Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity
It`s just like the fly in the glass jar scenarmino : Just imagine a glass jar [with a lid] with a fly flying around inside the jar. The jar is being accelerated towards it`s inevitable demise when it hits the sun. Does the fly stay in the same position in the jar, or is it pushed towards the end of the jar furtheest from the sun. Now I don`t know the answer to this one..but I sure wouldn`t like to bee the fly. Affectionately yours,...Don Collie jnr. - Original Message - From: Dr Bruce Griffiths [EMAIL PROTECTED] To: Discussion of precise time and frequency measurement time-nuts@febo.com Sent: Wednesday, May 30, 2007 12:49 PM Subject: Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity Bill Beam wrote: Not true. Very simple experiments will show occupants of the satellite that they are in a non-inertial reference frame. (Release a few test masses about the cabin and you will observe that they move/accelerate for no apparent reason, unless the satellite is in free fall which you'll know soon enough,) The experimenter must conclude that the satellite is undergoing acceleration due to the influence of an attractive (gravitational) field. Except when released at rest with respect to the satellites centre of mass the test masses will both drift towards the satellites centre of mass. The outermost test mass will have too slow an orbital speed to remain at the position it was released and the innermost test mass will have too large an orbital speed to remain at the position at which it was released. Bruce ___ time-nuts mailing list time-nuts@febo.com https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts -- No virus found in this incoming message. Checked by AVG Free Edition. Version: 7.5.472 / Virus Database: 269.8.1/822 - Release Date: 5/28/2007 11:40 AM ___ time-nuts mailing list time-nuts@febo.com https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity
Ulrich, Yes, but your use of the term system instead of frame of reference confused me instead of helping. It sounded like you were talking about different cases, instead of the same case under different viewpoints. I did not understand what you meant. That's OK, I believe I got it now. Thanks Didier KO4BB -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Ulrich Bangert Sent: Wednesday, May 30, 2007 12:37 AM To: 'Discussion of precise time and frequency measurement' Subject: Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity Didier, my first posting to that topic contained the semtences: Centrifugal forces are so called fictitious forces which are only observed from within accelerated systems. Normal physics is done in inertial systems. Is that not pretty much what you have found out after all? 73s and my best regards Ulrich, DF6JB -Ursprüngliche Nachricht- Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Im Auftrag von Didier Juges Gesendet: Mittwoch, 30. Mai 2007 02:35 An: Discussion of precise time and frequency measurement Betreff: Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity James, Where were you all week-end? Your explanations are so clear, it makes sense now. Thank you very much. I understand now that centrifugal forces are necessary to explain the behavior of objects when an accelerating frame of reference is used, but not necessary (actually counter-productive) to explain the behavior of the same objects when an inertial frame of reference is used. That solves my problem and the apparent contradiction that sometimes the centrifugal force is necessary and sometimes not, because I did not appreciate the effects of changing the frame of reference. Thanks a lot again. I had no idea time-nuts would drive me to brush-up on physics :-) Didier KO4BB James Maynard [EMAIL PROTECTED] wrote: The reason that the frame of reference matters is that gravity is indistinguishable from acceleration. (This is an assumption that Einstein made when deriving his general theory of relativity. It seems to work.) An inertial frame of reference is a non-accelerating frame of reference. In an inertial frame of reference, Newton's laws of motion work -- if you use Newton's gravitational relationship, that the gravitational force (weight) that each of two bodies exerts on the other is proportional to both their masses, and inversely proportional to the square of the distance between them. In an accelerating frame of reference (either linear acceleration, or rotational acceleration, or both) additional forces, technically called fictitious forces, must be introduced in order to explain the motions of bodies with Newtonian mechanics. The fictitious forces on a body are also proportional to the body's mass. (A body's mass is just a measure of its inertia: to accelerate at an acceleration a, a force F must be applied, and the mass m is just F/a.) .. ___ time-nuts mailing list time-nuts@febo.com https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts ___ time-nuts mailing list time-nuts@febo.com https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts -- No virus found in this incoming message. Checked by AVG Free Edition. Version: 7.5.472 / Virus Database: 269.8.1/822 - Release Date: 5/28/2007 11:40 AM ___ time-nuts mailing list time-nuts@febo.com https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity
On Wed, 30 May 2007 01:10:02 -0800, Bill Beam [EMAIL PROTECTED] wrote: Gentlemen: Those of you who have never taken a university physics course are excused for confusion over centripital/centrifugal/psudo forces. Some of you who did take a university physics class spent too much time asleep in class. I did and I paid attention and I didn't smoke anything I wasn't supposed to but I don't remember this aspect. This is a learning experience. John --- John De Armond See my website for my current email address http://www.neon-john.com Cleveland, Occupied TN *fas-cism* (fash'iz'em) n. A system of government that exercises a dictatorship of the extreme right, typically through the merging of state and business leadership, together with belligerent nationalism. -- The American Heritage Dictionary, 1983 ___ time-nuts mailing list time-nuts@febo.com https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity
John, I did and I paid attention and I didn't smoke anything I wasn't supposed to... Just in case you forgot to mention: What was your favourite drink at these times? I really enjoy being part of time-nuts for this exclusive combination of severe scientific stuff with humor like that which is not so easy to be found at other places. Regards Ulrich Bangert -Ursprüngliche Nachricht- Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Im Auftrag von Neon John Gesendet: Mittwoch, 30. Mai 2007 21:18 An: Discussion of precise time and frequency measurement Betreff: Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity On Wed, 30 May 2007 01:10:02 -0800, Bill Beam [EMAIL PROTECTED] wrote: Gentlemen: Those of you who have never taken a university physics course are excused for confusion over centripital/centrifugal/psudo forces. Some of you who did take a university physics class spent too much time asleep in class. I did and I paid attention and I didn't smoke anything I wasn't supposed to but I don't remember this aspect. This is a learning experience. John --- John De Armond See my website for my current email address http://www.neon-john.com Cleveland, Occupied TN *fas-cism* (fash'iz'em) n. A system of government that exercises a dictatorship of the extreme right, typically through the merging of state and business leadership, together with belligerent nationalism. -- The American Heritage Dictionary, 1983 ___ time-nuts mailing list time-nuts@febo.com https://www.febo.com/cgi- bin/mailman/listinfo/time-nuts ___ time-nuts mailing list time-nuts@febo.com https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity
