Relativity of rotating system(s)

In summary, according to general relativity, the equatorial bulge of the Earth is an equilibrium shape that is shaped by the combined effects of gravitational time dilation and velocity time dilation.
  • #1
S.R.Wilton
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I was just playing around with thoughts while laying in bed, and I need confirmation or rejection of a random idea that popped up in my head. I have a mild idea of relativity from various readings. I'm an undergraduate student studying engineering physics, so I have a decent background on Newtonian mechanics and basic quantum mechanical principles, but not so much of relativistic effects. I always thought of relativity and the speed of light in a strictly linear sense because it was straightforward (haha, pun intended >_<), but today was the first day I thought from a different perspective.

Do the atoms on the surface of Earth decay slower relative to the atoms in the core of the Earth due to the velocity difference?

If so, wouldn't that mean there is a "gradient of time" that exists on the planet? (i.e, aquatic animals on the bottom of the ocean age more quickly relative to animals on the surface of the planet; Other animals/matter move through time in between the speeds of everything on the outside of the planet and everything toward the center.)

If so, wouldn't that mean that our entire galaxy rotating about it's center has a space-time gradient in which all matter on the outside edge of the galaxy is aging slower relative to the matter on the inside of the galaxy toward the center?

If not, why not?

Maybe I'm just tired and not thinking clearly.Edit: Back awake again, just realized I completely neglected the gravitational effects of being at the center of the Earth as opposed to being on the surface of earth. Now I'm just confused. Matter at the center feels (approximately) the same amount of gravitational attraction from all sides of the planet as opposed to being drawn toward the center of mass of the planet (yeah? no?)... but it's still attracted to the sun and everything else anyway... so... I just realized I don't know anything about general relativity at all. How does all this factor into relativity? How would all this factor into the rate at which time moves relative to us? I'm just confusing myself.

Hopefully my confusion will lul me to sleep, and I can have dreams space-time makes sense to me.
 
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  • #2
S.R.Wilton said:
Do the atoms on the surface of Earth decay slower relative to the atoms in the core of the Earth due to the velocity difference?

As you remark later on in your posting, in such a case the overall relativistic effect has both a gravitational nature and a velocity difference nature.

There is actually a more interesting scenario: if you have atomic clocks on the poles, and on the equator, which at some point in time have been synchronized, will they remain in sync?

In 1905 Einstein wrote they wouldn't. The 1905 special relativity predicted a velocity time dilation. Clocks on the equator are circumnavigating the Earth's axis, and clocks on the poles are stationary with respect to the Earth's axis.

But later the Equivalence principle came into view. There is an equatorial bulge, and clocks on the poles are closer to the Earth's geometrical center; they are deeper in the gravitational well.

The Earth's equatorial bulge is an equilibrium shape. Over its entire lifetime the Earth's rotation rate has decreased, and its equatorial bulge has readjusted accordingly.

The counterpart of that equilibrium in terms of General Relativity is that in equilibrium state the same amount of proper time elapses everywhere.

Matter tends to move towards a region where a larger amount of proper time elapses. If the amount of proper time at the poles would differ from the amount of proper time at the equator, then the shape of the Earth will deform until equilbrium state is reached.

Everywhere on Earth clocks located at sealevel count the same amount of lapse of proper time. That is: at every point on the surface of the Earth the contributions of velocity time dilation and gravitational time dilation add up to the same amount. (If they would not add up to the same amount then the principle of equivalence would be violated.)

So on completion of General Relativity it turned out that the 1905 prediction was wrong. Clocks all over the Earth (at sea level) count the same amount of proper time. Actually, it's a bit more complicated than that. Spacetime curvature involves both space and time. The path of light through spacetime is affected about fifty-fifty. (In his very first exploration of a theory that implements the Equivalence principle, in 1907, Einstein predicted half the light bending of what the correct theory predicted.)

For things moving slower than light the contribution of space curvature is accordingly smaller. For motion of planets a theory that would fail to incorporate space curvature would only fail to predict Mercury's orbital precession.

Cleonis
 
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  • #3
Woah, that's pretty awesome. Thanks for the info! I really didn't know or understand anything about relativity at all until yesterday. I'm still having trouble conceptually grasping everything, but everything you said seems logical. What do you mean by "proper time" though?
 
  • #4
S.R.Wilton said:
What do you mean by "proper time" though?

I suppose that expression may sound as if I just invented it. But it's a standard expression in relativistic physics. Think of the word 'property'. For any traveller, the traveller's proper time is his own time, the amount of time that is measured by a clock that is co-moving with that traveller.

For instance, the satellites of the GPS system have clocks onboard that are so accurate that relativistic effects are clearly noticable. The 'proper time' of those satellites is the time as measured by those onboard clocks.

Cleonis
 
  • #5
Cleonis said:
The Earth's equatorial bulge is an equilibrium shape. Over its entire lifetime the Earth's rotation rate has decreased, and its equatorial bulge has readjusted accordingly.

The counterpart of that equilibrium in terms of General Relativity is that in equilibrium state the same amount of proper time elapses everywhere.
Is there any simple way to explain why this is? I don't understand why the equivalence principle should imply this about the equilibrium shape of a rotating planet.
Cleonis said:
Matter tends to move towards a region where a larger amount of proper time elapses. If the amount of proper time at the poles would differ from the amount of proper time at the equator, then the shape of the Earth will deform until equilbrium state is reached.
Why would matter tend to move towards a region with greater proper time? I suppose the inertia of matter makes it "want" to move on a geodesic path which is where proper time is really maximized, but no point on the surface of the Earth is moving on a geodesic. Why wouldn't the equilibrium shape an object assumes depend on the nature of whatever non-gravitational forces (in this case electromagnetism) are preventing the particles from moving on geodesics? Can the equilibrium shape be defined as a minimum of potential or some other quantity which would depend on the non-gravitational force? If the electromagnetic force between atoms worked differently--if it obeyed an inverse-cube law or something--I wonder if it would it still be true that a solid object would naturally assume a shape where clocks on the surface all tick at the same rate.

If these kinds of questions aren't easy to answer without doing a complicated mathematical analysis, don't worry about it, I was just curious.
 
