Relativity and simple algebra II

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

  1. ralfcis Registered Senior Member

    Relativity can be explained with basic algebra using only 2 formulae (the one for gamma (Y) and the relativistic velocity combo formula). These construct a simple graphical building block that can be joined by light signals to graphically show how relativistic phenomena such as time dilation, simultaneity, the muon experiment, the twin paradox can be explained through simple algebra.

    There are no rotated frames here, only three types of lines v, \(v_t \) and \( v_h \)(half speed relativistic velocity), their reciprocals and their behavior when multiplied by Y (gamma). The math also avoids square roots and squares in the equations to make them work using simple addition and subtraction. There is no clock sync, only light signals and the assumption atomic clocks all run at the normal clock rate within their frames thanks to the principle of relativity.
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  3. ralfcis Registered Senior Member

    The main equation (space and time velocity combination) is
    \( c^2 = v^2 +v_t^2\)
    also written as the gamma function
    \(Y = \frac{c}{ \sqrt{(c^2-v^2)} } \)
    \(v_t= c/Y \)
    \( v = c/Y_t\)

    \(v_t= c/Y \) is the velocity through time while v is the velocity through space. The faster you are observed through space, the slower your time rate is observed.

    People have great difficulty understanding what a velocity or rate of time means. Velocity through space is the rate of distance through time just as your rate of hot dogs eaten is the number of hot dogs you can eat in an hour. Velocity does not depend on distance. Velocity of time is how fast you observe a rate of time compared to your own such as fast forward or slow motion on your tv.

    Here is the relativistic velocity combo equation of two velocities through space:

    \(w =(v+ u) / (1 + vu/c^2)\)

    So here is a more universal universal equation that includes relativistic velocity combination with space and time velocity combination:

    \(c^2= ((v+u)^2+ v_t^2u_t^2) / ( 1 + vu/c^2)^2\)

    The main equation is also written as:

    \((ct')^2 = (ct)^2 - x^2 \) (Minkowski hyperbolic (difference of squares) form where hyperbolas intersect the same proper time for all velocity lines)

    \((ct)^2= (ct')^2+ x^2 \) (Epstein pythagorean (sum of squares) form where circles intersect the same proper time for all velocity lines)
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  5. ralfcis Registered Senior Member

    This is my equation sheet to cut and paste equations in future text:

    [ tex] c^2 = v^2 +v_t^2[/tex]
    [ tex]Y = \frac{c}{ \sqrt{(c^2-v^2)} } [/tex]
    [ tex]v_t= c/Y [/tex]
    [ tex] v = c/Y_t[/tex]
    [ tex]w =(v+ u) / (1 + vu/c^2)[/tex]
    [ tex]c^2= ((v+u)^2+ v_t^2u_t^2) / ( 1 + vu/c^2)^2[/tex]
    [ tex](ct')^2 = (ct)^2 - x^2 [/tex]
    [ tex](ct)^2= (ct')^2+ x^2 [/tex]
    [ tex] v = v_h(v_t/c +1) [/tex]
    [ tex] v_t = v_{ht}(v/c +1) [/tex]
    [ tex] v = 2c^2v_h / (c^2 + v_h^2)[/tex]
    [ tex] v_t = 2c^2v_{ht} / (c^2 + v_{ht}^2)[/tex]
    [ tex] v_h = c(c - v_{ht}) / (c +v_{ht}) [/tex]
    [ tex] v_ht = c(c - v_h) / (c +v_h) [/tex]
    [ tex] v_t = DSR(c+v) [/tex]
    [ tex] t_{ps} = xv_{ps}= xYv/(1+Y)[/tex]
    [ tex] v_{ht} =cDSR[/tex]
    [ tex]Y(c-v) = c/DSR[/tex]
    [ tex]v/(c-v) = 2cv_h / (c - v_h)^2[/tex]
    [ tex]v = c(1/YDSR -1)[/tex]
    [ tex]v = c(1/sqrt(1-1/Y^2))[/tex]
    [ tex]DSR= Y(1-v/c)[/tex]
    [ tex]1/DSR= Y(1+v/c)[/tex]
    [ tex]Y = 2Y_h^2 - 1 [/tex]
    [ tex]Y= (c^2 + v_h^2)/(c^2 - v_h^2) [/tex]
    [ tex]Y= (1/(1-v_h/v)) -1 [/tex]
    [ tex]Y = 2c^2v_h/v(c^2 - v_h^2) [/tex]
    [ tex](Yv)^2 = v^2/(1-v^2)[/tex]
    [ tex]Yv= 2Y_h^2v_h[/tex]
    [ tex]Yv= v_h(Y+1)[/tex]
    [ tex]t'=x(v_h +c)/c [/tex]
    [ tex]t = Yt'[/tex]
    [ tex]Yv =x/t'[/tex]
    [ tex]Yu/Yw=DSR_v[/tex]
    [ tex]vx/c^2=RoS[/tex]
    [ tex]t{pad}=t′−x/Yv[/tex]
    [ tex]t' =xDSR_o/Yv[/tex]
    [ tex]w_t= c / Y_w = c / (Y_vY_u(1 + vu/c^2)) [/tex]
    [ tex]Y_ww = (v+u)Y_uY_v [/tex]
    [ tex] cDSR = sqrt((c-v)/(c+v)) [/tex]
    [ tex]DSR_w= DSR_v* DSR_u [/tex]
    [ tex]DSR^2= (c-v)/(c+v) [/tex]
    [ tex]w = c(1-DSR_v^2 DSR_u^2 )/(1+DSR_v^2 DSR_u^2 )[/tex]
    [ tex]A = (v+u)[/tex]
    [ tex]B= (1+vu/c2)[/tex]
    [ tex]sqrt(c^2B^2-A^2) = c/YvYu =v_tu_t[/tex]

