# Relativity And Simple Algebra

relativity

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### #902 ralfcis

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Posted 07 August 2019 - 11:41 AM

Upon closer inspection, c is not a velocity vector in the equation (wrong it is a vector as shown in next post)

Y(v + c) = c/DSR

it is a scalar unit, it has no sign or direction. Hence v+c is not an expression of relative velocity and the equation will be of no use in my dissertation on the MMX. Fortunately I have two other equations that do handle relative velocity combination of velocities written as Yv. I have to be very careful of the math because a change in sign of v causes the answer to flip. For example, in the above equation v= -.6c has DSR =2 while v = .6c has DSR = 1/2. Subtraction results in division. Einy's equations don't have to worry about this because v is usually squared so its sign doesn't matter.

Edited by ralfcis, 25 December 2019 - 05:15 PM.

### #903 ralfcis

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Posted 11 August 2019 - 06:49 AM

I couldn't finish this, I will in a few days.

I must admit, after days of thinking I do not have a handle on this yet so I'm going to have to flail around a bit.

I think my last post was wrong, c is not a scalar value, it is indeed a velocity vector. Why do I suspect this? Well, the other two equations, that are based on Einstein's velocity combo law, yield the same results. Here they are again:

Yww = (v +u)(YvYu) where Yw = YvYu(1 + vu/c) and w =(v+ u)/(1 + vu/c2)

DSRw = DSRv * DSRu  where DSRw2 = (c-w)/(c+w)

The sign of the velocities is positive if the direction is positive where the distance increases in the direction of the vector. The velocity w is the relativistic combination of v + u.

So Yc = Yv +cDSR where cDSR = vht which is the half speed velocity through time.

For example for v = .6c

vt = .8c (vt is velocity through time = c/Y)

vh = 1/3c (half of .6c)

vht = .5c (half of .8c) (also vht = c(c-vh)/(c+vh))

Edited by ralfcis, 22 December 2019 - 02:50 PM.

### #904 ralfcis

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Posted 12 August 2019 - 11:09 PM

The math is leading to seemingly nonsensical results when plugging v=c into the above equations.

First Y can't equal infinity, it must always be cancelled out in any useful equation. I proved this was true previously in this thread when I did the math for the twin paradox with the impossible scenarios of the twin returning at -c or accelerating away at +c in post #551.

A brief summary of the result is that light travels at 2c with the velocity through space = c and the velocity through time also = c. Only for light is this true as a material object going near c would have its vt approach  0. Time does not stop for light. There is a discontinuity at v=c where the results for material objects are very different for light at v=c. The math further showed that as Alice approaches c away from Bob after travelling 3 ly, she will permanently age 2 yrs more than Bob. If she returned to Bob at -c, she will permanently age 2 yrs less than Bob. Relativity has no comment on these issues which it deems are out of bounds.

Normally the velocity through time is an observed illusion controlled by the equation

c2 = v2 +vt

where vt = c/Y

So at v = .6c, vt = .8c.

But during the time of relative velocity imbalance, which lasts from the time a change in relative velocity is made to the time it is received, the velocity through time is no longer an illusion and is no longer controlled by vt=c/Y but by vht=cDSR  (vht is the half speed of vt) . So what's happening is constant relative velocity has a constant Y but I don't know what form Y takes during an imbalanced relative velocity. Actually I do know what form it takes but I don't yet know the formula  relating cDSR in terms of Y. It's possible the formula is

Yv(c-v) = cDSR or Yv(c+v) = c/DSR

Some other clues are that the lines of proper simultaneity have a rate of changing slope only during the velocity imbalance period. This rate of slope change may provide another significant formula. Another clue is the time lengths of the light signals multiplied by DSR yield Y.

In the train rear example, the pink light signal from the middle of the platform to the rear of the train takes .625 yrs from t=0 where DSR = 2 in that direction and 1/2 for the yellow light signal direction which has a length of 2.5 yrs. So .625 * 2 = Y = 2.5 * 1/2. This mathematical pattern also appears in the formula

Yww = (v +u)(YvYu) where Yw = YvYu(1 + vu/c2) and w =(v+ u)/(1 + vu/c2)

When you set u=c then w =c because that's the combo result of any v +c.

