Re: Simple Sagnac


"sal" <pragmatist@xxxxxxxxxx> wrote in message
news:pan.2005.07.31.03.00.11.992321@xxxxxxxxxxxxx
| Between this and the 1500 line tome in another thread which I haven't
| yet touched there's a lot to respond to, and I'm going to have to take
| it in installments.
|
| The first thing, though, is that I'm confused by your objections. You
| always object to the derivation of the Lorentz transform, c+v, c-v,
| and all that -- no surprise there. But now you're saying the Lorentz
| transform itself isn't linear:

Yep.


|
| A (quotes pulled out of context):
| > "Since tau is a LINEAR function, it follows from these equations
| > that"
| >
| > I can't use italics here so I can only capitalize it to NON-linear
| > and correct the math, not the translation.
| >
| > Doesn't really matter, tau is not linear. tau(60,0,0,20) = 8+8
| > tau(60,0,0,4) = 8
| > tau(80,0,0,16) = 8

Yep. Focus in on

tau(60,0,0,20) = 8+8
tau(60,0,0,4) = 8

We can forget the y and z, both are zero.
We can forget x, both are the same.
tau(20) = 8+8, so tau(t) = 5t/4
tau(4) = 8, so tau(t) = 2t
The one way light time from engine to caboose is 4 seconds
stationary frame, 8 seconds train frame.


| and:
|
| A (again, somewhat out of context):
| > There is no linear relationship between t and tau and you've made
| > the right point.

Yep, the point you made was correct. The speed of light in the train
frame is zero, it goes from one end to the other and BACK AGAIN,
ending up where it started. Making the train shorter, even to the
infinitessimal, doesn't help. The light ends up where it started.
You've found the divide-by-zero in gamma, c = 0.



| and:
|
| > Are you 100% ABSOLUTELY CERTAIN as you possibly could be that the
| > cuckoo transforms are linear?
|
| To which I say, WTF???

Well, you would... it's your religion under attack.
|
| At risk of insulting your intelligence I will walk through what a
| "linear" transformation is.

You can insult all you want, I'm used to it. The seed of destruction of
your
faith is in your own words.

|
| A linear function of one variable is a constant times that variable,
| like this example:
|
| y = 3x
|
| A linear function of two variables is the sum of a constant times each
| of them. For example:
|
| z = 3x + 2y
|
| A NON-linear (polynomial) function involves higher order terms or
| cross-terms, such as, for example:
|
| z = 3x^2 + 2xy
|
| And just to be complete, a MULTI-linear function of multiple variables
| is linear in EACH of its arguments when the others are held constant
| ... such as, for example,
|
| z = 17 * x * y

I took linear algebra as part of my BA, thank you.



| Now, as to the Lorentz transform ... if we have two inertial
| (non-accelerating) frames of reference (i.e., coordinate systems), and
| one frame is moving at constant velocity V with respect to the other,
| and C is a constant,

YOU CAN STOP RIGHT THERE!
C isn't constant, it reverses direction; the tip of the ray ends where
it
started. Shrinking the distance doesn't help. c = 0 in the cuckoo
transforms.






then the value
|
| g = gamma = 1/sqrt(1-V^2/C^2)
|
| is a constant.

Yes, well, that was preceded by the caveat
"IF [we have two inertial...]| AND C is a constant" so
"g = gamma = 1/sqrt(1-V^2/C^2)" is nonsense, C =0.
Not that I'd insult your intelligence, I'm sure your IQ is in double
digits.



|
| Thus, the function
|
| tau = g * (t - V*x/C^2)
|
| is a constant times t plus a constant times x, and is, therefore a
| linear function of t and x.

There is no "Thus". C = dx/dt = 0.

|
| In the example you gave (which I have not included in this post, but
| which you've got right there on your screen in Google Groups if you
| need to refer back to it, so please don't object too much to the
| "grand snip") you specified v=3 and c=5.

Yep.

In that case, we had
|
| g = 1/sqrt(1-3^2/5^2) = 1.25

No we don't, we have g = 1/sqrt(1-3^2/0^2) = undefined,
because C = x'/tau - x'/tau = 0/tau = (32-32)/20.


|
| and, multiplying everything out, we have
|
| tau = 1.25 * t - 0.15 * x
|
| Do you agree that this is obviously a linear function of t and x?

Nope. Not a snowball's chance in hell. Write a paper, YOU found
the error in Einstein's paper. Well done.

|
| Now, quoting again, you claimed, apparently as an objection:
|
| > tau(60,0,0,20) = 8+8
| > tau(60,0,0,4) = 8
| > tau(80,0,0,16) = 8
|
| But you seem have plugged in _differences_ in "t" coordinate values on
| the left, and equated them with _differences_ in "tau" coordinate
| values on the right. Right?

I'll add one if that worries you.
The light reaches 60 at 5 units/second in 12 seconds.
The engine reaches 60 by travelling 60-32 =28 units at 3 units/second
= 28/3 seconds = 9 and a bit.
tau(60,0,0,12) = 9 and a bit.