    Important equations for the table in the next post:

    [ tex] v^2= (Y^2-1)/Y^2[/tex]
    [ tex] v_h = Yv/(Y+1)[/tex]
    [ tex] v_h = v/(1+v_t)[/tex]
    [ tex]DSR = Y(c-v)[/tex]
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  7. ralfcis Registered Senior Member

    I will provide verifications of the above formulas from the two main formulas upon request. However, I am very prone to arithmetic or cancelling terms errors so I use the following tables to check my work. I tend to favor using whole number rational fractions instead of long strings of decimal places. This has revealed beautiful sequences that relate the velocities.

    There's probably some way to format these tables properly but I don't have the time to find out.

    The first sequence is for multiples of .6c which is related to binary powers:

    v Y v_h v_t v_ht DSR w

    1 1 1 0 0 0 1+any
    511/513 513/64 31/33 64/513 1/32 32 1/3+255/257
    255/257 257/32 15/17 32/257 1/16 16 15/17+15/17
    63/65 65/16 7/9 16/65 1/8 8 3/5+15/17 or 7/9+7/9
    15/17 17/8 3/5 8/17 9/36 4 3/5+3/5
    7/9 9/(4sqrt2) 7/(9+4sqrt2) (4sqrt2)/9 sqrt2/4 4/sqrt2 3/5+1/3
    3/5 5/4 1/3 4/5 18/36 2 1/3+1/3
    1/3 3/(2sqrt2) 1/(3+2sqrt2) (2sqrt2)/3 (sqrt2)/2 2/sqrt2

    The next sequence is for multiples of .8c mixed in with .6c

    v Y v_h v_t v_ht DSR w
    3280/3281 3281/81 40/41 81/3281 1/81 81 40/41+40/41
    40/41 41/9 4/5 9/41 4/36 9 4/5+4/5
    35/37 37/12 5/7 12/37 6/36 6 4/5+3/5
    13/14 14/3sqrt3 13/(14+3sqrt3) 3sqrt3/14 1/3sqrt3 3sqrt3 4/5+1/2
    4/5 5/3 1/2 3/5 12/36 3 1/2+1/2
    1/2 2/sqrt3 1/(2+sqrt3) sqrt3/2 1/sqrt3 sqrt3

    Some miscellaneous numbers I calculated and used once:

    v Y v_h v_t v_ht DSR w
    24/25 25/7 3/4 7/25 1/7 7
    12/13 13/5 2/3 5/13 1/5 5
    77/85 85/36 77/121 36/85 8/36 9/2
    11/12 12/sqrt23 11/(12+sqrt23)
    8/17 17/15 1/4 15/17 3/5 5/3
    5/13 13/12 1/5 12/13 24/36 3/2
    12/37 37/35 1/6 35/37 5/7 7/5
    7/25 1/7 24/25 27/36 4/3
    16/65 1/8 255/257 15/17 17/15
    9/41 1/9 40/41 4/5

    21523360/21523361 21523360/21529922 6561/21523361 1/6561
    Last edited: Feb 7, 2021
  8. ralfcis Registered Senior Member

    Could James R repost his 2 questions from the former thread so I could quote reply to them so this thread would be equivalent to the last?