  • #6
A quick calculation shows that the proper time rates are *not* the same at the equator and the poles. Define the following:

c = speed of light = 3.00 x 10^8 m/s

v = equatorial velocity at Earth's surface = 465 m/s

M = mass of Earth = 5.97 x 10^24 kg

G = Newton's constant = 6.673 x 10^-11

r_eq = equatorial radius of Earth = 6.378 x 10^6 m

r_po = polar radius of Earth = 6.357 x 10^6 m

Then we have

[tex]\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex]

[tex]\phi_{eq} = \sqrt{1 - \frac{2 G M}{c^2 r_{eq}}}[/tex]

[tex]\phi_{po} = \sqrt{1 - \frac{2 G M}{c^2 r_{po}}}[/tex]

and the ratio

[tex]\frac{\phi_{eq}}{\phi_{po}}[/tex]

is what we should compare to [itex]\gamma[/itex] to see whether the effects on proper time rate of increased altitude and velocity cancel out. When I plug in numbers, I get:

[tex]\gamma = 1 + 1.2 * 10^{-12}[/tex]

[tex]\frac{\phi_{eq}}{\phi_{po}} = 1 + 2.3 * 10^{-12}[/tex]

So the speed-up due to increased altitude is almost twice the slow-down due to the equatorial velocity. Clocks will run faster at the equator than at the poles.
 
  • #7
JesseM said:
Why would matter tend to move towards a region with greater proper time? I suppose the inertia of matter makes it "want" to move on a geodesic path which is where proper time is really maximized,
This is a model of how geodesics deviate towards greater gravitational time dilation:
http://www.physics.ucla.edu/demoweb...alence_and_general_relativity/curved_time.gif
But you could also think of space time having a density gradient. And the geodesics behave like light rays in a medium with varying optical density: they deviate towards the denser region.

More Generally on the topic of gravity and rotation: I guess you would have to combine the Schwarzschild-Mertic:
http://en.wikipedia.org/wiki/Schwarzschild_metric
with the metric for rotating frames of reference:
http://books.google.de/books?id=DH7...ge&q=rotating frame space time metric&f=false
 
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  • #8
Is there any simple way to explain why this is? I don't understand why the equivalence principle should imply this about the equilibrium shape of a rotating planet.
Gravity is a fictitious force just like centrifugal force. They are on equal footing, except that the latter is easier to transform away. In a rotating frame, you add both potentials to get the total time dilatation. Fictitious force potentials are the weak field approximation of GR when you can express everything as a change in the time-time component only.
It's not a coincidence that potential energy goes with time dilatation.
 
  • #9
I've been wondering a few things for a while and this looks like the place to find some answers and more questions, so here's another:

An object moving relative to another at content velocity has equal claims to it being stationary and the other moving (at least in otherwise empty space).

Nothing (nothing in our three brane) can (truly) exceeded light speed as (not because) the velocity along ANY (maybe all, but at least the 3 extended) spatial dimension is shared with the time dimension (if it exists :eek:), thus the closer you get to light speed the slower time travels relative to the rest of the universe).

So if an object, a sphere, is rotating at close to light speed, and moving through space at close to light speed what would be the effects of time dilation relative to a stationary object.

Thanks
 
  • #10
PeterDonis said:
Clocks will run faster at the equator than at the poles.

Hi Peter,

I need to check: have you taken the Earth's equatorial bulge into account?

What I mean by that is the following: in the case of an oblate spheroid the center of gravitational attraction does not coincide with the geometrical center.

Here, with 'center of gravitational attraction' I mean the following: the point where all of the Earth's mass would have to be concentrated, so that it would exert exactly the same gravitational force as the entire Earth does.

The Earth's equatorial bulge is such that the equator is about 20 kilometers further away from the Earth geometrical center than the poles.
In the case of the Earth, with its equatorial bulge: for an object located at the equator, the center of gravitational attraction does not lie on the Earth's axis, but about 10 kilometers away from it.

(Actually, I have that figure of 10 kilometers on authority. I would love to be able to carry out the integration to compute the location of the Earth's gravitational center (as a function of latitudinal position) but my math ability isn't up to that task.)

Cleonis
 
  • #11
Cleonis said:
Hi Peter,

I need to check: have you taken the Earth's equatorial bulge into account?

Of course I did; look at the different values I used for the equatorial vs. the polar radius of the Earth.

Cleonis said:
What I mean by that is the following: in the case of an oblate spheroid the center of gravitational attraction does not coincide with the geometrical center.

Here, with 'center of gravitational attraction' I mean the following: the point where all of the Earth's mass would have to be concentrated, so that it would exert exactly the same gravitational force as the entire Earth does.

The Earth's equatorial bulge is such that the equator is about 20 kilometers further away from the Earth geometrical center than the poles.
In the case of the Earth, with its equatorial bulge: for an object located at the equator, the center of gravitational attraction does not lie on the Earth's axis, but about 10 kilometers away from it.

This is all wrong. An oblate spheroid is still symmetrical about one axis; in the case of the Earth that's the axis that runs through the North and South poles. The gravitational center of the Earth is the point on that axis that's equidistant between the poles. (This is all assuming, of course, that the Earth is a perfect oblate spheroid, which it isn't; but the variations are too small to matter here.) Also, the location of that center is not observer-dependent; it's at the same point for all observers, whether they're on the equator, at the poles, or orbiting far out in space.
 
  • #12
PeterDonis said:
An oblate spheroid is still symmetrical about one axis; in the case of the Earth that's the axis that runs through the North and South poles. The gravitational center of the Earth is the point on that axis that's equidistant between the poles. (This is all assuming, of course, that the Earth is a perfect oblate spheroid, which it isn't; but the variations are too small to matter here.) Also, the location of that center is not observer-dependent; it's at the same point for all observers, whether they're on the equator, at the poles, or orbiting far out in space.

Peter,

your comment above is very problematic.

The sphere is the only shape with the property that the center of gravitational attraction coincides with the geometrical center.

In the case of the Earth the fact that the center of gravitational attraction does not coincide with the geometrical center is what gives rise to the precession of the equinoxes. The precession of the equinoxes is a case of gyroscopic precession.

Let me discuss the example of a pole in a gravitational field.
Imagine a long pole in a gravitational field, aligned with the gradient in the field. Consider two halves of the pole; the first one closest to the source of the gravitational field, and the second one furthest away from the source. Evaluate the total gravitational force upon each section. Since gravity falls off with the square of the distance the half section closest to the source will be subject to stronger gravity than the half section furthest away from the source. The center of mass of the pole, however, will be at the geometrical center. Therefore in a gravitational field a pole will be subject to a torque.

The Earth's gyroscopic precession is due to the fact that the Earth is subject to a torque, arising from the Sun's gravity and the Moon's gravity.

