Symmetrical time dilation implies relativity of simultaneity

pervect
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This is something I've been meaning to write for a while, I finally got the time to do it, though the topic isn't currently "hot". I'm sure it will pop up again, though.

Imagine we have two observers, moving at a constant velocity relative to each other far away from any objects that might perturb their motion. We will call the two observers Adam (abbreviated as A) and Beth (abbreviated as B). Each of these observers carries an idealized clock - for the purposes of this thought experiment, a physical atomic clock is close enough to ideal as makes no difference.

One of the results of special relativity is that in Adam's frame of reference, Beth's clock runs slow - and vica versa. Formally, we can say the following. Along A's trajectory, which is also called A's worldline, every unique reading on A's clock corresponds to a unique event. The same applies to B's worldine, every clock reading defines a unique event. The concept of simultaneity means that we can define a 1:1 mapping between the events on A's worldine and the events on B's worldline that happen "at the same time" in A's frame of reference.

Example: Suppose we have a time dilation factor of 2:1. Then in A's frame of reference, the event on A's worldline where his clock reads 1 is simultaneous with the event on B's worldline where her clock reads 1/2.

We can write a general formula for this, for ease of exposition we will continue to assume the time dilation factor is 2:1 as in our example. The argument can easily be generalize to other time dilation factors, however. When A's time reads X, B's time reads X/2, where X is any number. Equivalently we can also say that when B's clock reads X, A's clock reads 2X using basic algebra.

Now, the concept of symmetrical time dilation implies that in B's frame of reference, A's clock is the clock that runs slow. So in B's frame of reference, when B's clock reads X, A's clock reads X/2 Or, equivalently, when A's clock reads X, B's clock reads 2X

Now, let's try to reconcile these facts. In A's frame of reference, when A's clock reads X, B's clock reads X/2. In B's frame of reference, when A's clock reads X, B's clock reads 2X. We note that 2X is not equal to X/2. So, what event on B's woldline is "really" simultaneous with the event on A's worldline where A's clock reads 1?

The answer is actually very simple. Simultaneity in relativity is relative, so both statements are true. One is true in A's frame of reference, the other is true in B's frame. Thus, "at the same time" in A's frame of reference is not the same concept (or mapping) as "at the same time" in B's frame of reference. In Newtonian physics, time is absolute, so the concept of "at the same time" is universal and not dependent on the observer. This is not the case with relativity, and symmetrical time dilation is just one of the consequences.

The concept of the relativity of simultaneity was popularized by Einstein in "Relativity: the special and general theory" in chapter 9, in a thought experiment involving a train. This is currently available online (and has been for some time) at http://www.bartleby.com/173/9.html.

Another paper on the topic that I find useful (I've had little reader feedback) is
"The challenge of changing deeply held student beliefs about the relativity of simultaneity", by Scherr, Shaffer, and Vokos. The abstract of this paper is on arxiv at http://arxiv.org/abs/physics/0207081. The full pdf text is usually available at http://arxiv.org/ftp/physics/papers/0207/0207081.pdf.
 
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Without affecting logic, it seems to me you multiply by 2 when you should divide by 2. When A's clock reads 1, B's clock reads 1/2, not 2; else B's clock would be faster not slower.
 
PAllen said:
Without affecting logic, it seems to me you multiply by 2 when you should divide by 2. When A's clock reads 1, B's clock reads 1/2, not 2; else B's clock would be faster not slower.
Should be fixed now. Let me know if I missed something :(.
 
pervect said:
This is something I've been meaning to write for a while, I finally got the time to do it, though the topic isn't currently "hot". I'm sure it will pop up again, though.
It does seem that there are rushes of people with the same problem. I've never quite worked out why that happens.

pervect said:
vica versa
Very minor, but shouldn't that be vice versa?
 
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A spacetime diagram showing this might be helpful.
A week ago, I posted a link to an interactive one in another thread.

The geometrical picture is that
"simultaneity according to an inertial observer"
is akin to
"tangency to a hyperboloid of constant timelike-interval from an event
[note that this defines what it means to be perpendicular to the radius]".

http://www.desmos.com/calculator/ti58l2sair

ti58l2sair.png
 
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Take the case of soon-not-to-be smiling Bob falling into a black hole. His companion, Alice observes daredevil Bob's plunge from the safety of an orbiting ship. She believes Bob never makes it to the event horizon. Bob, on the other hand, believes his harrowing journey was altogether too brief.
 
Let's say that there are two events A and B that occur at some non-zero time t on the y-axis of a frame S, and that another frame S' which had its origin coincide with S at t=0 is moving with a constant velocity v such that the motion of this frame is parallel to the x-axis of frame S. Now there will always be a symmetric time dilation between the two frames regardless of the direction of the velocity, but events A and B will be simultaneous in both frames. So any event that occurs on the y-axis of either frame is simultaneous in both frames. How does one understand the time dilation with these events?
@robphy I was a little confused with the hyperbola you have shown until I realized you'd plotted time on the x-axis instead of the y axis. I know it's arbitrary, but is it used often? I'm still (relatively!) new to this branch of physics and I'd like to practice "switched" spacetime diagrams as well if this is common in scientific literature. (Sideways light cones etc)
 
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Although relativity books have time running upwards, all other position vs time graphs ( in PHY 101) have time running horizontally to the right.

In the way I've drawn, it's less mysterious to novices and the slope is the velocity. In addition, with my signature convention (below), where I view time as more fundamental, it makes it easier to show the analogies with Galilean-Spacetime and Euclidean geometries...try the E-slider on the interactive graph. That's why I called the vertical axis, representing space, as y (and not x).
 
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Ibix said:
It does seem that there are rushes of people with the same problem. I've never quite worked out why that happens.
Memes?, well it is some kind of an explanation of why topics get popular both on internet and elsewhere.
http://en.wikipedia.org/wiki/Meme
 

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