20 seconds is the round trip time, 4
| seconds is the return time, and 16 seconds is the outbound time.
| Again, those are not coordinate _values_, they are coordinate
| _differences_,

Who gives a hoot?
The engine driver turned on a flashlight when he got to 80, the light
took 8 seconds to reach the caboose and 4 seconds to reach 60.
The caboose was at 60 when the light arrived.
½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c+v))
½[tau(0,0,0,t)+tau(0,0,0,t+32/2+32/8)] = tau(32,0,0,t+32/8)
and not
½[tau(0,0,0,t)+tau(0,0,0,t+32/2+32/8)] = tau(32,0,0,t+32/2)

Do you imagine tau(32,0,0,t+4) = tau(32,0,0,t+16) is LINEAR?
Not that I'd insult your intelligence, I'm sure your IQ is in
AT LEAST double digits.

and the coordinate transforms won't give you anything
| very helpful when applied to them; indeed, the values you give on the
| right are not all the values of tau corresponding to those arguments
| on the left! The middle one is wrong: tau(60,0,0,4) = -4, not 8.
|
| But what you needed were _differences_ in tau, not _values_ of tau at
| particular points. To obtain the differences in tau between two
| points on the path, you need to take the values of x and t for each of
| the points, plug them into the formula to obtain tau at each of the
| points, and subtract them.
|
| Do you understand that?
I understand that you don't want to insult my intelligence,
which makes you very polite idio.... err... person.



| That's what I did in my earlier post, and it worked out just fine for
| me. I'll walk through it once again, showing the transforms
| algebraically rather than just showing them as matricies, as I did
| previously.
|
| Again, you had three points on the path:
|
| Point 1: t=0, x=0 ... light ray emitted

Yep

| Point 2: t=16, x=80 ... light ray reflected at engine

Yep

| Point 3: t=20, x=60 ... light ray arrives at caboose

Err....
t = 12, x = 60. light ray arrives at 60. Engine has travelled
12*3+32 = 68 units. Caboose hasn't arrived yet.
ARE YOU ABSOLUTELY SURE, 100% CERTAIN that
the cuckoo transforms have any sense to them?
And quit pointing, it's rude. :-)



| The value of tau at each of the three points is:
|
| Point 1: 1.25*0 - 0.15*0 = 0
| Point 2: 1.25*16 - 0.15*80 = 8
| Point 3: 1.25*20 - 0.15*60 = 16

I'm getting tired of this pointless garba..err... algebra.
tau is a function and tau(60,0,0,12) is undefined, the light
hasn't reached the engine yet.
One might as well figure out tangent of 90 degrees = y/0.
Who taught you math anyway? Shoot the gorilla, it's not your fault.


|
| The "elapsed track time" for the outbound segment is
|
| dt = 16 - 0 = 16 seconds

Yep

|
| The "elapsed track time" for the return segment is
|
| dt = 20 - 16 = 4 seconds

Yep


| The "elapsed train time" for the outbound segment is
|
| dtau = 8 - 0 = 8 seconds

Which train, the "Galilean" or "Einstein Express"?
There is a difference, one is 32 units long and the other
has 8 cars added, 40 units long.

| The "elapsed train time" for the return segment is
|
| dtau = 16 - 8 = 8 seconds

| Is there any part of this that doesn't make sense to you? (Aside from
| possible algebra mistakes of which I am frequently guilty :-) )

It makes no sense to me, but since it was you that found the
divide-by-zero I'll give you the credit. C=0 because
x'/tau - x'/tau = 0, well done! Don't you know how to take
"yes" for an answer?



| Finally, is it possible that you are mixing up "linear" functions with
| "multilinear" functions?

What do you want, TWO tau() functions with the same name?
Ok, derive them.

½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))
½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c+v))

Be my guest.

| A linear function of several variables is as
| I described above: It is a linear combination of the arguments. A
| multilinear function, however, is a linear function of each of its
| arguments if you hold the other arguments constant. The Lorentz
| transforms, like all "linear" coordinate transformations, are linear
| functions. But they are not multilinear functions.
|
| Anyway, enough for now. I'll try to reply to specific parts of your
| post as well as the other tome some time in the next few days.

When you've wasted your time writing another page like Sagnac_1?
No data, doesn't understand Doppler... collects congratulations
from fellows of the Holey Church of Relativity. No accolades
from me, mate. Your page sucks. Shame I can't help you.
Androcles.


.

Relevant Pages

  • Re: Simple Sagnac
    ... > "Since tau is a LINEAR function, ... > There is no linear relationship between t and tau and you've made ... and the coordinate transforms won't give you anything ... The "elapsed track time" for the outbound segment is ...
    (sci.physics.relativity)
  • Re: Simple Sagnac
    ... In the frame of the train, ... > |> There is no linear relationship between t and tau and you've ... > |> cuckoo transforms are linear? ... > | A linear function of two variables is the sum of a constant times ...
    (sci.physics.relativity)
  • Re: Formulae for Latin squares of size 2^n
    ... If I want good linear transforms I'll use FFT or MDS based transforms. ... BBM Mixing isn't always ideal. ... designs for which we can prove amazing things [for the ...
    (sci.crypt)
  • Re: Clear writing about relativity
    ... why homogeneity of space and time imples linearity of tau. ... Hence tau() is not linear. ... But Einstein said it must be. ...
    (sci.physics.relativity)
  • Re: Clear writing about relativity
    ... which a prerequisite for relativity. ... Hence tau() is not linear. ... But Einstein said it must be. ...
    (sci.physics.relativity)
Quantcast