I Reconciling two observers in two frames

[Freixas]

I am not a physicist—not even close—just a guy who, for some crazy reason, decided to try to understand some of the basics of relativity. I’d like to understand them well enough to be able to explain them (correctly) to another lay person. I’m trying to see how much I could explain without relying on complex math formulas or those lovely diagrams with tilted space axes.

There are a gazillion tutorials and references about relativity on the web, but sometimes I have a question that all these seem to avoid or gloss over.

Let me introduce Bob and Alice, who are in different inertial frames. Bob sees Alice as moving away from him at ½ c. Alice would say the same about Bob. Both Bob and Alice have a light speed measurement device: a track, 1 light second (LS) long, with synchronized clocks on both ends that record when a light pulse passes by. The tracks are aligned along the axis of motion.

Based on the theory of relativity (and a lot of experiments), we know that both devices will time the pulse at exactly 1 LS/sec. This remains true even if Bob originates the pulse and has it travel through his track before reaching Alice and her track.

That’s how each sees their own track measurement. But how does Bob see Alice’s measurement?

My premise, and what I see implied in various articles, is that time dilation and length contraction allow each observer to see their own measurement at 1 LS/sec, while they also see the other’s measurement at 1 LS/sec, even though classical physics would say that the “moving” object should measure light at ½ LS/sec (since the object is moving at ½ c).

I thought I would test my understand by checking the numbers. I went to http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/tdil.html to use their handy calculators for time dilation and length contraction. For Bob, Alice’s track is only 0.8660254037844386 LS long. Alice is moving at ½ c, so from Bob’s point of view, a light pulse traveling down Alice’s track will take twice the time it would if the track weren’t moving.

Bob will see that the pulse travels 2*0.866… in 2*0.866… seconds, of course. Bob understands that Alice may view herself as the one who is not moving and that she perceives her track as 1 LS long. So, to my naïve way of thinking about it, time dilation will convert Bob’s 2 * 0.866… seconds into Alice’s 1 second and everyone will be happy—no matter how you measure it, light travels at c.

The time dilation is the inverse of the length contraction, but we are going to convert from Bob’s frame to Alices’, so diving by the inverse is the same as multiplying by the original number. Even without doing the math, I can sense trouble. The equation looks like 2 * 0.866… * 0.866… or 3 * 0.866, which is clearly not even close to 1 second.

What factor did I fail to take into account or what concept did I get wrong?

 

[Pencilvester]

The equation looks like 2 * 0.866… * 0.866… or 3 * 0.866, which is clearly not even close to 1 second.

What factor did I fail to take into account or what concept did I get wrong?

Right off the bat: $2 \cdot 0.866 \cdot 0.866 \neq 3 \cdot 0.866$

 

[PeterDonis]

Let me introduce Bob and Alice, who are in different inertial frames.

You mean at rest in different inertial frames, I take it. An object cannot be "in" only one frame; it's "in" all frames. It just can't be at rest in more than one frame.

My premise, and what I see implied in various articles, is that time dilation and length contraction allow each observer to see their own measurement at 1 LS/sec, while they also see the other’s measurement at 1 LS/sec

If you mean the output of each measuring device, as in, say, the number that comes up on its digital readout, yes, Bob and Alice will agree on what number appears on the digital readout of both of their devices. And for any given light pulse, both of their devices will read "1" (since you've chosen light-seconds per second as your unit of speed, which is a good choice). Ok so far.

What factor did I fail to take into account or what concept did I get wrong?

You left out relativity of simultaneity. Bob's and Alice's clocks are not synchronized the same way. If you are going to use time dilation and length contraction to do your calculations, you also have to take relativity of simultaneity into account; otherwise you will get wrong answers.

A simpler way to get the right answer is to not even think about time dilation and length contraction, but only about coordinates and Lorentz transformations. Assign coordinates to all of the relevant events in Bob's frame, then use the Lorentz transformation to obtain the coordinates of those same events in Alice's frame. Once you have the coordinates in a given frame, you can compute speeds in that frame using those coordinates.

Based on your scenario, here are the key events that I think you will need to assign coordinates to:

Origin: This is the event that has coordinates $(x, t) = (0, 0)$ in both frames. Usually it is taken to be the event at which the two observers (Bob and Alice in this case) pass each other. The implication is that at the instant they pass each other, they both set their clocks to read the same time, which is called time $0$.

B1: The event at which the light pulse passes the left edge of Bob's track.

B2: The event at which the light pulse passes the right edge of Bob's track.

A1: The event at which the light pulse passes the left edge of Alice's track.

A2: The event at which the light pulse passes the right edge of Alice's track.

The speed of light in Bob's frame will then be the rest length of Bob's track, divided by the time elapsed, in his frame, from B1 to B2. Similarly, the speed of light in Alice's frame will be the rest length of Alice's track, divided by the time elapsed, in her frame, from A1 to A2. Those are the two speeds that should come out to 1 (and which correspond to the speeds shown on the digital readouts of the two measuring devices).

Note that it is *not* going to be the case that the moving length of Alice's track in Bob's frame, divided by the elapsed time in Bob's frame from A1 to A2, will give 1. Nor will the corresponding "moving" calculation give 1 for Bob's track in Alice's frame.

Once you have the above coordinates and speeds calculated, it should be easier to see how time dilation, length contraction, and relativity of simultaneity come into play.

 

[Dale]

What factor did I fail to take into account or what concept did I get wrong?

Almost always it is the relativity of simultaneity that is forgotten.

 

[Freixas]

Right off the bat: $2 \cdot 0.866 \cdot 0.866 \neq 3 \cdot 0.866$

Duhhh! I used to do this in my math classes all the time. I'd get all the hard parts right, then take 2 + 2 and get 22. :-)

 

[PeterDonis]

Right off the bat: 2⋅0.866⋅0.866≠3⋅0.866

This is a valid point, but fixing it will not fix everything else in the OP.

 

[Freixas]

Bear with me for a short while longer.

Alice measures a light pulse through her track. It takes 1 second.

Bob measures a light pulse through Alice's track. It may not be the same light pulse at all. He observes that light travels 2*0.8660254037844386 light seconds in 2*0.8660254037844386 seconds. Great! c is c everywhere. Now, Bob wants to convert his time measurement into what he perceives Alice experiencing (he views her clock as running slower). So he divides his time by the time dilation factor of 1.1547005383792515 and gets 1.5 seconds. I was really hoping Bob would get 1 second. Bob thinks Alice sees light traveling slower than c.

