name123
- 510
- 5
Sorry that should be "uniform"jbriggs444 said:What does "unified" mean?
Sorry that should be "uniform"jbriggs444 said:What does "unified" mean?
Dale said:Please re read the definition I gave previously: "Local means close enough in space and time so that tidal effects are negligible. I.e. a uniform gravitational field". Is the gravitational field uniform in the region containing both clocks? Are tidal effects negligible?
Yes, but note that the metric will have a different form in these different charts. With a correct expression for the metric, all invariants will be correctly calculated.name123 said:If the satellites were labelled A and B and the mass C then would the clock comparison figure be correctly calculated no matter which you considered at rest?
A.T. said:The energy distribution affects the gravitational potential and thus the clock rates between distant points.
Gravitational time dilation is always between two separated clocks. With just one clock there is nothing to explain.name123 said:So considering just the distant clock
These two clocks are local:name123 said:(the clock in flat spacetime in distant space outside the shell and the clock in flat spacetime in flat spacetime inside the shell)
name123 said:once the radiation had passed the distant clock
The distribution of the energy in the shellname123 said:What energy distribution difference between 1 and 2 were you thinking would explain the affect?
A.T. said:Gravitational time dilation is always between two separated clocks. With just one clock there is nothing to explain.
So the gravitational energy causes the time dilation?Dale said:The distribution of the energy in the shell
Dale said:name123 said:Consider satellites A and B, going at different velocities, in free fall orbit around a massive body C at different altitudes for a million years, then being brought together and the clocks on them compared. Presumably the bringing them together would become less significant the longer they orbited, and the amount of time dilation due to curvature would be frame of reference independent, but what about the observed velocities from the mass being orbited's perspective? If the satellites were labelled A and B and the mass C then would the clock comparison figure be correctly calculated no matter which you considered at rest?
Yes, but note that the metric will have a different form in these different charts. With a correct expression for the metric, all invariants will be correctly calculated.
name123 said:Also regarding A, B and C being at rest, I am not quite clear how A or B could be considered at rest and not rotating, as while C (the mass) could be considered to orbit A or B, if it was so considered, would it not also need to be considered that it was rotating, for an explanation of why the same section of C was not always facing A (or B), and could it not be objectively measured that it was not rotating?
Also energy due to "stuff" like matter or light or whatever.name123 said:So the gravitational energy causes the time dilation?
https://en.wikipedia.org/wiki/Stress–energy_tensorname123 said:So the gravitational energy causes the time dilation?
Yes, completely.name123 said:If each had a ruler and a clock on board, when they were at rest together could they not all go by their own clock and ruler, and agree that their clocks and rulers were metrically equivalent?
The question is ambiguous. I cannot tell if you are asking about proper rotation or coordinate rotation.name123 said:Could you explain this please?
Dale said:The question is ambiguous. I cannot tell if you are asking about proper rotation or coordinate rotation.
Whether or not a given object is undergoing proper rotation is an invariant. It does not depend on the coordinates chosen. You can choose coordinates where a proper-rotating object is at coordinate-rest. In such coordinates there will be "fictitious forces" which will lead to the correct amount of proper rotation.name123 said:Proper rotation assuming that it what causes the proper acceleration that can be measured.
Dale said:Yes, completely.name123 said:If each had a ruler and a clock on board, when they were at rest together could they not all go by their own clock and ruler, and agree that their clocks and rulers were metrically equivalent?Dale said:name123 said:Consider satellites A and B, going at different velocities, in free fall orbit around a massive body C at different altitudes for a million years, then being brought together and the clocks on them compared. Presumably the bringing them together would become less significant the longer they orbited, and the amount of time dilation due to curvature would be frame of reference independent, but what about the observed velocities from the mass being orbited's perspective? If the satellites were labelled A and B and the mass C then would the clock comparison figure be correctly calculated no matter which you considered at rest?
Yes, but note that the metric will have a different form in these different charts. With a correct expression for the metric, all invariants will be correctly calculated.
Dale said:Whether or not a given object is undergoing proper rotation is an invariant. It does not depend on the coordinates chosen. You can choose coordinates where a proper-rotating object is at coordinate-rest. In such coordinates there will be "fictitious forces" which will lead to the correct amount of proper rotation.
Yes, as edited.name123 said:So A would agree with B about the expected differences between their clocksdue to gravitational time dilation. And the affect due to bringing the clocks together could be considered insignificant if the orbiting had gone on for long enough.
Dale said:I have no idea how to parse the other question. In GR, the proper time on any clock undergoing any motion in any frame in any spacetime is given by ##\int \sqrt{ g_{\mu \nu} dx^{\mu} dx^{\nu}}##
Dale said:Why don't you work through the first few chapters of Sean Carroll's lecture notes on general relativity first.
name123 said:Also I am not quite clear how A or B could be considered at rest and not rotating, as while C (the mass) could be considered to orbit A or B, if it was so considered, would it not also need to be considered that it was rotating, for an explanation of why the same section of C was not always facing A (or B), and could it not be objectively measured that it was not rotating?
Dale said:Whether or not a given object is undergoing proper rotation is an invariant. It does not depend on the coordinates chosen. You can choose coordinates where a proper-rotating object is at coordinate-rest. In such coordinates there will be "fictitious forces" which will lead to the correct amount of proper rotation.
Dale said:Yes, they are free
https://arxiv.org/abs/gr-qc/9712019
Working an example without the background would require a lot of effort on my part and result in very little gain for you. But reading the first couple of chapters of the lecture notes will result in a lot more gain for you.
Which is a good hint that they aren’t inertial.name123 said:Also from the perspective of the inertial frames A and B if there were distant stars, wouldn't they appear to be moving faster than the speed of light?
They are only locally inertial. And even locally it is only inertial if the object is not undergoing any proper rotation.name123 said:But the question was about from the frame of reference A or B which as I understand it are both inertial frames of reference.
Neither the stars nor the other objects are local.name123 said:Also from the perspective of the inertial frames A and B if there were distant stars, wouldn't they appear to be moving faster than the speed of light?
Dale said:They are only locally inertial.
In curved space-time there is no such thing as a global inertial frame of reference.name123 said:reference A or B which as I understand it are both inertial frames of reference
Rotating frames are not inertial.name123 said:So I then assume A relative to B and C in the scenario should be considered as non-inertial even though an accelerometer at rest with respect to A would measure no acceleration?
Yes, if you want to make any non-local measurements then you need to consider them to be non inertial.name123 said:So I then assume A relative to B and C in the scenario should be considered as non-inertial even though an accelerometer at rest with respect to A would measure no acceleration?
They are called "Christoffel symbols" (I know, it is a weird name). The lecture notes describe them in detail.name123 said:What is the fictitious force that would explain it?