# Relativity and simple algebra II

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

1. ### ralfcisRegistered Senior Member

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In keeping with my layering of several spacetime diagrams into one, the ct' axis is a line that represents 3 velocities and 3 different slopes.

It turns out v_t overlaps the same line as v and v' and the slope of each overlapping line is dependent on what coordinates you assign to the ct' axis.

Let's look at the point labelled ct'=1.

The coordinates of the velocity line at this point are (1.25,.75) where v=x/ct =.6c and the slope of the line is 1/v = 5/3. The velocity is x-axis/ct-axis.

The coordinates of the v'=Yv line are (1,.75) where v'=x/ct' = .75c and the slope of the line is 1/v' = 4/3. v' is the x-axis/ct'-axis.

The coordinates of the v_t=ct'/ct = c/Y line are (1.25,1) where v_t=ct'/ct = .8 and the slope of the line is 1/v_t = 5/4. v_t is the ct'-axis/ct-axis.

Normal algebra would not have labelled that point as 1, it would have only labelled it with the cartesian coordinates of (1.25,.75). I'm adding a 3rd ct'-axis which not only allows all facets of the main equation to be seen on the diagram but it allows both perspective views to be seen on one diagram instead of 2. It also avoids the mistakes relativity makes when switching from one perspective to the other as I'll discuss in the next post.

But let's go back to basic normal algebra for now. The simplest equation is y=x (which comes from the equation for a line y=mx+y_0 ). This equation would define the speed of light c on a spacetime diagram. c has no coordinates in relativity because c is actually made up of overlapping lines due to a cheat Minkowski added to his diagrams to make c appear the same from all perspectives (which it is anyway). The cheat he used was to take a normal frame rotation and flip the x'-axis up over the x-axis to make it look like the ct'-axis and x'-axis were rotated independently towards each other. Mathematically perfectly acceptable in order to force all c lines from every frame to overlap each other and have the same 45 degree angle on the Minkowski diagram. Epstein could have done the same thing on his diagrams except all frames would have the slope of c at 0 degrees. Epstein diagrams reveal a lot about the underlying mathematical constructs of relativity that do not necessarily follow the philosophical conclusions the Minkowski diagrams advocate.

In algebra, the equation y=1/x defines a hyperbola. The main equation of relativity creates straight lines and hyperbolas shifted 45 degrees from the simple equations of basic algebra. (This is not set in stone as the Epstein sum of squares version of the main equation uses circles that do not shift the velocity lines at 45 degrees or have an overlapping c line at 45 degrees.) The hyperbolas in relativity intersect the velocity lines at the same proper time. These hyperbolas connect a true, unseen, universal present at a distance and are not to be confused with the fake perspective present that Einstein invented in his clock sync method. In "ralfativity", I replace the hyperbolic lines with straight lines of 1/v_h lines of proper simultaneity of the Loedel half speed perspective (my own term).

Ralfativity only cares about the slopes of key velocity lines (v, v', v_t ,v_h ,v_ht, ) and their reciprocals which are lines of simultaneity. It only considers velocities separated into their coordinates of time and space and perspectives at the end. All clocks beat at the same universal rate within a frame and while earth may measure a ship going at .6c using earth clocks, the guy in the ship measures his v' speed at .75c using his own clock. The lines of perspective simultaneity (also called in relativity as now slices and x'-axes), arising out of Einstein's clock sync method, are illusory curiosities. They do not define a perspective present because the distance between points negates the very definition of a shared present.

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

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ralfcis:

Are you planning on responding to any of my posts, above - especially #58?

Okay. Could you please present the main postulates (assumptions) of the theory you have constructed?

Here are the ones from special relativity:
1. The laws of physics take the same forms in all inertial reference frames.
2. The speed of light in a vacuum is the same in all inertial reference frames.

Those two postulates are sufficient to derive all of special relativity, including things like the Lorentz factor; the Lorentz transformations; time dilation; length contraction; the relativity of simultaneity; relativistic mass, momentum and energy conservation; mass-energy equivalence, and more.

What are your core physical assumptions, which replace or alter the ones from special relativity?

Your ct' axis is a rotated axis, despite the fact that you claim your framework has no rotated axes.

What about an x' axis? How does an observer measure distances in the "primed" frame, in your system? Do effects like length contraction exist in your framework, or not? If not, how is the constancy of the speed of light maintained between frames (assuming that is an axiom)?

Not really. It is clear from the usual spacetime metric in flat space (which is special relativity) that the time coordinate/dimension is not on the same footing as space. On the other hand, the Lorentz transformations do transform time and spatial coordinates as a unit, which is why relativists speak of "spacetime".

