Symmetry Argument: Resolving the Twin Paradox Contradiction

[Gear300]

This is pretty much like the twin paradox question: if person A and person B were moving with a velocity v relative to each other and away from each other, then person B would observe a time dilation in person A's reference frame while person A would observe a time dilation in person B's reference frame...which seems contradictory. What would be the resolution to this apparent contradiction?

 

[Doc Al]

Where do you see a contradiction? (To compare measurements between frames, one must consider length contraction and clock synchronization as well as time dilation.)

 

[JesseM]

Person B observes A's clock is dilated in B's own reference frame (i.e. if you look at the event of A's clock reading 5 seconds and the event of A's clock reading 15 seconds, in B's frame the time elapsed between these two events is larger than 10 seconds), it doesn't really make sense to say "person B would observe a time dilation in person A's reference frame" since the convention is to say that each observer "observes" whatever is true in their own rest frame (note that this is actually different from what they see visually). But this is just a semantic issue, it is true that each one would observe the other one's clock to be running slow. This isn't a contradiction though, it can be understood in terms of the relativity of simultaneity, which says that different frames disagree about whether a pair of events at different positions in space happened at the same time or different times (whether the events are simultaneous or not). Suppose for example that you and I are moving at 0.6c relative to one another, and that when we cross paths, we both set our clocks to read 0, then continue moving apart at the same speed. In my frame, the event of my clock reading 10 years will be simultaneous with the event of your clock reading 8 years, so in my frame your clock is running slower than my own. But in your frame, the event of your clock reading 8 years is not simultaneous with the event of my clock reading 10 years, instead it is simultaneous with the event of my clock reading 6.4 years, so in your frame it is my clock that is ticking slower (likewise, in your frame the event of my clock reading 10 years is simultaneous with the event of your clock reading 12.5 years).

 

[Gear300]

Hmmm...it took me a while to get what you're saying...but to make sure of things...what you're saying is that because the notion of space-time is not universal as assumed in the equations, my assumption of being able to connect the two observations as if they were at the same time or in the same reference frame was not valid?

 

[JesseM]

Hmmm...it took me a while to get what you're saying...but to make sure of things...what you're saying is that because the notion of space-time is not universal as assumed in the equations, my assumption of being able to connect the two observations as if they were at the same time or in the same reference frame was not valid?

I don't know what you mean by "the notion of space-time is not universal". It's the notion of simultaneity that's not universal--two events which are assigned the same t-coordinate in one frame can be assigned different t-coordinates in another. In Newtonian physics it's already true that two events could have the same x-coordinate (or same y or z) in one coordinate system but different x-coordinates in another; relativity extends this to time coordinates as well.

In Newtonian physics, if you know the coordinates x,y,z,t of an event in one inertial frame, and you want to know the coordinates x',y',z',t' of the same event from the perspective of a different inertial frame whose origin is moving at velocity v along the x-axis of the first frame, then you'd use the Galilei transformation:

x' = x - vt
y' = y
z' = z
t' = t

But in relativity, you'd instead use the Lorentz transformation:

x' = gamma*(x - vt)
y' = y
z' = z
t' = gamma*(t - vx/c^2)

where gamma = 1/sqrt(1 - v^2/c^2). You can see that with the Galilei transformation the x coordinate of the event in the first frame can be different than the x' coordinate of the event in the second frame, but the t coordinate is always equal to the t' coordinate. On the other hand, with the Lorentz transformation, the t coordinate of an event in frame #1 can be different from the t' coordinate of the event in frame #2.

 

[Gear300]

So then what you're saying is that it only seems like a contradiction because we're used to thinking of these sorts of things as contradictions (and it would be a contradiction in a Newtonian system but not in a Relativistic system)?

 

[JesseM]

So then what you're saying is that it only seems like a contradiction because we're used to thinking of these sorts of things as contradictions (and it would be a contradiction in a Newtonian system but not in a Relativistic system)?

Yes, we're used to thinking there should be some truth about who is "really" older, but if simultaneity is relative than the answer will depend on your choice of reference frame. Keep in mind, though, that disagreements about simultaneity only happen for events at different locations in space, if the two of them got back together to compare ages at a single location, every frame would agree about who was older (but in order for them to get back together one of them would have to accelerate to turn around, so then you have a typical twin paradox type situation where the one that accelerates is always going to be the one who has aged less when they reunite.

 

[Gear300]

Yes, we're used to thinking there should be some truth about who is "really" older, but if simultaneity is relative than the answer will depend on your choice of reference frame. Keep in mind, though, that disagreements about simultaneity only happen for events at different locations in space, if the two of them got back together to compare ages at a single location, every frame would agree about who was older (but in order for them to get back together one of them would have to accelerate to turn around, so then you have a typical twin paradox type situation where the one that accelerates is always going to be the one who has aged less when they reunite.

