This post started out as a simple comment on the last sentence of thread post #73 by Doc Al. The actual result is more of a digression on some of the formal aspects of Special Relativity that, in the context of this discussion, either have only been sketched or have been described piecemeal.
Doc Al said:
\gamma L is not the distance L as measured by the train! It is the position coordinate of an event.
I agree with what you say here, but as stated (and presuming it
is meant as a warning about my previous post [#72]), it may obscure what is correct and what might be misleading about my post.
1) (To clarify for readers unfamiliar with the terminology)
Event is the term used for a given point in the mathematical space used to describe relativity. Specifying an event requires both space coordinate(s) and a time coordinate. The definition covering the current situation, as given by jcsd, is:
"What I've done is basically create a co-ordinate susyetm the co-ordinates for each event are (x,t) where x is the distance from the midpoint of the space station (i.e. x = 0 at M) and t is the difference in time from when the spaceship and M occupy the same spot (i.e. t = 0 as the spaceship arrives at M)."
Each reference frame describes the same set of events. However, different reference frames overlay different systems of coordinates onto these. The differing measurements determined by these different coordinate systems reflect the space and time distortions that occur when measurements are made in different reference frames.
2) The
Lorentz transformation is the mathematical operation that maps the coordinates of events in one reference frame to the coordinates in a second frame.
The ordinary Lorentz transformation maps event coordinates in the rest frame to event coordinates in some other inertial frame.
If an event has coordinates
(x, t) in the rest frame, then the general transform \Lambda to coordinates
(x', t') in a frame with relative velocity
u is:
(x', t') = \Lambda (x, t) = \left( \gamma (x - ut), \gamma(t - \frac{ux}{c^2}) \right)
If, however, what is known is the coordinates
(x', t') in the moving frame, then to find the coordinates
(x, t) in the rest frame what is required is the
inverse Lorentz transformation \Lambda^{-1}. This has the general form:
(x, t) = \Lambda^{-1} (x', t') = \left( \gamma (x' + ut'), \gamma(t' + \frac{ux'}{c^2}) \right)
3) Ok, I screwed up. I just realized that jcsd was talking about the spaceship scenario not the train scenario, so the way my previous post (#72) compares what I say to what jcsd said makes no sense, and I should probably edit it. (Without the comparison, the logic still holds - though with the caveats given below.)
4) Just to be clear: I'm discussing the train scenario.
In the embankment frame, coordinates are denoted:
(x_m, t_m)
where
xm is a spatial coordinate along an axis defined by the track, and
tm is a time coordinate.
For any time
tm in the embankment frame, the observer
M has coordinates:
M(t_m) = (x_M, t_M) = (0, t_m)
We also have two events - flash at
A, and flash at
B - that happen at
tm = 0. Coordinates for these are:
\textrm{flash}_A = (x_A, t_A) = (-L, 0)
\textrm{flash}_B = (x_B, t_B) = (L, 0)
(I've switched the signs so that the train moves toward positive
xm.)
The flashes occur at the same moment that the rear and front of the train pass (respectively)
A and
B. In other words, in the embankment frame the length of the train equals the distance from
A to
B.
In the train frame, coordinates are denoted:
(x_{m'}, t_{m'})
The axis along which the spatial coordinate
xm' is measured is defined by the track and thus coincides with spatial axis used in the embankment frame (it's coordinate system, however, is, of course, separate).
For any time
tm' in the train frame, the observer
M' has coordinates:
M'(t_{m'}) = (x_{M'}, t_{M'}) = (0, t_{m'})
The coordinates of
flashA and
flashB transformed into the train frame (as given by jcsd, but with signs reversed) are:
\textrm{flash}_{A'} = (x_{A'}, t_{A'}) = (- \gamma L, \frac{\gamma\beta L}{c})
\textrm{flash}_{B'} = (x_{B'}, t_{B'}) = (\gamma L, \frac{-\gamma\beta L}{c})
This is an example of the inverse Lorentz transformation as defined in item 2) above. (We consider the train to be at rest so we are going from a moving frame to a rest frame.) Calculating using the formula from item 2) we get:
This has been corrected. See post #99 below for the reasoning behind the use of -u.
<br />
\begin{equation*} \begin{split}<br />
(x_{A'}, t_{A'}) &=<br />
\Lambda^{-1}(x_A, t_A) =<br />
\Lambda^{-1}(-L, 0) \\<br />
&= \left( \gamma (-L + 0(-u)), \gamma (0 + \frac{(-u)(-L)}{c^2}) \right) \\<br />
&= ( -\gamma L, \frac{\gamma\beta L}{c} )<br />
\end{split} \end{equation*}<br />
The last point necessary for the set up is that both the observer M and the observer M' set the origin of their respective time axes to the moment that M' passes M.
5) So far quantities such as L and \gamma L are, as Doc Al said, "position coordinate of an event".
To get the spatial distance in one dimension we can use the formula:
\Delta x = |x_1 - x_0|
Thus the distances from the observer M to the points A and B are (in the embankment frame):
\Delta x_A = |x_A - x_M| = |-L - 0| =\ L
\Delta x_B = |x_B - x_M| = |L - 0| =\ L
and thus as I said in the previous post, "each half of the train measured in the embankment frame is also length L (so total length 2L)"
And the distances from the observer M' to the points A' and B' are (in the train frame):
\Delta x_{A'} = |x_{A'} - x_{M'}| = |-\gamma L - 0| = \gamma L
\Delta x_{B'} = |x_{B'} - x_{M'}| = |\gamma L - 0| = \gamma L
and so, again as in the previous post, "the rest length of each half of the train, i.e. the length measured in the train reference frame is indeed \gamma L"
I'll leave off for now, the third assertion of my previous post will have to wait for next time.
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