4 ms·
The lower gravity clock will run faster, but the experiment should give the same result, regardless of which frame it's running in. The same way that the caesiu
by incompatible 2y ago
The lower gravity clock will run faster, but the experiment should give the same result, regardless of which frame it's running in. The same way that the caesium-133 atom transition frequency is 9192631770 Hz, regardless of the gravitational field.
- deleted 2y ago[deleted]
- mikewarot 2y agoThe time shift of a Cesium beam atomic clock at different altitudes is a well established experimental result. With Optical Lattice clocks, they are much more stable, and the output is light, so a phase difference can be detected much, much quicker. According to [1], a difference of 1 centimeter can be measured. The quote of interest: An optical lattice clock with a frequency accuracy of 1 × 10^−18, which is currently the most accurate in the world, has a detectable gravitational potential equivalent to an elevation difference of approximately 1 cm. If you can access it, here's the related letter in Nature[2] from 2018. [1] https://www.rd.ntt/e/research/JN202304_21619.html https://www.rd.ntt/e/research/JN202304_21619.html [2] https://www.nature.com/articles/s41586-018-0738-2 https://www.nature.com/articles/s41586-018-0738-2
- incompatible 2y agoYes, nevertheless each clock in its own frame of reference is measuring 9192631770 Hz, or whatever. It's only with multiple communicating clocks that you'd observe the difference: that 9192631770 Hz in one place is different to 9192631770 Hz somewhere else.
- deleted 2y ago[deleted]
- magicalhippo 2y agoSo if you have three clocks, one on the floor, one on a table and one at the ceiling, you could tell that relative to the clock on the table, the bottom one runs slower than the one at the ceiling, right? Is the statement then that if the elevator is sufficiently tall, this difference disappears if the elevator is accelerated by a rocket in outer (flat) space, vs hanging in an elevator shaft?
- mikewarot 2y agoIn a gravity scenario, the clock closest to the "bottom" would run slowest. In an accelerating elevator scenario, the acceleration would be equal in all 3 clocks, and they would all run at the same rate.
- hnuser123456 2y agoSo a real universal UTC should have a unit of seconds that assumes a perfectly homogenous universe, with each user responsible for correcting for the depth of their local gravitational well/hill.
- nyrikki 2y agoNo, there is no 'real universal' time, just agreed upon reference frames. UTC uses TAI, which is a weighted average of global atomic clocks. Any physically realized time standards will have errors. Proper time is the only invariant form under GR.
- fanf2 2y agoUTC is defined on the rotating “geoid” which is the modern version of mean sea level. If you happen to be NIST (one of the USA’s principle time labs) or Schriever space force base (which hosts the ground-based reference atomic clocks for the GPS satellite fleet) you need to adjust the speed of your clocks slightly to compensate for being in Colorado 1 mile above sea level.
- toast0 2y agoUniversal Time would really need to be based on seconds passing at the center of mass of the universe. Each observer could run their own clock, and calculate adjustments from there. Although that might not be very convenient.
- hnuser123456 2y agoWhy would it have to be at the center of mass of the universe? I am going ahead and assuming that if every observer assumed that their local observable universe was homogenous, then in most global (co-moving) locations across the entire universe, we would have similar results, but I am eager for someone smarter's reply.
- hnuser123456 2y agoAlso, is the CMB not a reference frame?