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# Scary relativity

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### #1 deSitter

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Posted 25 November 2013 - 09:14 AM

Scary because there is some linear algebra here!

A square matrix as many of you know, is an arrangement of numbers that embodies in some contexts a transformation of a certain sort - for example when I rotate the coordinates in the plane,

x' = x cos a - y sin a
y' = x sin a + y cos a

This is more easily expressed (read brackets as a column vector, and parentheses as a row vector, and double brackets as a matrix with entries in "reading" order)

[x' y'] = M [x y]

where M is the square matrix [[ cos a, -sin a, sin a, cos a ]]

The determinant is the product of the diagonal entries minus that of the off-diagonal entries, here cos^2 a + sin ^2 a = 1.

OK let's introduce four basic matrices (with complex numbers as entries)

Sx = [[ 0, 1, 1, 0]]
Sy = [[ 0, -i, i 0 ]]
Sz = [[ 1, 0, 0, -1 ]]
S0 = [[ 1, 0, 0, 1 ]]

and let's make the matrix object

R = x Sx + y Sy + z Sz + t S0 = [[ z + t, x - iy, x + iy, -z + t ]]

The determinant is

|R| = (z+t)(-z+t) - (x-iy)(x+iy) = -x^2 - y^2 - z^2 + t^2.

You'll see where this is going eventually. Ask questions now.

-drl

### #2 Otto Piechowski

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Posted 25 November 2013 - 10:24 AM

For me, Danny, this math is over my head. I would need pictures to explain each step and a glossary (with pictures) to explain the words vector, row vector, column vector, and matrix.

BUT DO NOT DO THIS, at least at this time. Rick and Brentwood also asked you to explain the special relativity thing. It is quite likely they do understand simple math like this and I do not want to hold the three of you up. Later, perhaps y'all can draw pictures for me and stuff like that BUT NOT NOW, or perhaps, even at all.

Thank you for taking this on.

Otto

### #3 llanitedave

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Posted 25 November 2013 - 11:49 AM

Sounds like a job for numpy.

### #4 Rick Woods

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Posted 25 November 2013 - 12:08 PM

Rick and Brentwood also asked you to explain the special relativity thing. It is quite likely they do understand simple math like this and I do not want to hold the three of you up.

It's even more likely that at least one of them has no idea what any of it means. I feel like Homer Simpson when someone talks to him, and all he's hearing is "blah, blah, blah". It may as well be runes, or the Voynich manuscript.

Is this degree of immersion in the math really necessary to create an understandable analogy for the car-headlight thing? If so, well, thanks anyway, but there's nothing here for me.

### #5 PeterR280

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Posted 25 November 2013 - 12:10 PM

The Lorentz transformations would be much easier to follow.

### #6 The Mighty Mo

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Posted 25 November 2013 - 12:23 PM

Post deleted by The Mighty Mo

### #7 brentwood

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Posted 25 November 2013 - 12:35 PM

Nothing here for me either. I can understand some of Otto's posts more!
If the answer to the question, "What does an outside observer see when a vehicle travelling at just below c puts on its headlights" can only be answered with an equation, then I would say that there is no answer, it is still an unknown.

### #8 deSitter

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Posted 25 November 2013 - 01:06 PM

C'mon be patient! You have to see things before you can understand them! It's not a linear process. Ask questions!

-drl

### #9 dickbill

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Posted 25 November 2013 - 01:17 PM

why do you introduce imaginary numbers?

### #10 deSitter

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Posted 25 November 2013 - 01:26 PM

You'll see! It's nothing unreal - it's very real. You HAVE to be patient. It will all come down to a flag...

-drl

### #11 choran

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Posted 25 November 2013 - 01:56 PM

Will the simplified version come down to solving a right triangle with sides
ct, vt, and (hypotenuse) ct' and eventually solving for t'? I've seen that approach and it is pretty understandable, but requires pics.

### #12 brentwood

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Posted 25 November 2013 - 02:08 PM

Danny,a&#1082;&#1086; &#1072;&#1079; &#1074;&#1080; &#1077; &#1087;&#1086;&#1080;&#1089;&#1082;&#1072;&#1083; &#1074;&#1098;&#1087;&#1088;&#1086;&#1089; &#1074; &#1041;&#1098;&#1083;&#1075;&#1072;&#1088;&#1080;&#1103;, &#1097;&#1077; &#1084;&#1086;&#1078;&#1077;&#1090;&#1077; &#1076;&#1072; &#1075;&#1086; &#1088;&#1072;&#1079;&#1073;&#1080;&#1088;&#1072;&#1084;&#1077; &#1073;&#1077;&#1079; &#1076;&#1072; &#1080;&#1079;&#1087;&#1086;&#1083;&#1079;&#1074;&#1072;&#1090;&#1077; Google? &#1053;&#1077; &#1084;&#1086;&#1075;&#1072; &#1076;&#1072; &#1087;&#1086;&#1089;&#1090;&#1072;&#1074;&#1103; &#1089;&#1080; &#1087;&#1086;&#1089;&#1090; &#1074; Google &#1080; &#1076;&#1072; &#1089;&#1077; &#1087;&#1088;&#1077;&#1074;&#1077;&#1078;&#1076;&#1072;&#1090;.?

### #13 choran

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Posted 25 November 2013 - 02:27 PM

### #14 deSitter

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Posted 25 November 2013 - 02:43 PM

Will the simplified version come down to solving a right triangle with sides
ct, vt, and (hypotenuse) ct' and eventually solving for t'? I've seen that approach and it is pretty understandable, but requires pics.

Nope, 100% rigorous.

