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Why do clocks tick slower near massive objects like earth ?

by Abidal Sala
Tags: clocks, earth, massive, objects, slower, tick
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Abidal Sala
#1
May5-12, 04:26 AM
P: 30
So clocks _on_ earth are slower than the clocks in satellites because earth is massive? it doesn't make sense to me, what does the mass of an object have to do with the passage of time of other objects surrounding this massive object or that lie on it? I can't think of any reasonable explanation.. is there any scientific explanation for this?
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Matterwave
#2
May5-12, 04:45 AM
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Einstein's general relativity. Massive objects curve the space-time around them, giving different clock tick-rates depending on where the clocks are located.
tom.stoer
#3
May5-12, 07:57 AM
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P: 5,464
In GR the time-dilation can be calculated as follows

[tex]\tau_{ab} = \int_a^b d\tau = \int_a^b \sqrt{dx_\mu dx^\mu} = \int_{t_a}^{t_b}dt \sqrt{g_{\mu\nu}\,\dot{x}^\mu\,\dot{x}^\nu}[/tex]

First we look at the special case of SR with a moving observer with v>0 but in flat space. That means the metric is diag(1,-1,-1,-1) and we get

[tex]\tau_{ab} = \int_{t_a}^{t_b}dt \sqrt{1-v_iv^i} = (t_b - t_a)\,\sqrt{1-v_iv^i}[/tex]

When we do this calculation is a non-flat spacetime (e.g. due to the presence of a gravitating body) we must consider the full metric g. In many cases the metric is still diagonal and we get

[tex]\tau_{ab} = \int_{t_a}^{t_b}dt \sqrt{g_{00}+g_{ii}\,v^i\,v^i}[/tex]

(in all cases one has to sum over multiple indices)

The last formula shows that in a gravitational field even with vanishing velocity v the 00-component of the metric contributes to a gravitational time dilation.

One can e.g. use the Schwarzsschild solution

[tex]d\tau^2 = \left(1-\frac{R_s}{r}\right)\,dt^2 - \left(1-\frac{R_s}{r}\right)^{-1}\,dr^2 - r^2\,d\Omega^2[/tex]

to calculate the effect for a body at rest at "coordinate" radius r

[tex]\tau_{ab} = (t_b - t_a)\,\sqrt{1-\frac{R_s}{r}}[/tex]

Have a look at

http://en.wikipedia.org/wiki/Gravita..._time_dilation
http://en.wikipedia.org/wiki/Schwarz...zschild_metric

moatasim23
#4
May5-12, 08:49 AM
P: 78
Why do clocks tick slower near massive objects like earth ?

Suppose an observer is at the surface of earth holding a laser whose frequency is synchronized with that of his heart beat.Say for example 1 wave per heat beat.Now consider another person at some height observing the laser.He will observe a change in frequency as frequency is directly proportional to energy and laser will lose energy while moving up.So heart beat he sees comes out to be less per second.So he concludes heart beats at a slower rate.
Naty1
#5
May5-12, 09:36 AM
P: 5,632
it doesn't make sense to me,.....I can't think of any reasonable explanation.. is there any scientific explanation for this?
I agree....and all scientists until Einstein would have,too!!!

GR explains WHAT happens...and is confirmed by our observational measurements. Nobody knows WHY most stuff happens.....Why are you here asking this? Why am I here answering? Why does nature conform with so much of our 'artifical' mathematics??

The way the knowledge evolved is that it was experimentally determined that everyone sees the speed of light [absent gravity, in flat spacetime] as 'c'. Light always is going faster than you by 'c'....WOW is that curious!!! so scientists started with a theory of 'ether'....Lorentz and Fitzgerald worked on pieces of this puzzle....one fussed with length contraction the other time dilation which results from relative differences in velocity...It took an "Einstein' to put the pieces of the puzzle together in a coherent way.

Did you catch that?? Relative velocity ALSO causes changes in elapsed time!! So both differences in gravitational potential [GR] AND relative velocity [SR] cause time to tick at different rates!!! Now you should be 'doubly' confused....

edit: To add to the 'incomprehensibility' you should also consider that time is inextricably woven into space....as spacetime....so in our universe, space and time 'warp' in unison....when one changes so does the other....and both space and time are 'warped' by velocity and gravitational potential....