I don't understand. When you talk about the "moving observer" seeing a velocity, which velocity are you talking about? Not the moving observer's own velocity, surely, because the moving observer is always stationary relative to himself. I guess you're talking about the moving observer's measurement of the "stationary" observer's relative velocity, then.

If the two observers are moving at constant velocities at all times (or one is stationary and the other is moving at constant non-zero velocity), how could it be the case that each one measures the other's speed to be different?

If you're standing still on a road and I drive my car past you at 60 km/hr, then according to me I'm stationary in my car and you are moving backwards relative to my car at 60 km/hr.

I don't see how it could ever be the case that, as long as we're dealing with constant relative velocities, you see me driving at 60 km/hr, but I see you moving backwards at 75 km/hr, for instance.

So you're trying to replace relativity by an alternative theory in which distances are absolute, but time changes between frames, which makes the speed of light variable between frames.

Is that correct?

Would one of your postulates be "distances are absolute", then?

I don't think you understand the special theory of relativity very well. I agree with you that time does not behave like distance. That does not mean, however, that a concept like "spacetime" is meaningless. On the contrary, the Lorentz transformations show that such a concept is required.

Okay, but I'd prefer it if you could respond at least to post #58, first.

We're having a discussion here, aren't we? Or are you treating this more like a blog?

Your own diagrams use different time scales on the t and t' axes. Why is that?

We can avoid that confusion by talking about events in spacetime, as I said previously.

I don't really know what you mean. Maybe we can unpack that claim later.

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5. ### ralfcisRegistered Senior Member

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absolutely all of them but i figured you need more background. more is coming.

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

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ralfcis:

I'd also like to ask you a specific question. Consider the case of one observer (B) with (1 dimensional) space and time coordinates x and t, who is stationary in the x coordinate, and another "moving" observer (A) who moves with speed +v in the positive x direction in B's coordinates. Let A have coordinates x' and t', and let A be stationary in the x' coordinate. Assume that at time t=t'=0, the coordinate x=0 is coincident with x'=0.

Here's how Galileo or Newton would have defined the transformations between A and B's reference frames:

$x' = x - vt$
$t' = t$.

And here's how Einstein and Lorentz say the coordinates transform between the same two frames:

$x' = \frac{1}{\sqrt{1-(v/c)^2)}}\left( x - vt \right)$
$t' = \frac{1}{\sqrt{1-(v/c)^2)}}\left ( t - \frac{v}{c^2}x \right)$.

Could you please give the coordinate transformation equations that are used in your theory?

Here's my guess at them, based on what you have written so far:

$x' = x - vt$
$t' = \sqrt{t^2 - \left(\frac{x}{c}\right)^2}$

Note that the second equation can also be written as
$(ct')^2 = (ct)^2 - x^2$.

Are these the correct coordinate transformations between the two frames, in your system? If not, please provide the correct ones.

Last edited: Feb 13, 2021
8. ### James RJust this guy, you know?Staff Member

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If I need something, I'll ask for it.

Could you please respond to my posts? Maybe start with post #58, although a reply to post #64 would also be very helpful.

9. ### ralfcisRegistered Senior Member

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421
I will but not on your schedule, you need more background first because your questions show no understanding of what I'm saying because I have not yet given you a primarily adequate description of the elephant.

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

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Out of interest, assuming that the ralfcis transformation equations I have given above are correct, I have looked at a light signal propagating outwards from the origin. Consider a signal that is sent from x=0 at t=0, and received at distance x=c at time t=1, for instance. The speed of the signal is calculated as x/t, which in this case is c, of course.

What happens if we transform to a different frame, in which an observer moving at speed v relativity to the x coordinate?

The coordinates of the "transmission" event in the new frame are (x',t')=(0,0), no matter which set of transformations we use from post #64. The coordinates of the "receipt" event are:

Newtonian relativity

$x'=c-v,\qquad t'=1$

The calculated speed of light in the 'primed' frame is $x'/t'=c-v$. That makes sense, since in the Newtonian universe the "moving" observer must subtract his own speed from the speed of light in order to find the speed of light relative to him. The speed of light is not the same in all frames of reference in the Newtonian picture.

