Re: Simple Sagnac
- From: sal <pragmatist@xxxxxxxxxx>
- Date: Tue, 02 Aug 2005 11:03:20 -0400
You must be right; I must be stupid. Nothing else would explain my
wasting yet more time looking over this post of yours again. Be that
as it may, that's what I'm doing, and I saw a couple of items worth
additional comments...
On Mon, 01 Aug 2005 23:40:44 +0000, Androcles wrote:
>
> "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.
No, we can't forget the x -- if it's fixed, then it provides a
constant offset. The formula was
tau = 1.25 * t - 0.15 * x
When x = 60, this reduces to
tau = 1.25 * t - 9
With x held fixed at 60, when t=4, tau=-4. Got that? It's not 8,
it's -4. In the frame of the train, the event at those coordinates
had already happened before the train left the origin.
On the other hand, at time t=4 the light ray was actually at x=20
.... it hadn't gotten to x=60 yet, and wouldn't for quite a while.
The front of the train was at 32+3*4=44, so the engine hadn't gotten
to 60 yet, either. If it had there would have been a contradiction,
of course, since time in the engine increases after it leaves the
station, so it's a good thing it hadn't.
On the other hand, with t=20 with x fixed at 60, we find tau=16, as
you said.
> 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,
The _speed_ of light is, by your assumption, 5. The _velocity_ of a
particular ray of light, viewing the problem as having one dimension,
is +5 or -5, or, if you want to be pedantic, it's occasionally
undefined, as at the instant when the ray of light is being reflected
from the mirror.
In general you can't talk about either velocity or speed without
having two points separated by a space in which the object (ray or
otherwise) traveled nearly linearly. Any interval which includes
reflection from a mirror is one over which one can't talk sensibly
about the velocity of the ray of light.
> 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.
As I just said, you can't talk reasonably about the speed of the light
over the entire interval which includes the reflection.
You walk across the room, turn around, and come back. Your start and
end point are the same, so your velocity must have been zero.
Therefore you didn't move. Paradox? No, just confusion over how you
compute "velocity" of something.
> | 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.
Yes, I know that. But when you claim
tau = 1.25 * t - 0.15 * x
is not a linear function, you should not be surprised if I take that
to indicate that you may need some review.
> | 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 ray reverses direction; C is just a constant and has no
"direction".
> the tip of the ray ends where it
> started. Shrinking the distance doesn't help. c = 0 in the cuckoo
> transforms.
You defined c=5. Therefore it is constant.
> 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.
You said c=5.
> 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.
You said C=5, and that's what I've been using. Now you want it to be
0. Make up your mind.
If the ray travels at +c, and c=0, it'll take quite a while to get to
the engine, though.
> | 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.
So you finally agree, c=5. (But you just said c=0, didn't you?)
> 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.
You just agreed that c=5 in the previous paragraph, now you want it to
be zero again.
> | 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.
Again, you just said that linear function isn't a linear function.
Then what is it?
> | 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.
It doesn't "worry" me. But it's useless -- tau=1.25t+0.15x doesn't
relate _differences_ in tau to _differences_ in t, it relates their
_values_ at particular points. If you plug in differences instead of
actual values you'll get nonsense out, as you've shown very clearly.
> 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.
Right -- the engine gets there _before_ the light. The engine gets to
60 at t=9, the light gets to 60 at t=12.
The light doesn't catch up with the engine until it reaches 80, at
time t=16.
> tau(60,0,0,12) = 9 and a bit.
No, tau(60,0,0,12) = 6. The event at time 12, location 60, has time 6
in the train's frame.
> | 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?
Well, _you_ should, because you're the one plugging _differences_ in
track time into the coordinate transformation formula and then
complaining that it doesn't produce equivalent _differences_ in the
train coordinates.
> 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 I'm sure yours is too, else you never could have learned to type.
(But wait, come to think of it, there's somebody who claims she taught
her dog to type, and I'm not sure about his IQ, so maybe that
reasoning isn't airtight...)
As to tau, one must constrain it to be linear or the gedanken
experiment in E:1905 wouldn't be sufficient to let one derive it.
> | 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.
But did you understand what I said? Yes, or no?
> | 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.
Did you forget the mirror? The light passes x=60 _twice_ because its
path is "folded" by the mirror.
At t=12 the light hasn't caught up with the engine yet.
> 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.
You should keep going over it until you understand it.
> 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 gorilla was in physics class, not math class. And it was a
Newtonian gorilla; I don't even know if the same experiment works with
relativistic gravity but I suspect it doesn't.
> | 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.
This is not a Galilean train. On the Galilean train, SoL is C in only
one preferred frame.
This is an important point. Put down your glass, stop looking out the
window, and listen for a moment. C IS CONSTANT IN GALILEAN/NEWTONIAN
MECHANICS, TOO. The difference is that in relativity, SoL is always
measured to be C in inertial frames.
And in any system, "_speed_ of light" refers to its _magnitude_.
Light traveling at velocity -5 has a _speed_ of 5.
> | 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,
Then you should read it again, of course.
> 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?
Not unless it's a _sincere_ "yes".
> | 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.
God you're polite today, all those words and not one real cuss
word. ("sucks" doesn't count, it's only a contextual cuss word.)
I tried to make the page simple enough so that _anybody_ could
understand it. As you have pointed out, I clearly didn't succeed.
> Shame I can't help you. Androcles.
--
Nospam becomes physicsinsights to fix the email
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