Time Dilation in Special vs. General Relativity

[philipp]

Hi all,

I was trying to understand the time dilation in special and general relativity and after much time of "overthinking" I am pretty much stuck now. My problem is, that what seems to me to be the same premises apparently imply opposite things.



In special relativity, for two inertial reference frames moving relative to each other with velocity v, we have the following formula for time dilation:

T' = γ ⋅ T₀

where

• T' is the time measured in the moving reference frame
• T₌ is the proper time measured in the resting system
• γ = 1/√1 - (v²/c²) ≥ 1
• v is the relative velocity of the intertial reference frames
• c is the speed of light

We see that: T' ≥ T₀

We further know that "moving clocks run slow."
So in the resting reference frame, the time runs faster, meaning more time passes in the resting frame relative to the moving one.


So a smaller T (in this case T₀) ⇒ more time passing relative to the other frame




Now for the general relativity. Let's imagine a source of gravitaton, e.g. a planet, and T₁ being a time interval measured close to that planet and T₂ being a time interval measured further away. We have

T₂ = gh/c² ⋅ T₁ + T₁

where

• g is the gravitational acceleration
• h is the distance to the center of gravity (≈ hight above ground)
• c is the speed of light

We see that: T₂ ≥ T₁

We further know that "clocks close to gravitation run slow."
So for a place closer to gravitation the time runs slow meaning less time passes relative to a place further away.


So a smaller T (in this case T₁) ⇒ less time passing





Now how can this be? How can a smaller time interval once imply more time being passed and another time imply less time being passed. Where am I wrong?

Thanks in advance

[Strange]

You seem to have swapped what you are comparing between the two cases.

 

In the first case, the moving clock runs slow => therefore time appears to run slower for the moving frame or, equivalently, more time passes in the stationary frame (you).

 

In the second case, the clock in higher gravitational potential runs slow => therefore time appears to run slower for the high-gravity frame or, equivalently, more time passes in the low-gravity frame (you).

[philipp]

Thank you for your quick answer. Rverything you pointed out seems perfectly reasonable, but still:

 

 

for the special relativity:

 

stationary (T₀) = more time passing

moving (T') = less time passing

 

and T₀ (more time) < T' (less time)

 

 

 

for the general relativity:

 

high gravity potential (T₁) = less time passing

low gravity potential (T₂) = more time passing

 

and T₁ (less time) < T₂ (more time)

 

 

 

so:

T₀ (more time) < T' (less time)

T₁ (less time) < T₂ (more time)

 

 

I still can't see how this adds up..?

Edited May 26, 2016 by philipp

[imatfaal]

Because the man in the observatory - ie you - sees a clock in a rocket running slow and a clock on a spacestation a long way from the sun's gravitational potential run quickly; that is to say that you are T_prime in the first equation and T_one in the second.

 

The equations make sense but are confusing if you try to equate as one single observing position what would normally be denoted as T_zero in both equations. In my example of the scientist/you observing a clock on a speeding rocket and a clock on a very distant spacestation you are T in the relative velocity and T_zero in the Gravitational potential

 

T = T_zero (1/(1-v^2/c^2)^1/2)

 

T = T_zero(1/(1-2MG/c^2)^1/2)

 

edit - missed out sqrt. Thnx Dima

Edited May 27, 2016 by imatfaal

[Strange]

To explain a bit more (or confuse further) note that in the special relativistic case, the situation is symmetrical: you will see the other person's clock running slow but he will also see your clock running slow.

 

In the case of gravitational time dilation, if you see his clock running slow then he will see your clock running fast.

Edited May 26, 2016 by Strange

[MigL]

Maybe the equivalence principle can simplify things for you.

Consider that wherever you feel the force of gravity ( i.e. weight ), you can be considered to be accelerating.

A far away observer ( where space-time is nearly flat ) will measure your time to be dilated.

 

So both acceleration and relativistic velocity will produce time dilation.

 

And as has been pointed out, SR's time dilation is symmetrical between the two observers because velocity is relative.

GR's time dilation is not because acceleration is not relative

[DimaMazin]

Because the man in the observatory - ie you - sees a clock in a rocket running slow and a clock on a spacestation a long way from the sun's gravitational potential run quickly; that is to say that you are T_prime in the first equation and T_one in the second.