Cleonis
 
  • #13
Cleonis said:
Peter,

your comment above is very problematic.

The sphere is the only shape with the property that the center of gravitational attraction coincides with the geometrical center. ...

Not with the definition of "center of gravitational attraction" you've given. The "source point" of the Newtonian gravitational force of an object is the geometric center for any shape that has an axis of symmetry. For shapes that aren't spheres, there are additional multipole moments that come into play for things like precession, as you note, but those don't change the Newtonian force exerted on a small test object like a person standing on the Earth's surface, and that is the only factor that comes into play for determining the rate of lapse of proper time, which was the OP's original question.

Cleonis said:
Let me discuss the example of a pole in a gravitational field.

Here the pole isn't a small test object--it has enough extension to see the gradient in the gravitational field, so of course it's going to see effects that a small test object wouldn't see. But again, this is irrelevant to the OP's original question.
 
  • #14
PeterDonis said:
The "source point" of the Newtonian gravitational force of an object is the geometric center for any shape that has an axis of symmetry. For shapes that aren't spheres, there are additional multipole moments that come into play for things like precession, as you note, but those don't change the Newtonian force exerted on a small test object like a person standing on the Earth's surface, [...]

This issue will require further discussion, but it no longer fits/suits this relativistic physics thread. Tomorrow I will start a new thread in 'Classical Physics', where I will submit this issue to the forum. (As a name for that thread I think 'Center of gravitational attraction' is a good candidate, but I'm undecided yet.)

Needless to say, I'm convinced I made no errors.

Cleonis
 
  • #15
Hmm... looking at various web pages discussing quadrupole moments, I'm no longer sure I was correct that they have *no* effect on the force on a small test particle. However, I haven't been able to find any definite formulas. Does anyone know of any online references?
 
  • #16
Cleonis said:
This issue will require further discussion, but it no longer fits/suits this relativistic physics thread. Tomorrow I will start a new thread in 'Classical Physics', where I will submit this issue to the forum

I agree that this particular question belongs better in Classical Physics; however, the effects on proper time lapse still belong here, IMHO.
 
  • #17
Cleonis said:
Matter tends to move towards a region where a larger amount of proper time elapses.
I think you may have this the wrong way round, if by "a region where a larger amount of proper time elapses", you mean a region where proper time is faster or a region where the least time dilation occurs.

For example imagine a mine shaft at one of the poles with clocks at the top and bottom of the shaft. The upper clock will be running faster than the lower clock and observers at the top and bottom of the shaft would both agree this was the case. The lower observer would see more tick pulses from the upper clock compared to his local clock and vice versa. Since matter would fall from the top of the mine towards the bottom, it seems that saying matter tends to move towards a region where a lesser amount of proper time elapses, would be more accurate. Talking in terms of coordinate time might be even better, because technically the rate of proper time is always one second per second.

Now imagine a very rigid perfectly spherical spinning Earth. The clock at equator would be running slower at than the clock at the pole because they both experience the same amount of gravitational time dilation and the equatorial clock experiences an additonal amount of time dilation due to its velocity. The velocity component subtracts from the gravitational component to make the effective gravitational potential at the equator lower than the gravitational potential at the poles. Now if liquid is added to the surface of this rigid spinning sphere at one of the poles, it will move from the higher effective gravitational potential to the lower effective gravitational potential at the solid surface of the equator. If enough liquid is added the effective gravitational potential will be the same everywhere at the surface of the liquid (sea level) creating the classic bulged shape of a rotating planet. Clocks anywhere on the liquid surface will run at the same rate because they are all at the same effective gravitational potential.

S.R.Wilton said:
If so, wouldn't that mean there is a "gradient of time" that exists on the planet? (i.e, aquatic animals on the bottom of the ocean age more quickly relative to animals on the surface of the planet;

Taking the statement by PeterDonnis "So the speed-up due to increased altitude is almost twice the slow-down due to the equatorial velocity." at face value, it follows that animals at the bottom of ocean (at the poles or at the equator) age slower relative to animals on the surface.

S.R.Wilton said:
Now I'm just confused. Matter at the center feels (approximately) the same amount of gravitational attraction from all sides of the planet as opposed to being drawn toward the center of mass of the planet (yeah? no?)...

The important thing to remember in these sort of considerations is that gravitational time dilation is more closely related to the gravitational potential, than to the attraction or force of gravity. For example a clock at the centre of a Moon sized object will be running faster than a clock at the centre of an Earth sized object, because although the force of gravity is zero in both cases, the gravitational potential at the centre of the Moon sized object is greater than the gravitational potential at the centre of the Earth sized object.
 
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  • #18
After some further noodling with this, I find that I made some errors in my previous posts. The effect of Earth's oblateness does make enough of a difference to matter, as Cleonis says. Here's a modified calculation, properly accounting for the Earth's oblateness and consequent quadrupole moment.

I'll use the same definitions of variables as above, plus the following additional ones:

J_2 = Earth's dimensionless quadrupole moment = approximately 1.35 * 10^-3. (This is an upper bound of 2/5 of the Earth's oblateness of 1/297; the upper bound is 2/5 of that because 2/5 is the correct coefficient for the moment of inertia of a spheroid. I haven't been able to find a more accurate value, but the real value should be smaller than this upper bound--see below for comments on this.)

Let me now re-write things in terms of the Newtonian gravitational potential, [itex]\phi[/itex]. (I apologize for changing the use of the variable [itex]\phi[/itex] from my previous post, but the use I'm giving here is a standard one.) What I was using previously was the value of [itex]\phi[/itex] for a spherically symmetric source, namely

[tex]\phi = - \frac{G M}{r}[/tex]

However, for an oblate spheroid, like the Earth, the potential gains an additional term, and looks like this:

[tex]\phi = - \frac{G M}{r} + \frac{G M r_{eq}^2}{2 r^3} J_2 \left( 3 cos^2 \theta - 1 \right)[/tex]

where [itex]\theta[/itex] is zero at the poles and [itex]\pi / 2[/itex] at the equator. (The formula comes from the book Classical Mechanics, by Kibble & Berkshire, which can be found on Google Books using http://books.google.com/books?id=0a8dk0eDxgEC&pg=PA140&lpg=PA140".) Note that the effect will be to make the potential *less* negative at the poles and *more* negative at the equator.