Einsteinian special relativity

$x'=\frac{1}{\sqrt{1-(v/c)^2)}}(c-v), \qquad t'= \frac{1}{\sqrt{1-(v/c)^2)}}(1-v/c)$

The calculated speed of light in the 'primed' frame is $x'/t' = \frac{c-v}{1-v/c}=\frac{c-v}{\frac{c-v}{c}}=c$. That is, the "moving observer sees light as still travelling at 'c', which is consistent with the postulate that the speed of light is the same in all frames.

ralfcis relativity

$x'=c-v, \qquad t'=\sqrt{1-(c/c)^2}=\sqrt{1-1}= 0$

The calculated speed of light in the 'primed frame' is $x'/t' = \frac{c-v}{0} = \infty$.

It appears, that in ralfcis relativity, the speed of light is not the same in all frames. In this example it is actually infinite in the 'primed' frame.

More generally, if we consider propagation of the light from (x,t)=(0,0) for a general time, so that it reaches coordinates (ct,t) after time t, then ralfcis relativity says we would have, in the "moving" frame of the observer with velocity v:

$x'=ct-vt=(c-v)t, \qquad t'=\sqrt{t-(ct/c)}=0$

which would put the speed of light in the 'primed' frame at $x'/t'=\infty$.

So it appears that, if the transformations I have given are correct, ralfcis relativity has the speed of light as infinite in every frame except for one "special" frame.

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

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Okay. No problem. Maybe your "background" posts will answer some of my questions.

Bear in mind, though, that I'll most likely try to reply to you as your post, rather than letting a lot of new material bank up, so it could save time if you answered my specific queries first, because it would prevent some of the misunderstandings you say I have from persisting in future posts of mine.

12. ### ralfcisRegistered Senior Member

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421
Just glancing at your last post. If v=x/t is limited to c is v'=Yv=x/t' also limited to c?

It's going to take a considerable amount of time to answer your questions because it's almost as if you've mentally redacted the parts you don't understand and substituted your own reality instead. That's ok as most people try to understand things in the context of what they already understand. I appreciate your fervent desire to ask questions and gain understanding. No one has ever tried before.

So here we go, the last bit of background before I continue answering your questions in order.

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This is the basic graphical unit. It can be flipped, mirrored, paralleled and joined with other units using light signals and the Loedel lines of simultaneity. SR uses the spacetime path which has its own rules and can't be constructed from basic units or light signals.

In the basic unit, time dilation is explained without Einstein's clock sync method, without lines of perspective simultaneity, without length contraction, without perspective relativity of simultaneity, without time slowing, all excess baggage. The perspectives of the participants are derived from the Loedel perspective where proper time trumps perspective time. The flipped version is the muon experiment. The basic unit joined to the flipped version is the twin paradox. Two basic units joined in parallel is the pole in the barn paradox, the train in the station example, the Michelson Morley experiment. That just about covers all of Special Relativity except E=mc2.

So consider this Md is depicting a time trial race. The starting judge Bob fires a starting gun at the origin where he is co-located with the racer Alice. They each have the same atomic stop watch that flashes a light at regular intervals (regardless of anything else you might be thinking affects that interval). When the racer crosses the finish line 1.5ly away marked by a planet on the star charts, she will fire off a light signal. It this case it happens to correspond to her two year interval signal because her v= 3/5c relative to the judge (from the judge's perspective). (Or using her own on-board clock, her v'=3/4c because x=1.5 and her watch says 2 yrs have passed for her.) From the racer's perspective the judge is moving away from her at v=3/5c. But the judge has not moved anywhere from the starting line so his x=0 (he has only moved through time at v_t=c). He knows the racer has been moving away from him at 3/5c because the DSR he received showed the racer's clock was ticking at half the speed of his own.

The judge has only received the TV broadcast from the racer to determine the DSR but has not yet received the yellow finish line light signal. He will wait 4 yrs from the start of the race to receive that signal. The signal will have an encoded message," I have reached the finish line at the 1.5ly mark in two ticks of my atomic stopwatch". The judge knows the light signal took 1.5 of his years to reach him (inaccurate statement). So his time from his perspective for the end of the race is 4-1.5=2.5yrs. Time dilation is that the racer did it in 2 yrs. Notice no time has slowed! It's all a matter of relativity of proper simultaneity.

Relativity would have drawn a horizontal blue line of Bob's perspective line of simultaneity between those 2 time points. The only line of simultaneity I use is the green one for Loedel simultaneity. t=2 for the judge is simultaneous with t'=2 for the racer according the relativity's hyperbolic lines of proper time simultaneity generated by the main equation

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

Alice would also have received his pink line from t=2 at t'=4. Whoa but that pink line took 3 yrs to reach her and the yellow line only took 1.5 yrs to reach Bob. This is a false equivalency. Using Alice's stop watch, Bob's signal was Loedel simultaneous with her proper time of t'=2 sent at a distance of 1.5 ly away from her. At t'=4, the pink light travelled a total of 3 ly in two of her years according to her watch so the c' of this light was 3/2 = 1.5c.

