# Relativity and simple algebra II

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

1. ### QuarkHeadRemedial Math StudentValued Senior Member

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Oh my! What a bargain!

exchemist likes this.
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3. ### ralfcisRegistered Senior Member

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Ok so I'm going to finish off my last post before I tackle the questions which are legit and am grateful James R took the time to ask them.

The results of my last Md are shocking to anyone who thinks they know relativity.
The pink light signal, from Bob's perspective using his own clock, takes 3 yrs to travel 3 ly but if he was going according to a simulation of Alice's watch it would be 3ly in 2.4yrs.
From Alice's perspective, using her watch and her star charts (which were drawn on Earth using various techniques like the parallax view to measure distances), her thin red line of perspective simultaneity points to the signal starting at 2.5 yrs her time and ending 1.5 of her yrs later. So the pink light signal travelled 3 ly in 1.5 of her yrs. If she used Bob's simulated time, it would be 3 ly in 3 Bob yrs. I'm not including how length contraction handles this because in this math, I don't need it to.

But I've introduced a new perspective, the Loedel perspective that allows one to peer into proper time without the hysteresis of other perspectives. From that perspective, Alice and Bob both agree the 3ly were travelled in 2 proper years. But, but, those unfamiliar with what the true meaning of "c being the speed limit from all perspectives" would complain that my perspectives involve super-luminal speeds and they're all different values..

They confuse the clocks you use with perspective. They also confuse an absolute velocity c with the relative velocity to c. If I choose to say Alice's gamma velocity v'=Yv and her v=3/5c then her Yv is 3/4c. (For v=4/5c, the Yv=4/3c). This can't mean light doesn't beat her to her destination using her clock (and relativity of simultaneity of when the light signal started). But what is the speed of light using her clock in that case? Is it 2c from her perspective or is it 1.5c using proper distance over proper time (the Loedel perspective)? Or since see's moving at Yv=3/4c does her light also move at Yc =5/4c?

The answer is my math doesn't care about individual perspectives but only the proper perspective (which is an oxymoron). The true simultaneity of when the signals were sent was when both proper times were 2 and received at 4. Alice travels 1.5ly in 2 of her years which are also proper years which is Yv=3/4c and light travels 3ly in 2 proper years which is 3/2c (which is not Yc as I had thought for years). The pink light signal, from Bob's perspective, also travels 3ly in 2 proper years so this is how c is the same from all perspectives but not a universal value of c. The yellow and pink lines don't share the same value (the yellow's is 1.5ly/2 proper years) but the slopes of each have the same value of c.

Relativity goes to great lengths to maintain c is the relative velocity limit in any direction by keeping all slopes at c but it also reacts to relative velocities by altering the length of the light lines (pink is twice the length of yellow). (Without c as a limit of information, there would be no such thing as causality in this universe.)

This is all controlled by

$Y_u/Y_w=DSR_v$ as Y_u approaches infinity causing Y_w to approach infinity but the ratio approaches a finite value because the infinities cancel out as they must in physics.
I'll show the derivation of this most important equation in the next post. (w=u (+)v . I'm going to use the symbol (+) for relativistic combination from now on. )

I think besides TEX, the greatest failing of this forum is you can't go back and correct mistakes in past posts because they don't trust the users enough to be honest. That's what I miss about my previous forum which was the only one that allowed unlimited corrections. The quote feature should be enough to keep people honest. I'm still catching many mistakes and can't correct them in situ. It's ridiculous. Don't bother to reply, just fix it or not please.

Last edited: Feb 10, 2021
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5. ### DaveC426913Valued Senior Member

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Thing about discussion fora is that you can't willy-nilly go back and correct things people have already responded to - for obvious reasons. (I say "Yay for Science!" and then you respond with "I agree" and then I change my post to say "Yay for Nazis!")

You appear to be trying to use this forum as if it's a blog - where your words will stand for all time, to be read by future readers, not current readers.

No one here (or in any forum), having followed you 25 posts into a thread, is going to tolerate you saying "Oh, I changed all that. Go back and reread the first two dozen posts."

That's just not a workable scheme.

It's not broken. Your old forum was.

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7. ### ralfcisRegistered Senior Member

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No I'm cutting and pasting formulas and values from my tables and I'm still finding mistakes even after the reboot of this thread and have to remember to re-correct in the future. I guess that's what I'll have to do.