For a test object at rest in the field (such as one at the poles), the proper time lapse rate is given by

[tex]\frac{d \tau}{dt} = T = \sqrt{1 + \frac{2 \phi}{c^2}}[/tex]

However, for a test object moving in the field (which an observer on the equator is, since he's rotating with the Earth), the lapse rate is

[tex]\frac{d \tau}{dt} = T = \sqrt{1 + \frac{2 \phi}{c^2} - \frac{v^2}{c^2}}[/tex]

(This is actually an approximate formula; for very weak fields and small velocities, it's more than good enough.)

So the two values we need to calculate are:

[tex]T_{pole} = \sqrt{1 + \frac{2 \phi_{pole}}{c^2}} = \sqrt{1 - \frac{2 G M}{c^2 r_{pole}} \left( 1 - \frac{r_{eq}^2}{r_{pole}^2} J_2
\right)}[/tex]

[tex]T_{eq} = \sqrt{1 + \frac{2 \phi_{eq}}{c^2} - \frac{v^2}{c^2}} = \sqrt{1 - \frac{2 G M}{c^2 r_{eq}} \left( 1 + \frac{1}{2} J_2 \right) - \frac{v^2}{c^2}}[/tex]

Again plugging in numbers, I get:

T_pole = 1 - 6.963 * 10^-10

T_eq = 1 - 6.966 * 10^-10

Since T_pole is closer to 1, clocks will run faster at the *poles* than at the equator, by about 3 parts in 10^-13, *if* we use the value of J_2 given above. However, if the value of J_2 is somewhat smaller (as it probably is), T_pole and T_eq will in fact be equal (since the effect of *decreasing* J_2 will be to move T_eq *closer* to 1 and T_pole *further* from 1). And the actual value of J_2 *is* smaller than the upper bound I used above. Kibble & Berkshire, in their textbook, discuss how the "equipotential surface" of the Earth can be determined by including the effect of centrifugal force; their calculation is basically equivalent to the one I did above, where I included the term in v^2 in T_eq. (That v^2 term can be considered as a "potential due to centrifugal force" if you want to look at it that way.) Their contention is basically the same as what Cleonis was asserting in an earlier post: the surface of the Earth *is* an "equipotential surface" when the effects of centrifugal force are included. The value of J_2 they calculate that would support that conclusion is about 1.1 * 10^-3, which is indeed somewhat smaller than the one I used above, and is a perfectly reasonable value.

So, Cleonis, mea culpa: you were correct that, assuming the real value of the Earth's J_2 is the one that Kibble & Berkshire calculate, the proper time rates at the poles and the equator *will* be the same.
 
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  • #19
J2 is actually 0.00108263, see http://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html" [Broken]. It's indeed hard to find if you search for English keywords.
The equipotential surface is called the Geoid, and it deviates some 100 m from sea surface. It is used as a reference surface for global time (TAI, I think), because all clocks on it run at the same rate.
 
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  • #20
PeterDonis said:
for an oblate spheroid, like the Earth, the potential gains an additional term, and looks like this:

[tex]\phi = - \frac{G M}{r} + \frac{G M r_{eq}^2}{2 r^3} J_2 \left( 3 cos^2 \theta - 1 \right)[/tex]

where [itex]\theta[/itex] is zero at the poles and [itex]\pi / 2[/itex] at the equator. (The formula comes from the book Classical Mechanics, by Kibble & Berkshire, which can be found on Google Books using http://books.google.com/books?id=0a8dk0eDxgEC&pg=PA140&lpg=PA140".)

As I recall a more exhaustive formula also contains terms for J_4, J_6, etc., but the contributions from those terms are much smaller.


PeterDonis said:
the surface of the Earth *is* an "equipotential surface" when the effects of centrifugal force are included.

Certainly the Earth is an equilibrium shape. It's interesting to examine the nature of that equilbrium.

An object located on either of the poles is at the lowest gravitational potential of any location on the surface. Objects located on the equator have more gravitational potential energy than at the poles, since they are further away from the center of gravitational attraction, and they have kinetic energy (as they are circumnavitating the Earth's axis).

Imagine an ice-planet, the same size as the Earth, same rotation rate, same oblateness, with a perfectly smooth and perfectly frictionless surface. Place an object on the surface, co-rotating with the ice-planet. It will stay on that latitude.

- Imagine what would happen if there would be too much oblateness. Then at every latitude there would be more gravitational potential energy than kinetic energy, and released objects would tend to slide to the nearest pole.
- Imagine what would happen if there would be insufficient oblateness. Then at every latitude there would be less gravitational potential energy than kinetic energy, and released objects would tend to slide away from the central axis of rotation, towards the equator.

With exactly the right oblateness there is equilibrium. Along the surface the oblateness sets up a gradient in gravitational potential energy, from the equator down to each of the poles. Gradient in gravitatinal potential means there is a centripetal force, in this case precisely the required force for objects to remain co-rotating with the Earth.

Relativistic perspective
To a first approximation there is a correspondence between gravitational potential and rate of proper time. (Connecting rate of proper time with energy potential is only approximately correct, as spacetime curvature involves space too. The slower the velocity, the smaller the contribution from curvature of space.)

In terms of GR there is equilibrium at the surface when the rate of proper time is the same all over the surface. Over it's entire lifetime the Earth's rotation rate has decreased, and the Earth's oblateness has decreased in step with that. So if we would have found different rates of proper time for clocks around the world then that would have constituted a violation of GR.

Cleonis
 
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  • #21
Cleonis said:
Matter tends to move towards a region where a larger amount of proper time elapses.

kev said:
I think you may have this the wrong way round,

Yeah, you're right, I goofed.
Deeper in a gravitational well a smaller amount of proper time elapses (as compared to higher up in the gravitational well).

I should have replayed in my mind the visualization of clocks on a rotating disk. From a co-rotating point of view there appears to be a gravitational force away from the central axis. Hence close to the central axis is "up", and further away from the central axis is "deeper down in the gravitational well". But I didn't do the full visualization, and I goofed.

Annoyingly I'm too late to place an addendum in the original posting, so I can't correct the error there.

Cleonis
 
  • #22
Ich said:
J2 is actually 0.00108263, see http://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html" [Broken].

Thanks! With this value of J_2, both T_pole and T_eq are the same to four significant figures:

T_pole = T_eq = 1 - 6.971 * 10^-10

Cleonis said:
As I recall a more exhaustive formula also contains terms for J_4, J_6, etc., but the contributions from those terms are much smaller.

Yes. With the above value of J_2, discrepancies between T_pole and T_eq show up at about 1 part in 10^-14; this may be because of the higher order terms, or it may be because of the error bars around some of the input values (particularly the gravitational constant G).

