## Relativity and the proper usage of time dilation

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### Relativity and the proper usage of time dilation

There are frequent misrepresentations as to what relativity does and does not claim. For instance, there is a oft-misused (or at least misunderstood) equation, specifically the time dilation one:

$\Large \Delta t' = \gamma \Delta t$

I am writing here precisely what goes into that equation, so people can no longer misuse it on this board.

Let us start with the usual equations for converting a time and spatial event in one frame and see how those coordinates are

$\Large x' = \gamma (x - v t)$
$\Large t' = \gamma (t - v x / c^2)$

where the two frames are the primed and unprimed one. One frame (the unprimed one) we treat at rest and say the primed frame is moving in the positive x direction, with velocity v (v < c).

If we consider two things that occur at spacetime points (x1,t1) and (x2,t2), we can define the duration between the two occurrences as

$\Large \Delta x = x_2 - x_1$
$\Large \Delta t = t_2 - t_1$

We can do the same thing in the primed frame and have: (x'1,t'1) and (x'2,t'2), with

$\Large \Delta x' = x'_2 - x'_1$
$\Large \Delta t' = t'_2 - t'_1$

A trivial bit of algebra can combine the above equations and give the space and time interval conversion equations from special relativity:

$\Large \Delta x' = \gamma (\Delta x - v \Delta t)$
$\Large \Delta t' = \gamma (\Delta t - v \Delta x / c^2)$

Let's focus on the time equation, since we have described the length situation in other posts in this Expert Forum.

Specifically, let's consider how one gets the canonical time dilation equation.

$\Large \Delta t' = \gamma (\Delta t - v \Delta x / c^2) = \gamma \Delta t$

where the final equality sign is only valid for either: v = 0 or $\Delta x = 0$. Since the first possibility is when they have no relative velocity, only the second case is of interest.

We can interpret this case ($\Delta x = 0$) as the primed observer peeking at the unprimed observer's clock, but only if that clock is not moving at all in the unprimed observer's frame. It's as if the moving observer is trying to see the time experienced by the non-moving observer.

However, one could as easily ask what the primed observer understands to be the unprimed observer's position on a clock co-moving with the primed observer. In this case, both $\Delta t \neq 0$ and $\Delta x \neq 0$. Specifically, these two space time intervals are related via: $\Delta x = v \Delta t$. Putting that in the correct equation, reduces to

$\Large \Delta t' = \gamma (\Delta t - v \Delta x / c^2) = \gamma (\Delta t - v^2 \Delta t / c^2) = \gamma (1 - v^2 / c^2) \Delta t$
$\Large = \Delta t / \gamma$

Note that this says the following thing:

The primed observer says that his clock runs faster than a clock stationary with respect to an unprimed observer. The primed observer also says that his clock runs slower than a clock the unprimed observer says is comoving with the primed observer.

Thus it is completely imperative that you understand that the primed observer will observe different time intervals than the unprimed observer depending on both the relative velocity and on their position. While the unprimed observer will see the two clocks running at the same rate, the primed observer will see these two clocks running at different rates.

So, the real moral of this story is you really need to be careful when you use relativistic equations and do not use them incorrectly or outside their region of intended use.

The above discussion is adapted from an earlier series of posts:

http://www.sciencechatforum.com/bulletin/viewtopic.php?p=15726#15726 (and some surrounding posts.)

Lincoln
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