On Jan 23, 2006, at 5:12 AM, Stephen A. Lawrence wrote:
Just a couple brief comments.
Horace Heffner wrote:
On Jan 22, 2006, at 5:05 AM, Stephen A. Lawrence wrote:
Acceleration doesn't affect clocks. That's been verified
(can't cite references, sorry). A clock in a centrifuge
slows only as a result of the speed at which it's traveling,
not as a result of the centripetal force.
[HH]
This can not be consistent with relativity,
[SAL]
But it is. It's built into GR from the get-go.
[HH]
I thought Einstein's equivalence principle was built into
relativity.
The equivalence principle is built in. So is the principle of
relativity, and, as a consequence of the assumption that you can
change to any arbitrary coordinate system without affecting the
results, the lack of any local effect due to acceleration is built
in, too.
[snip enormous amounts, after reading -- thanks for the additional
explanation of the retardation comments]
The rate at which a clock is observed to tick does not depend on
whether the clock is _currently_ undergoing acceleration. That
has been both predicted and observed to be true, to the limits
of the experiments which have been done.
Then you have conclusively proved GR is based upon a false
assumption.
No, I haven't, because, as stated elsewhere, clocks in GR are
apparently affected by gravitational _potential_ but not by the
local intensity of the gravitational _field_.
When you accelerate, in SR, you find that distant clocks are
apparently affected by _your_ acceleration. _THAT_ is equivalent
to the GR clocks being affected by the gravitational potential.
The effects are identical.
You cannot separate the observations from the observer, and the
concept of observable properties of external things being affected
by changes within yourself (such as your acceleration) is a
consequence of that.
[ ... ]
Agreed - but only if you agree that clocks involving mass
actually change due to velocity alone. In other words, if m =
m0*gamma is purely due to appearances, i.e. due to retardation,
then the only effect left to cause a time difference upon
rejoining the clocks is acceleration. I don't think it is
generally accepted an more that m=m0*gamma is a real effect. I
definitely read that in some text.
Whether an effect is "real" or not is so slippery that I don't
think there's any definitive answer.
IMHO the Sagnac effect proves that time dilation is "real". In the
opinion of lots of other people it does not. If you spin a rigid
disk it will crack due to Fitzgerald contraction. I think that
proves the contraction is "real". Many other people think it does
not.
The trouble with m0*gamma is it's the total energy of the object,
and that's frame-dependent. Your point of view determines how big
it is. But does that mean it isn't "real"? I don't think so --
that's like saying kinetic energy isn't "real" because it's frame
dependent.
What acceleration _DOES_ do is affect _DISTANT_ clocks. When
YOU accelerate, clocks that are far, far away and toward which
you are accelerating seem to you to run _faster_.
Irrelevant. Who cares how the clocks appear during the journey.
I only care what happens when they come back together in the
same location in the same reference frame. Then all retardation
effects have cancelled because there is no retardation remaining.
Retardation explains what you see as you watch the other party, right?
By using a powerful telescope, you can actually watch the other
party's clock throughout the whole trip. By looking at the _size_
of the image, and the rate at which it's changing, you can see how
fast the other party is moving (relative to you) and how far away
the other party is. That picture-show which you can watch _MUST_
agree with the physical effects observed when you get home again
and put the clocks next to each other. If you can explain how that
happens you're probably as close to "understanding" this as you can
get.
The weird thing is that all the "effects of retardation" do _not_
cancel when you get home, and it's very hard to draw a line between
what was "real" and what was an illusion.
The weirdness you apparently see comes directly from the (assumed)
fact that the light signal travels at C relative to _both_ the
stationary and the moving parties.
* * *
In the "stationary" frame you can explain it all by using time
dilation -- you can, with the help of extra (stationary) observers
spaced out along the route, actually observe the traveler's clock
running _slow_. (Just assume time dilation is real and you're done!)
In the "moving" frame you've got a much bigger problem; just
exactly when does the stationary clock run _fast_? The answer is:
while you are accelerating.
[ ... ]
If the traveler accelerates in a blazing flash lasting a few
microseconds, and then IMMEDIATELY decelerates again, he'll
experience negligible time skew. On the other hand, if he
accelerates, coasts a long time, and then decelerates, he'll
experience a lot of time skew.
OK, but then this implies the clock is mass related, and
m=m0*gamma is a real, not a retardation effect.
Relativistic Doppler shift includes a term for gamma. The
emitter's motion changes the apparent frequency, _and_ the
emitter's different time base changes the apparent frequency, and
the two effects must be combined to obtain the total observed effect.
The "extra" mass of a moving body is (gama-1)*m0. At relativistic
speeds that's where most of the body's energy is. If that's not
"real", then most of the energy isn't "real", either.
The gravitational time dilation is due to the gravitational
potential, _not_ the local acceleration of the field.
I think this is not the only possible explanation. An
alternative explanation is the red shift is due to the effect of
gravity on the photon. Gravitons exchange momentum with photons,
but not virtual photons. If this were not true black holes would
not exist. In the case of a spherical shell object with a hole
in it, I think the red shift would occur at the surface as light
goes through the hole.
