Alfonso says...

On 23/03/11 13:50, PD wrote:

Time is not the reciprocal of frequency. Time is benchmarked by a
locally stationary reproducible process. See the NIST standards.

You are ducking the issue:

It's not a matter of ducking the issue, it's a matter of
it being very difficult to teach a course in physics on
Usenet, because you have no idea what kind of backgrounds
the students have.

a/The frequency of a transverse moving clock is reduced.
b/The time interval between ticks is increased (dilated means increased)

That's an incorrect statement, according to SR. What
is correct is this: Let F and F' be two inertial frames
such that a clock is at rest in frame F' and is traveling
at speed v as measured in frame F. Let e_1 and e_2 be events
taking place at this clock. (For example, e_1 might be a
tick of the clock, and e_2 might be the next tick).

Then the prediction of SR is:

T_12 = gamma T'_12

where T_12 is the time between events e_1 and e_2 as
measured in frame F, and T'_12 is the time between
those events as measured in frame F'.

This is different from your B because it is explicitly
frame-dependent. If the clock were measured from a
different frame, the time between ticks would be different.

c/ What the moving clock registers is reduced.

I have no idea what you mean by that.

What are the units of a/b/ and c/
a is 1/s or Hz
b is s the reciprocal of Hz
c is the number of ticks (unitless)

Which statement do you disagree with?

None of this has much directly to do with Doppler shifts.
A Doppler shift occurs in the following situation:

As before, let e_1 and e_2 be two events taking place
at a clock that is at rest in frame F'. Let there be
a second clock be at rest in frame F, some distance
from the first clock. Let e_3 be the event at which
a light signal produced at e_1 reaches the second
clock. Let e_4 be the event at which a light signal
produced at e_2 reaches the second clock.

If we let T_34 be the time between e_3 and e_4,
as measured in frame F, and T'_12 be the time
between e_1 and e_2, as measured in frame F',
then the Doppler shift is the ratio

T'_12/T_34

The prediction of SR, in the case in which the F'
clock is moving away from the F clock at speed v,
is

T'_12/T_34 = square-root((1-v/c)/(1+v/c))

(The ratio of the frequencies is the reciprocal
of this)

The prediction of SR, in the case in which the
F' clock is moving transversely (perpendicular
to the line between the two clocks) is:

T'_12/T_34 = square-root(1-(v/c)^2)

(The ratio of the frequencies is the reciprocal
of this)

You ask about units. Since it is a ratio, it
doesn't matter. If the clocks are identical
in construction, then we can let T'_12 =
the number of "ticks" of the F' clock between
e_1 and e_2, and let T_34 be the number of
"ticks" of the F clock between e_3 and e_4.

If instead of the F' clock being the sender
of the signals, it is the receiver of the
signals (that is, if signals are sent by
the F clock at events e_3 and e_4, and
the signals are received by the F' clock
at events e_1 and e_2) then the prediction
of SR is:

T'_12/T_34 = square-root((1+v/c)/(1-v/c))

for the case of the F' clock moving directly
away from the F clock, and

T'_12/T_34 = 1/square-root(1-(v/c)^2)

for the case of the F' clock moving transversely.

For the case of the F' clock moving *toward*
the F clock, instead of away, you just reverse
sign of v:

T'_12/T_34 = square-root((1+v/c)/(1-v/c))
for F' clock sending

T'_12/T_34 = square-root((1-v/c)/(1+v/c))
for F clock sending

--
Daryl McCullough
Ithaca, NY