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Relativity And Simple Algebra

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#902 ralfcis

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Posted 06 August 2019 - 11:15 AM

100% of the people reading my last post would agree that from the Earth's perspective, its time rate was 9.14 times the muons time rate. Earth aged .627 Earth pYrs from the muon's birth until its death in .068 muon pYrs. Even if the muon sent out a light signal to earth at its birth, Earth would not have been aware of the muon's birth until .624 Earth pYrs had passed. .003 Earth pYrs later the muon arrives so it could be argued that from the real time earth's perspective, the muon only aged .003 Earth years from birth until death. Relativity's definition of earth perspective time cannot be experienced in real time just like the causal time perspective (which is .068 pYrs for both) can't be experienced in real time. Only from the arrival of light signals from a known distance can these two perspectives be calculated after the fact. By then, they have no relevance to reality. The only thing that's real in real time is the muon's perspective of how much distance it travelled in it's own time at Yv.

 

The next example I'll do is the MMX. Hopefully I can answer the riddle why does v+c = c in the velocity combo law yet Yv + Yc = c/DSR in my velocity combo law and what does it mean to understanding the results of MMX where the velocity of the inferometer had no effect on c. 


Edited by ralfcis, 06 August 2019 - 11:18 AM.


#903 ralfcis

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Posted 07 August 2019 - 11:41 AM

Upon closer inspection, c is not a velocity vector in the equation

 

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.



#904 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 vx = .6c (vx and v are used interchangeably for velocity through space)

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

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

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



#905 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 = vx2 +vt

 

where vt = c/Y

 

So at vx = .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 causal 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, 14 August 2019 - 08:04 AM.


#906 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.

 

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

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

To check, plug vhx into the combo equation 

vx = (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

 

vx = vhx (1 + vt/c) (or vt = vht (1 + vx/c) because vx 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 vx -> c we see a pattern in the equation

 

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

 

So since that is true for vx, what is the formula for vhx using n.

 

vhx = n / ((n+1) + sqrt(2n+1)) 

vhx is always smaller number than vas vx -> c, the "infinities" of the gammas for vx and vhx 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, 15 August 2019 - 09:46 AM.


#907 ralfcis

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Posted 15 August 2019 - 04:45 PM

I find this unexpected but the formulas

 

vx = n / (n+1) and vhx = 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, 15 August 2019 - 08:00 PM.


#908 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 vx = .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 vx = 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.



#909 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 + vx/c) and its counterpart vhx = vx / (1 + vt/c)

 

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

 

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

 

As v-> c, F -> 1/2 when written as 1 / (1 + vx/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 + vhx2) / (c2 - vhx2) and DSR = (c-vhx) / (c+vhx)

 

Y*DSR =( ((c2 + vhx2) - 2cvhx ) / ((c-vhx)(c+vhx))) * ((c-vhx) / (c+vhx))

 

= 1 - 2cvhx / (c+vhx)

 

as vx -> c so does vhx 

So

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



#910 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.


#911 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 vx = c,  and vhx =c from the equation

vhx = vx / (1+vt/c) because vt =0 when vx = c.

 

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


Edited by ralfcis, 18 August 2019 - 06:45 AM.


#912 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 + vx/c) , not its counterpart vhx = vx / (1 + vt/c))

 

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


Edited by ralfcis, 18 August 2019 - 06:48 AM.


#913 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 vhx = 1/2 vx and vhx = vx when vx -> c. This contradiction can't be true under any circumstances and yet I tried to pass it off as true.


Edited by ralfcis, 18 August 2019 - 06:57 AM.


#914 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(B2-A2) = 1/YvYu

 

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, 21 August 2019 - 09:10 AM.


#915 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.





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