8. ### James RJust this guy, you know?Staff Member

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Thanks, ralfcis.

Something is wrong, then. It doesn't matter who is measuring the speed of light. They should always measure it to be c, in any reference frame. If you think Alice is measuring the speed as 1.25c, then you've made a mistake somewhere - on your diagram, in your calculations, or maybe in both.

You've got a different speed of light there for Alice, again. This time it is 2c. Again, without spending more time trying to track it down, I don't know where your error is, but there's no doubt there's an error. Alice's speeds of light are not even consistent in two different calculations you have made, let alone consistent with special relativity.

There's not much point in my trying to track down where you went wrong, seeing as you're leaving us anyway. Good luck in sorting it out.

I don't know what you mean by "proper years". A time interval for a light signal cannot be a "proper" time, because a proper time is, by definition, the time measured between two events that happen at the same location in space (in a particular reference frame). Light is always travelling, so the sending and receipt of a light signal never happens at a single location, in any frame of reference.

Yes! See above.

What's "perspective"?

I don't know what you mean by "the relative velocity to c".

If something has velocity v, does that mean it's "relative velocity to c" is just c - v? Or are you talking about something else?

What is this "gamma velocity", physically? Your use of v' and v is reminiscent of a transformation between two different reference frames, but velocities don't transform that way between frames, as you know.

You seem to be saying that Alice is travelling faster than light, in some frame. Relativity doesn't allow that.

Define "the proper perspective". Who has "the proper perspective"?

Do you think there's a preferred frame of reference that is "the proper perspective"? If so, which frame would that be?

You seem to be throwing the relativity out of the theory of relativity.

I don't understand what you mean by "true simultaneity". Simultaneity means that two events happen at the same time. Two spatially-separated events that are simultaneous in one frame can never be simultaneous in a different frame. That is a result of special relativity.

Yes. Length contraction is a standard, derived, result of special relativity in the special case where the length measurement in one of two reference frames is a "proper length" (i.e. a length measured in a frame where the object is at rest).

Special relativity limits the speed of information transfer to c.

I commented on that formula in a previous post, which you have not responded to.

Please explain why TEX is a great failing of this forum.

You can, but only for a limited time after you post them.

As DaveC said, if you could endlessly edit previous posts you could mess with the accuracy of the historical record of the conversation, to put a false spin on it. Also, for later readers, it might well make no sense. Posters would appear to be commenting on things that were no longer there, for example.

That may be a small part of the reason, but there are many other considerations.

For example, I've seen people on forums leave in a huff, trying to delete every single post they made on the forum in order to remove material they have come to regret posting, or just to destroy the record. I've also seen people edit old posts to fill them with pages and pages of junk, to try to "break" the forum. There are reasons for many of the controls we have in place here.

Can't you go back there? Did they ban you?

9. ### ralfcisRegistered Senior Member

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421
Let me go back to your previous questions first. This could take some time because I'd like to make sure each question is put to bed before I go on to the next.

Yes I can go back but I'll stay here for a while to see how it goes.

Last edited: Feb 11, 2021
10. ### ralfcisRegistered Senior Member

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421
Yes all inertial frames are equivalent and every inertial frame moves through time at the normal rate of v_t=c and the units of time are proper time. The main equation:
$(ct')^2 = (ct)^2 - x^2$ is of hyperbolas bound by +c and -c (timelike region of the light cone) that intersect all velocity lines at the same proper time.

Here is an Md showing the hyperbola that intersects all velocity lines at proper time =2. The green Loedel half-speed perspective or line of simultaneity intersects the same proper time for v=0 and v=3/5c.

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11. ### ralfcisRegistered Senior Member

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As I explained when Alice and Bob, the two participants, are engaged in constant relative velocity, their Doppler Shift Ratios are the same value and their time dilation is reciprocal. At v=3/5c (separating so v is positive) , DSR = 1/2 and Y=5/4 (or as I say v_t=4/5c). When Alice makes a change in velocity, let's say to -4/5c, her DSR =3 and Y changing to 5/3. But this is not reciprocal due to the delay of c. Bob's relative velocity remains at +3/5c indicated by his DSR of Alice still remaining at 1/2 and his Y at 5/4. This is an imbalance of relative velocity and if Alice has travelled out 3ly before she makes the velocity change, Bob won't see the balance restored until t=8 when the info of the change reaches him at c. At that point, his DSR and Y will match Alice's. Here's a sneak peek at various return velocities. Relativity doesn't recognize the effects of any velocity that does not result in a re-unification of Bob and Alice because of its definition of spacetime paths and proper time (proper present to be more precise) only exists during co-location ( the two clocks can't be separated). The flaw appears if Alice stops. It's obvious using math you can extrapolate infinite return time results in Alice ages 1 yr less than Bob as a result of her stop. My math is just bound by math and this Md is the result of any velocity change Alice makes even if she accelerates away from Bob at velocities greater than 3/5c.

