# Relativity and simple algebra

Discussion in 'Physics & Math' started by ralfcis, Jan 29, 2021.

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1. ### ralfcisRegistered Senior Member

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Relativity can be explained with basic algebra using only 2 formulae (the one for gamma 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.

I have a bunch of useful forms of the two main equations but I don't notice any easy way to put subscripts and superscripts here.

Last edited: Jan 29, 2021

3. ### DaveC426913Valued Senior Member

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15,410
Best way to display formulae is to use LaTeX.

[ TEX ] H_2O^2 [ /TEX ] = $H_2O^2$

5. ### mathmanValued Senior Member

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1,831
I presume you are discussing special relativity.

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10,288
H₂O²?

Wot dat?

8. ### DaveC426913Valued Senior Member

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15,410
Sorry. That was a typo. Here it is:

$H_2O^3$

It's an ice cube!

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

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375
Yes, it'll take some time to roll out the whole picture. It's mostly math and spacetime diagrams. However if you have specific questions I'll answer them.

10. ### ralfcisRegistered Senior Member

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375
Thank you I will try that.

11. ### ralfcisRegistered Senior Member

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375
The main equation (space and time velocity combination) is

c^2 = v^2 +v_t^2 also written as the gamma function
Y = c / sqrt(c^2-v^2) where ( v_t= c/Y and v = c/Y_t)

v_t 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.

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:

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

or (ct)^2= (ct')^2+ Yx^2 (Epstein pythagorean (sum of squares) form where circles intersect the same proper time for all velocity lines) [/tex]

The basic graphical building block will be shown next

I tried $v_t$. It doesn't work so my equations will be difficult to follow

Last edited: Jan 29, 2021
12. ### DaveC426913Valued Senior Member

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15,410
It looks fine.
I see this:
$v_t$

[ TEX ]Y = \frac{c}{ \sqrt{(c^2-v^2)} } [ /TEX ]

$Y = \frac{c}{ \sqrt{c^2-v^2} }$

Note: LaTeX will not be immediately visible after you save your post and view it. You must refresh the page to see it it rendered correctly.

Last edited: Jan 29, 2021
13. ### ralfcisRegistered Senior Member

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375
Ok one more time

The main equation (space and time velocity combination) is

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

$v_t$ 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.

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:

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

or $(ct)^2= (ct')^2+ x^2$ (Epstein pythagorean (sum of squares) form where circles intersect the same proper time for all velocity lines)

The basic graphical building block will be shown next

Last edited: Jan 30, 2021
14. ### ralfcisRegistered Senior Member

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375
Ok now I need to create a sheet of all my equations with the latex control script visible so I can cut and paste equations in subsequent posts. How do I make the script words like [tex visible like you did in your examples?

Last edited: Jan 30, 2021
15. ### ralfcisRegistered Senior Member

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375
[$h_2^0$]

Last edited: Jan 30, 2021
16. ### ralfcisRegistered Senior Member

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375
This is my equation sheet to cut and paste equations:
(Not all were successful, a 2nd page will be needed for corrections)

[ tex] [/tex]

[ 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_h_t(v/c +1)[/tex]
[ tex]v = 2c^2v_h / (c^2 + v_h^2) [/tex]
[ tex]v_t= 2c^2v_h_t/ (c^2+ v_h_t^2)[/tex]

[ tex]v_h = c(c-v_h_t) / (c+v_h_t) and v_h_t = c(c-v_h) / (c+v_h)[/tex]

[ tex]v/(c-v) = 2cv_h / (c - v_h)^2[/tex]

[ tex]Y = 2Y_h^2 - 1 = (c^2 + v_h^2)/(c^2 - v_h^2) = c/v_t = (1/(1-v_h/v)) -1 = 2c^2v_hv/(c^2 - v_h^2) = 2Y_h^2 v_h /v [/tex]

[ tex]v_h=Yv/(Y+1) [/tex]

[ tex](Yv)^2 = v^2/(1-v^2)[/tex]

[ tex]v= c sqrt(Y^2-1)/Y[/tex]

[ tex]v_t = c/Y = DSR(c+v) = v_h_t(1 +v/c)[/tex]

