# Relativity And Simple Algebra

relativity

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### #868 sluggo

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Posted 08 July 2019 - 09:38 AM

"If Bob drove 60 miles at 60 mph to see Alice, he would see her in 60 min. If he asked her to leave at the same time and drive to meet him, he would see her in 30 min. No one exceeded the speed limit, they only drove half the distance simultaneously.]"

But the relative velocity would be 120 mph. Alice's car is going 60 mph towards Bob. Bob's car is going 60 mph towards Alice. No one is exceeding the speed limit but their relative velocity is 120 mph. The gap between them is also closing at 120 mph which a person on the side of the road witnesses. The closing speed equals the relative velocity except for an insignificant relativistic effect. But with higher relativistic effect, the closing speed gets up to two times the relative velocity. Bob and Alice could  see their relative velocity is also 120mph with radar guns. I just don't see how two cars headed towards each other wouldn't crash at twice the speed as one would being parked and the other hitting him at 60mph. That's how head on collisions work. If what you're saying is true then we live in 2 different physical universes.

Agree with gap closing at 120. Alice moves 30 mi in 30 min, v=60. Bob moves 30 mi in 30 min, v=60.

Note, they move simultaneously

We have accounted for all objects in motion. Where is the thing moving at 120?

### #869 sluggo

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Posted 08 July 2019 - 09:56 AM

Also true is the fact that sound is not effected by the motion of the person hearing it.

Doppler shift results from relative motion of source and detector.

Most people have heard the variation from sirens.

### #870 ralfcis

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Posted 08 July 2019 - 10:11 AM

Where is the thing moving at 120?

The thing is called relative velocity. They are each moving at 60 mph relative to the road but they are moving at 120mph relative to each other. The closing speed, as viewed by someone watching them from the side of the road, is not a relative velocity. It does not depend on how close he is to them or even if he's directly in line with them. But relative velocity of each to him does depend on how far he is from the road because relative velocity is a vector. If he was watching two satellites closing at 120mph, their relative velocity to each other would be 120 mph but to him it would be near zero. No? None of this sounds familiar?

Edited by ralfcis, 08 July 2019 - 10:38 AM.

### #871 sluggo

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Posted 09 July 2019 - 10:43 AM

Ralf;

But with higher relativistic effect, the closing speed gets up to two times the relative velocity.

[Composition of velocities does not agree.

B velocity v, as measured by A, v = (b-a)/(1-ab), a = .9 and b=-.9.

v= -1.8/1.81 = .99.]

### #872 sluggo

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Posted 09 July 2019 - 10:46 AM

Ralf;

The bus is mostly empty now. A ride to nowhere.
This is where I get off.

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### #873 ralfcis

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Posted 09 July 2019 - 11:18 AM

You really are illiterate. I said closing speed can be 2c  not relative velocity which must always be less than c according to the velocity combo law. Where do you get the balls to think you can discuss relativity when you don't even know what relative velocity is? Please stay off the bus this time until you learn some basic terms at least. I wouldn't be surprised if no one here knows what velocity is let alone relative velocity because they let your total ignorance pass without a word.

### #874 ralfcis

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Posted 09 July 2019 - 11:34 AM

Since you're too lazy  to google questions here's a link for you

I know you have severe reading disabilities but you're going to have to try to join the letters into words, words into sentences and use that place holder between your ears to try to make sense of the sentences.

### #875 ralfcis

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Posted 09 July 2019 - 09:18 PM

I'm hoping everyone got off the bus with Sluggo because I'm going to go way off road. I'm going to try to derive new math for velocity addition based on Yv + Yu = ?. It should be a more linear way to add velocities than Einstein's equation and should require no hidden tricks to make things look like they apparently work. The math will look nothing like Einy's math and be based on none of his assumptions so relativists will have no hope of understanding it. They're pretty busy with trying to understand what relative velocity is so they may not have time to grasp the new math. Don't know how long it will take or when I'll start or if it'll be ultimately successful so this may be a rabbit hole you might want to avoid as I flail around.

### #876 ralfcis

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Posted 15 July 2019 - 10:06 AM

Ok equations and their derivations are coming. Here's what I have so far but it's not formally derived:

The first equation, as you may remember, is the relative velocity to c.

DSR = Yv(1-v/c) if v is negative.

example v= -.6c (Alice going back to Bob)

Yv = 1.25 (I've added the subscript because when I start adding velocities you need to know which Y belongs with which velocity)

DSR = 1.25 (1.6) =2

What's interesting in this equation is the sign of v controls the inverse of DSR.

if v=.6c (Alice going away from Bob)

DSR = 1.25 (.4) = 1/2 which is the inverse of the original DSR. (Doppler Shift ratio).

These tricks come in handy when you want to reduce complex expressions with squares, ratios and square roots into expressions of simple addition.

