jxbjxbjxb

I just read Einstein's own "physical reasoning" example for showing
the effect of Gravitational Time Dilation. It is:

Hang a clock on a string from the ceiling. On the bottom of this
clock is attached a beam emitter, pointing downard, which makes a
pulse of light every time the clock ticks.

Place another clock on the floor next to a long hole, such that the
hanging ceiling clock is above this floor clock.

At the very instant before the ceiling clock makes a tick, and thus
emits a pulse of light, cut the string. Because the clock starts to
fall just very slightly before it ticks/flashes, the distance it has
travelled is imperceptible, thus the clock "still exists in the
timeflow of the ceiling" as it emits a pulse of light.

As the clock ticks, pulses of light move downward toward the clock on
the floor.

At the very instant before the first pulse of light reaches the floor
clock, drop the floor clock into the hole. Because this clock starts
to fall just very slightly before it receives the first pulse of
light, the distance it has travelled is imperceptible by the time the
light reaches it, thus the floor clock is "still flowing in the
timeframe of the floor."

So now we have two falling clocks. Since the ceiling clock was
released before the floor clock, it is moving a bit faster, thus
gaining on the floor clock. As a result of this, the light pulses
will be doppler-shifted such that the floor clock perceives them as
being more frequently emitted than the ceiling clock claims it is
emitting them. Thus, the flow of time for the floor clock must be
slower than the flow of time for the ceiling clock, thus time is
dilated closer to the center of gravity, i.e. Time slows down as you
near gravity.

--

Now, this is Einstein's example for showing how gravity time dilation
works. You may see this spelled out on page 102 of Kip Thorne's book
"Black Holes and Time Warps", and probably in plenty of other books as
well. The problem I have with this example is:

Suppose you also placed a beam emitter on the top of the floor clock,
such that as it ticks, it also sends out a pulse of light upward,
toward the ceiling clock.

Since both clocks are moving toward each other relative to each other,
the Doppler shift would be observed identically by the ceiling clock,
thus suggesting that time also slows for the clock further away from
the center of gravity.

Because both clocks are free-falling in the same gravity, they may be
considered inertial frames of reference thus Special Relativity can
explain them, thus neither clock is ultimately experiencing slower
time. The exercise seems to ultimately have nothing to do with
gravity, so why did Einstein claim it was a demonstration of
Gravitational Time Dilation????

Jaxon

richard

Exactly the same flaw is present in each and every time dilation
argument that the man constructed, moreover illustrating this flaw
constitutes a good portion of the counter-arguments to the theory, to no
avail, the adherents simply cite provably incorrect sources while
turning a deaf ear.

Richard Perry

http://www.cswnet.com/~rper

perion

I always had the same objection to this scenario as an example of
gravitational time dilation. Hawking makes no reference to Doppler shifting
in the following ****ysis:
Energy is proportional to frequency so as light moves away from a
gravitational source, it loses energy and therefore it's frequency is
lowered - and vice versa for light moving toward a grav. source. Don't cut
either clock's strings - just suspend them. Energy is lost (freq. lowered)
as light moves from bottom clock to top one. Energy is gained (freq.
increased) as light moves from top clock to bottom one. Hence -
gravitational time dilation. What's Doppler shifting have to do with it???

It seems to me that introducing the relative motion of the two clocks in
freefall and (therefore a Doppler shift) doesn't show anything about
gravitational time dilation - as your counter-example seems to point out. It
just becomes an exercise in Special Relativity.

I'll be interested to see what others have to say.

Perion

jxbjxbjxb

Indeed, Kip Thorne says in his book:

"Einstein converted the problem of comparing the flow of time as felt
by the ceiling and the floor into the problem of comparing the ticking
rates of two freely falling clocks. His equivalence principle then
permitted him to compare the ticks of the falling clocks with the aid
of his SR laws."

Just makes no sense to me.

