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- Problem statement: Two rings rotate with equal and opposite angular (relativistic) velocity about a common center. Matt rides on one ring and Eve on the other and there's a point they meet and their clocks agree. At the moment they pass by one another, each asserts that is the other's clock which is running slow. Show that the next time they meet their clocks agree again.

Problem statement:

I understand that when they say 'there's a moment they meet and their clocks agree' they mean that their proper time is the same (let c=1):

$$d\tau_{M} = d\tau_{E} = dt\sqrt{1-(r\omega)^2}$$

Next time they meet ##d\tau_{M} = d\tau_{E}## is again true so above equation is valid but I think this is not enough to answer the question

Actually my books provides a method that I do not fully understand:

'We could consider a non-inertial coordinate system attached to one of them (let's say to Eve). Then Eve defines a surface of simultaneity by extending hypersurfaces orthogonal to her word line, the distance between hypersurfaces being at equal intervals of her proper time. At points where her hypersurfaces intersect Matt's world line, his proper time ##\tau_{M}## is read off and Eve can compute ##\tau_{E}## as a function of ##\tau_{M}##.

Eve's world line is defined by:

$$t= \gamma \tau_{E}$$

$$x = sin (\omega t) = sin (\omega \gamma \tau_{E})$$

$$y = cos(\omega t) = cos (\omega \gamma \tau_{E})$$

$$z = 0$$

Matt's world line is defined by:

$$t= \gamma \tau_{M}$$

$$x = -sin (\omega t) = sin (\omega \gamma \tau_{M})$$

$$y = cos(\omega t) = cos (\omega \gamma \tau_{M})$$

$$z = 0$$

Then one proceeds by determining the four vector ##w## which connects Eve's and Matt's world lines and which is orthogonal to Eve's 4-velocity.'

Using ##w \cdot u_{E} = 0## condition one gets that ##\tau_{M} = \tau_{E}## when ##sin(2\omega \gamma \tau_{E}) = sin(2\omega t)=0## (which means that whenever their world lines cross you get ##\tau_{E} = \tau_{M}##). Note I have not provided full details on how to get ##sin(2\omega \gamma \tau_{E}) = sin(2\omega t)=0##; I could do so if needed be.

What does defining a surface of simultaneity by extending hypersurfaces orthogonal to one's world line mean?

I am trying to figure out this method so further questions may be asked.

Thanks.

*Two rings rotate with equal and opposite angular (relativistic) velocity about a common center. Matt rides on one ring and Eve on the other and there's a moment they meet and their clocks agree. At the moment they pass by one another, each asserts that is the other's clock which is running slow. Show that the next time they meet their clocks agree again.*I understand that when they say 'there's a moment they meet and their clocks agree' they mean that their proper time is the same (let c=1):

$$d\tau_{M} = d\tau_{E} = dt\sqrt{1-(r\omega)^2}$$

Next time they meet ##d\tau_{M} = d\tau_{E}## is again true so above equation is valid but I think this is not enough to answer the question

Actually my books provides a method that I do not fully understand:

'We could consider a non-inertial coordinate system attached to one of them (let's say to Eve). Then Eve defines a surface of simultaneity by extending hypersurfaces orthogonal to her word line, the distance between hypersurfaces being at equal intervals of her proper time. At points where her hypersurfaces intersect Matt's world line, his proper time ##\tau_{M}## is read off and Eve can compute ##\tau_{E}## as a function of ##\tau_{M}##.

Eve's world line is defined by:

$$t= \gamma \tau_{E}$$

$$x = sin (\omega t) = sin (\omega \gamma \tau_{E})$$

$$y = cos(\omega t) = cos (\omega \gamma \tau_{E})$$

$$z = 0$$

Matt's world line is defined by:

$$t= \gamma \tau_{M}$$

$$x = -sin (\omega t) = sin (\omega \gamma \tau_{M})$$

$$y = cos(\omega t) = cos (\omega \gamma \tau_{M})$$

$$z = 0$$

Then one proceeds by determining the four vector ##w## which connects Eve's and Matt's world lines and which is orthogonal to Eve's 4-velocity.'

Using ##w \cdot u_{E} = 0## condition one gets that ##\tau_{M} = \tau_{E}## when ##sin(2\omega \gamma \tau_{E}) = sin(2\omega t)=0## (which means that whenever their world lines cross you get ##\tau_{E} = \tau_{M}##). Note I have not provided full details on how to get ##sin(2\omega \gamma \tau_{E}) = sin(2\omega t)=0##; I could do so if needed be.

**Question:**What does defining a surface of simultaneity by extending hypersurfaces orthogonal to one's world line mean?

I am trying to figure out this method so further questions may be asked.

Thanks.

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