Yes, it is true that “either twin could be considered to be moving at close to the speed of light”. And it is also true that so long as neither of them accelerates each will think that the other is ageing more slowly. But this is no paradox as there is no way for them to get together to see who is right. That can only happen if one or other of them turns around. And in that case whoever turns back towards the other will end up thinking that the other aged so much more quickly during the turn-around that when they get back together there will be a well-defined answer to the question of who is indeed younger and by how much. (They will not agree on the exact pattern of “when” the different ageing rates occurred, but since there is no absolute definition of “when” for events that do not happen at the same place, that disagreement is just a matter of having different points of view.)
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Moving Through Time
Time does not even exist without some frame of reference, as space and time are just ways of assigning coordinates to events and different observers may do so in different ways. And in terms of any particular frame of reference, motion through space is usually thought of as having different positions at different points in time. Since the position varies with the time I suppose this could be described by saying that we “move with time”. But the idea of motion through time is kind of vacuous since any choice of time coordinate obviously changes when it changes; so I guess we could always be described as moving with time through the same time at “speed” one. This might get more interesting for the case of different reference frames with different time coordinates – in which case one might be moving with the time of one frame through the time of another (with a rate that would correspond to the Lorentz time dilation factor corresponding to the relative speed between the two frames).
So whether you are moving through space or time depends on what coordinate system you are using. In terms of coordinates based on your own body you are always fixed in space and “move” only through time. But in terms of the coordinates of a frame that is moving with respect to you, you are moving through its version of space (with exactly the opposite velocity).
Pluto’s Status and Peers
One-way speed of light
Some authors (referenced in this Wikipedia article ) have alleged that The “one-way” speed of light, from a source to a detector, cannot be measured independently of a convention as to how to synchronize the clocks at the source and the detector – and that this somehow has important implications with regard to the interpretation and application of Special Relativity.
But I beg to differ.
Consider the scenario of two remote systems, B and C, that are stationary with respect to us who, let us say, are at position A. Now let a light signal from B be emitted towards both C and A, and let C, on receiving that signal immediately signal towards us at A. The time delay between our receipt of the two signals will be the light travel time from B to C plus the difference between the light travel times to us from B and C.
So, if we let [math]t_{XY}[/math] be the light travel time from X to Y, then the time difference between the direct and indirect signals starting at B is given by [math]\Delta t_B = (t_{BC} +t_{CA})-t_{BA}[/math].
And if we do the process in reverse for signals starting at C we get a time difference of [math]\Delta t_C = (t_{CB} +t_{BA})-t_{CA}[/math].
So, if we get the same time difference both ways, then we’ll know that [math](t_{BC} +t_{CA})-t_{BA} = (t_{CB} +t_{BA})-t_{CA}[/math],
which gives [math]t_{BC} -t_{CB} =2 (t_{BA}-t_{CA})[/math].
So in that case the difference between light travel times from B to C and from C to B is exactly twice the difference between the travel times from B to A and C to A.
Now let’s restrict to the case where B and C are separated by a distance [math]d[/math] and are both at the same distance [math]D[/math] from A, and let [math]v_{XY}[/math] be the speed of light in the direction from X to Y and [math]v_0[/math] be the speed in the perpendicular direction towards A from the midpoint of the line BC.
Then for the case of equal time differences both ways our previous result becomes [math]\frac{d}{v_{BC}} -\frac{d}{v_{CB}} =2 (\frac{D}{v_{BA}}-\frac{D}{v_{CA}})[/math].
In the case that A is midway between B and C (which is what the equidistance gives us in the case of one space dimension) we have [math]d=2D[/math] with [math]v_{BA}=v_{BC}[/math] and [math]v_{CA}=v_{CB}[/math], so the equal time differences condition is just that [math]\frac{d}{v_{BC}} -\frac{d}{v_{CB}} =2 (\frac{d/2}{v_{BC}}-\frac{d/2}{v_{CB}})[/math] – which is clearly always true even if the two one-way speeds are different.
For the case that A is not on the line BC, if we let [math]\omega=2\theta[/math] be the angle at A between AB and AC then [math]d=2D\sin{\theta}[/math] and the condition of equal time differences becomes [math]2D\sin{\theta}(\frac{1}{v_{BC}} -\frac{1}{v_{CB}}) =2D (\frac{1}{v_{BA}}-\frac{1}{v_{CA}})[/math].
If we let [math]v_{\theta}[/math] denote the speed of light in direction at angle [math]\theta[/math] from the right bisector of BC, and [math]T_{\theta}=\frac{1}{v_{\theta}}[/math], and if the equal time condition is observed in all directions, then we need to find a function [math]T_{\theta}[/math] satisfying the conditions [math](T_{\theta+\phi}-T_{-\theta+\phi})=\sin{\theta}(T_{\pi/2+\phi}-T_{-\pi/2+\phi})[/math]
OneWay Speed of Light Links
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GreenhouseSaturationLinks
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EulerLagrangeLinks
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