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12. ### ralfcisRegistered Senior Member

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1. Bob, Alice, Earth or the background grid can be participants.
2. The sign or magnitude of the relative velocity.
3. Acceleration which in my math is nothing but a sequence or average resultant change to the relative velocity or a handoff of clock information between an outgoing and incoming ship with no acceleration involved. Also called a frame jump by some.
4. Any participants you wish to include in a relative velocity.
5. The change is permanent until another change is made. This has to be done when the participants are separated because the permanent change is dependent on the delay of information between them. When I say information it's like this. If the sun magically disappeared (in the god's eye proper present), we would be orbiting nothing and getting light from nothing in our perspective present for 8 minutes until reality caught up with us. Relativity, unfortunately, defines reality as the perspective present and has no concept, except in the hyperbolas Md, of a god's eye proper present separated by distance.
6. No, you are implying absolute motion and a universal present with the words you used.
7. If more changes occur before the imbalance reaches Bob, they accumulate and propagate from their origin in the same way the first one propagated (from its origin at 3 ly e.g.) Relativity does not allow you to show how time progresses during a relative velocity imbalance. Here is how my math shows it in progress with how the Loedel perspective lines change slope during the period of imbalance:

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13. ### ralfcisRegistered Senior Member

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1. I have many equivalent equations for the DSR equation relativity uses but graphically it is defined like this:

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The interval between the light lines is how much time interval is broadcast and how much time interval is received. So the Bottom two pink light lines show Bob broadcasts 1 year but Alice receives it over 2 years so she observes Bob's TV broadcast at half speed slow motion or DSR = 1/2. The bottom two yellow light lines show Alice also broadcasts 1 year of her life and Bob sees it over two years so their DSR is reciprocal as is their Y because their relative velocity is reciprocal.

After Alice makes her velocity change she immediately receives Bob's year at twice the speed (in only half of her year) according to the spacing of the pink light lines. But the spacing of the Yellow light lines shows Bob still receives Alice's broadcast at DSR=1/2 until t=8 when Alice's change of velocity reaches him. Because the DSR's are different, Bob's relative velocity to Alice is .6c but Alice's relative velocity to Bob is -.6c hence a period of imbalance. Then Bob's DSR of Alice changes from 1/2 to 2 at t=8 which means their relative velocity is the same at -.6c. You'll notice that during the imbalance period, Alice's proper time falls behind Bob's until the imbalance is over and she ends up 2 proper years behind Bob permanently. Relativity can only make this call at co-location and can't see how Alice's proper time falls behind. It's because if the clocks are separated, different perspectives will read them differently but the underlying proper time relationship is revealed by the Loedel perspective and any other perspective can be deduced from that. Relativity defines perspective reality and does not delve into the calculatable proper time reality that is the source of the perspective realities.

14. ### ralfcisRegistered Senior Member

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Think of a clock 3 ly away that is broadcasting its clock face. You would see the time on that face 3 yrs in the past. If the clock was moving, the delay in clock info would be affected by the motion. But the motion would also affect the underlying time you are observing which the clock is recording. So the apparent rate of the clock info and the real time dilation affecting the clock you are observing is the apparent rate and the time rate that combine to make up DSR. So DSR can be written with gammas or (c-v), (c+v) terms. I'm going to lay out all my formulas for DSR and explain the significance of each. Note: time dilation is not due to time slowing but to the relativity of simultaneity of when the clocks are stopped and started but I'm trying to answer all these questions before I explain how that works mathematically.

15. ### ralfcisRegistered Senior Member

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I hope you understand this now because you seem stuck on units. Like I said, the DSR is like a slow motion or fast forward on your DVD player and the normal rate of time is your play button. I've explained how v_t is also a rate of time ct'/t and is related to the gamma factor, v_t=c/Y so v_t is not dimensionless.

