- #1
Grimble
- 485
- 11
One of the first expressions that I came across in studying SR was that it was Counter Intuitive.
In particular this expression was used to define the relationship between local time and the time measured by a 'moving clock'.
It seemed to be thought that newcomers to the theory would assume that, as the traveling clock would be seen to run slowly by a stationary observer, then an observer moving with the clock, would expect see the stationary clock run fast.
{But that would mean that the 'newcomer' would be imagining it all as a single frame of reference - (such as the clock in a moving car running fast compared to a stationary one, which would be true if that clock were, indeed, running fast)}.
But that is not what is being compared, is it?
Are we not comparing the measuring of a moving clock, compared with a stationary (local) one?
And, if so, then the reciprocal is measuring exactly the same measurement, i.e. the measurement of a clock (local to the stationary observer) moving relative to an observer local to the 'moving' clock.
So in each case it is the measurement of a clock moving relative to the observer making the measurement. So, at least to me, it is intuitive that the two measurements would show the same relationship.
This brings one to the question; if the two clocks are identical and the moving clock runs slow, which is the real time?
(And I suppose, the associated question; if something can be measured, is it there fore real?)
Now, my answer would be Yes they are real and therefore that both measurements have to be real.
But how can both measurements be 'real' if they are identical clocks showing different times?
The answer that I give has to be that both measurements are real and correct because they are not measuring the same thing.
It seems to be assumed that because we are dealing with clocks that are local to each of two frames of reference, and that because 'clocks' somehow define how we measure time in SR, that we are measuring the same interval.
But in one case we are measuring a stationary (local) clock and in the other we are measuring a moving clock. i.e. we are measuring it under different conditions.
Am I getting this right?
In particular this expression was used to define the relationship between local time and the time measured by a 'moving clock'.
It seemed to be thought that newcomers to the theory would assume that, as the traveling clock would be seen to run slowly by a stationary observer, then an observer moving with the clock, would expect see the stationary clock run fast.
{But that would mean that the 'newcomer' would be imagining it all as a single frame of reference - (such as the clock in a moving car running fast compared to a stationary one, which would be true if that clock were, indeed, running fast)}.
But that is not what is being compared, is it?
Are we not comparing the measuring of a moving clock, compared with a stationary (local) one?
And, if so, then the reciprocal is measuring exactly the same measurement, i.e. the measurement of a clock (local to the stationary observer) moving relative to an observer local to the 'moving' clock.
So in each case it is the measurement of a clock moving relative to the observer making the measurement. So, at least to me, it is intuitive that the two measurements would show the same relationship.
This brings one to the question; if the two clocks are identical and the moving clock runs slow, which is the real time?
(And I suppose, the associated question; if something can be measured, is it there fore real?)
Now, my answer would be Yes they are real and therefore that both measurements have to be real.
But how can both measurements be 'real' if they are identical clocks showing different times?
The answer that I give has to be that both measurements are real and correct because they are not measuring the same thing.
It seems to be assumed that because we are dealing with clocks that are local to each of two frames of reference, and that because 'clocks' somehow define how we measure time in SR, that we are measuring the same interval.
But in one case we are measuring a stationary (local) clock and in the other we are measuring a moving clock. i.e. we are measuring it under different conditions.
Am I getting this right?