Question

show that the circle x^2+y^2= a^2 in frame S appears to be an ellipse in frame S' which is moving with velocity v relative to S​

Answer 1

Explanation:

usual, prove it using the general heuristic that

X=YX=Y if and only if X⊆Y and X⊇Y.X⊆Y and X⊇Y.

In other words, take a generic element of A×(B−C)A×(B−C) and show that it also must be an element of (A×B)−(A×C)(A×B)−(A×C), and vice versa.

Answer 2

Answer:

Circle x^2+y^2= a^2 in frame S appears to be an ellipse in frame S' which is moving with velocity v relative to S​. This can be shown as,

Explanation:

Let in S-frame , equation be x²+y²=a²------------(i)

Using Lorentz transformation,

The change in x will be :

$x' =\frac{x-vt}{\sqrt{1-\frac{v^{2} }{c^{2} } } }$

$y' = y$

$z'=z$

putting the values in equation 1, we get

$(\frac{x'-vt'}{\gamma\sqrt{1-\frac{v^{2} }{c^{2} } } })^{2} + (\frac{y'}{\gamma}) ^{2} = (\frac{\gamma}{\gamma}) ^{2}$

⇒ $\frac{(x-h)^{2} }{a^{2}} +\frac{y^{2}}{b^{2}}= 1$

this is equation of ellipse.