Reconciling two observers in two frames

[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!

 

[Freixas]

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.

Thanks, PeterDonis! I appreciate all your time and effort.I apologize for any frustration you may have experienced.

I'll give a couple of examples of things you might find clear and that confused me: In your reply for example, you consider A2 and A3 as "events" even though, if people associate a location with an event, it is the location at which the event occurred. In physics perhaps you define it differently. Event A3 seems completely irrelevant anyway and I'm still scratching my head as to why it's there. Why do you give Bob's values as A1, A2 and A2 instead of B1, B2 and B3 (or: "why is there an 'A' prefix at all?"). This particular reply was detailed enough that I could figure out what you meant, but sometimes people provide answers with embedded assumptions and not enough detail to decode.

Among responses, "Go study the relativity of simultaneity" is a correct, but unproductive answer. "Use the Lorentz transform" misses the entire point since I was trying to cross-check the transform. Using it (properly) gives the correct answer, but without any insight. "You forgot to account for the fact that the entry and exit clocks don't appear synchronized to Bob" provides the insight and completely answers my question. It doesn't even require that many words. Credit goes to PeroK--we should put a gold star next to his reply. SlowThinker's calculations were icing on the cake since I was interested in cross-checking numbers and that was what he did.

I'm sure I phrased my question poorly. Putting up a diagram at the start would have been better. That PeroK was able to decipher my question and address the key point succinctly is amazing.

 

[Freixas]

I don't seem to be able to edit my response with the diagram, so here's the corrected diagram.

 

[PeroK]

In other words, if Alice has a row of synchronised clocks At rest relative to her at regular intervals; and Bob is moving relative to Alice in the direction of the row of clocks; then, in Bob's reference frame the clocks are closer together, but still regularly spaced, out of sync (with leading clocks lagging) and running slow (all at the same rate).

The Lorentz Transformation effectively encapsulates that difference between the two reference frames.

 

[Dale]

In your reply for example, you consider A2 and A3 as "events" even though, if people associate a location with an event, it is the location at which the event occurred. In physics perhaps you define it differently.

In physics an event is a location and a time.

 

[robphy]

To summarize @PeterDonis 's careful analysis of your original problem,
( https://www.physicsforums.com/threa...vers-in-two-frames.952195/page-2#post-6032350 )
here's a spacetime diagram on rotated graph paper
(for the case of $\beta=3/5$, which leads to easier arithmetic... then for the case of $\beta=1/2$).

By the way, $\gamma(1+\beta)=\sqrt{\frac{1+\beta}{1-\beta}}=k$ (the Doppler Factor).
For $\beta=3/5$, we have $k=2$.
For $\beta=1/2$, we have $k=\sqrt{3}\approx 1.73205$.

In my rotated graph paper diagrams,
the "light-clock diamonds" (the "ticks") have the same area
and the "aspect-ratio"="width"/"height"=$k^2$ (which is easier to see for $\beta=3/5$).

 

[SlowThinker]

"Use the Lorentz transform" misses the entire point since I was trying to cross-check the transform. Using it (properly) gives the correct answer, but without any insight.

You can work with various light clock scenarios to get qualitative insights (which clock is ahead or behind, what's shorter and what's longer, and even derive the Lorentz transformation itself), but since you came asking for actual numbers, there's really no way around LT.
It's not my first choice, I prefer to guess what the result is going to be, but LT is a reliable cross-check, and most importantly a quantitative tool.

I recommend that you spend some time looking at what are events A1 and A2, what are their x and t coordinates in Bob's frame and in Alice's frame, and what they mean.

 

[PeterDonis]

In your reply for example, you consider A2 and A3 as "events" even though, if people associate a location with an event, it is the location at which the event occurred.

"Event" in relativity means a point in spacetime--in a particular frame it is a point in space (a "location") at an instant of time. Note that all of my events have both $x$ and $t$ coordinates; that's how coordinates work in relativity, spacetime is 4-dimensional. (Two dimensions are left out here because they don't matter for this particular problem, so we just have one dimension of space and one of time.)

Event A3 seems completely irrelevant anyway and I'm still scratching my head as to why it's there.

You can leave it out, it's not really necessary. I just included it to illustrate how elapsed time on the clock that is actually co-located with Alice works.

Why do you give Bob's values as A1, A2 and A2 instead of B1, B2 and B3

Because the "A" events are the key ones to look at, since you said you wanted to figure out how the events on Alice's track look in Bob's frame. The "B" events were events associated with Bob's track, which you said you weren't really interested in. If you want to look at the "B" events, a similar analysis can be done for those.

"Go study the relativity of simultaneity" is a correct, but unproductive answer.

"You forgot to account for the fact that the entry and exit clocks don't appear synchronized to Bob" provides the insight and completely answers my question.

And, as I noted, the second statement ("clocks not synchronized") is saying the same thing as the first ("relativity of simultaneity"), just in different words. They're not two different concepts; they're just two different ways of stating the same concept. That's why I said that, if "clocks not synchronized" works better for you as a way to describe that concept, you should substitute that wherever you see "relativity of simultaneity". That is important because "relativity of simultaneity" is a much more common way of referring to this concept, so you need to be aware of what that phrase refers to.