clock rate

[xxe...@att.net]

I station a clock a few meters above the ground. From somewhere above, I drop an identical clock so that it passes by the stationary clock. At the instant it passes, as a matter of principle, which clock will have the faster time rate?


This is a question of physics in principle - so make it easy and consider a non-rotating Earth.

[dlzc]

Dear xxe...:

On Sunday, August 21, 2016 at 1:28:23 PM UTC-7, xxe...@att.net wrote:
> I station a clock a few meters above the ground.
> From somewhere above, I drop an identical clock
> so that it passes by the stationary clock. At
> the instant it passes, as a matter of principle,
> which clock will have the faster time rate?

Piece of cake.
The falling clock will see the not-falling clock, as "ticking more slowly", The not-falling clock will see the falling clock, as "ticking more slowly".

David A. Smith

[xxe...@att.net]

xxein: The logic of that is absurd. Can you show why you think that would happen?

[dlzc]

Dear xxe...:

On Sunday, August 21, 2016 at 4:54:47 PM UTC-7, xxe...@att.net wrote:
...

> The logic of that is absurd. Can you show why you
> think that would happen?

Sure. Simple relative motion. Either clock is at rest in its own frame, and the Earth's surface simply isn't accelerating enough to account for much time dilation.

The better question is, which clock sees the other clock as "ticking faster", longest.

David A. Smith.

[dlzc]

On Sunday, August 21, 2016 at 5:03:19 PM UTC-7, dlzc wrote:
...

Correction:

> The better question is, which clock sees the other

> clock as "ticking [slower]", longest.

David A. Smith.

[xxe...@att.net]

xxein: That's a silly question. Did that just come out of your typing or did you think about it first? Can you use different words to say what you mean?

[Tom Roberts]

As a matter of principle, both clocks tick at their usual time rates, and since
they are identical, they have identical rates. Earth rotating makes no difference.

If you want to ask how they observe (measure) the other clock's tick rate (not
time rate, as that is unobservable), or how some observer measures their tick
rates, then you must specify how the measurements are performed, as the result
depends on that, both in principle and in practice.

dlzc answered this latter question, not the one you actually asked.
And he assumed a specific measurement procedure, which he did not
describe.


Tom Roberts

[dlzc]

Dear tjrob137:

On Sunday, August 21, 2016 at 6:05:39 PM UTC-7, tjrob137 wrote:
...

>
> dlzc answered this latter question, not the one you actually asked.

He asked "which clock will have the faster time rate?" Each clock sees the other clock as slowed, so each clock is the "faster time rate", in its frame.

> And he assumed a specific measurement procedure, which he did not
> describe.

How about light pulses, each fraction of a second, detected say 1 meter (on that frame) above and below each clock, light pulses from the other clock. We could make each pulse a slightly different color, like a serial number. For the rest frame, the topmost mirror starts the counts, the bottom-most stops the counts. Vice versa for the falling clock.

Just so you can get an average clock rate for the interval...

David A. Smith

[David (Time Lord) Fuller]

xxe...@att.net wrote:
I station a clock a few meters above the ground. From somewhere above, I drop an identical clock so that it passes by the stationary clock. At the instant it passes, as a matter of principle, which clock will have the faster time rate?

the stationary clock a few meters above the ground will have a slower clock rate as the orbit of the electrons in orbit around the molecules that make up the clock will be eccentric from being pushed upwards

[David (Time Lord) Fuller]

the stationary clock a few meters above the ground will have a slower clock rate as the orbit of the electrons in orbit around the molecules that make up the clock will be eccentric from being pushed upwards

Just "inside" the sperically radial radiating electric charge of the atoms, would be seen as an orthogonal magnetic field.
The acceleration of gravity pushing upwards on the atoms would displace this electromagnetic field away from its 90 degree "happy place", slowing the atoms clock until the elections can compensate in their orbits back to the 90 degree happy place by lowering the energy density of spacetime by lowering the vacuum impedance.

[David (Time Lord) Fuller]

the stationary clock a few meters above the ground will have a slower clock rate as the orbit of the electrons in orbit around the molecules that make up the clock will be eccentric from being pushed upwards


Just "inside" the sperically radial radiating electric charge of the atoms, would be seen as an orthogonal magnetic field.
The acceleration of gravity pushing upwards on the atoms would displace this electromagnetic field away from its 90 degree "happy place", slowing the atoms clock until the elections can compensate in their orbits back to the 90 degree happy place by lowering the energy density of spacetime by lowering the vacuum impedance

The "falling unrestrained clock" is (maintaining a constant kinetic energy) by accelerating towards the surface relative to the increasing gravitational field.


((Vacuum impedance/(4pi/10^7))^2 / (acceleration of gravity))*velocity * mass of clock = (total kinetic energy of falling clock.)