That would make sense. The redshift "happens" in regions where the
field is non-zero, of course, which is exactly where the photons
would be interacting with it.
If the field is the gradient of the potential, then the places
where the field is strongest are also the places where the
potential, and degree of redshift, are changing most rapidly.
But I'm just talking about where and when you would "observe" a red-
shift. The "observed" redshift is a function of the gravitational
potential, but that's not the same as saying it's "caused" by it --
I should be more careful about how I say things...
It sounds like you are attributing a "real" effect to static
gravitational potential that should be matched by an equivalent
"real" effect from a static electromagnetic potential A. No such
effect exists to my knowledge. AFAIK, The only effects that
manifest as real are the result of changes in A, i.e in @a/@t.
It seems that way but it's not.
In relativity, the E field and the G "field" produce totally
different kinds of "forces". Let's see if I can dredge this out of
my memory....
The E field contains a heat-like component (heat is a force, too --
a candle increases the momentum of an object placed above it, so dP/
dt is nonzero in that case). Gravity is not a heat-like force, and
I'm failing completely to recall just what difference that makes in
this case.
More mundanely, charge is conserved; you drop a charged rock down
the hole, at the bottom of the hole it still has the same charge as
it had at the top of the hole. If you turn it into a beam of light
you need to figure out what to do with the charge -- you can't just
throw it away. Gravitational mass is apparently _not_ conserved,
not the same way; in particular, when you drop the rock down the
hole, it gains gravitational mass.
Oh well I'm just babbling at this point I should drop this line of
reasoning until and unless I look it up again...
Here's another cute example: A spherical chamber cut out of a
uniformly dense planet which was _offset_ from the center would
have a _uniform_ (but non-zero) G-field inside it.
It should have a g field due to the sphere having the radius from
the hole to the center.
If you work it out, it's a completely uniform field. Very strange.
(Easiest way to analyze it is to pretend the chamber is a separate
sphere of "negative mass" and just sum its "negative" field with
the field of an intact planet.)
I think any object held in that chamber would experience a
gravitational red shift proportional to the g at its location,
not to the gravitational potential.
I don't know if I understand you. Light should be redshifted as it
crosses the chamber, in proportion to the intensity of the field in
the chamber, right?
If that's what you're saying, I agree. Note again that since the
field strength is the gradient of the potential, that's equivalent
to saying the degree of redshift varies with the potential.
If what you were saying were true then objects in the center of
the universe (assuming here a big bang) should all be massively
red shifted, instead of vice versa.
Only if there's a gravitational field filling the universe,
pointing to the center. That's the only way you'll get a lower
gravitional potential at the center of the universe.
And if there is such a field, then there must be a redshift
associated with it, too.
No matter how you cut it, clock rate is a function of
gravitational field. If the effects of the gravitational
field differ from the effects of acceleration (this difference
at any point) then Einstein's fundamental assumption for GR is
violated and GR disappears in a flash! 8^)
I also have to question the validity of the tangential straight
rod approach you use. I could be missing something, but it
doesn't seem to account for how we would see the clock advance
as it passes behind the earth in the opposite direction.
You can't synchronize all the clocks on a rotating disk.
I didn't mention synchronization.
You can't synchronize all the clocks on the Equator. If you
try, you find there is a "date line" where two adjacent clocks
are out of sync. It's crossing the "date line" which causes the
hiccup.
Here again you are talking about how things appear in motion. I
just want to figure out in an intuitive way what accounts for
differences when clocks are brought back together.
Hmmm ... Consider again the laser-ring gyro. What causes the
fringe shift when you rotate it?
Signal velocity relative to the rim of the disk can be measured and
is constant.
If there is no velocity effect which does this, then what
remains except acceleration? Retardation is out of the picture.
Now that I can see some real data it would be good to look at the
effects of gravimagnetism, because these should modify the
expected values. It is probably going to take an FEA program to
do this right, and I just do not have the time right now.
What I *can* see from the airplane data is it can not be fully
analysed using only the earth's gravimagnetic field. It requires
quantifying the solar and/or galactic gravimagnetic field.
But would such effects not be swamped by the local influence of the
Earth?
Gravimagnetics, given the presence of the significant ambient
gravimagentic field, should enhance the time difference on the
*airplane* clocks. In other words it agrees qualitatively with
the time differences, but may be too small to make any
difference. It does mean the two east-west opposed orbit
satellites would have their orbital parameters affected, thus
their velocities, and thus their clocks.
It is interesting that polar satellites should veer left going
over the North pole and right going over the South pole, from the
point of view of a person in the satellite oriented feet down and
facing the direction of motion. Satellites going west-to-east
should experience a lower g value than those going east-to-west,
and the higher the velocity the lower the g value. This means
the g value at the surface should be, due to gravimagnetism,
slightly less at the equator than at the pole, and should cycle
in value over a 24 hour period, due to the earth's axis not
aligning with the ambient gravimagnetic field.
Hmmm ... Interesting.
