# Time dilation confusion

I'm just starting to learn about special relativity, and I'm a little bit confused about something. Take the example of an observer in $S$ on the ground observing a train move at constant velocity $v$ relative to $S$, an observer in $S'$ is on the train, and this observer in $S'$ flashes a light that reflects from the ceiling and returns to him in a time, $t'$ he measures.

I know that in general, if two frames $S$ and $S'$ are in relative uniform motion with respect to each other, and an observer in $S$ can see the clock of an observer $S'$ and the observer in $S'$ can see the clock of the observer $S$, then the observer in $S$ will see the clock of the observer in $S'$ run slower than his own, and vice versa. But I also know that $S'$ time is proper time, and so $t'\leq t$. This seems very strange to me. Observer $S$ sees his clock running faster, so intuitively, observer $S$ expects to measure less time for the event, but the time he measures for the light to return is no less than the time that observer $S'$ measures.

Am I understanding this correctly?

• You just managed to confuse me about special relativity... can you draw this scenario, please? I think you may actually discover the source of your problem, when you do. Sep 11, 2014 at 7:31
• "but the time he measures for the light to return is no greater than the time that observer S measures." By no greater do you mean equal to or less than?
– BMS
Sep 11, 2014 at 7:52
• @BMS Yes, I mean that the time observed by the observer in the train is $\leq$ than the time observed by the observer on the ground Sep 11, 2014 at 7:56

I agree with your statements up through the claim that $t'<t$. That's all fine.

Here I think is the issue you're running into:

The quantity $t'$ in the relationship above represents the time interval as measured in frame $S'$. It does not represent the number of ticks by the moving clock as measured in frame $S$. That's a subtle but important distinction, read it again.

• Oh jeez, I think I asked my question the wrong way: $S$ also sees his clock as ticking faster, but $S$ observes a larger time than does $S'$. I think that's what I meant to ask :) is there still no contradiction? Sep 11, 2014 at 7:58