Janus' method for non-linear velocity addition

Oh, ok, you've given me a phyti to ralfcis Md translation dictionary.

P.S. Not nice, just grateful when I can learn something new.

PPS. No big deal to watch movies in any order and still understand what's happening. It just puts the surprise plot twists at the beginning when I start from the middle.

 

Log in or Sign up to hide all adverts.

Ralph;

ct' = x', ct = x, so line 1 says: (x')^2 = (x)^2 - (x)^2 = 0

What's the meaning of that?

Your own form is the distribution of light energy within the light clock, obvious from a simple inspection.
The u component is the active part of the clock. The clock moves on the x axis, while the light moves simultaneously on the x and p axis, i.e. in 2 spatial dimensions. The x component compensates for the motion of the clock. The observer moving with the clock is only aware of the u component, i.e. his clock runs slower along with all other em processes in his frame.

You seem to want to reinvent SR with unnecessary ideas.
I recently saw Don Lincoln's video promoting the same metaphorical interpretation as Briane Green.
Expecting a Walt Disney film coming soon to a theater near you.

 

Log in or Sign up to hide all adverts.

"ct' = x', ct = x" Where do you get that? I have no idea what you're assuming here or the rest of what you're saying. You can apply the last equation all you want to the light clock because that's what your book might be telling you but I'm just setting up the math for a more important interpretation. Are you saying the 3 equations are not equivalent? They are and they're all forms of relativity's primary equation unless you have another one you favour. And how am I inventing unnecessary ideas which were already stated by Lincoln and Greene. Those ideas clash with your interpretation, so what. Contrary to wiki there are many interpretations of relativity so I'm not going to get into a battle with yours. Mine is either right or wrong mathematically, not philosophically.

PS. I've decided not to be dragged onto this merry-go-round so if you don't want to see things from another perspective, I'm going to have to put you on ignore. I only want to talk to 1 person here anyway.

 

Log in or Sign up to hide all adverts.

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:

Please Register or Log in to view the hidden image!

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.

 

I have calculated that the precession of Mercury is 40.3 " per century and the precession of the earth is 1.85 " per century under the Doppler effect.
I do n’t know what the measurement result of the precession of the earth is. Does anyone know?
I can calculate the precession of any planet, as long as you can tell me some basic information about it, including the distance to the perihelion, the distance to the perihelion, speed, eccentricity e.

I checked the information and according to GR calculations, Mercury ’s progress deviation is 0.41 "per hundred years.
I can now say with certainty that the precession deviation of Mercury is caused by the Doppler effect of the gravitational field!
http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node116.html

 

In order for the method to work for 2 ships at .6c leaving earth at 45 degree angles, I have to find what velocity through space of B intersects A's line of simultaneity at t=.8586 for B. I'm working backwards from the answer that the relative velocity between B and A is .512727c when you use the long formula or Janus' method to solve this problem. For .5127, Y =1.1647 so 1/Y= .8586. So I need to find a slope for v of B=.6c relative to earth that A's line of simultaneity intersects at t=.8586.

The closer the slope of v of B at .6c relative to earth gets to 0 , the closer the relative velocity of A to B = .7684c because that was the answer for A and B at .6c at 90 degrees to each other. My final answer must be .5127c. So I've got to move the slope of v of B at .6c to the right. But too far right reaching a slope of 3/5 for v of B and the relative velocity of A to B drops to zero as the angle between the two .6c velocity lines drops to 0 degrees. Somewhere between these two extremes is the slope of v of B that gives the correct answer of .512727c for the relative velocity between A and B.

I guess I can try random guesses for the slope of v of B=.6c until I find the right slope and maybe that'll trigger in me why that's the right slope and relate that to the 45 degrees between A and B.