Let's work it out by events as it may more clearly reveal my problem. Alice and Bob coordinate their clocks as they pass each other. The time reads 0. A pulse of light enters Alice's track at time 0. This is PeterDonis's A1 event.

We do not care about Bob's track, we only care about Alice's, as observed by both Bob and Alice. Events B1 and B2 are irrelevant.

At time 2*0.866..., Bob see's the pulse exit Alice's track, event A2. Alice sees the same event at time 1. I would be tempted to say that the time dilation is 1/(2*0.866...) which is 1.732050807568877. But the time dilation formula says 1.1547005383792515. Using the correct time dilation, Alice's clock reads 1.5 seconds when Bob sees the light pulse emerge from her track.

Issues related to simultaneity may provide the answer, but I need the dots connected a little more directly than just making that statement. I've gone through a ton of articles and tutorials, but most of them do what PeterDonis was trying to do: connect A events with B events. I even found one article that said basically that my calculations should lead to the result I expect, but didn't actually bother doing the calculations.

Maybe I don't know the point of time dilation. I thought it was to map the clock in one frame of reference to another. If Bob maps the pulse time to 1.5 seconds, but for Alice, the pulse exited half a second earlier, then what's the point of the mapping?

Let me try another way: Alice sends a video of the light pulse measurement to Bob. Bob receives the video and corrects it for the increasing light delay. He should see the clock start at 0 and stop at 1 and, if the time dilation factor is correct, he should measure this as taking 1.154... seconds. Being a physicist, he doesn't need to actually look at Alice's track--he just calculates the length compression, factors in the speed at which the track is moving and comes up with 1.7320... The numbers don't match.

Once again, since I'm prone to adding 2+2 to get 22, don't assume any of my calculations are actually correct. The time dilation factors come directly from the website I referenced (using cut/paste), but I wouldn't just accept those, either.

Thanks for your patience.

 

[PeroK]

What factor did I fail to take into account or what concept did I get wrong?

Suppose I tried to solve this problem using only time dilation (because I hadn't learned about length contraction). The numbers wouldn't work out. I'd say: but I'm taking time dilation into account, what's going wrong?

You would say to me: actually, there's this thing called length contraction that is important also. So, we try the problem again using our knowldege of time dilation and length contraction. But, still, the numbers don't work out.

So, we go and ask @PeterDonis and he says: there's also this thing called "relativity of simultaneity" and in fact, all three of these things: time dilation, length contraction and simultaneity are all neatly packaged up in the Lorentz Transformation.

We then learn about the Lorentz Transformation, try the problem again and - hey presto - the numbers work out!

 

[PeterDonis]

Bob wants to convert his time measurement into what he perceives Alice experiencing (he views her clock as running slower). So he divides his time by the time dilation factor of 1.1547005383792515 and gets 1.5 seconds.

Nope. Again, you are ignoring relativity of simultaneity.

I strongly suggest that you take the suggestion I gave in my previous post, and work this problem using coordinates and Lorentz transformations first, and then, once you've got the right answer that way, go back and figure out how time dilation and length contraction fit in.

To give you a hint, consider the events A1 and A2 that I described in my previous post, that correspond to the light signal passing the left and right ends of Alice's track (I'm assuming the light is traveling left to right). The time elapsed in Bob's frame between those two events is some time $T$. (I know you've calculated the numerical value of $T$, but that's actually a red herring; you can work the problem and get the speed of light in both frames without ever actually calculating this value!) So if we assume that event A1 is also the origin of coordinates (which implies that Bob and Alice pass each other right at that same point in space at that same instant), then we can assign coordinates $(0, 0)$ to A1 and coordinates $(T, T)$ to A2, since we know the light pulse travels at speed $1$ in Bob's frame.

Now, consider another event which I'll call A3. This event is the event on Alice's worldline that happens at the same time as event A2. So the time it takes for the light pulse to traverse Alice's track will be the time between events A1 [edit: corrected from A2] and A3. But of course I left out a key qualifier: the time in which frame? We are looking for the time in Alice's frame, right? But that means we also have to use the definition of simultaneity for Alice's frame! In other words, which event on Alice's worldline happens "at the same time" as A2 depends on which frame you choose; to properly pick out event A3, we need to use the definition of "at the same time" in Alice's frame. And that is different from the definition of "at the same time" in Bob's frame, which is the simultaneity convention you have (without realizing it) implicitly been using in your calculations based on time dilation.

The great advantage of the coordinates/Lorentz transformation method is that you don't have to keep track of any of this; everything automatically falls into place because of the way the Lorentz transformation works. That is why I am again strongly suggesting that you try working the problem using that method first, before trying to understand how time dilation fits into the picture.

 

[Freixas]

Ok, Perok's reply is just snarky.

PeterDonis, I appreciate your time. I went to the page I referenced and found a link to the Lorentz transformation (http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/ltrans.html). As I was setting things up, I ran the square root of .5 through my calculator. Hmmm... The length contraction formula on http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/tdil.html says that the contraction for an object moving at .5c should be the square root of .5 or .707106... But the calculator, and the value I was using, says .866...

If I use sqrt(.5)*2 light seconds for the length of Alice's track as Bob observes it (both contracted and moving at .5c), then that is also the seconds it takes for a pulse to pass through. Using the time dilation formula, to go from Bob's time to Alice's, I need to divide by 1/sqrt(.5) which just happens to equal 2*sqrt(.5) and I get 1 second.

So now it all seems to work exactly as I expected, but perhaps not how you expected. The web page calculators are the source of the error--I just figured it was me!

Just to check, the formula I used for length contraction is L = L₀ * sqrt(1 - v²/c²) = 1 light second * sqrt(1 - .5c²/c²) = sqrt(.5) light seconds. The time dilation is the inverse.

I keep thinking it can't be this easy.

 

[Freixas]

I got a little nervous, so I tried a different speed, .9c. The length of Alice's track becomes .3162277... light seconds. Because she's moving at .9c, it takes the burst 10*.316227 seconds to traverse the track. The inverse of .3162277 just happens to be 3.162277..., so again, we get 1 second.

 

[PeterDonis]

The length contraction formula on http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/tdil.html says that the contraction for an object moving at .5c should be the square root of .5 or .707106... But the calculator, and the value I was using, says .866...

$$
\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{1}{\sqrt{1 - 0.5^2}} = \frac{1}{\sqrt{1 - 0.25}} = \frac{1}{\sqrt{0.75}} = \frac{1}{0.866}
$$

So your first answer for the length contraction factor was right.

