Re: Principle of equivalence

On Apr 16, 7:43 am, Bryan Olson <fakeaddr...@xxxxxxxxxxx> wrote:
rbwinn wrote:
Bryan Olson wrote:
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.

So what are scientists objecting to?

Exactly what they said.

I said the same thing you did.

No, you did not. "If we measure the time on the cesium clock
in S when it hits the floor" is gibberish. The time "on the clock"
is what the clock reads, which is different from the time we
measure in S.

Well, you are being pretty technical, but I will give you this one.
We will do it this way. We put a clock on the floor and say, this
clock represents t'=t, or "time we measure in S", as you put it. Now,
again speaking very technically, the clock that hits the floor will
measure slightly less time than the clock on the floor because it fell
a few feet. I was just rounding off by saying that the two clocks
would read the same, since the amount of time involved would be
negligible.





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.

It does not confuse me.  Nothing changed in the experiment except the
path of the S' clock, which now falls straight down as seen by the
observer in S.  Galileo was right about what he would see.


The Lorentz equations disagree with this.
Flat-out wrong. Big red X through the rest.

Show the math, Bryan.  The S' clock hits at x=0.
            t'=(t-vx/c^2)gamma = t gamma
In order to agree with Galileo's result, t' would have to equal t.  If
you can get t' to equal t, let's see you do it.

Ah, a challenge. That clock, as you said "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." Thus in S, the velocity of
that clock is v=0.

             gamma = c / (c**2 - v**2)**(1/2)
             gamma = c / (c**2 - 0)**(1/2)
             gamma = c / (c**2)**(1/2)
             gamma = c / c
             gamma = 1

             t'=(t- v x/c^2) * gamma
             t'=(t- 0 x/c^2) * 1
             t'= t - 0
             t'= t

You challenged me to show "t' would have to equal t". Challenge met.

Well, you did exactly what I said you would do. You flipped frames of
reference and said the clock was in S. Figure it from S'.
If you can get it to work from S', I will have to say you have done
something.

[...]

Well, as I said, if you are correcting papers, just show the correct
math as you see it.

Done.

I have to give you an incomplete on this one.
Robert B. Winn
.

Relevant Pages

  • Re: time dilation
    ... train car to the floor, it falls a distance of y'. ... In the Lorentz equations, a clock in S', the frame of reference ... We are using cesium clocks as weights. ...
    (sci.physics.relativity)
  • Re: time dilation
    ... train car to the floor, it falls a distance of y'. ... In the Lorentz equations, a clock in S', the frame of reference ... than the clock in S if the picture is taken from S, ...
    (sci.physics.relativity)
  • 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 ...
    (sci.physics.relativity)
  • Re: Principle of equivalence
    ... 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 ...
    (sci.physics.relativity)
  • 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. ... in S when it hits the floor, scientists tell us that the cesium clock ... frame changes you will no longer be tempted to apply the Lorentz ...
    (sci.physics.relativity)