Time Dilation in Special vs. General Relativity

[swansont]

You can see this in fig 2 here

http://relativity.livingreviews.org/Articles/lrr-2003-1/fulltext.html

 

"The effects cancel at "

[Robittybob1]

You can see this in fig 2 here

http://relativity.livingreviews.org/Articles/lrr-2003-1/fulltext.html

 

"The effects cancel at "

I struggled to find that reference. I click the link and a page comes up but no figures are on it. Do I have to scroll through the document?

I've downloaded the document http://relativity.livingreviews.org/Articles/lrr-2003-1/download/lrr-2003-1Color.pdf

and the figures are showing up OK. thanks.

Edited May 29, 2016 by Robittybob1

[Janus]

I wasn't trying to correct Janus, I was just trying to work out which bit was the time dilation factor (so it is the bit under the square root sign. You say both t' and T start at 0 i.e. that is when t'=T=0, I presume. So to see how much time dilation there was I subtracted t'-T = TD (time dilation) [is t'-T = TD (time dilation) a valid way of looking at it?]. I then tried to look for the region around the Earth in which an orbiting satellite experiences gravitational time dilation (GTD) equal but opposite to the velocity dependent TD (VTD).

Since in #7 it is agreed that:

I was looking for the situation where instead of T(International Space Station)<T(GPS satellite)

both would have combined time dilation of zero, where T(New satelite)=T(on Earth surface).

You're looking for the situation where

 

[math]\sqrt{1-\frac{3GM}{rc^2}} = \sqrt{1-\frac{2GM}{r_e c^2}}[/math]

 

[math]1-\frac{3GM}{rc^2} = 1-\frac{2GM}{r_e c^2}[/math]

 

[math]-\frac{3GM}{rc^2} = -\frac{2GM}{r_e c^2}[/math]

 

[math]\frac{3}{r} = -\frac{2}{r_e}[/math]

 

[math] r= \frac{3r_e}{2}[/math]

 

Where r is the radius of the orbit and r_e is the radius of the Earth.

[Robittybob1]

You're looking for the situation where

 

.....

 

[math] r= \frac{3r_e}{2}[/math]

 

Where r is the radius of the orbit and r_e is the radius of the Earth.

Is this the same as saying it orbits at a height (from the mean Earth's radius) that is half the Earth's mean radius? (A whole and a half gives the 3/2 fraction. This is value I had proposed on another site but I was told I was wrong. I trust your result totally.

As above ""The effects cancel at [latex]\approx 9545 kilometers[/latex]" My spreadsheet result was 9556814 m (9557 km) so there is a small difference only. Thank you very much.

 

PS: I was using Earth radius = 6371000 m, that times 3/2 = 9556500 m. The result above was the first value that was after the change in sign so that would explain the small difference of the additional 314 meters for I was changing the velocity by 1 km/h increments.

Edited May 29, 2016 by Robittybob1

[Robittybob1]

If the second formula is the combined effect, what is the first one measuring? Since it hasn't got a velocity term in it I was suspecting it is the gravitational time dilation at that radius value.

Was that correct?

I made [math] t' = 1 [/math] and compared the result to the orbital speeds, and there was point where the results became extremely small and then changed signs. I think this must be the point where "the combined effect of gravitational and relative velocity time dilation effects" = zero.

In other words [math]T = 1 \sqrt{1-\frac{2GM}{rc^2}- \frac{v^2}{c^2}}=0[/math]

Which must mean [math]{1-\frac{2GM}{rc^2}- \frac{v^2}{c^2}}=0[/math]

The sqrt of zero is zero (that makes sense, and confirmed). So why would that expression become zero?

 

That can only happen when both fractions are equal and equal to -1/2. ie 1 - 1/2 - 1/2 = 0

Do you agree?

 

PS: No that is wrong. Both fractions just have to add to -1 they don't both need to be equal to -1/2.

 

What is the value of the fractions when ""the combined effect of gravitational and relative velocity time dilation effects" = zero""?

 

I'll look at this again tomorrow.

Note I am seriously wrong here. I will correct it later. If I wanted GTD and VTD to be equal but opposite this next equation needs to be equal to 1 not zero

[latex]T = 1 \sqrt{1-\frac{2GM}{rc^2}- \frac{v^2}{c^2}}=0 [/latex]

 

[latex]T = 1 \sqrt{1-\frac{2GM}{rc^2}- \frac{v^2}{c^2}}= 1 [/latex] I may need to rethink the algebra on this one sorry.

