Re: relativity of time


francisco wrote:
> concerning the following site:
>
> http://users.powernet.co.uk/bearsoft/LtClk.html
>
> Typical Anti-Relativty garbage, not worth your time.

Then you should have no problem addressing the issues it
raises with logical argument instead of disparaging remarks.

Your opinion of the website does explain how a relative
moving observer can alter the path between the mirrors
to permit the appliction of special propagation calculations.

Rain doesn't follow a longer path, nor take a longer period
of time to reach the ground, simply because you view it
from a moving vehicle.

Sue...

> __________________
> Janus
>
> The whole problem with the world is that fools and fanatics are always so
> certain of themselves, and wiser people so full of doubts.--Bertrand Russell
>
> "Sue..." <suzysewnshow@xxxxxxxxxxxx> wrote in message
> news:1125448209.388401.54560@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> >
> > francisco wrote:
> >> let observer S be at rest on the ground. let S carry a timing device,
> >> consisting of FD (a flashing lightbulb F attached to a detector D). let a
> >> mirror M be a distance L0 from FD, so that M - FD = L0. let M and FD be
> >> two
> >> points of a vertical line that represents L0, so that M is north of FD.
> >> let
> >> the bulb emit a ray of light that travels north to M from F. when the
> >> reflected light returns to D, the clock ticks and another flash is
> >> triggered. let delta t0 be the time interval between ticks according to
> >> S,
> >> so that delta t0 = 2L0/c.
> >>
> >> let observer S' be in a train moving horizontally east on a long,
> >> straight
> >> track. the train moves at constant speed v relative to S. let S' carry an
> >> identical timing device, consisting of FD and M, so that M - FD = L0. let
> >> M
> >> and FD be two points of a vertical line (a line that is perpendicular to
> >> the
> >> length of the track) that represents L0, so that M is north of FD. in
> >> other
> >> words, L0 is represented by a line (inside the train) that extends across
> >> the train . let the bulb emit a ray of light that travels north to M from
> >> F.
> >> and when the reflected light returns to D, the clock ticks and another
> >> flash
> >> is triggered. again, let delta t0 be the time interval between ticks
> >> according to S', so that delta t0 = 2L0/c.
> >>
> >> the time interval delta t0 is observed by either S or S' when the clock
> >> is
> >> at rest with respect to that observer.
> >>
> >> let's consider now the situation when S looks at the clock carried by S'
> >> on
> >> the moving train. .
> >>
> >> imagine that ABC is an isosceles triangle. let AC be the horizontal base
> >> of
> >> ABC, so that A and C are the base angles of ABC. let C be east of A . let
> >> B
> >> be the angle opposite AC. and let B be north of AC. let P be the midpoint
> >> of
> >> AC, so that BP is the perpendicular bisector of AC, and PC = AC/2.
> >>
> >> let the train move east, parallel to AC
> >>
> >> according to S, the following takes place:
> >>
> >> the bulb in the train emits a ray of light at A. a light signal (at that
> >> instant) alerts S that the bulb has emitted a ray of light at A.
> >>
> >> the ray of light is reflected at B (from the mirror M inside the train).
> >> a
> >> light signal (at that instant) alerts S that the ray of light is
> >> reflected
> >> at B.
> >>
> >> the ray of light is detected at C (at the detector D inside the train).
> >> and
> >> a light signal (at that instant) alerts S that the ray of light is
> >> detected
> >> at D.
> >>
> >> according to S, delta t is the time interval between the light signal at
> >> A
> >> and the light signal at C, and
> >>
> >> AC = v*(delta t).
> >>
> >> also, according to S, L = SQRT[(BP)^2 + (PC)^2], where BP = L0 and PC =
> >> AC/2
> >> = v*(delta t)/2
> >>
> >> so, L = SQRT{(L0)^2 + [v* (delta t)/2]^2}.
> >>
> >> S observes that the light beam travels a distance of 2L, so that delta t
> >> =
> >> 2L/c
> >>
> >> so, delta t = 2*SQRT{(L0)^2 + [v* (delta t)/2]^2}/c
> >>
> >> substituting for L0 from delta t0 = 2L0/c and solving delta t =
> >> 2*SQRT{(L0)^2 + [v* (delta t)/2]^2}/c for delta t gives the following:
> >>
> >> delta t = (delta t0)/SQRT(1 - v^2/c^2)
> >>
> >> the factor in the denominator of this equation is always less than or
> >> equal
> >> to 1, thus delta t is greater or equal to delta t0. that is, S measures a
> >> greater interval between ticks. the effect is called time dilation. the
> >> time
> >> interval delta t0, measured by S', who is at rest relative to the clock
> >> inside the train is called the proper time.
> >
> > << The result depends on the angle of the light pulse to the direction
> > of the two observers relative motion. The choice of the light clock and
> > its orientation arbitrarily selects a particular direction which gives
> > the correct result. We do not prove that the square of the longest side
> > of any triangle is equal to the sum of the squares of the other two
> > sides by drawing a right angled triangle, proving that it works in that
> > case and then generalising. That is mathematically unsound and it is
> > mathematically unsound to pick a particular direction for the light
> > pulse to travel in when the result depends on the direction. In short,
> > it is a mathematical fiddle. >>
> > http://users.powernet.co.uk/bearsoft/LtClk.html
> >
> > Sue...
> >

.

Relevant Pages

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