Ok, so what? How do I get from c'=1.5c the same result of 4-1.5=2.5yrs from Alice's perspective of Bob's t=2? First of all there is no c' in relativity. But using Yv, the light signal using Alice's watch must be faster than her Yv (the velocity according to her watch). If she was going at .8c, her Yv=4/3c, that does not mean she would suddenly be able to beat light, which should be c from all perspectives. v is not the same as v' and neither are c and c'. Nothing can exceed c and that rule is actually preserved by having factors for c when you apply the Y factor to v.

Secondly, Alice can be set as the stationary frame which would mean all the numbers that applied to Bob's perspective would now apply to Alice's. So why is Alice's c' so much higher than Bob's when there should be symmetry between the perspectives? The problem began in Bob's perspective with the assumption that the yellow light took 1.5 of Bob's years to return when in fact it took 2 of his proper years to return just like it took the pink signal 2 of Alice's proper years. The good thing about proper yrs is that they're not subject to Bob's or Alice's perspectives. The Loedel perspective allows a glimpse into proper time which is independent of perspective. Difficult to understand this fine point that a perspective allows a glimpse into proper time which is independent of perspective.

So if you want to depict both perspectives equally on one Md, Bob's equation should really be $(ct')^2 = (ct)^2 - (x/Y_0)^2$ where $Y_0$ =1 and Alices equation should really be $(ct')^2 = (ct)^2 - (x/Y_1)^2$ where $Y_1$ =5/4. c' represents the number of proper years light has travelled although I have not yet proved that mathematically, only empirically so far by doing examples. So for Alice, the light travel is also 1.5 yrs and 4-1.5= Alice 2.5 yrs when Bob sent the signal at t=2 which is the correct reciprocal of Bob's time dilation when his t=2.5, Alice's t'=2.

So plugging these values into the new equation for Alice $(ct')^2 = (ct)^2 - (x/Y_1)^2$:
$2^2 = 2.5^2 - (1.875/1.25)^2$ correct!

You can depict both perspectives on one Md if you alter the main equation in the way I showed.

Last edited: Feb 13, 2021
13. ### ralfcisRegistered Senior Member

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Let's flip the basic graphic unit to see how it would apply to the muon experiment

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I'll show the next example (sometime in the future after I answered all of James' questions) using real muon numbers, the above Md is to demonstrate the concept of time dilation from relativity of proper simultaneity.

The first consideration is that since no velocity change has occurred during the journey in from space, the proper times must be the same at the muon's crash landing. (Of course if this flipped graphic unit was joined to the basic one to show how the twin paradox works, the proper times at re-unification would not be the same.) So we count down the numbers from the apex to set the start of the muon's journey at t'=0, 3ly away from Earth. We draw the green line of the Loedel simultaneity to join both zeroes on the ct and ct' axes. The hyperbolic purple line from the main equation is also drawn in to join the zeroes. I've also added the blue lines of the stationary perspective simultaneity to make it easier to follow how relativity of simultaneity causes time dilation in this race to Earth. Again it has to do with the perspective of when the race starts as opposed to any notion of time slowing.

So the bottom blue line says the race began at t=-1 while it actually began at proper time simultaneity of t=t'=0 as seen by the Loedel perspective (of proper time). The length of the velocity line above 0 is the same length as the stationary perspective (both have t=t'=4 years). From Bob's perspective the total time in from the start at t=-1 is 5 of his years. So the time dilation is 5 for Bob, 4 for Alice and the proper relativity of simultaneity is 5-4=1.

In relativity there is no proper relativity of simultaneity only a perspective one. This is shown by Alice's perspective line of simultaneity intersecting at t=.8. So RoS in relativity is -1+.8 = 1.8. The formula for RoS is vx/c^2 so plugging in the values you get .6*3=1.8.

Anyhow Bob's view of the simultaneous start is one year more than the proper time start which is where time dilation really comes from, relativity of simultaneity and not of time slowing. The same can be applied to relativity's notion of length contraction. It's not due to any physical phenomenon of length actually shrinking in the direction of motion and then magically springing back when the motion stops. It's due to relativity of simultaneity. No where in my math of the muon example is time dilation or length contraction mentioned.

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

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ralfcis:

I am replying to your post #69 here. I won't look at your muon experiment one until we reach agreement on this one. If we cannot agree on this one, then I am afraid we will part company on the matter of relativity.

