# Relativity and simple algebra II

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

1. ### ralfcisRegistered Senior Member

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375
So lets take inventory of all the scenarios that are being piled up on me. Every single person is convinced I will not address them because I somehow can't and want to live in my own unchallenged self-consistent delusion. Again, I'm not anyone's employee here and you'll all just have to be patient. I'll answer all questions in whatever order I feel like. So don't go assuming I'm avoiding any of your tasks just because I haven't gotten around to them yet.

Right now I've been concentrating on whether my equation for length contraction has been wrong for years. I've found the source in one of Greene's videos where he says x=x'/Y
at 10:36
and in another where x=Yx' just like t=Yt'. For me, (t',x') are the moving rotated coordinates while (t,x) are the stationary cartesian coordinates. v=x/t , v'= x'/t' and if x'=x/Y and t'=t/Y then x'/t' =v' = v. There are no concepts like time contraction (t'= Yt") or length dilation (x=Yx') similar to time dilation (t=Yt') in SR taken from different perspectives as I showed in #116. Dilation and contraction mean the same thing even though two different words are used. Greene was inconsistent with his labels and since I never had to use length contraction I didn't notice I was using the wrong formula x'=Yx. Now why would I hate to be challenged if it points me to mistakes like these. Maybe SR uses a convention that both coordinate systems use x and t axes within their frame but the moving one gets converted to x' and t' axes when viewed from an outside perspective. If this is true then his formula x'=Yx would make sense but I just assigned the moving frame automatically with x' and t' axes. Maybe that's why no one has been understanding my x' and t' notation.

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

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35,782
ralfcis:

I'll show you how length contraction can be derived in SR, from the Lorentz transformation for position coordinates:
$x'=\gamma(x-vt)$

Notice that if we consider two events in spacetime, we can write two transformation equations relating the positions and times of the two events, where the subscripts indicate event 1 or event 2:
$x_1'=\gamma(x_1-vt_1)$
$x_2'=\gamma(x_2-vt_2)$
Subtracting the first equation from the second we have:
$x_2' - x_1'=\gamma[(x_2 - x_1)-v(t_2-t_1)]$

To measure the length of something (a spaceship, say), observers in the two reference frames need to determine the position coordinates at the two ends of the spaceship. The length is then $\Delta x' \equiv x_2' - x_1'$, or $\Delta x \equiv x_2 - x_1$. It can't be both of those, however, for reasons I'll get to in a moment. It follows that:
$\Delta x' = \gamma(\Delta x - v \Delta t)$,
where $\Delta x$ and $\Delta x'$ are distance intervals measured in the two frames, and $\Delta t$ is a time interval (the time difference between two events) in the frame that is using the $(x,t)$ coordinates.

To make this concrete, let's consider a specific example. Assume that Bob is stationary and Alice is moving at velocity $v$ in the $x$ direction. Bob uses coordinates $x$ and $t$, while Alice uses $x'$ and $t'$.

Assume that Alice's spaceship has a length $\Delta x_{12}'$, measured by her using rulers attached to the spaceship. To call this "the length" of her spaceship, Alice must measure the two events (1 and 2) that define the endpoints of the spaceship simultaneously, in her frame of reference. That means that we require $\Delta t_{12}' = 0$, while $\Delta x_{12}'$ is certainly non-zero. Since the length being measured here is stationary relative to Alice's rulers (but not Bob's), $\Delta x_{12}'$ is the rest length of the spaceship (also called its "proper length"), whereas in this case $\Delta x_{12}$ does not correspond to a rest length.

What we want to know now is: how long is Alice's spaceship, as measured in Bob's frame?

Bob is trying to measure the length of the spaceship as it flies past his rulers at speed $v$. But his procedure is the same as Alice's. He has to find the locations of the ends of the ship (i.e. coordinates of two events at the ends of the ship, which I will call $x_3$ and $x_4$), and he has to make sure that he measures those two position coordinates simultaneously in his own frame of reference. That is, Bob requires $\Delta t_{34}=0$ between the two events that define the length of the spaceship.

For Bob, we have the equation he needs to relate $\Delta x_{34}$ to $\Delta x_{34}'$, remembering that $\Delta x_{34}'$ is the rest length of the spaceship. It's this one, from above:
$\Delta x' = \gamma(\Delta x - v \Delta t)$.