The only 'sense' I can make of it is to think there is still a deeper underlying relationship that we do not yet fully understand.

"Realtivity whispers about spacetime issues which black holes seem to scream."
moatasim23
#6
May6-12, 01:28 AM
P: 78
Quote Quote by moatasim23 View Post
Suppose an observer is at the surface of earth holding a laser whose frequency is synchronized with that of his heart beat.Say for example 1 wave per heat beat.Now consider another person at some height observing the laser.He will observe a change in frequency as frequency is directly proportional to energy and laser will lose energy while moving up.So heart beat he sees comes out to be less per second.So he concludes heart beats at a slower rate.
I was in a hurry so couldnt give the full explanation.Now consider two persons whose heart beat at the same rate on the surface of the earth.Their heart beats are also synchronized with the frequency of light emitted from a laser.Say one wave per heart beat.Now if one person gets to some height and observes the laser emitted from the first observer then he observes a decrease in frequency as compared to that while standing on earth.This decrease in frequency is due to the decrease in energy, as light moves up, due to Gravitation.This decrease in frequency causes increase in wavelength of light.So the observer at height observes that heart beats at a slower rate of the person at ground bcz the wave emitted will be longer.
Zeron_X25
#7
May9-12, 09:37 PM
P: 7
Quote Quote by tom.stoer View Post
When we do this calculation is a non-flat spacetime (e.g. due to the presence of a gravitating body) we must consider the full metric g. In many cases the metric is still diagonal and we get

[tex]\tau_{ab} = \int_{t_a}^{t_b}dt \sqrt{g_{00}+g_{ii}\,v^i\,v^i}[/tex]

(in all cases one has to sum over multiple indices)

The last formula shows that in a gravitational field even with vanishing velocity v the 00-component of the metric contributes to a gravitational time dilation.

One can e.g. use the Schwarzsschild solution

[tex]d\tau^2 = \left(1-\frac{R_s}{r}\right)\,dt^2 - \left(1-\frac{R_s}{r}\right)^{-1}\,dr^2 - r^2\,d\Omega^2[/tex]

to calculate the effect for a body at rest at "coordinate" radius r

[tex]\tau_{ab} = (t_b - t_a)\,\sqrt{1-\frac{R_s}{r}}[/tex]
If I were to try and calculate the time dilation of a moving object near or on the surface of the Earth, do I have to convert the velocity to Schwarzschild coordinates? If so, how would I go about doing that?
Ich
#8
May10-12, 03:20 AM
Sci Advisor
P: 1,910
If you know the locally measured velocity, you have all you need. Just calculate the time dilation and multiply by it.
Zeron_X25
#9
May10-12, 10:17 PM
P: 7
Quote Quote by Ich View Post
If you know the locally measured velocity, you have all you need. Just calculate the time dilation and multiply by it.
So are you saying that I should calculate the velocity time dilation separately and then solve the Earth's time dilation and multiply the two numbers?
But wouldn't that make the time extremely small because of the gravitational time dilation term?
tom.stoer
#10
May11-12, 12:50 AM
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P: 5,464
Let's compare the gravitational time dilation of a clock located on earth and a clock located at some height h. Using the radius of the earth RE you calculate

[tex]\tau_{ab}(x) = (t_b - t_a)\,\sqrt{1-\frac{R_S}{R_E+x}}[/tex]

and compare x=0 and x=h.

[tex]\frac{\tau(0)}{\tau(x)} = \sqrt{\frac{1-\frac{R_S}{R_E}}{1-\frac{R_S}{R_E+x}}}[/tex]

The effect is tiny due to the small parameter RS/RE ≈ 8.9*10-3m / 6.6*106m.
Ich
#11
May11-12, 03:04 AM
Sci Advisor
P: 1,910
But wouldn't that make the time extremely small because of the gravitational time dilation term?
"Small time dilation" doesn't mean a small number to multiply by. It means that the "clock rates" are almost equal, t'/t=(1- a little bit). You multiply by the ratio (1- a little bit), not by a little bit. Look again at tom's formula, it is close to 1.
the_emi_guy
#12
May11-12, 04:06 PM
P: 589
Abidal,
Of course it does not make sense, but is the nature of modern science.
Did Newton's law of gravity make sense to his contemporaries? I'm holding an apple in my hand, I can make it weigh less by climbing a tree? That would have been completely counterintuitive in his era. Modern science is accepting things that agree with or explain our experiments even though they may seem contrary to our native senses.

