Relativity and simple algebra II

Discussion in 'Alternative Theories' started by ralfcis, Feb 6, 2021.

  1. James R Just this guy, you know? Staff Member

    Just for fun, here's how the formula for the spacetime interval is derived. Start with the Lorentz transformations:

    Now consider the quantity $(ct')^2 - x'^2$:

    $(ct')^2 - x'^2 = \gamma^2(ct - \frac{vx}{c})^2 - \gamma^2(x-vt)^2$


    $\gamma^2 = \frac{1}{1-(v/c)^2} = \frac{c^2}{c^2-v^2}$


    $(ct')^2 - x'^2 = \frac{c^2}{c^2-v^2}[(ct - \frac{vx}{c})^2 - (x-vt)^2]$
    $=\frac{c^2}{c^2-v^2}[(ct)^2 - 2vxt + (\frac{vx}{c})^2 - x^2 + 2vxt - (vt)^2]$
    $=\frac{c^2}{c^2-v^2}[(ct)^2 - x^2 + (\frac{vx}{c})^2 - (vt)^2]$
    $=\frac{c^2}{c^2-v^2}[(c^2-v^2) t^2 - x^2 (1 - (\frac{v}{c})^2)]$
    $=\frac{c^2}{(c-v)(c+v)}[(c-v)(c+v) t^2 - x^2 (\frac{(c^2 - v^2)}{c^2})$
    $=\frac{c^2}{(c-v)(c+v)}[(c-v)(c+v) t^2 - x^2 (\frac{(c-v)(c+v)}{c^2})$
    $=\frac{c^2}{(c+v)}[(c+v) t^2 - x^2 (\frac{(c+v)}{c^2})$
    $= (ct)^2 - x^2$.

    We see that the quantity 's', defined by
    $s^2 \equiv (ct')^2 - x'^2 = (ct)^2 - x^2$

    has the same numerical value in both frames of reference. In other words, s is a spacetime invariant quantity, which is really useful.

    We can extend this by considering not just coordinates in spacetime, but actual intervals (i.e. the differences between coordinate measurements for two spacetime events). It is easy to show that:

    $(\Delta s)^2 \equiv (c\Delta t')^2 - (\Delta x')^2 = (c\Delta t)^2 - (\Delta x)^2$

    which tells us that the "spacetime interval", $\Delta s$ between two events in different frames is a constant for any two given events, and it can be measured in any inertial frame.

    It turns out that the sign of $\Delta s$ is also really useful for determining whether it is possible for two given spacetime events to be causally connected to one another. A special case is that any two events that have $\Delta s=0$ can communicate by exchanging a light signal.
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  3. origin Heading towards oblivion Valued Senior Member

    Sure there is. Bob's odometer would show he traveled less distance than Alice's star chart said he did. Once Bob has returned to Alices frame they would both agree on the duration of a second and the length of a meter, but Bob aged less than Alice and Bob traveled a shorter distance than Alice measured during his trip.
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  5. James R Just this guy, you know? Staff Member

    In ralfcis's example, Alice is the one in the spaceship and Bob stays on Earth. So your post is correct as long as we swap A and B.

    Just to add: conceptually, we don't need to use spaceships. Alice could, in principle, drive a very fast car along a road at v=(3/5)c, and she could measure the distance covered using an actual odometer - i.e. standard car odometer. At the end of the trip, the odometer in her car would read the "contracted" trip distance, while the milepost signs on the side of the road would read Bob's "uncontracted" distance measurements.
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  7. ralfcis Registered Senior Member

    It's always the same example of the same thing unless I use the example of v=.8c.

    I will but things come up.

    I've explained this many times. Brehme used "your space in my time" without objection from relativists so it is appropriate to mix any perspective with any clock. Yv is Alice using her on-board clock, the only measurement device she has access to, to measure the time it took her to cross Bob's proper space. Yv=x/t'. Alice cannot use the parallax distance measurement technique while she is travelling because it would take 2 ships. She could remap the entire universe from her 3/5c perspective but that is not practical because it would have to be done for all velocities. So long as relative velocity is depicted correctly, you can have your choice of all other valid depictions and you choose the easiest and one that makes physical sense. You don't choose the perspective that you're stationary to an entire universe moving past you (no rule against that) just as you could choose the sun orbits the earth but why choose impractical perspectives that don't reflect reality? I choose to ignore the existence of length contraction for the reasons I gave. It's redundant so it means special relativity is not the simplest theory to explain the physics.

