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Question about calculating time dilation of clocks at sea level

  1. Aug 22, 2010 #1
    I am trying to understand how relativity explains that clocks all over the globe count at about the same rate at sea level. One of the papers that I have looked at is "The global positioning system, relativity, and extraterrestrial navigation" at http://www.nist.gov/customcf/get_pdf.cfm?pub_id=904814. The most common equation for the gravitational potential including the quadrupole term is given as eq.1.4 and the coordinate time increment for a clock is then given as eq. 1.5, see attached eqs.jpg.

    If I use these equations to compare two clocks at different latitudes using a spreadsheet, I find that they count the same regardless of latitude to within a fraction of a nanosecond per day. So far so good.
    However, if I attempt to back calculate the value of gravitational acceleration implied from equation 1.4, I get a value different than the actual measured value. This seems to be because the value for the gravitational potential in equation 1.4 is based on the distance r from a point mass (V= GM/r) rather than from an object like the earth where the mass is distributed over a large volume. The quadrupole term as far as I can tell corrects only for the earth's oblateness.
    So here is where I see a problem - based on the V from equation 1.4, we would get an implied gravitational acceleration at the north pole of 9.854 m/s2, whereas the actual measured value of g at the north pole is 9.832 m/s2. If you attempt to substitute the real values for g or GM/r into equation 1.5, you end up with a clock at the north pole countering slower than a clock at the equator by 92.59 ns per day, which is inconsistent with the data. So how does this calculation account for the fact that the earth is not a point mass? Thanks.
     

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  3. Aug 22, 2010 #2

    bcrowell

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    I think there must be two separate issues here.

    (1) If you look near the end of section 1 in the Ashby paper, they discuss two different potentials, one for the nonrotating frame (equation 1.4) and one for the rotating frame, which differs from it by adding a term involving [itex]\omega[/itex]. If you use 1.4 to calculate g, you'll get an acceleration relative to the fixed stars, as measured by an observer at the earth's surface who is at rest relative to the fixed stars. Terrestrial experiments actually measure g relative to the surface of the earth, so you would want to use the potential expressed in the rotating frame in order to reproduce that.

    (2) The discrepancy between 9.854 m/s2 and 9.832 m/s2 can't be explained by effect #1, which vanishes at the poles. I don't know how big effects like local geology are in computing g, but I could easily imagine that they would be this big. Note that although the GR calculations are finding times to incredibly high precision, the relativistic effects are such tiny corrections that one does not need a fantastically good model of the earth's field to get them to the desired precision. This is presumably why they can get away with ignoring geology.
     
    Last edited: Aug 22, 2010
  4. Aug 22, 2010 #3

    bcrowell

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    If your goal here is to check whether it's true that there is no relativistic time dilation between the equator and the poles, then I don't think the method you're using to check it will work. In general relativity, the potential is actually *defined* in terms of the relativistic time dilation. If there were a difference in clock rates between the open sea at the equator and the (summer-time) open sea at the north pole, then there are only three possibilities: (1) water is being blocked from flowing downhill (from high V to low V) as it should between these two locations, (2) general relativity is incorrect, or (3) the earth's field doesn't, to a good enough approximation, meet the technical requirement (a stationary field) that is needed in GR in order to define a gravitational potential.
     
  5. Aug 22, 2010 #4
    plenum:
    I'm pretty sure aspects of this have been discussed before...try some searchs if you don't get answers you want.....in past discussions one thing that I recall discussed was the diameter of the earth being larger at the equator than the poles, so "sea level" is a bit of a vague term. Also, mass density variations all over the planet distort the local passage of time.

    I did not look at your source reference to see how these might/might not be accounted for.
     
  6. Aug 22, 2010 #5
    Thanks for your response - I am assuming that it is correct that clocks all count at the same rate across the globe. I am just trying to understand how the relativistic terms all add up to give this result. Yes you are right, there is the centripetal term due to the earth's rotation, it is not in eq.1.4 but it is in equation 1.5 when the actual time increment is calculated. I did try adding that term in to the gravitational potential in one place in my spreadsheet, as you say, it makes no difference at the pole, but if you add it to the equatorial V equation then you get 9.821 m/s^2 as compared to 9.804 m/s^2 for V/r^2 whereas the actual g at the equator is 9.781 m/s^2 (makes it worse). I can't subtract it, since this centripetal term is basically Einstein's velocity effect term v^2/2c^2 and must make time count slower rather than faster. I want to understand this more so that I can figure out some issues around the Hafele and Keating experiment - there was an article in wireless world that got me thinking about this that I am still trying to make heads or tails of... see: http://www.wbabin.net/historical/murray13.pdf [Broken]
     
