Transformation equation of velocity contradicts stellar aberration?

[Happiness]

Let S be the frame where the Sun is at rest. Imagine light from the North Star reaches the centre of the Sun, and let's define the equatorial plane as the plane that is perpendicular to this light and cuts the Sun into two hemisphere.

Suppose a distant star A is on this equatorial plane and its light also reaches the centre of the Sun. Let the direction of the propagation of this light be our x axis. And let our y-axis be pointing in the direction of the North Star.

In other words, the light from star A is traveling in the positive x direction, while that from the North Star is traveling in the negative y direction.

Let S' be the frame that is moving with respect to S in the positive x direction at velocity v.

By the transformation equation of velocity,

$u_x^\prime=\frac{u_x-v}{1-u_xv/c^2}$ , $u_y^\prime=\frac{u_y\sqrt{1-v^2/c^2}}{1-u_xv/c^2}$ , $u_z^\prime=\frac{u_z\sqrt{1-v^2/c^2}}{1-u_xv/c^2}$

the light from the North Star and that from star A are still traveling in the negative y direction (or y', to be precise) and in the positive x direction (or x'), respectively, to an observer in S'.

This contradicts the observation of stellar aberration, where the two velocities are not perpendicular in S'.

What did I miss?

Are two perpendicular vectors still perpendicular when we change reference frames?

 

[BiGyElLoWhAt]

Conceptually, I would think that they would not be perpendicular in S'. The light from the north star should gain a x component due to the observers velocity in the x direction. I'm also not sure what you're doing with the transformation equations. What is u the velocity of?

 

[Happiness]

u and u' are the velocities in S and S', respectively.

$u_{North\ Star}$, velocity of light from the North Star in frame S = $\begin{pmatrix}u_x\\u_y\\u_z\end{pmatrix}$ = $\begin{pmatrix}0\\-c\\0\end{pmatrix}$

 

[BiGyElLoWhAt]

Ok, so transform $u_{North Star}$ to $u_{North Star}'$ and you'll see that in one frame, u', you have an x component from the orthogonality of the two velocities.

 

[Happiness]

No, it does not. Have you learned special relativity before?

 

[BiGyElLoWhAt]

-.-
$\left ( \begin{array}{c} 0 \\ -c \\ 0 \\ \end{array} \right )$
$\left ( \begin{array}{c} v \\ 0 \\ 0 \\ \end{array} \right )$
$u'=\frac{u-v}{ 1-\frac{uv}{c^2}} =\frac{ \left ( \begin{array}{c} 0 \\ -c \\ 0 \\ \end{array} \right ) - \left ( \begin{array}{c} v \\ 0 \\ 0 \\ \end{array} \right )}{1-\frac{uv}{c^2}}$
What you are trying to do is set up the transforms for when the velocity of the north star is along the x-axis (parallel to the velocity).

Also, no, I haven't learned SR. I have, however, studied it. Once I learn it, I will quit studying it.

 

[Happiness]

Oh yes, you are right! Now, I see the mistake. Thanks!

 

[BiGyElLoWhAt]

Think about it like this. If these 2 rays of light are going to meet at the center of the sun, then in their respective frames (S' and S'', where S'' has a velocity along the y-axis parallel to the velocity of the north star light coming towards the sun ), there must be a non orthogonal velocity component between the 2 frames in BOTH frames (S' and S'')

 

[BiGyElLoWhAt]

No problem, but do me a favor, and don't attack people that are trying to help you like that.

 

[Happiness]

Yes, don't take it to heart.

 

[DrGreg]

u&#039;=\frac{u-v}{ 1-\frac{uv}{c^2}}<br /> =\frac{<br /> \left ( \begin{array}{c}<br /> 0 \\<br /> -c \\<br /> 0 \\<br /> \end{array} \right ) -<br /> \left ( \begin{array}{c}<br /> v \\<br /> 0 \\<br /> 0 \\<br /> \end{array} \right )}{1-\frac{uv}{c^2}}

That's not correct for non-parallel velocities. See velocity-addition formula.

 

[BiGyElLoWhAt]

Ahh, true, I missed a 1/gamma on my v. Regardless, the point is that the velocities have non orthogonal components after you transform out of the S frame to S' or S''.