So then let's say that both person A and B were holding mirrors when they set their time to 0s, and that they held the mirrors so that one person could see his/her clock in the other person's mirror. If person B's mirror held the reflection of person A's clock, and if we had to choose between whether the reflection showed the time dilation of A or followed the same time intervals as B's clock, then would we choose the latter because the reflection is occurring right next to person B?

 

[JesseM]

So then let's say that both person A and B were holding mirrors when they set their time to 0s, and that they held the mirrors so that one person could see his/her clock in the other person's mirror. If person B's mirror held the reflection of person A's clock, and if we had to choose between whether the reflection showed the time dilation of A or followed the same time intervals as B's clock, then would we choose the latter because the reflection is occurring right next to person B?

If they're right next to each other then the delay between the light leaving A's clock and being reflected off the mirror and reaching B's eyes will be negligible, so the delay between B seeing his own clock reading 0 and seeing A's clock reading 0 will be negligible. But time dilation is about the rate each clock ticks over time, not about the time they show at a particular instant (like the instant when they are right next to each other)--if they watch the other guy's clock for any extended length of time, whether by looking in a mirror or by looking directly at the other guy's clock (I don't actually understand the point of the mirror in this thought-experiment), they will see it running slower than their own, due to a combination of time dilation and the Doppler effect (remember my point that what you 'observe' in your rest frame is different from what you see visually--the Doppler effect doesn't have anything to do with how fast a clock ticks in your frame, but it does effect how fast you see it ticking visually when you're watching it with your eyes).

 

[Gear300]

The reason I included the mirrors was because I was assuming the situation would be different if person B could view the behavior of person A's clock from a different point in space (if the source was a mirror stationary with respect to B). I was also thinking that person B could view his/her own clock from a moving source (person A's mirror). So what you're saying is that it doesn't matter whether the mirrors are there or not?

 

[JesseM]

The reason I included the mirrors was because I was assuming the situation would be different if person B could view the behavior of person A's clock from a different point in space (if the source was a mirror stationary with respect to B). I was also thinking that person B could view his/her own clock from a moving source (person A's mirror). So what you're saying is that it doesn't matter whether the mirrors are there or not?

If you have mirrors at different points in space far away from A or B that would be a different issue because the light from a given event would take some noticeable extra time to reach the mirror and then return to the eyes of A or B. But you said the mirrors were just being held by A and B, which makes them seem totally pointless--the light from a given tick of A's clock will reach B's mirror at almost the exact same time it reaches B's eyes since the mirror is right next to B's eyes (and the time for the light to bounce off the mirror into B's eyes is negligible since they're right next to each other), so why not just ask what B sees when he looks directly at A's clock, instead of asking what he sees when he looks in his own mirror at the reflection of A's clock? Unless you're asking what A will see when he looks at his own clock in B's mirror or something like that.

 

[Gear300]

Unless you're asking what A will see when he looks at his own clock in B's mirror or something like that.

That was actually the question I was originally intending to ask when I started this mirror question. Would A see a time dilation in his/her clock when looking at B's mirror?

 

[JesseM]

That was actually the question I was originally intending to ask when I started this mirror question. Would A see a time dilation in his/her clock when looking at B's mirror?

He will see his clock slowed down, but this will be purely due to the signal delays and not due to time dilation--he would see his own clock slowed down by the same amount in a Newtonian universe too (the signal delays have to do with the fact that the light from each successive tick of his clock has a greater distance to travel to reach the mirror and return to him than the light from the previous tick, since the mirror is constantly moving away from him, so there'll be a greater time between reflected ticks than there would be if he was watching his own clock in a mirror which was stationary relative to himself). On the other hand, if he watches B's own clock which is being held next to the mirror that holds the reflection of his clock, he'll see it running even slower than the reflected image of his clock, due to time dilation--this would not be the case in a Newtonian universe.

 

[Gear300]

He will see his clock slowed down, but this will be purely due to the signal delays and not due to time dilation--he would see his own clock slowed down by the same amount in a Newtonian universe too (the signal delays have to do with the fact that the light from each successive tick of his clock has a greater distance to travel to reach the mirror and return to him than the light from the previous tick, since the mirror is constantly moving away from him, so there'll be a greater time between reflected ticks than there would be if he was watching his own clock in a mirror which was stationary relative to himself). On the other hand, if he watches B's own clock which is being held next to the mirror that holds the reflection of his clock, he'll see it running even slower than the reflected image of his clock, due to time dilation--this would not be the case in a Newtonian universe.

I think I'm starting to get the bigger picture (starting, of course). Thanks for the replies.

One last question (don't worry...this is the last...hopefully). Let us say that we were both familiar with a particular movie and its run-time. Let us also say that there is a direct wireless network between my computer and your computer and that one of us hacked into the other's computer (we'll say that you hacked into my computer...whatever hits your preference). The frame I'm in is actually moving at 0.6c relative to your frame (but we're not observing this because we are looking at our computers). I'm actually watching a movie on my computer (the movie we're both familiar with), and because you've hacked into my computer, you're also watching it. Taking into account that my frame is moving at 0.6c relative to your frame, would you notice a slower run-time of the movie?

