# Stopping the clock

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From my understanding, it seems that stopping time would be like reaching light speed with matter (like .75c + .75c = .96c). Relative to Earth clocks we can almost stop time as we increase relativistic speeds close to C. We can also almost stop time relative to Earth if we could put a clock on a black hole. So I think that a black hole traveling at relativistic speed would just have a very very slow clock.

Is this in the ball park?

I have been looking for a gravity time calculator but so far have not found exactly what I want. The "space math" site has everything but what I want. I would like to compare earth clocks to different g forces. Would Gravitational potential vs G-force complicate things?

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...I have been looking for a gravity time calculator but so far have not found exactly what I want. The "space math" site has everything but what I want. I would like to compare earth clocks to different g forces. Would Gravitational potential vs G-force complicate things?

I don't know of anything absolutely ideal, like an online calculator or even a really clear explanation.

But I just now tried wikipedia "graviational redshift" and it seemed OK

http://en.wikipedia.org/wiki/Gravitational_red_shift.

It gives an approximate formula for z.

Let's consider the ratio of clock rates to be z+1

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it gives the ratio of frequencies between a distant clock, with no redshift at all from earth gravity, and a clock on the earth's surface.

You just have to plug in the earth mass for M and the earth radius for r. And find z + 1.

It is a simple enough formula.

And if you have two clocks one at the top of the building and one at the ground floor, then ground clock would be slower than top clock by the ratio

(ztop + 1)/(zground + 1)

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Down at the bottom of the wikip page they give a slightly more complicated more exact formula which basically says the clockrate ratio

1/(z+1) = sqrt(1 - 2GM/rc2)

It is tempting, as an example, to use google calculator and find by what ratio our earth surface clocks are slowed down compared with a clock out in space far from earth. (forget about other gravity fields for simplicity, like the sun etc.)

Try putting this in the google window and press return

sqrt( 1 - 2G*mass of earth/(radius of earth*c^2))

ugh! it says our earth clocks are slower by a factor of

0.99999999

only slightly slower and probably too small for the precision of the calculators. ugh bah and fiddlesticks (Phi objects to profanity).

I will try just the important number

(G * mass of Earth) / (radius of Earth * (c^2)) = 6.95453588 × 10^-10

In other words the redshift is approximately 0.7 parts per billion.

Our earth clocks are slower by about 0.7 parts per billion.

The great thing is that google calculator knows the mass of earth and radius of earth and Newton G etc etc so this is all very quick to do. Don't have to look up constants.

Edited by Martin
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...

Try putting this in the google window and press return

sqrt( 1 - 2G*mass of earth/(radius of earth*c^2))

ugh! it says our earth clocks are slower by a factor of

0.99999999

only slightly slower and probably too small for the precision of the calculators. ugh bah and fiddlesticks (Phi objects to profanity).

...

I will try just the important number

(G * mass of Earth) / (radius of Earth * (c^2)) = 6.95453588 × 10^-10

In other words the redshift is approximately 0.7 parts per billion.

Our earth clocks are slower by about 0.7 parts per billion.

But Martin, Is the 0-G from being in orbit even being considered? Or does it need to be considered? It doesn't look like 0-G is being considered when calculating time delays for the GPS system.

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"Further, the satellites are in orbits high above the Earth, where the curvature of spacetime due to the Earth's mass is less than it is at the Earth's surface. A prediction of General Relativity is that clocks closer to a massive object will seem to tick more slowly than those located further away (see the Black Holes lecture). As such, when viewed from the surface of the Earth, the clocks on the satellites appear to be ticking faster than identical clocks on the ground. A calculation using General Relativity predicts that the clocks in each GPS satellite should get ahead of ground-based clocks by 45 microseconds per day."

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I would have thought that the 0-G from being in orbit would be more significant then a floor and ceiling type calculation. I am a little confused now.

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When people talk about 0G they do not mean there is no gravity (well they might but they're wrong), they mean they are in freefall, normally with what they're in as well.. It is true that gravity is lower, which is what your quote says, but there is still gravity from the earth.

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When people talk about 0G they do not mean there is no gravity (well they might but they're wrong), they mean they are in freefall, normally with what they're in as well.. It is true that gravity is lower, which is what your quote says, but there is still gravity from the earth.

So a satellite has a balance of centrifugal force and centripetal force and is actually in a free fall. (1) Does true 0G exist anywhere in the universe (between galaxies)? So if our solar system was located closer to the black hole in the center of the Milky Way the clocks would be slower but our gravity would feel the same. (2) Artificial gravity from spinning would slow a clock, right?

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from the little bit of math i know the force of gravity decreases by the square of the distance so something that weighs 1 pound some distance from the surface of earth,weighs 1/4 pound when it's twice that distance 1/d^2. since 1 divided by a huge number is a tiny little number no matter how huge the big number gets (even 13 billion light years) this means earths gravity(for example) is still a force even at the edges of the observable universe.

i don't know about centrifugal force and time dilation it seems like you would need to be spinning really fast to slow a clock.

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Thanks moth, That's the kind of equation I can work with.l I did find

"an accelerating dragster or space shuttle. Spinning objects such as merry-go-rounds and ferris wheels are subjected to gravitational time dilation as an effect of their angular spin."

I feel much better about gravitational time dilation. My last question is how much will each additional artificial G force on earth (if location matters) slow a clock?