The web page calculators are the source of the error

Unfortunately not. See above.

 

[Freixas]

Man, you're fast. OK, back to the drawing board...

 

[pervect]

Man, you're fast. OK, back to the drawing board...

I haven't seen you acknowledge the various posters pointing out the issues with the relativity of simultaneity yet. As about 3-4 posters have mentioned, already, and I believe it's what you're missing.

I assume it's not a familiar term to you at all. But I don't see how we can give you a better answer if the answer to your question is something you're not familiar with other than to tell you the name of what it is. It seems premature to go into a longer explanation, so I'll just repeat the point that the "relativity of simultaneity", which I'm assuming you're not familiar with, is the answer to the discrepancy. If we get to the point where you want to know more about what that is, we can try to talk more, but until we've gotten to the point where you're looking at the issue of "Relativity of simultneity" seriously, I dont't think it makes much sense to write a lot about it.

 

[Freixas]

I haven't seen you acknowledge the various posters pointing out the issues with the relativity of simultaneity yet. As about 3-4 posters have mentioned, already, and I believe it's what you're missing.

I assume it's not a familiar term to you at all. But I don't see how we can give you a better answer if the answer to your question is something you're not familiar with other than to tell you the name of what it is. It seems premature to go into a longer explanation, so I'll just repeat the point that the "relativity of simultaneity", which I'm assuming you're not familiar with, is the answer to the discrepancy. If we get to the point where you want to know more about what that is, we can try to talk more, but until we've gotten to the point where you're looking at the issue of "Relativity of simultaneity" seriously, I don't think it makes much sense to write a lot about it.

Actually, I have seen and read quite a few articles on simultaneity. That doesn't mean that I understand them well or how to apply the principle here. Most of these deal with two people in different frames of reference arguing about the order of in which some events occurred. In this case, Bob observes things in his frame of reference and then does a calculation about what happens in Alice's frame. It may be equivalent to some other simultaneity thought experiment, but I have yet to work the steps out for myself.

I'll be honest: before I posted the question, I figured that simultaneity would be involved in the answer. Just knowing that doesn't help me actually get an answer.

I suspect the people on PhysicsForums get a lot of similar beginner confusion and don't want to work through all the steps yet one more time. I don't blame you guys at all—this is my problem to solve and any help is a bonus. PeterDonis's post which included event A3 includes some tantalizing bits and I probably need to think more about it. One question: I have no idea why "the time it takes for the light pulse to traverse Alice's track will be the time between events A2 and A3" instead of A1 and A3. Was that a typo?

Be patient. It may take me a while. Thanks to all who've made an effort to participate. If anyone still feels like helping, the most useful answer would be along the lines of "here's how the relativity of simultaneity addresses the issue", preferably broken into small steps. :-)

 

[PeterDonis]

I have no idea why "the time it takes for the light pulse to traverse Alice's track will be the time between events A2 and A3" instead of A1 and A3. Was that a typo?

Hm, you're right, the events on Alice's worldline are A1 and A3, so the time is the time between those two events in Alice's frame (which is just the elapsed time on Alice's clock between those two events). I've edited the previous post to correct the typo.

 

[jartsa]

@Freixas May I suggest that on a spaceship Alice throws a clock from the rear towards the front, and while the clock is still on the air Alice shoots it with a laser gun, which causes the clock to stop ticking and to deflect onto the nearby wall, on witch a mark is made by the collision.

Then Alice calculates what the speed of the laser beam was.

And Bob in another frame calculates that the distance to the mark is some amount shorter than what Alice says, and the ticking rate of the clock was some amount slower than what Alice says ... and so on and so forth.

(Alice knows the speed of the clock without measuring, because she has practiced throwing a clock at certain speed)

 

[Freixas]

I was reviewing some simultaneity examples. My favorite is the car driven at high speed into a garage (see https://www.physicsforums.com/threads/car-garage-paradox.283377/). There are light beams traveling towards the end of trains, distant explosions, etc. These all seem to have one observer who thinks two events occur simultaneously and one who thinks otherwise.

My case seems different to me. Everything I care about if from Bob's point of view. There are no arguments about whose clock is slower or whose lengths are contracted. Alice is passive. She is observed but does not get to observe. Her thoughts and opinions are not solicited. She doesn't get to decide when an event occurs--Bob will look at her track and her clock and make up his own mind. I don't think any simultaneity issues come into play.

Consider:

Bob sets up a powerful telescope to record the light pulse entering and exiting Alice's track. Alice also places a big clock right by the track exit and the telescope can resolve the numbers on it. Bob corrects the video for light delay, so he can play it back in "real time".

We have now set up a continuous mapping of Bob's world to Alice's world as seen by Bob. Again, we do not care at all about what Alice sees during any of this. A transformation is an equation that maps something from one reference to another. Bob can use the video to calculate how much slower he perceives her clock running. The time dilation calculation that Bob gets if my numbers are correct are different from what he gets if he uses the Lorentz time dilation transformation.

As we have seen, the likelihood that the Lorentz transformation is wrong is 0. The odds that I have miscalculated Bob's timing of the traversal is high. Honestly, I think that's where the problem lies and not in any simultaneity issues.

I send a pulse of light down a track moving at speed v in the same direction as the light pulse. The track's length is L light seconds. When do you think the pulse will emerge? The velocity and track length are relative to some "fixed" observer, of course, and the answer is what that observer would time.

 

[PeroK]

I don't think any simultaneity issues come into play.

To Bob, Alice's clocks are out of sync. Assume Alice has two clocks, one at either end of the track, and has synchronised them in her frame, so that she can measure the time the pulse is sent from one end and received at the other.

To make up some numbers:

In Bob's frame, when the clock at one end of the track reads $0s$, the clock at the other end reads, say, $-1s$. Let's assume the pulse starts at the clock reading $0s$ and takes $5s$ in Bob's frame to reach the other end. Alice's clocks are time dilated in Bob's frames and, say, only advance $4s$ during this experiment.

When the pulse reaches the second clock, Alice's clocks read $4s$ and $3s$. Therefore, in this case, Alice records
only $3s$ for the journey. Not the $4s$ that you are insisting on.

One of the standard texts on SR (by Morin) actually starts with the relativity of simultaneity. He derives this before he derives time dilation or length contraction. It seems to me that this is a good idea, as it hopefully prevents students from persisting with the "relativity is time dilation and length contraction and I don't think I need to bother about simultaneity issues".