 

[JesseM]

One last question (don't worry...this is the last...hopefully). Let us say that we were both familiar with a particular movie and its run-time. Let us also say that there is a direct wireless network between my computer and your computer and that one of us hacked into the other's computer (we'll say that you hacked into my computer...or whatever your preference). The frame I'm in is actually moving at 0.6c relative to your frame (but we're not observing this because we are looking at our computers). I'm actually watching a movie on my computer (the movie we're both familiar with), and because you've hacked into my computer, you're also watching it. Taking into account that my frame is moving at 0.6c relative to your frame, would you notice a slower run-time of the movie?

Whoever's computer is running the movie will see it playing at normal speed, while the one who hacked into the other one's computer will see it playing at an altered rate, due to a combination of signal delays due to motion (i.e. the Doppler effect) and genuine time dilation.

 

[Gear300]

Whoever's computer is running the movie will see it playing at normal speed, while the one who hacked into the other one's computer will see it playing at an altered rate, due to a combination of signal delays due to motion (i.e. the Doppler effect) and genuine time dilation.

Alright then (I was just checking the objectivity of relativistic phenomenon...in this case, if the two didn't observe the relative velocity; I was suspecting the time dilation but forgot to - or simply didn't - take into account the Doppler effect). Thanks again for all the replies.

 

[Rasalhague]

On the other hand, if he watches B's own clock which is being held next to the mirror that holds the reflection of his clock, he'll see it running even slower than the reflected image of his clock, due to time dilation--this would not be the case in a Newtonian universe.

Wouldn't B's clock always show a later time than A's reflected clock, albeit not later by as much as in a Newtonian universe?

The time shown by the reflected image of A's clock is the time shown by A's clock at the instant when the light left A's clock. Call the event of this light being emitted from A's clock A₁. Call the event of A receiving the reflected image of A's clock showing this time A₃.

Unless I'm mistaken, the time observed on B's clock should be equal to (1-(u/c)²)^1/2 times the time shown by A's clock at an event A₂ half way along A's worldline between events A₁ and A₃. (Half way because light travels at a constant speed, so its journey to the mirror will take as long as its journey back to A.) This time value will be less than that shown by A's clock at event A₂ because, although in A's rest frame A₂ is simultaneous with the event of the image being reflected from B's mirror, this event of reflection, call it B₂, will happen before A₂ in B's rest frame. This is the slowing down related to "time dilation".

The reason why I don't think A could ever observe B's clock showing an earlier time than the reflected image of A's own clock is because, for this to happen, B₂ would have to be simultaneous in B's rest frame with an event on A's worldline separated from the mutal origin by a smaller spacetime interval than A₁ is separated from the origin. But this couldn't happen unless perhaps B was traveling faster than light...

 

[JesseM]

Wouldn't B's clock always show a later time than A's reflected clock, albeit not later by as much as in a Newtonian universe?

Hmm, when I actually do the calculations below it seems you're right. But since both B's clock and the reflected image of A's clock read 0 when B was right next to A, this must mean that A sees B's clock continually ticking faster than the reflected image of his own clock, not slower as I said. But A will definitely see B's clock ticking slower than his own (non-reflected) clock by the amount predicted by the relativistic Doppler effect, so he must see the reflected image of his own clock slowed down (relative to his own non-reflected clock) by an amount even greater than would be given by either the relativistic Doppler effect or the classical Doppler effect. When I think about it, this makes sense though--just thinking in purely Newtonian terms, it makes sense that if you are watching your own clock in a mirror which is moving away from you, it should be slowed down by a factor greater than the ordinary classical Doppler effect, since light from successive ticks is not being emitted by the mirror at one tick per second, as there is also a Doppler-type delay for the light from these ticks reaching the mirror as a consequence of the fact that the mirror is continually getting farther and farther from your clock and so the light from each tick has farther to travel to reach it. So I wasn't wrong when I said that the amount that A sees his own reflected image slowed down would be the same in relativity as it is in Newtonian physics, I was just wrong when I imagined that the rate it would be slowed down in Newtonian physics would be given by the classical Doppler equation.

The time shown by the reflected image of A's clock is the time shown by A's clock at the instant when the light left A's clock. Call the event of this light being emitted from A's clock A₁. Call the event of A receiving the reflected image of A's clock showing this time A₃.

Unless I'm mistaken, the time observed on B's clock should be equal to (1-(u/c)²)^1/2 times the time shown by A's clock at an event A₂ half way along A's worldline between events A₁ and A₃. (Half way because light travels at a constant speed, so its journey to the mirror will take as long as its journey back to A.)