    Also I'd like to show an example of the pattern in the tables above which saves a lot of calculation.

    The velocity 511/513. \(512= 2^9\). 511 is one less, 513 is one more.
    513 is the numerator of Y and the denominator is \(64= 2^6\)
    \( v_h = Yv/(Y+1)\) which is the numerator of v divided by the sum of the denominator of v plus the denominator of Y
    so I made a math error.
    v_h is not 31/33 it's 511/(513+64). My arithmetic is just awful.
    v_t is just the reciprocal of of Y.
    \(DSR = Y(c-v) and v_{ht} = cDSR\) so DSR = \(32= 2^5 and v_{ht}\) is the reciprocal of DSR.

    v=511/513 is the combination of 1/3+255/257
    255/257 is the combination of 15/17+15/17
    15/17 is the combination of 3/5+3/5
    3/5 is the combination 1/3+1/3

    You can group other combinations into 7/9 and 63/65 because the increments are in 1/3c.
    The advantage of this is in having all perspectives in one spacetime diagram without having to reassign the stationary frame. For example, 15/17 c has a relative velocity to 3/5c of 3/5c because 15/17 = 3/5+3/5 in relativistic combination.
    Last edited: Feb 7, 2021
  9. James R Just this guy, you know? Staff Member

    Are you saying you can derive all of (special) relativity from your two formulae?
    Can you show how this is derived from your two formulae?

    Can you please give us two or three concrete examples of the application of this "more universal" equation, to show how it is applied?
  10. ralfcis Registered Senior Member

    That's a loaded question for me. I believe my list of equations is the result of only manipulating my two formulas but I'd have to go through years of derivations to confirm that's true. I deal in velocities. The velocity combo formula shows any velocity relative to c is still c without going into the idea that this must be true if time dilates and distance contracts. So my math supports relativity without the need for the latter discussion as I'll show. I also use the Brehme idea of your space in my time. It comes up when people say you can cross the universe without ageing much if you go fast enough. That's valid in relativity but it makes no mention that you'd need some kind of odometer to measure your perspective of the distance travelled. Also if you're going that fast using your watch, light must still be going faster from your perspective because it will beat you to your destination. Just as your Brehme velocity is Yv from your perspective and Earth's star charts, c must also have a factor for its Brehme velocity to beat you but, spoiler alert, it's not Y. Nevertheless c is still c from every perspective. My math also shows time does not slow in time dilation, it has to do with relativity of simultaneity depending on when the stop watches are started from each perspective. This is just a math exercise, philosophy can be debated but math can't be, it's either right or wrong.

    Yes, I'll go look it up and write it out if you insist but I won't be using tex to save me time. But it will take you less time to understand this equation works if you plug in values from the tables I've provided.

    I don't know if there's any application. It was an example that if c has 2 velocity components in the main equation then those two velocities are subject to the velocity combo equation which is the other main equation. And, vice versa, if you're adding 2 velocities through space in the velocity combo equation, you can have an equation that includes velocities through time. I ws just curious what the result would be that keeps c at c no matter what velocity you throw in the mix.

    Anyway, I'm not finished transcribing my previous thread which includes your 2 past questions. Could you cut and paste them here? Also my list of equations allows a cut and paste just by removing the space in the first [ tex]:
    so [ tex]c^2= ((v+u)^2+ v_t^2u_t^2) / ( 1 + vu/c^2)^2[/tex] beomes \(c^2= ((v+u)^2+ v_t^2u_t^2) / ( 1 + vu/c^2)^2\)
    Last edited: Feb 8, 2021
  11. ralfcis Registered Senior Member

    Ok I transcribed your questions from the previous thread:
    James R said

    What is \(Y_t\)?

    More importantly, what is \(v_t\) physically? What do you mean by a "velocity through time"? Clearly, vt has dimensions of a regular velocity, so what is it, exactly?