    They're using clocks that suffer from relativity of simultaneity to measure length so of course their length measurements are different.

    Assuming length contraction allows the math used for relativity to work. In fact, it is absolutely essential for it to explain the physics. I explain the same physics with different math that also allows me to predict extra measurable physical phenomena that relativity cannot. So the two agree but relativity calls certain scenarios indeterminate that my math is able to determine.

    It is physically measuring a velocity using your own clock, without the need for clock sync or referring to the other clock, and using proper time star charts from a common stationary perspective such as the earth, to understand why you seemed to cross those distances in less time than your relative velocity v would indicate. Have you not heard of people able to cross the universe in a very short amount of their time if their v is close enough to c? Relativity explains this using length contraction. I use Yv and the Brehme perspective of your space in my time.

    The question you should have asked is why is v a relative velocity and Yv is not. Bob's not going anywhere but his relative velocity to Alice is x/t = 3/5c. Alice's relative velocity to Bob is also 3/5 from her DSR reading. But she is going somewhere, 3ly as measured by her clock in 4 yrs. This is her Yv=x/t'=3/4c which Bob has no access to seeing. He sees Alice's clock ticking at half of his so he can't see her Yv. She can't see her motion contracting the space she's travelling. I choose a real clock and a real star chart over her non-existent odometer.

    Acceleration has nothing to do with it and relativity still works fine without any acceleration being involved at all. You refuse to believe this, google clock handoff.

    Relativity can't say Alice has any permanent age difference unless she returns and compares their clocks co-located. If she stops 3ly away, she's in the same stationary frame as Bob, and they can send light messages of their ages but this is not enough for relativity to conclude their permanent age difference because they are separated and other perspectives can see different results of permanent age diff between them.

    Again a clock handoff does not have to incrementally sweep the lines of simultaneity. I have seen the curved analysis trying to mitigate the acceleration and it is just dead wrong. Permanent age difference has nothing to do with acceleration but everyone uses the illogical conclusion that if constant velocity doesn't cause age difference then non-constant velocity must. I explain what is the cause of age difference and it's due to a mismatch of relative velocity.

    Yes I am on purpose because that formula applies to constant relative velocity. At least you know that but everyone I've ever met incorrectly assumes the results of reciprocal time dilation are the same as permanent age difference hence time dilation is the cause of permanent age difference. If Alice did not turn around and kept going, ct'=8, ct=10 and x=6 and the formula applies but this result is for time dilation and not permanent age difference. If you look at the Greene videos, the twin paradox is derived using different math but my math is simpler and applies to twin paradoxes where the twins never meet again.

    Non-sequitur. You've never used a ruler to measure length and you'd prefer a stopwatch to do so?

    Again, why make your life difficult to shoe horn in a philosophy that is not required? You say it is, I can show you in every example it isn't.

    In relativity, perspective is reality. The earth will orbit a removed sun for 8 minutes but the underlying and inescapable reality is the sun is gone. I could draw up a perspective reality for each planet but the universal reality is the sun is gone so why not do the math from that one perspective?

    No they don't if she shifts her subjective perspective to look from the Loedel perspective. All that changes is when each turn on their stopwatches to start timing an event. They don't observe time slowing, they observe a relativity of simultaneity of when timing starts or ends for an event.

    Yes I do, you just don't see it yet and you haven't received enough information to see it yet.
  8. James R Just this guy, you know? Staff Member


    This is the first time you've mentioned "Brehme" to me. I don't know who that is, or why he is authoritative. Regardless, I don't see why we need to rely on his authority.

    If you randomly mix space and time measurements made in different frames, you can certainly calculate some numbers, but they are unlikely to relate to anything that's meaningful physically.

    Alice and Bob are both in the same position as regards measuring devices and the like. You can't privilege one over the other. Whatever measuring devices Bob has, we can assume Alice has identical ones available to her. There are no preferred reference frames in relativity, so you can't even say that Bob was the one who was "really" stationary and Alice was the one who "really" moved.

    So what? Yv isn't physically meaningful. What is it the velocity of? Nothing, as far as I can tell.

    If Bob wants to use parallax, he needs to observe from two different locations in his frame. Similarly, if Alice wants to use parallax, she needs to observe from two different locations in her frame. They are on an equal footing. It is absurd to say that Alice can't use parallax but Bob can.