[xxe...@att.net]

xxein: OK. My question still stands with time rate but, if it's OK by you, let's thoroughly go through this then with maximum observable clock tick rate as a rough stand-in.

How accurately can a clock tick rate be measured (secs/sec)? On the same time scale, what length of time is needed to record this clock tick rate?

I really want to go through this with you and get whatever resolution we can. As a measure of what I am thinking I'll give you this:

***
Can we say that M = Gm/c^2 as in meters of mass?
Can we say that M can be used as a distance as in 2M (Schwarzschild radius)?

Can we say that sqrt(1-(2M/r)) is the time rate in gravity at r?
Can we say that sqrt(2M/r) is the escape velocity at r?
Can we say that sqrt(1-((v/c)^2)) is the time rate for a velocity?
Can we say that (v/c)^2 = 2M/r when v = escape velocity?
Is there a difference between v(velocity) and v(escape velocity)?
When we are stationary at r, are we traveling at escape velocity at r?
***

[Tom Roberts]

On 8/21/16 8/21/16 9:01 PM, dlzc wrote:
> Dear tjrob137: On Sunday, August 21, 2016 at 6:05:39 PM UTC-7, tjrob137
> wrote: ...
>> dlzc answered this latter question, not the one you actually asked.
>
> He asked "which clock will have the faster time rate?"

Right. So the question has to do with each clock's INTRINSIC time rate. There's
nothing at all about the other clock making any observation.

Note carefully the phrasing: he is asking about a clock's time
rate, NOT about a clock's time rate AS OBSERVED BY THE OTHER
CLOCK. But you keep discussing the latter.

> Each clock sees the other clock as slowed,

Perhaps, but IRRELEVANT as the question is not concerned with how other clocks
or observers might observe it.

> so each clock is the "faster time rate", in its frame.

No. See my earlier response.

This is about ENGLISH PHRASING. When you say "time rate in its frame"
you are not discussing anything but the clock itself, and for that
each clock has its usual time rate (tick rate).

You are implicitly trying to compare how each clock sees the other clock, when
the question, AND YOUR ANSWER HERE, does not do that.


Tom Roberts

[Alan Folmsbee]

On Sunday, August 21, 2016 at 7:07:02 PM UTC-10, xxe...@att.net wrote:
> On Sunday, August 21, 2016 at 9:05:39 PM UTC-4, tjrob137 wrote:

This answer uses the mass=area theorem

Gm/c^2 = meter/second^2 meter^2 second^2 / meter^2 = meter ::YES

Can we say that M can be used as a distance as in 2M (Schwarzschild radius)? No

Can we say that sqrt(1-(2M/r)) is the time rate in gravity at r? 

M/r = meters, SQRT(meter) not eq time rate of anything/

Can we say that sqrt(2M/r) is the escape velocity at r? No :: M/r = meters

Can we say that sqrt(1-((v/c)^2)) is the time rate for a velocity? 

No :: sqrt(1-((v/c)^2)) = 1 for units

Can we say that (v/c)^2 = 2M/r when v = escape velocity?  No

Is there a difference between v(velocity) and v(escape velocity)?  No

When we are stationary at r, are we traveling at escape velocity at r? Maybe.

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$

How accurately can a clock tick rate be measured (secs/sec)?

60 picosecond accuracy is easy to get with common electronics you can afford.

On the same time scale, what length of time is needed to record this clock tick rate? 2000 picoseconds

[Alan Folmsbee]

That is evaluated using the mass=area theorem.

(sqrt(Z) / accel )* v * mass = E

((1/sqrt(second) ) / meter/sec^2) * meter/sec * meter^2 = E

(second^(3/2) / meter) * meter/sec * meter^2 = E

(second^(1/2) * meter^2 = E

No , not energy, sqrt of angular momentum

(second^(1/2) * meter^2 = sqrt(angular momentum)

angular momentum = meter^4 / second

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$

At the instant it passes, as a matter of principle, which clock will have the faster time rate? 

If the falling clock has velocity more than 4.9 meters per second, each clock is locally observed to be the faster one. At less than 4.9 meters per second fall rate, gravitational time dilation becomes important. Then the freely falling clock appears to run faster from the perspective of both observers.

[David (Time Lord) Fuller]

Falling clock total energy = ( mass of clock*c^2) + (0.5*mass of clock* (velocity towards surface)^2)

Clock resting on surface total energy = ( mass of clock*c^2) - (mass of clock * (velocity of gravity)^2)

Total energy = clock rate.

Gravity is (increasing entropy) is (lower energy carrying capacity of spacetime)

Gravity is (lower vacuum impedance) is (lower speed of light) is (lower energy) is (redshifted energy density of spacetime).

Gravity is (redshifted spacetime) is a (less energetic photon)

[David (Time Lord) Fuller]

Falling clock feels No force, same as the photon in simply following curvature of spacetime,

The total energy of the falling clock (mass of clock* c^2 + ((mass of clock)* (velocity of clock relative to surface)^2) is constant.

