Time dilation equilibrium with speed and gravity

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Jimmyboy

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Does anyone know if there is some kind of ratio between the speed of an B object and the gravity of object A for time dilation to be at some kind of equilibrium. I will try to explain it clearer.. If a person (A) is on earth they will age slower than if a person (B) was in space due to the gravity of Earth... but the speed of an object can also effect time, so at what speed would person B have to travel relative to earth to be at some kind of time dilation equilibrium... it there some ratio between gravity and speed???
 
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csmyth3025

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Jimmyboy":3gywa61g said:
I'll try to answer your original post from December - I don't know enough about this to be sure I'm giving you the right answer, though. Hopefully, someone else will read this and correct any mistakes I make.

As I understand it, time for an atomic clock here on Earth at sea level passes a tiny bit slower than time for an atomic clock located in Denver at an elevation of 5,280 ft. More to the point of your question, I found the following passage [edited] in the Wikipedia article on GPS (global positioning system):

"...For example, the relativistic time slowing due to the speed of the satellite [is] about 1 part in 10^10 [and] the gravitational time dilation [on Earth] makes a satellite [clock] run about 5 parts in 10^10 faster than an Earth based clock..."
From this passage I'm guessing that the Earth's gravity causes a clock on its surface to run one-half second slow for every billion seconds counted by the clock in the GPS satellite - which, being in orbit around the Earth, could be considered a free-falling object in the Earth's gravitational field and, therefore, an inertial frame of reference. In this regard I believe the clock on the GPS satellite acts as if it's floating freely in space outside of any gravitational effects. One way to say this is that for every second the clock in the GPS satellite ticks off, the clock on the surface of the Earth ticks off 0.9999999995 seconds. A better way to say the same thing (for our calculation) is that for every second the clock on Earth ticks off a second, the clock in the GPS satelite ticks off 1/0.9999999995 seconds, or about 1.000000001 seconds.

If I understand your original question correctly, you want to know how fast a space ship which is not being effected by any gravitational field would have to go (in a direction away from Earth) in order for the clock on the space ship to run as slow as an identical clock on the surface of the Earth. There is a formula - called the Lorentz factor - by which this speed can be calculated. The symbol for the Lorentz factor is the lower case Greek letter Gamma - which looks like a curly "y". I'll just use y since I don't how to make greek letters on my computer:

y(delta t)=1/sqrt[1-(v^2/c^2)]
In this case (delta t)=the seconds ticked off by the clock in the space ship (which we'll say is one second to simplify matters), y=the Lorentz factor, [y(delta t)]= the seconds ticked off by the "stationary" clock, v=the velocity of the space ship relative to Earth, c=the speed of light (~300,000 km/sec), and sqrt means square root.

We already know that we want the "staionary" clock in an inertial frame of reference (outside the influence of gravity) to run faster, relative to our space ship, by the same amount as the GPS clock does, relative to the clock on the Earth. This amount - or factor - is 1.000000001 seconds for every second on our space ship. In other words y=1.000000001. Since we've said that (delta t)=1 second we can leave it out of the discussion. We also know that y=1/sqrt[1-(v^2/c^2)]. To make this formula work we know that the "one" in the numerator on the right side is going to have to be divided by a number that's just a tiny bit smaller than one. As the original formula for the Lorentz factor suggests, this number is the reciprocal of 1.000000001 or 0.9999999995. This number is equal to the denominator on the right side of the equation, which is sqrt[1-(v^2/c^2)].

The square of 0.9999999995=~0.999999999. Since this=[1-(v^2/c^2)], then (v^2/c^2)=0.000000001 or (1 x 10^-9). Multiplying both sides by c^2 (9 x 10^10 km^2/sec^2) we wind up with v^2=90 km^2/sec^2 and, taking the square root of both sides we have v=~9.4868 km/sec.

If my calculation is right, then the space ship would have to be travelling about 21,200 mph for the clocks on the ship to match the clocks on Earth.

I have to caution you again that I probably don't know any more than you do about how these things are calculated. I may be using this formula entirely wrong and my math probably sucks. It's the best I can do. I sure hope someone else looks at this for your sake.

Chris
 
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Jimmyboy

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Thanks Chris,

The answer sounds about right, after all there isnt much difference between time on a shuttle and time on earth, I came up with 774 km/s which doesnt sound right, using your formular. I would have thought Earths gravitation would have been in the formular, perhaps it is, maybe that is what the 1 is for for the numerator.
 
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SpeedFreek

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I remember seeing the OP and going away to think about it, but the more I thought about it the more complicated it got, and I forgot to come back and post something! I am not 100% sure of all of this myself, so again, take this on approval.

I started down the same avenue as Chris (nice post, btw), but then as I thought it through I had a realisation. It all depends on how accurate an approximation you want.

If we are going to use relativity, we also have to consider the problem from the view of the traveller. Time-dilation is symmetrical between observers in relative inertial motion. Each will calculate the other is "running slower" by the same amount. So then we have to consider the accelerations involved, during which time the relationship between the two observers will always be changing asymmetrically. So we can never have the twins remaining the exact same age as each other, but they might continue to age at a similar rate once they are both in inertial frames.

But then we have to consider how we are going to define time, for the purposes of this thought experiment.

Firstly, when in an inertial frame in orbit around the Earth, you are still within the Earths gravitational influence and so are still time-dilated relative to an observer as far away from any gravitational influence as possible. An inertial orbit is still subject to gravitational time dilation, and the same is true for any object receding from Earth.

As you recede at a constant speed, the difference in the gravitational potential between the two clocks will always be changing. The only place it stops changing is at infinity - as far away from the gravitational influence as possible. Of course, the effect becomes negligible after a certain distance, but you might want to take it into account.

