Question

The rest distance Patna and Delhi is 1000km. A non-stop train travels at 360$km{{h}^{-1}}$.
(a) What is the distance between Patna and Delhi in the train frame?
(b) How much time elapses in the train frame between Patna and Delhi?

Answer 1

Hint: Use the formula for contracted length in the frame of a moving body, i.e. $l'=l\sqrt{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}}$ and find the distance between Patna and Delhi in the train frame. Then use the formula $\text{time = }\dfrac{\text{distance}}{\text{speed}}$ and calculate the time taken in the train frame.

Formulae used:
$l'=l\sqrt{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}}$
$\text{time = }\dfrac{\text{distance}}{\text{speed}}$

Complete answer:
(a) According to the theory of relativity, the length contract with respect to a moving object. Let us understand this statement.
Suppose an object travels a distance l. With respect to the ground or a stationary observer, the distance travelled by the object will be equal to l. This distance is also called rest distance. However, if measured the length of the distance in the frame of the object (i.e. with respect to the object) then the length will be shorter than the actual length. This contracted length is called apparent length.
The apparent length l’ is given as $l'=l\sqrt{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}}$ ….. (i),
Where v is the speed of the object and c is the speed of light in vacuum, i.e. $3\times {{10}^{8}}m{{s}^{-1}}$
Now, we have to calculate the apparent distance that the train covers in the train frame.
Here, l = 1000km and v =$360km{{h}^{-1}}=360\left( {{10}^{3}}m \right){{\left( 3600s \right)}^{-1}}=100m{{s}^{-1}}$.
Substitute the values of l, v and c in equation (i).
$\Rightarrow l'=1000\sqrt{1-\dfrac{{{\left( 100 \right)}^{2}}}{{{\left( 3\times {{10}^{8}} \right)}^{2}}}}$
$\Rightarrow l'=1000\sqrt{1-\dfrac{\left( {{10}^{4}} \right)}{\left( 9\times {{10}^{16}} \right)}}$
$\Rightarrow l'=\left( 1000-56\times {{10}^{-12}} \right)km$
Therefore, the distance between Patna and Delhi in the train frame is $56\times {{10}^{-12}}$km (i.e. 56mn) less than the actual length of the distance.

(b) Let us calculate the time taken for the train to travel from Patna to Delhi in the train frame. For this, let us use the formula $\text{time = }\dfrac{\text{distance}}{\text{speed}}$.
Let the apparent time taken be t’.
Therefore, $t'=\dfrac{l'}{v}=\dfrac{\left( 1000-56\times {{10}^{-12}} \right)km}{100m{{s}^{-1}}}=\dfrac{\left( 1000-56\times {{10}^{-12}} \right)\times 1000m}{100m{{s}^{-1}}}=\left( {{10}^{4}}-0.56\times {{10}^{-9}} \right)s$

Note: As per the above question, it means that a moving body sees the lengths of surrounding objects shorter than the actual lengths. However, when we are in motion we see the length of the objects to be equal to the lengths when we see them by being stationary.
This does not mean that there is no length contraction when we move. The contraction of the length is so small that it cannot be seen by our naked eyes. This is because our speed is very very less than the speed of light in vacuum.