Cheers.
feynmanisbest
#13
May11-12, 04:18 PM
P: 6
Objects that are located in a stronger gravitational field pass through time more slowly. Science has not yet answered why that is, so far it has merely confirmed the fact that it is so.

Understanding why would probably require a deeper understanding of the nature of time, or of gravity, or of both, than we currently have.
PAllen
#14
May11-12, 05:01 PM
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Just to be clear that this effect is experimental, modern, identically constructed atomic clocks, detect difference in time rate between the floor and a table top, quantitatively consistent with predictions of GR.
Passionflower
#15
May11-12, 11:35 PM
P: 1,555
Quote Quote by PAllen View Post
Just to be clear that this effect is experimental, modern, identically constructed atomic clocks, detect difference in time rate between the floor and a table top, quantitatively consistent with predictions of GR.
Actually that is a good point to discuss:

Now I do not question the validity of GR but does experimentally comparing the clock rates at different stationary locations in a static gravitational field actually prove GR?

I do not think so, we could even prove this with only SR!

Two different stationary locations in a static gravitational field have different proper accelerations. We can calculate the clock rate difference with SR only and get a result in agreement with experiment!

The GR part is there but it is simply not measurable because our instruments are not accurate enough or we cold also say that the Earth is gravitationally too weak to measure.

Agree?
yuiop
#16
May12-12, 12:00 AM
P: 3,967
Quote Quote by Passionflower View Post
Two different stationary locations in a static gravitational field have different proper accelerations. We can calculate the clock rate difference with SR only and get a result in agreement with experiment!

The GR part is there but it is simply not measurable because our instruments are not accurate enough or we could also say that the Earth is gravitationally too weak to measure.
The SR part has acceleration proportional to 1/r while the GR equation is proportional to 1/(r^2)*1/√(1-2GM/(rc^2)) so there is a slight difference, but I guess (as you suggest) this tidal effect of GR is too small to detect in a tabletop experiment.
tom.stoer
#17
May12-12, 02:26 AM
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Quote Quote by Passionflower View Post
Now I do not question the validity of GR but does experimentally comparing the clock rates at different stationary locations in a static gravitational field actually prove GR?

...

The GR part is there but it is simply not measurable because our instruments are not accurate enough or we cold also say that the Earth is gravitationally too weak to measure.

Agree?
I guess that in the meantime we have experiments taking into account varying gravitational fields, e.g. on board of satellites.

But I disagree with the idea using SR to interpret the effect in constant gravitational fields. Yes, one could use proper acceleration ~ 1/rē from Newtonian gravity + SR to do the calculation but we know that we are cheating. And we know that in other cases the calculation will become wrong simply b/c the instantaneous 1/r potential explicitly violates SR. So I think this is another example where the combination of different theories (Newtonian gravity + SR) seems to work to some extent - but from a from an educational standpoint this may be missleading.
PAllen
#18
May12-12, 10:43 AM
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Quote Quote by Passionflower View Post
Actually that is a good point to discuss:

Now I do not question the validity of GR but does experimentally comparing the clock rates at different stationary locations in a static gravitational field actually prove GR?

I do not think so, we could even prove this with only SR!

Two different stationary locations in a static gravitational field have different proper accelerations. We can calculate the clock rate difference with SR only and get a result in agreement with experiment!

The GR part is there but it is simply not measurable because our instruments are not accurate enough or we cold also say that the Earth is gravitationally too weak to measure.

Agree?
Well you have to use SR plus principle of equivalence with GR definition of inertial frames. That is not really pure SR, though such analyses are useful in bridging from SR to GR, distinguishing frame dependent features (coordinate acceleration) from independent ones (tidal gravity)- in GR.

Without the GR definition of inertial frames, there is no basis in SR to say that free fall is the frame in which Minkowski metric applies.

Unambiguous is that these measurements measure what is defined as gravitational time dilation. How the latter factors into 'pure SR' versus GR is subject to interpretation. Pressed for precision, I would say this type of gravitational time dilation measurement is an accurate measure of the equivalence principle - which is not normally considered part of SR.


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