Discover the Science Behind Age Differences at the Equator vs. Poles

[Quantum1332]

Since relativity says that the faster you go, the slower time goes, then you are younger at the equator, since the equator spins faster than the poles. Therefore you are younger at the equator than at/near the poles. Am I right? Even if it is a very minute amount.

 

[Haelfix]

Younger, relative to whom?

Everyone ages (in their own laboratory frame) at identical rates, but its how you synchronize your clocks where special relativity makes a difference. Moreover at the velocities we are talking about, the difference is of order .00001 seconds or somesuch over a lifetime.

 

[pervect]

Since relativity says that the faster you go, the slower time goes, then you are younger at the equator, since the equator spins faster than the poles. Therefore you are younger at the equator than at/near the poles. Am I right? Even if it is a very minute amount.

No, there are additional effects due to GR that cause this not to happen. Though you are indeed moving faster on the equator, you are also further from the center of the Earth, because of the Earth's equatorial bulge. Higher clocks tick faster than lower clocks, and this counteracts the effect due to the velocity.

It turns out that alll clocks on the geoid (roughly speaking, the Earth's surface at sea level) click at the same rate. There is a rather simple energy conservation argument that suggests why this should be. If you displace a fluid element from any point on the Earth to any other point on the geoid (i.e. at sea level), it will take no energy (ignoring very small effects like departure of the sea level from equilibrium, tides from the sun and moon, etc.)

Therfore, there will be no net redshift or blueshift if you transmit a photon from one point on the geoid to another point - because the photon, like the matter, will neither gain or lose energy.

Thus it is not a coincidence that the blueshift due to the higher altitude is canceled out by the redshift due to the velocity - energy conservation makes it happen this way.

For a technical reference, see
http://relativity.livingreviews.org/open?pubNo=lrr-2003-1&page=node3.html

Thus, the very useful result has emerged, that ideal clocks at rest on the geoid of the rotating Earth all beat at the same rate. This is reasonable since the Earth's surface is a gravitational equipotential surface in the rotating frame. (It is true for the actual geoid whereas I have constructed a model.) Considering clocks at two different latitudes, the one further north will be closer to the Earth's center because of the flattening - it will therefore be more redshifted. However, it is also closer to the axis of rotation, and going more slowly, so it suffers less second-order Doppler shift. The Earth's oblateness gives rise to an important quadrupole correction. This combination of effects cancels exactly on the reference surface.

 

[nikolai]

Since relativity says that the faster you go, the slower time goes, then you are younger at the equator, since the equator spins faster than the poles. Therefore you are younger at the equator than at/near the poles. Am I right? Even if it is a very minute amount.

The speed is the same, the inertia is the difference, so not really.

 

[Haelfix]

Actually I was under the impression that departure from spherical symmetry breaks the geoid surface symmetry, but whatever its a small point, the Earth isn't quite spherical or elliptical, it has a certain equatorial bulge. I haven't done the full calculation so I am not entirely sure about it.

 

[pervect]

Actually I was under the impression that departure from spherical symmetry breaks the geoid surface symmetry, but whatever its a small point, the Earth isn't quite spherical or elliptical, it has a certain equatorial bulge. I haven't done the full calculation so I am not entirely sure about it.

I'm not sure which symmetry you think is broken?

If the Earth were actually a ball of fluid in equilibrium (or a solid core with a fluid-covered surface), all clocks on that fluid surface would run at the same rate.

The Earth does not actually have the shape described above, but I won't get into details of the (small) errors. The actual shape of the Earth is quite close to this "idealized" shape.

The equatorial bulge caused by the Earth's rotation does not "mess up" the constancy of rate of clocks on the Earth surface - it's an intergal part of why clocks on the Earth's surface run at the same rate.

http://www.physicstoday.org/vol-58/iss-9/p12.html
and the previously mentioned
http://relativity.livingreviews.org/open?pubNo=lrr-2003-1&page=node3.html

both discuss this. Note that Einstein himself in his original SR paper, like the OP in this thread, predicted that clocks at the poles would run at a different rate than clocks on the equator. It wasn't until much later when Einstein developed GR that he realized that the clocks would run at the same rate at both locations.

The detailed argument gets rather technical, but the short version is easy to understand. If two observers can exchange light signals without any red or blue shift, they will infer that their clocks run at the same rate.

Because the Earth is an equipotential surface, we can infer that there will be no red or blue shift in transmitted signals.

[add]
Here is a slightly more technical argument that addresses the problem. Consider the rotating fluid-covered Earth as a static problem in GR (static because the metric coefficients do not vary with time).

The force of gravity in the local coordinate system will be a vector, given by $\nabla g_{00}$.

The fluid has the property that the fluid surface is always perpendicular to the local gravitational field if it is in equilibrium.

Therfore the equilibrium fluid has the property that the fluid surface is at a constant value of $g_{00}$.

 

[Haelfix]

"Because the Earth is an equipotential surface"

Yes I thought we were talking about departures from this...

 

[pervect]

The geoid is actually defined to be an equipotential surface AFAIK. The "mean sea level" idea is actually a popularization to describe the idea in less techical terms.http://www.google.com/search?&q=define:Geoid&sa=X&oi=glossary_definition&ct=title

# The particular equipotential surface that coincides with mean sea level and that may be imagined to extend through the continents. This surface is everywhere perpendicular to the force of gravity.
www.eurofix.tudelft.nl/glossary.htm[/URL]

# The hypothetical surface of the Earth that coincides with sea level everywhere.
[url]www.maps-gps-info.com/maps-gps-glossary.html[/url]

# The equipotential surface in the gravity field of the Earth that coincides with the undisturbed mean sea level extended continuously through the continents. The direction of gravity is perpendicular to the geoid at every point. The geoid is the reference surface for geodetic leveling (surveying) and some inertial navigation systems.
earth-info.nga.mil/GandG/coordsys/definitions.htm

# That equipotential surface (a surface of equal gravity potential) which most closely matches mean sea level. An equipotential surface is normal to the gravity vector at every point
[url]www.rbf.com/cgcc/glossary.htm[/url]
[/quote]