 

The equations make sense but are confusing if you try to equate as one single observing position what would normally be denoted as T_zero in both equations. In my example of the scientist/you observing a clock on a speeding rocket and a clock on a very distant spacestation you are T in the relative velocity and T_zero in the Gravitational potential

 

T = T_zero (1/1-v^2/c^2)

 

T = T_zero(1/1-2MG/c^2)

T = T_zero (1/(1-v^2/c^2)^1/2)

T = T_zero(1/(1-2MG/rc^2)^1/2 )

Thank you for your quick answer. Rverything you pointed out seems perfectly reasonable, but still:

 

 

for the special relativity:

 

stationary (T₀) = more time passing

moving (T') = less time passing

 

and T₀ (more time) < T' (less time)

 

 

 

for the general relativity:

 

high gravity potential (T₁) = less time passing

low gravity potential (T₂) = more time passing

 

and T₁ (less time) < T₂ (more time)

 

 

 

so:

T₀ (more time) < T' (less time)

T₁ (less time) < T₂ (more time)

 

 

I still can't see how this adds up..?

T_{InternationalSpaceStation}<T_{on Earth surface}<T_{GPS satellit}

[Robittybob1]

T = T_zero (1/(1-v^2/c^2)^1/2)

T = T_zero(1/(1-2MG/rc^2)^1/2 )

T_{InternationalSpaceStation}<T_{on Earth surface}<T_{GPS satellit}

Is it possible to describe "<T" in terms of atomic clock tick rates?

Would this be equivalent to what you had said above:

The tick rate of atomic clocks on the International Space Station are slower than the tick rate of clocks on the Earth and both are slower than the tick rate of clocks on GPS satellites?

Edited May 27, 2016 by Robittybob1

[DimaMazin]

Is it possible to describe "<T" in terms of atomic clock tick rates?

Would this be equivalent to what you had said above:

The tick rate of atomic clocks on the International Space Station are slower than the tick rate of clocks on the Earth and both are slower than the tick rate of clocks on GPS satellites?

Yes.

[swansont]

In fact, the time dilation factor you calculate is the fractional change in the frequency, ∆f/f. What everyone tends to call time dilation is the integral of this, which isthe elapsed proper time.

[DimaMazin]

I think Philipp can use the equation for calculation of clocks indications on Earth and on its satellites

 

t₁gamma₁/(1- 2GM /( r₁c²))^1/2=t₂gamma₂/(1-2GM /( r₂c²))^1/2=t₃gamma₃/(1-2GM /( r₃c²))^1/2

Edited May 28, 2016 by DimaMazin

[Robittybob1]

I think Philipp can use the equation for calculation of clocks indications on Earth and on its satellites

 

t₁gamma₁/(1- 2GM / r₁c²)^1/2=t₂gamma₂/(1-2GM / r₂c²)^1/2=t₃gamma₃/(1-2GM / r₃c²)^1/2

[latex]t_1\gamma_1 / (1- 2GM / r_1c^2)^{1 / 2} = t_2\gamma_2 / (1-2GM / r_2c^2)^{1/2} = t_3\gamma_3 / (1-2GM / r_3c^2)^{1/2}[/latex]

 

Conclusion: seems to imply [latex]\frac{t_n\gamma_n}{r_n}[/latex] for any [latex]n[/latex] is a constant. Did I get that conclusion right?

 

[Thanks Mordred for the Latex correction.]

Edited May 27, 2016 by Robittybob1

[Mordred]

[latex]\frac {t_1\gamma_1}{(1-\frac {2GM}{ r_1c^2})^{1/2}}=\frac {t_2\gamma_2}{(1-\frac {2GM} { r_2c^2})^{1/2}}=\frac {t_3\gamma_3}{(1-\frac {2GM}{ r_3c^2})^{1/2}}[/latex]

 

 

Fixed the latex

 

DimaMazin please confirm this is correct to your post. As it looks incorrect

Edited May 27, 2016 by Mordred

[Robittybob1]

[latex]\frac{t_n\gamma_n}{r_n}[/latex] That [latex]\gamma[/latex] is a velocity dependent term. For satellites surely that is related to the orbital velocity? The ISS and GPS satellites orbit at different [latex]{r}[/latex] values and different velocities accordingly so their time dilations are different.

[latex]\gamma = \frac {1}{\sqrt{1-\frac{v^2}{c^2}}}[/Latex]

#7

 

TInternational Space Station<Ton Earth surface<TGPS satellite

Edited May 28, 2016 by Robittybob1

[DimaMazin]

[latex]\frac {t_1\gamma_1}{(1-\frac {2GM}{ r_1c^2})^{1/2}}=\frac {t_2\gamma_2}{(1-\frac {2GM} { r_2c^2})^{1/2}}=\frac {t_3\gamma_3}{(1-\frac {2GM}{ r_3c^2})^{1/2}}[/latex]

 

 

Fixed the latex

 

DimaMazin please confirm this is correct to your post. As it looks incorrect

Yes.You are correct.

