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Special relativity confusion

After reading many books out of interest I thought I understood the basics of relativity, but a few extracts from a book I am studying from have confused me.

I shall copy them.

'Imagine a 22nd century rocket race over a distance of 30 million kilometres. Light takes 100 seconds to travel this distance so you would expect a rocket travelling at half the speed of light to take 200 seconds. The rocket crew and the race officials would also expect a time of 200s if, like Newton and Galileo, they thought in terms of absolute motion. However, according to Einstein's theory of special relativity, the rocket crew's on-board 'clock' would show they arrive at the finishing line 173 seconds after passing the starting line. The race officials at the starting or finishing line would still record 200 seconds on their clocks.'

'A moving clock runs more slowly than a stationary clock.'

I am also aware of the formula for time dilation, which seems to tell me that the faster a person moves relative to an observer at rest, the longer the time recorded is for the person who is moving compared to the time measured by the person at rest.

Now the first extract says the time recorded by the rocket which moves relative to the people at rest is shorter. Surely this isn't consistent with the formula, which shows that as the person who is moving has a larger speed, the ratio of v/c is larger so 1-(this ratio)^2 becomes smaller. So dividing the time measured by the person at rest by 1-(this ratio)^2 gives a larger measured time for the mover. Nor can I see how the second extract is consistent with the formula, as a slow clock would record a smaller time between two events. The first two extracts seem consistent to me.

I am aware that you would age more slowly if you travelled faster, which is consistent with the first two extracts.

So why can I not piece together the equation with the explanations in writing. This annoys me. The seem to be saying the exact opposite... Thankyou if anybody can see what I am getting at.
(edited 10 years ago)
It took me a while to get my head around this when I first did it.

If you use the Lorentz factor to find the time dilation for 1 second of someone in a rocket travelling very fast, and get an answer of 2, what that tells you is that 1 second on the rocket is equivalent to 2 seconds for an outside observer. It's counterintuitive, because the equation gives you an increase in time when you're thinking of a shorter period of time.

Basically, if time t is the length of time between two events, then dilated time t' is that length of time from the perspective of someone moving at relativistic speed *as measured by someone who isn't*. Thus the stationary person thinks the moving person's clock takes 2 seconds to tick instead of one, and the moving person experiences less time.
(edited 10 years ago)
Reply 2
Original post by feebleFanfare
It took me a while to get my head around this when I first did it.

If you use the Lorentz factor to find the time dilation for 1 second of someone in a rocket travelling very fast, and get an answer of 2, what that tells you is that 1 second on the rocket is equivalent to 2 seconds for an outside observer. It's counterintuitive, because the equation gives you an increase in time when you're thinking of a shorter period of time.

Basically, if time t is the length of time between two events, then dilated time t' is that length of time from the perspective of someone moving at relativistic speed *as measured by someone who isn't*. Thus the stationary person thinks the moving person's clock takes 2 seconds to tick instead of one, and the moving person experiences less time.


How does this fit in with what you said then?

Einstein showed that if the time interval between two events measured by an observer at rest relative to the events is t', (called 'the proper time'), an observer moving at speed v relative to the events would measure a longer time interval, t, given by
t=t1v2c2t=\frac{t'}{\sqrt{1-\frac{v^2}{c^2}}}

So time goes more slowly for the observer moving fastest, but they measure a longer time - so is this because we are talking about measuring the time between two events occuring?
(edited 10 years ago)
Reply 3
The v in the equation is the speed of the observer not the observed.
Reply 4
Original post by Ferrus
The v in the equation is the speed of the observer not the observed.


Can't see where I have suggested otherwise...
Reply 5
Original post by fayled
Can't see where I have suggested otherwise...

The observer always sees time as being the same for anything travelling at the same speed as him, such as a watch in the same inertia frame. So for the observer the practical effect of time observing time as being faster for those things which are not co-moving.
Reply 6
Original post by Ferrus
The observer always sees time as being the same for anything travelling at the same speed as him, such as a watch in the same inertia frame. So for the observer the practical effect of time observing time as being faster for those things which are not co-moving.


Sorry could you rephrase that last sentence please?
Reply 7
Original post by fayled
Sorry could you rephrase that last sentence please?

Your watch records the same time even as you increase your velocity.
Original post by fayled
How does this fit in with what you said then?

Einstein showed that if the time interval between two events measured by an observer at rest relative to the events is t', (called 'the proper time'), an observer moving at speed v relative to the events would measure a longer time interval, t, given by
t=t1v2c2t=\frac{t'}{\sqrt{1-\frac{v^2}{c^2}}}

So time goes more slowly for the observer moving fastest, but they measure a longer time - so is this because we are talking about measuring the time between two events occuring?


That's not what the equation tells you. As the articles you quoted said, if a fast moving vehicle covers a certain distance an on-board clock would record a shorter time than one at the side of the track.

What the equation tells you is how much more slowly time runs as you speed up, effectively how much longer each second is. However, as Ferrus said, when you move quickly you still perceive time passing at the same rate, so the effect is a discrepancy between the two measurements.

Say person A observes from beside the tracks as a train moves from one place to another, and person B is on the train. Person A sees the train take 10 seconds to complete it's journey. Using the equation, we see that 10 seconds of proper time gives an answer of more than 10 - say 20 seconds. This tells us that if A watches 10 seconds pass on B's clock, they experience 20 seconds. It follows, then, that as A experiences 10 seconds B experiences less time than that. The reason (with regards to the equation) that it's this way round is that, as noted above, v in the equation is the speed of the observer.
Reply 9
Original post by feebleFanfare
That's not what the equation tells you. As the articles you quoted said, if a fast moving vehicle covers a certain distance an on-board clock would record a shorter time than one at the side of the track.

What the equation tells you is how much more slowly time runs as you speed up, effectively how much longer each second is. However, as Ferrus said, when you move quickly you still perceive time passing at the same rate, so the effect is a discrepancy between the two measurements.

Say person A observes from beside the tracks as a train moves from one place to another, and person B is on the train. Person A sees the train take 10 seconds to complete it's journey. Using the equation, we see that 10 seconds of proper time gives an answer of more than 10 - say 20 seconds. This tells us that if A watches 10 seconds pass on B's clock, they experience 20 seconds. It follows, then, that as A experiences 10 seconds B experiences less time than that. The reason (with regards to the equation) that it's this way round is that, as noted above, v in the equation is the speed of the observer.


Thankyou, I think this probably solves my problem then.
Reply 10
Original post by Ferrus
The v in the equation is the speed of the observer not the observed.


Oh, I just realised this is where I have been going wrong after reading everything again. Thanks.

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