Re: lorentz transformations with angle
- From: george_llew_jones@xxxxxxxxx
- Date: 23 May 2005 07:01:25 -0700
albert wrote:
> Hi all. I'm having some trouble figuring out how to come up with
> Lorentz transformations in 3 dimensions with angle. The usual
> transformations:
> x' = gamma (x - vt)
> y' = y
> z' = z
> t' = gamma (t - vx/c)
> are just one spatial dimension. I'm having trouble figuring out how
to
> extend this and come up with transformations in all three spatial
> dimensions with angle.
> Can anyone offer some suggestions on how to go about figuring this
out,
> or pointers to books/references, or whatever? I'd appreciate any
help
> you can
> offer.
> Thanks in advance.
Although I can see the general idea behind what you're asking, I'm not
sure what you mean by "angle" - angle between what and what?
Any general Lorentz transformation (that doesn't involve spatial
and/or time reflections) can be written as compositions of 2 specific
types of Lorentz transformations - rotations and boosts. The Lorentz
transformation that you have given above, and that is often used to
introduce Lorentz transformations, is a boost along the x axis.
Consider 2 inertial observers that are in uniform relative motion with
respect to each other, and whose spatial origins coincide at t = t' =
0.
In the situation given in your post the primed observer (origin) moves
along the unprimed x-axis, and the the unprimed observer moves along
the
x'-axis (in the negative direction). I think you're trying to say that
this doesn't have to be the case, that the motion can have an arbitary
orientation with respect to the axes.
As above, consider 2 inertial observers that are in uniform relative
motion with respect to each other, and whose spatial origins coincide
at
t = t' = 0, but now such that the direction of relative motion of the
primed has an arbitrary orientation with respect to the unprimed axes.
Now imagine slowing the relative speed down without changing the
direction of motion. If when the speed reaches zero the unprimed and
unprimed axes are aligned, then the original transformation was a
boost.
This doesn't have to be the case, though. When the speed reaches zero
the unprimed and unprimed axes might not be aligned. In this case, the
original transformation was a combination of the rotation necessary to
align the axes, and a boost.
This is where I become unsure as to what angle your're referring. Are
you referring to: 1) the arbitary orientation of the relative motion of
a boost; or, 2) the arbitary orientation of the relative motion of a
boost *and* the aritrary orientation of the primed and unprimed axes?
I'll work through a special case of 1).
First, a change to matrix notation (which looks best in a fixed width
font). The Lorentz transformation that you gave can be expressed as
_ _ _ _ _ _
| t' | |gamma -gamma*v 0 0 | | t |
| x' | = |-gamma*v gamma 0 0 | | x |
| y' | | 0 0 1 0 | | y |
|_z'_| |_ 0 0 0 1_| |_z_|
or, r' = B r, where r', B, and r are the appropriate matrices.
Now consider a boost, i.e., unprimed and primed axes aligned, such that
the direction of motion is in the unprimed x-y plane at an angle theta
with respect to the x-axis. Things now get a little complicated, so
first an outline of what happens: i) rotate such that the motion is
along the x-axis; ii) perform the standard boost in the x direction;
iii) perform the inverse roatation to "undo" the original rotation.
Now the details. Consider a double-primed frame that is at rest with
the
unprimed frame, and that has axes aligned such that the motion of the
primed observer is along the x''-axis. This is accomplished by the
transformation r'' = R r, where R is the rotion matrix
_ _
| 1 0 0 0 |
| 0 cos(theta) sin(theta) 0 |
R = | 0 -sin(theta) cos(theta) 0 |
|_0 0 0 1_|.
Now move to a triple-primed frame by using B (given above) to boost
along the x''-axis: r''' = B r''. The triple-primed frame is stationary
with respect the primed frame, but, because of the rotation R, the
axes of the primed and triple-primed frame are not aligned. Align the
axes by undoing the rotation:
r' = R^(-1) r'''
= R^(-1) B r''
= R^(-1) B R r,
where
_ _
| 1 0 0 0 |
| 0 cos(theta) -sin(theta) 0 |
R^(-1) = | 0 sin(theta) cos(theta) 0 |
|_0 0 0 1_|,
since cos(-theta) = cos(theta) and sin(-theta) = sin(theta).
Bottom line: the primed and unprimed frames are related by the Lorentz
transformation L with L = R^(-1) B R. I leave the matrix
multtiplication
as an exercise :-).
Regards,
George
.
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- From: albert
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