Cleonis said:
An object located on either of the poles is at the lowest gravitational potential of any location on the surface. Objects located on the equator have more gravitational potential energy than at the poles, since they are further away from the center of gravitational attraction, and they have kinetic energy (as they are circumnavitating the Earth's axis).

Viewed from a non-rotating frame, that's true. Viewed from a frame that's rotating with the Earth, however, the effect of centrifugal force appears as an additional "potential" whose value is just right to compensate for the increase in height as you move from the poles to the equator. The observed effects are the same either way.

Cleonis said:
So if we would have found different rates of proper time for clocks around the world then that would have constituted a violation of GR.

There is one additional assumption that's needed to deduce that the Earth's surface must be an equilibrium surface in this sense: we have to assume that the Earth's material acts sufficiently like a fluid to be able to equilibrate this way. That certainly seems to be a valid assumption for the Earth, but it might not be for objects with a lot less angular momentum and/or a lot less mass (asteroids, for example, are known to come in all sorts of weird shapes).
 
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  • #23
PeterDonis said:
There is one additional assumption that's needed to deduce that the Earth's surface must be an equilibrium surface in this sense: we have to assume that the Earth's material acts sufficiently like a fluid to be able to equilibrate this way. That certainly seems to be a valid assumption for the Earth, but it might not be for objects with a lot less angular momentum and/or a lot less mass (asteroids, for example, are known to come in all sorts of weird shapes).

If an Earth sized sphere was made of steel or titamium, it would be sufficiently rigid to prevent it deforming into the ideal oblate shape of spinning fluid massive body. If such a body was spinning, I wonder if it would still form a time equilibrium surface?

This possibility came to mind when it occurred to me that such an object would have stress in its surface at the poles and pressure in the surface at the equator, and stress and pressure contribute to the gravitational potential in GR. I wonder if anybody here can calculate that?
 
  • #24
kev said:
If an Earth sized sphere was made of steel or titamium, it would be sufficiently rigid to prevent it deforming into the ideal oblate shape of spinning fluid massive body. If such a body was spinning, I wonder if it would still form a time equilibrium surface?

If the body really was rigid enough to resist deformation, then no, it would not form a time equilibrium surface, by definition. The time equilibrium surface is the surface that the deformation would produce if not resisted, so obviously if the deformation is resisted, it won't produce that surface. Put another way, if deformation would change the surface if it weren't resisted, then the existing surface (when the deformation is resisted) can't be an equilibrium surface.

kev said:
This possibility came to mind when it occurred to me that such an object would have stress in its surface at the poles and pressure in the surface at the equator, and stress and pressure contribute to the gravitational potential in GR. I wonder if anybody here can calculate that?

By the method we've been using for determining the time equilibrium surface, only the Earth's mass and its rate of rotation are relevant. Internal stresses within the Earth will contribute to its externally measured mass, of course, but since we already know the externally measured mass, we don't need to go into those details. In essence, we're idealizing the Earth's exterior spacetime geometry as Schwarzschild spacetime with a central mass M equal to the mass of the Earth (so the proper time lapse rate of observers at rest is determined purely by the metric coefficient g_tt, which depends only on the radial coordinate r--thus, the difference in r between poles and equator comes into play here), and we're treating observers rotating with the Earth as moving within that geometry (which adds an additional time dilation factor dependent on their velocity, which is determined by the Earth's rate of rotation). If we wanted to be really pedantic, we could use Kerr spacetime instead, which would include the frame-dragging effects of the Earth's angular momentum, but those are too small to matter for determining the time equilibrium surface.

It is true, though, that to have a valid solution, we should be able to match an interior solution of the EFE along the boundary at the Earth's surface to the exterior solution we've been using. For that match to be valid, the surface should be an equipotential surface using the metric of the interior solution as well as that of the exterior solution. For the standard metric for relativistic stars, which is the one used in conjunction with the http://en.wikipedia.org/wiki/Tolman%E2%80%93Oppenheimer%E2%80%93Volkoff_equation" [Broken], the metric coefficients automatically match up at the boundary for a non-rotating star, since the metric looks just like the Schwarzschild metric except that m(r), the mass as a function of the radial coordinate r, appears instead of the total mass M. At the boundary, of course, they are the same, if the star is non-rotating (so that the boundary is at the same radial coordinate r for all values of the angular coordinates).

The question then is whether having the star be rotating, so that the radial coordinate r of the boundary depends on the "latitude" [itex]\theta[/itex], introduces any problem. I'm not sure, but I'm inclined to think not, for the weak field case we are considering. In other words, for the weak field case, I would think that for any point on the boundary, the interior metric coefficient should be governed by the total mass M for an object at rest--i.e., not rotating with the body--but using the actual radial coordinate r at that latitude [itex]\theta[/itex], just as we did for the exterior metric above. The extra correction for an object rotating with the body would then obviously be the same, since the velocity is the same, so that the time equilibrium surface would be the same for the interior metric as for the exterior. However, I haven't done a detailed calculation to confirm that this is true.

(P.S.--After writing the above, I found http://adsabs.harvard.edu/full/1967ApJ...147..317H", by Hartle & Sharp, which derives a metric for a rotating star. Unless I've misread something, it appears to support my conjecture above; their g_tt metric coefficient is just two terms, one corresponding to the Schwarzschild g_tt using the actual radial coordinate r at a given point on the boundary, and the other corresponding to the velocity correction for objects rotating with the star.)
 
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  • #25
Cleonis said:
An object located on either of the poles is at the lowest gravitational potential of any location on the surface. Objects located on the equator have more gravitational potential energy than at the poles, since they are further away from the center of gravitational attraction, and they have kinetic energy (as they are circumnavitating the Earth's axis).

PeterDonis said:
Viewed from a non-rotating frame, that's true. Viewed from a frame that's rotating with the Earth, however, the effect of centrifugal force appears as an additional "potential" whose value is just right to compensate for the increase in height as you move from the poles to the equator. The observed effects are the same either way.
I concur, but I also have an objection: the above statement is tautological.

Ultimately there is no such thing as 'viewing from the frame that is rotating with the Earth'. When the motion is mapped in a coordinate system that is co-rotating with the Earth, then the centrifugal term contains the angular velocity of the rotating coordinate system with respect to the inertial coordinate system. That is: using the rotating coordinate system hinges on relying on the inertial coordinate system as reference of motion.