[ tex]t'=x(v_h +c)/c [/tex]

[ tex]v' = Yv[/tex]
[ tex]t = Yt'[/tex]
[ tex]x'=Yx[/tex]

[ tex]Yv =x/t'[/tex]

[ tex]Yu/Yw=DSR_v[/tex]

[ tex] t'(1-Y) [/tex]
[ tex]t'=x/Yv[/tex]
[ tex]vx/c^2[/tex]

[ tex]t'=sqrt(t^2- x^2)[/tex]

[ tex]t' =xDSR_o/Yv[/tex]

[ tex]t_p_s= xv_p_s = xYv/(1+Y) [/tex]

[ tex]wt= c / Yw = c / (YvYu(1 + vu/c^2)) [/tex]

[ tex]Y_ww = (v+u)YuYv [/tex]

[ tex]v_h_t=cDSR [/tex]

[ tex] cDSR = sqrt((c-v)/(c+v)) [/tex]

[ tex]DSR_v= Yu/Yw[/tex]

[ tex]Y(c-v) = c/DSR[/tex]

[ tex]DSR_w= DSR_v* DSR_u [/tex]
[ tex]DSR_w^2= (c-w)/(c+w) [/tex]

[ tex]w = c(1-DSR_v^2 DSR_u^2 )/(1+DSR_v^2 DSR_u^2 )[/tex]

[ tex]DSR = Y(1-v/c)[/tex]

[ tex]t' = X(c+v_h)/c [/tex]

[ tex] t'(DSR - 1)[/tex]

[ tex]t' = X(-2v_h)/(c+v_h) [/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]

v Y $v_h$ $v_t$ $v_h_t$ DSR t'

1 1 1 0 0 0

3280/3281 3281/81 40/41 81/3281 1/81 81

40/41 41/9 4/5 9/41 4/36 9

255/257 65/16 15/17 16/65 1/8 8

24/25 25/7 3/4 7/25 1/7 7

35/37 37/12 5/7 12/37 6/36 6

12/13 13/5 2/3 5/13 1/5 5

77/85 85/36 77/121 36/85 8/36 9/2

15/17 17/8 3/5 8/17 9/36 4

4/5 5/3 1/2 3/5 12/36 3

3/5 5/4 1/3 4/5 18/36 2

1/2 1.155 .268 .866 .577 1.73

8/17 17/15 1/4 15/17 3/5 5/3

5/13 13/12 1/5 12/13 24/36 3/2

1/3 1.06 .17 .943 .707 1.41

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

0 0 1 1

21523360/21523361 21523360/21529922 6561/21523361 1/6561

Last edited: Jan 30, 2021
17. ### James RJust this guy, you know?Staff Member

Messages:
35,782
What is $Y_t$?

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

You haven't yet posted a derivation of this.

What does $t'$ indicate, in this equation, as opposed to $t$? I notice there are no $x'$ terms in the equation...

18. ### ralfcisRegistered Senior Member

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375
Ok problem solved with the equations that weren't coming out properly:

[ 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 = c/Y = DSR(c+v) = v_{ht}(1+ v/c)[/tex]

[ tex] t_{ps} = xv_{ps}= xYv/(1+Y)[/tex]

[ tex] v_{ht} =cDSR[/tex]

v Y [ tex]v_t[/tex] [ tex]v_h[/tex] [ tex]v_{ht}[/tex] DSR

19. ### ralfcisRegistered Senior Member

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375

$v_t = c/Y = ct'/ct$ 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.)

I'll answer each question in a separate post.

20. ### ralfcisRegistered Senior Member

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375
$v_t= c/Y and v = c/Y_t$

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.

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

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35,782
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?

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.

22. ### ralfcisRegistered Senior Member

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375
You haven't yet posted a derivation of this.

I did on another forum. I can cut and paste it here but it's long and not written in TEX. I have found that no one has ever taken the trouble to look at my math because it takes a lot of effort to do so. It's not like Good Will Hunting where mathematicians read math like it was a language. So I'd suggest if you doubt the formula, plug in some values and make sure the numbers come out correctly.

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