Relativity And Simple Algebra

[ralfcis]

I'm not really a fan of people squatting on my thread with their personal theories. This place is indeed a looney bin but it doesn't mean there's only one general ward without private rooms. I promise not to crash or trash yours.

[Mutex]

I'm not really a fan of people squatting on my thread with their personal theories. This place is indeed a looney bin but it doesn't mean there's only one general ward without private rooms. I promise not to crash or trash yours.

 

Rightly so, I probably would not be a fan of that either, however, my response and Flummoxed's response are the continuation of the question you asked.. 

 

 

"Is there a physical (non-time based) ruler that allows one to directly measure length contraction? "

 

So as light has a length value that can directly measure length contraction,  a discussion of why light has a length and why and how that is a function of the properties of the length of space and time at the creation of that photon, it is on topic and pertinent to your question. It's kind of the nature of how science is done.. 

[ralfcis]

This is a thread about relativity, not unrelated theories from bizarro world. See if I got into a debate with you, it would be about me shredding every one of your mystical opinions and that should be what you need to do on this thread to my theories. Unfortunately, you'd need to know algebra first and pick out my flaws mathematically instead of via Quija board.

[Mutex]

I guess I'm done with the PSX. I'm banned for lack of clarity. Usually I'm banned for being a dick or for trying to push a personal theory which is what I was partly trying to do (except I don't view pushing new math as pushing a new theory). But their heads are so far up their asses my posts look like hieroglyphics to them (except for Dale who is open-minded but has disappeared). I laid all my cards on the table and they had no idea what the numbers meant or what hand I was playing. How can people deal with the most complex math on the planet and not understand algebra? I can still interact somewhat but what's the point, they have no information for me. Even this place, a total cuckoo's nest, influenced my quest greatly. I changed my views a lot here which usually led me back to my original path which I hope is near the final final end. This last piece of the puzzle on how to use light signals to tell time was huge and a total surprise to me. I don't think there are any pieces left (although I've said that many times before).

 

I'm not seeing much relativity here in this post from you on this thread.. Just saying.. 

 

 

See if I got into a debate with you, it would be about me shredding every one of your mystical opinions and that should be what you need to do on this thread to my theories.

 

I might be a little pithy here, but it seems to me that much of what you post is "about me". Science is not about you, or your ability to shred others, I find that an adversarial approach to learning and understanding nature tends to put many offside.

 

Just saying, perhaps that has something to do with being banned or whatever from other science sites. 

 

 

Unfortunately, you'd need to know algebra first and pick out my flaws mathematically instead of via Quija board.

 

OK, Do you have flaws mathematically? I know algebra already, I guess that is why I am able to pick out some of your flaws, some flaws don't require algebra to pick out at all. BTW it's Ouija Board.

 

The best you can do is listen and try to understand other peoples opinions and ideas, that is what I did for you, I read all your posts (and there are SO many and they are big), I considered your arguments and your analysis, I did not disparage you or insult you in any way. I did participate in the discussion and engaged in a attempt at a debate.

 

It's up to you how you carry yourself and how you interact with other people, that often comes down to whether you want to talk with people or at people, do you want discourse and dialogue, or dogma and dissent..

 

Have a nice day...

[ralfcis]

Well you were right about one thing, it is Ouija. So what are the other flaws you've picked out? Is it the one where you say light has no frequency only wavelength and then your other theory explaining photons moving? When I was a kid, I saw the word "quantum" in a Superman comic. I repeated that word to my dad who thought I was very smart. Being able to parrot words out of a comic book is not intelligence.

Edited February 3, 2020 by ralfcis

[ralfcis]

Here's a sample of the awesome power of the Loedel velocity perspective: photos.app.goo.gl/TXD5j6HzvhvUDEvt9.

 

The 1/3c Loedel lines of simultaneity extend up from the original .6c velocity and then intersect the light line from the turnaround. Then new -1/3c Loedel  neon green lines of simultaneity (due to the new -.6c velocity at turnaround) extend to the -.6c velocity line. A new Loedel line of simultaneity (thick green line) extends back from there to the stationary perspective's point where the 1/3c lines of simultaneity started. This method can be employed from any original velocity to any subsequent one as shown in: photos.app.goo.gl/b9D3FSk3Dh8TYeVG9 and photos.app.goo.gl/DfRjhge4x7E3mVc68.