That said, as others have noted, you do appear to be particularly stubborn in this matter.

 

[Freixas]

That said, as others have noted, you do appear to be particularly stubborn in this matter.

Stubborn's my middle name. :-)

That said, this is the first reply that actually provides me with some insight on where simultaneity raises its evil head. Yes, I've been assuming one master clock for both of ends of this rather long track. I can sync Bob's clock with the time at the start of the track or with the time at the end of the track. Each clock runs slow to Bob and by the same amount, but if Bob looks at both, they don't appear to agree with each other. Alice would disagree with Bob.

Got it. Thanks! Now to run the numbers...

 

[PeterDonis]

I don't think any simultaneity issues come into play.

You are wrong.

We have now set up a continuous mapping of Bob's world to Alice's world as seen by Bob.

You have set up a way for Bob to see directly the readings on Alice's clocks. But in order to correctly predict what those readings will be (what readings Bob will see through his telescope), you must know how Alice's clocks are synchronized, because there are two of them at different spatial locations. That is where simultaneity comes in: Alice's clocks are synchronized according to Alice's notion of simultaneity, not Bob's.

I send a pulse of light down a track moving at speed v in the same direction as the light pulse. The track's length is L light seconds. When do you think the pulse will emerge?

When according to whose clock? According to whose simultaneity convention?

Please, please, please stop speculating and do what I suggested before: work the problem using coordinates and the Lorentz transformation. If you had done that when I first suggested it, you would already have solved your problem. It's not that hard and I don't understand why you keep resisting this obvious solution.

 

[SlowThinker]

The systematic approach is to use Lorentz transformation. If you take it for granted, good. It can be enlightening to derive it yourself, but you can do it another time.
I've tried to do that, but can't say I'm happy with the results.

Lets assign coordinates to the events B1, B2, A1, A2 in Bob's frame. Use shortcut $\gamma$=0.866
B1: x=0, t=0
B0: x=1, t=0.5 (end of Bob's track when Alice sees the pulse to start)
BE: x=1, t=0 (end of Bob's track when Bob sees the pulse start)
B2: x=1, t=1
A1: x=X, t=T where X is the distance between the ships, of little relevance, and T=X
A2: x=X+2/$\gamma$, t=T+2/$\gamma$ (I found these backwards to get 1 second in Alice's frame, 2/$\gamma$=8$\gamma$/3=2.309)

Now let's plug these into Lorentz transformation with v=0.5c: x:=$\gamma$(x-vt), t:=$\gamma$(t-vx) to get to Alice's frame.
B1: x=$\gamma$(0-0.5*0)=0, t=$\gamma$(0-0.5*0)=0
B0: x=$\gamma$(1-0.5*0.5)=3$\gamma$/4, t=$\gamma$(0.5-0.5*1)=0
BE: x=$\gamma$(1-0.5*0)=$\gamma$, t=$\gamma$(0-0.5*1)=-$\gamma$/2
B2: x=$\gamma$(1-0.5*1)=$\gamma$/2, t=$\gamma$(1-0.5*1)=$\gamma$/2
A1: x=$\gamma$(X-0.5*T)=$\gamma$X/2, t=$\gamma$(T-0.5*X)=$\gamma$T/2
A2: x=$\gamma$(X+2/$\gamma$-0.5(T+2/$\gamma$))=$\gamma$(X/2+1/$\gamma$)=$\gamma$X/2+1
t=$\gamma$(T+2/$\gamma$-0.5(X+2/$\gamma$))=$\gamma$(T/2+1/$\gamma$)=$\gamma$T/2+1

I'd appreciate if someone could check it, because between B1 and B0 I'm getting length contraction of 3$\gamma$/4=9/(16$\gamma$)=0.650 as seen by Alice.
Also Bob's clocks tick 1s between BE and B2 while Alice sees it as $\gamma$ seconds, i.e. she sees Bob's clocks tick faster than hers.
I've been looking at these results for some hours now but still can't find a mistake. Using v=-0.5 rather than 0.5 didn't seem to help.

It could make the calculations a bit simpler if we agree to set B2=A1=(0,0) i.e. X=T=0 from now on.

 

[Freixas]

You are wrong.

Beating a dead horse.

You have set up a way for Bob to see directly the readings on Alice's clocks. But in order to correctly predict what those readings will be (what readings Bob will see through his telescope), you must know how Alice's clocks are synchronized, because there are two of them at different spatial locations. That is where simultaneity comes in: Alice's clocks are synchronized according to Alice's notion of simultaneity, not Bob's.

For whatever reason, PeroK's explanation makes the error a bit more obvious, although I see you are both saying the same thing.

Please, please, please stop speculating and do what I suggested before: work the problem using coordinates and the Lorentz transformation. If you had done that when I first suggested it, you would already have solved your problem. It's not that hard and I don't understand why you keep resisting this obvious solution.

Hey, it sounds like I'm causing you to get stressed. Sorry! None of this is life-and-death, right? Now, as far as working the problem using the Lorentz transformation, you're assuming a few things:

• That I didn't. That's incorrect, I did and I have a spreadsheet to prove it.
• That using the Lorentz transformation would make everything clear to me. It didn't.
• That I know how to properly use the Lorentz transformation to solve the problem. Probably not: garbage in, garbage out.
• That it's not that hard. The Lorentz transformation that I saw uses coordinates (x,y,z,t) to calculate (x',y',z',t'). I probably made some errors in choosing x and t or in interpreting the result.

When I was posting something earlier, I thought I had mentioned my spreadsheet, but I must have edited it out. As far as I can tell, the spreadsheet results just look like nonsense.

 

[Freixas]

Lets assign coordinates to the events B1, B2, A1, A2 in Bob's frame. Use shortcut $\gamma$=0.866
B1: x=0, t=0
B0: x=1, t=0.5 (end of Bob's track when Alice sees the pulse to start)
BE: x=1, t=0 (end of Bob's track when Bob sees the pulse start)
...

I'm not going to check your math, but I will say that when I see "Bob's track" and "Alice sees", then I know whatever problem you are trying to solve is not the one I posed. In my problem, Bob's track lies forlorn and unused and Alice left on a much needed vacation to Tahiti (after launching "her" track, which is completely automated and has no observers or observation devices, not even imaginary thought experiment observers).

 

[pervect]

The systematic approach is to use Lorentz transformation.

