Re: Principle of equivalence
- From: rbwinn <rbwinn3@xxxxxxxx>
- Date: Thu, 17 Apr 2008 18:20:59 -0700 (PDT)
On Apr 17, 6:56�am, PD <TheDraperFam...@xxxxxxxxx> wrote:
On Apr 15, 10:55�pm, rbwinn <rbwi...@xxxxxxxx> wrote:
� �The basic problem with current interpretation of relativity can be
seen from considering two frames of reference S and S' and how they
relate to the principle of equivalence.
� �S is a set of coordinates at rest and S' is a set of coordinates in
motion in the x direction relative to S with a velocity of v. �Let S'
represent a train car in which we will drop a cesium clock at t'=t=0
from the roof. �In S we construct an apparatus on a floor level with
the train car floor at x=0 from which we will drop an identical cesium
clock from a height equal to the height of the train car roof.
This is different than what physicists mean by the principle of
relativity (which you call the principle of equivalence, which in turn
means something completely different). The principle of relativity
considers a SINGLE sequence of events and asks about the physical
description of that single sequence as seen from two reference frames,
rather than considering two similar sequences of events in two
different reference frames.
For example, let's take a SINGLE rock dropped from a SINGLE elevated
cup into a SINGLE bucket on the floor, and let's just do that drop
ONCE, but we'll look at that same sequence of events (from the event
marking the drop of the rock from the cup to the event of the rock
landing in the bucket) from reference frames S and S'.
We can arrange the instruments in S and S' such that x=0 and t=0 as
*measured* in S at the event of the release, and such that x'=0 and
t'=0 as *measured* in S' at the same event. The question then arises
whether x and x' will be the same when the ONE rock lands in the
bucket, and whether t and t' will be the same at that same event.
Galilean relativity says that t *measured* in the second event will be
the same as t' *measured*. Relativity says they will not. Then it is a
simple matter of making the measurement and seeing what is right. It
turns out that the *measured* t and the *measured* t' are not the
same, despite the precision of the clocks being more than adequate.
Now, if you'd like to suppose that there is a *mythical* but
*unmeasured* t' that is equal to t and that the *measured* time in S'
is something other than this mythical t', I suppose that's your
perogative. But Galilean relativity makes a statement about *measured*
t and t', as does special relativity. And the measured numbers bear
out relativity.
PD
Well, so what is the problem, PD? You drop a rock into a bucket of
water and "measure" the events according to what is seen by an
observer in S and an observer in S'. Event one is when the rock is
released and t'=t=0. Event two is when the rock hits the water in the
bucket. Both events are timed by a clock on the floor in S which
shows t'=t. The observer in S sees the rock hit the water when t=t.
The observer in S' sees the rock hit the water when t'=t.
The clock on the floor in S reads the same as seen by either observer
when the rock hits the water. What is the problem you see?
Robert B. Winn
.
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