Jaxon

dirk van de moortel

And just before it sends a pulse, drop it in the hole.
Just before the pulse reaches the ceiling clock, cut the
rope and the top clock will start falling.
Since the floor clock was dropped first, it will be
*receding* from the ceiling clock. As a result of this,
the light pulses will be doppler-shifted such that the
ceiling clock perceives them as being *less* frequently
emitted than the floor clock claims it is emitting them.
Thus, the flow of time for the ceiling clock must be
faster than the flow of time for the floor clock, thus
time is 'contracted' further away from the center of
gravity, i.e. time speeds up as you recede from gravity.
So you have the same result as before.

Dirk Vdm

harry

I look at the below example from the lab frame (ceiling or floor).

Also the velocity relative to my frame is imperceptible.
The frequency is therefore unaffected.


And so is the emitted light pulse almost unaffected by what happens to the
clock.
Moreover, the very slight frequency increase of both is identical, so there
is no Doppler shift between them at all.

Now I'm lost! There was only one light pulse coming from each clock,
unaffected by what happens with the clocks, right?!
On the other hand, if you consider light pulses as sent from an accelerating
clock, then the example is different, and Feynman described that in "six
not-so-easy pieces". But I had problems with his example too, and may come
back to that another time...

"Thus"? Hmm. No, if your description is correct, that example is simply faulty!


Again, with one pulse each and equal speeds, in the lab frame the
frequencies shifts are identical.

No! There is an increasing velocity difference between them. However, as I
pointed out, the two pulses are indeed emitted and compared in the same
inertial reference frame (moving at a very slight velocity relative to the ceiling/floor).


I fully agree.

Harald

dirk van de moortel

There are many light pulses coming from the ceiling clock
only. The floor clock is not sending any pulses.

Dirk Vdm

harry

Thanks Dirk, in that case the example is very different from how I
understood it!

Still the example doesn't make sense to me: as Jaxon pointed out, the
situation looks perfectly symmetrical and with light signals coming from
both clocks, both receivers should observe the same Doppler effect (I saw
your other reply, but you changed the experiment to the inverse experiment,
and changing v into -v).

Also, a constant Doppler rate difference can only be the result of a
constant speed difference, and in this example v changes, thus the frequency
changes, even at constant acceleration and gravitation.

Harald

dirk van de moortel

But light signals do not come from both clocks.
There are two completely different cases:
Case 1:
The ceiling clock is dropped first and sends signals down.
The floor clock is dropped later when it receives the
first signal.
The clocks approach each other.
The floor clock receives signals at a higher rate
than they are sent out.

Case 2:
The floor clock is dropped first and sends signals up.
The ceiling clock is dropped later when it receives the
first signal.
The clocks recede from each other.
The ceiling clock receives signals at a lower rate
than they are sent out.

No symmetry: they both give the same result (see below).
And obviously there is no way both clocks can be sending
signals to each other since one of them must be dropped
first. You cannot possibly have both experiments going
on at the same time.

Bu there *is* a constant speed difference as you can
easily verify:
In case 1:
Height of ceiling clock: h_c(t) = H - 1/2 g t^2
Height of floor clock: h_f(t) = -1/2 g (t-H/c)^2
Height difference h(t) = h_c(t) - h_f(t) = -gH/c t + junk
Relative velocity = v = dh/dt = -gH/c
(initially approaching, catching up and then receding) In case 2:
<exercise for you>
Relative velocity = v = gH/c (permanantly receding)

Dirk Vdm

harry

It does in HIS counterexample!


NO, I think that that is your example!
His Case 1-extended, as I understood it:

The ceiling clock is dropped first and sends signals down.
The floor clock is dropped later when it receives the first signal.
-> Then the floor clock also sends signals up
The clocks approach each other.
The floor clock receives signals at a higher rate than they are sent out.
-> Also the ceiling clock receives signals at a higher rate than they are
sent out.

That is normal Doppler, as used to illustrate the Twin paradox ("symmetrical time dilation").

Oops yes of course, sorry!

Harald