Trying to understand time dilation better

[ghwellsjr]

Thanks for your explanations.

You're welcome.

...
So IMHO we now have a more realistic specification for what “causes” a different aging. It does not refer any longer to the relative motion between the red and blue clocks, but it spells out the relative motion between the (black) rest clock and each of the moving (red and blue) clocks. However it refers to a peculiar IRF and I guess it can't be extended to any IRF. May be there is a better way to solve this issue...

DaleSpam already gave you the complete answer in post #28. Instead of concerning yourself with the black clock which happens to be at rest in a particular IRF, you want to use the Coordinate Time, represented by "t" in his equation. (They may be the same but you still don't want to restrict yourself to any particular clock at rest in an IRF.) In my diagrams, I use a simplified version of his equation which simply means that for the increment of the Coordinate Time grid lines, I determine (or specify) the speed of an object and calculate how much Time Dilation there is for that increment of time and that's how far apart I spread the dots.

Once I build a complete scenario in the defining IRF, I can use the Lorentz Transformation process to get to any other IRF moving at any speed within the range of +/- c. That is one reason why you don't want to conceptualize Time Dilation as dependent on or in relation to any other actual clock. It's related to the Coordinate Time of an IRF which may or may not have any clock at rest in it and even if it did, that clock would be subject to the same determination of Time Dilation as any other clock. For example, in the second and fourth diagrams on post #18, there is no clock at rest in those IRF's, and yet the determination of Time Dilation for all the clocks still works.

 

[rede96]

So just to check my understanding...

My twin and I are the same age. We both have very fast star ships and are challenged to a race to see who's star ship is the quickest. The race starts and ends on Earth and the first one back wins.

We both set off in our separate star ships to a star that is 10 ly's away. When we get there we must turn around and head back to earth.

I have a very special star ship that allows me to travel very close to the speed of light, so the time I actually experience on my journey is arbitrarily small, say just 1 day. However my twin is only traveling at say 0.5c so the time he experiences is longer.

When I arrive back on Earth I find that people I knew are nearly 20 years older then when I left. And, to my surprise I find my twin is already back on Earth and older than I am. So he won the race? But how can that be if I was traveling almost twice as fast?

Am I missing something or is that just how space-time works?

 

[ghwellsjr]

So just to check my understanding...

My twin and I are the same age. We both have very fast star ships and are challenged to a race to see who's star ship is the quickest. The race starts and ends on Earth and the first one back wins.

We both set off in our separate star ships to a star that is 10 ly's away. When we get there we must turn around and head back to earth.

I have a very special star ship that allows me to travel very close to the speed of light, so the time I actually experience on my journey is arbitrarily small, say just 1 day. However my twin is only traveling at say 0.5c so the time he experiences is longer.

When I arrive back on Earth I find that people I knew are nearly 20 years older then when I left. And, to my surprise I find my twin is already back on Earth and older than I am. So he won the race? But how can that be if I was traveling almost twice as fast?

Am I missing something or is that just how space-time works?

I'd be surprised too if your twin got back before you. Why do you think this? You are correct about your trip but your twin will take twice as long meaning he won't get back until 40 years after he left and he will have aged 34.6 years.

 

[rede96]

I'd be surprised too if your twin got back before you. Why do you think this? You are correct about your trip but your twin will take twice as long meaning he won't get back until 40 years after he left and he will have aged 34.6 years.

Ah right, of course. Thanks. And I wasn't thinking, but it has been a long day. :)

Cheers.

 

[Dale]

So how can we better qualify what causes (the physical conditions leading to) a different aging? What do we actually need to know in order to calculate the time gap?

The cause of differential aging is simply that different paths through spacetime have different intervals. This is no different nor any more surprising than the fact that different paths in space have different lengths. To determine the time elapsed on a timelike path through spacetime we calculate:

$$\tau=\int \sqrt{-g_{\mu\nu} dx^{\mu}dx^{\nu}}$$

This formula works for any timelike path, inertial or non-inertial, through any spacetime, curved or flat, using any coordinate system. It is the completely general explanation for differential aging. In an inertial frame in flat spacetime it simplifies to the equation I gave in 28.

 

[phyti]

Can we say that “only the average speed according to an IRF matters to the Time Dilation which determines the average value of the period assigned, in this IRF, to a relatively moving clock” (I assume all velocities are alongside the x-axis and that both the initial and final meeting points are collocated)?

Then the Differential Aging of both clocks at their second meeting point would depend on the difference between the average speed, across its journey, of each clock in respect to the same IRF (this difference being independent on the selected IRF, irrespective of the maximum speed reached by any of the clocks, of the number of direction changes alongside the x axis, of the magnitude of their accelerations and finally not directly dependent on their relative speed in respect to each other).

If you examine the expression for gamma, it is not linear but exponential, yeilding the familiar hyperbolic curve for time unit vs speed.

With respect to SR;
for distance: x = vt
dx/dt = v, increasing directly proportional to speed
for time: t = x/v
dt/dv = -x/v^2, decreasing inversely proportional to speed squared