Re: [math-fun] atomic clocks & gravitational time dilation
On 2014-02-11 08:08, Warren D Smith wrote:
On 2/9/2014 12:40 PM, Wouter Meeussen wrote:
in http://www.nist.gov/pml/div688/2013_1_17_newera_atomicclocks_3.cfm a atomic clock, using Y-Sr ?lattice?, is said to achieve 3E-18 accuracy & stability. This would allow to detect time dilation caused by a mere 4 cm altitude difference.
If guestimating correctly, this would be equivalent to the gravitational effect of a 19 kg mass at a horizontal distance of 10 cm.
--The time-dilating factor is (to first order) sqrt( 1 + 2*P/c^2 ) where P is the (negative) gravitational potential energy per unit mass of you, G=Newton gravit. constant, c=light speed.
Using P = -G * (19 kg) / (10 cm) one finds about 1 - 1.4*10^(-25). So you'd got that wrong by factor > million.
To make the time-dilate factor be 1-3*10^(-18), which is the accuracy claimed for the clock, we would want to lower the clock in Earth gravity field by height h where h = 3*10^(-18) * c^2 / g = 2.8 cm.
You might want to consider the fact that the ground under your feet rises and falls twice a day by about 50 cm due to the moon ("solid tide").
Is not the Sun almost equally culpable?
You don't normally notice this happening.
In 1961 I toured a lab with a gyro recording this while sitting on a concrete column down to bedrock. (Despite the difficulty of touring while sitting on a column.)
This would seem to suggest that any "true" time standard would have to be produced by a space-based clock, Earth-based clocks are inherently too inaccurate? Or software adjustment? Or what? The very meaning of "time" depends where you are on the Earth and it would seem we are reaching the limits of what one might call "usable accuracy." Further clock improvements would seem futile.
What if you want to record the close passage of a 19 kg child? --rwg PS, has anyone ever found a correct explanation of tides in a middle or high school text?
On Tue, Feb 11, 2014 at 2:41 PM, Bill Gosper <billgosper@gmail.com> wrote:
PS, has anyone ever found a correct explanation of tides in a middle or high school text?
http://www.ck12.org/earth-science/Tides/lesson/Tides-Basic/ -- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
This explanation from the text Mike sugggests is the same explanation I've long heard from reliable sources: ----- • Water on the side of Earth facing the moon is pulled hardest by the moon’s gravity. This causes a bulge of water on that side of Earth. That bulge is a high tide. • Earth itself is pulled harder by the moon’s gravity than is the ocean on the side of Earth opposite the moon. As a result, there is bulge of water on the opposite side of Earth. This creates another high tide. • With water bulging on two sides of Earth, there’s less water left in between. This creates low tides on the other two sides of the planet. ----- But I'm curious: * Are the tides on the sides of the earth nearest and farthest from the moon symmetrical? * If so, why (since the reasons given for those tides are different) ? --Dan On 2014-02-11, at 2:17 PM, Mike Stay wrote:
On Tue, Feb 11, 2014 at 2:41 PM, Bill Gosper <billgosper@gmail.com> wrote:
PS, has anyone ever found a correct explanation of tides in a middle or high school text?
http://www.ck12.org/earth-science/Tides/lesson/Tides-Basic/
-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
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Gravitational force goes down as r^2, so the effect on the sunward side should be slightly different from that on the other side. This is, of course, neglecting the interaction of the water with the irregular topography of the earth (cf Bay of Fundy) and rotational effects, which are larger. Actual tide tables are very local and wildly different in both amplitude and timing over a distance of tens of miles or less. Far from symmetrical. --R -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Dan Asimov Sent: Tuesday, February 11, 2014 5:35 PM To: math-fun Subject: Re: [math-fun] atomic clocks & gravitational time dilation This explanation from the text Mike sugggests is the same explanation I've long heard from reliable sources: ----- . Water on the side of Earth facing the moon is pulled hardest by the moon's gravity. This causes a bulge of water on that side of Earth. That bulge is a high tide. . Earth itself is pulled harder by the moon's gravity than is the ocean on the side of Earth opposite the moon. As a result, there is bulge of water on the opposite side of Earth. This creates another high tide. . With water bulging on two sides of Earth, there's less water left in between. This creates low tides on the other two sides of the planet. ----- But I'm curious: * Are the tides on the sides of the earth nearest and farthest from the moon symmetrical? * If so, why (since the reasons given for those tides are different) ? --Dan On 2014-02-11, at 2:17 PM, Mike Stay wrote:
On Tue, Feb 11, 2014 at 2:41 PM, Bill Gosper <billgosper@gmail.com> wrote:
PS, has anyone ever found a correct explanation of tides in a middle or high school text?