Edited May 30, 2016 by Robittybob1

[Janus]

Note I am seriously wrong here. I will correct it later. If I wanted GTD and VTD to be equal but opposite this next equation needs to be equal to 1 not zero

[latex]T = 1 \sqrt{1-\frac{2GM}{rc^2}- \frac{v^2}{c^2}}=0 [/latex]

 

[latex]T = 1 \sqrt{1-\frac{2GM}{rc^2}- \frac{v^2}{c^2}}= 1 [/latex] I may need to rethink the algebra on this one sorry.

The formula as stated is the time dilation factor as measured by our far distant observer (someone who is much, much further from the mass M than r and at rest with respect to M.) For this observer there is no situation for which gravitational time dilation and time dilation due to relative velocity "cancel out" . Clocks closer to M(lower in the gravity field) than he is run slower and clocks with a non-zero velocity relative to him also run slower. The two effects can only compound each other and never work against each other. It is only an observer closer to the center of M than r that can see the two effects offset each other. The clock higher than he is runs faster due to gravitational time dilation, but runs slower due to its relative velocity.

[Robittybob1]

The formula as stated is the time dilation factor as measured by our far distant observer (someone who is much, much further from the mass M than r and at rest with respect to M.) For this observer there is no situation for which gravitational time dilation and time dilation due to relative velocity "cancel out" . Clocks closer to M(lower in the gravity field) than he is run slower and clocks with a non-zero velocity relative to him also run slower. The two effects can only compound each other and never work against each other. It is only an observer closer to the center of M than r that can see the two effects offset each other. The clock higher than he is runs faster due to gravitational time dilation, but runs slower due to its relative velocity.

So do you disagree with Swansont's post http://www.scienceforums.net/topic/95329-time-dilation-in-special-vs-general-relativity/page-2#entry922497?

I am having trouble and I must admit I haven't got the problem licked yet.

I found a chart in Wikipedia that might help.

https://commons.wikimedia.org/wiki/File:Orbit_times.svg

This graph shows what I have been trying to calculate. The "net orbital time gain line" meets zero line at just below 10,000 km from the Earth's center. This is the r value in the equations. Considering the Earth's radius is 6371 km and I estimated the satellite height of 3185 km that adds to 9558 km from the Earth's center. These figures match the chart.

 

Have I misunderstood what the chart is showing?

Associated with the article https://en.wikipedia.org/wiki/Error_analysis_for_the_Global_Positioning_System

Edited May 30, 2016 by Robittybob1

[swansont]

So do you disagree with Swansont's post http://www.scienceforums.net/topic/95329-time-dilation-in-special-vs-general-relativity/page-2#entry922497?

I am having trouble and I must admit I haven't got the problem licked yet.

I found a chart in Wikipedia that might help.

https://commons.wikimedia.org/wiki/File:Orbit_times.svg

This graph shows what I have been trying to calculate. The "net orbital time gain line" meets zero line at just below 10,000 km from the Earth's center. This is the r value in the equations. Considering the Earth's radius is 6371 km and I estimated the satellite height of 3185 km that adds to 9558 km from the Earth's center. These figures match the chart.

 

Have I misunderstood what the chart is showing?

Associated with the article https://en.wikipedia.org/wiki/Error_analysis_for_the_Global_Positioning_System

 

 

What I posted was not from the perspective of a far distant observer, it was from the perspective of someone on earth, which is the question you asked.

 

A distant observer is only going to see effects that slow the clocks — deeper in the gravitational well and moving faster. Someone on earth will see effects of opposite sign, since the gravitational effects speed the clocks up.

[Janus]

So do you disagree with Swansont's post http://www.scienceforums.net/topic/95329-time-dilation-in-special-vs-general-relativity/page-2#entry922497?

I am having trouble and I must admit I haven't got the problem licked yet.

I found a chart in Wikipedia that might help.

https://commons.wikimedia.org/wiki/File:Orbit_times.svg

This graph shows what I have been trying to calculate. The "net orbital time gain line" meets zero line at just below 10,000 km from the Earth's center. This is the r value in the equations. Considering the Earth's radius is 6371 km and I estimated the satellite height of 3185 km that adds to 9558 km from the Earth's center. These figures match the chart.

 

Have I misunderstood what the chart is showing?

Associated with the article https://en.wikipedia.org/wiki/Error_analysis_for_the_Global_Positioning_System

As already explained, Swansont's post was in regard to an observer on the surface of the Earth. For such a observer what would be the orbital radius need to be for a clock in orbit to tick at the same rate as his own.

 

In post #28, I come at it from a slightly different angle. I assumed a far off observer looking at both the orbiting clock and one resting on the surface, and looked for the orbit which had the two ticking at the same rate as each other (but not at the same rate as his own clock.) He sees both clocks running slow by the same factor( as long as he is at rest with respect to the surface clock.). From his perspective, both clocks run slow due to gravitational time dilation, with this effect being greater on the lower surface clock. But the higher orbiting clock also has an added time dilation due to its relative motion, and this makes up for the difference in gravitational time dilation, so the two clocks run at the same rate.