Yes, I agree. The Lorentz transform provides a map from Bob's view of the experiment to Alice's view of the experiment, because it provides a map from when (and where) events happen in Bob's frame to when (and where) event's happen in Alcie's frame. Or vica-versa. This is done by a mathematical relationship that takes the time and place (t,x,y,z) that an event happens in one frame and provides the time and place (t', x', y', z') that it happens in another frame.

So it seems like something the OP should be interested in. However, I have noticed that anything involving the phrase "Lorentz transform" seems likely to get ignored when it's mentioned to a non-physicist. Explaining in detail what the Lorentz transform is doesn't seem to help - perhaps the (hopefully non-techincal) overiview of what it does will help, perhaps not.

 

[PeterDonis]

That I didn't. That's incorrect, I did

Then you can post your work here.

The Lorentz transformation that I saw uses coordinates (x,y,z,t) to calculate (x',y',z',t'). I probably made some errors in choosing x and t or in interpreting the result.

If you post your work, we can help you to see where it went wrong (I'm assuming it did since it seems like it didn't give you the answer that both Alice's and Bob's devices will measure $c = 1$). But instead you seem to have posted some different calculation that just uses time dilation, which as you have seen, certainly won't give you the right answer.

 

[PeterDonis]

it sounds like I'm causing you to get stressed

No, it's just somewhat frustrating to see you continuing to pursue a calculation that is already known to be wrong.

It's a bit more frustrating when you say you did do a calculation using the right method, but that's not the calculation you posted in this thread. If you're here for help, then help us help you: show us the calculation you did using the Lorentz transformation, which is the method that everybody here agrees is the right method, so we can help you see where it went wrong.

 

[SlowThinker]

I know whatever problem you are trying to solve is not the one I posed.

Well I did my best to analyze the problem exactly as you described it.
Perhaps it's time to show us your spreadsheet.

 

[Dale]

and Alice left on a much needed vacation to Tahiti (after launching "her" track, which is completely automated and has no observers or observation devices,

Umm, so what is there to reconcile then? I thought the whole point of the thread was that you were having trouble “Reconciling two observers in two frames”. With that gone I am not getting what you want to understand.

 

[Freixas]

Well I did my best to analyze the problem exactly as you described it.
Perhaps it's time to show us your spreadsheet.

Appreciate it! I am trying to add a correction for what people have pointed out. Let's say Bob's and Alice's clocks sync when the entry end of track passes by Bob. The exit end would have passed by Bob some time ago.

Alice thinks the clocks on both ends of her track are perfectly sync'ed (she made sure before she left for Tahiti). Bob doesn't see things that way: the exit clock is off by some amount. The exit end is 1 light second away and the track is moving at .5c. What is the time discrepancy that Bob see's on the exit clock relative to the entry clock?

Anyway, that is what I trying to figure out now.

 

[Freixas]

Umm, so what is there to reconcile then? I thought the whole point of the thread was that you were having trouble “Reconciling two observers in two frames”. With that gone I am not getting what you want to understand.

Two frames, one observer watching what is happening in both frames and then trying to use formulas to confirm what he saw and discovering that his formulas don't give the same answer as what he sees. Said observer then appealing to PhysicsForum to figure out why. PhysicsForum people pointing out the missing factor, observer working on correcting his calculations and thinking maybe he should chuck it all and join Alice in Tahiti.

This was always the problem; any assumption about it involving Alice's observations is incorrect.

 

[Dale]

Two frames, one observer

In SR problem descriptions an observer is just another way to identify a reference frame. Observers don’t do anything other than identify frames, so this distinction doesn’t affect anything. Certainly is isn’t grounds for dismissing anyone’s response above

 

[Freixas]

No, it's just somewhat frustrating to see you continuing to pursue a calculation that is already known to be wrong.

It's a bit more frustrating when you say you did do a calculation using the right method, but that's not the calculation you posted in this thread. If you're here for help, then help us help you: show us the calculation you did using the Lorentz transformation, which is the method that everybody here agrees is the right method, so we can help you see where it went wrong.

I somehow missed this reply. Well, if you really want to look at the spreadsheet and fix it, I'm happy to post it. I tried to label things so they make sense, but you never know. "Relative speed" is a multiplier I used to calculate how long it takes for light to traverse an object X light seconds long when said object is moving at velocity v away from the light.

The spreadsheet is prior any attempts to correct for the entry/exit clocks not being synchronized from Bob's point of view.

 

[Freixas]

In SR problem descriptions an observer is just another way to identify a reference frame. Observers don’t do anything other than identify frames, so this distinction doesn’t affect anything. Certainly is isn’t grounds for dismissing anyone’s response above

Fair enough. Let's say Alice's job is to verify that her entry/exit clocks are sync'ed from her perspective. She can then go off duty. I believe everything else is just Bob's observations. Alice might disagree with his observations if she were around, but she's not and we're only concerned about why Bob can't properly map his calculations of what he should see with what he actually sees. There's no paradox or anything, it's just Bob (me, really) trying to make the numbers work.

 

[SlowThinker]

I believe everything else is just Bob's observations. Alice might disagree with his observations if she were around

If you're not switching between frames, you can't possibly run into problems. Whatever is the distance D between points where the pulse enters and exits Alice's track, it will take D/c seconds for light to travel between those points.
What is D is questionable, naïvely I'd say D=2$\gamma$=1.732, my calculations show it's 2/$\gamma$=2.309 .
In fact I used Lorentz transformation from https://www.mathstools.com/section/main/Lorentz_Transformation#.W1yoeIqLvmo where they define incorrectly $\gamma$ as 1/$\gamma$. I'll try to redo the calculations with correct $\gamma$.

 

[Dale]

Let's say Alice's job is to verify that her entry/exit clocks are sync'ed from her perspective. She can then go off duty.

Sure, but if someone else solves the problem with Alice on duty then it is a valid solution for the Alice off duty problem too. Alice being on or off duty doesn’t change any of the calculations.

we're only concerned about why Bob can't properly map his calculations of what he should see with what he actually sees.

And since his calculations of what he should see involve the frame formerly set up by Alice, there is nothing wrong with saying things like “Alice observes” or “according to Alice”. It is the same thing.

My concern is your dismissal of some valid responses for an invalid reason. I would encourage you to go back and revisit those responses with the understanding that it is the same problem.

 

[Dale]

If you're not switching between frames, you can't possibly run into problems.

He is switching between frames. He is just confused about the traditional role of an “observer” in SR.

