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Re: Variable speed of light (was: Relativity conundrum)



Pentcho Valev wrote:

Bob LaMontagne wrote:

Pentcho Valev wrote:

Michael Burns-Kaurin wrote:

I find that the best solution to relativity conundrums is to apply the
Lorentz transformations.

Let the light enter the train at the origin in both reference frames. The
train moves in the +x direction according to the track frame. The light
moves across the track in the +y direction in the track frame. The width
of the train is w, same in both frames.

In track frame, time of light leaving train is simply w/c. Location of
that event is x=0, y=w, t=w/c.

Now plug those numbers into the Lorentz transformations to find the
spacetime locations for the leaving-train event in the train frame.

x' = gamma * (0 - v*t) = - gamma * v * w / c
y' = w
t' = gamma *(t - (v/c^2)x) = gamma * w/c

You obtain t' > t so your argument could be regarded as reductio ad absurdum.
If clocks on the train run slower, as textbooks say, then t' < t.

Pentcho

t' > t is actually correct in this case - if you correctly interpret t' and t.
They measure the time between the same to events, light entering the train and
light leaving the train. That interval has to be greater for the person in the
train because the light travels a longer distance.

Still this is reductio ad absurdum since, by using the same method, we can obtain
contradictory results. In another experiment, the events are again "light enters
train" and "light leaves train". However in the rest frame the beam approaches the
train obliquely so that, in the train frame, it runs perpendicular to the train. In
this case Lorentz equations give

x' = 0 ; x = vt ; t' = (1 / gamma)*t

Clearly, t' < t whereas in the original experiment the analogous time intervals were
found to obey t' > t. These are absurd results in my view. Lorentz equations predict
that, in the train frame, some processes are slower than in the rest frame whereas
others are faster. To call this "time dilation" is both wrong and misleading.
"Absurd time deformation" seems to be the most suitable name.

Pentcho

Not at all. You are comparing apples and oranges. Let's be specific:

1. For events located at the same point in the train frame (ticks of a
train clock, for instance), the time interval delta t' (as measured by
the train observer) is less than the time interval delta t (the temporal
separation of the same events measured in the station frame). The
quantity known as the spacetime interval is the same in both frames.
Here

c^2 delta-t^2 - delta-x^2 = c^2 delta-t'^2 - delta-x'^2
= c^2 delta-t'^2

The square root of this quantity is the spacetime interval that is
"timelike". If the quantity above were negative we would multiply
through by -1 before taking the square root, resulting in a interval
that is "spacelike".

Notice that delta-t' is the minimum temporal interval, since
delta-x'=0. This partially corresponds to looking at a vector with x-
and y-components in the plane. A rotation of coordinates can let you
have different components x' and y', but the length of the vector (x^2 +
y^2)^(1/2) remains constant. One choice of coordinate systems gives the
maximum y' when x' is zero, another gives the maximum x when y is zero.
This is only an analogy, since in Minkowski spacetime there's a minus
sign in finding the spacetime interval instead of a plus in finding
spatial distances. But any spacelike interval can be minimized (in the
frame where the events are simultaneous, dt=0) and any timelike interval
can be minimized (in the frame where the two events are coincident --
occur at the same position, dx=0).

2. Similarly, for events located at the same point in the station frame
(ticks of a station clock, for instance), the time interval delta t (as
measured by the station observer) is less than the time interval delta
t' (the temporal separation of the same events measured in the train
frame). The quantity spacetime interval is the same in both frames:

c^2 delta-t'^2 - delta-x'^2 = c^2 delta-t^2 - delta-x^2 = c^2 delta-t^2

3. For events that are not located at the same point in either frame the
temporal components do not vanish in either frame.

4. In your example you have both x- and y- separations, so the formula
above needs to be generalized:

c^2 delta-t^2 - (delta-x^2 + delta-y^2)
= c^2 delta-t'^2 - (delta-x'^2 + delta-y'^2)

Letting the events be the entrance and exit of the light beam, we will
have delta-x = 0, and delta-y = delta-y' =/= 0. This simplifies down to

c^2 delta-t^2 = c^2 delta-t'^2 - delta-x'^2

This is the same as in case 2 above, so it should not be surprising that
we get delta-t' > delta-t, _not_ vice versa.

I quote the last part of your reaction again in order to interleave my
comments:

These are absurd results in my view.

They only appear contradictory when applied incorrectly.

Lorentz equations predict
that, in the train frame, some processes are slower than in the rest frame whereas
others are faster.

Is the "process" occurring at a fixed point in some reference frame? If
so, the time intervals in the process will appear _smallest_, i.e., time
flowing slowest in that reference frame. Not all processes will appear
slower in the train frame, simply because not every process (series of
events in spacetime) occurs at the same (stationary) spatial point in
the reference frame of the train! No contradiction.

Incidentally, the invariance of the spacetime interval is the concept
which generalizes the equation you brought up in a previous message (c =
x/t = x'/t' = c'), expressing the spatial distance travelled by the
light beam as (delta-x^2 + delta-y^2)^(1/2), setting it equal to c
delta-t, then squaring both sides to get rid of the square roots:

(delta-x^2 + delta-y^2)^(1/2) = c delta-t

==> (delta-x^2 + delta-y^2) = c^2 delta-t^2

==> c^2 delta-t^2 - (delta-x^2 + delta-y^2) = 0

Similarly, in the primed frame,

c^2 delta-t'^2 - (delta-x'^2 + delta-y'^2) = 0

So clearly the expression on the left-hand-side are is invariant, the
same in any inertial reference frame. (Turns out this is true for any
pair of events, not just events on the world-line of a light beam.)
This only invokes Einstein's 2nd principle (that light travels at the
same rate as measured in any inertial frame), so I trust that you will
like it better than derivations that rely on the Lorentz
transformations! ;-)

To call this "time dilation" is both wrong and misleading.
"Absurd time deformation" seems to be the most suitable name.

I hope that the above remarks help to show the consistency and logic of
the relativistic viewpoint.

Ken