asymmetric effect of SR

[bwanaaa]

Haefele and Keating flew an atomic clock from the west coast to the east coast in 1971 and then they flew it back. The east going clock lost 59 +- 10 ns and the west going clock gained 273 +- 7 ns. Why are the times not similar since the jet flew at roughly the same speed? If taking account of the jet stream, the speed variation is not 500%. And taking account Earth's rotational velocity, the speed delta of the plane cannot account for 500% change in time dilation/contraction due to SR (special relativity). What am I missing?

 

[Revolution 11]

Correct me if I am wrong, but they flew around the world twice, not just from the west to east coast and vice-versa. Would that explain the 500% difference?

The jet moved at the same speed relative to the jet (or the air around the jet). But to the Earth's perspective, the jet moved faster when going eastward, causing a loss in time.

Also, it seems although both planes were roughly at the same altitude in the experiment, the slight difference in altitude caused the westward plane to gain 35 ns throughout the flight. Experimentally, time dilation can currently be measured on a difference of 33 cm.

 

[bwanaaa]

you are correct. i shoud've wiki'd first
..................pred grav (GR)... pred (SR)............pred sum............obs
eastward.....144±14..............−184 ± 18..........−40 ± 23..........−59 ± 10
westward.....179±18...............96±10...............275±21..............273±7

 

[Revolution 11]

No problem. Hope you don't mind if I introduce a new question related to yours.

So the Earth is not a sphere due to its rotation. I get that. But how is the gravitational field equal at sea level at all latitudes then? How is the time dilation equal at sea level?

I figured that the reduced speed at the poles is compensated by the smaller radius of the Earth, but I don't think that is accurate. I looked at a few papers talking about this and they mention gravitational equipotentials and something about how fluids follow equipotentials (sea level) but there was a lot of math behind it and I have no background in physics.

 

[Ventanni]

The Earth's gravitational field is much more uniform that, say, that of the Moon's, but it's not equal at all levels. An individual will experience a slightly higher level of gravity at the top of a mountain than they will at sea level, which means the rate of time dilation will be slightly different.

I guess the question is, what were the flight paths of these flights?

 

[Revolution 11]

Wait, I thought that elevation determined gravitational strength so shouldn't the mountain have less force than at sea level? The mountain would have more mass underneath it but the distance from the center of Earth would be greater so gravity has a net decrease.

 

[bwanaaa]

yes, revolution 11, these sources agree with you
http://www.nytimes.com/2003/12/23/science/q-a-altitude-and-weight.html
http://wiki.answers.com/Q/Would_an_object_weigh_more_at_the_top_of_a_mountain_or_at_sea_level
http://answers.yahoo.com/question/index?qid=20080813143602AAbDeP2

but getting back to my SR asymmetry question.
after double thinking it, why should there be an asymmetry?
shouldnt the westward and eastward trips just give the same amount of time dilation-after all, the atomic clocks were travelling relative to the surface of the earth-who cares how he earth is rotating because the atmosphere in which the jet travels is also rotating.

so shouldnt the result be the same negative number regardless of the direction the jet travels, N,S,E,W?

 

[Revolution 11]

Ah that's a common mistake that I made as well. When you look at a solid (non-hollow?) object's gravity, the gravity acts as a single point source at the center of the object's mass.

In this case, your frame is wrong. The speed is relative to the Earth's core, not the surface. Eastward would be faster because you get the rotational speed to help as well.

 

[bwanaaa]

yes but the earth clock is not at the core, it's on the surface. Anyway, the negative and positive signs do not make sense. Consider what this kind of result this analysis would give to a plane flying to the north pole from the equator? The earth's rotation should not be germane.

We are considering the aircraft velocity relative to a fixed point on the earth's surface. And the same kinematic time dilation should be found regardless of the direction the plane travels.