-drl

### #15 Crow Haven

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Posted 25 November 2013 - 03:04 PM

I haven't the math background or understanding of relativity to be able to see how this turns out. Can light be slowed down or appear slowed down from an observer's perspective at the side of the road? I have a hard time with the idea of a light beam traveling at a foot a minute or looking like it is proceeding this slowly. I thought that the light sent from the headlights would just travel at the "speed of light" as soon as turned on from the car. That it would appear instanteous from the observer in the car and from the observer at the side of the road it would also, regardless of how fast the car moved. Is there enough time and distance between the car and when the light is switched on to make it look like the light comes on before the car appears? Sorry if this sounds really stupid. I'm totally confused by it, but am hoping there will be a way to describe what actually would appear to happen for the observer at the side of the road that I can understand.

### #16 choran

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Posted 25 November 2013 - 03:07 PM

Heck! Was hoping I could follow.

### #17 Neutrino?

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Posted 25 November 2013 - 03:09 PM

Tough crowd.

You'll see where this is going eventually. Ask questions now.

-drl

SL(2,C) to SO+(1,3) to PSL(2,C) to M ?

### #18 Rick Woods

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Posted 25 November 2013 - 03:14 PM

Maya, it doesn't sound stupid to me; or else, I'm equally stupid. I can't help feeling that there must be a simple analogy that would give people like us at least a rough picture of the situation.

Danny, I appreciate your attempting to explain the whole thing, and I will follow it and try to understand. But I really hope this ends up with something on the order of "think of it like this..."
But I have no questions now, beyond "what does it mean?"

### #19 Rick Woods

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Posted 25 November 2013 - 03:16 PM

Danny,a&#1082;&#1086; &#1072;&#1079; &#1074;&#1080; &#1077; &#1087;&#1086;&#1080;&#1089;&#1082;&#1072;&#1083; &#1074;&#1098;&#1087;&#1088;&#1086;&#1089; &#1074; &#1041;&#1098;&#1083;&#1075;&#1072;&#1088;&#1080;&#1103;, &#1097;&#1077; &#1084;&#1086;&#1078;&#1077;&#1090;&#1077; &#1076;&#1072; &#1075;&#1086; &#1088;&#1072;&#1079;&#1073;&#1080;&#1088;&#1072;&#1084;&#1077; &#1073;&#1077;&#1079; &#1076;&#1072; &#1080;&#1079;&#1087;&#1086;&#1083;&#1079;&#1074;&#1072;&#1090;&#1077; Google? &#1053;&#1077; &#1084;&#1086;&#1075;&#1072; &#1076;&#1072; &#1087;&#1086;&#1089;&#1090;&#1072;&#1074;&#1103; &#1089;&#1080; &#1087;&#1086;&#1089;&#1090; &#1074; Google &#1080; &#1076;&#1072; &#1089;&#1077; &#1087;&#1088;&#1077;&#1074;&#1077;&#1078;&#1076;&#1072;&#1090;.?

Brentwood, is the above what you intended to post? Looks like a lot of ASCII to me.

### #20 Otto Piechowski

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Posted 25 November 2013 - 03:20 PM

ask questions: OK; now mind you, I'm not being sassy or trying to be stupid; I just have no more familiarity with this math than I do with a language foreign to me.

What is a "square matrix"?

What do
x' = x cos a - y sin a
y' = x sin a + y cos a
mean?

And what is the meaning of these things in relation to the paradox of 1+1=1 at near light speeds?

I mean, I can do the simple equations of t=t(0) X the-square-root of 1 - v^2/c^2; the same for Length and the same for mass when mass(0) is divided by the-square-root of 1 - v^2/c^2. I can do it. I don't understand why these do what they do. I don't understand why, from the perspective of an outside observer, the combined velocity of headlight beam and the gal-in-the-spaceship-already-traveling-at-near-c is still just less than c? I would like to understand why this is the case. And as one of our colleagues here said, it would be nice if there was a real world example that explained this phenomenon; not just gave an analogical similarity like that silly higgs-boson-like-a-snow-field thing.

But, as I am writing, I am afraid I am getting to a really clear question.

Danny, even though this is special relativity and not quantum mechanics, is this an area of relativity which, like quantum mechanics, it cannot be explained with analogies from the ordinary world of fried eggs, baseball, and why the landlord can't get the furnace working?

Otto

### #21 Crow Haven

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Posted 25 November 2013 - 03:28 PM

I can't help feeling that there must be a simple analogy that would give people like us at least a rough picture of the situation.

I sure hope so, Rick.

A square matrix? Otto, I picture a square of sedimentary rock embedded with fossilized sea shells... It's bad when you haven't the math training....

### #22 dickbill

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Posted 25 November 2013 - 03:30 PM

...

### #23 brentwood

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Posted 25 November 2013 - 03:31 PM

The page of gibberish I posted above was a google translation of a couple of questions in Bulgarian. It looked all right when I posted it, but it was in Cyrillic script so maybe that's why it came through as just numbers.
The point that I was trying to make is the same as Otto's, why post in a format that is unintelligible to most on here.
That maths makes no sense to me, I can't even get enough of it to ask any questions!

### #24 Crow Haven

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Posted 25 November 2013 - 03:34 PM

...

Problem is, it's pretty sad too!

### #25 choran

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Posted 25 November 2013 - 03:36 PM

I'm no expert, but the idea is that -- by postulate --ALL observers will always measure the speed of light as the same, regardless of their state of relative motion. SO, since that number c is FORCED to remain a constant, and since
velocity=distance/time, TIME is forced to change. Two postulates (constancy of light speed, and same rules of physics in all systems moving at constant velocity relative to one another) must be accepted, and the rest follows. If you accept the postulates, you must swallow that time is dilated.
My brief summary based on what I'm sure is an incomplete understanding.

It's all quite counterintuitive and hard (impossible?) to swallow.

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