Solve Velocity 4-Vector Homework

[MikeLizzi]

Homework Statement

Given
Space Time Coordinate of object <t, x, y, z> = <0, 0, 0, 0>
Velocity of object as Vector <betaX, betaY, betaZ> = <.866, 0 ,0>
Velocity of target reference frame as Vector <betaX, betaY, betaZ> = <-.866, 0 ,0>

Transform velocity of object to the target reference frame using the 4-vector technique.

Homework Equations

The Attempt at a Solution

[/B]Following directions from a source on the internet...

Velocity as a 4-vector (tau, vx, vy, vz)
= <1, betaX, betaY, betaZ>
= <1, 0.866, 0.0, 0.0>

magnitude of 4-vector
= Sqrt( 1 + betaX * betaX + betaY * betaY + betaZ * betaZ )
= 1.3228590249909473

Velocity as a 4-vector normalized (tau, x, y, z)
= <1/magnitude, betaX/magnitude, betaY/magnitude, betaZ/magnitude>
= <0.7559384493044097, 0.6546426970976188, 0.0, 0.0>

Transform the 4-vector where L is the Lorentz Transform populated for the target velocity:
v' = Lv

resulting in v'(tau, x, y, z)
= <2.6454852575216505, 2.6183403845739543, 0.0, 0.0>

? I should be getting a vx value somewhere around .98c. Totally lost.

 

[PeroK]

If you think you're totally lost, I'm lost trying to figure out what you're doing. First, I suggest stating what the 4-vector technique is. Is that the Lorentz transformation of 4-vectors? Perhaps you could state the relevant transformation equations?

 

[PeroK]

To give a start and also help you with some latex. If a particle has a velocity $u$ in the x-direction with gamma factor $\gamma_u$, then its four-velocity is:

$(u_{\mu}) = (\gamma_u, \gamma_u u, 0,0)$

 

[MikeLizzi]

To give a start and also help you with some latex. If a particle has a velocity $u$ in the x-direction with gamma factor $\gamma_u$, then its four-velocity is:

$(u_{\mu}) = (\gamma_u, \gamma_u u, 0,0)$

That helps a lot. I am going to spend some time working the problem (and my latex).

 

[MikeLizzi]

That helps a lot. I am going to spend some time working the problem (and my latex).

I think I got it!
It took a couple days to learn latex but PeroK's hint and seeing the formulas clearly made me realize what I was doing wrong.

Here's the latex by the way (still working on format)

The Lorentz Transformation is defined as

$$\lambda=
\begin{bmatrix}

\gamma & -\gamma\beta_x & -\gamma\beta_y & -\gamma\beta_z \\
-\gamma\beta_x & 1+\frac{(\gamma -1)\beta_x^2}{\beta^2} & \frac{(\gamma -1)\beta_x\beta_y}{\beta^2} & \frac{(\gamma -1)\beta_x \beta_z}{\beta^2} \\

-\gamma\beta_y & \frac{(\gamma -1)\beta_x\beta_y}{\beta^2} & 1+\frac{(\gamma -1)\beta_y^2}{\beta^2} & \frac{(\gamma -1)\beta_y\beta_z}{\beta^2} \\

-\gamma\beta_z & \frac{(\gamma -1)\beta_x\beta_z}{\beta^2} & \frac{(\gamma -1)\beta_y\beta_z}{\beta^2} & 1+\frac{(\gamma -1)\beta_z^2}{\beta^2}

\end{bmatrix}
$$

Where v is the velocity of S with respect to S' and
$$\beta_x = v_x/c$$
$$\beta_y = v_y/c$$
$$\beta_y = v_y/c$$
$$\beta^2 = \beta_x^2 + \beta_y^2 + \beta_z^2$$
$$\gamma = \frac{1}{\sqrt{1-\beta^2}}$$

The velocity 4-vector in reference frame S' is defined as
$$u'_\mu=
\begin{bmatrix}
\gamma_{u'} \\
\gamma_{u'} u'_x \\
\gamma_{u'} u'_y \\
\gamma_{u'} u'_z
\end{bmatrix}
$$
Where u' is the 3-vector velocity of an object with respect to the S' reference frame and gamma is the gamma value using u'.

The velocity 4-vector in reference frame S is defined as
$$u_\mu=
\begin{bmatrix}
\gamma_{u} \\
\gamma_{u} u_x \\
\gamma_{u} u_y \\
\gamma_{u} u_z
\end{bmatrix}
$$
Where u is the 3-vector velocity of an object with respect to the S reference frame and gamma is the gamma value using u.

Thanks PeroK !