Generalized Galilean transformation

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Pushoam
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Homework Statement


Write the Galilean coordinate transformation equations
for the case of an arbitrary direction for the relative velocity v of one frame with respect to the other. Assume that the corresponding axes of the two frames remain parallel. (Hint: let v have componentsvx, vy, vz.)

Write down the equivalent matrix equation.

Homework Equations


Consider a frame S' moving with uniform velocity v with respect to another inertial frame S.
Then
x'=x- vx t, $$
$$y'=y- vy t, $$
$$z'=z- vz t, $$
$$t' = t

The Attempt at a Solution


The matrix formulation is
$$\begin{pmatrix}
x' \\
y' \\
z'\\
t'
\end{pmatrix} =\begin{pmatrix}
x& -v_x \\
y & -v_y \\
z & -v_z\\
0&1\end{pmatrix}
\begin{pmatrix}
1 \\
t\\

\end{pmatrix}
$$
Is this right?
 
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Pushoam said:
Is this right?
I don't think so. Your relevant equations are fine. For a coordinate transformation one expects something of the form $$
\begin{pmatrix}
x' \\
y' \\
z' \\
t'
\end{pmatrix} = M\begin{pmatrix}
x \\
y \\
z \\
t
\end{pmatrix}$$
(compare with the matrix for a simple rotation in 3D, e.g. around the z-axis)
 
Then,
##\begin{pmatrix}
x' \\
y' \\
z'\\
t'
\end{pmatrix} =\begin{pmatrix}
1&0&0& -v_x \\
0&1&0& -v_y \\
0&0&1 & -v_z\\
0&0&0&1\end{pmatrix}\begin{pmatrix}
x \\
y\\
z\\
t

\end{pmatrix}##
Is this correct?
Can you please tell me how to write the matrices side by side?
 
Last edited:
I think this is what the exercise composer meant, yes.
And the inverse transformation matrix looks the same, except that the minus signs are now plus signs. Good exercise to check that ##M^{-1}M = MM^{-1} = {\mathbb I}##
Pushoam said:
Can you please tell me how to write the matrices side by side?
But you did that already in post #1 !

Generally: right-click on a formula and pick show math as ##\TeX## commands :smile:
 
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Thank you.
Pushoam said:
Can you please tell me how to write the matrices side by side?
I got it. I need to write" ## "before and after the text command for matrices.