When Y_t is 0, Y is infinite and vice versa. Zero in relativity is not a natural number, it is 1/infinity. Since v and v_t are symmetrical, you can get around the infinities of the gammas if you consider the v_t version of an equation with the v version and vice versa. Relativity gives infinities a pass but by grouping Y with other variables, you can get all infinities to cancel out mathematically. This is why my math can derive finite answers that relativity can't because some of its results are infinite when they're really not.

I'll keep answering your questions as time permits.

Last edited: Feb 12, 2021
16. ### ralfcisRegistered Senior Member

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I'd just like to add why I can have proper time as an underlying reality is because I have to math that allows light signals to pass proper time information while relativity does not. It instead chose to incorporate light information into Einstein's clock sync method which generates perspective lines of simultaneity. All clocks on a line of perspective simultaneity are set to the same time using Einstein's clock sync method. But this is an artificial present because 2 clocks separated by distance do not define any kind of present. Anyone separated from you does not share your present just because your clocks are synced to the same time.

17. ### James RJust this guy, you know?Staff Member

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ralfcis:

Thanks for your replies. I'll respond in more detail in the following posts. But here I'll make a few general comments.

The first thing I want to say is that it is good that you're aware that what a particular observer "sees" at any particular time depends on when light reaches him or her, often from distant points in space. Due to the finite speed of light, there are "signalling delays" that mean that the observer never sees distant changes instantly.

Because of these signalling delays, it is very important in relativity to distinguish, in any given frame of reference, the time that some event actually happens from the time that some particular observer in spacetime "sees" that event happen. For example, consider if the sun were to suddenly explode as you read this. The spacetime event "sun explodes" would be happening right now (call it t=0), but at a point in space 150 million km from where you are sitting (call it x=150,000,000 km). You would only "see" the sun exploding once light has enough time to travel that distance of 150 million km, which takes about 8.5 minutes. So the spacetime event "ralfcis sees the sun explode" happens here, where you are sitting (x=0), at t=8.5 minutes. It is vitally important to appreciate that the spacetime event "sun explodes" is a different event to the event "ralfcis sees sun explode". They have different spacetime coordinates, even though they both occur in the same frame of reference (i.e. your reference frame as you sit in your chair on Earth).

The main equations of special relativity, the Lorentz transformations, tell us how to translate the spacetime coordinates of a single spacetime event between two different frames of reference, where those two frames are moving at constant relative velocity. In general, both the space and the time coordinates of a given event change when you change reference frames, in a precisely specified way.

I'll probably be referring to this in future posts.

The second thing I want to say is that your graphs look very complicated, in general, although some of them are very helpful. I assume that you calculate the graphs from the equations, rather than basing the equations on the graphs. Is that correct? I think that in some instances it might be easier to talk about a situation if you reduced the graph to just the necessary lines required to make your point. So, for example, if you have only two relevant observers, Alice and Bob, then just show Alice and Bob's worldlines (at a single relative speed), along with a few worldlines for light signals exchanges between them. Some of your graphs have worldlines for a whole bunch of different relative velocities. Also, in some cases you are adding in lines of simultaneity and things like that, which is useful, but your lines are not always clearly labelled. I should also tell you that I have a form of colour blindness, so referring to various lines on your graphs purely by colour isn't always helpful to me. The same will no doubt apply to some other viewers, too.

The third general comment I want to make is about Doppler shifts. The Doppler effect happens even in a non-relativistic universe, of course, since it is due to aspects of wave motion where the source and detector are in relative motion to one another. In relativity, however, part of the observed Doppler shift is an effect of time dilation between the two different reference frames (the source and observer frames), which is conceptually separate from their "pure" relative motion.

Lastly, I have a question, which I have asked before, but I'm still a little confused. The question is: do you believe that the theory of special relativity is "flawed" or incorrect in some way, or is your work simply an attempt to reformulate it from a different set of starting equations?

The reason I'm confused is that, on the one hand, I think you've said that your results agree with the relativistic results (i.e. calculated numbers etc. agree), but on the other hand, you seem to be constantly referring to "problems" with the relativistic calculations or results. So, I can't tell whether you think there are errors in relativity, or not. I also can't tell whether you think your own work is "correcting" errors in relativity, "improving" on it, or what. Could you perhaps explain to me what you're trying to do?