Bob just chooses two events at the two ends of the spaceship that occur simultaneously in his frame ($\Delta t_{34}=0$). Then:
$\Delta x_{34}' = \gamma \Delta x_{34}$

This equation tells us that, if Bob measures the length of the spaceship to be $\Delta x_{34}$, Alice measures a different distance $\Delta x_{34}'$ in her frame, between the same events. The Lorentz factor, $\gamma$ is a number greater than 1, so $\Delta x_{34}' > \Delta x_{34}$. In other words, Alice measures a greater distance interval than Bob, between the events that Bob is using to measure the length of the spaceship. However, Alice agrees with Bob that the two events in question were located at the two ends of her spaceship. She can use the same two positions to measure the same distance simultaneously in her frame, so we can identify the distance between events 3 and 4 in Alice's frame, $\Delta x_{34}'$, as the being the same as the "proper length" of her spaceship. Note, however, that Alice will need to use different events to actually measure the length directly, because if she wants to measure the length of her ship she needs to measure both ends simultaneously (as was the case with events 1 and 2).

To summarise, then, we know that $\Delta x'_{34}=\Delta x'_{12}$, and that $\Delta t'_{12}=0$.

Bob measures a certain length for the spaceship, and Alice measured a longer length. The length of Alice's ship is contracted, in Bob's frame.

Let's take a closer look at Bob's events 3 and 4.

Here's the time interval transformation between the event coordinates, which I haven't mentioned yet. This also comes from the Lorentz transformations, in the same way as the distance interval transformation, derived above:
$\Delta t' = \gamma (\Delta t - \frac{v}{c^2}\Delta x)$

We know that events 3 and 4 are used for Bob's measurement of the length of the spaceship, and for those two events $\Delta t_{34}=0$. So:
$\Delta t_{34}' = -\gamma \frac{v}{c^2}\Delta x_{34}$.

Now $\Delta x_{34}$, Bob's measurement of the length of the ship, is certainly not zero. Therefore, we conclude that, for events 3 and 4, which Bob is using to measure the length, $\Delta t_{34}=0$ but $\Delta t_{34}'\ne 0$. In other words, the events Bob is using to measure the length are not simultaneous in Alice's frame of reference.

So what about events 1 and 2, which Alice is using to measure the rest length of her ship? For those two events, we have $\Delta t_{12}'=0$ (Alice's frame). Then, using the equation above again, we have, for those two events:
$\Delta t_{12}' = \gamma (\Delta t_{12} - \frac{v}{c^2}\Delta x_{12})=0$

which can be re-arranged to find $\Delta t_{12}$:
$\Delta t_{12} = +\frac{v}{c^2}\Delta x_{12}$

We can plug that expression back into:
$\Delta x' = \gamma(\Delta x - v \Delta t)$.

to find a relationship between $\Delta x_{12}'$ and $\Delta x_{12}$, once again:
$\Delta x_{12}' = \gamma(\Delta x_{12} - v \Delta t_{12})$
$\Delta x_{12}' = \gamma(\Delta x_{12} - \frac{v^2}{c^2}\Delta x_{12})$
$\Delta x_{12}' = \gamma(1 - (\frac{v}{c})^2)\Delta x_{12}$
$\Delta x_{12}' = \gamma \frac{1}{\gamma^2}\Delta x_{12}$
$\Delta x_{12}' = \frac{1}{\gamma}\Delta x_{12}$

Again, remembering that $\Delta x_{12}'$ is the rest length of the spaceship, as measured by Alice, we see that $\Delta x_{12}$ (the distance between events 1 and 2, as measured by Bob) is longer than the length Alice measures.

You might then object: "But doesn't that mean that Bob would say that Alice's ship has undergone 'length expansion' rather than 'length contraction'?" The answer is: no, it doesn't mean that. The distance interval between events 1 and 2, in Bob's frame, is not a length. Bob requires that the endpoints of lengths in his frame be measured simultaneously in his frame, which requires $\Delta t=0$. Clearly, we don't have $\Delta t=0$ for events 1 and 2 (see the expression for $\Delta t_{12}$, above, in terms of $\Delta x_{12}$, both of which are non-zero. Between the times that events 1 and 2 occur, in Bob's frame, Alice's spaceship moves a certain distance, in Bob's frame, so that the distance between those two events ($\Delta x_{12}$) is not the length of the spaceship, in Bob's frame.