Whether it's explicitly there or implicitly, the reference of motion is always the inertial coordinate system. Hence my claim that ultimately 'viewing from a rotating frame' is not viable.You refer to 'another way', in the phrasing 'The observed effects are the same either way'. Let me elaborate on what I think those two different ways are.

Taking the example of an observer in a pilot training centrifuge, pulling G's: the observer can interpret his sensation as an outward tug, which he may refer to as 'centrifugal force'. When he does that the observer is reversing cause and effect. What is actually happening is that a centripetal force is exerted on the observer, and acceleration gives rise to inertial effects, which are physically sensed by the observer.
If the observer reverses cause and effect he thinks the centrifugal force comes first, pressing him into the seat of the pilot training centrifuge. And as remarked, the observed effects are the same either way.

As we know, the centrifugal force interpretation does not encounter discrepancies since the physics of inertia is indeed invariant under reversal of the direction of time. Acceleration and deceleration are indistinguishable, and as expressed by the principle of relativity, acceleration and deceleration are one and the same thing.

<Addendum>
Correction: acceleration and deceleration being indistinguishable is not the right example here. Anyway, I think the key issue of reversal of cause and effect is clear enough.
</Addendum>

When someone refers to 'viewing from a rotating frame', then the shift of perspective is not just the 'inertial versus rotating', but there is also the element of reversing cause and effect.

Cleonis
 
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  • #26
Cleonis: I'm not sure that I disagree with you, but I'm not sure I'd put things quite the way you do. After all, we live our everyday lives in a frame rotating with the Earth, and we have no trouble relating our everyday motions to that frame. So let me re-state what I was saying in a little more detail:

Viewed from a frame rotating with the Earth, the "fixed stars" appear to be rotating about the Earth. What we call "Earth's angular velocity of rotation" from an inertial frame, we call the "angular velocity of the fixed stars" from the "Earth rotating" frame--we can measure this angular velocity by measuring the time it takes for particular stars to go from one meridian passage to the next. This rotation of the fixed stars gives rise to an extra gravitational potential which depends on the angular velocity at which we observe the fixed stars to rotate. So we can make predictions about what the equipotential surface of the Earth will be based on measurements that we can make entirely within the "Earth rotating" frame; there is no need to refer anything to an inertial frame.

Cleonis said:
When someone refers to 'viewing from a rotating frame', then the shift of perspective is not just the 'inertial versus rotating', but there is also the element of reversing cause and effect.

I agree that's true in the centrifuge example you gave; the actual force in that example is the force exerted by the seat of the centrifuge on the pilot, which acts inward, not outward. I'm not so sure it's true for the "Earth rotating" frame as I described it above.
 
  • #27
I'd suggest looking at http://adsabs.harvard.edu/abs/2005gr.qc...1034D

In this paper we show how the student can be led to an understanding of the connection between special relativity and general relativity by considering the time dilation effect of clocks placed on the surface of the Earth. This paper is written as a Socratic dialog between a lecturer Sam and a student Kim.

This is on arxiv, but the above URL shows that it's also been published in print (papers on arxiv that haven't been published in print haven't been peer reviewed and are best avoided).

The arxiv link is http://arxiv.org/abs/gr-qc/0501034
 
  • #28
PeterDonis said:
Viewed from a frame rotating with the Earth, the "fixed stars" appear to be rotating about the Earth. What we call "Earth's angular velocity of rotation" from an inertial frame, we call the "angular velocity of the fixed stars" from the "Earth rotating" frame--we can measure this angular velocity by measuring the time it takes for particular stars to go from one meridian passage to the next. This rotation of the fixed stars gives rise to an extra gravitational potential which depends on the angular velocity at which we observe the fixed stars to rotate. So we can make predictions about what the equipotential surface of the Earth will be based on measurements that we can make entirely within the "Earth rotating" frame; there is no need to refer anything to an inertial frame.

I think the underlying question is:
"Is it possible at all to formulate any theory of motion without referring to inertial frames?" I will argue that the answer to that is 'no'.

I want to make a comparison. For that I'd like to recount the story of Roemer's hypothesis. By the year 1670 and later, observations of Jupiter's largest moons were accurate enough to predict events like Jupiter's moon's eclipses with a precision of minutes. Assuming that the period of the Moons themselves were constant, then the observations suggested that the Moons were consistently too early during the time of the year that Earth is closest to Jupiter, and too late when Earth is farthest away. Roemer suggested this was an apparent effect, arising from a finite speed of light.

Until then it was not known whether the speed of light was infinite or finite. Roemer's hypothesis gave weight to the notion of a finite speed of light.

Philosophy of science
Poincaré, in an essay on philosophy of science, used the story of Roemer's hypothesis as an example. Poincaré discussed that at the time scientist could have opted to make adjustments to their theory of motion, so as to preserve a notion that light moves instantly. Finding such adjustments would have been difficult, but not impossible. With enough mathematical ingenuity adjustments could have been devised of the theory of motion, to account for the anomalies in the orbits of Jupiter's moons.

Of course, that would have resulted in an ugly theory of motion. For the "adjusted" theory of motion would be doing a job that it shouldn't have to do in the first place. The "adjusted" theory of motion would not just be theory of motion, but also a cosmetic device, to hide the fineteness of the speed of light.
The significance, argued Poincaré, is the illustration that in science we regard that theory as superior that is among the candidates the simplest in how the world is conceptualized. Other candidates can be devised, with effort and ingenuity, but those strained theories are cumbersome, so much so that scientists reject them intuitively.


PeterDonis said:
Viewed from a frame rotating with the Earth, [...] we call the "angular velocity of the fixed stars" This rotation of the fixed stars gives rise to an extra gravitational potential which depends on the angular velocity at which we observe the fixed stars to rotate. [...]

Well, I do grant that with enough effort and mathematical ingenuity it's possible to come up with a scheme in which rotation of the assembly of stars relative to the Earth is construed to give rise to the gravitational potential that you refer to.
But it seems to me that such a scheme does not constitute any gain. It will just be a strained scheme, arduously doing a job that is naturally done by the concept of inertial coordinate systems. Yes, I do grant that such a scheme can be made self-consistent, but I reject it on the grounds that it's doing things the hard way, while the natural way is right there, and has been there all along. It's not on equal par. It does not stand side by side with the inertial coordinate system concept as an interexchangable interpretation.

There is a world of things to add to the above discussion, but I have to stop this posting somewhere. Attempts to cast GR as a theory that relativises rotation are dead ends, I think. At the same time, I've seen it argued well that GR relatives acceleration - but that is another story.