    \(c^2= ((v+u)^2+ v_t^2u_t^2) / ( 1 + vu/c^2)^2\) You haven't yet posted a derivation of this.

    What does t′ indicate, in this equation, as opposed to? I notice there are no x′ terms in the equation...

    \(v_t\)=c/Y=ct'/ct is the velocity or rate of time through time. Relative to what? Some absolute standard of time?

    That doesn't make sense. c is a velocity with dimensions of distance over time.

    Can you please give a specific example of the permanent slowing of time and show your calculations of that. What effect does a permanent slowing of time have on the rest of the universe?

    "the faster you are observed through space, the slower your time rate is observed." By whom?

    You're not making the "relativity" part very clear in what you're writing.

    Do you understand reference frames? Do your equations or derivations use the idea of different reference frames?

    Are you attempting to re-derive special relativity here, or is this a new theory of your own with its own postulates, supposed to replace special relativity? It's not clear to me.
  12. ralfcis Registered Senior Member

    \(v_t=c/Y and v=c/Y_t\) There is a symmetry between velocity through space and velocity through time that arises from the main equation \( c^2 = v^2 +v_t^2\)

    For example if v=3/5c , Y = 5/4 but there is a symmetry between velocity through space and the one through time. So if v = 3/5c , \(v_t=4/5 and Y_t=5/3\).

    You may find this view unorthodox but I swear it came about from a discussion with Don Lincoln on the SPCF forum I was on years ago.
  13. ralfcis Registered Senior Member

    \(v_t\)=c/Y=ct′/ct is the velocity or rate of time through time. Now all clocks in their frames beat at the normal rate c through time. It's like looking at video in fast forward or slow motion. There are 3 types of rates through time in relativity. One the permanent slowing of time during an imbalance of relative velocities when one participant initiates a change which takes time for the other to register the change, another is the Doppler Shift Ratio which is partly an apparent rate caused mostly by the rate of info delay from a moving clock. The third is \(v_t\) which satisfies the main equation so that the faster you are observed through space, the slower your time rate is observed. As you can see from the equation, it's basically reciprocal time dilation.

    For example if v=3/5c, \(v_t\)=4/5, DSR = .5c and if the participant was to turn around after 3 ly, she would age 2 yrs less when the two re-unite at Earth (i.e. the twin paradox.)
  14. ralfcis Registered Senior Member

    c2=w2+w_t2 where
    w=v+u/(1+vu/c2) =c2(v+u)/(c2+uv)
    and by symmetry
    w_t= =c2(v_t+u_t)/(c2+u_tv_t)
    v_t =c/Y_v and u_t=c/Y_u
    so w_t=c/Y_w=c/(Y_vY_u(1+vu/c2)

    so c2=c4(u+v)2/(c2+uv)2 +c6/(Y_v2Y_u2(c2+vu)2)

    cancelling a bunch of c2 and resubbing v_t =c/Y_v and u_t=c/Y_u you get

    c^2= ((v+u)^2+ v_t^2u_t^2) / ( 1 + vu/c^2)^2

    easier if you group the terms
    \(A = (v+u)\)
    \(B= (1+vu/c2)\)
    \(sqrt(c^2B^2-A^2) = c/YvYu =v_tu_t\)
  15. ralfcis Registered Senior Member

    Oh yes you said you didn't see x' mentioned anywhere. I deal with the main equation that has 3 axes in it: ct', ct, x. I superimpose coordinates on the ct' axis that represent the coordinates for 3 different velocities: v, v' and vt. I deal in velocities and only break them up into their time and space components at the end if requested. I don't use coordinate rotation to create ct' and x' coordinates. The x' axis is the same as the line of simultaneity from the 'moving' perspective which is 1/v with a v slope= ct/x. Again this math deals primarily in velocities, not their space and time constituents (except at the end).
  16. ralfcis Registered Senior Member

    We are all stationary in our own frames and while our velocities through space in our own frames are zero, according to the main equation I posted, our velocities or rates through time are c. I did give examples of rates through time but for example if a ship was leaving earth at v=3/5c and both the earth and the ship were broadcasting a TV signal, it would appear that each would see the other moving in slow motion at .5c or half the normal rate that they would be moving in their own frames. If the ship was coming back, both would eventually see each other moving at double the normal rate in fast forward at 2c. Their clocks would appear to be moving double fast forward to each other while in their own frames, their clock rates would be normal at c.