What all of this means is that Bob and Alice can't use the same two events to measure the "length" of Alice's spaceship in their respective frames. Bob needs to choose two events that have $\Delta t=0$, while Alice needs to choose two events that have $\Delta t'=0$. At least one of the two events that Alice uses to measure the length must be different from either of the events Bob is using to determine the length of the ship.

But we have derived the length contraction relationship of SR unambiguously. It is:

$\Delta x_{34}' = \gamma \Delta x_{34}$

Recalling that $\Delta x_{34}' = \Delta x_{12}'$, which is the rest length, we see that the apparent length in a frame that is not measuring the rest length is contracted, compared to the rest length. So, we can drop the subscripts and just write:

$L = L_{rest}/\gamma$

where $L_{rest}$ is the rest length and $L$ is the measured length (of any object) in some "other" frame in which the measured distance is not at rest in the coordinate system.

In using this formula, we need to be very careful about which frame is measuring the rest length of a particular object (or distance), and which is the "other" frame. That will vary depending on which object(s) we're measuring the lengths of, and their states of motion. Note, also, that in some situations, neither of the two frames of interest will measure the rest length of a given object, in which case we might need to look at the more general Lorentz transformations.

Last edited: Feb 28, 2021

5. ### ralfcisRegistered Senior Member

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375
I choose the easiest frame, the one that actually reflects real physically possible scenarios, to be my stationary reference frame. I assign it (t,x) coordinates. I don't use another frame, I have a velocity line with units that are labelled with t'=Yt units. The 1/slope of that line can be expressed as v=x/t or Yv=x/t' or $v_t=t'/t$. So the coordinates of that line can be expressed as (t,x), (t',x) or (t',t). There is no x' axis so there is no slope or coordinate relationship to x'. This is nothing like how SR handles coordinates.

I was under the impression that SR had another frame that used (t',x') coordinates and they were Alice's dilated/contracted coordinates from Bob's perspective. From Alice's perspective of Bob those coordinates became (t",x"). (For example if Bob's t=5, Alice's corresponding t'=4 and she saw Bob's t"=3.2 from her perspective and if Bob's perspective of Alice's x was 3, her perspective of that x=3 was x'=2.4 and her perspective of Bob's perspective of that x'=2.4 was x"=1.6.) But no, the ' notation has nothing to do with contraction/dilation, it has only to do with who's the stationary perspective. The formula for length contraction from Bob's perspective is x=x'/Y and from Alice's perspective is x=Yx' because the context of x has changed. I find that extremely confusing as the formula for x' is always the same if I always associate the ' notation with the contraction/dilation of the moving frame. It's lucky for me that I never have to switch contexts or worry about the formula for x' but I do finally see my ' notation is nothing like SR's and has caused years of confusion.

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

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35,782
ralfcis:

All frames reflect real physically possible scenarios. There are no preferred reference frames in relativity. We know exactly how to translate events from one frame to any other frame where we need them.

Also, note that all frames agree that what happens in spacetime happens in spacetime. It is impossible for an event to happen in one frame and not happen in some other frame. It can happen at different coordinates, of course - at a different time coordinate and/or space coordinate - but everybody agrees that what happens happens. In the Minkowski picture, events are fixed points in spacetime.

Your velocity line is just a world line in the x,t coordinates, as I have explained. You label it with the time coordinates t' in the frame in which the object or person to whom that world line applies remains at rest.

You're using two frames, even if you don't recognise it as such.

The coordinates of an event must have both spatial components (usually 3, but here we're just considering the direction of relative motion, x) and a time component. So (t', t) does not specify an event in spacetime (it's missing a spatial coordinate). As for (t',x), that is a mixed set of coordinates from two different frames: Alice's clocks and Bob's rulers. That's not usually what we're interested in, in practice. As you sit in your chair there, you don't measure time using some clock on a distant planet speeding away from Earth at close to the speed of light; you use clocks that are stationary on Earth. You also use the rulers you have at hand.

