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# Relativity And Simple Algebra

relativity

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### #1225 ralfcis

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Posted 20 March 2020 - 09:07 AM

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

### #1226 ralfcis

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Posted 03 April 2020 - 02:34 PM

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...mage2-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.

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...mage3-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.
1.
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 by ralfcis, 03 April 2020 - 02:46 PM.

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