 

Notice linearly progressing Loedel age difference only occurs during the time of relative velocity imbalance (whose duration is the light delay from the turnaround to the ct axis) between the two worldlines. For constant relative velocity there is no change in Loedel age difference. This explains the invariant age difference at re-unification. What's relativity's year by year mathematical explanation of the progression of age difference after the turnaround? Instantaneous lump sum at the turnaround or at the re-unification point? Rindler metric? Not a valid question? Actually I already know that relativity has no answer for this.

 

If anyone has questions about the algebra involved I'm open.

Edited February 5, 2020 by ralfcis

[ralfcis]

Here's a muon-like example using the Loedel perspective. photos.app.goo.gl/fhSGp1WVdvRiNoh17 A ship comes in at .6c and can make various velocity changes at 3 ly away from earth. The swing in Loedel simultaneity lines occurs between 4 and 6 Earth time. No change in velocity results in no Loedel age difference. A stop results in the ship ageing 1 yr more but a velocity change to .8824c results in the ship ageing .8 yr less than Earth for the final Loedel age difference.

[ralfcis]

Here's an Md of the hysteresis of simultaneity at the top and bottom triangles of a return trip at .6c. photos.app.goo.gl/rFd42uvoNWJ9r1EP7 The age difference from sample perspective velocities from -c to +c is the same for both top and bottom if you you subtract the Loedel age difference which is constant and without hysteresis during constant relative velocity. This is a constant age difference from which all other perspectives of the age difference between Bob and Alice can be calculated using a formula for the hysteresis. Hence, the perspective age differences are unimportant. This is huge but those PhD's on the PSX won't even glance at this. 

 

Hysteresis can be compared to a camera's aperture and shutter. The aperture is wide open at +c and -c perspectives but closed at the Loedel perspective where Alice and Bob both agree on their proper times. (There is a reciprocal fuzz of time between the two when the aperture is open.) This is a window into a proper time present from which all perspectives begin. Hence, this is the basis of a common reality unlike Einstein's idea that the perspectives are individual realities and a common reality is the illusion. I think he chose the wrong perspective on how to perceive time. This is the math that proves it.

Edited February 10, 2020 by ralfcis

[ralfcis]

I've figured out the formula for the hysteresis of simultaneity around the Loedel perspective. For the example above, A=B=2 And you're trying to find the resultant intersections on the velocity lines b or a. d = Y of ( relative velocity - perspective velocity) / (Y of perspective velocity). So b=Ad and a=B/d.  For example if B =2 and we want the value of a for a perspective velocity of +7/9c (Y = 1.599), Y of (.6c - 7/9c) = Y of (-1/3c) = 1.06066. So b= 2(1.06066/1.599) = 1.33. So the age difference between Alice and Bob from the +7/9c perspective of A-b = 2-1.33 = .67yrs. 

Edited February 11, 2020 by ralfcis

[ralfcis]

Wow, so they, on the PSX, hold a reopen vote on my Loedel age difference question but some unknown person deleted all my math supporting my argument just before the vote. No record of it ever having existed and all my avenues to repost the math have been closed. Do they really fear anyone was going to look into my math and actually be convinced? Only in movies are mathematicians able to read anothers math as if it were text or 006 hieroglyphics. No one knows algebra anymore. 

[ralfcis]

Here's a nice little article that shows relativity doesn't need to be interpreted as Minkowski spacetime:

 

http://www.timeone.ca/losing-time/#sthash.yEsvwfhl.gg3iXMb9.dpbs

 

I've been arguing this on the PSX but their minds are absolutely closed shut ignoring other mathematical interpretations:

 

There is an obsession on wikipedia of whether some physical ruler could be held up to something long whizzing past and being able to determine that the moving object has physically shrunk due to length contraction. There is all kinds of supposed evidence that the results of certain particle collisions do not make sense if the particles hadn't have contracted before the collision. In a previous question, I had hoped Raskar photography, with its femtosecond frame resolution would be that physical ruler but the frame resolution is not enough to detect length contraction.