Before I get into the details, I want to tell you the main problem that I'm seeing in your analysis. It is that, while you take into account the different rate at which A's clock ticks relative to B's (i.e. time dilation), you do not take into account that A's measurements of distance are also different from B's. The reason you end up concluding that light signals travel faster than c in A's frame is because you're using the wrong distance units in A's frame. Specifically, instead of using A's rulers, you want A to use B's rulers. You can't do that and get consistent results.

First, I don't know where you get v'=Yv from. That is an incorrect expression for transforming velocities between reference frames. In your example in post #59, things are very simple: B says that A's velocity is (3/5)c, and A says that B's velocity is -(3/5)c.

You also need to be careful to distinguish the velocity of some object or person inside a reference frame from the velocity of the frame itself. This is what the special relativistic velocity addition formula does - it allows us to transform velocities of objects between two reference frames. I should also note that, according to special relativity (hereafter referred to as SR), the speed of anything that travels at velocity c in one frame must be c in every other frame.

To me, it looks like, most of the time, your "Loedel lines" are equivalent to SR's lines of simultaneity, although in several posts above you have made errors when drawing those lines. The ones in post #59 look okay, though.

I should point out that the "Loedel lines" on your graph look like* SR lines of simultaneity for Alice (A). Bob's (B's) lines of simultaneity are horizontal lines on that graph. SR lines of simultaneity, in general, connect points with the same time coordinate in a given frame.
---
* P.S. Further down, you will see that I discovered that your "Loedel lines" are not the same as SR lines of simultaneity, after all. I'm not sure what they are, now.
---

You have not yet responded to tell me whether you agree with my definition of "proper time" or to note that yours is different, so for the moment I will ignore parts of your posts that use that term, until you explain how you define it.

From B's perspective, A has to travel 1.5 ly at a speed of (3/5) light years per year, which will take 2.5 years.
From A's perspective, she travels nowhere (she is stationary in her own frame). Instead, the "finish line" travels towards her at a speed of (3/5)c. The initial distance to the "finish line" is 1.2 ly, according to her own rulers, so it will take 2 years according to her clocks.

Notice that I have used SR length contraction to calculate the "contracted" distance for the trip in A's frame of reference. A must use rulers in her reference frame to measure distances; she can't use B's rulers, because B measures distances differently than she does. This is a consequence of the need to measure the two endpoints of any length interval simultaneously, or else the "length" measurement makes no sense. Since A and B do not agree on which sets of events are simultaneous, similarly they do not agree on the measured lengths of objects - or other distance interval measurements.

Your graph, though, doesn't show A's length measurements. You have shown the time axis in A's frame, in effect - it is the same as her worldline, which you have drawn. Distances in her frame must be measured along lines of simultaneity.

I don't know what your v' is. In A's frame of reference, her speed is zero, because her frame of reference moves with her (obviously). In A's frame, Bob (B) moves with a speed of -0.6c. In B's frame, A moves with a speed of +0.6c and B's speed is zero. Neither of them sees the other, or themselves as travelling at 0.75c.

You can see above how A could calculate B's speed in her frame. She sees B moving away from her, to a final distance (when the finish line, which is also moving in her frame, reaches her) of 1.2 ly. She measures, on her clock, that this takes 2 years, so she can calculate B's speed as (1.2 ly)/(2 years) = 0.6 ly/year = 0.6 c.

Correct. Remember, x is B's coordinate. A must use coordinate x', in her own frame. B does not move in his own x coordinate, but he certainly moves in A's x' coordinate - from, say, x'=0 to x'=-1.2 ly.

I agree. B sees A's regular "tick" signals coming at half the rate, because A is moving away from him and there is a Doppler shift.
Similarly, in her frame, A see's B's signals coming at her at half the rate, because B is moving away from her (at the same speed).

Almost, but not quite right. The encoded message from A will be "I have reached the finish line at the 1.2 ly mark (according to my own stationary rulers) in two ticks of my atomic stopwatch."

I don't know why you say it is inaccurate. Your graph shows that, correctly. We can read off the times from the graph axis. The axes are B's reference frame.

Yes.

If B's watch tells him the race took 2.5 years and A's tells her that the race took 2 years, then it seems bizarre of you to claim that time was running at the same rate for both A and B. If it was, why would their watches end up showing different times? We're assuming these are well-calibrated atomic clocks, aren't we? What went wrong with them?

Correct. B's lines of simultaneity are horizontal on your graph. A's are similar to, but apparently not the same as your "Loedel" lines, on that graph (see below).

No!

SR doesn't have "hyperbolic lines of proper time simultaneity". SR's lines of simultaneity are similar to (but not the same as) your "Loedel lines" - certainly straight lines on your graph in this example.

