I Two rings rotating about a common center

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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: 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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The easiest way to prove this is to argue by symmetry. Rotate by 180 degrees about an axis in the plane of the ring. The problem is identical, but you have swapped Matt and Eve. Therefore by symmetry their reading must be identical.
 
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The easiest way to prove this is to argue by symmetry. Rotate by 180 degrees about an axis in the plane of the ring. The problem is identical, but you have swapped Matt and Eve. Therefore by symmetry their reading must be identical.
Yes, I see what you mean. Actually my book provides two methods: by symmetry and the one I have posted.

Now I am trying to understand the later.
 
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Yes, I am not really convinced by that method. Rotating reference frames are notoriously difficult to do correctly. I suspect that the method they suggest does not lead to a valid coordinate system.
 

jbriggs444

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What does defining a surface of simultaneity by extending hypersurfaces orthogonal to one's world line mean?
Sounds like: "Consider a tangent inertial frame at an event on the world line with x=y=z=t=0 at the event. The set of events at t=0 per this frame defines a hypersurface of simultaneity".

This leads to a coordinate system for at least some portion of the region of interest. You get the t coordinate from the simultaneous clock time on the object and x, y and z coordinates per the tangent inertial frame.

If intuition serves, the problems arise far to the outside of the orbital circle where the hypersurfaces of simultaneity might sweep backward in remote proper time, leading to coordinate singularities and events that are multiply mapped by the coordinate system.
 

Ibix

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If you imagine a (2+1)d Minkowski diagram, Matt and Eve's worldlines are spirals in opposite directions lying in the surface of a cylinder. At any event on Eve's worldline you can draw a plane that is Lorentz-orthogonal to Eve's worldline at that event. That plane intersects Matt's worldline at exactly one event, and ##w## is the straight line in that plane that joins Eve's event to Matt's.

That's what I think the author is describing. However, (s)he appears to be finding ##w## by letting it be a line between some chosen event on Eve's worldline and then choosing an event on Matt's worldline such that ##w## is orthogonal to Eva's four velocity. As others have commented, I'm not at all sure that the coordinate system implied by the description is valid - but the specified line does exist. Although it's not well defined when the worldlines cross.

The method seems enormously over-complicated. You just need to equate the coordinate expressions to find the events where they meet and then solve for the proper times. In fact, in this case, equating the two coordinate time expressions gives you that ##\tau_E=\tau_M## at all times (including the meetups) in this frame. That won't hold so generally in other frames, but must be true at the meetups because they are single events.
 
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Sounds like: "Consider a tangent inertial frame at an event on the world line with x=y=z=t=0 at the event. The set of events at t=0 per this frame defines a hypersurface of simultaneity".
Thanks for your clarification.

Does tangent inertial frame imply that the world line of (let's say) Eve is orthogonal to the set of events at ##t=0##?
 
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At any event on Eve's worldline you can draw a plane that is Lorentz-orthogonal to Eve's worldline at that event.
Mathematically we can express that condition as ##w \cdot u_{E} = 0##, isn't it?
 
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The method seems enormously over-complicated.
I agree. However, for fun's sake, I am studying it. Let me post both the exercise and solution (note we're dealing with the second method):

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