Yes, that's correct. When A's own clock reads A3, he will be seeing the reflected image of his own clock reading A1, and he will be seeing B's clock reading A2*sqrt(1 - v²/c²), where A2 = (A3 - A1)/2. So if B is moving away from A at constant velocity v, then at time t=A1, B will be at a distance of v*A1 (in A's frame). Then since B continues to move away at velocity v while the light from A at A1 moves toward him at velocity c, the time needed for the light to catch up with B will be (v*A1)/(c - v). That will be the time interval between A1 and A2, which means the actual time of A2 will be given by A1 + (v*A1)/(c - v) = (c*A1 - v*A1)/(c - v) + (v*A1)/(c - v) = c*A1/(c - v). And since we know B's clock reads A2*sqrt(1 - v²/c²) at the moment the light from A1 reaches it, that means B's clock reads A1*c*sqrt(1 - v²/c²)/(c - v) = A1*sqrt(c² - ²)/(c - v) = A1 * sqrt((c + v)*(c - v))/(c - v) = A1 * sqrt(c + v)/sqrt(c - v). This number will always be greater than A1, so A will always see B's time as ahead of the time on the reflected image on his own clock.

Just to pick a numerical example, if B is moving away from A at 0.6c, and A1 is the event of A's clock reading 10 seconds, then at this moment B is 6 light-seconds away, so it takes another 15 seconds for the light to catch up with B, at which time B's clock reads 25*sqrt(1 - 0.6²) = 25*0.8 = 20 seconds, which is in fact equal to 10*sqrt(c + v)/sqrt(c - v) = 10*sqrt(1.6)/sqrt(0.4) = 20. So in this case B appears to be ticking twice as fast as the reflected image of A.

 

[Rasalhague]

I suppose the closer B's speed (in A's rest frame) to c, the smaller the difference between the time A sees on clock B and the time A sees on the reflected clock A, and the slower they'll both be compared to A's clock as A views it directly, approaching the limit at c of infinite slowness, i.e. not ticking at all.

 

[JesseM]

I suppose the closer B's speed (in A's rest frame) to c, the smaller the difference between the time A sees on clock B and the time A sees on the reflected clock A, and the slower they'll both be compared to A's clock as A views it directly, approaching the limit at c of infinite slowness, i.e. not ticking at all.

It's true that both clocks will approach infinite slowness in their visual ticking rate in the limit as B's velocity in A's frame approaches c, but it's not right that the difference in times between B and the reflected image of A gets smaller as you approach c...according to the formula I found, when A sees the reflected image of his own clock showing time A1, then at the same moment the image of B's clock will show time A1*sqrt(c + v)/sqrt(c - v), and that factor of sqrt(c + v)/sqrt(c - v) approaches infinity in the limit as v approaches c. So although the images of the two clocks are approaching infinite slowness, they are approaching it at different rates, and the ratio between their two rates of apparent ticking is getting larger and larger. It's in the opposite limit as v approaches 0 that they approach ticking at the same apparent rate, unless there's a mistake in my derivation.

 

[Rasalhague]

Maybe it's just my algebra skills failing me, but I don't understand how you get to

$$A_{2} - A_{1} = \frac{v \cdot A_{1}}{c - v}$$

I got

$$A_{2} - A_{1} = \frac{v \cdot A_{1} + v \cdot \left(\frac {v \cdot A_{1}}{c} \right)}{c}$$

where v * A₁ is the distance that B has traveled when the light sets out from clock A, and v * A₁ / c is the time it takes the light to travel this distance, and v * v * A₁ / c is the distance B travels during that time.

Subtracting A₁ from each side,

$$A_{2} = A_{1} \left(1 + \frac{v}{c} + \frac{v^{2}}{c^{2}} \right)$$

So the time shown by clock B is

$$B_{2} = A_{1} \left(1 + \frac{v}{c} + \frac{v^{2}}{c^{2}} \right) \sqrt[]{1 - \frac{v^{2}}{c^{2}}}$$

And

$$\lim_{v \rightarrow c} = A_{1} \left(1 + \frac{v}{c} + \frac{v^{2}}{c^{2}} \right) \sqrt[]{1 - \frac{v^{2}}{c^{2}}} = 0$$

which agrees with what we'd see on a Minkowski spacetime diagram if we took the vertical and horizontal axes as those of A's rest frame and let the axes of B's rest frame approach the light cone from either side as B's speed in A's rest frame approaches c. Whereas

$$\lim_{v \rightarrow 0} = A_{1} \left(1 + \frac{v}{c} + \frac{v^{2}}{c^{2}} \right) \sqrt[]{1 - \frac{v^{2}}{c^{2}}} = A_{1}$$

which makes sense, since when there's no velocity, B doesn't move away from A, so the light takes no time to reach it, and there will be no time dilation:

$$A_{1} = A_{2} = B_{2} = A_{3}$$

And of course, when B has no velocity, emission, reflection and reception all take place at the same time and in the same place.

 

[JesseM]

Maybe it's just my algebra skills failing me, but I don't understand how you get to

$$A_{2} - A_{1} = \frac{v \cdot A_{1}}{c - v}$$

I got

$$A_{2} - A_{1} = \frac{v \cdot A_{1} + v \cdot \left(\frac {v \cdot A_{1}}{c} \right)}{c}$$

where v * A₁ is the distance that B has traveled when the light sets out from clock A, and v * A₁ / c is the time it takes the light to travel this distance, and v * v * A₁ / c is the distance B travels during that time.