I think you're possibly confusing SR curves that has a constant value for the quantity $(ct)^2 - x^2$, which obviously has both space and time components, not just time. That quantity is essentially similar to what, in SR, is called the "spacetime interval". Note carefully: spacetime interval, which is not the same as time interval.

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

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(continued...)

Correct.

Why should there be equivalency? The yellow line shows a light signal sent to B (who is stationary) from a fixed point in space that is 1.5 ly from him. Since the speed of light is c, the signal must take 1.5 years to reach him.

On the other hand, the pink line shows a light signal sent to A, who is continuously moving away from the light source at a speed of 0.6c. The light must "catch up" to A as she moves away from it, and that takes extra time. By the time it gets to her, she is 3 light years away from the source, as measured by B. We can see that clearly on your graph. At the time that B sent that signal, A was 1.2 ly away from him, according to your graph, but she didn't stay at that distance; she moved further away from him.

Again, you're ignoring the fact that A's frame doesn't use B's rulers.

Let's do this using the Lorentz transformations, for clarity. There are two events of interest:
Event 1. B sends the pink signal. This occurs at (x,t)=(0,2 years).
Event 2. A receives the pink signal. This occurs at (x,t)=(3 ly, 5 years).

Note: coordinates here are given in B's frame of reference and can be clearly read from the graph.

Now, from SR, we have $\gamma = 1.25$, due to the speed at which B's frame moves according to A.

Consider event 1. Using the Lorentz transformations, we find:
$x' = \gamma (x-vt) = (1.25)(0-(0.6)(2)) = -1.5$ ly
$t' = \gamma (t-vx/c^2) = (1.25)(2 - 0) = 2.5$ years.

Your graph wrongly indicates that the pink signal was sent at 2 years on A's clocks. In fact, it was sent at 2.5 years, according to A's clocks. That means the green line on your graph is not a line of simultaneity for A. Your "Loedel lines", whatever they are, aren't lines of simultaneity at all, but something else.

I would appreciate it if you explained what your "Loedel lines" are. Previously I assumed they were lines of simultaneity.

The Lorentz transformation makes it clear that when the pink signal is sent (i.e at the time, on A's clocks, when the signal is sent), A is 1.5 ly away from B, as measured using A's rulers. But that corresponds to when A was at a distance of 1.875 ly, as measured by B's rulers.

Thinking about it, maybe the mistake you've made is to draw the line where B measures the distance as 1.5 ly, rather than drawing it where A measures the distance as 1.5 ly. In other words, it might all come back to your error in assuming that A and B share the same rulers.

And for event 2:
$x' = (1.25)(3 - (0.6)(5))=0$
$t' = (1.25)(5 - (0.6)(3)) = 4$ years.

This makes perfect sense. Notice that the receipt of the pink signal happens at x'=0 in A's frame. A is always at x'=0 in her own frame, of course, so if she is receiving the signal, it must occur at x'=0.

The receipt time, according to A's clock, is 4 years, and you apparently agree with that.
---
That brings us to the speed that A calculates for the pink light signal.

According to A, that light signal had to travel 1.5 ly to reach her, and it took 4-2.5 = 1.5 years, so its speed was 1 light year per year in her frame, as expected from SR.

I am still not clear on whether you are trying to replace SR by your own theory, with your own theory having a variable speed of light. Is that what you're trying to do?

Here, you say that the speed of light should be c from all perspectives. But above, you just calculated a c'=1.5c. So which is correct?

Notice that my SR calculation shows a speed of c in both frames of reference: A's and B's. It appears to be only your theory that calculates different speeds for the light in different frames.

I don't understand how you justify applying a Y factor to any v. What are you doing when you do that?

No! Alice's clocks run at different rates from Bob's clocks; you agree with this, apparently. Alice's rulers also are not the same lengths as Bob's rulers; you appear to be consistently ignoring this.

Because you made some mistakes. Because you're trying to use a theory of relativity of your own making - one that is not consistent with SR or with real-world experiments.

Define "proper year".

You're clearly using the term "proper time" in a completely non-standard way. Please define it.

Difficult to understand until you post your definition of "proper time".

Last edited: Feb 14, 2021
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16. ### ralfcisRegistered Senior Member

Messages:
421
Take a look at this video course from Brian Greene:

https://www.youtube.com/playlist?list=PLj6DWzIvBi4PFDXCCV1bNhVUgDLTwVbFc
It talks about much of what I mentioned about special relativity that I don't think you've seen before. It's where I learned relativity before I discovered my own math to predict the same things with a completely different philosophy.
Also take a look at this highly rated post I made on the Physics Stack Exchange:

https://physics.stackexchange.com/q...ing-twin-also-come-back-permanently-length-co

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

Messages:
39,286
Why add more things for me to look at?