 

[SlowThinker]

Correction where $\gamma$ is $\frac{1}{\sqrt{1-v^2/c^2}}$. Now it makes sense.

Lets assign coordinates to the events B1, B2, A1, A2 in Bob's frame. Use shortcut $\gamma$=1.155
B1: x=0, t=0
B0: x=1, t=0.5 (end of Bob's track when Alice sees the pulse to start)
BE: x=1, t=0 (end of Bob's track when Bob sees the pulse start)
B2: x=1, t=1
A1: x=X, t=T where X is the distance between the ships, of little relevance, and T=X
A2: x=X+2/$\gamma$, t=T+2/$\gamma$ (I found these backwards to get 1 second in Alice's frame, 2/$\gamma$=1.732)

Now let's plug these into Lorentz transformation with v=0.5c: x:=$\gamma$(x-vt), t:=$\gamma$(t-vx) to get to Alice's frame.
B1: x=$\gamma$(0-0.5*0)=0, t=$\gamma$(0-0.5*0)=0
B0: x=$\gamma$(1-0.5*0.5)=1/$\gamma$, t=$\gamma$(0.5-0.5*1)=0
BE: x=$\gamma$(1-0.5*0)=$\gamma$, t=$\gamma$(0-0.5*1)=-$\gamma$/2
B2: x=$\gamma$(1-0.5*1)=$\gamma$/2, t=$\gamma$(1-0.5*1)=$\gamma$/2
A1: x=$\gamma$(X-0.5*T)=$\gamma$X/2, t=$\gamma$(T-0.5*X)=$\gamma$T/2
A2: x=$\gamma$(X+2/$\gamma$-0.5(T+2/$\gamma$))=$\gamma$(X/2+1/$\gamma$)=$\gamma$X/2+1
t=$\gamma$(T+2/$\gamma$-0.5(X+2/$\gamma$))=$\gamma$(T/2+1/$\gamma$)=$\gamma$T/2+1

Between B1 and B0 I'm getting length contraction of 3$\gamma$/4=1/$\gamma$=0.866 as seen by Alice.
Also Bob's clocks tick 1s between BE and B2 while Alice sees it as $\gamma$ seconds, so she sees Bob's clocks tick slower than hers.
And obviously, the speed of light over Alice's track (between A1 and A2) is 1 in both frames.

 

[bahamagreen]

I am not a physicist—not even close—just a guy who, for some crazy reason, decided to try to understand some of the basics of relativity. I’d like to understand them well enough to be able to explain them (correctly) to another lay person. I’m trying to see how much I could explain without relying on complex math formulas or those lovely diagrams with tilted space axes.

Am I the only one who views a non-physicist wanting to explain correctly to another non-physicist something developed by some of the smartest physicists in history which actual working physicists devote years of university study to grasp as sounding like a request to assist the blind in leading the blind? I'm amazed at the indulgence through two pages, this thread not being closed shortly after pointing to the LT in post #3.

 

[robphy]

The systematic approach is to use Lorentz transformation.

Yes, I agree. The Lorentz transform provides a map from Bob's view of the experiment to Alice's view of the experiment, because it provides a map from when (and where) events happen in Bob's frame to when (and where) event's happen in Alcie's frame. Or vica-versa. This is done by a mathematical relationship that takes the time and place (t,x,y,z) that an event happens in one frame and provides the time and place (t', x', y', z') that it happens in another frame.

So it seems like something the OP should be interested in. However, I have noticed that anything involving the phrase "Lorentz transform" seems likely to get ignored when it's mentioned to a non-physicist. Explaining in detail what the Lorentz transform is doesn't seem to help - perhaps the (hopefully non-techincal) overiview of what it does will help, perhaps not.

One of my favorite quotes (I bolded the sentence below):

( "Lapses in Relativistic Pedagogy ", AJP 62, 11 (1994) http://dx.doi.org/10.1119/1.17728 )

"I couldn't agree more with Mallinckrodt that the Lorentz transformation doesn't belong in a first exposure to special relativity. Indispensable as it is later on, its very conciseness and power serve to obscure the subtle interconnnectedness of spatial and temporal measurements that makes the whole business work. Only a loonie would start with real orthogonal matrices to explain rotations to somebody who had never heard of them before, but that's how we often teach relativity. You learn from the beginning how to operate machinery that gives you the right answer but you acquire little insight into what you're doing with it."

Since the OP said

I am not a physicist—not even close—just a guy who, for some crazy reason, decided to try to understand some of the basics of relativity. I’d like to understand them well enough to be able to explain them (correctly) to another lay person. I’m trying to see how much I could explain without relying on complex math formulas or those lovely diagrams with tilted space axes.

and (in my opinion) has been relying on complex math formulas...
...maybe it's time to try those "lovely [spacetime] diagrams with tilted space axes".

"A spacetime diagram is worth a thousand words."

My $0.02.

 

[Freixas]

Correction where $\gamma$ is $\frac{1}{\sqrt{1-v^2/c^2}}$. Now it makes sense.

Lets assign coordinates to the events B1, B2, A1, A2 in Bob's frame. Use shortcut $\gamma$=1.155
B1: x=0, t=0
B0: x=1, t=0.5 (end of Bob's track when Alice sees the pulse to start)
BE: x=1, t=0 (end of Bob's track when Bob sees the pulse start)
B2: x=1, t=1
A1: x=X, t=T where X is the distance between the ships, of little relevance, and T=X
A2: x=X+2/$\gamma$, t=T+2/$\gamma$ (I found these backwards to get 1 second in Alice's frame, 2/$\gamma$=1.732)

Now let's plug these into Lorentz transformation with v=0.5c: x:=$\gamma$(x-vt), t:=$\gamma$(t-vx) to get to Alice's frame.
B1: x=$\gamma$(0-0.5*0)=0, t=$\gamma$(0-0.5*0)=0
B0: x=$\gamma$(1-0.5*0.5)=1/$\gamma$, t=$\gamma$(0.5-0.5*1)=0
BE: x=$\gamma$(1-0.5*0)=$\gamma$, t=$\gamma$(0-0.5*1)=-$\gamma$/2
B2: x=$\gamma$(1-0.5*1)=$\gamma$/2, t=$\gamma$(1-0.5*1)=$\gamma$/2
A1: x=$\gamma$(X-0.5*T)=$\gamma$X/2, t=$\gamma$(T-0.5*X)=$\gamma$T/2
A2: x=$\gamma$(X+2/$\gamma$-0.5(T+2/$\gamma$))=$\gamma$(X/2+1/$\gamma$)=$\gamma$X/2+1
t=$\gamma$(T+2/$\gamma$-0.5(X+2/$\gamma$))=$\gamma$(T/2+1/$\gamma$)=$\gamma$T/2+1

Between B1 and B0 I'm getting length contraction of 3$\gamma$/4=1/$\gamma$=0.866 as seen by Alice.
Also Bob's clocks tick 1s between BE and B2 while Alice sees it as $\gamma$ seconds, so she sees Bob's clocks tick slower than hers.
And obviously, the speed of light over Alice's track (between A1 and A2) is 1 in both frames.

