Time Dilation in the ISS: 2 Seconds in a Year

[Aristarchus_]

Why do you need to convert into %? It's pretty pointless.

You have still not answered my question in what frame 1 year is measured in. Is it 1 year on Earth or 1 year on ISS?

I am not sure, I am supposed to compare the two. For example how much younger or older is the twin in ISS after a year in comparison to the one on Earth

 

[PeroK]

You have still not answered my question in what frame 1 year is measured in. Is it 1 year on Earth or 1 year on ISS?

That can't be important, given we are dealing with tiny difference in proper times.

 

[Aristarchus_]

That's definitely wrong. I get about $0.01 s$ taking into account only the speed of the ISS.

I miscalculated than, can you provide your calculation?

 

[malawi_glenn]

That can't be important, given we are dealing with tiny difference in proper times.

It's not important here, but its not all about getting the correct answer.

 

[PeroK]

I miscalculated than, can you provide your calculation?

It's in a spreadsheet. That said I don't like your method. Do you know how to use the binomial theorem for the square root?

 

[malawi_glenn]

I miscalculated than, can you provide your calculation?

It has already been pointed out that in your formula, there is a factor of 2 wrong in the calculation of the speed. Have you tried again after that?

 

[Aristarchus_]

It has already been pointed out that in your formula, there is a factor of 2 wrong in the calculation of the speed. Have you tried again after that?

I think I figured out the mistake in the calculation. Now I will brainstorm a bit...

 

[malawi_glenn]

Do you know how to use the binomial theorem for the square root?

My Texas TI-82 manage fine just cranking in the entire calculation at once

 

[Aristarchus_]

My Texas TI-82 manage fine just cranking in the entire calculation at once

Another issue is getting 4.18*10^{-9} % as the percentage of how much faster time elapses on ISS than on Earth...

 

[jbriggs444]

Another issue is getting 4.18*10^{-9} % as the percentage of how much faster time elapses on ISS than on Earth...

Excuse me? What leads you to believe that the proper time on the ISS elapses more rapidly than that on Earth?

Time dilation describes how moving clocks appear to run slow (tick less elapsed proper time) compared to reference clocks in an inertial frame. In the context of Newtonian gravity and special relativity, there is only one inertial frame here.

 

[PeroK]

I would tend to use $\Delta t_s, \Delta t_o$ for the proper times on the surface and in the ISS orbit. A useful approach is to rewrite this as:
$$\Delta t_o = \Delta t_s \sqrt{1- \frac{v^2}{c^2}} = \Delta t_s \sqrt{1- \frac{GM}{rc^2}}$$Then expand that term to first order using a binomial expansion.

Let me show you a technique that you need to be familiar with. It's used all over physics and applied maths:
$$\Delta t_o = \Delta t_s \sqrt{1- \frac{GM}{rc^2}} \approx \Delta t_s(1 - \frac 1 2 \frac{GM}{rc^2})$$That's called a first order binomial expansion. From that you get:
$$\Delta t_s - \Delta t_o = \Delta t_s(\frac{GM}{2rc^2})$$And, of course, if you want to use $v$, we have:
$$\Delta t_s - \Delta t_o = \Delta t_s(\frac{v^2}{2c^2})$$Messing around with percentages is not the way to go.

 

[malawi_glenn]

Let me show you a technique that you need to be familiar with

I teach this problem for my 17y old students at high school. They don't bareley know what a derivative is at this point.

Now if OP have taken a university level class on calculus, then of course it is a good idea to invoke Mac Laurin expansion. However, modern pocket calculators can deal with this problem just fine. For lower speeds, Mac Laurin expansion is needed though.

 

[jbriggs444]

However, modern pocket calculators can deal with this problem just fine. For lower speeds, Mac Laurin expansion is needed though.

At ten digits of .999999999... before the significant digits start, one is starting to tickle the precision boundaries of 64 bit floating point (16 digits or so). I'm with @malawi_glenn. We're OK in this case but need to be aware of the limitations of our tools.

You can get complete loss of precision when dealing with numbers like 0.99999999999999998 (sixteen digits of nines) which become indistinguishable from one.

The ideas that taking a square root halves the difference between the original value and 1 and the idea that taking the reciprocal inverts the difference between the original value and 1 are pretty useful and may be accessible without calculus.

$\sqrt{0.99} \approx 0.995$
$\sqrt{1.01} \approx 1.005$
$1/0.99 \approx 1.01$
$1/1.01 \approx 0.99$
$1.01^2 \approx 1.02$
$0.99^2 \approx 0.98$

 

[PeroK]

I teach this problem for my 17y old students at high school. They don't bareley know what a derivative is at this point.