    If she can't do it, nobody can do it. Is Bob, sitting on Earth, absolutely stationary? No. Earth moves around the Sun. The Sun moves around the galaxy. The galaxy moves relative to other galaxies. And so on and so forth. You're essentially claiming that nobody anywhere can make a map.

    All of them make physical sense.

    They do reflect reality. There are no preferred reference frames in relativity. Alice's frame, in which the Earth moves away from her and Proxima Centauri approaches her, is just as valid a Bob's frame, in which Alice moves between a "stationary" Earth and Proxima Centauri.

    When you ignore it, you get wrong answers, like those speeds of light that you obtain that are greater than or less than c.

    Your theory fails to accurately describe reality. SR succeeds where you fail.

    You should state your postulates. SR only has two, and I've given them to you. Everything else is derived from those. So, what is your alternative theory derived from?

    Try to be consistent, at least. Just above, you said you "ignore" length contraction, but here you admit that "of course their lengths measurements are different".

    Can you see that you can't acknowledge that length measurements are different in different frames and, at the same time, expect to get correct answers if you ignore that?

    It's not assumed; it's derived from the postulates.

    I assume you're read up on relativity. You will have seen how length contraction is derived. Right?

    I mean, if you're going to criticise a theory, the first step is to make sure you fully understand it. Have you done that?
    So far, I have pointed out a number of errors in your calculations, but you have not responded.

    Which "extra physical phenomena" can you predict, that relativity cannot? Please give one example, at least.

    Please give one example of something that relativity calls "indeterminate" but your math is able to determine. Be sure to choose something that is physically testable, at least in principle.

    To measure a velocity, one needs two things: a measurement of distance and a measurement of elapsed time. It is impossible to measure a velocity using a clock by itself.

    You are yet to define "proper time" or "proper distance", as you understand them, and I won't be commenting on those until you do.

    You're not being very specific about reference frames and the like in what you've written here, so for now I'm not going to try to unpack this.

    Her clock can't measure the distance. Clocks only measure time. She needs some rulers if she wants to measure the distance she travels.

    That is an error. In Alice's frame, Bob travels at -3/5c, while she stays still. In Bob's frame, Alice travels at 3/5c, while Bob stays still. Neither of them measure anything as travelling at 3/4c.

    Yv is just a randomly calculated quantity. It is not physically meaningful.
  9. James R Just this guy, you know? Staff Member


    Sure she can. She can pull out a ruler and measure something that she sees flying past her, for instance.

    Alice has access to just as many "real clocks" as Bob does. If Bob makes a star chart using his own rulers, then distances on the chart will necessarily be distances that Bob measured using his rulers. Alice could make her own chart using identical methods to Bob, but she would have to use her rulers, not Bob's.

    You have not yet responded to my comment about odometers, above. If Bob can make a working odometer, so can Alice, in principle. Actually, though, odometers are problematic, because they are invariably attached to a vehicle. There's no such thing as an odometer that measures distances while it remains stationary. Probably you ought to forget the whole complication of the odometer, and just use the idea of regular rulers instead.

    You wrote about permanent time differences between Alice and Bob which are discovered only after Alice and Bob are brought to rest relative to one another. That requires acceleration.

    I agree with you that it is not necessary for Alice and Bob to ever meet again once Alice flies away. If you're not interested in comparing their clocks side-by-side and at rest, then nobody needs to accelerate.

    Please don't try to tell me what I refuse to believe. Not before you've asked me, anyway.

    You seem to be hung up on "permanent" age differences, for some reason. I agree that whenever Alice moves and then stops moving relative to Bob, there will be permanent time differences, due to Alice's accelerations. It doesn't matter whether Alice physically returns to Bob's location. She doesn't have to. She could, for example, just stop some distance away from Bob.

    You're wrong. That would be a situation in which relativity could draw conclusions about their "permanent age difference". Also, once they were at rest relative to one another, every other frame would agree about their "permanent age difference" in the "stationary" frame.

    If "everyone" except for you reaches one conclusion and you reach a different one, that should be food for thought, at the very least. What are the chances that you're the real expert?

    If Alice never stops moving relative to Bob, then they are never again both stationary in Bob's frame. They can compare their respective clock readings in various ways (e.g. by sending each other light signals carrying the time), but they don't share the same notion of simultaneity, so while they can say things like "Your clock read X at the same time my clock read Y" they will need to keep in mind that "at the same time" doesn't mean the same thing for both of them.