http://www.ck12.org/earth-science/Tides/lesson/Tides-Basic/
-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
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My understanding is that there is a lag between the strongest gradient of the gravitational force and the high tides. Below is for Honolulu this coming Valentine's Day. The simple model would predict high tides at roughly noon and at midnight. Coastal features can also complicate things, I believe. Day High Low High Low Moon Sunrise Sunset Fri 14 3:59 AM HST / 1.94 ft 10:39 AM HST / 0.14 ft 4:05 PM HST / 1.09 ft 9:44 PM HST / 0.03 ft Full Moon 7:02 AM HST 6:29 PM HST BC -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Dan Asimov Sent: Tuesday, February 11, 2014 3:35 PM To: math-fun Subject: [EXTERNAL] Re: [math-fun] atomic clocks & gravitational time dilation This explanation from the text Mike sugggests is the same explanation I've long heard from reliable sources: ----- * Water on the side of Earth facing the moon is pulled hardest by the moon's gravity. This causes a bulge of water on that side of Earth. That bulge is a high tide. * Earth itself is pulled harder by the moon's gravity than is the ocean on the side of Earth opposite the moon. As a result, there is bulge of water on the opposite side of Earth. This creates another high tide. * With water bulging on two sides of Earth, there's less water left in between. This creates low tides on the other two sides of the planet. ----- But I'm curious: * Are the tides on the sides of the earth nearest and farthest from the moon symmetrical? * If so, why (since the reasons given for those tides are different) ? --Dan On 2014-02-11, at 2:17 PM, Mike Stay wrote:
On Tue, Feb 11, 2014 at 2:41 PM, Bill Gosper <billgosper@gmail.com> wrote:
PS, has anyone ever found a correct explanation of tides in a middle or high school text?
http://www.ck12.org/earth-science/Tides/lesson/Tides-Basic/
-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
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On 2014-02-11 14:34, Dan Asimov wrote:
This explanation from the text Mike sugggests is the same explanation I've long heard from reliable sources:
----- • Water on the side of Earth facing the moon is pulled hardest by the moon’s gravity. This causes a bulge of water on that side of Earth. That bulge is a high tide.
• Earth itself is pulled harder by the moon’s gravity than is the ocean on the side of Earth opposite the moon. As a result, there is bulge of water on the opposite side of Earth. This creates another high tide.
• With water bulging on two sides of Earth, there’s less water left in between. This creates low tides on the other two sides of the planet. -----
But I'm curious:
* Are the tides on the sides of the earth nearest and farthest from the moon symmetrical?
I assumed they were not, and asked this same question when we were camping at Steep Ravine, and thought I had circumstantial evidence that I was correct that they were asymmetrical (I thought higher tide closer to the moon). There were tide tables posted, and I notice that the variance between the two high tides per day varied by more than the corresponding tide varied from day to day. The high high-tide seems to vary from the low high-tide by more than the variation in high high-tides for 2 consecutive days. I guessed that the low high-tide was approximately the same as the antipodal tide at the same time as the high high-tide. But my friend (and my son) didn't accept my argument (that the low high-tide corresponded to/approximated the antipodal tide at the time of the high high-tide) at all, saying the tidal system was far more complicated than I was modelling it. In particular, they said that the sun's measurable influence on tides was enough to make my argument unconvincing. They were probably right. I had planned on finding tide tables from antipodal points on the earth; I never looked. I tried just now, and did not find anything useful, although I did come across a middle-school/high-school lesson (distinct from Mike Stay's) that seems far less bogus than anything I learned about tides when I was a kid: pumas.nasa.gov/files/01_25_11_1.pdf It implies that the 2 tides are asymmetric. I suspect there are more reasonable middle-school level explanations of tides than I expected (I'm pleasantly surprised). There's probably good coverage of other topics, too, so I should alter my expectations (and stop complaining). Oh well. My favorite aspect of this lesson, and the biggest distinction between this lesson and my own elementary school experience (long, long, ago) is exemplified by these two quotes: "By studying tide tables for various places, you can see that the behavior of lunar tides is not nearly as simple as the analogy and the physical/mathematical calculations in the DISCUSSION section demonstrate. Additional complications include: ..." "Technical note: The spring analogy should not be pushed too far ..." If I had teachers who could admit that (or be aware that their analogies had limitations), then I would have been much less frustrated as a kid, and forgiven them much more significant simplifications.
* If so, why (since the reasons given for those tides are different) ?
--Dan
On 2014-02-11, at 2:17 PM, Mike Stay wrote:
On Tue, Feb 11, 2014 at 2:41 PM, Bill Gosper <billgosper@gmail.com> wrote:
PS, has anyone ever found a correct explanation of tides in a middle or high school text?