Oh, and one last thing: in a number of instances, you are using various technical terms in non-standard ways, or using terms that are non-standard. That means that I can't always immediately work out what you're referring to, and that means that I need to ask what may seem like "stupid" questions from time to time. Please be patient with me if I ask you to define a term, because I want to make sure we both understand each other correctly. There's no point in talking past each other. Please ask me to clarify if I say something that doesn't seem clear, too.

Last edited: Feb 12, 2021
18. ### James RJust this guy, you know?Staff Member

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Okay. I'm going to work through your latest posts now, ralfcis. There's a lot of stuff to read. If you want to skip to what's most important, it's probably post #58, below.

That doesn't comport with my understanding of what a reference frame is. In standard texts on relativity, a reference frame can be conceptualised as being like a literal 3-dimensional "frame" of rulers in a 3-d cubic lattice (it is actually imaginary, of course). At each intersection of the lattice, we can imagine there is a clock. For a single reference frame, all the clocks at all points of the lattice can be synchronised so that, no matter where they are in space, they all tick at the same rate in that frame and read the same time.

Since a reference frame encompasses all of time, it doesn't make much sense to me to say that the frame "moves through time" at any rate. I would prefer to say that the clocks in the frame can be used to record the time that any event happened in that frame. All we need to do is to choose the clock at the point in the frame that is nearest to where the event occurred and read off the time coordinate from the clock.

Also, the term "proper time", as I explained previously, has a very specific meaning in special relativity. The "proper time" is the time measured by a clock in a reference frame that is attached to some moving object. A clock on a spaceship, for example, measures the "proper time" in a reference frame that keeps its "rulers" stationary relativity to spaceship at all times. Or, to look at it the other way around, the clocks in a particular reference frame measure the spaceship's proper time only if the spaceship never moves (spatially) relative to that frame. i.e. the spaceship stays at one "intersection" of the frame's rulers at all times (in that frame).

I don't understand what t, x and t' are in that equation. Are they time and space coordinates for a particular set of events? If so, which set of events? I'm also puzzled as to why x' does not appear in that equation. Is it a special case in which x'=0 for some reason, rather than being a more general equation?

It reminds me of the "spacetime interval" in special relativity, which is the quantity (in one dimension):

$s^2 = (ct)^2 - x^2 = (ct')^2 - x'^2$

The usefulness of this quantity is that for any single event in spacetime, s has the same value, no matter whether we calculate it in the 'primed' frame (using x' and t' coordinates) or the 'unprimed' frame (using x and t coordinates).

That seems sort of okay to me. When you talk about the "proper time", here, you mean the proper time of the "moving" spaceship (i.e. the observer who has v=3/5c in the "stationary" frame)?

In that case, assuming the 'primed' frame is "attached" to that spaceship (i.e. the spaceship is stationary in the 'primed' frame), then we set x'=0, and then the standard relativistic equation reduces to the "special case" that is your "main equation", quoted above. The x and t coordinates are the coordinates on the axes of your graph.

Last edited: Feb 12, 2021
19. ### James RJust this guy, you know?Staff Member

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I'm still not certain what your "Doppler Shift ratio" is. You most often seem to write it like this:
$DSR_v = Y_u/Y_w$

but I don't know what the subscripts v, u and w refer to. It seems like there are three reference frames involved in that equation. Is that correct? If so, why do we need three frames to calculate a Doppler shift? Can't we just use the source and observer frames, if we want to do that?

Are you saying that people in different frames of reference do not always agree on the value of the "DSR"? You seem to be saying that Alice can change the DSR, while it doesn't change for Bob, or something like that.

It could just be that it's a measure that is frame-dependent, and I guess that's okay, but I feel like I'm missing why it is useful.

The potential red flag for me in your statement is the word "When". You haven't specified which frame of reference you're using when you say "When Alice makes a change of velocity...". Are we using Alice's clocks to say when her velocity changes? Or Bob's clocks? Or the clocks in some third frame of reference?

Clearly, the clocks in different frames tick off time at different rates, and so mostly don't display the same time simultaneously at the same location in space. Also, different frames have different notions of simultaneity, as you know, so the set of other events that are simultaneous to the event "Alice makes a change in velocity" are different for Alice, compared to Bob, say. In particular, any two spatially-separated events cannot be simultaneous for both Alice and Bob.