And, as usual, you have a blind spot to the coordinate x', probably because it hasn't occurred to you that it can have any value other than zero.

But you're wrong when you say your way of handling coordinates is "nothing like" how SR does it. You borrow directly from SR. You use SR's Lorentz factor to convert between t and t' coordinates. You use SR's Minkowski diagrams (only you apparently call them "Loedel diagrams"). You use SR's invariant spacetime parameter, without calling it that, in the special case where x'=0. Mostly, all this borrowing is the core of what you call "your" theory.

I'm not even sure you have any theoretical basis in your version of relativity for deriving the Lorentz factor $\gamma$ (gamma), or Y as you call it. So far, you've been unable to specify any actual postulates of your "alternative theory", and I haven't seen any derivations of any of the equations you use, from first principles.

That's a hopeless muddled description. Bob and Alice only need (x,t) and (x',t') coordinate frames to translate the coordinates of spacetime events between their frames. They don't need a third or fourth set of coordinates.

It should be clear to you from reading my post about length contraction that whether a given observer sees a greater or lesser distance between a pair of events in spacetime depends on whether those events are simultaneous in one of the two frames, and if so, in which of the two frames they are simultaneous. The coordinate of some event that has t'=4 in Alice's frame might correspond to t=5 or t=3.2 or some completely different value of t, in Bob's frame, depending on the spatial coordinate of that event in Alice's frame.

Remember: the length contraction and time dilation formulas are special cases of the more general Lorentz transformations. They only apply when one of the two frames is measuring a rest length or a proper time interval. It is possible - indeed it is often the case - that neither of the two frames is measuring a rest length or a proper time interval between the two events of interest, in which case there is no option but to use the full Lorentz transformations.

You're not alone in finding it confusing. Lots of people reach peak confusion not when they start learning about relativity, but when they've learned a certain amount of relativity, so that they're starting to come to grips with it but not fully across it yet. If they give up at that point, then they can be forever lost. You have to push through the peak confusion phase, pay careful attention to definitions and other things, and gradually you start to really understand relativity. It takes effort.

One thing to notice straight away is that, at constant relative velocity, the "famous" effects of relativity are reciprocal. Bob sees Alice's clocks as running slower than his (i.e. Alice's clocks "dilate" from Bob's point of view), but Alice sees Bob's clocks as running slower than hers (i.e. Bob's clocks "dilate" from Alice's point of view). Bob sees Alice's length contract; Alice sees Bob's length contract.

This means you can't blindly write down a formula like $x'=\gamma x$ and pretend that's the formula for length contraction, where $x$ is always the "stationary" frame and $x'$ is always the "moving frame". For starters, you need to realise that you shouldn't even strictly be writing $x$ or $x'$ when you're talking about a length or distance, because any length always has two endpoints, which are two events in spacetime, which means you're dealing with two position coordinates, not one. Strictly, you should be writing $\Delta x' = \gamma \Delta x$, which I was careful to do when I explained length contraction in my earlier post. But even then, you're not done, until you've identified whether $\Delta x$ and $\Delta x'$ are lengths at all, in their respective frames and, if so, which one is a rest length.

8. ### ralfcisRegistered Senior Member

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375
I'm just going to ignore what you've written because this thread is not about SR. I don't think you've understood 1 thing I've said but now I understand I misinterpreted what the ' means in SR thanks to you. In my math the formula is x=Yx' not x'=Yx (not that I use either) because x is for the cartesian coordinates unless, of course, I want to look at the cartesian coordinates from the other perspective then the formula is x'=Yx". The ' in my math stands for how many times Y is used. So x=Yx'=YYx". Glad we got this cleared up.

Anyways I consider perspectives as illusions whereas SR considers them the only reality except when clocks are co-located and perspective disappears. So to change all SR's formulas to change with perspective seems very error prone if you're not constantly mindful of the perspective and which form of the formula to apply. No thanks, my way is consistent and blind to all this and I only work with proper distance and proper time from which you can calculate any perspective you want. What is horrible is that it took me so many years to find out about this bad choice (among many from my perspective) in SR.