    This is just an example of an apparent rate through time and how things might look if we could actually see time dilation in real time which is not possible. It can only be calculated after the fact and is not the same rate as DSR shows us in real time. These are not absolute, they are relative and due to relative velocity between the two participants.
  17. ralfcis Registered Senior Member

    c is also a rate of time as well as a rate of distance however as a rate of time it appears pretty slow to us unlike c velocity through space.
  18. ralfcis Registered Senior Member

    Permanent time difference due to the twin paradox.

    If Alice goes out at 3/5c and turns around at 3ly, there will be a stretch of time when Bob will not be aware that Alice has changed their relative velocity. Relative velocity is the cause of time dilation which is normally reciprocal because the relative velocity is reciprocal. But a change of relative velocity made at a distance apart means there is an imbalance in relative velocity because Bob continues to see Alice going away from him at .6c for a time while Alice sees immediately sees Bob approaching him at -.6c which are two different relative velocities. So time will be permanently affected between the two by Alice aging 2 years less than Bob when they re-unite and that 2 yr difference will continue until one of them makes another velocity change.
  19. ralfcis Registered Senior Member

    Gravitation also causes a permanent slowing of time like in the movie Interstellar based on the physics stylings of Kip Thorne. The closer they moved to the black hole, the faster they aged their loved ones at home (but actually the slower they aged relative to Earth). It's relative between two participants when in fact both frames aged at the same proper velocity of time c because they are stationary relative to their own frames.
  20. ralfcis Registered Senior Member

    By the one who is not you observing your frame.
  21. ralfcis Registered Senior Member

    I'm just presenting a new math framework based on 2 equations and a single algebraic/geometric unit that can re-construct relativity from simple algebra. It's a math exercise, not a philosophical one. It's like Euclid's geometry based on real geometric shapes was superseded by Hilbert's geometry which didn't require that (not implying that I'm either of them).
  22. ralfcis Registered Senior Member

    Ok let's begin with relative velocity.

    When Alice takes off from Earth at 3/5c, she has no on-board speedometer that reads .6c. Her DSR (Doppler Shift Ratio) reading relative to Earth is .5 and this is what tells her she's moving .6c relative to Earth. Her relative velocity to other ships moving would be different relative velocities. The thing about relative velocities to everything is they are just as valid if you made them all relative to one common thing such as the Earth.

    Consider all cars on Earth being equipped with radar guns instead of speedometers. You could measure your velocity relative to every car. You could assume, despite the Earth's momentum, that you can't tell if your car is moving relative to a stationary Earth or your wheels are spinning the entire planet underneath them like a giant hamster cage. You could extend that view to thinking, from your perspective, the sun orbits the Earth and the entire universe speeds past a proton in the LHC. With adjustments to gravity and acceleration effects, your calculations would be pretty close to those of a more realistic perspective of relative velocity.

    Before I continue, consider this very important fact: space (and substances like water or glass) house an electromagnet medium with permittivity and permeability properties that define the speed of the electromagnetic wave that propagates using that medium. Space is made up of a distributed capacitance and inductance that can be compared to a material medium's inertia and elasticity that allows it, via Hooke's law, to allow propagation of mechanical waves. The equation for EM wave propagation is very similar to Hooke's law for mechanical wave propagation.

    The important difference is material has no relative velocity to a vacuum's electromagnetic medium. Why? Because c travels through its medium and if you were able to push the medium you'd be able to add that push to c but you can't because c is the speed limit. If you blasted a vacuum bottle into space, the relative velocity of the bottle to the vacuum it contains is the same as the bottle's relative velocity to the vacuum it's speeding through. It's as if the enclosed bottle's behaving like an open ended cylinder. Since light is an EM wave travelling through an EM medium, the bottle also has no velocity relative to light.

    The Fizeau experiment suggests that the EM medium of water limits the speed of light to .75c. This means if you blasted a water bottle into space, the velocity of the bottle could be added to the medium and hence to the light that travels through that medium. It would act as any wave through a medium moving relative to an outside observer. The medium has a relative velocity to an outside observer and that would be added to the speed of light in that medium so long as that addition does not violate c as the speed limit.

    Anyway, relativity uses spacetime diagrams (Minkowski, Epstein or Loedel) to try to depict relative velocity. However they are incomplete without considering each perspective as stationary so two diagrams are needed to depict relative velocity. I have gotten around this and can provide both perspectives in a single diagram because I don't use coordinate rotation in my math.