In a message dated 5/30/2007 13:06:23 Pacific Daylight Time, [EMAIL PROTECTED] writes: Regards Ulrich Bangert Hi Ulrich, I am still using plotter daily, easier to use than Stable32. Some comments on Plotter: it cannot directly read-in RS-232 output from the 53132A (such as I posted yesterday). This is because of the Comma the HP unit inserts in the numbers. Any chance you can make Plotter comma compatible? Also, Plotter always comes up in scientific notation on the vertical scale, I always have to set it to #.6 mode manually. Any chance to make it come up in normal notation? thanks, Said ** See what's free at http://www.aol.com. ___ time-nuts mailing list time-nuts@febo.com https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity
On Thu, 31 May 2007 01:52:34 +1200, Dr Bruce Griffiths wrote: Bill Beam wrote: Assume satellite in circular orbit. (Not really necessary.) Assume test mass's released at rest wrt satellite center of mass. Inner test mass released closer to Earth and outer released farther from Earth. Also assume no air currents, no relativity, no luminiferous ether, no static, no s- -t. It helps if this problem is solved in a proper (Earth based) inertial frame and to consider the total energy (kinetic plus potential) of the test masses. But there are no strictly inertial frames based on the Earth. The earth rotates around its axis (neglecting precession, nutation etc), it also orbits the sun which in turn ... An actual test of these predictions would be somewhat expensive to carry out. The damping due to the air in the shuttle or ISS (as well as a host of other small effects) would tend to damp out such motion. The question is how quickly? This contradicts the last assumption stated above. Clearly a satellite based frame is non inertial and therefore Newtons laws of motion are not valid. Gentlemen: Those of you who have never taken a university physics course are excused for confusion over centripital/centrifugal/psudo forces. Some of you who did take a university physics class spent too much time asleep in class. Regards, Bill Beam NL7F ___ time-nuts mailing list time-nuts@febo.com https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts Bill Beam NL7F -- No virus found in this outgoing message. Checked by AVG. Version: 7.5.467 / Virus Database: 269.8.0/818 - Release Date: 5/25/2007 12:32 PM ___ time-nuts mailing list time-nuts@febo.com https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity
Bill Beam wrote: Assume satellite in circular orbit. (Not really necessary.) Assume test mass's released at rest wrt satellite center of mass. Inner test mass released closer to Earth and outer released farther from Earth. Also assume no air currents, no relativity, no luminiferous ether, no static, no s- -t. It helps if this problem is solved in a proper (Earth based) inertial frame and to consider the total energy (kinetic plus potential) of the test masses. But there are no strictly inertial frames based on the Earth. The earth rotates around its axis (neglecting precession, nutation etc), it also orbits the sun which in turn ... An actual test of these predictions would be somewhat expensive to carry out. The damping due to the air in the shuttle or ISS (as well as a host of other small effects) would tend to damp out such motion. The question is how quickly? This contradicts the last assumption stated above. Yes, but if one wishes to experimentally test the predictions it is not always practical to use an SV with an internal vacuum. The question is really could this be done on the ISS or shuttle or would the effects of the internal atmosphere disturb/damp the motion too quickly? In other words what would actually happen to 2 such test masses within the space shuttle, for example? The other question is how large would the interior of the SV have to be to avoid the test masses colliding with internal surfaces? The other point that in practice the frames in which virtually all measurements are made are non inertial. Sure one can correct the results to an Inertial frame if one can find/identify one that is inertial to a sufficient approximation. However this is an elusive target which keeps shifting around as the precision of measurement increases. General relativity surely indicates that the concept of an Inertial frame has a strictly local existence/validity? Bill Beam NL7F ___ time-nuts mailing list time-nuts@febo.com https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity
Said, it cannot directly read-in RS-232 output from the 53132A (such as I posted yesterday). This is because of the Comma the HP unit inserts in the numbers. please send short file and I will look what I can do. Also, Plotter always comes up in scientific notation on the vertical scale, I always have to set it to #.6 mode manually. Any chance to make it come up in normal notation? This is more severe. Not that it were a problem to change the default scale to whatever. But: The scientic notation FITS ALL while #.6 fits only YOU. Perhaps I think about a way to store such things in the ini-file. Best regards Ulrich -Ursprüngliche Nachricht- Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Im Auftrag von [EMAIL PROTECTED] Gesendet: Mittwoch, 30. Mai 2007 22:43 An: time-nuts@febo.com Betreff: Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity In a message dated 5/30/2007 13:06:23 Pacific Daylight Time, [EMAIL PROTECTED] writes: Regards Ulrich Bangert Hi Ulrich, I am still using plotter daily, easier to use than Stable32. Some comments on Plotter: it cannot directly read-in RS-232 output from the 53132A (such as I posted yesterday). This is because of the Comma the HP unit inserts in the numbers. Any chance you can make Plotter comma compatible? Also, Plotter always comes up in scientific notation on the vertical scale, I always have to set it to #.6 mode manually. Any chance to make it come up in normal notation? thanks, Said ** See what's free at http://www.aol.com. ___ time-nuts mailing list time-nuts@febo.com https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts ___ time-nuts mailing list time-nuts@febo.com https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity
On Tue, 29 May 2007 16:31:40 +1200, Dr Bruce Griffiths wrote: Ulrich, Didier Talking about forces, gravitational fields etc makes no physical sense if the observer's reference frame isn't specified. For an observer in/on a satellite orbiting about the Earth with their reference frame fixed with respect to the satellite. There is no gravitational field, whatever methods chosen to measure a gravitational field (within the satellite) will always produce a null result. Not true. Very simple experiments will show occupants of the satellite that they are in a non-inertial reference frame. (Release a few test masses about the cabin and you will observe that they move/accelerate for no apparent reason, unless the satellite is in free fall which you'll know soon enough,) The experimenter must conclude that the satellite is undergoing acceleration due to the influence of an attractive (gravitational) field. Just because NASA calls it 'microgravity' doesn't make it true. It means NASA is wrong. Weightlessness is not the same as zero-g. Pendulum clocks fail to work, given an initial push they will just rotate around the pivot, provided the pivot suitably constrains the motion of the pendulum (ie a shaft running in a set of ball or roller bearings or similar and not a knife edge pivot). If, however the satellite acts as a rigid body and has a large enough diameter then it would be possible for an observer on the satellite to detect a gravitational field gradient. Therefore, you must conclude that somewhere inside the satellite g is not zero. If the satellite is large enough and orbits close enough to the Earth, this gravitational field gradient would tear the satellite apart. For an observer located on the Earth however the motion of the satellite can be accurately described by Newtonian mechanics where the centripetal pull of gravity acts on the satellite causing it to have a centripetal (radial) acceleration as it orbits the Earth. Bruce Regards, Bill Beam (PhD, physics 1966, past tenured Associate Professor of Physics) Bill Beam NL7F -- No virus found in this outgoing message. Checked by AVG. Version: 7.5.467 / Virus Database: 269.8.0/818 - Release Date: 5/25/2007 12:32 PM ___ time-nuts mailing list time-nuts@febo.com https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity
Bill Bill Beam wrote: On Tue, 29 May 2007 16:31:40 +1200, Dr Bruce Griffiths wrote: Ulrich, Didier Talking about forces, gravitational fields etc makes no physical sense if the observer's reference frame isn't specified. For an observer in/on a satellite orbiting about the Earth with their reference frame fixed with respect to the satellite. There is no gravitational field, whatever methods chosen to measure a gravitational field (within the satellite) will always produce a null result. Not true. Very simple experiments will show occupants of the satellite that they are in a non-inertial reference frame. (Release a few test masses about the cabin and you will observe that they move/accelerate for no apparent reason, unless the satellite is in free fall which you'll know soon enough,) The experimenter must conclude that the satellite is undergoing acceleration due to the influence of an attractive (gravitational) field. Just because NASA calls it 'microgravity' doesn't make it true. It means NASA is wrong. Weightlessness is not the same as zero-g. Only, if you insist on sticking to Newtonian physics with all its attendant problems. Pendulum clocks fail to work, given an initial push they will just rotate around the pivot, provided the pivot suitably constrains the motion of the pendulum (ie a shaft running in a set of ball or roller bearings or similar and not a knife edge pivot). If, however the satellite acts as a rigid body and has a large enough diameter then it would be possible for an observer on the satellite to detect a gravitational field gradient. Therefore, you must conclude that somewhere inside the satellite g is not zero. A finite gradient doesn't imply that the field itself is nonzero, except of course towards the extremeities of the satellite. Regards, Bill Beam (PhD, physics 1966, past tenured Associate Professor of Physics) Bill Beam NL7F Bruce ___ time-nuts mailing list time-nuts@febo.com https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity
On Tue, 29 May 2007 22:27:42 +1200, Dr Bruce Griffiths wrote: Bill Bill Beam wrote: On Tue, 29 May 2007 16:31:40 +1200, Dr Bruce Griffiths wrote: Ulrich, Didier Talking about forces, gravitational fields etc makes no physical sense if the observer's reference frame isn't specified. For an observer in/on a satellite orbiting about the Earth with their reference frame fixed with respect to the satellite. There is no gravitational field, whatever methods chosen to measure a gravitational field (within the satellite) will always produce a null result. Not true. Very simple experiments will show occupants of the satellite that they are in a non-inertial reference frame. (Release a few test masses about the cabin and you will observe that they move/accelerate for no apparent reason, unless the satellite is in free fall which you'll know soon enough,) The experimenter must conclude that the satellite is undergoing acceleration due to the influence of an attractive (gravitational) field. Just because NASA calls it 'microgravity' doesn't make it true. It means NASA is wrong. Weightlessness is not the same as zero-g. Only, if you insist on sticking to Newtonian physics with all its attendant problems. This discussion began as a classical problem. The relativistic effects are many orders of magnitude smaller than Newtonian (v/c=2.6e-5). For example: A test mass released on the Earth side of the satellite cabin will advance in its own orbit a few mm/sec faster than one released on the far side due to purely classical differences in orbits. Easily observable without even using a timepiece. Once your feet leave the ground, not even Newtonian mechanics is intuitive. Who would have thought that 'putting on the brakes' to leave orbit would cause a satellite to speed up Pendulum clocks fail to work, given an initial push they will just rotate around the pivot, provided the pivot suitably constrains the motion of the pendulum (ie a shaft running in a set of ball or roller bearings or similar and not a knife edge pivot). Run the numbers - depends on how hard the push. Consider sheeparding of material in Saturn rings by small moons. If, however the satellite acts as a rigid body and has a large enough diameter then it would be possible for an observer on the satellite to detect a gravitational field gradient. Therefore, you must conclude that somewhere inside the satellite g is not zero. A finite gradient doesn't imply that the field itself is nonzero, except of course towards the extremeities of the satellite. Of course it does. If g=0 everywhere in the neighborhood of a point then the gradient is zero. Else, what is the meaning of gradient? Grad not zero implies field not uniform implies not(field zero everywhere). Regards, Bill Beam (PhD, physics 1966, past tenured Associate Professor of Physics) Bill Beam NL7F Bruce ___ time-nuts mailing list time-nuts@febo.com https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts Bill Beam NL7F -- No virus found in this outgoing message. Checked by AVG. Version: 7.5.467 / Virus Database: 269.8.0/818 - Release Date: 5/25/2007 12:32 PM ___ time-nuts mailing list time-nuts@febo.com https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity
Didier Juges wrote: Bruce, A lot of the statements that have been made lately on this subject kind of make sense to me in a way taken in isolation, but they do not all agree with each other, and that makes me uncomfortable. Example: I do not understand why the frame of reference would matter when you talk about gravity field. There is a gravity field or not, and the frame of reference should not matter. I understand that the frame of reference matters when you talk about displacement, velocity or acceleration. But the magnitude of a field, or a force, does not depend on the observer as it is static, or maybe a better term would be absolute or self-referenced? The reason that the frame of reference matters is that gravity is indistinguishable from acceleration. (This is an assumption that Einstein made when deriving his general theory of relativity. It seems to work.) An inertial frame of reference is a non-accelerating frame of reference. In an inertial frame of reference, Newton's laws of motion work -- if you use Newton's gravitational relationship, that the gravitational force (weight) that each of two bodies exerts on the other is proportional to both their masses, and inversely proportional to the square of the distance between them. In an accelerating frame of reference (either linear acceleration, or rotational acceleration, or both) additional forces, technically called fictitious forces, must be introduced in order to explain the motions of bodies with Newtonian mechanics. The fictitious forces on a body are also proportional to the body's mass. (A body's mass is just a measure of its inertia: to accelerate at an acceleration a, a force F must be applied, and the mass m is just F/a.) If the frame of reference has linear acceleration (relative to an inertial frame of reference), bodies within that frame of reference will experience a fictitious force that is proportional to their masses and to the acceleration of the frame of reference. Viewed from the frame of reference of a car that is accelerating away from a stop light, the passengers are pressed back in their seats by a force proportional to the acceleration of the car and to their masses. This fictitious force disappears when you view the situation from the an inertial frame of reference. Viewed from that point of view, the seats are pressing forward on the passengers to cause them to accelerate with the car. Viewed from a rotating frame of reference, we have other fictitious forces: centrifugal force and Coriolis force. Both of these are proportional to the mass of the body on which they act -- when viewed from the rotating frame of reference. Both vanish if you view the situation from a non-rotating frame of reference. Sometimes - usually, even - it's simpler to view the problem from an inertial frame of reference. Sometimes, though, it's easier to look at the problem in an accelerating frame of reference. If you do that, you account for the frame of reference's acceleration by introducing fictitious forces. Now, it makes sense that an object immersed in gravity fields from several larger objects may not be able to tell the difference between multiple fields, and a unique, net field (in the sense of Newton's net force), at least as long as the gradient is small enough that it cannot be observed within the dimensions of the object. So if the net field is zero and the gradient small enough to be ignored, the object will behave the same as if there were no field. When you say within the dimensions of the object I assume that you are looking at the problem from the frame of reference of the object. That's natural if you are, for example, in an orbiting satellite, such as the International Space Station. Viewed from an inertial frame of reference, the ISS is following an orbit determined by the vector sum of the gravitational forces (from earth, moon, sun, etc.) that act upon it. Viewed from the frame of reference of the space station, however, these forces add to zero. However, for an observer on earth, a satellite is in the gravity field of earth (let's assume all other gravity fields from the sun and other planets are negligible), which is not zero at the altitude of the satellite, Even an observer on earth is on an accelerating frame of reference. (The earth rotates on its axis.) ... yet for an observer on the satellite, the net field appears to be zero. Where is the counter-field coming from? And why can't we observe it from earth? How can the field be different when observed from different points? For an observer on the satellite (in the satellite's frame of reference), the counter-field is created by the fictitious forces due to the satellite's acceleration. For example, centrifugal force due to the satellite's gravitational acceleration towards the center of mass of the earth. Could it be that the effect of the gravity field (with is
Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity
Finally, something that makes sense! Thanks, James Maynard. The idea that the centripetal force that balances the gravitational force is fictitious was not popular when I was educated, before 1960. But centripetal force goes away if gravity goes away. The orbiting object continues in a straight line because no forces are causing acceleration. When there is gravity, and an object falls around the Earth, the velocity vector is not constant - it rotates 360 degrees for each orbit of the Earth. An additional acceleration is required to make that happen, hence centripetal force. Gravity and centripetal force must balance if the object is to keep falling in an orbit, which does not have to be circular. If the orbit is not circular then the object's velocity magnitude changes to match its altitude. Centripetal force also goes away if radial motion goes away. The space shuttle has rocket engines that can reduce the radial motion so that the altitude falls low enough to start atmospheric braking. Note that great forces are required to change the angle of the velocity vector. A shuttle can not drive around the sky like an aircraft (when it is in space) but it does have some control of altitude. Bill Hawkins -Original Message- From: James Maynard Sent: Tuesday, May 29, 2007 11:26 AM To: Discussion of precise time and frequency measurement Subject: Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity Didier Juges wrote: Bruce, A lot of the statements that have been made lately on this subject kind of make sense to me in a way taken in isolation, but they do not all agree with each other, and that makes me uncomfortable. Example: I do not understand why the frame of reference would matter when you talk about gravity field. There is a gravity field or not, and the frame of reference should not matter. I understand that the frame of reference matters when you talk about displacement, velocity or acceleration. But the magnitude of a field, or a force, does not depend on the observer as it is static, or maybe a better term would be absolute or self-referenced? The reason that the frame of reference matters is that gravity is indistinguishable from acceleration. (This is an assumption that Einstein made when deriving his general theory of relativity. It seems to work.) An inertial frame of reference is a non-accelerating frame of reference. In an inertial frame of reference, Newton's laws of motion work -- if you use Newton's gravitational relationship, that the gravitational force (weight) that each of two bodies exerts on the other is proportional to both their masses, and inversely proportional to the square of the distance between them. In an accelerating frame of reference (either linear acceleration, or rotational acceleration, or both) additional forces, technically called fictitious forces, must be introduced in order to explain the motions of bodies with Newtonian mechanics. The fictitious forces on a body are also proportional to the body's mass. (A body's mass is just a measure of its inertia: to accelerate at an acceleration a, a force F must be applied, and the mass m is just F/a.) If the frame of reference has linear acceleration (relative to an inertial frame of reference), bodies within that frame of reference will experience a fictitious force that is proportional to their masses and to the acceleration of the frame of reference. Viewed from the frame of reference of a car that is accelerating away from a stop light, the passengers are pressed back in their seats by a force proportional to the acceleration of the car and to their masses. This fictitious force disappears when you view the situation from the an inertial frame of reference. Viewed from