Einstein's clock

[chinglu]

Cordin to Einstein, if you are moving relative to each other, you disagree on the timing of events.

So, for example, Saturn and Mercury are moving relative to each other, therefore, they will disagree on the timing of events.

Now, we have a clock that humans have used for thousands of years, the revolution of the earth around the sun. So, that is a clock.

Now, will observers on Saturn and Mercury disagree on the earth's position since they must disagree on time given the earth's position is a timing device?

 

[AlphaNumeric]

They will also disagree on the distances one another travels because of their relative motion and different locations in the Sun's gravitational field. My hack detection sense is telling me you're trying to go somewhere with this...

 

[RJBeery]

Cordin to Einstein, if you are moving relative to each other, you disagree on the timing of events.

So, for example, Saturn and Mercury are moving relative to each other, therefore, they will disagree on the timing of events.

Now, we have a clock that humans have used for thousands of years, the revolution of the earth around the sun. So, that is a clock.

Now, will observers on Saturn and Mercury disagree on the earth's position since they must disagree on time given the earth's position is a timing device?

Do you mean disagreement on the instantaneous position of Earth or the relative speed of the passing of "years"?

 

[brucep]

Cordin to Einstein, if you are moving relative to each other, you disagree on the timing of events.

So, for example, Saturn and Mercury are moving relative to each other, therefore, they will disagree on the timing of events.

Now, we have a clock that humans have used for thousands of years, the revolution of the earth around the sun. So, that is a clock.

Now, will observers on Saturn and Mercury disagree on the earth's position since they must disagree on time given the earth's position is a timing device?

Your problem is set in the weak field so any difference in tick rate ratio between

dTau_mercury/dt_earth, dTau_saturn/dt_earth, or dt_shell saturn/dt_shell mercury is exceedingly miniscule becoming irrelevant for weak field experiments which don't require nano second accuracy.

The weak field is modeled by the Schwarzschild metric solution to the EFE. We can do that in the weak field because all the big things like the Sun and all the Planets are ~ spherically symmetrical ~ non rotating and deltas associated with relativistic effects ARE weak field miniscule. The following equation is derived from the effective potential of the Schwarzschild equation of motion. This equation includes a component for gravitational time dilation and time dilation associated with relative motion.

dTau/dt = (1-3M/r)^1/2

For your example lets use geometric units so the ratio will be dimensionless

dt_earth = 1 [tick]

M_sun = 1477 meter

r_ mean mercury orbit = 5.790905E10 meter

r_mean saturn orbit = 1.43344937E12 meter

dTau_mercury/dt_earth = (1-3M_sun/r_mean orbit mercury)^1/2 = .999999961

dTau_saturn/dt_earth = .9999999998

dt_shell mercury/dt_shell_saturn = .999999961 / .9999999998 = .999999961....

Weak field miniscule. How would the great physicists hope to 'find that' without Maxwell? It's amazing what mathematical physicists have accomplished.

Maybe you should pick an example where most the action takes place in the strong field.

I forgot to pay homage to the great experimental physicists, astro physicists, and teachers.

 

[chinglu]

They will also disagree on the distances one another travels because of their relative motion and different locations in the Sun's gravitational field. My hack detection sense is telling me you're trying to go somewhere with this...

I am trying to determine whether the two observers will agree on the position of the earth in space.

So, when it is at some position and then returns to that position again, they can sync clocks on some common earth position for the next round, no?

This would be an agreed upon earth year.

 

[chinglu]

Do you mean disagreement on the instantaneous position of Earth or the relative speed of the passing of "years"?

I mean position within epsilon.

 

[chinglu]

Your problem is set in the weak field so any difference in tick rate ratio between

dTau_mercury/dt_earth, dTau_saturn/dt_earth, or dt_shell saturn/dt_shell mercury is exceedingly miniscule becoming irrelevant for weak field experiments which don't require nano second accuracy.