If w = c then Yw = Y= Y

So plugging in

Yv(c+v) = c/DSR

into the above equation and cancelling out Yw and Yu you get

c=c/DSR which is no longer an equation.

So something went wrong and it's probable that Yw, Yu and Yc are not equal when u=c. Just like in the example of Alice turning around at -c or changing velocity to +c away from Bob, Yc is not infinity and Yu has a finite value that is not equal to the finite value of Yw. I don't know yet but my math senses say this is where to look.

The length of the light signals changes depending on direction (and hence the sign of DSR) and the depiction of the relative velocity. In the Minkowski depiction, light takes longer to reach something receding from it than it does something going towards it. The very definition of relative velocity says these are two different combined relative velocities yet the MMX found no difference. Just as DSR gives Alice the power to go many times c through time during the velocity imbalance period, so too could DSR give light the same power to adjust its velocity through time mathematically to accommodate velocities relative to it to keep c constant.

The vt for c is manifest in its frequency which is affected by the DSR. It's possible that when light is blue shifted or red shifted, these are the manifestations of c having its vt adjusted by a velocity relative to it so that c remains constant.

Edited by ralfcis, 22 December 2019 - 02:54 PM.

### #905 ralfcis

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Posted 15 August 2019 - 09:45 AM

Ok this is a tough problem to solve and I love it. So far the problem with the equation is that infinity +1 is not equal to infinity and therefore can't cancel each other out as v approaches c. Let me show you what I mean in a numerical example.

v = 3280/3281 c. This is a velocity through space quite close to c.

vh = 40/41c. This is the half speed velocity through space.

To check, plug vh into the combo equation

v = (40/41 + 40/41) / (1 + 402/412) = (2*40) / (412 +402)/41) = (41* 2 * 40) / (412 +402) = 3280/3281. It checks

vt = 81/3281c The observed velocity through time.

Plug these into the formula

v = vh (1 + vt/c) (or vt = vht (1 + v/c) because v and vt are two sides of the same coin and interchangeable but not how relativity defines this.)

3280/3281 = 40/41 ( 1 + 81/3281) = 40/41 (3281 + 81) / 3281

So 3280 = 40/41 (3281 + 81). It checks

But as v -> c we see a pattern in the equation

v -> oo/(oo + 1) where oo is infinity. Whatever number close to c you choose, v can always be written as v = n/(n+1). Yeah, mind blown relativists. I guess Einy missed this relationship between c and infinity.

So since that is true for v, what is the formula for vh using n.

vh = n / ((n+1) + sqrt(2n+1))

vh is always smaller number than vas v -> c, the "infinities" of the gammas for v and vh have finite differences which can be exploited (the infinite parts can be cancelled out) to have valid equations when v=c. Now I need to show how this happens.

Edited by ralfcis, 22 December 2019 - 02:58 PM.

### #906 ralfcis

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Posted 15 August 2019 - 04:45 PM

I find this unexpected but the formulas

v = n / (n+1) and vh = n / ((n+1) + sqrt(2n+1))

work for any number (not just integers).

I was expecting it to work only for large integers. This means I'm not seeing a very fundamental underlying math and, like fluke-artist Einstein, have stumbled across something huge by accident without yet understanding it.

Edited by ralfcis, 22 December 2019 - 02:59 PM.

### #907 ralfcis

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Posted 15 August 2019 - 06:20 PM

Since vht = cDSR and vt is interchangeable with v, let's re-write the formulas in a more useful form.

vt = m / (m+1) and vht = m / ((m+1) + sqrt(2m+1))

Replace DSR with vht in the formula for velocity combination

DSRw = DSRv * DSRwhere w = c(1-DSRw2) / (1+DSRw2)

vhtw = vhtv * vhtu

So if v = .6c, vt = .8c and thtv = .5c

and if vu = .8c, v= .6c and thtu = 1/3c

so vhtw = 1/6 c = c DSR

Solving for combined velocity through space w we get 35/37c which is correct.