When you say "relativity velocity" here, which reference frame is measuring those relative velocities? Are you using Bob's frame, Alice's frame, the "average" frame which sees both Bob and Alice moving, or what?

It might be better, if you want to be specific, to talk about particular events in spacetime, because every observer agrees on which events happen; only the spacetime coordinates of events change between frames. As soon as you start talking about the velocity of something or somebody in a particular frame of reference at a particular time, it becomes difficult to translate that statement to some other frame, because the other frame obviously doesn't share the same notion of what is simultaneous, and it uses different clocks and rulers to measure velocities.

Here, you talk about what Bob "sees", which brings in issues of light signalling delays. Maybe what you're saying about what Bob sees is correct, but that doesn't raise any issues for me in terms of relativity, as far as I can tell.

I don't really understand what you mean by "effects of any velocity ...". Which physical effects are you trying to "recognise"?

Aren't you using the same definitions of "spacetime paths" as relativity? You're using Minowski diagrams. On the whole, I agree that they are correct from the point of view of special relativity, as far as I can tell. So, I'm not seeing a problem.

I have previously defined "proper time", above. Only one frame can ever measure the "proper time" of an object - the one attached to the object, in which the object never moves (spatially).

I'm not seeing a flaw.

I don't understand what "infinite return time" means.

There is a potential complication, in that when Alice turns around she is accelerating, by definition. As she does that, her velocity relative to Bob is changing. On your graphs, the "turn around" is modelled as a sharp point, indicating, in effect, an unphysical infinite momentary acceleration. While being the "explanation" for the "twin paradox" effect we see when Alice and Bob reunite, that turnaround doesn't affect any of the other reasoning we take away from the graphs.

20. ### James RJust this guy, you know?Staff Member

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38,033
Okay. Your "participant" seems to be roughly equivalent to "observer", but sometimes you seem to use it for "reference frame" (as when you refer to the "background grid"). I think it is important to distinguish the two, because a single observer can only ever be at a single position in space at a given time, where a reference frame is a system of coordinates that covers the whole of space and time.

Actually, the acceleration during the turnaround in the "twin paradox" scenario causes an effective "time gap" as the moving observer's notion of simultaneity shifts around due to her change of velocity.

I don't see how a "handoff of clock information" during a period of constant relative velocity (no acceleration) could cause any such "gap", unless you're talking about a physical adjusting or resynchronisation of the clocks in one frame.

That would suggest that it is temporary, not permanent. Just a matter of language; not that important, I guess.

I don't think so. I think the issue comes not from signalling delays, but from the change of reference frame during acceleration. If Alice turns around in her spaceship, then her "outgoing" frame (when she has velocity +v relative to Bob) is obviously different from her "return" frame (when she has velocity -v relative to Bob). At the turn around point, Alice must alter her frame from one to the other. In reality, she must pass through an infinity of frames in between as she does that (with velocities in the range from +v to -v). In the process, her notion of which spacetime events are simultaneous in her frame changes continuously.

Relativity doesn't define "reality" or "perspective present", to my knowledge. Those are not terms I have seen in any standard text on relativity.

When you say relativity has no idea of "god's eye" perspective, isn't that exactly what a reference frame (extending through the whole of space and time) is?

No, I'm not implying that. There is no absolute motion or absolute standard of rest in relativity, and no "absolute present". You and I appear to agree on the relativity of simultaneity and on relative motion, so far at least.

I think it is best to be specific and talk about events in spacetime. Those can be "changes", but they are changes that have specific space and time coordinates in every frame. Talking about "changes" in the abstract risks leading to ambiguity.

As for "propagation", we agree that things like light signals (or information carried by light signals) propagate through space, but they do that in a very well-defined way. Your graphs clearly show how information about distant "changes" propagates via light signals. I don't think the changes themselves propagate.

I think that at least some of the difficulties we're having here are more to do with use of precise language than anything else.

I'd say your graphs show how time progresses. One of the axes is time! You can draw a graph in whatever reference frame you need. The only problem with the "twin paradox" scenario is that if you're going to draw a Minkowski diagram for Alice, you're going to need two - one for each of her two frames, before and after the turnaround.

In post #49, can I ask you what the green lines represent on your top graph? You say they are "Loedel perspective lines", but I don't know what that means.

They clearly aren't lines of simultaneity in Alice's reference frame. So what are they?