Last edited: Feb 28, 2021
9. ### James RJust this guy, you know?Staff Member

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35,782
That's a mistake, for three reasons:
1. Your own theory is mostly borrowed from SR.
2. If you understood SR better than you currently do, you'd also be better equipped to challenge your own ideas, which is what good scientists always try to do, first.
3. By refusing to learn anything new, you're consigning yourself to irrelevancy, in terms of being a credible academic source. You just make yourself look like a crank, and you're probably smart enough that you don't have to put yourself in that box.
Still, I'll take it into account that I'm wasting my time with you, and scale back my postings in response. I have other things to do with my time, if you're not interested in what I have to say.

I am confident I have a very good understanding of your ideas, by now. The main problem is that you don't really understand what the equations you are using are actually telling you about the physics of the situations you are examining. You also seem to have a philosophical objection, in that you won't commit to the numerical values that come out of the equations being "real"; instead you seem to want to pretend that they are some kind of illusion, in all but one "preferred" frame.

The notation I have been using just uses (x,t) for one reference frame and (x',t') for the other. The primes (') are just to distinguish the coordinates in one from the coordinates in the other. I assumed you would already be familiar with this standard notation.

Regarding your x=Yx' versus x'=Yx thing: I just wrote a whole post explaining that it's not one or the other: it depends on what the x and x' represent, in the two frames. I also cautioned you about the difference between events and lengths. In a previous, lengthy post, I derived length contraction from scratch, and discussed the probable source of your confusion about which way around the formula goes (it can go either way).

It is now obvious that either you didn't bother reading my posts, or it all flew over your head, or you were determined not to learn anything regardless of what I wrote, or some combination of those. So, I'm wasting my time on you.

I have already addressed that false claim several times in this thread. Now that it is clear you are simply ignoring what I tell you, I see I am wasting my time.

The formulas don't change. You just have to remember that no one frame is preferred above any other.

It's easy to make errors in relativity if you aren't careful about frames and events, or if you don't really understand the theory properly. There's nothing error-prone about the theory itself. It is thoroughly tested and confirmed.

Clearly there are a lot of things you're blind to.

Are we done here, then?

10. ### ralfcisRegistered Senior Member

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375
Yes we've come to an agreement I prefer my math and will continue showing what it can do. We've also come to an agreement I didn't understand SR's ' notation and was using the wrong formula for length contraction as a result. I only ignore the parts of your posts I already thoroughly understand about SR which allowed me to see a part I was wrong about. That part does not mean I don't understand any of it and it's irrelevant to my math anyway as I don't need length contraction.

11. ### ralfcisRegistered Senior Member

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375
Ok, I wanted to avoid any lengthy discussions on the math behind SR but I'm going to have to present how coordinate frame rotation is done in SR and why I don't use any frame rotation despite what James seems to think. I'll also show how SR arrives at the reverse-Minkowski diagram for Alice's round trip flight at .6c and why I use non-rotated time units and no length contraction in my reverse-Minkowski (where Alice switches reference frames with Bob). I actually don't do reverse_Md's, I can switch my perspectives on a single Md if I needed to but don't. SR absolutely needs frame switching because its reality is based on perspective, mine is not. Happy reading everyone.

12. ### ralfcisRegistered Senior Member

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It all starts with two blank sheets of Cartesian graph paper that will be placed on top of each other and rotated. Each will have a vertical axis labelled ct and a horizontal axis labelled x . The rotated sheet will have the labels ct' and x' to differentiate the sheets. At no time will either square Cartesian coordinates be crushed sideways into any Minkowski rhombic coordinates. However as one sheet is rotated joined at the origin to the other, it's Cartesian squares will get larger and larger until they get infinitely large once a 45 degree rotation (signifying light speed) is achieved.

This frame rotation and you can see Alice's red c line for her frame and Bob's blue c line for his frame. They don't overlap meaning c is not the same from perspectives other than the frame in which it exists. So how did Minkowski make this frame rotation conform to Einstein's SR philosophy? Most people believe he didn't do a frame rotation, he did 2 separate axis rotations of ct' from ct and x' to a place that was symetrical around Bob's c line such that Alice's c line overlapped it. Notice the x' axis rotation was not even symmetrical with the ct' axis rotation. Also notice and axis rotation is not the same thing as a frame rotation despite what everyone seems to think.

