Re: The Nanometre Twin

Sue... says...

On Nov 8, 12:49 pm, stevendaryl3...@xxxxxxxxx (Daryl McCullough)
wrote:

Whatever. Do you have a proof of inconsistency, or not?
If you do, present it. If you don't, then retract your
claim.

Your twin's clock is at 1/3f

What are you talking about? Each twin sends out
a message at the rate of 1 per day, as measured
by his own clock. Where in the world are you
getting 1/3 f? I thought you said that you
understood the derivation of the relativistic
Doppler effect!

The factor of 1/3 is not the rate of either clock.
It is the rate that *images* from one clock are
received by the other twin, as measured in the
coordinate system of the receiving twin.

The rate of Dennis' clock, as measured by Sue's
coordinate system, is not 1/3, it is 1/gamma =
3/5 (for the case we are talking about, which is
v = 4/5 c).

You obviously do *not* know how the relativistic
Doppler shift is computed, if you think that the
Doppler shift factor is the rate of the clocks
involved.

Here's a derivation of the factor of 1/3, from
the point of view of Sue's coordinate system:

1. At time t, Sue sends out one signal.
2. At that time, Dennis is a distance of vt away.
3. The signal arrives at some time t1 = t/(1-(v/c)).
For the particular value v = 4/5 c, we have
t1 = 5 t.

To compute t1, we solved the following equation:

c (t1 - t) = v t1

The left side gives the x-coordinate of the light signal
at time t1, and the right signal gives the x-coordinate
of Dennis at time t1.

4. As measured by Sue's coordinate system, Dennis' clock
is running slow by a factor of 3/5.

5. So the time on Dennis' clock when the signal arrives
is not 5t, but 3/5 * 5 t = 3 t.

6. So the conclusion, from the point of view of Sue's
coordinate system, is that a signal sent on the first day
(as measured by her clock) will arrive on day 3 (as measured
by Dennis' clock). A signal sent on the second day (by
her clock) will arrive on day 6 (by Dennis' clock). Etc.

The factor of 1/3 is a *combination* of the time dilation
of Dennis' clock and the finite speed of light.

How about from Dennis' point of view? How does he account
for the factor of 1/3? From his point of view, he is the
one at rest, and Sue is moving. Well, he computes things
as follows:

1. Sue's clock is running slow, by a factor of 3/5.
2. When her clock says time t, the time according to
Dennis' coordinate system is t' = 5/3 t.
3. So Sue sends out her first signal on t' = 5/3 day,
the second signal on t' = 10/3 day, etc.
4. For a signal sent by Sue at time t', it arrives at
some time t1' given by t1' = (1+(v/c)) t'. For
the particular value of v, 4/5 c, we get:
t1' = 9/5 t'.

How was this computed? Well at time t', Sue is a
distance of vt' away from Dennis. Since light travels
at speed c, it takes time delta-t' = vt'/c for the
light to travel that distance. So the light arrives
at a time vt'/c after it is sent. So the light arrives
at time t' + vt'/c = t' (1+(v/c)).

5. Recall that the first signal is sent at time t' = 5/3 day,
according to Dennis' coordinate system. It arrives at time
t1' = 9/5 t' = 9/5 * 5/3 day = 3 days. The second signal
is sent at time t' = 10/3 day and arrives at time t1' = 6 days.
Etc. From the point of view of Dennis' coordinate system,
the signals from Sue arrive at the rate of one every three days.

Both Sue and Dennis agree that signals from Sue arrive once
every three days according to Dennis' clock.

--
Daryl McCullough
Ithaca, NY

.

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