The weak field is modeled by the Schwarzschild metric solution to the EFE. We can do that in the weak field because all the big things like the Sun and all the Planets are ~ spherically symmetrical ~ non rotating and deltas associated with relativistic effects ARE weak field miniscule. The following equation is derived from the effective potential of the Schwarzschild equation of motion. This equation includes a component for gravitational time dilation and time dilation associated with relative motion.

dTau/dt = (1-3M/r)^1/2

For your example lets use geometric units so the ratio will be dimensionless

dt_earth = 1 [tick]

M_sun = 1477 meter

r_ mean mercury orbit = 5.790905E10 meter

r_mean saturn orbit = 1.43344937E12 meter

dTau_mercury/dt_earth = (1-3M_sun/r_mean orbit mercury)^1/2 = .999999961

dTau_saturn/dt_earth = .9999999998

dt_shell mercury/dt_shell_saturn = .999999961 / .9999999998 = .999999961....

Weak field miniscule. How would the great physicists hope to 'find that' without Maxwell? It's amazing what mathematical physicists have accomplished.

Maybe you should pick an example where most the action takes place in the strong field.

I forgot to pay homage to the great experimental physicists, astro physicists, and teachers.

Yes, I wanted both a high speed differential and a gravity differential.

But, I am trying to sync earth years.

The question is whether they can agree on an earth position and then agree it is again there sometime later. In other words, earth does not occupy 2 different places in space regardless of the observer.

 

[AlphaNumeric]

The Earth doesn't occupy two different places in relativity. Everyone sees the Earth be at a single place and moving consistently. Their points of view are then linked by the transformation rules of relativity. Is this going to be another attempt by you to claim relativity says an object is at two places at once, like the light sphere thing? You completely failed to grasp relativity then, please don't let this be a repeat.

 

[przyk]

Cordin to Einstein, if you are moving relative to each other, you disagree on the timing of events.

So, for example, Saturn and Mercury are moving relative to each other, therefore, they will disagree on the timing of events.

Now, we have a clock that humans have used for thousands of years, the revolution of the earth around the sun. So, that is a clock.

Now, will observers on Saturn and Mercury disagree on the earth's position since they must disagree on time given the earth's position is a timing device?

Your question doesn't make much sense. Neglecting a bunch of subtleties, the answer is that observers on Saturn and Mercury could both decide to use the Earth's orbit as a clock if they wanted. It's just that their own, local clocks wouldn't agree on how long an Earth year was.

It gets more subtle than this because different observers in relativity don't share the same concept of simultaneity, so there isn't a simple correspondence between what's on one observer's clock compared with what's on another's clock at the "same time", because different observers in relative motion don't agree on what "same time" means over spatial distances. If you're taking GR into account, curvature makes this even worse.

 

[origin]

chinglu, I guess you think it is time to once again prove that relativity is wrong, I am afraid like every other time you will get smacked down and leave.

 

[brucep]

Yes, I wanted both a high speed differential and a gravity differential.

But, I am trying to sync earth years.

The question is whether they can agree on an earth position and then agree it is again there sometime later. In other words, earth does not occupy 2 different places in space regardless of the observer.

Read what I wrote down, the equation derived from the effective potential of the Schwarzschild equation of motion, accounts for BOTH gravitational and relative motion time dilation. It gives the same result as if you did them individually and summed the results together. If you don't understand how to transform from a coordinate frame to the invariant proper frame then you should be learning that first.

 

[Aqueous Id]

I am trying to determine whether the two observers will agree on the position of the earth in space.

Whether or not they are in agreement depends on whether or not they correct for relativistic effects, though they would tend to be miniscule without speeds near c.

So, when it is at some position and then returns to that position again, they can sync clocks on some common earth position for the next round, no?

If they are trying to be correct, they can be in agreement moment by moment, simply by agreeing to use a real time application of the Lorentz transformation.

This would be an agreed upon earth year.

Theoretically two ET observers could agree on what time it is anywhere on Earth by agreeing to apply Lorentz in real time, and by picking an Earth time zone (e.g. GMT) and by correctly arriving at the local diurnal position, i.e., the precise moment of midnight for that zone. In practice it would be extremely difficult to achieve with accuracy but they could employ an atomic clock, a model for the two worlds they are tracking (Earth and the other observer) and there are some robust algorithms, like the Kalman filter, to reduce drift.

