the jogger's frame of reference, the event that the finger snap begins and the event that the finger snap ends are (assumed to be) co-located. Thus, in this frame

$$\Delta \bar x = 0$$

According to the Lorentz transformation, the elapsed time between the events in your frame of reference is (setting $c = 1$)

$$\Delta t = \gamma\left(\Delta \bar t + \beta \Delta \bar x \right) = \gamma \Delta \bar t$$

This is time dilation - the elapsed proper time, $\Delta \tau = \Delta \bar t$ is less than the elapsed coordinate time in a relatively moving frame of reference.

and then dividing by gamma again to get back to frame A (because the gammas are the same) doesn't give you the same number.

This would be a misapplication of the Lorentz transformation since, in your frame, the two events are separated in space and thus, $\Delta t$ is not an elapsed proper time.

Let the distance between the two events in your frame be $\Delta x$. By the Lorentz transformation, this is

$$\Delta x = \gamma \left(\Delta \bar x + \beta \Delta \bar t \right) = \gamma \beta \Delta \bar t$$

Now, given $\Delta t$ and $\Delta x$, we have, by the Lorentz transformation

$$\Delta \bar t = \gamma \left(\Delta t - \beta \Delta x \right)$$

But the right hand side is just

$$\gamma \left(\gamma \Delta \bar t - \gamma \beta^2 \Delta \bar t \right) = \gamma ^2 \left(1 - \beta^2 \right)\Delta \bar t = \Delta \bar t$$

as desired.