 

This led to this question and indeed the parallax measurement does present a physical ruler, independent of time, to measure reciprocal length contraction. The physical ruler is the distance to a star and due to relative velocity, the ships can be viewed as the stationary frame and the distance to the star is whizzing past them. The parallax view, which is independent of time, does indeed measure the distance to a far off star as having length contracted, from the ships' perspective, due to the velocity of the ships approaching that star. But the fact that it's impossible that the velocity of a ship can shrink space all the way to the star and physically pull it towards them must mean that length contraction is an illusion of perspective caused by relativity of simultaneity. The star would only look closer due to a trick of time, not due to a length physically contracting.

 

Here's the subsequent exchange from a knowledgeable parrot:

 

Assume the origin of the coordinate is system is at the Sun, the earth is orbiting the Sun in the xy plane, the star is at a distance L along the z-axis, and the radius of the earth’s orbit is r. When the earth is at y=-r, an astronomer on earth points a little arrow at the star on the celestial sphere. When the earth is at y=+r, the astronomer must rotate the little arrow in the y z plane by θx so the arrow points to the star’s new location on the celestial sphere. If an observer is now boosted by λ toward the star, we can calculate the new angle and boost parameters that the boosted observer sees done to the arrow.

 

Sorry his 4-vector math did not copy and paste:

 

 

Θ′=⎡⎣⎢⎢⎢1000010000coshλ−sinhλ00−sinhλcoshλ⎤⎦⎥⎥⎥⎡⎣⎢⎢⎢000000−θx00θx000000⎤⎦⎥⎥⎥⎡⎣⎢⎢⎢1000010000coshλsinhλ00sinhλcoshλ⎤⎦⎥⎥⎥=⎡⎣⎢⎢⎢000000−θxcoshλθxsinhλ0θxcoshλ000θxsinhλ00⎤⎦⎥⎥⎥=⎡⎣⎢⎢⎢000000−θxγθxβγ0θxγ000θxβγ00⎤⎦⎥⎥⎥0]

Therefore, the new parallax angle is θx′=γθxθx′=γθx. The new observer could understand this as

 

 

θx′2rL′L′=γθx=γ2rL=Lγθx′=γθx2rL′=γ2rLL′=Lγ

So the boosted observer is correct in saying the new parallax angle is the diameter of the orbit divided by the Lorentz contracted distance L′ to the star.

 

Ok that is the correct and factual answer to my OP. But if I may go a bit further, is the contraction an illusion of perspective or is the star physically pulled forward by the line of contracted space between the moving ship and the star (which was the new part of my question which was removed). – ralfcis yesterday    

 

I don't think "illusion" or "star is physically pulled" are good concepts as to how L behaves. You may be led to unworkable conclusions like fictitious forces acting on the star. Here is an example to make the contraction of L seem more real to you. Your spaceship is going close to c toward the star. Via redshift of lines you see the star approaching you at close to c. Via the parallax measurement you conclude the star is the contracted L' away. What makes L' even more real is that it takes you ~L′/c seconds of your life to get to the star …. no illusion! – Gary Godfrey yesterday

 

Sure but anything you ascribe to length contraction can be ascribed exclusively to time effects. The point of the question is whether there's a length-only yardstick. The parallax measurement is that but it's undermined by the effects of relativity of simultaneity (RoS) otherwise fictitious or illusory forces causing actual shrinkage are being invoked as you said. Slowed time is also illusory as RoS affects the relative start of time measurement, not its rate. I'm pretty sure no one agrees with my last statement even though I have the math to support it. – ralfcis yesterday   

 

This feels like I'm jumping into an existing discussion you are having to which I can't add much. My only 2 cents is that physics seeks to mathematically copy what we measure with a minimal set of ideas. If more than one word story can decorate that math, fine, but I wouldn't choose to get lost in philosophy seeking which is really true. – Gary Godfrey yesterday