Hey, thanks! Can I ask a few clarifying questions?

B0 is a bit confusing for me because I had Alice running the timing track and Bob just observing it. Therefore, Bob says that the end of Alice's track is 1/$\gamma$ at t=0. Bob, however, can see Alice's clocks, both the one at the start of the track and the one at the end. Alice believes both clocks are synchronized. Bob sees Alice's entry clock read 0. What does he read for the exit clock? How did you derive 0.5 seconds?

You have B0: x=$\gamma$(1-0.5*0.5)=1/$\gamma$. No matter how I add parentheses to that formula, I can't make the math work out. (1-(.5*.5)) = (1 - .25) = .75. Or ((1-.5)*.5) = (.5*.5) = .25.

You generalized A1. It might be a little easier to follow if X and T are also 0.

I should probably draw picture of what I'm thinking. Thanks for your time and effort.

 

[robphy]

Programming note:

Numerical calculations are so much easier
if v=(3/5)c (so gamma=5/4, Doppler=2)
or v=(4/5)c (so gamma=5/3, Doppler=3)...
rational numbers! No calculator needed!

When v=(1/2)c, we have gamma=2/sqrt(3) and Doppler=sqrt(3)...
which is mind-numbing to me when expressed as a decimal approximation.
In addition, this numerical complication makes the possibly already-confusing physics a little harder to see.

 

[SlowThinker]

Hey, thanks! Can I ask a few clarifying questions?

B0 is a bit confusing for me because I had Alice running the timing track and Bob just observing it. Therefore, Bob says that the end of Alice's track is 1/$\gamma$ at t=0. Bob, however, can see Alice's clocks, both the one at the start of the track and the one at the end. Alice believes both clocks are synchronized. Bob sees Alice's entry clock read 0. What does he read for the exit clock?

For entry of Alice's track, you want to find point that has $t_B$=0 in Bob's frame, and $x_A$=$\gamma$X/2 in Alice's frame (as found previously).
$\gamma (x_B-0.5 t_B)=x_A=\gamma X/2$
$\gamma (t_B-0.5 x_B)=t_A$
Plug $t_B$=0 into first to get $x_B$, then plug both into second to get $t_A$.
We get $x_B=X/2$, $t_B=0$ in Bob's frame, or as seen by Alice $x_A=\gamma X/2$, $t_A=-\gamma X/4$.
Having seen this, you could probably have guessed the Bob's x, but Lorentz transformation gets you there every time.

For exit of Alice's track, solve for $t_B$=0, $x_A$=$\gamma$X/2+1.
We get $x_B=X/2+1/\gamma$, $t_B=0$ in Bob's frame, or as seen by Alice $x_A=\gamma X/2+1$, $t_A=-\gamma X/4-0.5$.

So Bob sees Alice's track's entry at X/2 when the pulse starts, Alice's clock showing $-\gamma X/4$.
At the same time Bob sees Alice's track's exit $1/\gamma$ farther and the exit clock showing 0.5 seconds less than the entry clock which is the answer to your questions.
Clearly, those numbers have no relevance to the original problem of tracking the light pulse.

How did you derive 0.5 seconds?

Similar to above, solving for $x_A=1$, $t_B=0$. You can do this all day long.

You have B0: x=$\gamma$(1-0.5*0.5)=1/$\gamma$. No matter how I add parentheses to that formula, I can't make the math work out. (1-(.5*.5)) = (1 - .25) = .75. Or ((1-.5)*.5) = (.5*.5) = .25.

It's not straightforward but it uses the fact that $\gamma=\sqrt{4/3}$. So $3/4\sqrt{4/3}=\sqrt{3/4}=1/\sqrt{4/3}$.

I agree that Lorentz transformation itself doesn't offer much insight, but now that you have coordinates of some interesting points, you can start to see how it all works.

 

[PeterDonis]

if you really want to look at the spreadsheet and fix it

No, I want to look at your calculations in the form of equations and numbers giving the coordinates you assigned to events in Bob's frame and the Lorentz transformation from Bob's frame to Alice's frame, posted directly into the thread.

 

[Freixas]

Thanks to everyone who has helped! SlowThinker, I appreciate your taking the time to step through the calculations.

After a lot of noise, I have to say my original calculations were off by .5 seconds. Having the exit clock off by -0.5 (from Bob's point of view), fixes it.

Rather late to the game, I'll add a diagram of the problem. Assume the clocks are synchronized (from Alice's point of view) and run continuously. It is just an accident (or careful planning) that the entry clock reads zero when it passes by Bob and the light burst enters. The exit clock doesn't stop, but Bob notes the reading as the light exits.

Timing and positions from Alice's point of view aren't included (except the clock, of course). We can say that at the Bob's t₀, x'₀ = 0 and t'₀ =0. At exit, x'₁ = 1, t'₁ = 1, assuming I got it all correct (which, I realize) is a big assumption.

@robphy: You may be right about the spacetime diagrams, but (crazy me), I actually wanted to see the numbers work out. And I learned something by trying to calculate the numbers that I might not have picked up on with all those diagrams. Personally, I'd close this thread and consider it done, but I wouldn't mind hearing if there are errors on the diagram.

 

[Freixas]

No, I want to look at your calculations in the form of equations and numbers giving the coordinates you assigned to events in Bob's frame and the Lorentz transformation from Bob's frame to Alice's frame, posted directly into the thread.

Actually, if you open the spreadsheet and use View Formulas, you'll see the equations as equations. Some of the variables have an underscore in front of them to keep Excel happy (e.g. v/x looks like _v/_x).

But SlowThinker has cranked through it all for me and showed me how to fix the 0.5 seconds too much I was getting, so no need to bother. Thanks.

 

[Dale]

Numerical calculations are so much easier
if v=(3/5)c (so gamma=5/4, Doppler=2)

This is my preference. I always use v = 0.6 c. It is easier to draw accurate spacetime diagrams than larger values.