The epsilon of error is probably similar to that of time difference of arrival systems. In the simplest of these, you get an "elliptical error probable" (EEP) meaning the correct position is probably somewhere within an ellipse. The axes of the ellipse are due to the relative system errors in the three sites (you, other guy, earth). In this case you would start by trying to construct a simple model that does only that and then try to account for orbital and axial motion of each observer.

As it turns out, the systemic errors will eat your lunch, and completely obliterate the small relativitistic error as the EEP grows with each imprecisely modeled feature. Besides, imagine trying to definitively nail down "the center of the earth", or of your planet of observation. In practice there are ungodly algorithms just to do that, and they will swallow your relativistic errors as well.

In short, it's an utterly impractical question, without accounting for all the models NASA used in successfully dropping the Mars rovers so close to the landing zones. Their success demonstrates the ability to get close enough despite all the lack of precision. Obviously, if you want to do something bad enough, you're going to create workarounds to all known obstacles.

 

[chinglu]

The Earth doesn't occupy two different places in relativity. Everyone sees the Earth be at a single place and moving consistently. Their points of view are then linked by the transformation rules of relativity. Is this going to be another attempt by you to claim relativity says an object is at two places at once, like the light sphere thing? You completely failed to grasp relativity then, please don't let this be a repeat.

No, I don't think observers think the earth is in 2 different places.

 

[chinglu]

Your question doesn't make much sense. Neglecting a bunch of subtleties, the answer is that observers on Saturn and Mercury could both decide to use the Earth's orbit as a clock if they wanted. It's just that their own, local clocks wouldn't agree on how long an Earth year was.

It gets more subtle than this because different observers in relativity don't share the same concept of simultaneity, so there isn't a simple correspondence between what's on one observer's clock compared with what's on another's clock at the "same time", because different observers in relative motion don't agree on what "same time" means over spatial distances. If you're taking GR into account, curvature makes this even worse.

Yes, so let's assume they have frequency clocks on their respective planets.

Now, like with GPS, they can adjust these clocks to agree on an earth year.

So, how is it possible that two observers at different speeds and different gravity potentials can agree on time?

How does this fit within relativity that claims there is no absolute time and all event force time disagreements?

Here, we can sync clocks in two different frames and GPS has proven as such.

How can that be?

 

[chinglu]

Read what I wrote down, the equation derived from the effective potential of the Schwarzschild equation of motion, accounts for BOTH gravitational and relative motion time dilation. It gives the same result as if you did them individually and summed the results together. If you don't understand how to transform from a coordinate frame to the invariant proper frame then you should be learning that first.

Yea, but can they adjust their clocks as does GPS satellites before launch?

Then, these two frames agree upon and absolute standard, the revolution of the earth around the sun.

How can frames agree upon an absolute standard of time?

 

[chinglu]

Whether or not they are in agreement depends on whether or not they correct for relativistic effects, though they would tend to be miniscule without speeds near c.


If they are trying to be correct, they can be in agreement moment by moment, simply by agreeing to use a real time application of the Lorentz transformation.


Theoretically two ET observers could agree on what time it is anywhere on Earth by agreeing to apply Lorentz in real time, and by picking an Earth time zone (e.g. GMT) and by correctly arriving at the local diurnal position, i.e., the precise moment of midnight for that zone. In practice it would be extremely difficult to achieve with accuracy but they could employ an atomic clock, a model for the two worlds they are tracking (Earth and the other observer) and there are some robust algorithms, like the Kalman filter, to reduce drift.

The epsilon of error is probably similar to that of time difference of arrival systems. In the simplest of these, you get an "elliptical error probable" (EEP) meaning the correct position is probably somewhere within an ellipse. The axes of the ellipse are due to the relative system errors in the three sites (you, other guy, earth). In this case you would start by trying to construct a simple model that does only that and then try to account for orbital and axial motion of each observer.

As it turns out, the systemic errors will eat your lunch, and completely obliterate the small relativitistic error as the EEP grows with each imprecisely modeled feature. Besides, imagine trying to definitively nail down "the center of the earth", or of your planet of observation. In practice there are ungodly algorithms just to do that, and they will swallow your relativistic errors as well.

