A stationary observer sees a particle moving north at velocity v very close to the speed of light. Then the observer accelerates eastward to velocity v. What is its new total velocity relative to the observer?

I ask because while the particles total velocity will be higher its velocity northward will be lower. It is counterintuitive that accelerating a particle in one direction will decrease its velocity in another.

Will its momentum in the north direction also be lower? I am sure it wont but why not? (because gamma increases?). How does the math work out?

If v is very large I know what the new velocity must look like and none of the equations below gives such an answer. The new v will be close to c and its northward component will be close to v/sqrt(2)

These equations from Wikipedia are all I know about velocity addition.

$u_x = \frac{u_x' + v}{1 + \frac{v}{c^2}u_x'}, \quad u_x' = \frac{u_x - v}{1 - \frac{v}{c^2}u_x},$

$u_y = \frac{\sqrt{1-\frac{v^2}{c^2}}u_y'}{1 + \frac{v}{c^2}u_x'}, \quad u_y' = \frac{\sqrt{1-\frac{v^2}{c^2}}u_y}{1 - \frac{v}{c^2}u_x},$

edit: The northward velocity does become much smaller but the proper velocity and therefore the momentum remains the same. The particles northward velocity decreases exactly as much as the particles clock slows down.

  • $\begingroup$ That is not another way to think of it, the two scenarios are different. In the first scenario, you are remaining in your inertial frame, in the second you are changing the inertial frame. You must formulate your question in one frame and seek the result there, or at least transform the question also. $\endgroup$
    – Umaxo
    Dec 28, 2020 at 14:10
  • 1
    $\begingroup$ Please note that the current question (v8) is not equivalent to the first version (v1). $\endgroup$
    – Urb
    Dec 28, 2020 at 15:20
  • $\begingroup$ OK I see what I did. The new velocity is not toward the north-west. It will be almost due west. So the velocity toward the north is greatly reduced. I still want to know the momentum $\endgroup$
    – R. Emery
    Dec 28, 2020 at 15:40
  • $\begingroup$ It's not just that. In (v1) the particle is accelerated until it points north-east (i.e. until it forms a $45^\circ$ angle with both the x and y axis). In (v8) you are asking for a velocity transformation that is not going to give the same x and y components. $\endgroup$
    – Urb
    Dec 28, 2020 at 15:47
  • 1
    $\begingroup$ I did not explain well. It is possible to accelerate the particle to the east for some time and make it move exactly north-east, but then the x-component of the final velocity will not be $v$ (the initial velocity), neither the y-component will remain $v$. Both (v1) and (v8) have answers, but they are different. $\endgroup$
    – Urb
    Dec 28, 2020 at 16:16

3 Answers 3


As you said, momentum in the y direction will be conserved. From this condition we can obtain the relationship between $v_y$ and $v_x$ at any given moment (indexes $i$ and $f$ indicate initial and final states):

$p_{yi}=m \gamma_i v_{yi}$

$p_{yf}=m \gamma_f v_{yf}$

So that:

$v_{yf}=\frac{\gamma_i }{\gamma_f }v_{yi}$

where $\gamma_i=1/\sqrt{1-v_{yi}^2/c^2}$ and $\gamma_f=1/\sqrt{1-(v_{yf}^2+v_{xf}^2)/c^2}$

That is, $v_{yf}$ is smaller than $v_{yi}$.

  • $\begingroup$ I think your fraction is upside down $\endgroup$
    – R. Emery
    Dec 28, 2020 at 19:41
  • $\begingroup$ I dont think so, which one? $\endgroup$
    – user65081
    Dec 28, 2020 at 20:04
  • $\begingroup$ The first one.. $\endgroup$
    – R. Emery
    Dec 28, 2020 at 20:12
  • $\begingroup$ the first fraction comes from making the first two equal to each other because the initial and final momentums are the same in the y direction $\endgroup$
    – user65081
    Dec 28, 2020 at 20:40

A stationary observer sees a particle moving north at velocity v very close to the speed of light. Then the observer accelerates eastward to velocity v. What is its new total velocity relative to the observer?

In the following gamma is the factor by which the y-speed of the particle slows down (time dilates), when the observer changes its speed from zero to v.

Does 'relative to the observer' mean 'according to the observer'?

