Re: Question about light clock and derivation of time dilation


"Daryl McCullough" <stevendaryl3016@xxxxxxxxx> wrote in message news:dame5u0dhg@xxxxxxxxxxxxxxxxxx
> sue jahn says...
>
> >> sue jahn says...
> >>
> >> >Then if you view rainfall from a moving vehicle and the
> >> >path appears diagonal, would you use the diagonal distance
> >> >to compute the speed the rain is falling?
> >>
> >> If I wanted to know the velocity of the rain relative
> >> to my vehicle, then yes. If I wanted to know the velocity
> >> relative to the ground, then no.
> >I didn't mention wind.
>
> Neither did I.
>
> >> >Light moves perpendicular between parallel mirrors unless
> >> >there is a moving dielectric between the mirrors.
> >>
> >> That's not correct. The angle of reflection is equal to
> >> the angle of incidence. If a light beam hits a mirror
> >> at an angle of 80 degrees, then it will bounce off at
> >> 80 degrees.
> >
> >So how many times do you have to add 180 to 90 to
> >get to a number that is not divisable by 90?
>
> What are you talking about?
<<if light hits a mirror at 90 degrees,
it will bounce off at 90 degrees. >>
Your words

Sue...
>
> >> In the case of a light clock, you have, in the frame
> >> of the mirrors, the angle is 90 degrees. In the frame
> >
> >Your gaze may jump across an absract frame of
> >reference. The light most certainly will not.
>
> What are you talking about? A frame of reference
> is used for *describing* the path of the light
> beam. The light itself doesn't care anything about
> frames of reference. The same path looks perpendicular
> in one frame of reference and diagonal in the other
> frame of reference. But in *both* frames of reference,
> light obeys the same laws, which specify that it travels
> at c in all directions.
>
> >> in which the mirrors are moving, the angle is given
> >> by cos(theta) = v/c.
> >>
> >> >Your eye's motion, relative to the mirrors does not
> >> >create such a mechanism does it?
> >>
> >> Eyes have nothing to do with it.
> >Exactly.
>
> Then why did you bring them up? Nothing I said
> had anything to do with eyes.
>
> >In a frame in which
> >> the mirrors are moving at speed v, then the location
> >> of one mirror is given as a function of t by
> >>
> >> x = vt
> >> y = 0
> >>
> >> The location of the other mirror is given by
> >>
> >> x = vt
> >> y = L
> >>
> >> The path of the light pulse going from the first
> >> to the second is given by
> >
> >No by the principle of relativity.
>
> The principle of relativity just says that the laws
> of physics are the same in any inertial frame. Those
> laws say that if light hits a mirror at 90 degrees,
> it will bounce off at 90 degrees. If it hits at 80
> degrees, it will bounce off at 80 degrees. The
> principle of relativity does *not* say that if
> the light hits at angle 90 degrees in one frame,
> then it hits at angle 90 degrees in every frame.
>
> >The departure angle is still 90 degrees.
>
> No, it's not. Where did you get the idea that light
> must depart from a mirror at 90 degrees? It can depart
> at any angle whatsoever. The angle is frame-dependent
> in Special Relativity (just as it is in Galilean relativity).
>
> Rather than getting into the technical details of how
> one makes a directional beam of light, let's make things
> simple, and just assume that the light source is a bulb,
> which produces a spherically symmetric wave front. So
> there are tiny little pulses of light going off in every
> direction. Each and every little pulse is travelling in
> a straight line at speed c. Now, if this spherically
> symmetric collection of light pulses is created at one
> mirror, then the only pulse that will make it to the
> other mirror are those that are travelling in exactly
> the right direction. Specifically, the only pulse that
> will make it to the other mirror is one travelling at
> an angle given by
>
> cos(theta) = v/c
>
> All other pulses travelling in other directions will be
> lost. But this particular pulse will bounce off at exactly
> the same angle theta, and will travel in exactly the right
> direction to reach the original mirror. From then on, this
> pulse will go from one mirror to the other, back and forth,
> indefinitely. The time for going from one mirror to the
> other will be (as I've already said)
>
> T = L/c 1/square-root(1-(v/c)^2).
>
> --
> Daryl McCullough
> Ithaca, NY
>


.

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