Solve velocity addition problems using 4-vector

[Natchanon]

Homework Statement

An inertial frame F' moves in the x direction with speed v = 0.9c relative to another frame F. A traveler is moving with velocity u' = 0.5c i + 0.5c j in the frame F'. Determine the traveler's velocity in the frame F.

Relevant Equations

u_parallel = (u_parallel' + v)/(1+vu_parallel'/c^2)
u_perpen = u_perpen'/gamma(1+vu_perpen'/c^2)

I used the two equations above to solve for u_x and u_y and got u = 0.987c, where u_parallel = u_x and u_perpen = u_y. I wonder if I can use velocity four-vectors to solve this problem. Modify η^'μ = Λ^μ_νη^ν so we can use it for velocity addition?

 

[mitochan]

Hi.
As a byproduct of recent study in another thread, in perpendicular case, the below shown multiplication rule of gamma or time component of four velocity seems to work.
$$\gamma=\gamma_1\gamma_2$$
where S_1 is moving against S with speed $v_1$ corresponding $\gamma_1$ in say x-direction, S_2 is moving against S_1 with speed $v_2$ corresponding $\gamma_2$ in say y-direction, and $\gamma$ is corresponding to velocity of S_2 against S with direction $(v_1, v_2/\gamma_1,0)$.
I will be glad if you check it works or not.

 

[PeroK]

Hi.
As a byproduct of recent study in another thread, in perpendicular case, the below shown multiplication rule of gamma or time component of four velocity seems to work.
$$\gamma=\gamma_1\gamma_2$$
where S_1 is moving against S with speed $v_1$ corresponding $\gamma_1$ in say x-direction, S_2 is moving against S_1 with speed $v_2$ corresponding $\gamma_2$ in say y-direction, and $\gamma$ is corresponding to velocity of S_2 against S with direction $(v_1, v_2/\gamma_1,0)$.
I will be glad if you check it works or not.

More generally, assume a particle is moving with velocity $u$ in a reference frame, S_1, that is moving at velocity $v$ with respect to another frame, S_2, (assume along the x-axis). Then, the gamma factor of the particle in S_2 is:

$\gamma_2 = \gamma \gamma_1 (1 - \frac{vu_x}{c^2})$

The simplest way to show this is to consider the time component of the 4-velocity of the particle:

In S_1 we have $u^0 = \gamma_1 c$

In S_2 we have $u_2^0 = \gamma_2 c$

And, these are related by the Lorentz Transformation:

$u_2^0 = \gamma(u^0 - \frac{vu_x}{c^2})$

Putting these together gives the required result.

In the special case where $u_x = 0$, we have $\gamma_2 = \gamma \gamma_1$

 

[mitochan]

Very exciting. Thanks.

RHS of your $\gamma_2$ formula coincides with inner product of non dimensional four-vectors,
$U^\mu V_\mu$ in (+---) convention, where
$$U^\mu=(\gamma_u, \gamma_uu_x/c, \gamma_uu_y/c, \gamma_uu_z/c)$$
$$V^\mu=(\gamma_v, \gamma_vv_x/c, \gamma_vv_y/c, \gamma_vv_z/c)$$
Is it right to ascribe the origin of term $uv_x$ in velocity addition rule to this four vector inner product ?

Doing substitution in your last formula I got
$$\gamma_2=\gamma(\gamma_1-\frac{vu_x}{c^2})$$
It does not coincide with your $\gamma_2$ formula. wrong where ?

 

[PeroK]

Very exciting. Thanks.

RHS of your $\gamma_2$ formula coincides with inner product of non dimensional four-vectors,
$U^\mu V_\mu$ in (+---) convention, where
$$U^\mu=(\gamma_u, \gamma_uu_x/c, \gamma_uu_y/c, \gamma_uu_z/c)$$
$$V^\mu=(\gamma_v, \gamma_vv_x/c, \gamma_vv_y/c, \gamma_vv_z/c)$$
Is it right to ascribe the origin of term $uv_x$ to this four vector inner product ?

Doing substitution in your last formula I got
$$\gamma_2=\gamma(\gamma_1-\frac{vu_x}{c^2})$$
It does not coincide with your $\gamma_2$ formula. wrong where ?

Yes, the formula generalises further to any two velocities, using the inner product.

I used different notation for the frames. That's the only reason it's different from yours.

 

[mitochan]

And, these are related by the Lorentz Transformation:

Thanks. I just restate it to confirm my understanding.

There are three IFRs, say $S,S_1,S_2$.
In S Origin of $S_1$ is moving with four velocity $v_1^\mu$.
In S Origin of $S_2$ is moving with four velocity $v_2^\mu$.
In $S_1$ Origin of $S_2$ is moving with four velocity
$$\Lambda(-v_1)^\mu_\nu v_2^\nu$$
In $S_2$ Origin of $S_1$ is moving with four velocity
$$\Lambda(-v_2)^\mu_\nu v_1^\nu$$
where $\Lambda(v)$ is matrix of Lorentz transformation with four velocity v. These two should have opposite sign but same magnitude in space components.