It's correct that v*A1 is the distance from A to B at the moment the light sets out from clock A, but why were you calculating the time it takes the light to travel that distance? You want to figure out the time it takes the light to catch up to B, and B is still moving outward as the light is approaching it, so the light will have to travel a greater distance than that before it catches up with B. Since B is moving away at velocity v while the light is moving towards it at velocity c, the "closing speed"--the rate that the distance between them is shrinking--must be c - v, and since the initial distance was v*A1, the time for the distance between them to shrink to 0 must be v*A1/(c - v). You could also calculate this a different way by saying that if we set A1 to have time coordinate t=0 in this frame, and define A's position coordinate as x=0 with B traveling in the +x direction, then the light's position as a function of time will be x(t) = ct, while B's position as a function of time will be x(t) = vt + v*A1, so then we can find the time t when the light catches up with B by setting these to equal to each other, giving ct = vt + v*A1, which works out to t = v*A1/(c - v).

 

[Rasalhague]

Since B is moving away at velocity v while the light is moving towards it at velocity c, the "closing speed"--the rate that the distance between them is shrinking--must be c - v, and since the initial distance was v*A1, the time for the distance between them to shrink to 0 must be v*A1/(c - v). You could also calculate this a different way by saying that if we set A1 to have time coordinate t=0 in this frame, and define A's position coordinate as x=0 with B traveling in the +x direction, then the light's position as a function of time will be x(t) = ct, while B's position as a function of time will be x(t) = vt + v*A1, so then we can find the time t when the light catches up with B by setting these to equal to each other, giving ct = vt + v*A1, which works out to t = v*A1/(c - v).

Okay, I understand your explanations of how to derive

$$t = A_{2} - A_{1} = \frac{v \cdot A_{1}}{c - v}$$

but I'm still unsure of where I went wrong in that first step.

It's correct that v*A1 is the distance from A to B at the moment the light sets out from clock A, but why were you calculating the time it takes the light to travel that distance? You want to figure out the time it takes the light to catch up to B, and B is still moving outward as the light is approaching it, so the light will have to travel a greater distance than that before it catches up with B.

That's why I didn't think that v*A₁ alone was the distance the light had to travel. If I had, I would have written simply

$$A_{2} - A_{1} = \frac{v \cdot A_{1}}{c}$$

But I knew this wasn't the case. Rather, I reasoned that this was part of the distance the light had to travel. I added to v*A₁ what I thought was the rest of the distance.

$$A_{2} - A_{1} = \frac{v \cdot A_{1} + v \cdot \left(\frac{v \cdot A_{1}}{c} \right)}{c}$$

The second term

$$\frac{v \cdot \left(\frac{v \cdot A_{1}}{c} \right)}{c}$$

was my attempt at representing the rest of the distance. That's to say, the distance that clock B travels from where it is in A's rest frame when the light is emitted - to where it is when the light is reflected.

I reasoned that, since v * A1 / c is the time it takes the light to travel from A to where B is in A's rest frame when the light is emitted, the distance B travels during this time would be v * v * A1 / c (that is, the speed of B multiplied by the time it took light to reach the place where B was when the light set out). I concluded that the distance the light had to travel from A to B was therefore

$$v \cdot A_{1} + v \cdot \left(\frac{v \cdot A_{1}}{c} \right)$$

and the time it would take the light to travel this distance, I thought would be

$$A_{2} - A_{1} = \frac{v \cdot A_{1} + v \cdot \left(\frac{v \cdot A_{1}}{c} \right)}{c}$$

But apparently not, if this gives a different result from your derivations. Presumably I've made a wrong assumption somewhere about how to calculate one (or more) of those times or distances...

 

[JesseM]

That's why I didn't think that v*A₁ alone was the distance the light had to travel.

I understood that, but I didn't understand why it was necessary to calculate it even as an intermediate step.

But I knew this wasn't the case. Rather, I reasoned that this was part of the distance the light had to travel. I added to v*A₁ what I thought was the rest of the distance.

$$A_{2} - A_{1} = \frac{v \cdot A_{1} + v \cdot \left(\frac{v \cdot A_{1}}{c} \right)}{c}$$

The second term

$$\frac{v \cdot \left(\frac{v \cdot A_{1}}{c} \right)}{c}$$

was my attempt at representing the rest of the distance. That's to say, the distance that clock B travels from where it is in A's rest frame when the light is emitted - to where it is when the light is reflected.

I reasoned that, since v * A1 / c is the time it takes the light to travel from A to where B is in A's rest frame when the light is emitted, the distance B travels during this time would be v * v * A1 / c (that is, the speed of B multiplied by the time it took light to reach the place where B was when the light set out).