that point of view, the seats are pressing forward on the passengers to cause them to accelerate with the car. Viewed from a rotating frame of reference, we have other fictitious forces: centrifugal force and Coriolis force. Both of these are proportional to the mass of the body on which they act -- when viewed from the rotating frame of reference. Both vanish if you view the situation from a non-rotating frame of reference. Sometimes - usually, even - it's simpler to view the problem from an inertial frame of reference. Sometimes, though, it's easier to look at the problem in an accelerating frame of reference. If you do that, you account for the frame of reference's acceleration by introducing fictitious forces. Now, it makes sense that an object immersed in gravity fields from several larger objects may not be able to tell the difference between multiple fields, and a unique, net field (in the sense of Newton's net force), at least as long as the gradient is small enough that it cannot be observed within the dimensions of the object. So if the net field is zero and the gradient small enough to be ignored, the object will behave the same as if there were no field. When you say within the dimensions of the object I assume
Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity
Aargh! Please change Centripetal force also goes away if radial motion goes away. to Centripetal force also goes away if angular motion goes away. -Original Message- From: Bill Hawkins [mailto:[EMAIL PROTECTED] Sent: Tuesday, May 29, 2007 12:17 PM To: 'Discussion of precise time and frequency measurement' Subject: RE: [time-nuts] FW: Pendulums Atomic Clocks Gravity Finally, something that makes sense! Thanks, James Maynard. The idea that the centripetal force that balances the gravitational force is fictitious was not popular when I was educated, before 1960. But centripetal force goes away if gravity goes away. The orbiting object continues in a straight line because no forces are causing acceleration. When there is gravity, and an object falls around the Earth, the velocity vector is not constant - it rotates 360 degrees for each orbit of the Earth. An additional acceleration is required to make that happen, hence centripetal force. Gravity and centripetal force must balance if the object is to keep falling in an orbit, which does not have to be circular. If the orbit is not circular then the object's velocity magnitude changes to match its altitude. Centripetal force also goes away if radial motion goes away. The space shuttle has rocket engines that can reduce the radial motion so that the altitude falls low enough to start atmospheric braking. Note that great forces are required to change the angle of the velocity vector. A shuttle can not drive around the sky like an aircraft (when it is in space) but it does have some control of altitude. Bill Hawkins -Original Message- From: James Maynard Sent: Tuesday, May 29, 2007 11:26 AM To: Discussion of precise time and frequency measurement Subject: Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity Didier Juges wrote: Bruce, A lot of the statements that have been made lately on this subject kind of make sense to me in a way taken in isolation, but they do not all agree with each other, and that makes me uncomfortable. Example: I do not understand why the frame of reference would matter when you talk about gravity field. There is a gravity field or not, and the frame of reference should not matter. I understand that the frame of reference matters when you talk about displacement, velocity or acceleration. But the magnitude of a field, or a force, does not depend on the observer as it is static, or maybe a better term would be absolute or self-referenced? The reason that the frame of reference matters is that gravity is indistinguishable from acceleration. (This is an assumption that Einstein made when deriving his general theory of relativity. It seems to work.) An inertial frame of reference is a non-accelerating frame of reference. In an inertial frame of reference, Newton's laws of motion work -- if you use Newton's gravitational relationship, that the gravitational force (weight) that each of two bodies exerts on the other is proportional to both their masses, and inversely proportional to the square of the distance between them. In an accelerating frame of reference (either linear acceleration, or rotational acceleration, or both) additional forces, technically called fictitious forces, must be introduced in order to explain the motions of bodies with Newtonian mechanics. The fictitious forces on a body are also proportional to the body's mass. (A body's mass is just a measure of its inertia: to accelerate at an acceleration a, a force F must be applied, and the mass m is just F/a.) If the frame of reference has linear acceleration (relative to an inertial frame of reference), bodies within that frame of reference will experience a fictitious force that is proportional to their masses and to the acceleration of the frame of reference. Viewed from the frame of reference of a car that is accelerating away from a stop light, the passengers are pressed back in their seats by a force proportional to the acceleration of the car and to their masses. This fictitious force disappears when you view the situation from the an inertial frame of reference. Viewed from that point of view, the seats are pressing forward on the passengers to cause them to accelerate with the car. Viewed from a rotating frame of reference, we have other fictitious forces: centrifugal force and Coriolis force. Both of these are proportional to the mass of the body on which they act -- when viewed from the rotating frame of reference. Both vanish if you view the situation from a non-rotating frame of reference. Sometimes - usually, even - it's simpler to view the problem from an inertial frame of reference. Sometimes, though, it's easier to look at the problem in an accelerating frame of reference. If you do that, you account for the frame of reference's acceleration by introducing fictitious forces. Now, it makes sense that an object immersed in gravity fields from several larger objects may not be able to tell the difference between
Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity
Bill, in general I would underwrite every single sentence of your explanation with the exception of The idea that the centripetal force that balances the gravitational force is fictitious was not popular when I was educated, before 1960. because gravitation IS the centripetal force for the satellite's motion. In this case the right word would have beeen indeed centrifugal. Centripetal forces are REAL forces and are the source of the permanent falling. While forces are one of the very first things that pupils are confronted with in learning physics they are by no means trivial and can be tricky to an high extend. If you would like to dive even deeper into this subject consider the following question: If I stand on the floor of my flat, clearly no acceleration is to be noticed on my body although it is clear that earth attracs me with my weight force (being much too high since years). If no acceleration is to be noticed at my body then a second force must be there that balances the gravitational force, and in this case it is really a BALANCE. Since I stand on the floor the floor must be the source of that force. Big question: HOW does it manage to exhibit this force to my body? Regards Ulrich Bangert Regards Ulrich Bangert -Ursprüngliche