Solve velocity addition problems using 4-vector

  • Thread starter Natchanon
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  • #1
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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 = Λμνην so we can use it for velocity addition?
 

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  • #2
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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.
[tex]\gamma=\gamma_1\gamma_2[/tex]
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.
 
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  • #3
PeroK
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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.
[tex]\gamma=\gamma_1\gamma_2[/tex]
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##
 
  • #4
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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
[tex]U^\mu=(\gamma_u, \gamma_uu_x/c, \gamma_uu_y/c, \gamma_uu_z/c)[/tex]
[tex]V^\mu=(\gamma_v, \gamma_vv_x/c, \gamma_vv_y/c, \gamma_vv_z/c)[/tex]
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
[tex]\gamma_2=\gamma(\gamma_1-\frac{vu_x}{c^2})[/tex]
It does not coincide with your ##\gamma_2## formula. wrong where ?
 
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  • #5
PeroK
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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
[tex]U^\mu=(\gamma_u, \gamma_uu_x/c, \gamma_uu_y/c, \gamma_uu_z/c)[/tex]
[tex]V^\mu=(\gamma_v, \gamma_vv_x/c, \gamma_vv_y/c, \gamma_vv_z/c)[/tex]
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
[tex]\gamma_2=\gamma(\gamma_1-\frac{vu_x}{c^2})[/tex]
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.
 
  • #6
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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
[tex]\Lambda(-v_1)^\mu_\nu v_2^\nu[/tex]
In ##S_2## Origin of ##S_1## is moving with four velocity
[tex]\Lambda(-v_2)^\mu_\nu v_1^\nu[/tex]
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.
 
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  • #7
PeroK
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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
[tex]\Lambda(-v_1)^\mu_\nu v_2^\nu[/tex]
In ##S_2## Origin of ##S_1## is moving with four velocity
[tex]\Lambda(-v_2)^\mu_\nu v_1^\nu[/tex]
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.
 
  • #8
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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

[tex]\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)

[/tex]
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

[tex]

\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) [/tex]



[tex]\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)[/tex]


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.
 
  • #9
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126
(Complement)

I can set z-axis be the direction of boost without losing generality.
For general direction of boost Lorentz transformation be

[tex]\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}
[/tex]
where
[tex]\alpha=\frac{\gamma-1}{v^2}=\frac{\gamma^2}{\gamma+1}[/tex]
##v, v_x, v_y, v_z## are dimensionless velocity, divided by c.
 
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