(Total energy of falling clock ) > ( total energy of stationary clock)

All lengths of all binding forces of all particles of stationary clock elongated by velocity of gravity
(c-velocity of gravity)/c = time dilation of stationary clock relative to falling clock.

Total Energy of clock at event horizon of blackhole is 1 (ONE)

Are you ONE Herbert ??

https://www.youtube.com/watch?v=IsTQ1cM-Hpg

[Tom Roberts]

On 8/22/16 8/22/16 - 12:06 AM, xxe...@att.net wrote:
> On Sunday, August 21, 2016 at 9:05:39 PM UTC-4, tjrob137 wrote:
>> On 8/21/16 8/21/16 3:28 PM, xxe...@att.net wrote:
>>> I station a clock a few meters above the ground. From somewhere above,
>>> I drop an identical clock so that it passes by the stationary clock. At
>>> the instant it passes, as a matter of principle, which clock will have
>>> the faster time rate? This is a question of physics in principle - so
>>> make it easy and consider a non-rotating Earth.
>>
>> As a matter of principle, both clocks tick at their usual time rates, and
>> since they are identical, they have identical rates. Earth rotating makes
>> no difference.
>

> xxein: OK. My question still stands with time rate

Not as a question of physics. If you want to indulge in theology or other
speculations about unmeasurable/unobservable quantities, don't use a science
newsgroup.

In physics, time is what clocks measure, because in any experiment testing any
theory's usage of "time", a clock is used to measure it.

> How accurately can a clock tick rate be measured (secs/sec)?

This depends on many aspects of the measurement. NIST has a clock that is
precise to two parts in 10^18, with accuracy known to a few parts in 10^16.

> On the same
> time scale, what length of time is needed to record this clock tick rate?

Again it depends on many details of the measurement. And the clocks.

> *** Can we say that M = Gm/c^2 as in meters of mass? Can we say that M can be
> used as a distance as in 2M (Schwarzschild radius)?

Yes. (Well, close enough -- there are caveats required to make this valid.)

> Can we say that sqrt(1-(2M/r)) is the time rate in gravity at r?

No.

> [...]

No to all the other unlikely things you say here. You need to LEARN something
about modern physics before you can reasonably expect to discuss it.


Tom Roberts

[David (Time Lord) Fuller]

As 99% of the clocks mass is Binding Energy, the stationary clock's clock rate is slowed to compensate for (increased binding energy Needs) due to binding forces needing to travel a longer distance

Binding energy contributes about 99% of the mass of a proton. The mass of a proton is about 940 MeV (these are the units most commonly used in particle physics). By comparison, the "bare mass"[1] of an up quark is around 2 MeV, and the bare mass of a down quark is about 5 MeV. A proton has two up quarks and one down quark as 'valence quarks', which combined contribute only about 10 MeV. The rest of the mass, about 930 MeV, comes from binding energy.

[mlwo...@wp.pl]

W dniu wtorek, 23 sierpnia 2016 17:21:06 UTC+2 użytkownik tjrob137 napisał:

> In physics, time is what clocks measure, because in any experiment testing any
> theory's usage of "time", a clock is used to measure it.

Almost, but not quite. In physics, time is what idiot physist
imagines clock measure.
Real clocks measure t'=t. Check at GPS.

[xxe...@att.net]

xxein: I am not driving your father's Oldsmobile nor do I wish to. Do you think the notion of a curved space-time is the proper model to explain gravity? I don't.

What would've happened if Albert Einstein thought he needed to rely upon somebody else to instruct him in how to think of physics?

[Tom Roberts]

On 8/28/16 8/28/16 8:48 PM, xxe...@att.net wrote:
> Do you think the notion of a curved space-time is the proper model to explain
> gravity? I don't.

I haven't the foggiest notion of what "the proper model" is. But I do know
that curved spacetime is the best model we have to date, and it has proven
itself to be an excellent model.

BTW YOU don't have any notion about "the proper model", either,
even though you seem to think that you do. I know this because
it is quite clear that you are ignorant of the experiments.

> What would've happened if Albert Einstein thought he needed to rely upon
> somebody else to instruct him in how to think of physics?

He DID have such instruction -- that's what an education is. NOBODY with such an
education would ever "rely" on it to tell them how to think. So this question is
just one more instance of you revealing your own lack of education. And also
displaying how STUPID is your approach.

People like you, who imagine they can "think it up" all on their own, are
HOPELESSLY out of touch with reality. In particular, a good education is NOT
"handcuffs" like you seem to think; it is, rather, LIBERATING.

There are three types of people in the world: those who don't
learn from their own mistakes, those who do learn from their
own mistakes, and those who learn from the mistakes of others.
You seem determined to remain in the first group. The whole
point of an education is to elevate one to the third group.


Tom Roberts