21. ### James RJust this guy, you know?Staff Member

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Continuing at post #50, I should start by saying that your diagram there seems to be accurate and is very useful, ralfcis, so that's good.

I agree. This is consistent with special relativity that says that when Alice and Bob are travelling at constant relative velocity, each sees the other's clocks (whatever they are) as running slower than their own, in this case by a factor of 2.

Yes.

If you consider it from a Doppler shift perspective, before the turnaround Alice was moving away from Bob, so received his regular signals at a lower rate, and after the turnaround she is moving towards Bob, so receives his signals at a higher rate.

When you say she "immediately receives", though, that isn't quite true. She receives a signal from Bob at her t'=4, and then doesn't receive another signal from Bob until t'=4.5, half a year later.

Alice immediately knows about her own change in velocity because she is on the spot when it happens. Bob, on the other hand, has to wait for information about Alice's change of velocity to reach him, because he's far away from her when she changes velocity. In other words, there is a light signalling delay that Bob has which does not apply to Alice.

You are correct that Bob doesn't find out about Alice's velocity change until his t=8.

You're not being specific enough about reference frames and times.

For Bob, for instance, as we can see from your graph, at time t=4.9 years, Alice's velocity is +0.6c according to Bob. At time t=5.1 years, Alice's velocity is -0.6c according to Bob. That information can be read easily from your graph.

Now, it is perfectly reasonable to say that at time t=5.1 years, Bob didn't know that Alice's velocity was -0.6c (in Bob's frame). At that time, Bob might well have believed that Alice was still moving away from him at +0.6c at that time, even though that wasn't the case in fact. But Bob's beliefs, right or wrong, can't affect the reality of what Alice was doing at that time, 3 light years away from Bob. We know that Bob has no way of instantly knowing or seeing what Alice is doing 3 light years away. He has to wait 3 years to see anything that happens 3 light years away. In other words, from his perspective at t=5 years, Bob has to wait until t=8 years until he can know anything about what is happening 3 light years away from him. And, sure enough, he receives a light signal from Alice at t=8 which lets him know she turned around 3 years earlier (according to his clocks).

There is no problem with this. Can you see a problem?

You might argue that, if Bob was drawing a Minkowski diagram of Alice's motion, we would be unable to do it "on the fly". He could only update it bit by bit, as he received signals from Alice. But there's no problem with that, either. Nobody can see into the future. We can only ever deal with the information we have at the time, unless we make assumptions.

t=8 is the first time Bob has information about Alice's new state of motion at t=5, which allows him to calculate a new DSR (if he wants to do that). There's no problem with that.

The entire "imbalance period" you refer to happens, in effect, at the turnaround point in Alice's journey. At that point, her frame of reference has to change from one moving at +0.6c to -0.6c (in Bob's frame). All other parts of her journey are constant velocity, relative to Bob, so signals are sent and received at regular intervals during those parts of the journey.

That's not correct. We could, for example, draw lines of simultaneity for Alice on your graph, and see the "time gap" in her idea of simultaneity at the turnaround point.

Incidentally, you have drawn some green lines on your diagram, like the "Loedel lines" in your previous one. What are those? What do they tell us? They are not lines of simultaneity for Alice. If you think they are, you have made an error, because all Alice's lines of simultaneity in each "half" of her trip should be parallel to one another (she is travelling at constant speed in each half of her journey).

Yes. That's the relativity of simultaneity. A standard, well-understood effect in special relativity. And we also have the signalling delays I talked about.

I don't know what that is. It almost sounds like you believe this is some preferred frame of reference "underlying" Bob and Alice's perspectives. But special relativity has no preferred frames. I thought we agreed on that.

I'm unfamiliar with the term "perspective reality".

Relativity has no problem with calculating "proper times". The times you have labelled on Alice's world lines are proper times for her journey. Bob does not measure proper time for that trip, because Alice is constantly changing spatial coordinates in Bob's reference frame (as is clear from your graph).

Last edited: Feb 12, 2021
22. ### James RJust this guy, you know?Staff Member

Messages:
38,033
A few loose ends...

I think we're in agreement on this. You're just talking about the regular relativistic Doppler shift. Right?

Okay. I can understand if you sometimes want to use units where c=1, in effect. The confusion only comes when you aren't consistent - putting factors of c into some equations, while assuming c=1 in others.

