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Length Contraction and Time Dilation 

 

Length Contraction

Material bodies contract when they are moving and this shortening is only in the direction of the motion. If the length of the body is L0, when it is at rest i.e. u = 0, then when it moves with the speed u along its length, then new length L is given by L  = L0 (1 u2/c2)1/2

To understand this, consider a spaceship. Suppose, initially spaceship is at rest and for an observer outside the spaceship, the length of the spaceship is L0 = 50 meters. For the traveller inside the spaceship, length of the spaceship as seen from inside is say, L01 = 49.8 meters which is less than L0 due to some thickness of front and rear part of the spaceship. Now, suppose spaceship is moving with velocity u along the direction of its length. Suppose the observer outside the spaceship and at rest, has some divine vision so that he can see the spaceship, even if spaceship is hundreds of miles away from him after some time. Suppose, through his divine vision, he can also see inside the spaceship, even if it is far away from him. For this observer, who is  outside the spaceship, the length of the spaceship, which is now moving with the velocity u will appear to be L  = L0(1 u2/c2)1/2. Thus, if spaceship is moving with the velocity equal to half the velocity of light i.e. u = c/2, then to that observer, the length of the spaceship will appear to be equal to L, where L = L0 (1 u2/c2)1/2

or L = L0 (1 1/4)1/2   (u = c/2 implies u/c = 1/2)

or L0 (3/4)1/2                                                                  

or .866L0

or .866 50 = 43.3 meters

Thus, to an observer outside the spaceship and at rest, the length of the moving spaceship, seems to be less than when the spaceship was at rest. For the traveller inside the spaceship, the length of the spaceship as seen from inside will still be L01 = 49.8 meters that is traveller does not feel any difference in the length of the spaceship.

 

Now suppose, the spaceship is moving with the velocity nearly equal to that of light i.e. v = .99c for example. Then, to the observer outside the ship and at rest, the length of the spaceship will appear to be equal to L, where L = L0 (1 u2/c2)1/2

or L = L0 (1 .992)1/2   (u = .99c implies u/c = .99)

or L = L0 (1 .9801)1/2  

or L = L0 (.0199)1/2                                                   

or L =.141L0

or .141 50 = 7.05 meters

For the traveller inside the spaceship, the length of the spaceship as seen from inside will still be L01 = 49.8 meters.

 

Now suppose, the spaceship is moving with the velocity u = .9999c for example. Then, to the observer outside the ship and at rest, the length of the spaceship will appear to be equal to L, where L = L0 (1 u2/c2)1/2

or L = L0 (1 .99992)1/2   (u = .9999c implies u/c = .9999)

or L =.0141L0

or .0141 50 = .705 meters or 70.5 centimeters.

For the traveller inside the spaceship, the length of the spaceship as seen from inside will still be L01 = 49.8 meters.

 

If spaceship were travelling with the velocity of light i.e u = c, then, to the observer outside the ship and at rest, the length of the spaceship will appear to be equal to L, where L = L0 (1 u2/c2)1/2

or L = L0 (1 1)1/2   (u = c implies u/c = 1)

or L = 0.

For the traveller inside the spaceship, the length of the spaceship as seen from inside will still be L01 = 49.8 meters.

Note that, as per theory of relativity it is impossible for a body to move with the velocity of light, thus the length of the spaceship can never appear to be zero to an outside observer.

 

 

Time Dilation

If Δt time interval is passed in a clock moving with the velocity u, then in the same type of clock, which is at rest, Δt / (1 u2/c2)1/2 time interval will have passed.

Suppose an observer outside the spaceship has a digital stopwatch, which is reset so that it shows the time as 00:00:00:000 that is 0 hour, 0 minute, 0.000 second. Suppose the traveller in the spaceship has the same type of digital stopwatch, which is also reset.

Now suppose, spaceship is moving with the velocity u and the traveller, who is inside the spaceship, presses the start button of his stopwatch at some time. The observer outside the spaceship and at rest, who is observing the traveller through his divine vision, presses the start button of his stopwatch at the same time, so that both the stopwatch start simultaneously.

Now suppose, some time later, the observer at rest, sees the time elapsed in the stopwatch of the traveller and suppose, he finds the stopwatch of the traveller is showing 00:00:14:000 i.e. 0 hour, 0 minute 14.000 seconds have elapsed in his stopwatch and at the same time, he stops his stopwatch and then sees the time elapsed in his stopwatch, then he will find that more than 14 seconds have in his stopwatch.

If spaceship were moving with the velocity u = c/2 and if 14 seconds (Δttraveller in spaceship  = 14) had passed in the stopwatch of the traveller inside the spaceship (i.e. the stopwatch of the traveller was showing 00:00:14:000), then in the stopwatch of the observer, time elapsed will be

Δtobserver at rest = Δttraveller in spaceship  / (1 u2/c2)1/2

or  Δtobserver at rest  = 14 seconds / (1 1/4)1/2       (u = c/2 implies u/c = 1/2 or u2/c2 = 1/4)

or  Δtobserver at rest  = 16.156 seconds i.e. stopwatch of the observer will show 00:00:16:156 

 

If spaceship were moving with the velocity u = .99c and if 14 seconds (Δttraveller in spaceship  = 14) had passed in the stopwatch of the traveller inside the spaceship, then in the stopwatch of the observer, time elapsed will be

Δtobserver at rest = Δttraveller in spaceship  / (1 u2/c2)1/2

or  Δtobserver at rest  = 14 seconds / (1 .992)1/2       (u = .99c implies u/c = .99)

or  Δtobserver at rest  = 99.232 seconds i.e. stopwatch of the observer will show 00:01:39:232  (1 min, 39.232 sec)

 

If spaceship were moving with the velocity u = .9999c

Δtobserver at rest  = 14 seconds / (1 .99992)1/2       (u = .9999c implies u/c = .9999)

or  Δtobserver at rest  = 989.968 seconds i.e. stopwatch of the observer will show 00:16:29:968  (16 min, 29.968 sec)

 

Thus time itself will appear to be slower in a moving spaceship. In other words moving clock ticks slower.

