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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 $0$ ?

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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.

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 at least is 0?

Yes, this implies that all relative velocity must either be zero or be in the perpendicular direction.

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You can use the rest frame of the "object" as one of your reference frames. In your example, you could say that $\mathbf{v}_s$ is the velocity of the object's frame with respect to the unprimed frame, and $-\mathbf{v}_s'$ is the velocity of the primed frame with respect to the object's frame (note the minus sign.) From here, it is straightforward to apply the velocity addition rule as you understand it to find the velocity of the primed frame with respect to the unprimed frame.

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  • $\begingroup$ This is exactly what I tried to do, and in regards to a particular question, I managed to get the right answer. However, the problem is, the velocity addition rule is slightly different with respect to parallel and perpendicular components. Now, the object has both $x$ and $y$ component of velocity just like the relative velocity between primed and unprimed frames. How do I guess which one is parallel and which one is perpendicular with respect to objects rest frame ? $\endgroup$ Sep 28, 2021 at 17:37
  • $\begingroup$ What I did was, relative to the velocity of the object along the $x$ direction of some rest frame, the velocity of $s,s'$ frame along that same direction in that frame is parallel. So I found the relative velocity in the $x$ direction between the two frames. I did the exact same thing in the $y$ direction. $\endgroup$ Sep 28, 2021 at 17:39
  • $\begingroup$ @NakshatraGangopadhay: You would need to use the more general velocity-addition equation that allows for the frames not to be moving in the $x$-direction relative to each other. Wikipedia has it, along with a brief derivation. I don't believe that you can separate out the transformations the way you did, since the velocity transformation rules are not linear. $\endgroup$ Sep 28, 2021 at 17:41
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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

If you know the velocity of the object with respect to frame s' (call it $v_{s'}$), then you know the velocity $s'_v$ of frame $s'$ with respect to the object: $s'_v = -v_{s'}$. This pair-wise reciprocity of velocities will always hold true.

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?

No, not really. This formula is only valid when all velocities are parallel. Essentially, the way relativistic velocity addition works is that you have a formula for composing parallel velocities and a different one for composing perpendicular velocities. The general formula, for two arbitrary velocities, is derived by splitting the velocity vectors into parallel and orthogonal components (as can be seen in this article, in the "General configuration" section: https://en.wikipedia.org/wiki/Velocity-addition_formula ).

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  • $\begingroup$ Yeah, this is what I thought of doing, but the problem with this is, it first assumes that the primed frame is moving in the $x$ direction relative to unprimed frame. The particle has both parallel and perpendicular velocities in both the frames, but what happens when the frames themselves are moving in an arbitrary direction and not along the $x$ axis as shown in the derivation from wikipedia. $\endgroup$ Sep 28, 2021 at 17:31

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