Maybe this is legal in math, but the more likely way Minkowski achieved his goal of manipulating Einstein's philosophy into his diagrams was that he flipped the x' axis over the x axis (shown as the dotted blue line in the first diagram) after the frame rotation. This is perfectly legal mathematically as a mirror image of the x' axis is the same as the original x' axis.

An axis rotation has no mathematical obligation to have the unit spacing expand on the rotating line whereas a frame rotation has the squares expand as one frame rotates with respect to the other. The length units have no choice but to expand with the time units so length dilation is tied to time dilation and is not a separate physical phenomenon. I only use axis rotation and it allows me freedom in my reverse-Md's as I'll show now.

Here is a combined Md with it's reverse-Md in SR. The units of the reverse Md are tied to the original Md and Alice's on-board clock in the reverse-Md becomes the standard for the universe replacing Bob's clock. This is not true in my reverse-Md's. I can keep Bob's clock as the standard for both perspectives which is not allowed in SR.

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13. ### ralfcisRegistered Senior Member

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reverse-Md/Md using SR frame rotation

reverse Md/Md using axis rotation

Notice the space yrs are the same as the Bob/Earth yrs. Alice's yrs in the reverse-Md side are Bob/Earth yrs * Y and there is no difference in distance travelled between reverse and regular Md so there is no length contraction as in the SR version of this diagram.

Space yrs also change in duration in the SR version between the two perspectives. Space yrs are the clock of the empty space frame in the reverse-Md where Alice remains for 4 yrs after Earth takes off from her. In SR, space yrs = Alice's proper time which is the universal perspective clock rate. But in my reverse-Md, Bob/Earth years remain the universal perspective clock whether Alice or Bob are in charge of the reference frame. I've also included the lines of perspective simultaneity in both diagrams so you can see they make far more sense in my axis rotation version.

Last edited: Mar 2, 2021
14. ### ralfcisRegistered Senior Member

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#227

All my latest posts are in response to #227.
This is mostly hilarious, "but for some reason I don't like drawing Alice's space axis." I also don't like to include poodles in my doodles because they are not needed to solve any problem. Yes, SR is absolutely dependent on having the concept of physical length contraction for its math to work according to its philosophical assumptions. Most importantly, both length contraction and considering both perspectives is absolutely critical for SR to explain the twin paradox. SR is very big on the reciprocity of both length contraction and time dilation so explaining how the twin paradox does not fit in with that reciprocity is an even bigger deal. It calls on the hocus pocus of reciprocity exists only during constant relative velocity hence that must lead to the conclusion that non-constant relative velocity must be the cause of twin paradox permanent time difference. The only problem is the twin paradox exists without any acceleration in the clock handoff scenario.

The reciprocity is not confined to time dilation Alice vs time dilation Bob or length contraction Alice vs length contraction Bob but most importantly time dilation Alice vs length contraction Bob. In the twin paradox, Alice ages permanently less than Bob but in return Alice has permanently shrunk the record of how far she has covered Bob's distance. Hence she may have aged 2 yrs less than Bob on the journey but in order to do so she must have shrunk the 3 ly distance Bob has measured to only 2.4ly she must have travelled in order for SR to work. The proof of the time difference is the reading on the co-located clocks. Co-located means when Bob and Alice re-unite or pass by each other on Earth, the proper times on the clocks can be compared in the same space if only for an instant. They do not have to be in the same inertial frame which is irrelevant to solving the twin paradox.

Now the problem is there is no way to have a ruler of Alice's entire path and compare it co-located with the distance Bob says she travelled. You can't even take a chunk of that empty space and compare it to Bob's chunk of empty space. There is no way to tell that Alice's path remains shorter once the velocity disappears but there is a way to tell she permanently aged less than Bob and that info does not disappear once the velocity goes away. So if there's no way to prove it, why would I include length contraction in any of my analysis or problem solving? Why would I keep the need of solving the twin paradox having to consider both perspectives. These are added complications that do not make the theory as simple as possible which is enough reason to throw out the entire theory.

Ok, now that I have the correct formula for length contraction straightened out, I can re-write the main equation from Alice's perspective with Bob in the stationary frame as

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

et voila I have re-shifted length contraction back as a time only effect. More precisely since I do not set Alice as a stationary frame but as a moving reference frame from the ct' axis, the formula becomes

$(Yct")^2 = (Yct')^2 - x^2$.