Nachricht- Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Im Auftrag von Bill Hawkins Gesendet: Dienstag, 29. Mai 2007 19:23 An: 'Discussion of precise time and frequency measurement' Betreff: Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity Aargh! Please change Centripetal force also goes away if radial motion goes away. to Centripetal force also goes away if angular motion goes away. -Original Message- From: Bill Hawkins [mailto:[EMAIL PROTECTED] Sent: Tuesday, May 29, 2007 12:17 PM To: 'Discussion of precise time and frequency measurement' Subject: RE: [time-nuts] FW: Pendulums Atomic Clocks Gravity Finally, something that makes sense! Thanks, James Maynard. The idea that the centripetal force that balances the gravitational force is fictitious was not popular when I was educated, before 1960. But centripetal force goes away if gravity goes away. The orbiting object continues in a straight line because no forces are causing acceleration. When there is gravity, and an object falls around the Earth, the velocity vector is not constant - it rotates 360 degrees for each orbit of the Earth. An additional acceleration is required to make that happen, hence centripetal force. Gravity and centripetal force must balance if the object is to keep falling in an orbit, which does not have to be circular. If the orbit is not circular then the object's velocity magnitude changes to match its altitude. Centripetal force also goes away if radial motion goes away. The space shuttle has rocket engines that can reduce the radial motion so that the altitude falls low enough to start atmospheric braking. Note that great forces are required to change the angle of the velocity vector. A shuttle can not drive around the sky like an aircraft (when it is in space) but it does have some control of altitude. Bill Hawkins -Original Message- From: James Maynard Sent: Tuesday, May 29, 2007 11:26 AM To: Discussion of precise time and frequency measurement Subject: Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity Didier Juges wrote: Bruce, A lot of the statements that have been made lately on this subject kind of make sense to me in a way taken in isolation, but they do not all agree with each other, and that makes me uncomfortable. Example: I do not understand why the frame of reference would matter when you talk about gravity field. There is a gravity field or not, and the frame of reference should not matter. I understand that the frame of reference matters when you talk about displacement, velocity or acceleration. But the magnitude of a field, or a force, does not depend on the observer as it is static, or maybe a better term would be absolute or self-referenced? The reason that the frame of reference matters is that gravity is indistinguishable from acceleration. (This is an assumption that Einstein made when deriving his general theory of relativity. It seems to work.) An inertial frame of reference is a non-accelerating frame of reference. In an inertial frame of reference, Newton's laws of motion work -- if you use Newton's gravitational relationship, that the gravitational force (weight) that each of two bodies exerts on the other is proportional to both their masses, and inversely proportional to the square of the distance between them. In an accelerating frame of reference (either linear acceleration, or rotational acceleration, or both) additional forces, technically called fictitious forces, must be introduced in order to explain the motions of bodies with Newtonian mechanics
Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity
Bill Hawkins wrote: Finally, something that makes sense! Thanks, James Maynard. Thank you. However, in the following paragraphs you should use the term centrifugal rather than centripetal. A centripetal force is directed towards the axis of rotation. A centrifugal force is directed outward, away from the axis of rotation. I have edited your reply, with my changes indicated in [bracketed] text. The idea that the [centrifugal] force that balances the gravitational force is fictitious was not popular when I was educated, before 1960. But centripetal force [that is, the satellite's weight, mg] goes away if gravity [g] goes away. The orbiting object continues in a straight line because no forces are causing acceleration. When there is gravity, and an object falls around the Earth, the velocity vector is not constant - it rotates 360 degrees for each orbit of the Earth. An additional acceleration is required to make that happen, hence centripetal force. [Right. Here, the centripetal force is the gravitational force, the satellite's weight. The fictitious centrifugal force that balances the satellite's weight is only present when you view the problem from the frame of reference of the orbiting satellite.] Gravity and [centrifugal] force must balance if the object is to keep falling in an orbit, which does not have to be circular. If the orbit is not circular then the object's velocity magnitude changes to match its altitude. Centripetal force also goes away if radial motion goes away. I would say, rather, that the centripetal force, mg in this case, causes the satellite's velocity to change its direction. When viewed in the non-inertial frame of reference of the satellite, the corresponding fictitious centrifugal force also goes away, because the satellite is not accelerating in a direction perpendicular to its velocity. The space shuttle has rocket engines that can reduce the radial motion so that the altitude falls low enough to start atmospheric braking. Note that great forces are required to change the angle of the velocity vector. A shuttle can not drive around the sky like an aircraft (when it is in space) but it does have some control of altitude. Bill Hawkins [I should also edit part of my previous post, as indicated in the bracketed text below.] -Original Message- From: James Maynard Sent: Tuesday, May 29, 2007 11:26 AM To: Discussion of precise time and frequency measurement Subject: Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity Didier Juges wrote: Bruce, A lot of the statements that have been made lately on this subject kind of make sense to me in a way taken in isolation, but they do not all agree with each other, and that makes me uncomfortable. Example: I do not understand why the frame of reference would matter when youtalk about gravity field. There is a gravity field or not, and the frame of reference should not matter. I understand that the frame of reference matters when you talk about displacement, velocity or acceleration. But the magnitude of a field, or a force, does not depend on the observer as it is static, or maybe a better term would be absolute or self-referenced? The reason that the frame of reference matters is that gravity is indistinguishable from acceleration. (This is an assumption that Einstein made when deriving his general theory of relativity. It seems to work.) An inertial frame of reference is a non-accelerating frame of reference. In an inertial frame of reference, Newton's laws of motion work -- if you use Newton's gravitational relationship, that the gravitational force (weight) that each of two bodies exerts on the other is proportional to both their masses, and inversely proportional to the square of the distance between them. In an accelerating frame of reference (either linear acceleration, or rotational acceleration, or both) additional forces, technically called fictitious forces, must be introduced in order to explain the motions of bodies with Newtonian mechanics. The fictitious forces on a body are also proportional to the body's mass. (A body's mass is just a measure of its inertia: to accelerate at an acceleration a, a force F must be applied, and the mass m is just F/a.) If the frame of reference has linear acceleration (relative to an inertial frame of reference), bodies within that frame of reference will experience a fictitious force that is proportional to their masses and to the acceleration of the frame of reference. Viewed from the frame of reference of a car that is accelerating away from a stop light, the passengers are pressed back in their seats by a force proportional to the acceleration of the car and to their masses. This fictitious force disappears when you view the situation from the an inertial frame of reference. Viewed from that point of view, the seats are pressing forward
Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity
-Ursprüngliche Nachricht- Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Im Auftrag von Didier Juges Gesendet: Montag, 28. Mai 2007 13:53 An: time-nuts@febo.com Betreff: [time-nuts] FW: Pendulums Atomic Clocks Gravity Ulrich, I am quite familiar with the cannon analogy. If I may use this analogy too, please consider the following: There must be a force balancing the force of gravity, otherwise the satellite would not cease from accelerating under gravity alone. Gravity exerts a force on the satellite which tends to make it fall towards earth. This is the Centripetal force. Inertia due to the mass of the satellite makes it resist this motion, and the tangential speed makes it miss the earth. Centrifugal force is the name we give to that resistance. When the satellite is in a stable orbit, it does not accelerate because both forces exactly balance each other. For the reason you pointed out, in a closed system the sum of forces must be zero, so there must be a force balancing the gravity force. So I see we agree. If there was no rotation, that force would not exist and the satellite would accelerate (under gravity alone) towards earth. Dont be confused by terminology. The terms centrifugal and centripetal are just names given to other forces, not actual forces by themselves. The centripetal force is due to gravity (but is could be electromagnetic, or anything else. In a centrifuge, it would be the force exerted by the rotating arm), the centrifugal force is due to mass, radius and speed. 73, Didier KO4BB -Original Message- From: Ulrich Bangert [mailto:[EMAIL PROTECTED] Sent: Monday, May 28, 2007 5:03 AM To: 'Didier Juges' Subject: AW: [time-nuts] Pendulums Atomic Clocks Gravity Didier, I am an physicist, not an engineer. Let me use an experiment of thought that Bill Hawkins has already used in the discussion: Assume an cannon mounted in an certain height with the barrel mounted tangetial to earth's surface. Fire an bullet and see it fall to earth after an certain time of flight. Now use more gun powder and see the the bullet fall to earth later. Use a BIG amount of powder and see the bullet leave earth's gravity completely. Between the extremes: Drop to surface and leaving earth's gravity completely there is one powder loading that brings the bullet into an circular orbit at the height of the cannon. The bullet never stops to fall to earth. However the motion towards earth's cencer is compensated by the fact that an tangential motion ALSO means to depart from the center of the body that you move tangential to. 73 and my best regards Ulrich, DF6JB -Ursprüngliche Nachricht- Von: Didier Juges [mailto:[EMAIL PROTECTED] Gesendet: Montag, 28. Mai 2007 02:02 An: [EMAIL PROTECTED] Betreff: Re: [time-nuts] Pendulums Atomic Clocks Gravity Ulrich, Please go ahead, I am all ears... (in all seriousness, I am not a physicist, just an engineer) If earth attracts the satellite and the satellite attracts earth, how come the satellite and earth don't get together? What is keeping them apart? When you say the gravity forces are of opposite direction, this is correct. The gravity applied by earth to the satellite causes a force vector directed towards the earth, the gravity applied by the satellite to earth is a force vector of equal magnitude and directed from earth to the satellite. The external result is null (as a system, there is no loss of force, action = reaction). The same holds true for centrifugal forces. The satellite affects the orbit of earth in proportion of their respective mass, so the satellite causes earth to move around it's theoretical orbit (if there was no satellite). The earth movement is very small (could not be measured for an artificial satellite, but but could certainly be calculated, the effect of the moon on earth's orbit can certainly be measured) but it causes an equal and opposite centrifugal force on earth, which balances the force exerted on the satellite. So I believe there are 2 sets of forces (gravity and centrifugal), and each set has a resultant that is null, as seen from the outside. However, at the level of earth and the satellite, the gravitational attraction is equal and opposite to the centrifugal force. I did not know physics cared if we used inertial system concepts or accelerated systems concepts (I do not know the difference). If I follow your theory, the speed of the satellite around the earth has no effect on gravity, so the satellite should stay where it is regardless of speed, but it does not! Please explain this to me. I agree that as long as the distance between a satellite and earth remains constant, the forces must balance each other. But if it's not centrifugal force that is balancing gravity, what is it? Thanks in
Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity
Ulrich, Your comparison with the linear motion is not valid. While you push the body, it accelerates. The energy spent giving the body increased speed (due to the excess force applied to the body while there is no counter force) is stored in kinetic energy. Once you stop pushing, the body moves straight at constant speed and there is no more force either way, the state of motion does not change. Going back to the satellite, I believe we agree in principal but you are hung up on the first law definition. The corollary to the first law is that objects resist change to their state of motion. They resist that change via inertia. Inertia causes real forces to be developed. I agree it's not a fundamental force like gravity or electromagnetism, but it is necessary to keep the system in equilibrium, otherwise what would be holding the satellite at a constant distance while it is being pulled towards earth? Example: a car is moving at 1m/s. You stand still on the driveway (your speed: 0m/s.) The car hits you. Because the car is much heavier (err: has more mass) than you, you now move at close to 1m/s and the car has just slowed down a bit (m * v stays the same). The car has exerted a force on you that gave you acceleration. You have exerted a force on the car that caused the car to change it's speed, you imparted on the car a certain acceleration in the opposite direction. The only way the car could feel acceleration is because a force was exerted upon it. The force that was exerted on the car came from the inertia of your body. It is a real force. Now, imagine the driveway was itself on a sliding plane moving at 1m/s in the opposite direction to the car. In fact, the car was not moving but you were. It makes no difference, once you realize the only difference is where the reference it. Two forces were developed. Which one you call