Yes, by the time the light travels the distance of v*A1, B has traveled an additional distance of v*v*A1/c. But that still isn't the distance B will be when the light catches up to it, because it will take some extra time for the light to cover that extra distance of v*v*A1/c (specifically, an extra time of v*v*A1/c*c = v^2*A1/c^2), and by that time B will have traveled even farther away (an additional distance of v*v*v*A1/c*c). Then when the light travels that small additional distance (taking a time of v*v*v*A1/c*c*c = v^3*A1/c^3), B will have itself moved away some even smaller additional distance, so it'll take an even smaller additional bit of time for the light to cross that distance--this is basically just a version of Zeno's paradox (the one about Achilles and the tortoise). Zeno's paradox is resolved in calculus because an infinite series with ever-decreasing terms can have a finite sum, so similarly you could calculate the total time until the light reached B by adding an infinite series of ever-decreasing times v*A1/c + v^2*A1/c^2 + v^3*A1/c^3 + v^4*A1/c^4 + ..., using the formula given here--the formula says that for the infinite series a + ar + ar^2 + ar^3 + ..., the sum is a/(1 - r). In this case we have a = v*A1/c and r = v/c, so the sum should be (v*A1/c)/(1 - v/c), and multiplying numerator and denominator by c gives v*A1/(c - v), same as what I found using the simpler argument about closing velocity.

 

[Rasalhague]

Zeno's paradox is resolved in calculus because an infinite series with ever-decreasing terms can have a finite sum, so similarly you could calculate the total time until the light reached B by adding an infinite series of ever-decreasing times v*A1/c + v^2*A1/c^2 + v^3*A1/c^3 + v^4*A1/c^4 + ..., using the formula given here--the formula says that for the infinite series a + ar + ar^2 + ar^3 + ..., the sum is a/(1 - r). In this case we have a = v*A1/c and r = v/c, so the sum should be (v*A1/c)/(1 - v/c), and multiplying numerator and denominator by c gives v*A1/(c - v), same as what I found using the simpler argument about closing velocity.

Aha, of course, good old Zeno! Thanks Jesse, I see it now.

 

[Rasalhague]

I was trying to write some equations summarising what we've found. First, the time interval in A's rest frame between the light leaving A's clock and being reflected off B's mirror:

$$t(A_{2}) - t(A_{1}) = \frac{v \cdot t(A_{1})}{c-v},$$

as you've derived in three different ways now! From this, we get the time on A's clock at the instant which is simultaneous in A's rest frame with the reflection event:

$$t(A_{2}) = t(A_{1}) + \frac{v \cdot t(A_{1})}{c-v}.$$

We multiply this by sqrt(1-(v/c)^2) to find the time shown on B's clock at the instant of reflection:

$$t\left(B \right) = t\left(A_{1} \right) \cdot \sqrt[]{\frac{c+v}{c-v}}.$$

So far so good, I think. Now I reasoned, by symmetry, that the current time shown by A's clock (the time when the image of B's clock together with the reflected image of A's own clock arrive) should be the time shown by A's reflected clock, t(A₁) plus twice the twice the interval t(A₂) - t(A₁), thus:

$$t(A_{3}) = t(A_{1}) + \frac{2v \cdot t(A_{1})}{c-v} = t\left( A_{1}\right) \cdot \left(1 + \frac{2v}{c-v} \right).$$

Twice the interval because it should take the light as long to return from B to A as it takes to get from A to B. Unless I'm mistaken (and somewhere in one of these steps I must be...), solving for t(A₁) gives

$$t\left(A_{1} \right) = t\left(A_{3} \right) \cdot \frac{c-v}{c+v}.$$

But when I substitute for t(A₁) in the expression for the time A sees on B's clock, I get

$$t\left(B \right) = t\left(A_{3} \right) \cdot \sqrt[]{\frac{c-v}{c+v}}.$$

But that can't be right because

$$t\left(A_{3} \right) \cdot \sqrt[]{\frac{c-v}{c+v}} \leq t\left(A_{3} \right) \frac{c-v}{c+v} \leq t\left(A_{3} \right),$$

and I thought we'd established that t(B), the time A sees on B's clock, could never be less than t(A₁), the time A sees on A's own reflected clock.

 

[JesseM]

I was trying to write some equations summarising what we've found. First, the time interval in A's rest frame between the light leaving A's clock and being reflected off B's mirror:

$$t(A_{2}) - t(A_{1}) = \frac{v \cdot t(A_{1})}{c-v},$$

as you've derived in three different ways now! From this, we get the time on A's clock at the instant which is simultaneous in A's rest frame with the reflection event:

$$t(A_{2}) = t(A_{1}) + \frac{v \cdot t(A_{1})}{c-v}.$$

We multiply this by sqrt(1-(v/c)^2) to find the time shown on B's clock at the instant of reflection:

$$t\left(B \right) = t\left(A_{1} \right) \cdot \sqrt[]{\frac{c+v}{c-v}}.$$

So far so good, I think. Now I reasoned, by symmetry, that the current time shown by A's clock (the time when the image of B's clock together with the reflected image of A's own clock arrive) should be the time shown by A's reflected clock, t(A₁) plus twice the twice the interval t(A₂) - t(A₁), thus:

$$t(A_{3}) = t(A_{1}) + \frac{2v \cdot t(A_{1})}{c-v} = t\left( A_{1}\right) \cdot \left(1 + \frac{2v}{c-v} \right).$$