Let's check the results using m.

if w = 35/37, w= sqrt (c2 - w2) = 12/37

wt = 12/37 = m / (m +1)

so m=12/25 and (m+1) = 37/25

Plugging these into the top formula we get

vhtw = (12/25) / ((37/25) + sqrt (24/25 +1)) = (12/25) / ((37/25) + 35/25) = 1/6 and that is indeed the half speed of w12/37 which is the corresponding velocity through time for w = 35/37 with a DSR of 1/6.

Looks like an even simpler pattern is emerging:

If w =d/e and wt = f/e, then vhtw = f / (e+d)

So for the example of v = 3280/3281, vt = 81/3281 so vhtw = 81/ (3281 + 3280) = 1/81 which is correct.

Since we know what answers to expect, it should be easy to work backwards to derive the formal math.

Edited by ralfcis, 22 December 2019 - 03:00 PM.

### #908 ralfcis

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Posted 16 August 2019 - 10:42 AM

So my math flailing has led me full circle back to these two equations:

vht = vt / (1 + v/c) and its counterpart vh = v/ (1 + vt/c)

Since vht = c DSR and vt = c/Y, the first equation can be re-written

Y*DSR = 1 / (1 + v/c) = F which I'll call the F factor.

As v-> c, F -> 1/2 when written as 1 / (1 + v/c)  but Y*DSR do not seem to approach 1/2. Y -> oo and DSR ->0 which is indeterminate.

How do you prove Y*DSR -> 1/2 and are not indeterminate?

Well Y = (c2 + vh2) / (c2 - vh2) and DSR = (c-vh) / (c+vh)

Y*DSR =( ((c2 + vh2) - 2cvh ) / ((c-vh)(c+vh))) * ((c-vh) / (c+vh))

= 1 - 2cvh / (c+vh)

as v -> c so does v

So

Y*DSR = 1 - 2cc / (2c) = 1/2 not indeterminate when v=c.

So I need to figure out how to use this result to scrub out the infinities in my velocity combo equations which are

Yww = (v +u)(YvYu) where Yw = YvYu(1 + vu/c2) and w =(v+ u)/(1 + vu/c2)

or DSRw = DSRv * DSRwhere w = c(1-DSRw2) / (1+DSRw2)

Edited by ralfcis, 22 December 2019 - 03:02 PM.

### #909 ralfcis

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Posted 16 August 2019 - 01:15 PM

Turns out to be a very neat little proof. We want to see as u->c what is the combined velocity w of v+u

c/DSR= Yv (c+v) with u=c

Yww = (v +c)(YvYu)

Yww =cY/ DSRv

From DSRw = DSRv * DSRwe get

DSRDSRDSRu

So w =  cY/ Yw DSR= c Yu DSR/ (Yw DSRw)

We know as any velocity approaches c, its Y DSR approaches 1/2 so the final result is

w = c.

It shows that cancelling out zeroes and infinities in an equation just won't work, they have to be finite real numbers. Specifically 1/oo does not equal zero as so many people think and the gamma function does not equal infinity when v=c. oo is not a real number, it cannot be mathematically manipulated as a value. It is like a math cancer that must be surgically removed from any equation.

Now we have to get back into what all this means for the MMX. (Michelson Morley experiment)

Edited by ralfcis, 16 August 2019 - 01:23 PM.

### #910 ralfcis

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Posted 17 August 2019 - 08:21 AM

I know on physics forums, math is viewed with disdain and suspicion just like anything people don't understand. It's probably impossible to comprehend how a term Y that goes to infinity can be multiplied by a term DSR that goes to zero and the answer is 1/2. Those of you who know how to type in numbers on a calculator keyboard can verify for themselves the answer is correct. Let's use a velocity very close to c.

v=21523360 / 21523361 c = .99999995c

Y = 21523361 / 6561 = 3280.5

DSR = 1/6561

Y*DSR = 3280.5 / 6561 = .50000001

No matter how close you get to c, the result for Y*DSR (infinity times 0) is 1/2.

Now I'd think relativists with PHD's in physics would understand this. There are no infinities in physics that don't approach a finite number somehow. In this case as Y approaches infinity, DSR tempers that with its approach to 1/infinity and the result is their product approaches 1/2.

PS.  vht = 1/2 vt when v = c,  and vh =c from the equation

vh = v / (1+vt/c) because vt =0 when v = c.