    It may be simpler, but that's no good if it's wrong.

    There are no preferred frames or "perspectives" in relativity. It might be easier to do the math in one frame rather than another. That's no problem at all, because we know exactly how to translate results derived in one frame to any other frame.

    So, go right ahead. Do your math in whichever frame you like. But make sure you don't start mixing different frames (e.g. using distances from one frame and times from a different one). That way only leads to error.

    When you say "she shifts her perspective to the Loedel perspective" or similar, are you saying that she should use a different frame, other than her own rest frame? Aren't you then just introducing a third frame into the mix, alongside the original two?

    Remind me, though, because I'm not understanding yet: what is the "Loedel perspective"? Is that your "average velocity" (or "half velocity") frame, or something else?

    Well, I've been waiting for quite a while now - days - for you to give me information I have expressly asked you for. I'll wait. But I'm especially eager to see your responses to posts #58, and #71/72.
    Last edited: Feb 16, 2021
  10. ralfcis Registered Senior Member

    OMG more? That's great but I have a huge backlog to get to starting at post #27 and then there'll be more questions breeding like rabbits and then you'll be screaming at me why I'm not answering quickly enough. It's probably the same questions over and over, I'll keep repeating the same answers and you'll keep repeating the same rebuttals that my answers don't agree with the philosophy of relativity which I point out that they don't and I'll never get to complete the presentation of the math behind my counter explanation of relativity. This could take months. I've already written the whole thing in 77 pages but the purpose of this thread was to edit out all the dead ends and mistakes of which there were many as the math evolved. At this rate it could take a year to get the whole story. Fine I'll continue with the questions but I'll probably be marking many with "already answered" and it'll be up to you to understand why.

    I appreciate the interest though. Thank you.
    Last edited: Feb 16, 2021
  11. origin Heading towards oblivion Valued Senior Member

    Thanks for catching that. I probably messed up because I haven't read ralfcis's posts I just read your replies. No need for me to wade through all his errors when you are willing to do it.
  12. origin Heading towards oblivion Valued Senior Member

    The problem isn't that your answers don't agree with 'the philosophy of relativity', the problem is that your answers don't agree with experimentation and observation. That means your answers are wrong.
  13. ralfcis Registered Senior Member

    Show me where they don't agree with experimentation all I'll show you where you don't understand the difference between math, experimental results and the theory that explains those results.
  14. origin Heading towards oblivion Valued Senior Member

    Here you go.
  15. ralfcis Registered Senior Member

    Well that's great, case closed then, length contraction is measurable. Except it's not. I once proposed an experiment using Raskar photography (1 trillion frames per second) and a very long high speed bullet except the frame duration was still way too long to measure length contraction if it existed. So your car scenario is absolutely impossible to quote as an experimental result so stop.
  16. ralfcis Registered Senior Member

    So what. You've produced a list of experimental results. It's your job to point out where my math contradicts those results.
  17. James R Just this guy, you know? Staff Member

    I've been trying to make your life easier. I've repeatedly suggested to you that you concentrate on responding to my posts #58, #71/72. If you can respond to those, then you might very well cover issues I have repeated in other posts.

    It's up to you what you respond to, of course, but so far our disagreement seems to boil down to a few matters, like your ignoring of length contraction, your inappropriate mixing of measurements from different reference frames, your misconstruing of the spacetime interval for a time interval, and your "Loedel lines", which look almost like lines of simultaneity but which are not.

    There's been no screaming so far. That's good, because I have had a number of past interactions with people who believe they have found flaws in relativity who have got very upset and insistent when I have pointed out their errors. So far, you've been quite reasonable, although you have yet to try to address the main objections I have raised to your analysis.

    The same questions will keep coming up until you answer them. There's no point, for example, in your talking about "proper time" any more, until we have reached an agreement about what "proper time" is. I have already posted my definition. You have yet to post yours, but it is clear from a number of your posts that your definition is different from the one I'm familiar with.

    All I have done, really, is to show you how special relativity does the maths to analyse the scenarios you have used as examples. I have also pointed out where relativity disagrees with your analysis. Philosophy hasn't really come into it.