    Last edited by a moderator: May 4, 2017
  7. Aug 22, 2010 #6
    Naty1 - thanks, I did check some other posts, I think maybe I am stuck at a deeper level - I already have taken into account the earth oblateness with the quadrupole term and the rotation term, I actually get the correct answer to within 60 meters of the reference ellipsoid which matches the claims I have seen in other papers that it is good to within 100 meters. I am just concerned about the mismatch on the gravitational accelerations - there probably is a good explanation for this but I haven't been able to find any mention of it.
     
  8. Aug 22, 2010 #7

    bcrowell

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    Plenum1, is there something wrong with explanation (2) in my post #2?

    [EDIT] OK, I checked, and it turns out that geology gives variations on the order of 3x10-4 m/s2, which is much smaller than the 2x10^-2 m/s2 discrepancy you found.

    The effect of the centrifugal term is to decrease the observed value of g, not increase it. The ground is falling out from under you, so it takes a falling object longer to hit the ground. To an observer in the rotating frame, this is the same as reducing g. If you subtract rather than adding, then it sounds like you're bang on.
     
    Last edited: Aug 22, 2010
  9. Aug 22, 2010 #8
    bcromwell-
    Sorry, should have replied about that geology comment - I don't think that geology is the source of the difference in g's, since the calculation based on equation 1.5 gives an excellent correlation with the positional height of the reference ellipsoid to within a maximum of about 60 meters at 45 degrees latitude. The problem is more with how this result can be obtained so precisely using an equation that assumes the earth is a point mass. I am wondering if it has something to do with the quadrupole potential constant J. If you look at the discussion on page 140 of the link below:
    http://books.google.ca/books?id=0a8...=onepage&q="quadrupole " earth oblate&f=false
    you will find that J is determined by assuming that the earth at sea level is an equipotential surface and then back-calculating to find a constant J that makes equation 1.5 give the correct answer. I don't know if J has ever been determined experimentally. It just seems strange to me that an equation using V~=GM/r and g ~= GM/r^2 would give the right answer for the surface of the earth, when the actual calculation of g on earth requires something more like:
    g = 9.80616*(1-0.00259*COS(RADIANS(2*latitude in degrees))*(1-0.000000314*height)
    which is only good to about 1000km height as well.
    See: http://joseph-bartlo.net/supp/geohdiff.htm for an example.
    So I am still puzzled... : )
     
  10. Aug 22, 2010 #9

    bcrowell

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    Are we straight now about the sign error at the equator? Are you sure there are no similar sign errors in your calculation at the pole? If you still can't figure out what's wrong at the pole, please post the numbers you're using for M, J2, a1, and r at the pole, and I'll see what I get from them.
     
  11. Aug 22, 2010 #10
    bcrowell-
    Okay - had to check over the signs in my calculation. Yes, you would think the centrifugal force should affect g that way - but as far as I can tell that's not the way that Ashby figures it in his equation - he puts it in as a centripetal force instead, pointing in the opposite direction. If you look at his equation 1.6, the gravitational, quadrupole and centripetal terms all have the same sign and add to 6.96928E-10 at the equator (which is exactly the result in my spreadsheet). This is the factor by which a clock at the equator is counting more slowly than a clock at a much more distant r. If I were to reverse the sign in the spreadsheet and treat this as a factor decreasing the gravitational potential, then the g at the equator would be almost right, but at the pole it would still be way off, and worse still, at the equator, the clock now reads fast by a whopping 207.86 ns per day (which amounts to 2X the centripetal amount). The equation doesn't balance (time difference between two clocks = 0) unless it adds to the gravitational potential. The e-book link also seems to treat it as the same sign at the equator. Your right, I would have thought to do it the other way too, but if I do then I don't get the same answers that they cite in the paper! It is very confusing.
    Here are my values:
    M = 5.9790 E 24
    j2 = 0.00108368
    a1 = 6378137 metres
    r (pole) = 6356752 metres
    Thanks
     
  12. Aug 22, 2010 #11

    bcrowell

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    If I differentiate Ashby's expression for V in the nonrotating frame, I get [itex]g=GM(1/r^2-3kP/r^2[/itex], where [itex]k=J_2a_1^2[/itex] and [itex]P=3(\cos^2\theta-1)/2[/itex].