Twice the interval because it should take the light as long to return from B to A as it takes to get from A to B. Unless I'm mistaken (and somewhere in one of these steps I must be...), solving for t(A₁) gives

$$t\left(A_{1} \right) = t\left(A_{3} \right) \cdot \frac{c-v}{c+v}.$$

But when I substitute for t(A₁) in the expression for the time A sees on B's clock, I get

$$t\left(B \right) = t\left(A_{3} \right) \cdot \sqrt[]{\frac{c-v}{c+v}}.$$

Yes, all that looks right.

But that can't be right because

$$t\left(A_{3} \right) \cdot \sqrt[]{\frac{c-v}{c+v}} \leq t\left(A_{3} \right) \frac{c-v}{c+v} \leq t\left(A_{3} \right),$$

That inequality isn't right--(c - v)/(c + v) will be a fraction that's smaller than 1, and the square root of a fraction smaller than 1 is greater than the original fraction, not smaller (for example, the square root of 0.25 is 0.5). So, the inequality should be:

$$t\left(A_{3} \right) \geq t\left(A_{3} \right) \cdot \sqrt[]{\frac{c-v}{c+v}} \geq t\left(A_{3} \right) \frac{c-v}{c+v}$$

 

[Rasalhague]

That inequality isn't right--(c - v)/(c + v) will be a fraction that's smaller than 1, and the square root of a fraction smaller than 1 is greater than the original fraction, not smaller (for example, the square root of 0.25 is 0.5).

Aha, yes, of course! Thanks again. My tired brain was thinking of squaring fractions when it should have been taking square roots. (I guess to be thorough, I'd also have to specify that these are all positive square roots.)

And thanks Gear300 for posing such an interesting question!

 

[Gear300]

Alright...apparently, I hit another wall.

Let us say a space station X is stationary relative to where you are and has a clock that is synchronized with your time. Because ss X is quite a distance (2 light years) from where you are, if you were to observe its time using a telescope, the reading would be some time interval C off. You are in a star cruiser and are preparing for take off. Time T_L is the take off time - at that time, you record the time in ss X as T_L + C. The space cruiser will travel at 0.6c and will not decelerate at ss X but instead pass through it (somewhat like through a tunnel) with a constant velocity. The tunnel will have a clock displayed for everyone to read while passing through (its a pretty big tunnel, resulting in a pretty big clock). In the frame of the space station, its inhabitants have been looking at the clock for quite a while; when you're passing through, you read the clock. Using the same argument as before, wouldn't you read a different value on the clock compared to what the inhabitants (who you are also zooming past) read?

 

[hamster143]

Let us say a space station X is stationary relative to where you are and has a clock that is synchronized with your time. Because ss X is quite a distance (2 light years) from where you are, if you were to observe its time using a telescope, the reading would be some time interval C off.

Precisely 2 years off, because it takes 2 years for the light from space station's clock to reach your eye. But, since your clocks are synchronized and you're motionless wrt the space station, you just add 2 years to correct for light propagation and conclude that time on the station = time shown by your wristwatch.

Time TL is the take off time - at that time, you record the time in ss X as TL + C.

At the moment of takeoff in your reference frame, the clock on the space station reads TL, the clock as you see it reads TL - 2 years.

The space cruiser will travel at 0.6c and will not decelerate at ss X but instead pass through it (somewhat like through a tunnel) with a constant velocity.

At the moment you pass by the station, its clock will read TL + 2 / 0.6 = TL + 3.33 years. Your own clock will read TL + (2 / 0.6) * sqrt(1-0.6*0.6) = TL + 2.67 years.

 

[Gear300]

Precisely 2 years off, because it takes 2 years for the light from space station's clock to reach your eye. But, since your clocks are synchronized and you're motionless wrt the space station, you just add 2 years to correct for light propagation and conclude that time on the station = time shown by your wristwatch.

At the moment of takeoff in your reference frame, the clock on the space station reads TL, the clock as you see it reads TL - 2 years.
At the moment you pass by the station, its clock will read TL + 2 / 0.6 = TL + 3.33 years. Your own clock will read TL + (2 / 0.6) * sqrt(1-0.6*0.6) = TL + 2.67 years.

I edited the post a bit...but altogether, the time you read on the space station's clock would appear the same compared to what the inhabitants read, right?...even if the space station clock was going slower than your clock in your frame?

 

[hamster143]

If you're at the same point as the inhabitants of the station, you'll see exactly the same reading as them.

 

[Gear300]

If you're at the same point as the inhabitants of the station, you'll see exactly the same reading as them.

I see...the point of my confusion is that while moving with a relative velocity of 0.6c, you would observe slower time intervals for the space station clock compared to your clock's time intervals...but when you read the time while in the space station, it turns out to have a larger reading than yours. Would this be because the star cruiser had to accelerate before it hit 0.6c?