    If you are trying to propose a replacement theory for relativity, then as long as your theory can explain all the experiments as well as - or better than - relativity can, it will be worth taking a look at it. If it's mathematically simpler than relativity, all the better. On the other hand, if it is not self-consistent or it conflicts with real-world experiments, then it's not going to be much use to anybody.

    My impression is that you're kind of piggy-backing on relativity anyway. Your Y is essentially the same as relativity's Lorentz factor, for instance, but that factor is derived in relativity, whereas you simply present it as a given, so that we can't tell where it comes from in your theory. And then, you also have things like v'=Yv, which doesn't appear to be a physically meaningful quantity, and which does not appear in relativity.

    Since there are clear differences between your equations and relativity's, it would be good to know what your starting postulates or axioms are, for starters.

    Got a link? Can I see your 77 page presentation somewhere online?

    Okay. Good.
  18. James R Just this guy, you know? Staff Member

    It doesn't really matter even if length contraction is not easily measured in a lab. It is impossible to take it out of relativity while leaving everything else intact. The constancy of the speed of light in all inertial frames, for instance, requires that length contraction occurs, as long as we accept time dilation. You could try tossing out any two of those three things - time dilation, length contraction and the constancy of the speed of light - but you can't have any two of those without also having the third. And if you don't have all three of them, then you won't be able to account for experimental results, such as the ones origin linked to, above.

    For now, it is sufficient that I have shown that your math does not even produce self-consistent results. Given that, there's no need to compare your predictions to experiment.
  19. James R Just this guy, you know? Staff Member

    Just quickly ...

    Regarding the odometer thing, again, I have thought about that some more, and I'm not 100% sure I was right that it would read a "contracted" length along a road. I think it probably would, but it's actually a complicated situation, since odometers are calibrated by knowing the circumference of the car tyres, assuming they are circular, assuming they don't slip on the road, etc. etc. From the point of view of somebody on the roadside watching a car drive past, for instance, the tyres are not circular, but elliptical. Even for somebody sitting in the car, different points on the tyre are travelling at different speeds relative to the road.

    It's a lot easier, conceptually, to imagine two sets of identical regular rulers: one set laid out end-to-end along the roadway, and the other set glued end-to-end and fixed to the top of the moving car, with both sets extending out to infinity in the direction of travel.
  20. ralfcis Registered Senior Member

    Here's the link

    Yes rulers hence my Raskar photography method:

    I mean to start writing every night but my neighbors get in the way and invite me over and that has to stop. Why #27 is important is because defining a "participant" is a great place to start and it will take a very large post to define. Whether you depict the scenario in Loedel, Minkowski, reverse-Minkowski or Epstein all must agree and you will see Epstein is not particular about how c is viewed from different perspectives but still agrees and is convertible with all the other spacetime diagram types.

    PS. I did mention Brehme in post #7 and since then referred to it as "your space in my time" and also mentioned relativists accept it.


    (I have not read this yet, I`m very busy.)
    Last edited: Feb 17, 2021
  21. James R Just this guy, you know? Staff Member

    I found this article on "proper velocity", also called "celerity":

    "proper velocity (also known as celerity) w of an object relative to an observer is the ratio between observer-measured displacement vector $x$ and proper time $\tau$ elapsed on the clocks of the traveling object: $w = dx/d\tau$.​

    The relationship between proper velocity and coordinate velocity is:
    $w = \gamma \frac{dx}{dt} = \frac{dx}{d\tau}$

    This might be helpful in unpacking what the ralcis "Yv" is, since in ralfcis relativity, v is a coordinate velocity and Y is $\gamma$.

    It should be noted, though, that it would be incorrect to write $v'=\gamma v$ and claim that $v'$ is the velocity of something in the frame of the moving object.

    $w=\gamma v$ is the velocity of the object, calculated using the initial coordinate frame's rulers and the moving frame's clocks. In contrast, $v$, the coordinate velocity, uses both the rulers and clocks of the frame of the observer watching the object flying past, and a "correct" $v'$ should rightly use both the rulers and clocks of the moving frame.
  22. ralfcis Registered Senior Member