From this I can derive an intermediate equation on my way to linearizing the velocity combo law:

Yvv=2Yh2v

An example of this is if you want to express .6c relative to a stationary earth as two ships leaving earth in opposite directions at 1/3c which is the half speed vh of v=.6c. Notice I try to make my math accessible unlike 006 which tries to obfuscate what he's doing like frauds tend to do.

So putting in these numbers we get

1.25(.6) = 2(1.06)2 /3 and it checks out if any of you know how to use a slide rule. (sarcasm)

So the above formula shows how to add 2 half velocities, the next shows how to add any velocities:

Yww = (v+u)YvYu where Yw=YvY(1+vu/c)

For example v=.6c and u=.8c. What is the combination w of these two velocities?

v=.6, Yv=1.25=5/4

u=.8. Yu=5/3

YvY= 25/12

YvY(1+vu/c) = 37/12

w=((1.4)25/12)/(37/12)=.9459

If we use Einy's formula

w= (v+u)/(1+vu/c2) =(1.4)/(1.48) = .9459

This is some extremely powerful math. Let's do another example.

v=1/3c and u = 1/2c

Yv= 1.06066 and Yu = 1.1547

YvY= 1.22474

YvYu (1+vu/c) = 1.428868

w=((.833)1.225)/1.428 = .7142c

Let's check that against the old formula:

w=c(v+u)/(c+vu) = .833/1.167=.7142c

You might look at this and say, hey man, the old equation looks simpler. Figuring out w is not the goal, Yww is the goal because velocity expressed in terms of Yv is not subject to a limit, velocities expressed as Yv can be linearly added according to the formula above and then converted into Einy's expressions of v limited by c. This new math will have powerful implications later.

Yww =YuYvv +YuYvu where Yw=YuYv(1+vu/c) .

I'm not really interested in any critiques from math illiterates if you're thinking on commenting.

Edited by ralfcis, 09 August 2019 - 07:05 AM.

### #877 ralfcis

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Posted 15 July 2019 - 11:48 AM

I just noticed DSR is independent of how you depict relative velocity. For example, DSR is the same between two opposite velocities at 1/3 c as it is for .6c relative to a stationary reference frame. This is not true for Y in Einy's theory. So converting the above equations using DSR instead of Y would make them truly based on relative velocity independent of a reference frame. DSR is based on light signals which are independent of perspective lines of simultaneity. Causal time is also based on independence from a reference frame. I'm going to explore where this leads next. Does anyone need more detailed clarification?

Edited by ralfcis, 15 July 2019 - 11:48 AM.

### #878 ralfcis

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Posted 15 July 2019 - 10:20 PM

Ok I've been skipping ahead to the answers without doing the formal math derivation because I just don't have the time. It's not like anyone here is going to notice. Plus when you know the answer it's easier to work backwards from that. So the answer is if you want the sum of two velocities you use these two formulas:

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

So let's do an example

Alice is leaving earth at .6c and Bob is leaving earth in the same direction at .8c. What is their combined relative velocity?

We know from the 2nd formula that DSRv = 1/2 and DSRu = 1/3

so plugging that into the 1st formula we get DSRw =1/6

Now we want to solve for w in the 2nd formula and we get

1/36c +1/36w = c - w

w=c(1-1/36)/(1+1/36)

w = 35/37c = .9459c

This is the same result as the old formula got for the relativistic combination of .6c and .8c but in a much simpler way arithmetically. No more Y, no more time dilation or relativity of simultaneity or lines of perspective simultaneity (aka length contraction) need be calculated. Since light lines define the DSR and can be used to sync clocks, Einy's clock sync method also ends up in the trash. There's also a new clock start method, using the causal line of simultaneity, for the start of the spacetime path for when the participants do not start co-located. This new velocity combo law, along with the abolition of Einy's spacetime path rules, will make quick work of calculating age difference without the requirement of co-location at the end of the path. Pretty damn neat if I do say so myself even if no one here has a clue of what I'm talking about. Now I just gotta sit here and wait to be discovered.

Edited by ralfcis, 09 August 2019 - 07:45 AM.

### #879 ralfcis

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Posted 17 July 2019 - 05:02 PM

Now that I have the new math on which to base the workings of the universe, I have to put it into practice on correctly solving all the relativity examples like MMX, train/station, muon, GPS, light clock, ladder paradox  and most importantly the twin paradox which relativity fails at miserably (does anyone remember why it fails?).

### #880 ralfcis

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Posted 20 July 2019 - 09:14 AM

One of the most important questions that I've never heard a satisfactory answer for is how can two clocks run at different rates relative to each other but still run at the same rate independently? Relativists say time is affected by velocity yet it is not affected at all within the moving frame, it still ticks at the normal rate from that perspective. That must mean that the space between frames experiences some sort of time distortion between the two frames. When 1 frame is observing the other, the space between them must be distorting the observation of the normal rate of time that is going on within the observed frame.