Perhaps 17 is a good age to lose one's mathematical virginity!

 

[vela]

I teach this problem for my 17y old students at high school. They don't barely know what a derivative is at this point.

They don't need to know how to derive the expansion, just how to use it.

You could also point out it's just a generalization of the binomial theorem, which they should learn about in algebra, to real exponents, i.e.,
$$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots$$ with $n=1/2$.

 

[malawi_glenn]

They don't need to know how to derive the expansion, just how to use it.

You could also point out it's just a generalization of the binomial theorem, which they should learn about in algebra, to real exponents, i.e.,
$$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots$$ with $n=1/2$.

We have a different system, they learn binomial formula when they are 18 in swedish high school. But only for positive integer coefficients.

If "new" math is not needed, no need of invoking it.

 

[malawi_glenn]

At ten digits of .999999999... before the significant digits start, one is starting to tickle the precision boundaries of 64 bit floating point (16 digits or so). I'm with @malawi_glenn. We're OK in this case but need to be aware of the limitations of our tools.

You can get complete loss of precision when dealing with numbers like 0.99999999999999998 (sixteen digits of nines) which become indistinguishable from one.

The ideas that taking a square root halves the difference between the original value and 1 and the idea that taking the reciprocal inverts the difference between the original value and 1 are pretty useful and may be accessible without calculus.

$\sqrt{0.99} \approx 0.995$
$\sqrt{1.01} \approx 1.005$
$1/0.99 \approx 1.01$
$1/1.01 \approx 0.99$
$1.01^2 \approx 1.02$
$0.99^2 \approx 0.98$

One can also use factorization of conjugate and calcluate in this order
$\dfrac{1}{\sqrt{1+v/c}}t_0\dfrac{1}{\sqrt{1-v/c}}$
And factorization of conjugate i teach them when they are 16 y old

 

[PeroK]

Just to clarify the overall situation here. If you look online there seems to be a lot of misinformation about this, with confusion over the significance of gravitational and velocity-based time dilation.

First, we use reference time $\Delta t$ for a clock at rest far from the Earth. From GR we have the dilated time of an object in a circular orbit of radius $r$:
$$\Delta t_o = \Delta t \big (1 - \frac{3GM}{rc^2} \big )^{\frac 1 2} = \Delta t \big (1 - \frac{3GM}{2rc^2} \big )$$This neatly splits into a velocity-based component:
$$\Delta t_o = \Delta t \big (1 - \frac{GM}{rc^2} \big )^{\frac 1 2} = \Delta t \big (1 - \frac{GM}{2rc^2} \big )$$And a gravitational component:
$$\Delta t_o = \Delta t \big (1 - \frac{2GM}{rc^2} \big )^{\frac 1 2} = \Delta t \big (1 - \frac{GM}{rc^2} \big )$$We see from this that, overall, the gravitational component is twice as significant as the velocity-based component. However, the Earth clock has velocity-based and gravitational time dilation components as well. The velocity component is negligible, as the Earth's rotational speed is less than 10% of the ISS orbital speed. But, the gravitational component is:
$$\Delta t_e = \Delta t \big (1 - \frac{2GM}{Rc^2} \big )^{\frac 1 2} = \Delta t \big (1 - \frac{GM}{Rc^2} \big )$$And, this is crudely approximately the same as the ISS gravitational component, as $R \approx r$. Which means that overall we can consider only the speed of the ISS relative to Earth to get a good approximation. I get a figure of $0.01s$ per year.

Note that for the second approximation (taking gravity into account), I get only $0.009s$ per year difference between the ISS and Earth.

I must confess I couldn't find anything on line that agrees with this. And, the OP's quoted answer seems to be $0.013s$ per year. But, I don't get that answer.

 

[PeroK]

PS actually Wikipedia agrees with me:

https://en.wikipedia.org/wiki/Time_dilation#Time_dilation_caused_by_a_relative_velocity

 

[malawi_glenn]

@Aristarchus_ have you covered time dilation effects due to gravity in your class? Have you covered MacLaurin series in any class prior to this class?

And, the OP's quoted answer seems to be 0.013s per year.

The answer is supposed to be 0.007 sec

4.2*10^{-9} is the correct answer for the percentage of time dilation in relation to Earth

Pretty hard for me to understand what the answer OP has been given. These two numbers are not consistent.

 

[Aristarchus_]

@Aristarchus_ have you covered time dilation effects due to gravity in your class? Have you covered MacLaurin series in any class prior to this class?

Pretty hard for me to understand what the answer OP has been given. These two numbers are not consistent.

No, just the Lorentz formula for dilation