 

For the surface observer, the higher orbiting clock runs fast due to the difference in gravitational time dilation, but also has a relative motion time dilation that slows it down. The two effects cancel out so that he measures the orbiting clock as running the same speed as his own. Same end result for these two particular observers.

[Robittybob1]

 

He sees both clocks running slow by the same factor( as long as he is at rest with respect to the surface clock.).

Just seeing if I misunderstand you on this one point only. (The rest of it, I think I got it.)

Breaking down the quoted portion:

"He" refers to far off observer looking at both the orbiting clock and one resting on the surface (of the rotating Earth). Can that part "(of the rotating Earth)" be added or not?

"both clocks" refers to the one on the surface and the one in orbit 9558 km from the center of the Earth.

"he is at rest with respect to the surface clock" So does this just occur only for an instant, for if he was to maintain that he would need to be orbiting at a height where the orbital speed is the same as the rate of the surface of the Earth (I don't know at what height (x) that occurs).

Is that possible and would that equation hold if the distant observer was orbiting at that height (x)?

If two clocks are ticking at the same rate e.g the one on the surface and the one in orbit 9558 km from the center of the Earth, will they always appear to be ticking at the same rate to everyone else? (Not to their own clock's tick rate but the tick rate signal received from the other two, or is this always dependent on the relative velocity being zero to the "surface clock".)

Edited May 30, 2016 by Robittybob1

[swansont]

Rotation of the earth ultimately doesn't matter; the dilation from the rotation is balaced by the gravitational dilation from the deformation of the earth. The only local effect is from elevation.

[Robittybob1]

Rotation of the earth ultimately doesn't matter; the dilation from the rotation is balaced by the gravitational dilation from the deformation of the earth. The only local effect is from elevation.

I had sort of understood this as well, with the time dilation on the geoid being the same no matter where it was measured. Elevation then is like bumps on the surface (land height above sea level, mountains etc) is that what you meant?

Edited May 30, 2016 by Robittybob1

[swansont]

Yes, I am using the usual definition of elevation

[Robittybob1]

You're looking for the situation where

 

[math]\sqrt{1-\frac{3GM}{rc^2}} = \sqrt{1-\frac{2GM}{r_e c^2}}[/math]

 

[math]1-\frac{3GM}{rc^2} = 1-\frac{2GM}{r_e c^2}[/math]

 

[math]-\frac{3GM}{rc^2} = -\frac{2GM}{r_e c^2}[/math]

 

[math]\frac{3}{r} = -\frac{2}{r_e}[/math]

 

[math] r= \frac{3r_e}{2}[/math]

 

Where r is the radius of the orbit and r_e is the radius of the Earth.

Since at this radius [math] r= \frac{3r_e}{2}[/math] the absolute value of the gravitational time dilation (GTD) equals the absolute value of the velocity dependent Time dilation (VTD) time on this satellite will still be different than at the surface of the Earth (geoid). I think previously I may have thought just because they cancelled each other out they may have been the same as on the geoid.

 

What can we say about the time dilation at the geoid? We are moving and we are in a gravitational field, they may not be equal but I have no idea how we could calculate that. Do we just add these [math]\sqrt{1-\frac{3GM}{rc^2}} and \sqrt{1-\frac{2GM}{r_e c^2}}[/math] where [math]{r} = {r_e} [/math] ?

Edited June 1, 2016 by Robittybob1

[swansont]

Since at this radius [math] r= \frac{3r_e}{2}[/math] the absolute value of the gravitational time dilation (GTD) equals the absolute value of the velocity dependent Time dilation (VTD) time on this satellite will still be different than at the surface of the Earth (geoid). I think previously I may have thought just because they cancelled each other out they may have been the same as on the geoid.

 

What can we say about the time dilation at the geoid? We are moving and we are in a gravitational field, they may not be equal but I have no idea how we could calculate that. Do we just add these [math]\sqrt{1-\frac{3GM}{rc^2}} and \sqrt{1-\frac{2GM}{r_e c^2}}[/math] where [math]{r} = {r_e} [/math] ?

The overall dilation value on the geoid is taken to be zero, as that is the reference point. We only care what the total effect is on a satellite, because that's what affects the clock reading.

[Robittybob1]

The overall dilation value on the geoid is taken to be zero, as that is the reference point. We only care what the total effect is on a satellite, because that's what affects the clock reading.

What I was trying to understand is whether at the height 9558 km where the two effects cancel and at the geoid time runs at the same rate. If the overall dilation rate at both places is zero does that mean also they run at the same rate?