Plug in vh = 21523360 / 21529922 c into the velocity combo equation for the example above to check for yourself.

Edited by ralfcis, 25 December 2019 - 05:52 PM.

### #911 ralfcis

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Posted 17 August 2019 - 10:49 AM

I just saw another crazy result

vhx = vx / (1+1/Y) since vt = c/Y

(F is not equal to 1 / (1+1/Yv) ,F only applies to the equation vht = vt / (1 + v/c) , not its counterpart vh = v / (1 + vt/c))

My previous conclusion that Yc =1 has not been verified by the math above. The product of YcDSRis a finite value (= DSRv =1/2) but what Yc and DSRc are individually has not been verified.

Edited by ralfcis, 25 December 2019 - 05:40 PM.

### #912 ralfcis

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Posted 18 August 2019 - 06:17 AM

Sorry I've made an equation error in my last 3 posts which I will correct in red. I should have realized something was wrong when I concluded vh = 1/2 v and vh = v when v -> c. This contradiction can't be true under any circumstances and yet I tried to pass it off as true.

Turns out in post 917, if you change the value of v as u -> c, you don't get Yu/Yw -> 1/2, you get Yu/Yw -> DSRv.

Edited by ralfcis, 25 December 2019 - 05:48 PM.

### #913 ralfcis

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Posted 21 August 2019 - 09:06 AM

I have an opposite view to the science philosophers who can't do a lick of math that infest "physics" forums. Their perspective is that the math can be mistaken but the philosophy can't be so long as it agrees with observable and measurable phenomena. The foundation of the philosophy can also extend into explaining currently unmeasurable phenomena such as dark matter which is discussed at great length, for no productive reason, in "physics" (i.e. science philosophy) forums.

The science history buffs may differ but I think Einstein was a turning point in science. Math was previously a tool to quantize and fit observation, Einstein showed observation can follow from the math. I think his philosophy is the wrong interpretation of the math so I've been primarily focused on math exploration and the math is pretty weird.

The first sign of weirdness is subtraction becomes division in that a negative velocity causes DSR to flip.

Another sign is the velocities through space and  time can be exchanged in equations.

What's even weirder is that velocities and gammas can be exchanged in equations. Here's how that works:

Certain terms keep appearing so I am going to assign them letters:

A = (v+u)

B= (1+vu/c2)

sqrt(c2B2-A2) = c/YvYu =vtut

The last term shows a conversion between velocities and their gammas and a mathematical relationship between a square root subtraction of squares and an inverted multiplication. These terms and the new math where complex operators such as sqrts, squares, multiplication and division can all be expressed as addition and subtraction, can make complex relationships quite simple.

The complex relationships I need to reveal are:

1. an equation for Y in terms of constituent gammas only

2. an equation for w in terms of constituent gammas only

3. The complimentary expressions using time and space subscripts interchanged.

I think I have those answers but need to do extensive checking and derivations. Once I understand the interrelationships of the operators of this new math , I will have a tool on how to explain the emergence and significance of the MMX.

Edited by ralfcis, 31 August 2019 - 02:46 PM.

### #914 ralfcis

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Posted 21 August 2019 - 09:17 AM

PS. If anyone is interested in shedding their math illiteracy, I will answer mathematical questions mathematically, But if you insist on philosophy or history discussions, I can't help you.

### #915 ralfcis

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Posted 25 August 2019 - 08:39 PM

The math turns out to be very cool. It loops back on itself like a sphere that is completely self-enclosed and no matter what direction you take, you always end up at the same spot. It's like being on a planet at the edge of the universe yet in any direction you look, you're still at the 3D center of the universe (just like you're always at the 2D center of the surface of the earth. So here goes:

c2 = v2 + vt2   The main equation that defines space and time

So  vt = sqrt (c2 -v2 )

Yvt = 1 / sqrt(1- (c2 -v2 ) / c) = c/v

so v = c/Yvt  just like vt = c/Yv as expected because time and space terms are interchangeable in the equations.

We plug this result into relativity's velocity combo equation

w = (v+u) / (1 + vu/c2) = c (Yvt + Yut) / (1+ YvtYut)

It looks like gammas and velocities are also interchangeable in the equations but it doesn't mean you can swap t's and x's however you like.