It all depends on why you want to know the answer - i.e. what you want to do with the information, how you want to apply it to other problems.

At first approximation, we might be able to find a relative speed between a spaceship and Earth where the amount of time-dilation is relatively small, but we will never be able to find a circumstance where it is zero.

There is no orbit where the relative motion balances out the gravitational time dilation, and no other inertial frame that remains at the same gravitational potential.



And then, depending on how want to use the answer, we might run headlong into the relativity of simultaneity. As I said earlier, it depends on how we want to define time between the two frames. Is the other frame ever coming back? Will they always be moving at the same relative speed?
 
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Jimmyboy

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Thanks speedfreak

These two posts have giving me abit of a headache although I do understand it (and i did ask for it)!
I suppose it makes you wonder in the future if and when intersteller space travel is possible, and time dilation becomes a problem when space travellers have family and such, that cannot go on the trip with them, that maybe some kind of accelerator could be used on earth to compensate the phenomenon..???
 
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MeteorWayne

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A relevant discussion of GPS time corrections...while the orbit is higher (hence slower and further out of the earth's gravity well) the concepts are the same.

And don't worry, this stuff is supposed to make your brain hurt :)

Special and general relativity

According to the theory of relativity, due to their constant movement and height relative to the Earth-centered, non-rotating approximately inertial reference frame, the clocks on the satellites are affected by their speed. Special relativity predicts that the frequency of the atomic clocks moving at GPS orbital speeds will tick more slowly than stationary ground clocks by a factor of , or result in a delay of about 7 μs/day, where the orbital velocity is v = 4 km/s, and c = the speed of light. The time dilation effect has been measured and verified using the GPS system.

The effect of gravitational frequency shift on the GPS system due to general relativity is that a clock closer to a massive object will be slower than a clock farther away. Applied to the GPS system, the receivers are much closer to Earth than the satellites, causing the GPS clocks to be faster by a factor of 5×10^(-10), or about 45.9 μs/day. This gravitational frequency shift is also a noticeable effect.

When combining the time dilation and gravitational frequency shift, the discrepancy is about 38 microseconds per day; a difference of 4.465 parts in 1010. Without correction, errors in position determination of roughly 10 km/day would accumulate. In addition, because GPS satellite orbits are not perfectly circular, their elliptical orbits cause the time dilation and gravitational frequency shift effects to vary with time. This eccentricity effect causes the clock rate difference between a GPS satellite and a receiver to increase or decrease depending on the velocity orbital altitude of the satellite.

To account for the discrepancy, the frequency standard on board each satellite is given a rate offset prior to launch, making it run slightly slower than the desired frequency on Earth; specifically, at 10.22999999543 MHz instead of 10.23 MHz. Since the atomic clocks on board the GPS satellites are precisely tuned, it makes the system a practical engineering application of the scientific theory of relativity in a real-world environment. Placing atomic clocks on artificial satellites to test Einstein's general theory was proposed by Friedwardt Winterberg in 1955.
 
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csmyth3025

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This discussion prompts me to ask a question I've been wondering about:

If two satelites, both with atomic clocks, are in different orbits - one higher and one lower - will the clock on the satellite in the lower orbit run slower than the clock on the satellite in the higher orbit due to being "deeper" in the gravitational well of the Earth (after adjusting for their relative velocities).

I ask this question because of the concept that an object in orbit is essentially free falling in a gravitational field. As I understand it, the physical laws for an object that's free falling in a (uniform) gravitational field are indistinguishable from those of a object that's freely floating in space outside any gravitational influence (an inertial frame). The relative size of the satelites compared to the Earth would, I think, allow them to be treated as point-like objects not subject to tidal effects.

Chris
 
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SpeedFreek

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csmyth3025":ydyeievx said:
If two satelites, both with atomic clocks, are in different orbits - one higher and one lower - will the clock on the satellite in the lower orbit run slower than the clock on the satellite in the higher orbit due to being "deeper" in the gravitational well of the Earth (after adjusting for their relative velocities).
Yes, the clock in the lower orbit will "run slower".

csmyth3025":ydyeievx said:
I ask this question because of the concept that an object in orbit is essentially free falling in a gravitational field. As I understand it, the physical laws for an object that's free falling in a (uniform) gravitational field are indistinguishable from those of a object that's freely floating in space outside any gravitational influence (an inertial frame).
The laws of physics are also indistinguishable between inertial frames of reference in relative motion, but there will be time-dilation between those frames.
 
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MeteorWayne

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csmyth3025":3jtahpyx said:
This discussion prompts me to ask a question I've been wondering about:

If two satelites, both with atomic clocks, are in different orbits - one higher and one lower - will the clock on the satellite in the lower orbit run slower than the clock on the satellite in the higher orbit due to being "deeper" in the gravitational well of the Earth (after adjusting for their relative velocities).

I ask this question because of the concept that an object in orbit is essentially free falling in a gravitational field. As I understand it, the physical laws for an object that's free falling in a (uniform) gravitational field are indistinguishable from those of a object that's freely floating in space outside any gravitational influence (an inertial frame). The relative size of the satelites compared to the Earth would, I think, allow them to be treated as point-like objects not subject to tidal effects.

Chris
Please read my post above yours. General and special relativity have two different effects that partially cancel out.

I think it clearly answered your question.
 
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Jimmyboy

Guest
"To account for the discrepancy, the frequency standard on board each satellite is given a rate offset prior to launch, making it run slightly slower than the desired frequency on Earth; specifically, at 10.22999999543 MHz instead of 10.23 MHz. "

Meteor could you explain this please, does this refer to how fast the onboard satellite computers operate?
 
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MeteorWayne

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No, though it does have a veryveryvery minor effect on the computing rate.

It refers to the master clock frequency.
 
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