Re: Principle of equivalence

rbwinn 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.
Now the train car comes by at velocity v and at t'=t=0 both
cesium clocks are dropped. If we measure the time on the cesium clock
in S when it hits the floor, scientists tell us that the cesium clock
in S' will still be in the air at the time the S clock hits and will
register a time of
t'=(t-vx/c^2)gamma at that moment.

Scientists keep objecting to your imprecise statement there. In S,
the clock at rest hits the floor-level apparatus before the clock
on the train hits the floor.

Scientists tell us that the
experiment comes out the same in S and S' because when the S' clock
does hit, it registers the same time that the S clock registered when
it hit. Then if you looked at the experiment from the frame of
reference of the train car, the S' clock would hit first, and the S
clock would still be in the air and would register less time than the
S' clock.

Fair enough.

Now we look at the experiment from the perspective of Galileo,
who discovered the principle of equivalence. We will make one change
in the experiment.

You'd be better off going one step at a time. Changing both to
Galileo's understanding of physics and changing the experiment has
confused you.

When the S' clock is released at the roof of the
train it will be sent opposite the direction of the motion of the
train with a speed relative to S' that is equal to the speed of the
train. The clock will therefore drop straight down relative to S,
falling alongside the clock in S. According to Galileo, both clocks
will hit the floor at the same time, and both will read t'=t.

O.K.

The Lorentz equations disagree with this.

Flat-out wrong. Big red X through the rest.

There are two events for
the S' clock. It is released at t'=t=0, and it hits at x=0 at a time
of t'2. According to the Lorentz equations, the clock hits at x=0 at
a time of
t'2=(t-v(0)/c^2)gamma =t gamma. The Lorentz equations show the S'
clock still in the air when the S clock hits even though it is falling
right beside it, and Galileo sees the two clocks hit the floor at the
same time.

The Lorentz transform applies to uniform motion. You accelerated the
clock, then applied the Lorentz transforms as if the clock were still
at rest in S'.

Looks like one of the issues that confused you was calling it "the
S' clock". Give it a name, such as "clock b", so that when its rest
frame changes you will no longer be tempted to apply the Lorentz
transform as if it were still at constant x in its former rest frame.

Scientists say, If this is a problem, it is just a tiny
problem.

Well, I expect you'd lose more that a couple points on this homework.

Don't feel too bad about that. We all get stuff wrong. What makes
kooks kooks is that upon correction they cling to their
misapprehensions all the more fervently.

Hope that helps. But I know which way to bet.


--
--Bryan

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Relevant Pages

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  • Re: Principle of equivalence
    ... 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. ... ?If we measure the time on the cesium clock ... in S when it hits the floor, scientists tell us that the cesium clock ...
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    ... clock from a height equal to the height of the train car roof. ... in S when it hits the floor, scientists tell us that the cesium clock ... reference of the train car, the S' clock would hit first, and the S ... frame changes you will no longer be tempted to apply the Lorentz ...
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