For example when ct= 5 from Bob's perspective, ct' = 4 from Alice's. Now we want to see Alice's reverse perspective of Bob. So in the 2nd equation Y=5/4, ct'=4, ct"=3.2 and x=3. So plugging those values into the 2nd equation and everything works out without any length contraction.

Enough for now, I've spent my whole day posting.

15. ### ralfcisRegistered Senior Member

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They go away because they don't physically exist.

clock hand off scenario

Take a look, this is what SR actually says.

After turnaround to re-unification she would see Bob's clocks at double rate of hers due to DSR but due to time dilation she would calculate (not see) them to be running at .8 of her normal rate.

AA

You have no idea what I said, you're just providing irrelevant rebuttals.

The muon experiment yields no permanent time difference between the clocks at co-location like the twin paradox does. We're talking past each other with you having no translator of what I'm saying.

16. ### ralfcisRegistered Senior Member

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#235 on page 12 coming up next then I'll continue on the questions from page 7. It looks like it'll take some time to catch up with the present but I want to be tediously thorough.

17. ### ralfcisRegistered Senior Member

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#235
His course is my only source of knowledge on standard SR, I'm just seeing if your knowledge aligns with that. Generally people just parrot SR and have no deep understanding of it. The ones I've met who do have a deep understanding, there's no confusion between us on concepts (but terminology is another matter).

I posted 2 links to his views on past present and future. If this is what SR believes then I think it's insanity.

You'd have to be able to understand my objections before you could determine if I understand SR or not. I don't think there's any way you could coherently repeat my argument back to me.

Yes you definitely caught me there so you win that point.

I'll go back and read but 10 times sounds like a pretty big exaggeration.

Yes finally but you're acting like you explained this to me when it was I who put the time in to explain it to you.

I asked for the permanent age difference between them after the period of velocity imbalance ends. If that wasn't clear I hope it is now. That's a specific time when Bob receives the news of Alice's last change of velocity. The answer from SR is that the age difference is indeterminate because the clocks do not re-unite at any time defining a valid spacetime path. Yet they also do not co-locate if Alice stops 1ly away from Bob and you seem to think that permanent age difference is calculable because she is in the same frame as Bob and light signal messaging is sufficient to establish permanent age difference between the two ignoring the fact that there is no universal agreement from other perspectives about their age difference. This answer you will not find in basic courses on SR. If you don't know this then it really makes me question what kind of training you received in SR.

Last edited: Mar 3, 2021
18. ### Neddy BateValued Senior Member

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2,127
It is not surprising that trying to do SR 'your own way' has made you so confused. You cannot even see the similarity between the cosmic muon experiment and the twin scenario. Let me summarise:

One stay-home muon sitting on the surface of the earth lives about 2.2 microseconds and then decays. Then another stay-home muon sitting on the surface of the earth lives about 2.2 microseconds and then decays. Repeat this for 10 consecutive stay-home muons living and decaying in a row, and you have 22 microseconds total. During that time, one cosmic muon traveling at 0.995c relative to the earth (gamma=10) lives for 22 microseconds earth time before decaying. In its own frame, the cosmic muon only lived 2.2 microseconds, as normal.

Likewise, one stay-home twin (human being) sitting on the surface of the earth lives about 90 years then dies. Then another stay-home human being sitting on the surface of the earth lives about 90 years then dies. Repeat this for 10 consecutive stay-home humans living and dying in a row, and you have 900 years total. During that time, one cosmic twin traveling at 0.995c relative to the earth (gamma=10) lives for 900 years earth time before dying. In their own frame, the cosmic traveling twin only lived 90 years, as normal.

To make the above into the exact twin scenario, let the traveling twin leave at birth and travel out and back at 0.995c to arrive back home at biological age 90 just before he dies, so that they can visit the 810 year old grave of their dead 90 year old sibling (for a total of 900 earth years).

19. ### ralfcisRegistered Senior Member

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375
Time dilation is not the same thing as the twin paradox. 99.9999% of people don't know this. Time dilation is reciprocal, the twin paradox is not. The twin paradox has nothing to do with time dilation. The muon experiment is 100% reciprocal time dilation.

