# Relativistic velocity, perpendicular acceleration and momentum

• I
Gold Member
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?

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

• PeroK, etotheipi and em3ry
Gold Member
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)

Gold Member
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!

Last edited:
Homework Helper
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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: $$v = \frac{d}{\gamma \tau} \ \ \text{and} \ \ v' = \frac{d}{\gamma' \tau}$$

• em3ry
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Gold Member
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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##.

• em3ry
• 