 

Wise words. Yes an existing discussion where everyone is telling me an iron bar is forcibly being shrunk by the energy imparted to get to the velocity or by quantum effects ellipsizing spherical electron orbits in the iron atoms. My math says the RoS between a travelling object relative to a stationary observer will cause the observer to start his stopwatch later than when his perspective said the worldline began:

 

http://photos.app.goo.gl/4b6UpRjw7U29pMgC6 . The blue perspective starts his stopwatch 1.8 sec after he determines the race started but both agree the race took 3.2s of stationary time. – ralfcis yesterday    

 

RoS = vx/c^2. It is like the distance between them has a time value of 1.8s in my example which is like a head start in the race, related to these equations (ct')^2=(ct)^2 - x^2 (or (ct")^2 = (ct')^2 - (x/Y)^2 from the moving perspective). The RoS is not the time dilation rate of time but a lump sum of time that is the difference between race start and timing start. v'=Yv=x/t'=6/4c which is the distance moved / proper time on the moving ship. That's why it takes less ship's time in your ship to reach the star not because length contraction is real. – ralfcis 9 hours ago   Delete

 

As you said, math trumps philosophy but the simplest math wins. – ralfcis 9 hours ago   

 

Then of course he disappears because his closed mind prevents him from understanding what I've written. A real expert would have come across my reasoning before and if it was wrong he would have had a ready made answer to what I've said. Instead my math proofs are deleted in silence anonymously.

 

Anyway the Loedel perspective gives the clearest understanding of how relativity of simultaneity works without worrying about perspectives.

 

https://photos.app.goo.gl/q4yT4DXHtqBiZToF7

 

Notice both perspectives show 5s of travel with a delayed timing start of 1s. Clear as a bell.

Edited February 16, 2020 by ralfcis

[ralfcis]

Since my comments that involved a lot of math have disappeared silently and anonymously on the PSX once in the past on my Loedel question before a reopen review, I'm going to address Gary Godfrey's answer here. There seems to be a confusion between Galileo's principle of relativity and the illusion of perspective relativity. When we get in the car and drive down the road, we can't say we're stationary and our wheels are turning the Earth beneath us like a giant treadmill. Similarly, we can't say the earth is revolving around a proton in the LHC. The sun does not orbit the Earth daily. It's orbit does not have a sinusoidal component that gives us our seasons nor does the Earth's axis tilt like a metronome to give us the same. These are all proven illusions of perspective, not examples of the principle of relativity.

 

Now distant stars that do not red/blue shift appear relatively motionless to us. So if we turn on a ship's engine to approach them and a blue shift begins, it does not mean we can assume the star with the entire universe has now coincidentally decided to move towards us and we are just treading water against the flow. That blue shift is an illusion of perspective and is not due to the principle of relativity. Even if we turn off the ship's engine and coast towards the star can we claim we now don't know who's actually moving between the two of us. To say the ship is the one moving relative to the star is not a claim of absolute motion because absolute motion would involve the entire universe.

 

So how do we tell what's real and what's illusion? Persistence. There is no persistence in the blue shift or length contraction to the star once the ship returns to Earth but there is persistence in the time difference between the ship's clock and the Earth's clock. This persistence is not due to time dilation which is also the same illusion of perspective as length contraction and relativite velocity. The reciprocal time dilation does not change on the outbound or inbound journey. The persistent reality of the time difference is due to something else which I was trying to mathematically explain but gets deleted without any reasons why.

Edited February 16, 2020 by ralfcis

[ralfcis]

Finally I got a nibble from someone with physics credentials after all these years:

 

Your word "illusionary" seems synonymous with any measurement of the rod's length not done in the rod's rest frame. The idea that the rest frame length of the rod is invariant under boosts is well known. You are correct, the invariant length of the rod (or invariant distance to the star) does not change just because we look at it from a boosted frame. In this sense, measurements of the rod's length in other reference frames is illusionary, and we must return to the rest frame of the rod to measure its "real" length. I am still thinking about your idea of Persistence. – Gary Godfrey 44 mins ago

 

Yes, thank you, I wish I could shove your comment in the face of everyone who says otherwise. The rest frame length is invariant. So frustrating when everyone else I've ever come across disagrees. I do have a problem without knowing the correct terms to use as my Loedel age difference post shows. – ralfcis 30 mins ago    

 

Here is a link to a good portion of my math explaining why the time effect persists. physics.stackexchange.com/questions/526395/… . The 2nd half of the math has been erased and I can't repost it until my ban expires in 5 months but I do have it.