    The closest relativity can come to depicting it in a single diagram is using the Loedel diagram. 3/5c leaving Earth could be depicted as two half speed 1/3c ships leaving Earth in opposite directions. A more precise description is that a single ship would leave Earth at 1/3c and somehow the Earth would blast away at 1/3c from a faceless point in space from where the ship and Earth once stood together. If you wanted to maintain Earth as the common background frame, the depiction of the ships leaving it at 1/3c would translate to a .6c relative velocity depiction as 1 ship being the stationary frame, the other leaving it at .6c and the Earth between the two leaving the blast off point at half speed 1/3c. Although a spacetime diagram is supposedly about the two ships, Earth is also a participant and so is this non-physical cartesian coordinate background which is the true reference frame in empty space.

    The Earth is travelling at what I call the Loedel or half speed \(v_h\) velocity through space. Wiki defines it as an intermediate perspective between two equal velocities of opposite sign. (I'm paraphrasing because this is difficult to find on Wiki.) It is completely ignored by relativity but is extremely important to this math exercise. If you want I can provide how all this post looks using spacetime diagrams but I've found no one knows how to read them so they get intimidated or suspicious I'm trying to pull a fast one on them. Plus they take a lot of work.
  23. ralfcis Registered Senior Member

    No further questions so I guess I've lost everyone? Ok let's get deeper into the math.

    Please Register or Log in to view the hidden image!

    Here is the Loedel diagram of Alice leaving Bob on Earth at .6c. It would not be the same diagram as Earth being the reference frame and Bob and Alice leaving Earth at 1/3c in opposite directions. In fact the reference frame is empty space here represented by the half speed Loedel velocity \(v_h\). Nothing is referenced to this stationary reference frame and everything is referenced to the Earth/Bob velocity 0f -1/3c. There are no rules in math where a reference frame needs to be stationary. Breaking this rule means I can set any velocity axis as the reference simultaneously on one spacetime diagram.

    It does alter the main equation slightly though. It introduces gammas with subscripts: \(Y_o\) for the reference frame and \(Y_1\) for the subsequent frame depending on what perspective you choose. If you want to look from Bob`s perspective the main equation is


    from Alice's perspective it's


    What I'm doing mathematically is layering the spacetime diagrams into one diagram. Relativity doesn't do this but again this thread is a math presentation that could be applied to relativistic physics to arrive at the same answers.

    You may also notice that I've layered 2 separate scenarios for Alice, one where she turns back to Bob at t'=2 and the other where she just keeps going.

    I've developed a quick formula for calculating \(v_h\) but it's only applicable to Minkowski diagrams where the reference frame is stationary. It doesn't strictly apply to a Loedel diagram where \(v_h=0\)) always. There's no need to develop a more general formula as I'll show Minkowski mostly doesn't get into the trouble of having empty space as a stationary reference frame.


    Notice the thin green curved line which represents the hyperbolas generated by the main equation. They intersect all the velocity lines at the same time on their universally accurate atomic clocks which all beat at the same time rate within their frames. Notice the thick green line which is the line of simultaneity from the Loedel perspective. It also intersects the two blue velocity lines at the same proper time. It's not physically correct to call the Loedel line of perspective simultaneity the line of proper simultaneity but I use it as such mathematically. It seems to me that the hyperbolas generated by the main equation suggest proper time simultaneity but I've never gotten a straight answer to this question.

    The two thin red lines are Bob and Alice's perspective lines of simultaneity. If I added more perspectives, their lines of simultaneity would look like a hysteresis eye around the thick green line which is the only one that joins 2 proper times as endpoints. The Loedel line of simultaneity (which I've tried to coin) is not discussed in relativity but to me, mathematically, it has significant attributes especially that it provides an unambiguous perspective of proper time.

    Notice there are two light lines emanating from Bob and Alice signalling each other simultaneously from the Loedel perspective at t'=2. I will show how light lines are used to stitch together the simple graphical construct that will be the mathematical building block of graphically representing all the physical phenomena of relativity. But for now, I will show how the light signals are affected when this Loedel diagram is depicted as Minkowski and reverse-Minkowski diagrams.
    Last edited: Feb 8, 2021

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