20. ### Neddy BateValued Senior Member

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2,127
There is no twin "paradox". I assume you mean time dilation is not the same thing as differential aging, which is true.

If the cosmic muon that I described above were to decelerate to be at rest with the earth just a short moment before it decays, then we would have a formerly-cosmic muon with an age of 2.2 microseconds sitting right next to 10 already consecutively-decayed earth-muons (for a total of 22 microseconds). In the earth frame, the creation of first of the 10 already consecutively-decayed earth-muons would have been simultaneous with the creation of the cosmic muon.

I know that you don't recognise Einstein-synchronised clocks in your 'way' of doing 'SR', but Einstein-synchronised clocks are a valid simultaneity convention in SR, and using them, we would be able to say that those two muon-creation events were simultaneous in the earth frame, whether you like it or not.

Likewise, if the traveling twin that I described above were to decelerate to be at rest with the earth just a short moment before they died, then we would have a formerly-traveling twin with an age of 90 years visiting the the 810 year old grave of their dead 90 year old sibling (for a total of 900 earth years). This is almost perfectly analogous, except for the traveling twin doing a turnaround in the middle of the trip. But of course that could be remedied by using the simple one-way twin scenario that I gave you in post #260, in which case it would be perfectly analogous.

Last edited: Mar 3, 2021
21. ### ralfcisRegistered Senior Member

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375

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22. ### Q-reeusValued Senior Member

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4,184
The muons both decay in flight and some undecayed ones finish up terminating in a detector. Both of those factors make the situation nonreciprocal but it's not immediately obvious why.
In #237 analysis of circular relative motion is death to your version of relativity. You probably never bothered to look up the article linked to there. I excerpt again here:

"And as per p684 in http://www.lightandmatter.com/lmn.pdf, there has long been adequate experimental proof of nonreciprocal time dilation for physical objects in circular motion."

Conveniently - muons in that experiment! This time check that lead out. Your claim time dilation is purely owing to nonsimultaneity fails totally there, and in the scenario I gave in #237. Stop and think it through for once. Circular motion is the key. But you must be prepared to use it to unlock your current thinking.

23. ### ralfcisRegistered Senior Member

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375
page 7 #128

Loedel diagrams are not Minkowski diagrams at all. Minkowski have a stationary frame and a moving frame that apply to the two participants Bob and Alice. When Alice takes over the stationary frame, the reverse Md doesn't look like the Md when Bob is in control. As I explained the Loedel diagram tries to depict true relative velocity in that both participants are moving away from each other equally. The reverse Loedel is the same as the Loedel. I did not make this up as you keep implying. Loedel diagrams are part of relativity.

The Loedel lines of simultaneity come from the half speed perspective between Bob and Alice and just like Bob and Alice's lines of simultaneity, all points on the Loedel lines have the same time except when they intersect another velocity line. The Loedel lines happen to intersect the same proper times on the Bob and Alice velocity lines. This is to depict how there is no proper time age difference between Bob and Alice during constant relative velocity. Bob and Alice's lines of simultaneity have a perspective hysteresis around the Loedel line. I don't know why I have to keep repeating the same answers over and over. You seem to be waiting for me to trip up and give you a different answer.

No. Proper time observed within a frame is independent of outside perspective. The Loedel lines of simultaneity are not the same thing as a line of proper simultaneity because proper time is independent of outside perspective. The Loedel lines are still a perspective simultaneity that gives us a peek at what the proper times are to tell us of any proper time age difference between the clocks. You just refuse to read what I'm saying and keep substituting your own words here.

No. Proper time ticks at the same rate of c within every frame. It's not an illusion, it's actually the principle of relativity (look it up). The Earth observer's observations of the participant's proper times have no relevance to him. The light signals sent out during those proper times and received at the other proper times is what's important to the participants establishing any proper age difference between the two of them.

You can't be serious. That is Bob and Alice's perspective simultaneity once again not the Loedel perspective simultaneity that allows us to peek at the proper times of Bob and Alice which are the same during constant relative velocity. I have no way to make this any clearer. Bob and Alice pre-agree to send their light signals when their clocks hit 2. They don't know what the other clock's time is in real time or even perspective time.

Last edited: Mar 4, 2021