Relativistic velocity, perpendicular acceleration, and momentum [closed]

Question

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.

Answer 1

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 $v_{yi}$.

Answer 2

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 ... )

Answer 3

https://www.physicsforums.com/threads/relativistic-velocity-perpendicular-acceleration-and-momentum.997831/

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.