Re: Question about light clock and derivation of time dilation

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?

>> 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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