Lag of a clock coming back to its initial position (Twins Paradox)

[Aleberto69]

Thank you very much.

I can't imagine how to set up the equations for calculating:
1) the new transformation coordinates
2) velocity of light in the NIRF
3) $ds$ in the NIRF

the NRFI I was immagining :
$v(t)= k~~~~~~~ when~~~~~~t<t_0$
$v(t)= k-(t-t_0)~~~~~~~when~~~~~~t_0<t<t_0+2k$
$v(t)= -k ~~~~~~~when~~~~~~t>t_0+2k$

and let assume k=0.8c

 

[Aleberto69]

Objects with constant acceleration can be described by the Rindler coordinates. Objects in a circular motion which experience centripetal acceleration are described by the Born coordinates. Both of these are coordinate charts on Minkowski spacetime.

TY very much,
could you suggest a textbook that address those kind of topics? SR applied to NIRF, flat space time, Born coordinates...

 

[Aleberto69]

In any coordinate system, you can compute elapsed time by using the expression:

$\tau = \int \sqrt{\sum_{\mu \nu} g_{\mu \nu} \frac{dx^\mu}{ds} \frac{dx^\nu}{ds}} ds$

where $\mu$ and $\nu$ range over the indices for the different coordinates, and where $g_{\mu \nu}$ are the components of the metric tensor in whatever coordinate system you are using.

An example of a noninertial coordinate system where it is possible to do the calculation is the "Rindler coordinates" $X$ and $T$, which are related to the usual $x$ and $t$ through:

$x = X cosh(gT/c)$
$t = \frac{X}{c} sinh(gT/c)$

or the inverse:

$X = \sqrt{x^2 - c^2 t^2}$
$T = \frac{c}{g} tanh^{-1}(\frac{ct}{x})$

The significance of this coordinate system is that it is the natural coordinate system to use for someone traveling in a rocket that is accelerating so as to give an apparent gravity of $g$ at all times (in the rear of the rocket). Objects at "rest" aboard the rocket will have a constant value of $X$. The coordinate $T$ is the time as measured by a clock in the rear of the rocket.

For these coordinates, $g_{TT} = \frac{g^2 X^2}{c^4}$, $g_{XX} = -\frac{1}{c^2}$. So the proper time for a path is given by:

$\tau = \int \sqrt{\frac{g^2 X^2}{c^4} (\frac{dT}{ds})^2 - \frac{1}{c^2} (\frac{dX}{ds})^2} ds$

For the particularly simple case of $X =$ constant (a clock that is at "rest" on board the accelerating rocket), this simplifies enormously to:

$\tau =\frac{g X}{c^2} T$

which shows that the proper time for a clock is proportional to its location $X$ (higher clocks run faster).

The odd thing about the Rindler coordinate system is that the point $X=0$ isn't actually the rear of the rocket. Instead, the rear of the rocket is at $X = \frac{c^2}{g}$, which means that for a clock in the rear, $\tau = T$, as I said.

TY very much,

could you suggest a good textbook which address those calculation and deduce the formula you have used?
TY very much in advance

 

[Mentz114]

TY very much,
could you suggest a textbook that address those kind of topics? SR applied to NIRF, flat space time, Born coordinates...

Have a look at this http://math.ucr.edu/home/baez/physics/Administrivia/rel_booklist.html
I have not come across a SR textbook which includes Born coordinates but I have only ever read two.

 

[Nugatory]

could you suggest a good textbook which address those calculation and deduce the formula you have used?
TY very much in advance

Any general relativity textbook will cover these mathematical techniques in detail, because that's how we work with the complicated and hard-to-visualize coordinate systems needed for curved four-dimensional spacetime. Often they start by showing how the formulas we know from special relativity can be derived using this machinery. For example, the appalling integral at the top of stevendaryl's post 35 above actually works just fine with ordinary Minkowski coordinates; in these coordinates all the $g_{ij}$ and $\frac{dx^i}{ds}$ are constants so you can move them out of from under the integral sign, and the integral evaluates to the familiar $\Delta{s}^2=-\Delta{t}^2+\Delta{x}^2+\Delta{y}^2+\Delta{z}^2$

Sean Carroll's online lecture notes on GR will get you started.

 

[Aleberto69]

Have a look at this http://math.ucr.edu/home/baez/physics/Administrivia/rel_booklist.html
I have not come across a SR textbook which includes Born coordinates but I have only ever read two.

TY very much... readings for half a century or more:-) !

 

[PAllen]

The classic work for special relativity without relying on inertial frames and the Lorentz transform is:

https://books.google.com/books/abou...BAJ&printsec=frontcover&source=kp_read_button

The Lorentz transform is not even mentioned until chapter 6, after all major SR pehenomena (Doppler, time dilation, aberration) have already been treated in generality, without any reference to inertial observers.

