Difference between revisions of "Time dilation"

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Dividing by <math>[1-(v^2/c^2)]</math> yields
 
Dividing by <math>[1-(v^2/c^2)]</math> yields
  
<math>(t_2)^2=\dfrac{(t_1)^2}{1-v^2/c^2}}</math>
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<math>(t_2)^2=\dfrac{(t_1)^2}{1-v^2/c^2}</math>
  
 
Finding the root of both sides finally yields
 
Finding the root of both sides finally yields
  
 
<math>\dfrac{t_1}{\sqrt{1-v^2/c^2}}=t_2</math>
 
<math>\dfrac{t_1}{\sqrt{1-v^2/c^2}}=t_2</math>

Latest revision as of 00:19, 31 January 2021

In special relativity, the time dilation that will be experienced can be expressed with the formula: $\dfrac{t_1}{\sqrt{1-v^2/c^2}}=t_2$

where $t_1$ is the "proper" time experienced by the moving object.

$v$ is the relative velocity the ovject is moving to the observer.

$c$ is the speed of light; it can be simply expressed as 1, but then the velocity will have to be given in terms of $c$.

$t_2$ is the time the events appear to occur, with the time dilation added.

This formula can be derived using a simple example. If there was a light clock that consisted of two mirrors with a light beam bouncing between them, the distance between the mirrors would be $ct_1$. If, the clock was inside a moving vehicle, than the distance the clock travels would equal $vt_2$. Finally, to an observer of the light clock, the light beam, instead of bouncing up and down, would move diagonally. The distance it appears to travel is $ct_2$. This forms a triangle, and using the Pythagorean theorem, it would obtain:

$c^2(t_2)^2=c^2(t_1)^2+v^2(t_2)^2$

Subtracting by $v^2(t_2)^2$ yields

$c^2(t_2)^2-v^2(t_2)^2=c^2(t_1)^2$

Dividing by $c^2$ yields

$(t_2)^2[1-(v^2/c^2)]=(t_1)^2$

Dividing by $[1-(v^2/c^2)]$ yields

$(t_2)^2=\dfrac{(t_1)^2}{1-v^2/c^2}$

Finding the root of both sides finally yields

$\dfrac{t_1}{\sqrt{1-v^2/c^2}}=t_2$

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