In short, it's an utterly impractical question, without accounting for all the models NASA used in successfully dropping the Mars rovers so close to the landing zones. Their success demonstrates the ability to get close enough despite all the lack of precision. Obviously, if you want to do something bad enough, you're going to create workarounds to all known obstacles.

Since the frames cannot disagree on the position of the earth in some grid, then either they disagree on the speed of the earth or they do not.

If they disagree on the speed of the earth, then on the next cycle, they must place the earth at 2 different positions in a Cartesian grid which is impossible.

So, whether they had local clocks or not, the earth's revolution around the sun seems to provide an absolute standard of time on which they can agree.

But, there can be no absolute time in relativity.

So what gives here?

 

[Aqueous Id]

Since the frames cannot disagree on the position of the earth in some grid, then either they disagree on the speed of the earth or they do not.

Huh?

If they disagree on the speed of the earth, then on the next cycle, they must place the earth at 2 different positions in a Cartesian grid which is impossible.

Double-huh?

So, whether they had local clocks or not, the earth's revolution around the sun seems to provide an absolute standard of time on which they can agree.

Huh-cubed?

But, there can be no absolute time in relativity.
So what gives here?

A failure to understand what the Lorentz transformation means. If you can wrap your head around it, none of the foregoing questions will bother you anymore.

If relativity is too difficult for you, you might start with "time difference of arrival" (TDOA) which is a good stepping stone to learning relativity. (It's the backbone of GPS.)

 

[przyk]

How does this fit within relativity that claims there is no absolute time and all event force time disagreements?

Relativity claims that there are a whole family of space/time coordinate mappings, related by Lorentz transforms, that are all equally good in flat spacetime as far as the laws of physics are concerned. That doesn't prevent you from simply picking one of them and using it as a standard for everyone. It's just that in many cases, an observer using someone else's reference frame is likely to be a pain because it doesn't correspond with how they'd naturally like to measure space and time and simultaneity. This isn't much of an issue within the solar system because the gravitation is relatively weak and the speeds involved are much less than the speed of light, but it could be a nuisance at higher speeds and/or in a stronger gravitational well.

 

[Motor Daddy]

So, how is it possible that two observers at different speeds and different gravity potentials can agree on time?

Because all observers agree the speed of light is c, which means all observers agree on the length of the path that light travels in a duration of time. Since all observers agree on the length of that path, t=l/c, where l is the length of the path that light traveled. All observers agree on c, t, and l. PERIOD!

 

[Aqueous Id]

Because all observers agree the speed of light is c, which means all observers agree on the length of the path that light travels in a duration of time. Since all observers agree on the length of that path, t=l/c, where l is the length of the path that light traveled. All observers agree on c, t, and l. PERIOD!

That isn't true in relativistic motion, which requires application of the Lorentz transforms to account for where and when everybody is at any moment of observation.

However, under the low velocities described here, relativistic effects are small and can be neglected.

That leaves calculating where everybody is by a model, since all three are orbiting their own axes at various rates, at various lat/long coordinates on planets of various radii, and these planets are in some arbitrary phase of their elliptical orbits around the sun, traveling at various orbital velocities.

It turns it that the calculations involved basically use the same 3D rotation transforms as the Lorentz transformation for relativity. The programmer who gets to code this simulator will be grateful for that much.

It's kind of like relativity in a way - where you are with respect to me also depends on what time it is for you, since your planet will have rotated so many degrees about its axis and that amount changes your distance from me. On the other hand, the observer from the third planet is looking at you from a different angle. You may have moved away from me and towards him. So all of that has to be calculated correctly.

The way this is unlike relativity is that, due to everybody's low velocities, we can neglect relativistic effects and normally get away with it. That means we can all run the same calculation and get the same results at the same time which is why we'll be in agreement.

We'll just never be in exact agreement. There will be that miniscule relativistic error. There will be larger errors in our ability to pinpoint ourselves precisely in space, due to uncertainties we can't precisely measure (where the exact center of as of each planet is, for example).

But those are all differences as a practical matter, such as a simulation. Theoretically, yes, we have a way to always be in agreement. We'll just never be able to precisely implement it. This is why GPS is good - really good - but not exact.