Can't you deal with the questions I have asked you, first?

I'm happy to discuss Brian Greene's video after you have responded to my previous posts, especially posts #58, #71 and #72.

You said you wanted some feedback on your work from somebody who understands Minkowski diagrams. But now that I've given you a lot of feedback, you seem uninterested. Why is that?

Later. After we've discussed the posts above that you haven't responded to.

Last edited: Feb 15, 2021
18. ### Michael 345New year. PRESENT is 72 years oldlValued Senior Member

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13,077
Looks to me the above is a hook to a

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19. ### ralfcisRegistered Senior Member

Messages:
421
Look I'm working on the answers starting with #27. In the meantime I wanted you to see where I got some of my ideas specifically on the definition of length contraction. There is no measuring device for length in relativity besides a clock that is subject to "time dilation" (via relativity of simultaneity). So if your clock is affected so are your measurements of length contraction. The two are the same thing. There is no separate ship odometer or ruler or crunching of matter in relativity but there are methods separate from using a clock, such as the parallax method and super nova candle method, to measure proper distance. Even though time can be permanently altered in the twin paradox, there is no permanent form of length contraction that survives re-unification of the participants. Here's a quick video from Greene:

20. ### originHeading towards oblivionValued Senior Member

Messages:
11,876
Of course there is. For the twin "paradox" the traveling twin will say that less time has passed and the distance traveled was less than that of the earth bound twin. The time difference and the distance difference are both 'permanent'.

21. ### ralfcisRegistered Senior Member

Messages:
421
Constant velocity only results in reciprocal time dilation, not permanent age difference. If Alice leaves at 3/5c and passes by Proxima Centauri 3ly away, she has not aged 1 yr less than Bob just because time dilation says t=5 yrs for Bob and t'=4 yrs for Alice. From Alice's time dilation perspective, Bob has only aged 3.2yrs when her t'=4. She has no odometer reading that says she's only travelled 2.4ly but she does have an on-board clock that says she has travelled 4 yrs. It's relativity that concludes the distance she travelled must have contracted to maintain a constant c from all perspectives. Or she can trust her clock and her star charts and Yv= x/t' = 3/4c while v from Bob's perspective is 3/5 c and Alice can verify that her relative velocity to Bob is also 3/5c because her measurable DSR = .5. Relative velocity v is not the same thing as Yv, only v as a relative velocity is subject to being limited by c but Yv is not technically a relative velocity the same way v is.

So if Alice stops at Proxima Centauri, I say she can say she has permanently aged 1 yr less than Bob by comparing light messages between them but relativity can't say that because their atomic clocks are not co=located and still subject to other perspectives. So Alice has to turn back to Bob and if she does it at 3/5c, their atomic clocks will show she has aged 2 yrs less than Bob.

Not correct. The reason Alice aged 2 yrs less is because the earth twin didn't travel any distance at all so it's because she travelled all of the distance and because of the main formula:

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

notice the minus sign, distance subtracts from time in a spacetime path. So according to the formula, ct'=8 for Alice, ct=10 for Bob and x=6 (roundtrip of 3ly). 3ly not 2.4 ly! There is no record of any length contraction. It's exactly the same length as the star charts. There is also no physical odometer that recorded any length contraction. There is no physical evidence that the space between earth and proxima centauri was contracted by the ship which somehow had the power to draw an entire star system closer to earth. All we have is the philosophy that if c is the same from all perspectives, a conclusion that agrees with the MMX, then length contraction is the only possible explanation. I have another explanation that also does not violate the results of the MMX.

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

Messages:
39,286
You already responded to #27.

You can go through all my posts in order if you like, but I'm trying to save you some time by pointing you towards posts #58, #71 and #72. Up to you.

Wrong. In principle, somebody could use a wooden ruler - or, say, a whole bunch of wooden rulers connected end-to-end if a much longer ruler was needed.

I already explained what a reference frame is to you, in some detail, along with how observers in that frame can take measurements of events in spacetime. I guess you'll get to replying to that post later (?)

I don't have much interest in quibbling with you. A measurement of length of obviously not the same thing as a measurement of time. However, I agree with you that time dilation and length contraction are both effects that derive from the same cause - namely the nature of spacetime itself, or from the postulates of special relativity, if you prefer.