Edited February 16, 2020 by ralfcis

[ralfcis]

Non one reads this thread anyway so I dump my conversations from the PSX here before they delete them there.

 

I should have included that the invariant length dsds is the length ds2=dx2+dy2+dz2−dt2ds2=dx2+dy2+dz2−dt2 of the 4-vector (dx,dy,dz,dt)(dx,dy,dz,dt) between two spacetime points. In the rod's rest frame take these two spacetime points to be at the ends of the rod at the same time (so dt=0dt=0). Then ds=Lds=L. In a boosted frame the 4-vector between these same two spacetime points will be (dx′,dy′,dz′,dt′)(dx′,dy′,dz′,dt′) and ds′=ds=Lds′=ds=L still. But dt′≠0dt′≠0. So the observer chooses a different spacetime point for one end of the rod such that dt′=0dt′=0, measures the new length ds′=L′ds′=L′, and reports a Lorentz contracted L'. – Gary Godfrey 12 hours ago

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  @Gary Godfrey So yeah, whether 4-vector or Lorentz transforms, space and time are interchangeable and there's always a proper perspective and a boosted one. That's why my Loedel perspective is so cool because it simulates a common proper length and time perspective for both participants simultaneously. If you want to chat I'll make a room but if not they'll delete all I'll put into it if no one responds. Or you can wait 5 months until the brief interlude before my next banishment. – ralfcis 3 hours ago   

  
  
  
  
  
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  My point is as soon as you accept all perspectives are equally real, you ignore the underlying objective reality of proper time and proper space from which all perspectives are derived and hence illusory. The Loedel perspective and its beautiful math gives you a window into the underlying reality. Heresy I know and it's a shame I can only present the vast amount of math in small erasable chunks. Math is not a personal theory, it's a language to convey meaning as I lack the terms to use the relativistic language. I hope Chris will open that chat room now. Makes it easier to delete my stuff.

  
  

   

  
  

  If space and time are interchangeable in equations, why is there no instance of persistence in length as there is for time? People argue the persistence exists because c would be exceeded in near c travels to distant stars without length contraction. This is false because v=x/t not x/t'. Travel is at Yv =x/t', which is the invariant distance in the time on your spaceship, not in Earth time; no violation of c, no evidence of length persistence. I remember this being a Breme or Epstein concept but I see nothing on wiki about this.

  
  
  
  
  

Edited February 17, 2020 by ralfcis

[ralfcis]

I was ready to just give up on my quest to overturn relativity as I concluded no one understands relativity enough or understands anything I'm saying. But I reviewed my initial questions on the PSX (actually I've only been asking the same question under different guises) and found Dale not only knows his stuff but has been understanding and answering my questions all along. I just didn't know enough until today to recognise this. I just have to go to another forum he's on because the PSX is just not the place to discuss things with all the trolls and moderators on it. I tend to drive people crazy so I hope I can get his stamp of approval before that happens.

[ralfcis]

It seems that Willo is using a math method I have never seen before and am wondering if anyone else has.

