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Relativity of rotating system(s) 
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#1
Oct1009, 10:28 PM

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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 spacetime 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 spacetime makes sense to me. 


#2
Oct1109, 05:04 PM

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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 fiftyfifty. (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 


#3
Oct1109, 06:53 PM

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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
Oct1109, 07:01 PM

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Relativity of rotating system(s)
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
Oct1109, 07:36 PM

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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
Oct1109, 11:33 PM

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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 speedup due to increased altitude is almost twice the slowdown due to the equatorial velocity. Clocks will run faster at the equator than at the poles. 


#7
Oct1209, 02:54 AM

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http://www.physics.ucla.edu/demoweb/...urved_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 SchwarzschildMertic: http://en.wikipedia.org/wiki/Schwarzschild_metric with the metric for rotating frames of reference: http://books.google.de/books?id=DH7j...metric&f=false 


#8
Oct1209, 03:38 AM

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It's not a coincidence that potential energy goes with time dilatation. 


#9
Oct1209, 06:26 AM

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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 :o), 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
Oct1209, 04:50 PM

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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
Oct1209, 05:04 PM

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#12
Oct1209, 05:24 PM

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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
Oct1209, 06:02 PM

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#14
Oct1209, 06:42 PM

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


#15
Oct1209, 07:10 PM

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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
Oct1209, 07:14 PM

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#17
Oct1209, 07:22 PM

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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. 


#18
Oct1209, 10:30 PM

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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 boundsee below for comments on this.) Let me now rewrite 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 this link.) 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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