 

[hamster143]

I see...the point of my confusion is that while moving with a relative velocity of 0.6c, you would observe slower time intervals for the space station clock compared to your clock's time intervals...but when you read the time while in the space station, it turns out to have a larger reading than yours. Would this be because the star cruiser had to accelerate before it hit 0.6c?

There's a difference between what you SEE and what the time on the station IS in your reference frame.

When your cruiser takes off, you SEE the reading equal to TL - 2 years. (Agreed?) When you arrive at the station, you see the reading of TL + 3.33, even though only 2.67 years have passed by your clock. So if you were to look at the station through the telescope during the trip, you'd SEE their clock ticking FASTER than yours ... but that would be because you're moving toward them - basic Doppler effect. Even though their clock is 'really' ticking slower than yours.

 

[JesseM]

Alright...apparently, I hit another wall.

Let us say a space station X is stationary relative to where you are and has a clock that is synchronized with your time. Because ss X is quite a distance (2 light years) from where you are, if you were to observe its time using a telescope, the reading would be some time interval C off.

These problems are easier to follow if you use specific numbers...if the space station X is 2 light years away from you in your frame, and its clock is synchronized with yours in this frame, then obviously when you look through the telescope it will be 2 years behind!

You are in a star cruiser and are preparing for take off. Time T_L is the take off time - at that time, you record the time in ss X as T_L + C.

It would be T_L - C, not + C...when you look through a telescope you don't see things as they will be in the future, but rather how things were in the past, since the light from past events is just reaching you now. So, just to have some specific numbers to work with, say that T_L = 10 years...in this case, at that moment you see the synchronized clock on the station reading 8 years in your telescope. But keep in mind that there's a difference between what you see visually and what's actually happening in your frame...in your frame immediately before takeoff, when you are still at rest relative to the station, both your clock and the station's clock "actually" read 10 years since they are synchronized in this frame, in spite of the fact that you only see the station clock reading 8 years. On the other hand, once you are in motion at 0.6c relative to the station you have a different rest frame, and because of the relativity of simultaneity it's no longer true in this second frame that the station clock was synchronized with your clock the moment after takeoff...specifically, the relativity of simultaneity says that if two clocks are synchronized and a distance D apart in their own rest frame, then in a frame where they are moving at speed v, they will be out-of-sync by vD/c^2. So from the perspective of a frame moving at 0.6c relative to the station, at the moment before you took off your clock and the station clock were out-of-sync by (0.6c)(2 light years)/c^2 = 1.2 years, so if your clock read 10 years at that moment the station clock read 11.2 years at the same moment (though this frame would still agree with the prediction that you'd see the station clock reading 8 years at the moment your clock read 10 years). Assuming the acceleration was brief, the moment after takeoff (when you come to rest in this frame) your clock still reads 10 years and the station clock still reads 11.2 years in this frame.

Also, note that in this frame the distance between the takeoff point and the station is less than 2 light years due to Lorentz contraction, instead it's only $$2*\sqrt{1 - 0.6^2}$$ = 2*0.8 = 1.6 light years away. And once your ship accelerates and comes to rest in this frame, the station is moving towards you at 0.6c, so it'll take 1.6 light years/0.6c = 2.666... years to reach you. Since you were at rest in this frame, your clock ticked forward by 2.666... years in this time, so your clock reads 10 + 2.666... = 12.666... years when you pass the station. On the other hand, the station's clock was slowed down by a factor of 0.8 so it only ticked forward by 0.8*2.666... = 2.1333... years during this time, so when you pass the station the station's clock reads 11.2 + 2.1333... = 13.333... years.

As it turns out, this is exactly the same thing you'd predict if you instead analyzed the situation from the perspective of the station's frame. In the station frame, at the moment of takeoff both your clock and the station's clock read 10 years, and the distance is 2 light years, so your ship will take 2/0.6 = 3.333... years to reach the station. The station's clock is ticking at normal speed in this frame since it's at rest, so the station's clock will read 10 + 3.333... = 13.333... years when you pass it. On the other hand, your clock was slowed down by a factor of 0.8 in this frame, so your clock only ticks forward by 0.8 * 3.333... = 2.666... years during the journey, so your clock will read 10 + 2.666... = 12.666... years when you pass the station. Thus you can see that both frames predict that your clock reads 12.666... years and the station's clock reads 13.333... years when you pass the station, despite the fact that they disagree about whose clock was ticking slowly during the journey. In general, it is always true in relativity that different frames will agree on local events like what two clocks read at the moment they pass one another; if they didn't, you could get genuine physical contradictions (imagine one clock was connected to a bomb that would explode when the clock reached a certain time--if different frames could disagree on what time that clock read when it passed a second clock, they could disagree on whether the bomb would go off when it was right next to the second clock and destroy it, or whether it would go off when it was far away from the second clock, leaving the second clock unharmed!)