 

[Mentz114]

The classic work for special relativity without relying on inertial frames and the Lorentz transform is:

https://books.google.com/books/abou...BAJ&printsec=frontcover&source=kp_read_button

The Lorentz transform is not even mentioned until chapter 6, after all major SR pehenomena (Doppler, time dilation, aberration) have already been treated in generality, without any reference to inertial observers.

Reading through the contents I get the impression that everything relevant is covered. There's a whole chapter on rotation !
Tempting ...

 

[stevendaryl]

TY very much,

could you suggest a good textbook which address those calculation and deduce the formula you have used?
TY very much in advance

Here's a long-winded derivation.

If the rear of a rocket has acceleration $g$ in a frame in which it is (momentarily) at rest, then it has acceleration $g_v = \frac{g}{\gamma^3}$ in a frame in which it is traveling at $v$. You can figure this out by using the inverse Lorentz transforms:

1. $x' = \gamma (x - v t)$
2. $t' = \gamma(t - \frac{vx}{c^2})$

Now, in the primed frame, assume that position of the rocket is given by: $x' \approx \frac{1}{2} g (t')^2$. ($\approx$ means I'm ignoring terms of order $(t')^3$ or higher.) So in this frame, the rocket is accelerating from rest at rate $g$. In the unprimed frame, assume that $x \approx vt + \frac{1}{2} a t^2$. In this frame it's accelerating from speed $v$ at rate $a$. We want to solve for $a$. Plugging things into our equations, we have:

1. $\frac{1}{2} g (t')^2 \approx \gamma \frac{1}{2} a t^2$. This implies that $a = g (\frac{t'}{t})^2 \frac{1}{\gamma}$.
2. $t' \approx \gamma (t - \frac{v^2}{c^2} t) = \gamma(1-\frac{v^2}{c^2}) t = \frac{t}{\gamma}$. this implies that $\frac{t'}{t} \approx \frac{1}{\gamma}$

Putting 1 and 2 together gives us: $a = g \frac{1}{\gamma^3}$

Since $a = \frac{dv}{dt}$, this means $\frac{dv}{dt} = g \frac{1}{\gamma^3}$. We can invert this to get: $\frac{dt}{dv} = \frac{1}{g} \gamma^3$

So we integrate to get: $t = \int \frac{1}{g} \gamma^3 dv = \frac{1}{g} \int (1-\frac{v^2}{c^2})^{-\frac{3}{2}} dv$

To do this integral, we do a variable substitution: $v = c tanh(\frac{gT}{c})$, so $dv = g sech^2(\frac{gT}{c}) dT$ ($tanh$ is hyperbolic tangent, $sech$ is hyperbolic secant.) Then $1-\frac{v^2}{c^2} = 1 - tanh^2(\frac{gT}{c}) = sech^2(\frac{gT}{c})$. So the integral becomes:

$t = \int \frac{1}{sech(\frac{gT}{c})} dT = \int cosh(\frac{gT}{c}) dT = \frac{c}{g} sinh(\frac{gT}{c})$

We can integrate $v$ to get $x$, and we find:

$x = \int v dt = \int c tanh(\frac{gT}{c}) cosh(\frac{gT}{c}) dT = c \int sinh(\frac{gT}{c}) dT = \frac{c^2}{g} cosh(\frac{gT}{c})$

(To simplify the expression, I've chosen the origin of the x-axis so that the rocket is initially at $x = \frac{c^2}{g}$ rather than $x=0$)

We can also compute:

$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = cosh(\frac{gT}{c}) = \frac{gx}{c^2}$
$\gamma v = cosh(\frac{gT}{c}) c tanh(\frac{gT}{c}) = c sinh(\frac{gT}{c}) = \frac{gt}{c}$

And we show that $x$ and $t$ obey the equation: $x^2 - c^2 t^2 = \frac{c^4}{g^2}$

All right. Now, let's try to figure out a coordinate system that can be used on board the rocket. We already have a time coordinate, $T$, but what's its significance? Well, since the proper time for a clock at the rear of the rocket is given by: $\tau = \int \sqrt{1-\frac{v^2}{c^2}} dt$, we can change variables from $t$ to $T$ using $t = \frac{c}{g} sinh(\frac{gT}{c})$. So $dt = cosh(\frac{gT}{c}) dT$. Using our formula for $v$: $v = c tanh(\frac{gT}{c})$ so $\sqrt{1-\frac{v^2}{c^2}} = \sqrt{1-tanh(\frac{gT}{c})^2} = sech(\frac{gT}{c}) = \frac{1}{cosh(\frac{gT}{c})}$. So our integral for proper time becomes:

$\tau = \int dT = T$

So $T$ is just the time on the rear clock.

