Suppose a particle travels with velocity $\textbf u=(c/3,c/4)$ as perceived in an inertial frame $S$. Let $S'$ be another inertial frame at speed $c/5$ in the direction of the positive $x$-direction (as perceived in $S$). In $S$, the particle's trajectory makes an angle of $\alpha=\arctan(3/4)\approx 36.9^{\circ}$ with the $x$-axis. The question is: what is the angle $\alpha'$ made by the particle's trajectory with the $x'$-axis as viewed in $S'$?

My first method was to use $4$-velocities. If $U$ is the $4$-velocity of the particle in $S$ and $U'$ is the $4$-velocity of the particle in $S'$, then because the $4$-velocity is a $4$-vector, it would transform via a Lorentz transformation as: $$U'=\Lambda U$$ Using this method, it can eventually be shown that the angle $\alpha'=\arctan(\frac{3\sqrt{6}}{4})\approx 61.4^{\circ}$.

My second method was to use the invariance of the Minkowski inner product. If the notation $U_{AB}$ denotes the $4$-velocity of inertial frame $A$ as perceived in inertial frame $B$, then I expected: $$U_{S'S}\cdot U_{PS}=U_{S'S'}\cdot U_{PS'}$$ and $$U_{S'P}\cdot U_{SP}=U_{S'S}\cdot U_{SS}$$ where $P$ denotes the particle's inertial frame. This gave me a system of equations for the components of the particle's velocity $\textbf u'$ as perceived in $S'$, which, when I solved it, yielded an angle of $\alpha'\approx 59.9^{\circ}$.

What is the reason for the discrepancy between these two methods?

  • $\begingroup$ Let's get this straight first: When you write $u = \left(c/3, c/4\right)$ what is it that you mean? Is this the velocity in a two-dimensional Cartesian coordinate system? In which case "4-velocity" is a bit misleading, in a 2+1-dimensional theory. ;) $\endgroup$
    – kricheli
    Dec 11, 2022 at 18:59
  • 1
    $\begingroup$ Yes, sorry that's what I meant. $\textbf u=(c/3,c/4)$ is the velocity vector as viewed in the two-dimensional Cartesian coordinate system of inertial frame (S). $\endgroup$ Dec 12, 2022 at 19:18
  • $\begingroup$ U is 4-velocity , thus $~U_\mu\,U^\mu=c^2~$ is invariant at any frame. I can't see it ? $~U^\mu=\gamma(u)\,[c,\vec u]^T~$ $\endgroup$
    – Eli
    Dec 13, 2022 at 17:58
  • $\begingroup$ I reckon you might've set up the wrong equation for $U_{S'P}$. Did you by any chance use something like $U_{S'P} = \gamma(c, -v_x, -v_y, 0)$, where $v_x$ and $v_y$ are the variables to solve for? That would yield the (wrong) 59 degree answer. $\endgroup$ Dec 13, 2022 at 21:08
  • $\begingroup$ @Nihar Karve I believe that was indeed what I did, however I'm not sure why that would be incorrect? Could you show me the correct way to set it up and I can then give you the bounty. $\endgroup$ Dec 14, 2022 at 21:37

1 Answer 1


Let $U_{PS'}=\gamma(c, \mathbf v_{PS'})=\gamma(c, v_x, v_y, 0)$ be the 4-vector of interest. Then $\alpha$, the variable we are solving for, is $\tan^{-1}(v_y/v_x)$ and is correctly obtained by method 1 (a straightforward Lorentz boost) to be $61.4^\circ$.

This discrepancy in the second method likely arises from writing $U_{S'P}=\gamma(c, -v_x, -v_y, 0)$, i.e. $\gamma(c, -\mathbf v_{PS'})$. But this is incorrect, though non-intuitive, since it clearly works in the Newtonian case.

It remains true even in special relativity that if frame K is related to frame L by a pure boost of velocity $\mathbf v$, denoted $B(\mathbf v)$, then frame L is related to frame K by a pure boost of velocity $-\mathbf v$. This is to say $B(\mathbf v)^{-1}=B(-\mathbf v)$, as one would expect intuitively (e.g. in the $U_{SS'}$ equation).

However while relating frames $P$ and $S'$, there are two different Lorentz boosts at play here, $P\overset{B(\mathbf u_1)}\longleftarrow S\overset{B(\mathbf u_2)}\longrightarrow S'$. To get the 4-velocity of $S'$ in the frame of $P$ we need to "end up at $P$" so the chain is $S'\to S\to P$, corresponding to a total Lorentz transformation of $B(\mathbf u_1)B(-\mathbf u_2)$ applied to the 4-velocity of $S'$ in its own frame, i.e. $(1, 0, 0, 0)$. To get the 4-velocity of $P$ in the frame of $S$, the combined Lorentz transformation is $B(\mathbf u_2)B(-\mathbf u_1)$ applied on $(1, 0, 0, 0)$.

In Newtonian mechanics $B(\mathbf u_1)B(-\mathbf u_2) \overset{(*)}= B(\mathbf u_1-\mathbf u_2) =-B(\mathbf u_2-\mathbf u_1) \overset{(*)}= -B(\mathbf u_2)B(-\mathbf u_1)$, so the relative velocities are negatives of each other. But the key difference in special relativity is that Lorentz boosts do not commute, so $B(\mathbf a)B(\mathbf b) \ne B(\mathbf b)B(\mathbf a)$. Thus the equations marked with $(*)$ do not hold and the chain of logic is broken: the relative velocities $\mathbf v_{PS'}\ne-\mathbf v_{S'P}$ in relativity.

In fact if you compute it correctly, $\mathbf v_{S'P}$ has the same magnitude as $-\mathbf v_{PS'}$; it's just rotated by some angle. This is a general phenomenon known as Wigner rotation that arises due to this non-commutativity of general Lorentz boosts: two successive boosts in different directions cannot be represented by only a boost with the "combined" velocity, a rotation must also come along for the ride.

However the principle behind the second method remains correct: we can still use the invariance of the inner product across inertial frames. Instead of the 2nd equation in the OP, we should use the following:

$$U_{S'S}\cdot U_{PS}=U_{S'S'}\cdot U_{PS'}\tag{1}$$ $$U_{SS}\cdot U_{PS}=U_{SS'}\cdot U_{PS'}\tag{2}$$

These set up a system of equations whose answer matches the first method, as verified by WolframAlpha.


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