 

https://photos.app.goo.gl/LXPNvN1ANViDVpoE8
Janus,
In this running competition, I got two different answers from other special relativists.
1. It is always unclear.
2. They reach the platform at the same time.
He said: "As A.T. points out, observers at rest or only moving parallel to y in the Earth frame (such as Earth and your platform) will regard this race as a draw. Observers with any motion in the x direction in the Earth frame (such as Newton and Einstein in your example) will get different results for who won. It isn't clear from what you've said which frame your astronomer was using. He may have made a mistake, or interpreted your question differently, or he may be using a different frame from the one you think he is. "
https://www.physicsforums.com/threads/who-will-win-this-running-race.986704/

The owner of physicsforumsis very impulsive, as long as he sees you say something he does not agree with, then you will be blocked.

Please Register or Log in to view the hidden image!

I have got at least 5 different answers. Is there a standard answer for special relativity?

Please Register or Log in to view the hidden image!

Please Register or Log in to view the hidden image!

Please Register or Log in to view the hidden image!

 

Ok here's what I got graphically:

https://photos.app.goo.gl/5U7gCDR7a1ynsSGA9

The slope of the velocity of B=.6c leaving at a 45 degree angle to A is ~.14c. I'm not sure what that means but it intersects the hyperbola that represents t=.8586 for all velocities and A's line of simultaneity to give the correct answer that A to B's relative velocity = .5127c according to the method I've used. In order to get a more precise answer for .14c, I'd have to figure out the slope of a line where t=.8586 intersects the .8586 hyperbola. Then I'd have to see the significance of the answer and how it relates to all the working bits and pieces of Janus' solution and how it relates to 45 degrees between two .6c velocity lines. I'm no math wizard and I might not even be on the right path to understanding what's happening using only relativity.

PS. The answer seems to make sense because if v of B is zero leaving earth and A's is .6c, that means the relative velocity between A and B is close to the .5127c answer we're looking for. B may be leaving earth at .6c (instead of zero) but the 45 degree angle between A and B would shave their relative velocity down to the .6c range.

 

Alright I figured out the algebra.
t'=.8586
plugging that into
(ct')^2 = (ct)^2 - x^2
we get .7372 = t^2 - x^2

t= .8 +.6x from where A's line of simultaneity intersects t'=.8586
(As a side note .6x is the relativity of simultaneity from A's perspective)

We end up with a quadratic equation

.96x -.64x^2 = .0972

Now I don't remember how to solve these so I used various values of x until I got close enough so x is ~.11

So v =x/t = .11/.866 = .127c (not 1.4) which allows the line of simultaneity to intersect that velocity line at .8586 which yields a Y of 1.1647 which yields a velocity of .5127 of B =.6c at a 45 degreee angle relative to A at .6c. I still don't see any deeper meaning of what .127c means. My goal is to not work backwards from an answer but to work forwards to get an answer. Yes I have that but it's not elegant and insightful. I usually get a great rush when math reveals some hidden meaning. No rush this time.

When written in words, the quadratic equation expresses this meaning (where u is the velocity through time and v is the velocity through space):

2uvx - (ux)^2 = (t'^2 -(t of A)^2) t of A is where A's line of simultaneity intersects the t axis.
I may need to check this formula with other examples to make sure I got the variables correct.

I hope I've convinced some of you how useless a wiki understanding of relativity is.

 

I need to work on massaging this formula so it applies to all angles, not just 45 degrees. This will give me the insight to what's going on.

 

The special relativity makes people go crazy. Looking for truth in contradictions, the final conclusion is still contradictory.

Please Register or Log in to view the hidden image!

 

The final form is
(ux +v)^2 = (some mixture of sin@, YofB, v of B)^2 which I haven't figured out yet.
Tony, I forgot to mention I put you on ignore. Your endless unsubstantiated sniping against relativity is pointless so go do it on your own thread, not mine. Continue to dance like non one's watching because no one is.

 

I'm sorry, I also think I'm wrong. After all, you have also put a lot of effort into this field. Don't keep in mind, brother.