For our spatial coordinate, we want it to be the case that the length of the rocket stays the same, as viewed by someone on board the rocket. Otherwise, the rocket would either be ripped apart or crushed. Let $L$ be the length of the rocket as measured in the rocket's accelerated frame. We're going to assume that this is also the length of the rocket as measured in the inertial coordinate system in which the rocket is (momentarily) at rest. So let's suppose that the rocket has traveled for a while, and the rear of the rocket is at location $x$ at time $t$. The rocket is traveling at speed $v$. Let $e_1$ be the event (moment in space and time) when the rear of the rocket is traveling at speed $v$. Let $e_2$ be the event at the front of the rocket that is simultaneous with $e_1$ in the instantaneous rest frame of the rocket.

• Let $(x,t)$ be the coordinates of $e_1$ in the Earth frame.
• Let $(x', t')$ be the coordinates in the instantaneous rest frame of the rocket.
• Let $(x_2, t_2)$ be the coordinates of $e_2$ in the Earth frame.
• Let $(x_2', t_2')$ be the coordinates in the rocket frame.

Since $e_1$ and $e_2$ are simultaneous in the rocket frame, that means: $t_2' = t'$.

Since the rocket has length $L$ in its own rest frame, this means: $x_2' = x' + L$

Now, we can use the inverse Lorentz transformations to find $x_2$ and $t_2$:

$x = \gamma (x' + v t')$
$t = \gamma(t' + \frac{vx'}{c^2})$
$x_2 = \gamma (x_2' + v t_2') = \gamma(x' + L + v t') = x + \gamma L$
$t_2 = \gamma(t_2' + \frac{v x_2'}{c^2}) = \gamma(t' + \frac{v}{c^2} (x' + L)) = t + \frac{\gamma vL}{c^2}$

But we know from above for our constant acceleration rocket, $\gamma = \frac{gx}{c^2}$. So

$x_2 = x (1 + \frac{gL}{c^2})$

We also know that $\gamma v = \frac{gt}{c}$. So

$t_2 = t + \frac{gt}{c} \frac{L}{c^2} = t (1 + \frac{gL}{c^2})$

Now, we see an amazing, or possibly not amazing relationship between $x_2$ and $t_2$:

$(x_2)^2 - c^2 (t_2)^2 = x^2 (1+ \frac{gL}{c^2}) - c^2 t^2 (1+\frac{gL}{c^2})^2 = (x^2 - c^2 t^2) (1+\frac{gL}{c^2})^2$

But $x^2 - c^2 t^2 = \frac{c^4}{g^2}$, so

$(x_2)^2 - c^2 (t_2)^2 = \frac{c^4}{g^2} (1+\frac{gL}{c^2})^2 = (\frac{c^2}{g} + L)^2$

Now, remember that I told you that the initial location of the rear of the rocket was at $\frac{c^2}{g}$? The initial location of the front of the rocket was $\frac{c^2}{g} + L$. Letting $X_{rear}$ be the initial location of the rear of the rocket, and $X_{front}$ be the initial location of the front of the rocket, then we find that
$x^2 - c^2 t^2 = \frac{c^4}{g^2} = (X_{rear})^2$
$(x_2)^2 - c^2 (t_2)^2 = (X_{front})^2$

What this means is that the front and rear are following the same sort of accelerated trajectory, but with different accelerations:

$x = X_{rear} cosh(\frac{gT}{c})$
$x_2 = X_{front} cosh(\frac{gT}{c})$

In general, if a section of the rocket starts off at location $X$, then its position later will be given by:

$x = X cosh(\frac{gT}{c})$

with the time given by:

$t = \frac{1}{c} X sinh(\frac{gT}{c})$

The interesting thing about this is that even though the acceleration experienced by the rear of the rocket is $g_{rear} = \frac{c^2}{X_{rear}}$, the acceleration experienced by the front of the rocket is smaller: $g_{front} = \frac{c^2}{X_{front}}$. The proper time computed earlier for a clock in the rear of the rocket is: $\tau = T$. But the proper time for the front of the rocket is $\tau = \frac{g T X_{front}}{c^2}$. So clocks in the front run faster than clocks in the rear. (proper time is only equal to $T$ at the rear of the rocket).

 

[hutchphd]

One last point:

Clocks A, B together and clock C some distance away. All at rest initially and synchronised.

Clock B accelerates. Using only time dilation arguments, clocks A and C must read the same (in B's frame). It doesn't matter whether you integrate using $dt$ or sum using $\Delta t$. After B accelerates, using only time dilation, clocks A and C must remain synchronised in B's frame.

But, clocks A & C are not synchronised in B's frame. So, if you change your IRF you cannot simply think about time dilation.

Perhaps I can diminish confusion slightly with something I didn't understand until recently. Add clock D also some distance away to this example. Allow it to be attached ti B and undergo acceleration along with B If there is acceleration along the axis BD these clocks will no longer be synchronized in their new inertial frame after acceleration ceases. This is a direct result of them being accelerated. . .