- #1
Inderjeet
- 2
- 0
I recently saw the derivation of length contraction in Special Relativity . At the end , it said
x' = (x - vt) γ(gamma)
x = (x' + vt') γ(gamma)
Where γ(gamma) is Lorentz transformation . It is = 1/√(1- v²/c²)
Then derivation continued , with expansion of x' = (x + vt)γ
As t = 0 in this case
We end up with , x'(√(1- v²/c²) ) = x
As √(1- v²/c²) is always between 1 and 0 , x ≤ x'
Thus , length of an object is contracted of any other observer ( if the speed of object is near the speed of light)
Now the question is , If we take x = (x' + vt') γ instead of x' = (x + vt)γ
We end up with x = (x')γ
and we take x' = (x + vt)γ (as we did above)
We end up with x' = (x)γ
How is this possible that
x = (x')γ and x' = (x)γ
If i have watched the wrong derevation , please send me the link of correct derevation .
x' = (x - vt) γ(gamma)
x = (x' + vt') γ(gamma)
Where γ(gamma) is Lorentz transformation . It is = 1/√(1- v²/c²)
Then derivation continued , with expansion of x' = (x + vt)γ
As t = 0 in this case
We end up with , x'(√(1- v²/c²) ) = x
As √(1- v²/c²) is always between 1 and 0 , x ≤ x'
Thus , length of an object is contracted of any other observer ( if the speed of object is near the speed of light)
Now the question is , If we take x = (x' + vt') γ instead of x' = (x + vt)γ
We end up with x = (x')γ
and we take x' = (x + vt)γ (as we did above)
We end up with x' = (x)γ
How is this possible that
x = (x')γ and x' = (x)γ
If i have watched the wrong derevation , please send me the link of correct derevation .