# Error in Manipulating $4$-Velocities?

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?

• 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. ;) Dec 11, 2022 at 18:59
• 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). Dec 12, 2022 at 19:18
• U is 4-velocity , thus $~U_\mu\,U^\mu=c^2~$ is invariant at any frame. I can't see it ? $~U^\mu=\gamma(u)\,[c,\vec u]^T~$
– Eli
Dec 13, 2022 at 17:58
• 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. Dec 13, 2022 at 21:08
• @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. Dec 14, 2022 at 21:37

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.