Well, according to the observer the new x and y speeds of the particle are -v and v/gamma, so the total new speed he calculates, using the Pythagorean formula, like this:
$$v = \sqrt{(-v_x)^2+(v_y/\gamma)^2 }$$

And the direction of the new velocity vector is:

$$\arctan(\frac{v_y/\gamma} { -v_x} )$$

(north is positive y-axis , east is positive x-axis)

(It would be quite pointless for the observer to use the 'relativistic velocity addition' here. Observer would have to say something like: "according to me the old me moves now at velocity u, and the old me is still saying that the particle moves at velocity v' so according to the current me the velocity of the particle is ... )

  • 1
    $\begingroup$ Correct. Just remember that $v_x=v_y=v$. $\endgroup$ Dec 28, 2020 at 18:41
  • $\begingroup$ So momentum in the y direction is not conserved? $\endgroup$
    – user65081
    Dec 28, 2020 at 19:25
  • $\begingroup$ @Wolphramjonny Momentum in the y direction must be conserved. So new y momentum must be: new y-speed * gamma * mass. Is it perhaps unclear what gamma means here? Well it's the gamma related to the velocity change of the observer. $\endgroup$
    – stuffu
    Dec 30, 2020 at 2:37
  • $\begingroup$ It does not seems that your momentum on y is conserved, see my answer $\endgroup$
    – user65081
    Dec 30, 2020 at 2:43
  • $\begingroup$ @Wolphramjonny Well I guess my answer is a mess which only I can understand ... Now I added an explanation what I mean by gamma. And I upvoted the better answer too :) $\endgroup$
    – stuffu
    Jan 1, 2021 at 18:23


We have a particle moving with a velocity $\vec{v} =v \vec{e}_x$ wrt. to an inertial frame $\Sigma$ and we want to know its velocity in an inertial frame $\Sigma'$ which moves with a velocity $\vec{w}=w \vec{e}_y$ where v and w are equal.

It's easier to work first with four-vectors. The particle's four-velocity components wrt. to $\Sigma$ are:

$ \begin{align} u^{\mu}&=\gamma_v (1,\beta_v,0,0) \\ &=(\gamma_v,\gamma_v \beta_v,0,0) \end{align} $

Its clock is ticking at $\gamma_v$
Its velocity is $\beta_v$
Its proper velocity is $\gamma_v \beta_v$

The components wrt. $\Sigma'$ are given by a Lorentz boost in $y$-direction with velocity $\vec{w}$, i.e., for the four-vector components you have

$ \begin{align} u^{\prime \mu}&=[\gamma_w (u^0-\beta_w u^2),u^1,\gamma_w(-\beta_w u^0 + u^2),u^3] \\ &= \gamma_v (\gamma_w,\beta_v,-\beta_w \gamma_w,0).\\ &= (\gamma_v \gamma_w,\gamma_v \beta_v,-\gamma_v \beta_w \gamma_w,0). \end{align} $

The first thing to note is that the resulting vector does not point halfway between $\vec{e}_x$ and $\vec{e}_y$. It points almost directly toward $\vec{e}_y$

The three-velocity thus is

$ \begin{align} \vec{v}'&=c \vec{u}'/u^{\prime 0} \\ &=\frac{c}{\gamma_v \gamma_w} (\gamma_v \beta_v,-\gamma_v \beta_w \gamma_w,0) \\ &=(v/\gamma_w,-w,0). \end{align} $

Its clock is ticking at $\gamma_v \gamma_w$
The $e_x$ component of its velocity is $v/\gamma_w$
The $e_x$ component of its proper velocity is $\gamma_v \beta_v$

The second thing to note is that the $e_x$ component of its velocity slowed down by $\gamma_w$ even though the observer accelerated perpendicular to that component. Exactly the amount its internal clock slowed down! (As measured by the observer).

The third thing to note is that even though this component of its velocity decreased neither its proper velocity nor its momentum in that same direction changed at all.

  • $\begingroup$ I wonder what will be the confusion in this site when even a small number of users post answers to their own questions and accept them as the best ones. $\endgroup$
    – Frobenius
    Jan 1, 2021 at 23:35
  • $\begingroup$ Well first of all thats not my answer. I was given that answer at Physics forums. Hence the link at the top. Secondly if they dont want people to answer their own questions then they can easily make it so they can't do that. $\endgroup$
    – R. Emery
    Jan 1, 2021 at 23:48

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