Cleonis
 
  • #29
Cleonis said:
Well, I do grant that with enough effort and mathematical ingenuity it's possible to come up with a scheme in which rotation of the assembly of stars relative to the Earth is construed to give rise to the gravitational potential that you refer to.

...

Attempts to cast GR as a theory that relativises rotation are dead ends, I think.

It's already known that GR predicts frame-dragging effects inside a rotating shell of matter. The http://en.wikipedia.org/wiki/Frame-dragging" [Broken] discusses it in the section on the Lense-Thirring effect inside a rotating shell. These frame-dragging effects will appear as "centrifugal potential" to observers that are not rotating with the shell.
 
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  • #30
Cleonis said:
Well, I do grant that with enough effort and mathematical ingenuity it's possible to come up with a scheme in which rotation of the assembly of stars relative to the Earth is construed to give rise to the gravitational potential that you refer to.
...
Attempts to cast GR as a theory that relativises rotation are dead ends, I think. At the same time, I've seen it argued well that GR relatives acceleration - but that is another story.

Here is a thought experiment that might help address the issue. Consider a hypothetical rotating Earth with a single geostationary sattelite in these 3 scenarios.

1) The Earth is surrounded by a shell of distant stars. It may be possible that the Earth could be considered stationary and the sattelite is suspended by a peculiar gravitational field generated by the rotating mass of the the distant stars possibly by the Lense-Thirring effect metioned by Peter Donnis in the previous poat.

2) The Earth is orbited around its equator by a single small and very distant star. In this scenario the mass of the single small distant star orbiting the non rotating Earth can not account for the equatorial bulge of the Earth and nor can it account for the artificial stationary sattelite suspended above the Earth.

3) A single small distant star is stationary above the North pole of the Earth. Again the single star can not account for effects mentioned in scenario 2.

Scenarios 2 and 3 are very different and yet the observered effects on the equatorial bulge of the Earth and the suspended sattelite are the same. This is very difficult to account for if we insist or rotation being relavistic. It can also be noted that in scenario 2 the single distant star orbiting the non rotating Earth is greatly exceeding the speed of of light. We might explain this away by stating the distant orbiting star is stationary with respect to spacetime that is co-rotating with the star. We have now defined an absolute form of motion because it is defined relative to the spacetime which is now being treated as a substance rather than just a vacuum. This spacetime rotating about the Earth generates the gravitational effects that account for the equatorial bulge of the Earth and also account for how the geo stationary artificial sattelite is suspended above the non rotating Earth. We have now defined the spacetime vacuum as a substance having mass or energy that generates its own gravitational field indepently of any classical massive bodies. By defining the distant star as stationary with respect to the co-rotating spacetime vacuum (to avoid the distant star exceeding the speed of light) it is only consistent that all motion is defined relative to the spacetime vacuum. We have now gone full circle, because by this new definiton of how motion is defined, we can no longer consider the Earth to be stationary because it is rotating with respect to the spacetime vacuum. The attempt to consider the Earth as non rotating becomes self defeating.
 
  • #31
PeterDonis said:
It's already known that GR predicts frame-dragging effects inside a rotating shell of matter. The http://en.wikipedia.org/wiki/Frame-dragging" [Broken] discusses it in the section on the Lense-Thirring effect inside a rotating shell. These frame-dragging effects will appear as "centrifugal potential" to observers that are not rotating with the shell.

About the thought experiment of frame-dragging effects inside a rotating spherical shell:

It's only purpose is to explore a specific implication of GR. I don't know any details, I just assume the inside space of the spherical shell must be at least the size of a planet, and the thickness of the shell must be at least the diameter of the inside space.

The rotating spherical shell scenerio is physically impossible; such a shell would collapse under its own self-gravitation. Presumably this physical impossibility is tolerated because it doesn't impact the thought experiment's actual purpose.

In 1917, in the course of corresponding with Einstein, Thirring commenced calculations to obtain an approximation for the effects inside a rotating spherical shell. Frame dragging effects were found, but they fell far short of complete frame dragging inside. The result was an unwelcome surprise to Einstein, for at the time his expectation was that GR was an implementation of a strong version of Mach's principle. To pass as an expression of Mach's principle the inside frame dragging effects would have had to be far stronger than what was found. (See also from http://www.tc.umn.edu/~janss011/" [Broken])

In the same period, around 1918, other evidence surfaced that GR isn't an implementation of Mach's principle (and in particular not an implementation of strongly demanding versions of Mach's principle.)

(Of course this doesn't diminish the importance of the concept of frame dragging in its own right. I avidly followed the evaluations of the Gravity Probe B experiment as they trickled out.)

If GR isn't an implementation of a strong version of Mach's principle, then is it perhaps an implementation of a comparatively weak version of Mach's principle?
For instance, what if the distribution of inertial mass in the universe can be thought of as bringing forth the spacetime itself? Then the existence of inertial mass and GR-spacetime can be thought of as mutually dependent; GR-spacetime allowing existence of inertial mass, and inertial mass bringing forth GR-spacetime.

It has been pointed out that in order to obtain the Schwarzschild solution from the Einstein Field Equations the following condition is imposed: that towards spatial infinity the solution must approach asymptotically to Minkowski spacetime.
It has been argued (and I find it compelling) that if GR would be an implementation of a Mass/spacetime mutual dependence, then the GR equations would not allow a solution like the Schwarzschild solution. The Schwarzschild solution describes a infinite universe with a single lump of inertial mass, and inertia everywhere.

I find this reasoning compelling. It appears to me that GR is not an implementation of a weak version of Mach's principle either. In itself this does not exclude versions of Mach's principle, it just means that GR doesn't seem to be an instrument that can help in assessing whether our Universe is in some form a Machian universe.

Cleonis

<Addendum>
Reading the article by Herbert Pfister http://philsci-archive.pitt.edu/archive/00002681/" [Broken] made me realize that while it's tempting to write about what I find compelling, I'm just out of my depth.

Pfister writes that in 1913 Einstein had performed similar evaluations as Thirring's, on the basis of what today is referred to as the 'Entwurf theory', a version of GR that Einstein in 1913 regarded as the finished GR. So Einstein could not have been surprised by Thirring's results.
</Addendum>
 
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  • #32
Cleonis said:
It has been argued (and I find it compelling) that if GR would be an implementation of a Mass/spacetime mutual dependence, then the GR equations would not allow a solution like the Schwarzschild solution. The Schwarzschild solution describes a infinite universe with a single lump of inertial mass, and inertia everywhere.