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in my last post i said the distance from the surface of the earth when i should have said center of the earth but i think you get the idea.

i'm just not sure if it applies when you're underground.it seems like your weight would decrease underground until you reached the center where you would be weightless.

i don't think artificial gravity would cause time dilation but thats just my opinion. there are people here who know.

i guess it's a question of wording. it would be interesting to find out if 1G in a rotating artificial gravity chamber caused the same amount of Lorentz contraction and dilation as 1G on earth, but when i try the math i get confused about that vx/c^2 term in the time part of Lorentz equation.

Edited by moth
expanding
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in my last post i said the distance from the surface of the earth when i should have said center of the earth but i think you get the idea.

I caught that too I was like "a ball 1 meter off the ground can't be 1/4 its weight an extra meter higher."

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So a satellite has a balance of centrifugal force and centripetal force and is actually in a free fall. (1) Does true 0G exist anywhere in the universe (between galaxies)? So if our solar system was located closer to the black hole in the center of the Milky Way the clocks would be slower but our gravity would feel the same. (2) Artificial gravity from spinning would slow a clock, right?

The use of centrifugal force means you are in a rotating coordinate system, not an inertial one. You still have a gravitational potential.

Spinning a clock will slow it. There was an experiment that did this by measuring the frequency shift of emitted radiation of material in a centrifuge.

Measurement of Relativistic Time Dilatation using the Mössbauer Effect by Champeney, Isaak and Khan

http://www.nature.com/nature/journal/v198/n4886/abs/1981186b0.html

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that makes sense to me now, acceleration is acceleration. gravity only seems special because it's not obvious whats doing the pushing.

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So a satellite has a balance of centrifugal force and centripetal force and is actually in a free fall.

That's only true for a perfectly circular orbit, and there is no such thing as a perfectly circular orbit in the real world. Moreover, centrifugal force isn't real. It is best to forget that concept. Try explaining elliptical or hyperbolic orbits with it.

from the little bit of math i know the force of gravity decreases by the square of the distance so something that weighs 1 pound some distance from the surface of earth,weighs 1/4 pound when it's twice that distance 1/d^2.

Careful! That is distance from the center of the Earth, not the surface. The Earth has an equatorial radius of 6378 kilometers. Grind through the math and you will see that the gravitational acceleration at 340 kilometers above the surface of the Earth (low Earth orbit) is about 90% of the acceleration at the Earth's surface.

since 1 divided by a huge number is a tiny little number no matter how huge the big number gets (even 13 billion light years) this means earths gravity(for example) is still a force even at the edges of the observable universe.

Correct. The force gets immeasurably small pretty quickly, however.

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The use of centrifugal force means you are in a rotating coordinate system, not an inertial one. You still have a gravitational potential.

...

Spinning a clock will slow it...

You say a lot with a few words. It is becoming more clear to me why "gravitational potential" is used often.

It seems the curvature of space is the key to clock speed and not just G forces that you feel. A merry-go-round warping space/time.

Merged post follows:

Consecutive posts merged
Moreover, centrifugal force isn't real. It is best to forget that concept.

How about that apparent centrifugal force or maybe inertia force.

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maybe centrifugal force is warped spacetime pushing us away from the center

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it would be interesting to find out if 1G in a rotating artificial gravity chamber caused the same amount of Lorentz contraction and dilation as 1G on earth, but when i try the math i get confused about that vx/c^2 term in the time part of Lorentz equation.

In another thread there were several that agreed that it was.

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It seems our local gravity does not have a large influence on our clocks

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(G * mass of Earth) / (radius of Earth * (c^2)) = 6.95453588 × 10^-10

Because of our Earth's mass and radius, clocks are slower by about 0.7 parts per billion. 60.5 microseconds per day. .022 seconds year compared to way out in space.

(G * mass of the Sun) / (radius of the Sun * (c^2)) = 2.12324397 × 10-6

Because of our Sun's mass and radius, Clocks on the surface would be slower by about 2.1 parts per million. .1728 seconds day. 63 seconds year compared to clocks way out in space.

(GPS satellite - Altitude 12,000 miles) Our Earth clocks are slower by 45 microseconds per day.

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I was trying to figure out how to use these formulas to calculate the 45 microseconds difference in time between the GPS satellite and the Earth surface. I can not even tell what unit of measurements are being used for "radius of Earth" or "mass of Earth". I would also like to know the time difference between the Sun's surface and the Earths surface. Can anyone help?

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check this if you haven't already

http://relativity.livingreviews.org/open?pubNo=lrr-2003-1&page=node1.html

i've been slowly working my way through this overview of g.p.s. and G.R., the math is over my head but there are enough words and results to get a good idea what's going on.

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Mass is probably kg and length (radius) metres, these are the SI units.

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Mass is probably kg and length (radius) metres, these are the SI units.

That was my first thought but...

radius of Earth = 6,378.1 kilometers

(G * mass of Earth) / (radius of Earth * (c^2)) = 6.95453588 × 10^-10

Of course. If I use meters instead of kilometers.

(G * mass of Earth) / ((6 378.1 * (10^3)) * (c^2)) = 0.695453588 nanometers.

Can I blame it on being old?

Thanks

moth, Thanks for the link. I understand what is happening but like you, I have a little trouble with the math. Even with the simple math sometimes as you see above.

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decimal points are slippery little suckers. fortunately, you can always catch mistakes later(but hopefully before anybody reads your post).

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decimal points are slippery little suckers. fortunately, you can always catch mistakes later(but hopefully before anybody reads your post).

It seems I would have easily seen a decimal point out of place. Not sure what I did at this point.

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