Let's work out what wt looks like in terms of ut and vt .

This equation previously derived:

Yw = YvYu(1+vu/c2)

wt = c / Yw = (vt ut) / c2 (1+vu/c2) = (vt ut) / (1+1/(YvtYut))

Notice the addition of v and u for w becomes a multiplication of vt and ut for wt.

I think that's a complete enough picture for the conversion operations. Now I need to see what happens when u is set to c. Can I get around the infinity of Yto see what happens to the velocity addition of c and v and show another explanation for the MMX result.

Edited by ralfcis, 25 December 2019 - 08:56 AM.

### #916 ralfcis

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Posted 26 August 2019 - 08:00 AM

So let's multiply these two equations together as good things always happen when you express velocities in terms of Yv.

w = (v+u) / (1 + vu/c2)

Yw = YvYu(1+vu/c2)

so YwwYvYu(v+u)

When v= u= 3/5,  w=15/17, Yw= 17/8, YvYu = 25/16, (v+u) = 6/5.

Let's boost u towards c by setting it to u= 4/5 and keeping  v= 3/5 for now.

so  w=35/37, Yw= 37/12, YvYu = 25/12, (v+u) = 7/5.

Let's double u to 40/41c

so  w=323/325, Yw= 325/36, YvYu = 205/36, (v+u) = 323/205.

Let's re-write YwwYvYu(v+u)

to w(v+u) =  YvYu Yw

As u -> c , the equation becomes

c/(3/5c+c) = 5/4 Yu Yw

So YuYw -> 1/2 (again!)

Turns out in post 917, if you change the value of v as u -> c, you don't get Yu/Yw -> 1/2, you get Yu/Yw -> DSRv.

Let's check the numbers for u = 40/41c

(323/325 / (3/5 + 40/41)) 4/5 = .5046

So infinity over infinity is limited to Yu/Yw =DSRv and the gamma of the combined velocities when one of them is c is Yw =Yu/DSRv  that of the velocity going to c.

Even though both w and u velocities are technically c this result shows there's a difference between c's. c can morph into relative multiples of c and still remain c from each perspective.

We've already seen an example of going many times c (through time) in the twin paradox but now we're going to see how this also happens in the MMX. Sad no one else can share in the beauty of how this multilayered lotus flower is opening up.

Edited by ralfcis, 25 December 2019 - 05:46 PM.

### #917 ralfcis

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Posted 26 August 2019 - 12:35 PM

Important

Another fn mistake. If you change the value of v as u -> c, you don't get Yu/Yw -> 1/2, you get Yu/Yw -> DSRv. Infinity over infinity does not tend to a fixed fraction but one that is highly dependent on v. That makes way more sense and shows how deeply interrelated the math is.

So I need to bring in the formula:

c/DSRv = Yv(c+v) and work it into the equation:

c (v+c) =  YvYu Yw  =  c/(c/Yv DSR)

So  DSRv Yu Yw    Yes! Correct!

This formula is the holy grail. It explains why the length of light signals change due to their direction. In the train station example, the light lines towards a train going at .6c towards the station were only half as long as the light lines going from train to station within the train. The light signal inside the train corresponds to the light signal inside the relatively stationary tube for the MMX and the light signal from the platform to the train corresponds to the moving tube relative to the aether in the MMX.

Since Y = c/vthe light signal in the moving tube at .6c has double the velocity through time as the stationary tube which erases the time effects of the tube's velocity on the total relative velocity of the light signal towards the end of the moving tube. I'm just spitballing, I still need to mathematically verify what i just said.

Too bad all the forumorons got off the bus but, in truth, the trip wasn't going to change them anyway. Oh, and I may be premature but Einstein's explanation for the MMX really looks kind of clumsy now. That's not going to help me but I really enjoy sticking it to relativists.

Edited by ralfcis, 25 December 2019 - 05:30 PM.

### #918 ralfcis

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Posted 28 August 2019 - 02:48 PM

I had decided that I was going to spend all day today explaining the physical relevance of the math I just presented but it's 4:46 PM and I haven't even started yet. I hope I can get to it sometime tonight because it's some really exciting stuff.