http://www.ck12.org/earth-science/Tides/lesson/Tides-Basic/
-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
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I think this link is a typical incomplete and rather misleading explanation of tides. It doesn't even suggest the solid tides that earth experiences, much less hint at an explanation for them. Nor does it suggest or begin to explain the asymmetrical ocean tides. It's all gravitational gradient, experienced equally by solid earth and liquid ocean. Tidal forces stress solids and liquids equally. The reason ocean tides are much bigger than solid earth tides: the tidal gradient has a component normal to earth's surface (which lifts or lowers), and a component tangent to earth's surface (which causes water, but not the semi-rigid earth, to flow). If it were not for the tidal-induced water flows (away from the tidal "equator" and toward the tidal "poles"), ocean and earth tides would be almost indistinguishable and we wouldn't notice them. Re symmetry of near/far tides: IIRC, the ratio of solar : lunar tidal effects is about 4 : 9. Also, tidal effects are proportional to d^3, where d is the earth-moon (or earth-sun) distance. Given the much farther earth-sun distance, the 4/13 solar tidal near/far component is pretty close to symmetrical. The much close earth-moon distance makes a stronger tidal gradient, and (I think) is the biggest factor in asymmetrical daily tide. I get a back-of-envelope estimate of about 1.066 near vs. far for the 9/13 lunar tidal effect, using .25M miles as rough earth/moon distance and 4K miles as rough earth radius. I make no claim having a good explanation for K-12 level. And I welcome any clarification or correction to any of the above. Thanks, - Scott -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Dan Asimov Sent: Tuesday, February 11, 2014 2:35 PM To: math-fun Subject: Re: [math-fun] atomic clocks & gravitational time dilation This explanation from the text Mike sugggests is the same explanation I've long heard from reliable sources: ----- * Water on the side of Earth facing the moon is pulled hardest by the moon's gravity. This causes a bulge of water on that side of Earth. That bulge is a high tide. * Earth itself is pulled harder by the moon's gravity than is the ocean on the side of Earth opposite the moon. As a result, there is bulge of water on the opposite side of Earth. This creates another high tide. * With water bulging on two sides of Earth, there's less water left in between. This creates low tides on the other two sides of the planet. ----- But I'm curious: * Are the tides on the sides of the earth nearest and farthest from the moon symmetrical? * If so, why (since the reasons given for those tides are different) ? --Dan On 2014-02-11, at 2:17 PM, Mike Stay wrote:
On Tue, Feb 11, 2014 at 2:41 PM, Bill Gosper <billgosper@gmail.com> wrote:
PS, has anyone ever found a correct explanation of tides in a middle or high school text?
http://www.ck12.org/earth-science/Tides/lesson/Tides-Basic/
-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
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Also, tidal effects are proportional to d^3, where d is the earth-moon (or earth-sun) distance.
1/d^3 , perhaps? In layman's terms: there is a squeeze effect around the (tidal) equator, of the same order of magnitude as the stretch effect along the (tidal) poles; yet the former is routinely neglected in popular accounts purporting to explain tides. WFL On 2/11/14, Huddleston, Scott <scott.huddleston@intel.com> wrote:
I think this link is a typical incomplete and rather misleading explanation of tides. It doesn't even suggest the solid tides that earth experiences, much less hint at an explanation for them. Nor does it suggest or begin to explain the asymmetrical ocean tides.
It's all gravitational gradient, experienced equally by solid earth and liquid ocean. Tidal forces stress solids and liquids equally. The reason ocean tides are much bigger than solid earth tides: the tidal gradient has a component normal to earth's surface (which lifts or lowers), and a component tangent to earth's surface (which causes water, but not the semi-rigid earth, to flow). If it were not for the tidal-induced water flows (away from the tidal "equator" and toward the tidal "poles"), ocean and earth tides would be almost indistinguishable and we wouldn't notice them.
Re symmetry of near/far tides: IIRC, the ratio of solar : lunar tidal effects is about 4 : 9. Also, tidal effects are proportional to d^3, where d is the earth-moon (or earth-sun) distance. Given the much farther earth-sun distance, the 4/13 solar tidal near/far component is pretty close to symmetrical. The much close earth-moon distance makes a stronger tidal gradient, and (I think) is the biggest factor in asymmetrical daily tide. I get a back-of-envelope estimate of about 1.066 near vs. far for the 9/13 lunar tidal effect, using .25M miles as rough earth/moon distance and 4K miles as rough earth radius.
I make no claim having a good explanation for K-12 level. And I welcome any clarification or correction to any of the above.
Thanks, - Scott
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Dan Asimov Sent: Tuesday, February 11, 2014 2:35 PM To: math-fun Subject: Re: [math-fun] atomic clocks & gravitational time dilation
This explanation from the text Mike sugggests is the same explanation I've long heard from reliable sources:
----- * Water on the side of Earth facing the moon is pulled hardest by the moon's gravity. This causes a bulge of water on that side of Earth. That bulge is a high tide.
* Earth itself is pulled harder by the moon's gravity than is the ocean on the side of Earth opposite the moon. As a result, there is bulge of water on the opposite side of Earth. This creates another high tide.
* With water bulging on two sides of Earth, there's less water left in between. This creates low tides on the other two sides of the planet. -----
But I'm curious:
* Are the tides on the sides of the earth nearest and farthest from the moon symmetrical?
* If so, why (since the reasons given for those tides are different) ?
--Dan
On 2014-02-11, at 2:17 PM, Mike Stay wrote:
On Tue, Feb 11, 2014 at 2:41 PM, Bill Gosper <billgosper@gmail.com> wrote:
PS, has anyone ever found a correct explanation of tides in a middle or high school text?
http://www.ck12.org/earth-science/Tides/lesson/Tides-Basic/
-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
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participants (8)
-
Bill Gosper -
Cordwell, William R -
Dan Asimov -
Fred Lunnon -
Huddleston, Scott -
Michael Greenwald -
Mike Stay -
Richard E. Howard