 

@Willo I've decoded what you tried to tell me into this Md. photos.app.goo.gl/JA8iwFX7yboAvYfdA. I've never seen this format and it's not used in popsci relativity. This is the format used photos.app.goo.gl/Cp7FjpQ3Rmhp9VYB6 t'=t/Y and t''=t'/Y=t/YY. The ' represents how many times gamma is used. t(E)=t′(E)=t′′(E)=0. t(F)=5, t′(F)=4, t′′(F)=3.2 from Alice's perspective outbound to t''(F)= 6.8 from Alice inbound. t(G)=10, t′(G)=8, t′′(G)=10. We don't use the same language so you'll never be able to read what I write. Brian Greene's course did not use your format at all. – ralfcis Mar 2 at 19:42   

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  @willo If your answer is the method professional relativists use, it sacrifices physical reality in order to get rid of the notion of equations changing due to perspective. The reality is Alice's journey is not 2 separate worldlines of inbound and outbound, it is 1 worldline of outbound to inbound. Alice's or Bob's perspectives do change the equations plus I even dispute the popsci version about Alice's perspective. It is not photos.app.goo.gl/Cp7FjpQ3Rmhp9VYB6 . It is photos.app.goo.gl/M1Ncn5LVD9padAds9 according to my math which no one has yet disputed (or even looked at). – ralfcis Mar 3 at 13:39    Delete

  
  
  
  
  

[ralfcis]

Sorry I'm about to get banned from the sciforms because the idiots that infest this forum also infest that one and I have no interest in engaging clueless morons anymore. So I'm dumping my discussion about non-linear velocity combination here. 

 

The method I used for the "right angle" scenario works because You don't have to account for length contraction or the relativity of simultaneity.

 

http://www.sciforums.com/attachments/image2-gif.3199/

For example, In the following diagram, we show A and B's velocity as measured from the Earth on top, and then the same scenario as seen from the rest frame of A.

In this case, B always remains directly "above" E at all times. If we were to put a clock (Cb) directly above E and at rest with respect to it, B would move on a straight line between the two. In others words everything in side the dotted box moves to the left as a system from A's perspective. A clock at E and the clock Cb are in sync according to both A and anyone traveling in that box. So if B takes 1 sec to travel 1 sec according to E and Cb (leaves E when both clocks reads 0 and arrives at Cb when both clocks reads 1 sec), According to A, B also leaves E when both clocks read 0 and arrives at Cb when both clocks reads 1 sec. However, according to A, both the clock at E and Cb are time dilated. If the relative velocity between A an E is 0.866c, then according to A, it take 2 sec for the Clock at E and clock Cb to tick off one sec, and thus it takes 2 sec between B leaving E and arriving at Cb. Since the vertical distance between E and Cb is the same for A as it is for E, , this means that the vertical component of B's motion as measured by A would be half that as measured by E. Now it is just a matter of adding the velocity components to get the resultant velocity.

As far as the idea of "velocity" dilation goes. In this case, it is just the consequence of time dilation. Velocity is distance/time. The vertical distance remains the same for both frames, so the measured vertical velocity follows the time dilation.
when the 2 velocities are not at right angles, you have to account for length contraction and relativity of simultaneity, which add extra complications to the calculation.

 

Thank you Phyti for teaching me something very important. I always assumed math was a universal language that can be easily read by mathematicians like in the movies. I'm not claiming to be a mathematician but I assumed since I can read my own math, every mathematician could read it. (I now see no one has ever tried so I have no reason to be so frustrated.) Conversely, I should be able to read your math format. You get the right answer so I should be able to work backwards from that and understand your motivation but I see now it's not so easy. I can't even follow your Minkowski diagrams unless I painstakingly convert them into the format I'm comfortable with. I don't understand most of the lines you draw in and what purpose they serve. I'm not saying your format is wrong, it's just that I can't understand it without putting a lot of effort in. (I totally failed to understand the last one and the accompanying 2 sentences in the length contraction thread.) We just don't speak the same math language.

Now so far as this thread goes, I'm not looking for the right answer, I'm looking for the meaning behind the answer. I try to use the long formula to get the right answer but I have a lot of trouble plugging in numbers and getting the right answer. (For example with two ships leaving earth at a 45 degree angle at .6c, my answer is .513c between them but I'm not confident that's correct.) In the method I've pasted together to try to solve that problem, I think angles in normal life, which use Pythagoras as a sum of squares, are not the same angles as relativity uses which use hyperbolic Pythagoras as a subtraction of squares but I don't know if what I'm saying is true. I have a very simple way of calculating two ships leaving earth at .6c at 0,90 and 180 degrees but my answer at 45 degrees does not match .513c so I'm stuck figuring out why not. I fear that because my math is so different, no one is going to bother to point out the mistake in my reasoning. I'll get the usual answer to stop thinking and just use the wiki formula. I've learned, though, that to get anyone to look at my math, I'll have to describe it in painstaking detail.