BTW, here is an article that refutes the H&K paper claiming they doctored the results
http://www.anti-relativity.com/hafelekeatingdebunk.htm

 

[Revolution 11]

Hmm, I can't really explain the kinematic time dilation then. But there is plenty of evidence that the H&K results are valid, within a reasonable error range, and have been independently verified by multiple teams. BTW I would not ever trust any site that has "anti-relativity.com" in its name. Bias much?

EDIT: Actually, you can't use the surface as a calculation starting point because it is not a proper inertial frame. The center of the Earth is still the correct inertial frame since the surface clock is still moving. The Earth is moving as well but that does not apply until you make the Sun the inertial frame.

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/airtim.html

 

[bwanaaa]

yes I actually came across that reference before my third post and the sentence i do not understand is
"The problem encountered with measuring the difference between a surface clock and one on an aircraft is that neither location is really an inertial frame. If we take the center of the earth as an approximation to an inertial frame, then we can compute the difference between a surface clock and the aircraft clock. "

The need to establish a third frame of reference-the center of the earth-is confusing. Why is neither the surface clock's nor the airborne clock's frame of reference valid? Because the earth's surface is moving? But that's a major point of the einstein equivalence principle-that moving frames of reference have the same physics.

I understand the equation for the time difference calculation:
Ta-Ts=To[(2Rwv+v^2)/2c^2]
where Ta is airborne time, Ts is surface time, To is center of the earth time
R is earth radius, w is earth angular velocity, v is plane velocity, c is light speed
And then they go on to use the following approximation:
Ta-Ts=-Ts[(2Rwv+v^2)/2c^2]
The difference being they use surface time instead of center of the earth time. They state:
Note that the "earth center" time has been replaced by the surface time in this expression. This is a valid approximation in this case since the time difference is many orders of magnitude smaller than the time itself, and this allows us to model the difference between two measurable times.

They do not explain why they introduce a negative sign. But I think they chose to do that to make time dilation a positive number. Still, how can they conclude by using Ts when they started the whole analysis saying that surface time was not a proper inertial frame?

And another confusing aspect of using the center of the earth as the reference frame is if you change the flight plan a little bit. Suppose the plane flies along a line of longitutude instead of latitude? Since v is orthogonal to earth angular velocity, is it measured as 0? A flight to south pole is the same as a flight to north pole in terms of time dilation? I wonder if that expt has been done?

 

[MrDudeMan]

yes but the earth clock is not at the core, it's on the surface. Anyway, the negative and positive signs do not make sense. Consider what this kind of result this analysis would give to a plane flying to the north pole from the equator? The earth's rotation should not be germane.

We are considering the aircraft velocity relative to a fixed point on the earth's surface. And the same kinematic time dilation should be found regardless of the direction the plane travels.


BTW, here is an article that refutes the H&K paper claiming they doctored the results
http://www.anti-relativity.com/hafelekeatingdebunk.htm

Signed deltas make perfect sense. A positive delta would be an increase in time relative to the static frame of reference and a negative sign would be decrease in time. The delta is relative to the length of a second in the static, or original, frame of reference.

 

[bwanaaa]

This article answers my question:

http://www.conspiracyoflight.com/Hafele/HafeleKeating.html

Apparently, the reference clock should be at the earth's pole to approximate an inertial frame of reference. This is as close as we can get to a 'true' inertial observer. They say:
"We first define our non-rotating inertial observer in stationary space, who is stationary with the CMBr.(cosmic microwave background)"

The sagnac effect and ring lasers are explained here:
http://www.ringlaser.org.nz/content/about_us.php

So it seems that correct calculations of special relativity require a 'true' inertial frame of reference, certainly not one that is spinning/rotating/revolving. But since everything in the universe is spinning, that's hard to come by. Even if you had ring lasers on your spaceship to make sure you were not spinning with respect to the CMBr, you still wouldnt have a 'true' inertial frame. Space is curved so even when you think you are going in a straight line, you are not- you are moving in a giant, gentle circle.