If you were to observe the broadcast signal of a frame leaving you at .6c, you'd see the TV picture in slow motion at half speed the normal rate of motion. This does not mean time itself has slowed for the people inside the ship. No, they are moving through time at the normal rate. What is happening is relativistic Doppler shift, the frequency of the TV signal is getting stretched so the rate of information coming to you is reduced.  Plus since relativistic doppler shift includes Y, time itself is slowed. (edited)

The question remains is time dilation really the result of time slowing? Time dilation is always the same whether the frames are approaching or receding while DSR is dependent on the sign of the velocity. You would see the rate of time double in the TV signal of a ship approaching at .6c but the time dilation would still be at 80% the normal rate. This agrees with the formula that the rate of time through time vt=c/Y.

But what does this mean really as you and the occupants in the ship are still ageing at the same rate of 1 year per year. The DSR doesn't really give you an idea that observed time is slowing somehow for the approaching ship as the TV signal looks like they're going at double speed fast forward. You'd have to relate Y in terms of DSR to mathematically see what's happening:

Y = cDSR/(c-v)

This formula tells us that if Y is the slowing of time itself, then it can be indirectly observed in the  time rate of the Doppler shift ratio.

Let me tell you about a parallel dimension Earth where clocks were never invented and time was told by the position of the sun in the sky via sundials. The earth rotates at 1000 mph relative to the sun's position in the sky. If a plane took off west from Ottawa at 1000 mph, his sundial would read the same time throughout his journey. Einstein would have maybe concluded that the guy is going an infinite distance in zero time. Einstein's definition of time was time is what clocks measure. Well this guy's clock is saying time has stood still while he is in fact ageing at the normal rate. So sorry, Einy, time and what clocks measure are two different things.

But is this what's happening with reciprocal time dilation (my sundial example was not reciprocal). Is Y about relating differing clock readings or are those clock readings being affected by the magical time warping space between two frames experiencing normal rates of time? I'll have the answer in my next post.

Edited by ralfcis, 20 July 2019 - 10:06 PM.

### #881 ralfcis

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Posted 20 July 2019 - 09:57 PM

OOps made a slight mistake  and corrected it in red above. There are 3 terms: doppler shift, relativistic doppler shift and doppler shift ratio. The formula for relativistic doppler shift is

1/f = c(c-v)/f0 sqrt(c2-v2)

(f0/f)2 = c2(c-v)2/(c2-v2) = c2(c-v)2/((c-v)(c+v)) = c2(c-v)/(c+v)

which means (f0/f)= DSR2/c

This means DSR is not a plain old doppler shift but has Y  and time dilation built right in.  So the TV picture being received is partly affected by plain old doppler shift and by time dilation in the transmission of clock info between frames. So part of the DSR reflects a real effect on time rate due to velocity.

Edited by ralfcis, 20 July 2019 - 10:09 PM.

### #882 ralfcis

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Posted 22 July 2019 - 09:27 AM

Ok I've found the answer: there are two velocity combo laws. c can't be exceeded not just by the summation of velocities between  frames but also in the summation of velocities that make up c. The formula is

c2 = vx2 + vt2

c is the velocity everything goes at and it is made of two component velocities: the velocity through space vx and the velocity through time vt. The summation of the square root of the squares of those velocities can't exceed c just like the sum of two velocities through space can't exceed c. Of course the speed limit of c is just an assumption that explains the experimental facts and is supported by math.

So it looks like there's only 1 velocity and since everything goes at that velocity and since that velocity is also the maximum velocity then all relative velocities are c and only c. By the standard velocity combo law, any vx relative to c will be c. So how do we make vx stand on its own without being erased by c? We multiply both sides of the top equation by Y.

Y2c2 = Y2vx2 + Y2vt2

Since vt = c/Y (derived from the top equation) and t=Yt' we have Yvx = x/t' and Yvt =c.

What does this mean in English? It means that if Alice sees her clock at 4yrs, which is the proper instantaneous present time for both her and Bob, and finds herself 3 light years from earth, it means her Yvx = 3/4. Bob who gets a message from Alice 3 yrs later also knows it took her 4 yrs to reach her destination since they aged at the same proper time rate during the journey. Even though time passed at the same rate for both, Alice's perspective of how much time passed for Bob is only 3.2 Bob yrs and Bob's perspective of how much time passed for Bob is 5 Bob yrs. When you properly convert type of year with perspective the answer is universally 4 proper time years.

So Yvis the correct, universal interpretation of velocity through space and, as shown in the previous posts, is not erased by c.

Yc -Yv= cDSR where vx is positive if separating. (I have trouble remembering the sign of vwhich is important because it causes DSR to flip.)