 

If the observer outside the spaceship and at rest saw that  the traveller inside the spaceship is having a dinner, then all the motions e.g. taking food in the spoon, thrusting the spoon in the mouth, pulling the spoon out of the mouth, chewing the food, drinking a glass of water appear to be slower than normal (whereas to the traveller inside the spaceship, everything moves at a normal rate) that is not only do the length shorten but also the clocks slow down. That is when the clock inside the spaceship records 1 second elapsed, as seen by the traveller inside the spaceship, the same type of clock near the observer outside the spaceship and at rest shows 1 / (1 u2/c2)1/2 seconds, which is more than 1 second.

 

Suppose traveller has the divine vision and he is now seeing the observer who is outside the spaceship and at rest. For the traveller, time is moving with the same normal speed in both the cases i.e. when spaceship is moving along at a uniform speed as well as when the spaceship is at rest.

Now if the observer is having a dinner, then to the eyes of the traveller, all the motions (e.g. taking food in the spoon, thrusting the spoon in the mouth, pulling the spoon out of the mouth, chewing the food, drinking a glass of water) of that man (who was observer and is now being observed by the traveller inside the spaceship) appear to be faster than normal, while to the man (who was observer and now being observed by the traveller inside the spaceship), everything moves at a normal rate. Thus for the traveller, velocity of time outside the spaceship appear to be increased by the factor 1 / (1 u2/c2)1/2.  If 14 seconds passed inside spaceship, which is moving with the speed u = c/2,  14/ (1 u2/c2)1/2 or 16.156 seconds will have passed outside the spaceship.

 

Just as the length of the spaceship appear to be contracted to the observer outside the spaceship and at rest, similarly the distance between the spaceship and the observer (who is now being observerd by the traveller) will appear to be contracted to the traveller in the spaceship as explained in the next example.

 

 

Length Contraction & Time Dilation

Suppose the spaceship is 100 light seconds away from the observer and is moving with velocity u = .8c towards the observer.

(one light second is the distance travelled by light in one second i.e. 1 light second = 3 108  meter/sec 1sec = 3 108 meter).

When the spaceship is 100 light sec away, suppose the traveller inside the spaceship resets & starts his stopwatch and at the same time, the observer at rest does the same.

When the spaceship reaches to the observer, time elapsed in the stopwatch of the observer will be distance / velocity = 100 light second / .8c

= (100 3 108 meter) / (.8 3 108 meter/second)

=125 seconds.

Thus the stopwatch of the observer will show 00:02:05:00 i.e. 2 min, 5 sec.

Now from Δtobserver at rest = Δttraveller in spaceship  / (1 u2/c2)1/2

Δttraveller in spaceship  = Δtobserver at rest (1 u2/c2)1/2

As the spaceship is moving with the velocity u = .8c, thus (1 u2/c2)1/2 = (1 .82)1/2 = .6

Thus Δttraveller in spaceship  = Δtobserver at rest .6 = 125 .6  = 75 seconds that is, for the traveller, he completed the journey ( which was of 100 light second for the observer outside the spaceship and at rest), when in his stopwatch 75 seconds had passed that is, for the traveller, in actual, 75 second time interval had passed inside the spaceship during his journey and during these 75 seconds, moving with the velocity u = .8c, the spaceship will travel .8c 75 sec = 60 light sec (or 60 3 108 meter).

Thus the traveller inside the spaceship will say, journey was of 60 light seconds (which he completed in 75 seconds while moving with the velocity u = .8c). But the observer at rest will say, journey was of 100 light seconds (which he completed in 125 seconds moving with the velocity u = .8c).

This contradiction disappears, if we consider the length contraction that is as the spaceship was moving towards the observer, the distance between the spaceship and the observer, as seen by the traveller will be L = L0 (1 u2/c2)1/2 = 100 light second  (1 .82)1/2 = 100 light second  .6 = 60 light second.

Thus, for the observer at rest, this journey was of 100 light second, which the traveller completed in 125 seconds moving with the velocity u = 0.8c.

For the traveller, this journey was of 100 (1 u2/c2)1/2 light second, which the traveller completed in 125 (1 u2/c2)1/2 seconds while moving with the velocity u = 0.8c. 

As u = .8c implies (1 u2/c2)1/2 =.6, thus for the traveller, this journey was of 100 .6 = 60 light seconds, which the traveller completed in 125 .6 = 75 seconds while moving with the velocity u = 0.8c. 

 

KINDLE STORE LINKS

SENTENCE PRACTICE - 1

SENTENCE PRACTICE - 2

  SENTENCE PRACTICE - 3

Useful Facts About Daily Life 

   RAQ

Length Contraction and Time Dilation

 

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