Do you know of any satellite that has taken advantage of this effect at this orbital radius?

[swansont]

What I was trying to understand is whether at the height 9558 km where the two effects cancel and at the geoid time runs at the same rate. If the overall dilation rate at both places is zero does that mean also they run at the same rate?

That's precisely what it means. There is no net time dilation. Time in each frame will run at the same rate.

 

 

Do you know of any satellite that has taken advantage of this effect at this orbital radius?

I'm not aware of any. Google might know.

[Robittybob1]

That's precisely what it means. There is no net time dilation. Time in each frame will run at the same rate.

 

 

 

I'm not aware of any. Google might know.

I tried Googling it once before and only got 1 result, and that was just confirming the height. I was stuck on the search terms to use that might help me with the search. There doesn't seem to be a name for this particular orbit. I'll try again tomorrow. Thanks.

Edited June 1, 2016 by Robittybob1

[imatfaal]

Do you know of any satellite that has taken advantage of this effect at this orbital radius?

 

Not as far as I can tell. A quick look at a table of satellites shows only a couple of satellites with semi-major axes even close (ie within a couple of thousand km). There are lots of low eath orbit with figures in the 1400 and below and lots of mid earth orbits over 20,00km; but only a handful in between - none close to your figure

[Robittybob1]

 

Not as far as I can tell. A quick look at a table of satellites shows only a couple of satellites with semi-major axes even close (ie within a couple of thousand km). There are lots of low eath orbit with figures in the 1400 and below and lots of mid earth orbits over 20,00km; but only a handful in between - none close to your figure

I'm thinking if it has the ratio of being exactly half the Earth's radius above the mean radius of the Earth and having the exact time dilation factor of the geoid it struck me as being a good orbit to have a GPS type system but it isn't being used.

Why wouldn't someone use it? Are we missing something about these time dilations?

 

I realise if we had two synchronous clocks and put one in a satellite and put that satellite into orbit at that radius the times would not stay the same during the journey, but even so they could then be reset if needed.

[Janus]

I'm thinking if it has the ratio of being exactly half the Earth's radius above the mean radius of the Earth and having the exact time dilation factor of the geoid it struck me as being a good orbit to have a GPS type system but it isn't being used.

Why wouldn't someone use it? Are we missing something about these time dilations?

 

I realise if we had two synchronous clocks and put one in a satellite and put that satellite into orbit at that radius the times would not stay the same during the journey, but even so they could then be reset if needed.

Putting a GPS satellite at an orbital radius of 1.5 Earth radii, would be trading a small adjustment to the GPS clocks for a much bigger headache. At their present altitude they orbit the Earth twice a day. Thus twice a day, any satellite has exactly the same relative position relative to the surface of the Earth. This makes it relatively easy to calculate your position. (At 1:00 you know that three satellites (a,b and c) be in certain positions relative to the Earth, and by determining their positions relative to You you know where you are.)

 

At an altitude where the clocks run at the same rate on the surface, the orbital period is 2 hr 35 min. Since this does not fit evenly into a day, you can't tell where the satellite is at any given time unless you also know how many times it orbited since its last known position. The complicates the working out of your position and increases the chances of errors sneaking in.

[Robittybob1]

Thanks Janus. You always surprise me with your extensive knowledge on the subject. Could you tell us what you do/did to be so clued up?

@ Mods - are we allowed to continue discussing how the GPS system works? [i recall a lot of it revolves around need to understand time dilation.]

[Strange]

I think it is slightly subtler than that (but it is many years since I worked on GPS systems).

 

When the GPS receiver finds a first satellite (which can take quite a while because it has to scan a wide range of frequencies looking for possible signals), the first thing it does is download the data (almanac?) with information about the current positions of all the satellites. It can then use this to find other satellites that should be visible much more quickly (because it knows their relative speeds and can do a rough compensation for Doppler effects).

 

The satellites need to be updated regularly with the positions (and functional status) of all the other satellites in the network. With the high orbits, this doesn't have to be done very often. If they were orbiting every three hours, they would be updating almost constantly. And the receiver would struggle to track satellites as they flew past and would be constantly having to look for new ones (it needs a minimum of four to calculate location, more is better, but some will be blocked by buildings, etc).

[swansont]

There's also the issue of using GPS satellites for "common view" time transfer. An orbital period of 2.5 hours means you get to "see" the satellite for maybe 30-60 minutes, depending on your field of view, and the common time for two people to see the satellite would be even shorter. That limits the precision of the measurement.

 

The clocks on the satellites are adjusted for their altitude, so that one can quasi-synchronize with the GPS clocks. It's not that big of a deal.