I have not been well prepared to do detailed calculation admitting general directions for v_1 and v_2.

(edit)
Say S_1 and S_2 are Lorentz transformed from S. Transformation between S_1 and S_2 should include boost and rotation. In $\Lambda(-v)$ minus means space inversion without change of time component.

 

[PeroK]

Thanks. I just restate it to confirm my understanding.

There are three IFRs, say $S,S_1,S_2$.
In S Origin of $S_1$ is moving with four velocity $v_1^\mu$.
In S Origin of $S_2$ is moving with four velocity $v_2^\mu$.
In $S_1$ Origin of $S_2$ is moving with four velocity
$$\Lambda(-v_1)^\mu_\nu v_2^\nu$$
In $S_2$ Origin of $S_1$ is moving with four velocity
$$\Lambda(-v_2)^\mu_\nu v_1^\nu$$
where $\Lambda(v)$ is matrix of Lorentz transformation with four velocity v.
These two have same magnitude but opposite sign in space components.

That may be true, but that's not what I said in post #3.

 

[mitochan]

Homework Statement: An inertial frame F' moves in the x direction with speed v = 0.9c relative to another frame F. A traveler is moving with velocity u' = 0.5c i + 0.5c j in the frame F'. Determine the traveler's velocity in the frame F.
Homework Equations: u_parallel = (u_parallel' + v)/(1+vu_parallel'/c^2)
u_perpen = u_perpen'/gamma(1+vu_perpen'/c^2)

I wonder if I can use velocity four-vectors to solve this problem.

Thanks to the teachings in above posts. Let me try answer it.

Suppose Lorentz transformation the matrix of which is

$$\left( \begin{array}{cccc} A'^0 \\ A'^1 \\ A'^2 \\ A'^3 \\ \end{array} \right)= \begin{pmatrix} \gamma_u & 0 & 0 & -\gamma_u u \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\gamma_u u& 0 & 0 & \gamma_u \\ \end{pmatrix} \left( \begin{array}{cccc} A^0 \\ A^1 \\ A^2 \\ A^3 \\ \end{array} \right) = \left( \begin{array}{cccc} \gamma_u A^0 - \gamma_u u A^3 \\ A^1 \\ A^2 \\ -\gamma_u u A^0 + \gamma_u A^3 \\ \end{array} \right)$$
where velocity is non dimension for divided by c. I can set z-axis be the direction of boost without losing generality. x, y components do not change.

Let A be four velocity of a moving body

$$\left( \begin{array}{cccc} A^0 \\ A^1 \\ A^2 \\ A^3 \\ \end{array} \right) = \left( \begin{array}{cccc} \gamma_v \\ \gamma_v v_x \\ \gamma_v v_y \\ \gamma_v v_z \\ \end{array} \right)$$
$$\left( \begin{array}{cccc}A'^0 \\ A'^1 \\ A'^2 \\ A'^3 \\\end{array} \right)= \left( \begin{array}{cccc}\gamma_u \gamma_v (1-uv_z) \\ \gamma_v v_x \\ \gamma_v v_y \\ \gamma_u \gamma_v (-u+v_z) \\\end{array} \right)= \ \gamma_u \gamma_v (1-uv_z)\left( \begin{array}{cccc}1 \\ \frac{v_x}{\gamma_u (1-uv_z)} \\ \frac{v_y}{\gamma_u (1-uv_z)} \\ \frac{ -u+v_z}{{1-uv_z}} \\\end{array} \right)$$Usual expression of velocity addition rule appears.
Numerator comes from Lorentz transformation of z component.
Denominator comes from Lorentz transformation on t component.
Product $u v_z$ originates from Lorentz transformation of four velocity. One from Lorentz transformation coefficient and the other come from four velocity component.
x and y components of four velocity are invariant by Lorentz transformation, but actual speed is reduced by denomination factor $\gamma_u (1-uv_z)$. It is due to change of $\gamma$ from $\gamma_v$ to $\gamma_u (1-uv_z)\gamma_v$ by the Lorentz trarnsformation.

 

[mitochan]

(Complement)

I can set z-axis be the direction of boost without losing generality.

For general direction of boost Lorentz transformation be

$$\Lambda=\begin{pmatrix} \gamma & -\gamma v_x & -\gamma v_y & -\gamma v_z \\ -\gamma v_x & 1+\alpha v^2_x & \alpha v_x v_y & \alpha v_x v_z \\ -\gamma v_y & \alpha v_x v_y & 1+\alpha v^2_y & \alpha v_y v_z \\ -\gamma v_z & \alpha v_x v_z & \alpha v_y v_z & 1+\alpha v^2_z \\ \end{pmatrix}$$
where
$$\alpha=\frac{\gamma-1}{v^2}=\frac{\gamma^2}{\gamma+1}$$
$v, v_x, v_y, v_z$ are dimensionless velocity, divided by c.