    Good, I've never heard of celerity before but I like the term proper velocity. Your v'=x'/t' but since your x'=Yx and your t'=t/Y, your \(v'= Y^2x/t = Y^2v\) (probably not if one of my perspectives for x' or t' is wrong).
    I'm not claiming that at all. I've already said v=x/t is a relative velocity and Yv =x/t'=x/(t/Y) is not. Can you not see the difference between the equations? You won't accept any number of papers that Brehme velocity or celerity or proper velocity or Yv is valid in relativity because you've been unaware of it and anything you've been unaware of can't be valid. I won't call Yv as v' anymore to avoid confusion with the definition of v' in relativity, ok? I mostly call it Yv because it sounds cool like a superhero's name and because it is clearly different from v. I'll stick with Y not \(\gamma\) because I have to be consistent with my spacetime diagrams.
    Last edited: Feb 17, 2021
  23. ralfcis Registered Senior Member

    Qusetion from #27
    No, v is the relative velocity of Bob and Alice relative to each other. But since there are the background frame and the Earth frame as 2 additional participants, there are other relative velocities depicted. (Save your comments until the end.) I may draw Alice leaving earth at 3/5c and Bob at 0c but their relative velocity is 3/5 even if, in the Loedel diagram, each of their velocities are labelled as -1/3c and +1/3c their relative velocity is 3/5c. I'm sick of having to explain what relative velocity is on every forum. Relative velocity can't be depicted as both labelled 3/5c. But the good news is, since all valid depictions are equivalent, you can choose the one that most fits the physics even though you can also choose the one where the entire universe whizzes past you just because your screaming down the highway in your car. The universe is not really, physically whizzing past you. It is always stationary relative to you without invoking a preferred frame or absolute motion as so often comes out of the mouths of those who have zero understanding of relativity. In relativity, Alice and Bob take turns being the stationary frame and yes I understand there is no such thing as stationary in relativity. But since I can choose a depiction that represents all others, I choose the Earth frame as stationary without implication of a preferred frame.

    Relativity is stuck on the example of 2 astronauts floating past each other in a deep featureless space with no distance markers and being unable to determine who is actually moving. It's makes no difference so move on from that. You have no choice but to draw relative velocity as absolute velocity so no more niggling comments that there's no such thing as absolute velocity. All spacetime diagrams have at least 4 participants, not just 2 as is shown in relativity's useless 2 astronaut example. Here is a reverse-Minkowski of Alice returning to Earth with Alice as the stationary frame (except she's not stationary and neither is the earth).

    Please Register or Log in to view the hidden image!

    Purists will say I've drawn it backwards (look at the signs on the x-axis) and what are these blue and red numbers? They're to show that once Alice is deemed the stationary frame, she is not actually on Earth past time 0. Earth has taken off from her, not she from it. She occupies a blank space where Earth used to be. This blank space is the background cartesian coordinates and it is one of the participants. Even though the entire universe is whizzing past Alice, space is not because, as I showed in my vacuum bottle example, you can't move a vacuum, there is no relative velocity to space and, by the transitive property, to the light that propagates through it using it as a non-material electromagnetic medium.

    The blue numbers on the ct-axis, borrow Earth's clock when it was the stationary frame. The red numbers on the ct-axis, are her formerly moving clock which she sets as the stationary clock for her stationary universe. In standard relativity, she would choose her red clock for the ct-axis and the blue numbers for the ct'-axis. What the drawing then shows is after occupying empty space for 4 red years she takes off to return to Earth which is speeding away from her at 3/5c. Her relative velocity to Earth was -3/5c in the regular Minkowski diagram so this means she must chase earth at 15/17c (relative to the blank space) in order to maintain the same relative velocity of -3/5c to the speeding Earth/Bob. She will re-unite with Bob when her red numbered clock is 8 and Bob's blue numbered clock is 10 just like in the regular Minkowski diagram. I'm doing this red and blue time numbering trick to show no change is required on the x-axis to accommodate the clock choice. Alice travels 6 ly no matter what.

    The important thing here is a participant is someone who can set his time on his clock as the reference time for everyone else (the other 3 participants). Look back at the Loedel conversion to Minkowski. If the Earth had been set as the background cartesian coordinates in the Loedel, it would have been a depicted as a 1/3c frame speeding in between Bob and Alice in the Minkowski Diagram. That's why the Loedel background was empty space so that the Earth could be the stationary frame with Bob in the Minkowski depiction. So when converting from one depiction to the next in relativity, one must always take into account all 4 participants, not just 2.

    Next up, the equivalent Epstein depiction to show the assumptions of relativity are irrelevant to correctly depicting the physics. The Minkowski diagram was doctored to support the assumptions of time dilation and length contraction, the Epstein was not.
    Last edited: Feb 17, 2021

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