    IMO the centrifugal term is 0 at the pole and [itex]-\omega^2 r[/itex] at the equator. This agrees in sign with Ashby's 1.5 (centrifugal term opposite in sign to the oblateness term). Note that Ashby's 1.5 is missing a parenthesis somewhere.

    The value of M you were using was different from the one I found on Wikipedia, and when I used yours, I didn't get good agreement with the observed values of g. When I changed it to the WP value, I got good agreement. When I calculate GM using WP's value of M, I get 3.9869*10^14 N.m2/kg, which agrees to 4 sig figs with Ashby's value.

    The details of the calculation are given below. The results, in units of m/s2, are:

    pole: theory=9.834, exp=9.832[1]
    equator: theory=9.783, exp=9.789[1], 9.780[2]

    The experimental values are from [1] http://en.wikipedia.org/wiki/Gravity_of_Earth and [2] http://en.wikipedia.org/wiki/Earth . I don't know what's up with the discrepancy between the equatorial values given in the two WP articles. The discrepancies between theory and experiment are bigger than geological variations, but I'm not sure they indicate anything other than the limited precision of this model.

    ---- rintintin ~ $ calc -x -e "M=5.9736 10^24 kg; k=(0.00108368*6378137^2)(1 m2); r=6356800 m; theta=0; P=(3(cos theta)^2-1)/2; w=2pi/(24*3600 s); o=GM(-3kP/r^4); ; g=GM/r^2+o" setting e to 1.60217648999999*10^-19 C
    setting G to 6.67427999999999*10^-11 N.m2/kg2
    setting k to 8.9875517873682*10^9 N.m2/C2
    setting kB to 1.38064999999999*10^-23 J
    setting c to 2.99792458*10^8 m/s
    setting h to 6.62606895999999*10^-34 J.s
    setting hbar to 1.05457162825177*10^-34 J.s
    M = 5.97360000000003*10^24 kg
    k = 4.40847868422846*10^10 m2
    r = 6.3568*10^6 m
    theta = 0
    P = 1
    w = 7.27220521664304*10^-5 s-1
    o = -0.0322921060222458 N/kg
    g = 9.83421573044428 N/kg
    ---- rintintin ~ $ calc -x -e "M=5.9736 10^24 kg; k=(0.00108368*6378137^2)(1 m2); r=6378100 m; theta=pi/2; P=(3(cos theta)^2-1)/2; w=2pi/(24*3600 s); o=GM(-3kP/r^4); c=-w^2r; g=GM/r^2+o+c"
    setting e to 1.60217648999999*10^-19 C
    setting G to 6.67427999999999*10^-11 N.m2/kg2
    setting k to 8.9875517873682*10^9 N.m2/C2
    setting kB to 1.38064999999999*10^-23 J
    setting c to 2.99792458*10^8 m/s
    setting h to 6.62606895999999*10^-34 J.s
    setting hbar to 1.05457162825177*10^-34 J.s
    M = 5.97360000000003*10^24 kg
    k = 4.40847868422846*10^10 m2
    r = 6.3781*10^6 m
    theta = 1.5707963267949
    P = -0.5
    w = 7.27220521664304*10^-5 s-1
    o = 0.0159314486627943 N/kg
    c = -0.0337305618948195 m/s2
    g = 9.78291931811831 m/s2
     
    Last edited by a moderator: Apr 25, 2017
  13. Aug 23, 2010 #12
    bcrowell,
    okay - that was somewhat miraculous - I went through your calculation - thank you for spending so much time on it! I can see the difference between your method and mine is primarily an extra 3 in the quadrupole term and a change of sign in the centripetal term- did you use calculus to get the extra 3? By your method:
    V = GM(1 - 3*J*a^2*(3*cos^2(theta)-1)/(2*r^2) + (-w^2*r) is that correct?
    Thanks.
     
  14. Aug 23, 2010 #13

    bcrowell

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    Yes, g=dV/dr (at the equator and the poles, where g is purely radial).

    No, the integral of r is r2/2, not r.
     
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