It might be worth mentioning that the Kerr solution for a black hole is distinctly non Machian too, because it defines rotation of the black hole without any reference to any other mass in the universe. The black hole is simply spinning with respect to spacetime.
 
  • #33
Cleonis said:
The rotating spherical shell scenerio is physically impossible; such a shell would collapse under its own self-gravitation.

Wouldn't that depend on the specific characteristics of the shell?

Cleonis said:
It has been pointed out that in order to obtain the Schwarzschild solution from the Einstein Field Equations the following condition is imposed: that towards spatial infinity the solution must approach asymptotically to Minkowski spacetime.
It has been argued (and I find it compelling) that if GR would be an implementation of a Mass/spacetime mutual dependence, then the GR equations would not allow a solution like the Schwarzschild solution. The Schwarzschild solution describes a infinite universe with a single lump of inertial mass, and inertia everywhere.

If you take the solution in isolation, yes. But there's nothing stopping you from patching a portion of a Schwarzschild solution--say one with a central mass equal to that of the solar system, and taking the r coordinate out to a couple of light-years, which would be some 10^16 times the Schwarzschild radius--into a tiny portion of a global solution something like an FRW model. The boundary condition of asymptotic flatness for the Schwarzschild solution would just be a result of the fact that, on a distance scale of a light-year or two, the spacetime geometry of the universe as a whole *is* flat, because that scale is much too small to see any curvature effects. (Similar remarks would apply to the Kerr solution.)

Of course, if the universe as a whole is not closed, then we've just pushed the problem of what determines the boundary conditions out to the universe as a whole. That's why some physicists like models in which the universe is closed, so that there is no boundary condition required. (Hartle and Hawking's "no boundary" proposal in cosmology goes a step further by saying that the universe is closed in the time dimension, roughly speaking, as well as in the space dimensions, so there's no boundary condition required in time either.)
 
  • #34
S.R.Wilton said:
Do the atoms on the surface of Earth decay slower relative to the atoms in the core of the Earth due to the velocity difference?

I'm not sure this particular question from the original post was ever adequately addressed.

Granted we are probably more concerned about the gravitational rather than the velocity difference, I'm still wondering whether this effect is significant enough to effect radioactive decay in the Earth's interior detectable over billions of years.

Despite any simplified calculations I've attempted, I'm not sure exactly what sort of gravitational potential a rock half-way-to-the-center-of-the-earth might be experiencing.

I'm particularly interested in long lived isotopic systems (like Lu-Hf, U-Pb, Rb-Sr, etc.) within the Mantle (so from around 7-70 km deep down to about 2855 km below the surface). Since in Geologic Time we're dealing with billions of years, I was wondering if these systems would need to be calibrated against the effect of relativity on their "apparent" rates of decay at depth.

For example, "mantle plumes" (which rise like lava lamp bubbles from the core-mantle boundary) are thought to be the source of hotspot volcanism (like that in Hawaii). These magmas have a distinctive isotopic signature, which is very different from magmas sourced from the upper mantle (like those of the mid-Atlantic Ridge). This is mostly due to the continued extraction of partial melts from the upper mantle, etc. due to plate tectonics over the past few billion years.

But assumptions about the lower mantle's composition are based off of measurements from undifferentiated meteorites (which were decaying far from a pronounced gravitational field) and on models from known decay rates combined with the age of the Earth (as calculated at the surface). Thus, we expect an undisturbed mantle to have a certain Lu-Hf ratio today by taking the starting ratio (from chondritic meteorites) and calculating what it ought to be today (using decay rates and time).

Specifically, would the dilation of time at ~3,000 km deep within the Earth be significant enough to affect model estimates of what today's isotopic ratios ought to be (in a closed system, and assuming meteorites do give a reliable starting ratio for 4.5 billion years ago)?

Thanks in advance for any help!
 
  • #35
hoiland said:
Specifically, would the dilation of time at ~3,000 km deep within the Earth be significant enough to affect model estimates of what today's isotopic ratios ought to be (in a closed system, and assuming meteorites do give a reliable starting ratio for 4.5 billion years ago)?
No.

Clocks at the center of the Earth do tick slower than do clocks at sea level. A tiny, tiny bit slower.

The scale factor is 1 - 6.969290134×10-10. Suppose a pair of ideal clocks were synchronized 4.5 billion years, with one kept at sea level for the next 4.5 billion years and the other magically placed and kept at the center of the Earth for the next 4.5 billion years. Today those two clocks would differ a grand total of a bit over 3 years. That of course is orders of magnitude smaller than the uncertainty in that 4.5 billion year figure.
 

1. What is the relativity of rotating systems?

The relativity of rotating systems, also known as the principle of relativity, is a fundamental concept in physics that states that the laws of physics should be the same for all observers in uniform motion. This means that the laws of physics should not change depending on the observer's frame of reference, even if they are rotating.

2. How does the relativity of rotating systems relate to Einstein's theory of relativity?

The relativity of rotating systems is a key component of Einstein's theory of relativity. It is an extension of the principle of relativity, which states that the laws of physics should be the same for all observers in uniform motion. Einstein's theory of relativity also includes the principle of equivalence, which states that the effects of gravity are equivalent to the effects of acceleration. This means that the laws of physics should be the same for observers in a rotating reference frame and observers in a non-rotating reference frame experiencing the same acceleration.

3. What is the Coriolis effect and how does it relate to the relativity of rotating systems?

The Coriolis effect is a phenomenon that occurs in rotating systems, where objects appear to be deflected from their expected path due to the rotation of the system. This effect is a direct result of the relativity of rotating systems, as the Coriolis force only appears when observing objects from a rotating reference frame. This effect is important in understanding the behavior of objects in systems such as the Earth's atmosphere and ocean currents.

4. Can the relativity of rotating systems be observed in everyday life?

Yes, the relativity of rotating systems can be observed in everyday life. For example, the Coriolis effect can be observed in the rotation of hurricanes and the direction of water draining in a bathtub. Additionally, the effects of relativity can be observed in the behavior of gyroscopes and pendulums, which are used in many everyday devices such as smartphones and navigation systems.

5. How does the relativity of rotating systems impact our understanding of the universe?

The relativity of rotating systems has greatly impacted our understanding of the universe, particularly in the field of cosmology. Einstein's theory of relativity, which includes the relativity of rotating systems, has helped us understand the behavior of objects at high speeds and in strong gravitational fields. This has led to advancements in our understanding of the structure of the universe, such as the theory of black holes and the expansion of the universe.

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