action and which one is reaction is irrelevant. It's only a matter of reference. Because you change the reference does not make the force go away. We agree on what is happening, we don't agree on what to call it. Since you know a lot more about this than I do, I will accept your statement that centrifugal forces (or more generally inertial forces) are fictitious, but only because you insist. As long as I can predict their effect and calculate their magnitude, that's all this engineer is interested in :-) 73, Didier KO4BB -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Ulrich Bangert Sent: Monday, May 28, 2007 8:23 AM To: 'Discussion of precise time and frequency measurement' Subject: Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity Didier, let us consider the more easier case of an linear motion. Imagine an body that can glide on an surface without any friction. Now you take a finger of your hand and press it on one side of the body so that it moves horizontally. Clearly your finger exercises an force on the body that makes it accelerate. And in your finger you feel an force into the opposite direction. The key question is about the physical reality of this force in the opposite direction. How big is it? I guess, you would argument that it has the same magnitude as the force that you apply with your finger but has the opposite direction, right? Now you have TWO forces. If they have the same magnitude but opposite direction their vectors add to zero and I am almost sure you would argument that this makes the sum of forces zero for the system. Now, that you have shown that the sum of forces is zero you are in the ungraceful position in that you must explain why the body IS ACCELERATING at all. According to F=m*a a non-zero F is necessary to generate a non-zero a. How do you explain? Please note that in physics there is one substantial thing that one must know about forces and counterforces: They never affect on the SAME body but always on DIFFERENT bodies. In case you do not believe take the next textbook and read it after. For the above experiment this means: Since BOTH forces that you are talking about affect on the same one body one of the forces CANNOT be the counterforce to the other. If this is so then lets search for the counterforce for the force that you apply with your finger. In order to be able to execute this force your feet or other parts of your body execute an force into opposite direction to the surface of earth. These two forces are counterforces to each other because they a) have opposite directions b) have same magnitude c) apply to different bodies. Now that we have found the counterforce that makes the sum of forces zero for the system you need to find the counterforce to your inertial force and you will find none. Perhaps it is helpful for your understanding that one of the definitions for fictitious forces is that no counterforce belonging to them can be found. And if no counterforce can be found they have NO physical reality because otherwise the rule of the sum of forces is violated
Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity
Ulrich, Didier Talking about forces, gravitational fields etc makes no physical sense if the observer's reference frame isn't specified. For an observer in/on a satellite orbiting about the Earth with their reference frame fixed with respect to the satellite. There is no gravitational field, whatever methods chosen to measure a gravitational field (within the satellite) will always produce a null result. Pendulum clocks fail to work, given an initial push they will just rotate around the pivot, provided the pivot suitably constrains the motion of the pendulum (ie a shaft running in a set of ball or roller bearings or similar and not a knife edge pivot). If, however the satellite acts as a rigid body and has a large enough diameter then it would be possible for an observer on the satellite to detect a gravitational field gradient. If the satellite is large enough and orbits close enough to the Earth, this gravitational field gradient would tear the satellite apart. For an observer located on the Earth however the motion of the satellite can be accurately described by Newtonian mechanics where the centripetal pull of gravity acts on the satellite causing it to have a centripetal (radial) acceleration as it orbits the Earth. Bruce ___ time-nuts mailing list time-nuts@febo.com https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
Re: [time-nuts] FW: Pendulums Atomic Clocks Gravity
Neville Michie wrote: Hi All, I am still having difficulty getting my head around the gravity point. Now I accept, in principle, that due to relativity an intense gravity field will slow a clock. My problem is visualising where you will find this field. At the centre of this planet gravity (from the planet) is zero. This comes about by an elegant piece of calculus that shows that everywhere inside a hollow sphere the gravity is zero. So a clock in the centre of Earth runs at the same rate as one on the surface? or does it run faster because the one on the surface has the planetary gravity acting on it? I think that the one inside the Earth runs faster. But when you are between the Earth and Moon at a point where gravity forces are neutral we should have the same rate as centre of planet? Now the way to measure gravity is to measure the force on a test mass. If there is zero force there is zero gravity, except when you are in orbit. This can be tested with 3 crossed gyroscopes that show your angular velocity. If your angular velocity is negligible then the magnitude of the gravity field is proportional to the force on a test mass. Unless you are in free fall accelerating towards a mass. I guess my question really is can you know that you are in a zero gravity field so your clock is running at the fastest rate? Or is relativity relative. Does relativity only show up when there are two frames of reference being compared, so there is no ultimate reference frame with the fastest clock? Or can any clock have a single relativity correction applied to it? How do you measure the gravitational potential at any site? (ie the scalar quantity that would be used to correct your clock). Has anyone got a clear answer? cheers, neville Michie Neville You are somewhat astray its not the gravitational field but the change in gravitational potential that red or blue shifts an atomic clock as it is moved from one location to another. The same gravitational potential difference can result for a small change in altitude in a strong gravitational field as a larger change in altitude in a smaller gravitational field. Also the gravitational field within a spherical shell is only zero when the density distribution on the shell is spherically symmetric. This is important in determining the gravitational field at the centre of a body that has mascons of different density to their surrounding matter they are embedded within. A clock doesn't necessarily run fastest when the gravitational field is zero, it runs fastest when the gravitational potential is highest. Locations of zero gravity and maximum gravitational potential do not necessarily coincide. For example an atomic clock located at the centre of the Earth runs slower than one located on the surface (ignoring the effects of rotation). However the gravitational field at the centre of the Earth is near zero whilst on the surface it is non zero. Bruce ___ time-nuts mailing list time-nuts@febo.com https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