[Robittybob1]

If we were to check this against the real life speed of the satellites at these heights/velocities what value do we give to [latex]T_1, T_2..... T_N[/latex] etc. Like is it [latex]0 {=} T_0[/latex]? Or better [latex] T_0 {=} 0[/latex]

Edited May 28, 2016 by Robittybob1

[DimaMazin]

If we were to check this against the real life speed of the satellites at these heights/velocities what value do we give to [latex]T_1, T_2..... T_N[/latex] etc. Like is it [latex]0 {=} T_0[/latex]? Or better [latex] T_0 {=} 0[/latex

If you know indication of one clock then you can calkulate simultaneity of any clock(if you know heights/velocities).It is real value.

[swansont]

Yes.You are correct.

 

Since gamma is a function of r for a circular orbit, perhaps you can derive how these are indeed constant. It's not obvious that they are.

[Robittybob1]

 

Since gamma is a function of r for a circular orbit, perhaps you can derive how these are indeed constant. It's not obvious that they are.

If we put real figures to the formula DimaMazin gave us? http://www.sciencefo...ty/#entry922079.

I think it will result in the OP seeing how the time dilation equation work and so will I. I just get confused how to put time in as a number. Maybe it just be a difference in time from [latex] T_0 {=} 0[/latex], then [latex] T_n [/latex] will either positive or negative or also zero.

[Janus]

The time dilation factor for a clock in circular orbit ( as measured by a distant observer) is found by:

 

[math]T = t' \sqrt{1-\frac{3GM}{rc^2}}[/math]

 

where r is the orbital radius.

 

This is what you get when you take the combined effect of gravitational and relative velocity time dilation effects found by.

 

[math]T = t' \sqrt{1-\frac{2GM}{rc^2}- \frac{v^2}{c^2}}[/math]

 

And substitute [math] \sqrt{\frac{GM}{r}}[/math], the orbital velocity at r, for v

Edited May 28, 2016 by Janus

[Robittybob1]

The time dilation factor for a clock in circular orbit ( as measured by a distant observer) is found by:

 

[math]T = t' \sqrt{1-\frac{3GM}{rc^2}}[/math]

 

where r is the orbital radius.

 

This is what you get when you take the combined effect of gravitational and relative velocity time dilation effects found by.

 

[math]T = t' \sqrt{1-\frac{2GM}{rc^2}- \frac{v^2}{c^2}}[/math]

 

And substitute [math] \sqrt{\frac{GM}{r}}[/math], the orbital velocity at r, for v

Thanks Janus

Just one clarifying comment regarding [math]T = t' \sqrt{1-\frac{3GM}{rc^2}}[/math]

the [math]T [/math] is the "time dilation factor" so what units will this have?

[math] t' [/math] What and where is that measured? I take it as a reading from a clock beside the distant observer, but if it is a clock do we put a period of time into the equation e.g. 1 second or 60 seconds? How does one put a number to this [math] t' [/math] when making a calculation?

Sorry for being a bit thick about this, but I'll get it soon.

Can I just say [math] t' = 1 [/math] second so [latex]T = t' \sqrt{1-\frac{3GM}{rc^2}}[/latex] becomes [latex]T = \sqrt{1-\frac{3GM}{rc^2}}[/latex]?

Edited May 28, 2016 by Robittybob1

[Robittybob1]

The time dilation factor for a clock in circular orbit ( as measured by a distant observer) is found by:

 

[math]T = t' \sqrt{1-\frac{3GM}{rc^2}}[/math]

 

where r is the orbital radius.

 

This is what you get when you take the combined effect of gravitational and relative velocity time dilation effects found by.

 

[math]T = t' \sqrt{1-\frac{2GM}{rc^2}- \frac{v^2}{c^2}}[/math]

 

And substitute [math] \sqrt{\frac{GM}{r}}[/math], the orbital velocity at r, for v

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.

Edited May 29, 2016 by Robittybob1

[swansont]

Thanks Janus

Just one clarifying comment regarding [math]T = t' \sqrt{1-\frac{3GM}{rc^2}}[/math]

the [math]T [/math] is the "time dilation factor"

 

No, T is the time. It will have units of time. The part other than the t' is the time dilation factor — the change in the fractional frequency ∆f/f. It is unitless.

 

The assumption is that at the start, t'=T=0

[Robittybob1]

 

No, T is the time. It will have units of time. The part other than the t' is the time dilation factor — the change in the fractional frequency ∆f/f. It is unitless.

 

The assumption is that at the start, t'=T=0

Could you check my edit of the post prior to yours please? It is not quite along your line but similar. I might be wrong, but tomorrow I'll recheck it.

[Robittybob1]

 

No, T is the time. It will have units of time. The part other than the t' is the time dilation factor — the change in the fractional frequency ∆f/f. It is unitless.

 

The assumption is that at the start, t'=T=0

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:

 

T(International Space Station)<T(on Earth surface)<T(GPS satellite) [parentheses added by me]

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).

Edited May 29, 2016 by Robittybob1