I'm not really sure what this is about. In special relativity we have
$\gamma = \frac{1}{\sqrt{1-(v/c)^2}}$.
and it looks to me like you're using Y as a substitute for gamma.

When v=0, gamma is 1. When v=c, gamma is infinite. I'd have to look back to find out how you're defining Y_t, but I don't think we need it for anything anyway right now, so I'm not going to going looking unless you tell me it is important for some reason.

Where are infinities cropping up? Since no material object or reference frame can ever have v=c, Y should never be infinite.

Can you please give me an example of where relativity would produce an infinite answer for any physical (i.e. measurable) quantity?

I don't see a problem. For example, Alice can send signals to Bob at any time about what her clocks say, and vice versa. Why do you say that it is impossible to pass "proper time information" in standard relativity?

No. The synchronisation of clocks in a single reference frame is something we assume has been done before we start using them for anything. The procedure is straightforward.

The relativity of simultaneity is a direct result of the Lorentz transformations. If you want to trace their origin back, it goes right back to the constancy of the speed of light in all frames and to the postulate of the constancy of the laws of physics in inertial frames. It has nothing to do with clock synchronisation, per se.

Well, yes. It is always possible to synchronise clocks in a single reference frame.

What? Sure they do. Two separated clocks that tick at the same rate can be used to define a "present". Events that happen at either clock at the same displayed time on the nearest clock happen simultaneously, assuming the clocks were previously synchronised.

That's wrong. You're most likely on the opposite side of the world to me, but we can agree on what is happening "now", with reference to our separate clocks (as long as we account for time zone differences). There are no relativity issues (ignoring the rotation of the Earth and on on, which causes problems for reasons unrelated to special relativity). As long as we're effectively at rest with respect to one another, we can agree on anything that is happening "now", no matter where it is happening, or how widely we're separated. Our clocks are synchronised (in practice, our respective times are synchronised to Greenwich Mean Time, or Universal Time).

Last edited: Feb 12, 2021
23. ### ralfcisRegistered Senior Member

Messages:
421
I see the problem here. It's like a blind man trying to understand an elephant with only a single nose hair from the trunk and having experience with mostly ostriches. Even with a full description it would still be difficult to piece together the enormity of this animal. So I'll give you another piece of the description.

The mathematical framework for relativity is quite different than what I'm presenting here. There are no Lorentz transform equations because there are no coordinate frame rotations. There are 3 axes ( ct, ct' and x) plotted on a Cartesian coordinate system. Because the two time axes ct and ct' look to be in distance units, relativists declare that time is a 4th spatial dimension. This is not true if you algebraically derive relativity's equations from the main equation which defines there is only one velocity in the universe (c) that has velocity through space (v) and velocity through time vt components. The faster you are observed to go through space, the slower you are observed to go through time. The relativistic combo equation similarly limits the combination of any two relative velocities through space to c.

$c^2 = v^2 +v_t^2$ also written as the gamma function $Y = \frac{c}{ \sqrt{(c-v)(c+v)} }$ ( $v_t= c/Y$ ).

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Ratios of the various axes define different types of velocity and the reciprocals of those velocities represent different types of lines of simultaneity (not shown and not important for now).

What is important is the 1/slope of the .6c line represents the velocity (x/ct) the stationary observer sees with his clock while x/ct' = Yv =.75c is the velocity the moving observer sees using his own clock. The distance common to both is invariant and proper and does not require any assumption of length contraction or reference to the stationary clock.

Using this info about Yv, algebraically multiply both sides of $Y = \frac{c}{ \sqrt{(c^-v^2)} }$ by v

$Yv = \frac{cv}{ \sqrt{(c-v)(c+v)} }$ =x/ct'

and v=x/ct and you get the main equation as

$(ct')^2 = (ct)^2 - x^2$

which represents all 3 axes.

It could have just as easily been written as

t'2 = t2 - x2 /c2 which would have led to the conclusion that all distance axes were really time axes according to relativist's logic. The time axis and length axes cannot be joined into the concept of spacetime because time does not behave like a distance axis and their lame logic does not make it so.

I will next show how the simple spacetime diagram I showed here can be combined with light signals to define the basic mathematical building block for relativity. Time slowing during constant relative velocity is an illusion. What causes that illusion is a mismatch in how start and end times are recorded differently from each perspective. Relativity of simultaneity causes the illusion of time dilation because perspective observations are illusion.