We know that the rules of relativistic addition are more complicated than those for simple classical cases. For example, we can have two frames of reference, and if we know the motion of the frames themselves, and the velocity of an object in one frame, we can use the formula to get the velocity of the object in the other frame.
However, imagine we know the velocity of an object with respect to both the frames, and we want to find out the velocity of them frames relative to one another. This can also be done using the formula.
We know $$s'_s = \frac{s'_v+v_s}{1+\frac{s'_vv_s}{c^2}}$$
Here, $s'_s$ is the velocity of $s'$ relative to $s$. $s'_v$ is the velocity of frame $s'$ relative to the object and $v_s$ is the velocity of the object relative to frame $s$.
My question is, suppose the motion of the object is in $2$ dimensions relative to both the frames. Do we use the same above formula for both directions separately? Normally, when we have the object moving along two dimensions, but the frame moving about one dimension, our formula is slightly modified and the Lorentz factor enters for finding velocity in those other directions. The formula becomes something like : $v_s' = \frac{v_s}{\gamma (1+\frac{s'_vv_s}{c^2})}$. I don't know exactly how it'll look like, because I'm used to calculating the velocity of an object (say) in the $y$ direction of a frame, which is moving in the $x$ direction relative to our original frame.
But, if we don't know how the two frames are moving relative to each other, and are only given the velocity of an object in each of these frames, how can we find the relative velocity of the two frames, in more than one dimensions ?
However, I don't know how to apply the formula here, since we don't know in the first place how the frames are moving relative to each other.
Suppose the velocity of an object along a particular direction is the same for both the frames. Does that mean the component of velocity of both the frames in that direction is the same.
For example, let $v_s = \frac{c}{\sqrt{2}}(\hat{i}+\hat{j})$ and $v_s' = \frac{c}{\sqrt{2}}(-\hat{i}+\hat{j})$
Since the velocity of the particle in the $y$ direction is the same with respect to both frames, can we say that the relative velocity of the frames in the $y$ direction atleast is $0$ ?
Best Answer
The easiest way is to use four-vectors. If an object has three-velocity $\mathbf{u} =\begin{bmatrix} u_x \\ u_y \\ u_z \end{bmatrix}$ in one frame, its four-velocity is $$\gamma_u\begin{bmatrix} c \\ u_x \\ u_y \\ u_z\end{bmatrix}$$
The general form of a Lorentz transformation from frame $S$ to frame $S'$ is $$\begin{bmatrix} \gamma&-\gamma\beta_x&-\gamma\beta_y&-\gamma\beta_z\\ -\gamma\beta_x&1+(\gamma-1)\frac{\beta_x^2}{\beta^2}&(\gamma-1)\frac{\beta_x \beta_y}{\beta^2}&(\gamma-1)\frac{\beta_x \beta_z}{\beta^2}\\ -\gamma\beta_y&(\gamma-1)\frac{\beta_x \beta_y}{\beta^2}&1+(\gamma-1)\frac{\beta_y^2}{\beta^2}&(\gamma-1)\frac{\beta_y \beta_z}{\beta^2}\\ -\gamma\beta_z&(\gamma-1)\frac{\beta_x \beta_z}{\beta^2}&(\gamma-1)\frac{\beta_y \beta_z}{\beta^2}&1+(\gamma-1)\frac{\beta_z^2}{\beta^2}\\ \end{bmatrix}$$ where $\begin{bmatrix} \beta_x \\ \beta_y \\ \beta_z \end{bmatrix}$ is the velocity of $S'$ relative to $S$. Since the four-vector transforms by matrix multiplication, and both four-vectors are given, it is then a straightforward task to solve for the relative velocity in the matrix.
Yes, this implies that all relative velocity must either be zero or be in the perpendicular direction.