 http://www.sciforums.com/attachments/image3-gif.3236/

You have A and B starting at the same point and heading off at 0.6c at a 45 degree angle to each other. Imagine a box with one corner at the starting point, and the other where B ends up after 1 sec ( the diagonal of the box will be 0.6 ls long in the box frame, and each side ~0.424 ls) You can also imagine clocks at all four corners.

The top two images show how the start and finish look in the box's rest frame.


The lower two images show the same events as determined by A's rest frame, where A is at rest and the "box" is moving to the left at 0.6 c. According to A both he and B where adjacent to the left bottom clock when it read 0, He also agrees the B arrives at the right upper clock when it reads 1 sec. However, according to A, the box is not a square, being length contracted in the left-right dimension to 0.424 X 0.8 ls = 0.34 ls. Also, all four clocks, due to the Relativity of simultaneity do not all read the same. The clocks on the right side of the box read ~0.25 sec ahead of the left side clock.
This means than in order for B to leave the left lower clock when it reads 0 and arrive at the upper right clock ( the dotted green arrow indicates B's path with relation to the box) when it reads 1 sec, only 0.75 sec tick off on the box clocks according to A. Due to time dilation 0.75/.08 = ~-0.94 sec pass for A in this time.
In that 0.94 sec the box moves ~0.56 ls to the right. B, in moving from left to right relative to the box will displace 0.34 ls to the right, and end up ~0.22 ls to the left of A. It will also be 0.424 ls "above" A.
Thus the distance from A to B along the green arrow is sqrt(-0.22^2+.424^2) = 0.478 ls.
This separation occurred over 0.94 sec as measured by A which gives a relative velocity between A and B of 0.509c A bit off from from what you got with the long equation, but close considering that a lot of the values I used above were rounded out to only 2 significant digits. For example, The time offset due to the relativity of simultaneity is closer to 0.2556 sec rather than 0.25 sec.

 

Okay, here's how I'd go about visualizing it.

 

I expanded all your numbers and the final answer works out to .5127c. Your last equation of the bottom 1st paragraph should be 0.75/.8=~.94 for anyone else following along. Everything makes sense except I don't understand how the box is a rest frame but inside it, B is going at .6c but A is going at .424c relative to the bottom left corner I assume. I've never seen movement inside a rest frame but I can see if there were, the Newtonian relative velocity formula would apply but not the relativistic combo formula as the velocities are not relative but are closing speeds. Am I interpreting this new trick correctly? If so, this shows any angles between the velocities are handled outside of relativity. I didn't see that coming.

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   I was approaching this problem by seeing a math pattern for two ships leaving earth at .6c at 0,90 and 180 degrees but couldn't establish any pattern for any other angles. I'll write out the approach in detail. Thanks for your answer..

 

 

Janus58, I'm going to butter you up before I get into this. I've only met 2.5 other people like you before, those with a clear (non-wiki) understanding of relativity. I don't think you're a hobbyist because it's difficult to maintain such a clear understanding without being a professional or a teacher in it. So my math is going to be different, it uses uncommon forms of relativistic math. This, as you know, is the primary equation of relativity (I assume the BB code editor is for math script but I'll pass on using it).

(ct')^2 = (ct)^2 - x^2

It has a famous form of

Y= c / sqrt((c-v)(c+v))

but I use my own form

c^2 = v^2 + u^2

where u is the rate of time through time = c/Y = ct/t' which I call the velocity through time.
(For those who can't visualise what a velocity through time looks like, just press fast forward or slow motion on your DVD player. The rate of time you're watching is different from your normal rate of time which is going at the velocity of c through time. c through time is 0 observed velocity through space and c through space is 0 observed velocity through time (which can't be reached))
Therefore c is the sqrt of the sum of squares of 2 velocity components at right angles to each other, v (the velocity through space) and u (the velocity through time). Everything moves at this composite velocity of c but, just like any velocity addition in relativity, the faster you are observed to move through space, the slower you are observed to move through time because no combination of any type of velocities can exceed c.