I can't comment until you define "proper distance". I have asked you many times to do that, now, but you still haven't done it. The only thing I know so far is that whatever you think "proper distance" is, it isn't the way that it is usually defined by those (like me) who understand special relativity.

I agree with you that the elapsed time on different clocks can be permanently altered as a result of relative motion. I also agree with you that whenever two observers come to rest relative to one another, they will agree on all length measurements, so that in that sense length contraction is no "permanent". Nothing important turns on that.

I'm not going to watch any videos until you have responded, at least, to post #58 and/or #71/72.

Greene agrees with me, by the way, about special relativity.

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

Messages:
39,286
Agreed.

For Alice, subjectively, the trip only took 4 years, so she only aged biologically 4 years. Bob, on the other hand, says the trip took 5 years, but to him Alice's clocks (including her biological age) appeared to be ticking slowly as she travelled, so that she only aged 4 years biologically while Bob aged 5 years biologically. From Alice's point of view, Bob travels away from her to a distance of 2.4 ly at a speed of 3/5c while she remains at rest in her frame, so the trip takes 4 years. But in Alice's frame, Bob's clocks tick slow, so Bob only apparently ages 3.2 years during Alice's 4 year trip.

Correct. We are in agreement.

She could measure the distance in any number of different ways, in principle. If Bob can measure the distance as 3 ly, then Alice could use exactly the same procedure (whatever it is) to measure the distance as 2.4 ly.

Yes.

Yes.

You're wasting time posting "new" examples of the same thing. You should instead spend your time addressing the comments I put to you regarding your scenario in posts #71 and #72.

I don't know what your Yv is supposed to represent, and it is clear that you are mixing frames inappropriately whenever you write something like x/t', rather than x'/t'.

It appears to be the case, though you will not confirm, that you think different observers all agree on length measurements. They do not. Not if SR is correct. You seem scared to say whether you are claiming that SR is incorrect or not. It would be good for you to clarify your position. Are you proposing an alternative theory to SR? Are you pointing our what you believe to be errors in SR? Or what?

What is Yv? Why is it useful?

Why do we need to worry about Yv? What is it, physically?

If Alice stops, then she has accelerated. In that case, the symmetry between Alice and Bob's frames is broken, because Alice accelerates while Bob does not. That acceleration can certainly result in a "permanent" discrepancy between their clock readings, and thus their ages.

You say that "relativity can't say that", but it can and it does. There's a myth that SR can't handle accelerations. It can, but the maths is more complicated since we have to approximate a non-inertial frame of reference by a whole bunch of approximately-inertial frames, then add up cumulative effects of all the frame changes. In practice, it requires integration. However, fortunately we can avoid all of that mess by looking at a spacetime diagram like the one I analysed in posts #71 and #72, above.

The way you're using that formula doesn't really make much sense, seeing as it is, essentially, a formula about the SR spacetime interval. I think that, probably, you don't understand where it comes from or what it actually means. Also, clearly, you don't understand that you're using a special case of the formula, in which x'=0. Obviously, in this specific example, Alice remains at x'=0 at all times, but Alice travels in Bob's x coordinates. So, if you're going to calculate some kind of spacetime interval for Alice's trip, of course you're going to have Alice's trip distance being non-zero in Bob's x coordinates, while it is zero in Alice's x' coordinates.

But none of that is very important, because you're using that formula incorrectly anyway. While the answer you pull out of it makes some kind of sense, that's more by accident than by design.

I could walk you through what that formula is really doing, but showing you actual spacetime intervals in the two frames, but I won't do that until you've responded to previous posts of mine, especially #58 or #71/72.

Rulers typically don't keep records. A stopwatch, on the other hand, records elapsed time.

Bob's star charts are drawn up based on measurements of distance made in Bob's reference frame. Alice could just as well draw up her own star charts while she is moving relative to the stars, but her distances would all be contracted in the direction of her motion, compared to Bob's distances in that direction.

It's probably better if you think about it this way: when Alice changes her state of motion with respect to the star system, she doesn't alter anything about the star system itself. She doesn't "draw it closer to Earth". Instead, she changes her own perception of space and time. She "rotates" her perception in a strange way - one that is described precisely by the Lorentz transformations of SR.

You apparently agree that when Alice moves, her perception and records of time change, compared to Bob's. But you're not claiming that Alice magically altered Bob's clocks. So why are you claiming that Alice magically alters Bob's distances? We're talking about the same thing, just with distance instead of time. Alice alters herself when she moves. She doesn't change the world around her. She doesn't change spacetime.

No you don't. Different observers measure different speeds of light if we use your "explanation", which is contrary to actual experiments, including the Michelson-Morley experiment.