 

[PeterDonis]

Let's say Bob's and Alice's clocks sync when the entry end of track passes by Bob.

Ok, then we can call this the origin of coordinates in both frames, so it has coordinates $(x, t) = (0, 0)$ in both frames.

Alice thinks the clocks on both ends of her track are perfectly sync'ed (she made sure before she left for Tahiti). Bob doesn't see things that way: the exit clock is off by some amount. The exit end is 1 light second away and the track is moving at .5c. What is the time discrepancy that Bob see's on the exit clock relative to the entry clock?

Anyway, that is what I trying to figure out now.

It looks like @SlowThinker has already given a solution that satisfies you, but I'm going to go ahead and work it the way I would work this type of problem. Actually I'll work it two ways.

The first way is to work things first from Alice's frame and then transform back to Bob's frame. This way is simpler, as we'll see, but it does assume that the light's speed is $1$ in Alice's frame, whereas you seemed to want to start with assuming that in Bob's frame. That's what the second way will show.

First way: since the entry end of the track is at $x = 0$ in both frames when Bob passes Alice, and since it's motionless in Alice's frame, the entry end of the track is always at $x' = 0$ in Alice's frame. (I'll use primed coordinates for Alice's frame and unprimed coordinates for Bob's frame.) Since the track is 1 light second long, the exit end of the track is always at $x' = 1$ in Alice's frame. So if the light is moving at speed $1$ in Alice's frame, then the events A1 and A2 (from my previous definitions) have coordinates $(0, 0)$ and $(1, 1)$ in this frame. And since Alice is co-located with the entry end of the track, event A3 (from my previous definitions) will have coordinates $(0, 1)$ in this frame. So obviously the light's elapsed travel time is 1 second in this frame, and it traveled distance 1, for speed 1.

Now, with all these coordinates, we can Lorentz transform to Bob's frame. Since this is an inverse transform, the equations are $x = \gamma \left( x' + v t' \right)$ and $t = \gamma \left( t' + v x' \right)$. So the coordinates of the three events in Bob's frame turn out to be:

A1: $\left( 0, 0 \right)$

A2: $\left( \gamma(1 + v), \gamma(1 + v) \right)$

A3: $\left( \gamma v, \gamma \right)$

Note that events A2 and A3 are not simultaneous in Bob's frame; A3 happens first. Note also that, by construction, the clock at the entry end of the track (which is also the clock co-located with Alice) reads 1 at event A3, while the clock at the exit end of the track reads 1 at event A2.

Finally, note that, in Bob's frame, the light pulse travels $\gamma(1 + v)$ light seconds in $\gamma(1 + v)$ seconds, so it has speed 1.

Second way: event A1 has coordinates $(x, t) = (0, 0)$ in Bob's frame. Since the light pulse travels from A1 to A2, event A2 must have coordinates $(T, T)$ in Bob's frame, where $T$ is a time (or distance, either one will work) that we have to calculate.

We calculate $T$ by figuring out the worldline of the exit end of the track in Bob's frame. The exit end of the track is traveling at speed $v$ in this frame, while the light pulse is traveling at speed $1$. And since the track is length contracted in this frame, the exit end of the track must pass through an event I'll call A4, which has coordinates $(1 / \gamma, 0)$ in Bob's frame (i.e., at time zero in this frame, the time when the entry end is co-located with Bob, the exit end is the length contracted length $1 / \gamma$ further along). So the worldline of the exit end of the track in Bob's frame satisfies the equation

$$
x = \frac{1}{\gamma} + v t
$$

We then simply set $x = t = T$ in the equation above to obtain for event A2:

$$
T = \frac{1}{\gamma(1 - v)} = \gamma(1 + v)
$$

So we have event A2 with coordinates $\left( \gamma(1 + v), \gamma(1 + v) \right)$.

We can now Lorentz transform the coordinates of event A2 to Alice's frame. The transformation equations are $x' = \gamma \left( x - v t \right)$ and $t' = \gamma \left( t - v x \right)$ (note the minus signs this time since this is a "forward" transform from the unprimed to the primed frame--this is the version you usually see in textbooks). This gives:

A2: $(x', t') = \left( \gamma^2(1 + v)(1 - v), \gamma^2(1 + v)(1 - v) \right) = \left( 1, 1 \right)$

So in Alice's frame, the light pulse travels 1 light-second in 1 second, for a speed of 1.

To illustrate the "clock offset", we Lorentz transform the coordinates of event A4 to Alice's frame. This gives:

A4: $(x', t') = \left( 1, - v \right)$

In other words, the clock at the exit end of the track reads time $- v$ at event A4, which is the event at which, in Bob's frame, the exit end is located at time $t = 0$, i.e., at the same time (in Bob's frame) as the pulse passes the entry end of the track. This illustrates the relativity of simultaneity: events A1 and A4 happen at the same time in Bob's frame, but not in Alice's frame; in Alice's frame, event A4 happens first, $v$ seconds before the pulse enters the track.

Since the "clock offset" terminology seems to work better for you, one way to help yourself figure out problems like this is to think "clock offset" whenever you see "relativity of simultaneity". They're two ways of viewing the same thing.

 

[SlowThinker]

I wouldn't mind hearing if there are errors on the diagram.

Alice's track should be 1/$\gamma$ long with the correct definition of $\gamma$ ($\gamma\geq 1$). Bob sees Alice's track shortened.
Also Bob's $t_1$ is correctly 1.732 but that's 2/$\gamma$ not 2$\gamma$.
I'm not sure if it's just a mistake or some deeper misunderstanding.

 

[Freixas]

Alice's track should be 1/$\gamma$ long with the correct definition of $\gamma$ ($\gamma\geq 1$). Bob sees Alice's track shortened.
Also Bob's $t_1$ is correctly 1.732 but that's 2/$\gamma$ not 2$\gamma$.
I'm not sure if it's just a mistake or some deeper misunderstanding.

No, that's me making one of my stupid mistakes. I've been correctly contracting the length and calculating t₁ using numbers and not symbols. As I am new to this, I got it in my head when doing the diagram that .866... was $\gamma$. My spreadsheet correctly uses 1/$\gamma$ for length contraction. Once I got the first one wrong, Bob's t₁ came along for the ride.

If those are the only errors, I'm happy. I hate to leave the wrong diagram up--I'll see if I can fix it and edit the post.

Thanks so much!