There are 3 main ways to graphically depict relative velocity as rotations of cartesian coordinates: Minkowski, Epstein and Loedel diagrams. The Epstein is a true rotation that represents a circular sum of squares relationship between the 2 coordinate systems whereas Minkowski is a hyperbolic difference of squares representation. Imagine two sheets of graph paper; one with ct and x axes and the other with ct' and x' axes and you rotate one sheet over the other pivoting at the common origin. The Epstein diagram rotates the ct/x sheet clockwise over the ct'/x' sheet whereas the Minkowski diagram rotates the ct'/x' sheet clockwise over the ct/x sheet. (Seems like this small difference should not create such vastly different graphical representations.) Minkowski also folds its x' axis over the x axis so all representations of c overlap the same 45 degree line which is graphically convenient to make c look the same from all perspectives. The Epstein diagram doesn't use this trick so things get messy with the slope of c. Just a note to all the Wiki readers, the Minkowski rotation is described as a clockwise rotation of ct' and a counter-clockwise rotation of x' so my mathematical interpretation is in quite a different form but still valid. I go further in my own Minkowski-hybrid diagram in that I need no rotation or mirror flip of the x-axis (meaning length is mathematically invariant) to get the correct results for any relativistic problem. (It's just a different form of math and it works so it's valid.) I also use Loedel perspectives and Loedel simultaneity (non-relativistic terminology) on my Minkowski-hybrid extensively to expose the underlying objective reality of proper time and proper space as opposed to the subjective "reality" of perspectives. But more on future threads if there are any.

All this will become clearer when I show my approach to solving the two ships leaving earth at different angles problem. Hopefully when I get to the bottom I'll be able to solve the 45 degree problem only using relativistic concepts without Pythagoras.

 

Before I get banned from here permanently, I'd like to try to finish my discussion with Janus58. Here is the Md I'll be discussing:

 

https://photos.app.goo.gl/ByaAVJU8UwiPN1bc6

A is the rightmost .6c velocity line. The earth timeline relative to A is the vertical line with black numbers for B=0 and red numbers for B=.6c at 90 degrees to A. Why is that? Let's discuss the black numbers first.

The black numbers are A's perspective of B's velocity through time. If B's velocity through space is 0, due to it being the stationary frame, B's velocity through time is c/Y. As a result, A's line of simultaneity from t'=1 intersects B's t=.8. Since Y=t/t', Y=1.25. Through the formula:

v^2 = (Y^2 -1) / Y^2 , v=.6c.

Now we get to the red numbers when B = .6c at 90 degrees to A. We know it's 90 degrees because B has no velocity through space component relative to A but it does have a velocity through space component relative to earth. A velocity through time of c yields the black numbers but a velocity of time for the velocity through space of B = .6c yields red numbers that are 1/Y of the black numbers. Now A's line of simultaneity intersects the earth time at t=.64 instead of t=.8 when v of B was 0. This means Y=1.5625 which yields a relative velocity of .7684 between A and B at .6c at 90 degrees away from earth. This is the correct answer.

For my 3rd example, I have A and B each at .6c separating from each other at 180 degrees. This is now a Loedel diagram and A's line of simultaneity intersects B t=8/17 so Y=17/8 which yields a relative velocity of 15/17c = .88235c which is the correct answer.

All this makes sense so far as only manipulating the velocity through time. As soon as you invoke velocity through space of B relative to A, I believe the numbers on the x axis will be affected but I haven't figured out how and will discuss the reasoning in the next post if there is one.

PS. I forgot to include the 0 degree example when A and B are leaving earth in the same direction at .6c. Since they overlap, A's line of simultaneity at t'=1 intersects B's t=1. Y =1/1 =1 which corresponds to v=0 between A and B.

Edited April 3, 2020 by ralfcis