Re: Simple Sagnac
- From: sal <pragmatist@xxxxxxxxxx>
- Date: Sat, 30 Jul 2005 23:00:13 -0400
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:
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
and:
A (again, somewhat out of context):
> There is no linear relationship between t and tau and you've made
> the right point.
and:
> Are you 100% ABSOLUTELY CERTAIN as you possibly could be that the
> cuckoo transforms are linear?
To which I say, WTF???
At risk of insulting your intelligence I will walk through what a
"linear" transformation is.
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
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, then the value
g = gamma = 1/sqrt(1-V^2/C^2)
is a constant.
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.
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. In that case, we had
g = 1/sqrt(1-3^2/5^2) = 1.25
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?
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? 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_, 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?
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
Point 2: t=16, x=80 ... light ray reflected at engine
Point 3: t=20, x=60 ... light ray arrives at caboose
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
The "elapsed track time" for the outbound segment is
dt = 16 - 0 = 16 seconds
The "elapsed track time" for the return segment is
dt = 20 - 16 = 4 seconds
The "elapsed train time" for the outbound segment is
dtau = 8 - 0 = 8 seconds
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 :-) )
Finally, is it possible that you are mixing up "linear" functions with
"multilinear" functions? 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.
--
Nospam becomes physicsinsights to fix the email
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