 

I have to admit defeat. z (.126c) is the velocity line that represents B=.6c at 45 degrees to A. z intersects A's line of simultaneity at t'=.85855 z's time which is 1/Y for the .512727c realative velocity between A and B. I did all sorts of manipulation of .85855 to see where 45 degrees was hiding in it like Janus' box technique but I couldn't find it. I did come up with a formula for Y of z:

Yofz^2 = 1 + x / (u^2 * (1/x +2v/u -x)) where v= .6c, u = .8c

z=x/t = (Yofz^2 - 1) / Yofz^2 and v=(t-u)/x.

t' = t/Yofz and x works out to .109 (I found a quadratic equation solver on google)

but I couldn't identify a 45 degree angle in any of this so no bridge between working backwards from the correct answer to a starting point that would lead to a correct answer. I think this is my first time not being able to solve a math problem that must have a solution.

 

So I woke up this morning and said to myself, what if I plugged my z=.126c result into the velocity combo formula where v=.6c.

w=(v+z)/(1+vz/c2)

So w = (.6 - .126) / (1 - .6*.126)

Turns out this gives the correct answer of .5127c. This means

(v+z)/(1+vz/c2) = the long form formula sqrt (z2+v2+2zv cos@ - (vz sin@/c)2) / (1+vz cos@/c2)

and that will give me an idea of how to express z in terms of trig angles. I'll do that later.

 

Correction: (v-z)/(1-vz/c^2) = the long form formula sqrt (A^2+B^2-2AB cos@ - (AB sin@/c)^2) / (1-AB cos@/c^2) where z=.126 and A=B=v=.6c and @=45 degrees.

Also correction:
Yofz^2 = t^2 / ((t-x)(t+x)) where t = u+vx

Still flailing

 

After much flailing I came up with a simplified equation that almost solves the 45 degree angle problem but I don't yet know what it really means and it's not applicable to other angles and velocities.

Answer = (v - v cos@ Answer) / (1 - v^2 cos @)

solving for Answer with v=.6c and @=45, we get Answer = .51294957 c but it does not exactly match Janus's answer of .5127272658692739c so the simplified equation must be wrong but deceptively close. This result did not come from a rigorous derivation, just from a pattern recognition within the equations for this one example but it's not correct so I give up. Too hard.

 

One of the things about working with 45 degree angles is that the cosines and sines of 45 degree multiples are equal to either - 1/sqrt(2) or 1/sqrt(2), which also can be expressed as (+/-) sqrt(2)/2 (0.707196781...)
Thus for example, by choosing the right substitutions, and making both u and v the same magnitude, you can get:

W = v sqrt(2- sqrt(2) - v^2/2)/(1-v^2/sqrt(2))

for the 45 degree scenario.

 

Yes that is the first step I took to solving the problem I'm trying to understand how in the combination of v=.6 and z=.126 (which is derived from an intersection on A's line of simultaneity), z represents B cos 45. So in what you wrote above

w= .3822/(1-.25855)

w in the combination of v and z is

w= (v-z) / (1 - vz)

So I try to force vz times a factor x to equal .25855 and worry about the top later.

so vzx=v^2 cos 45 so zx= v (cos45 )

Now we deal with the top part and we need to figure out a 2nd factor for zx which turns out to be w. plugging these into the formula for w

w = (v- vw cos 45)/ (1- v^2 cos 45) =

rewriting to solve for w

w =v / (1 + v cos45 -v^2 cos45) = .51294957

The only place for the error to have crept in was the 2nd factor was not quite equal to w but the error is so small I don't understand what's causing it.

Since I don't know where either fudge factor came from, I can't understand how z=.126 represents B cos 45 which would give me insight on how intersections on A's line of simultaneity represent B's angle leaving earth relative to A. Relative velocity seems to be a magnitude in 3D space that has some kind of internally expressed angle. I'm trying to understand that mathematically. Many of the numbers from your method pop up in this method as well but I'm not getting the full connection. I haven't run up against a problem like this for the rest of my math for relativity so it makes me worried there are gotcha's that I haven't run up against. This should be straight forward and yet I can't solve this problem.

 

Yup sorry, the 2nd fudge factor = (v-w(1-v^2 cos 45))/(v cos 45) =.5133401666 and it's not related to the real w and that's why my answer was wrong. I made a bad assumption to believe there was a pattern without checking it. So I still don't see how z=.126 relates to B cos 45.