[SOLVED] proper and improper time interval paradox

Sep 2009
409
155
india
please help me clear this paradox.
suppose, in my frame an event (1) takes place at A and after time t another event(2)occurred at a nearby place B.Both A and B are in my field of view, so i record the time interval in a single clock in my hand. this is t and it must be a proper time interval.
Now suppose a clock moves from A to B with a velocity L/t where L is the distance between A and B in my frame. This clock is moving with respect to me, so it runs slower, so the time interval noted in this clock will be t/gamma. however, this clock is moving wrt me, so it is the improper time interval.
conclusion: proper time int(t) greater than improper time int(t/gamma) ,which IS DEFINITELY WRONG .
 

physicsquest

PHF Helper
Feb 2009
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474
Maybe the confusion is in the idea of interval.

When there is "time dilation" the clock will read less.

The time interval increases, hence that clock reads less.

Let us look at it with an analogy in space. Let there be an object which you measure with a scale to be 10 cm long. Now let there be another scale which has been heated and so has expanded such that the 1cm mark is now actually at 2 cm (i know this is exaggerated) . This scale thus has a space dilation, i.e. its spatial interval has increased. However when you measure the length of the object with this scale, it will read 5 cm!

Thus a "length interval dilation" leads to a lower reading for length.
 
Sep 2009
409
155
india
sorry, but i am still not gathering the concept. i understand the space dilation, but cannot find the analogy with the time dilation. please save me.(joke)
 

physicsquest

PHF Helper
Feb 2009
1,426
474
Let us try again.

Take two identical robust pendulum driven second clocks.

The period for such a clock is given by the well known formula

T = 2 pi Sqrt (L/g) and is equal to 1 sec

Heat one of them so that the length of its pendulum becomes 4 L

(remember they are very robust).

If we plug this back into the formula we find that the new period T' = 2 T.

This means that the heated pendulum takes twice as long to complete one oscillation. Time has dilated for it, the interval between identical positions of the bob has doubled.

When the normal bob completes 1 osc, the heated one has completed only half. The hot one takes 2 secs to complete one osc. Its time period, which is an interval has increased.
Suppose 10 secs elapse .The normal clock reads 10 while the heated one reads only 5.

Thus the increase in period leads to a lower reading.

In effect the paradox is because the interval and reading are being mixed up.
 
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Pmb

PHF Hall of Fame
Apr 2009
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please help me clear this paradox.
suppose, in my frame an event (1) takes place at A and after time t another event(2)occurred at a nearby place B.Both A and B are in my field of view, so i record the time interval in a single clock in my hand. this is t and it must be a proper time interval.
Now suppose a clock moves from A to B with a velocity L/t where L is the distance between A and B in my frame. This clock is moving with respect to me, so it runs slower, so the time interval noted in this clock will be t/gamma. however, this clock is moving wrt me, so it is the improper time interval.
conclusion: proper time int(t) greater than improper time int(t/gamma) ,which IS DEFINITELY WRONG .

You have the wrong idea of what proper time is. A proper time interval is the time measured on a clock which is present at both events. It is therefore impossible for you to record the time intereval between two events which are spatially seperated and call that a propertime interval
 
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Sep 2009
409
155
india
thank you. now the problem is clear to me. my concept was wrong. thanks again.