So we see that Yv is an entire time domain where c is also multiplied by Y. This domain of x/t' treats c as any other velocity, since it must be multiplied by Y, but still does not violate that in the x/t domain, c is still limited to c. Yc might be 1.25 c but c is still c.

I didn't previously derive the above formula so this is how it's done:

Y2 = c2/(c2-v2) = c2(c-v)/((c+v)(c-v)2) = c2DSR2/(c-v)

so Y(c-v) = cDSR and Y(c+v) = c/DSR. Any questions math wizzes? (sarcasm)

What's weird about this equation is that it avoids the infinities that crop in when plugging c into any other velocity combo equation. That must be significant but I don't know how yet.

Another neat thing about using equations with Yv is that Yvt=c in them. This means that while Alice is covering massive distances in little of her time, the equation Yvt=c shows that time is passing at the normal rate for her while she's doing it.

Since I'm a graphical guy I'll show you how all this is represented on an STD. Any of this clicking with anyone out there yet?

Edited by ralfcis, 09 August 2019 - 07:54 AM.

### #883 ralfcis

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Posted 22 July 2019 - 09:51 AM

Oh I forgot to mention a very important point. I said everything moves at c but that's really relativity's schtick. What's really happening is everything is transmitting and receiving information at c. The universe goes to great mathematical trouble tweaking the information time rates to ensure this. It makes sure all the information transmitted is received so there is a conservation of information. The transfer of information from things separated from you within your frame is delayed but the time rate of information transfer is still vt=c. The further things are away, the greater the backlog of info that is not reaching you. You are seeing them in the further and further past. If they are receding from you, that backlog of info is increasing at a rate less than vt=c but if they are approaching you, the rate of information transfer is faster than c. This is the doppler shift ratio at work to get rid of the backlog faster.

But what would happen if the transfer rate ran out of backlog information to send? Would you get info from the future to fill the pipe? No. This is what happens in the twin paradox but it is taken care of for part of the journey back, less information is being created by forcing the returning twin to travel faster through time during the imbalance of relative velocity between the two frames. I went into great detail of how this happens previously but that discussion can't be understood by relativists and non-relativists alike because both sides are equally closed-minded.

PS. Here's a teaser for the more advanced among you (sarcasm again). Understanding the nature of DSR and Y wrt time lies in understanding the mathematical relationship between vx, vt and half speed vx and vt.

Edited by ralfcis, 22 July 2019 - 10:02 AM.

### #884 ralfcis

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Posted 27 July 2019 - 09:07 AM

I've spent many days analyzing these two STD's and have come to the conclusion that it's impossible, in relativity, to differentiate .6c between two ships leaving earth at 1/3c and 1 ship leaving earth at .6c. These STD's are not equivalent depictions of .6c relative velocity even though they look mathematically equivalent.

https://photos.app.g...dkCDU91bDDsWRW6

https://photos.app.g...nzUQWMZ173jdcg8

The problem is DSR is the same in both STD's so you get the same value for Y relative to each other even though 1/3c has a different Y relative to earth than .6c does. The problem stems from the timeline for earth in the .6c depiction and in the 1/3c depiction, they can't both be stationary. This means other depictions of relative velocity are also not equivalent and depend on setting the stationary frame as the one that's more massive. So the following relativistic ideas are wrong mathematically:

1. The sun does not revolve around the earth, nor does the earth revolve around satellites.

2. The LHC does not revolve around protons.

3. The entire universe moving past a stationary (to what?) spaceship is not equivalent to a spaceship moving through space.

4. Earth does not approach a muon at near c

These are idiotic ideas that even Einstein eventually realized were idiotic.

So when you do the analysis of pure relative velocity it is only valid in a grey featureless universe between two equally massive objects. Otherwise, in the real world, you must anchor the more massive object, such as earth, as the stationary frame. This does not imply an absolute frame but one that you can use to make meaningful relativistic calculations against. With pure relative velocity you can't tell who's moving how fast, how far and in what time but in the real world there are distance markers that are set without using time to determine them in 0 relative velocity frames. This means different depictions of relative velocity will have the participants end up in different points in space. When they do, you can tell the difference between two ships separating at 1/3c and one separating from the earth at .6c.

Now, how this affects me is I can't see a neat way to use light signals to determine time using pure relative velocity alone. Einy defined perspective time using light sync'd clocks. That idea is just stupid for the many reasons I've given before. I had hoped that the universality of atomic clocks would not need his pre-sync'd clock method but could be used to tell what the relative time was on the fly. Yes this can be done but it's not as neat (based only on pure relative velocity) as I had hoped. It requires what looks like mathematical fudging based on anchoring the stationary frame to the more massive object.

Edited by ralfcis, 27 July 2019 - 09:10 AM.