Relativistic Velocity, Perp. Accel., Momentum: Explained

[em3ry]

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 of the particle toward the north-west 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 won't but why not? (because gamma increases?). How does the math work out?

 

[vanhees71]

If I understand you right, you have a particle moving with a velocity $\vec{v} =v \vec{e}_x$ wrt. to an inertial frame $\Sigma$ and you want to know its velocity in an inertial frame which moves with a velocity $\vec{w}=w \vec{e}_y$.

It's easier to work first with four-vectors. The particle's four-velocity components wrt. to $\Sigma$ are
$$u^{\mu}=\gamma_v (1,\beta_v,0,0).$$
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
$$u^{\prime \mu}=[\gamma_w (u^0-\beta_w u^2),u^1,\gamma_w(-\beta_w u^0 + u^2),u^3]= \gamma_v (\gamma_w,\beta_v,-\beta_w \gamma_w,0).$$
The three-velocity thus is
$$\vec{v}'=c \vec{u}'/u^{\prime 0}=\frac{c}{\gamma_v \gamma_w} (\gamma_v \beta_v,-\gamma_v \beta_w \gamma_w,0)=(v/\gamma_w,-w,0).$$

 

[em3ry]

So the momentum in the north direction ($e_x$) becomes $\gamma_v \gamma_w \cdot v / \gamma_w$? The same as it was before? (Times the mass of course and c I guess)

 

[em3ry]

So its velocity in the north direction slowed down by $\gamma_w$. Exactly the amount its internal clock slowed down! (As measured by the observer)

Relativity is weird!

 

[PeroK]

So its velocity in the north direction slowed down by $\gamma_w$. Exactly the amount its internal clock slowed down!

The coordinates in the N direction are the same for both frames, as this direction is perpendicular to the relative motion. If we imagine the particle traveling a distance $d$ in the N-direction while its internal clock (proper time) advances by $\tau$, then the N-component of the particle's velocity is directly related to the particle's time dilation as measured in the two frames.

In other words, if $u_N, u'_N$ is the N-component of the particle's velocity in the two frames and $\gamma, \gamma'$ is the particle's gamma factor in the two frames, then the speed in the N-direction as measured in the two frames is: $$v = \frac{d}{\gamma \tau} \ \ \text{and} \ \ v' = \frac{d}{\gamma' \tau}$$

 

[PeroK]

So the momentum in the north direction ($e_x$) becomes $\gamma_v \gamma_w \cdot v / \gamma_w$? The same as it was before? (Times the mass of course and c I guess)

That follows automatically, as four-momentum is a four-vector: $p^x = p'^x$.

Where: $p^x = \gamma m v^x$ and $p'^x = \gamma' m v'^x$.

 

[vanhees71]

For a boost perpendicular to the velocity in $\Sigma$ all there is, is time dilation for the velocity component in the original direction, which explains the rescaling with $1/\gamma_w$ of this component.