"Jesse Mazer" wrote in message
...
Androcles wrote:
"Jesse Mazer" wrote in message
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Androcles wrote:
"Jesse Mazer" wrote in message
...
Androcles wrote:
"Jesse Mazer" wrote in message
...
Androcles wrote:
"Jesse Mazer" wrote in message
.. .
Androcles wrote:
Do me a favour.
Einstein wrote
"light is always propagated in empty space with a definite
velocity c which is independent of the state of motion of the
emitting body"
Reference :
http://www.fourmilab.ch/etexts/einstein/specrel/www/
Then he wrote:
"But the ray moves relatively to the initial point of k, when
measured in the stationary system, with the velocity c-v, so
that x'/(c-v) = t."
I don't know what the context of this is. I assume he's not
talking about how fast the light is moving in a given frame, but
rather how fast the light is moving away from some other object,
as seen not in the object's own frame but in a frame where the
object itself is moving at velocity v. In this case, although
light will still travel at c in this frame, the distance between
the light ray and the object moving at velocity v will be seen to
grow at the rate (c-v) in this frame. In the object's own frame,
though, the distance between itself and the light ray would grow
at the rate c, as relativity predicts.
You didn't reply to this part of my post--can you provide me with
the context of that statement by Einstein?
I've given you the reference.
Ah, I didn't notice that you were quoting from a website, sorry. But
that's a pretty long article, which section did you get the quote
involving the equation x'/(c-v) = t from?
Section 3. It wouldn't refer to it as an "article" though. It's the
original
paper that Einstein wrote in 1905 creating special relativity.
I see my interpretation was correct, then. He is saying that you have
two reference frames, K and k, and that in K's reference frame, the
distance between the origin of k and a ray of light is growing at the
rate (c-v).
For some duration of time.
For another duration of time the rate is v+c.
These durations are not equal.
Time is not a vector, it has no additive inverse.
This does not mean that the ray of light is moving at velocity (c-v)
in k's reference frame; in both K and k, the light is moving at
velocity c.
I don't believe in magic. If c = c+v and c = c-v, then v = 0.
If V = (c+v)/(1+v/c) then use that to derive the LTs.
If the system of equations is linear as Einstein claims, it shoud be
no trouble.
You are one of those that starts at the Lorentz
Transforms, proceeds to lecture on what you imagine I'm not aware
of, then conclude you are right.
You didn't really read my post, so how do you know this?
I didn't read it in depth, no. I quickly glanced down and saw some
equations I recognised as wasn't prepared to comment on, since they
cannot be derived in any sensible manner.
They can be derived from the fact that all the fundamental laws of
physics display the property of "Lorentz invariance".
Bull****.
This is a mathematical property which can be verified simply by
examining the equations. Do you understand what it means to say a
given equation shows Lorentz invariance? If not, I can go into more
detail.
Go on then, explain circularity .... err .... Lorentz invariance to me.
However, if it is true that all the fundamental laws of nature obey
Lorentz invariance,
it must be true that if different observers in motion with relation
to one another all use the same procedure to define the coordinates of
events in their frame--building a network of rulers and clocks which
are at rest with respect to themselves, and synchronizing the clocks
using the assumption that light moves at the same speed in all
directions in their frame--
It seemeth impossible for it to be, for if I walk away from a candle set
by the wall and you walk toward the same candle, we then have some
motion between us. If we then divide that motion equally between us, and
impart it to the candle such that it appeareth to be at rest upon the
floor upon which we walk, how then doth the light divide it's motion
between us, that we may both observe it to be the same?
And if I place another candle on the other side of us beside the
opposing wall, such that it is at rest with respect to the first candle,
how then shall I determine the light I approach and the light I recede
from, that they be of equal motion to me?
And yet one more candle, be it placed between us in the room, at centre,
that we may both recede from it and still determine that the motion of
light from all candles be the same?
I am but a simple man, and sorely perplexed by these deliberations, for
it is truly astounding that these assertions be true. Explain this
wondrous concept any sensible way you can, but gently, for it shall
surely confound me and cause my head to ache.
then the Lorentz transformation equations will indeed be the correct
way to transform measurements made with one set of rulers and clocks
into measurements made with another set.
I do not see any justification for magic or incorrect assumption.
From your comments I gather you probably believe that existing laws
are not really fundamental, and that when we find the real fundamental
laws they will not be Lorentz-invariant...
Quite so, the Lorentz Transforms do not exist.
but would you at least agree that *if* the fundamental laws are
Lorentz-invariant, then the Lorentz transformation will be the correct
way to transform between measurements on different observers' rulers
and clocks?
If you can use
1/2[tau(0,0,0,t)+tau(0,0,0,t+x'/V+x'/V)] = tau(x',0,0,t+x'/V)
since "the velocity of light c cannot be altered by composition with a
velocity less than that of light. For this case we obtain V =
(c+w)/(1+w/c) = c."
to derive the Lorentz Transforms, I'll take another look.
Go ahead. I'm amenable to reason.
Explain to me how the two velocities of light, c-v and c+v, measured
over a single distance, are really only one, c.
In fact, I didn't assume the Lorentz transformations were correct
without argument, I pointed out that all the current known laws of
physics are Lorentz-invariant,
Sorry, but I do not agree the laws of physics are illogical.
Are you saying there is something inherently illogical about the idea
that I will see rulers shrink as they move faster, and clocks slow
down?
Yes.
v = 3, t = 16, c = 5.
(Fixed font needed now)
S[----------]M moving frame, t = 0.
O---32----
Sam and Mike are carrying a ladder. You are going to shrink the ladder.
S[-----------]M moving frame, t = 16.
O---48----|----32---|X
Start mapping O in the stationary frame to S,
and X in the stationary frame to M. Linearly.
S[-----------]M moving frame, t = 20.
O----60----|----32---|X
E
Now show that half of 92 is 60 and half of 20 is 16, because the light
has
traveled from S to M and back again at speed 2 out and 8 back in the
moving frame, and from O to E via M (at t=16) in the stationary frame.
Again, you are free to believe there is a true Absolute Space and that
only rulers and clocks at rest in this space measure distance and time
correctly, and rulers and clocks moving in Absolute Space are
"objectively" shorter and slower. Even if you don't believe this is
how things actually work, are you saying it is *logically impossible*
that they could work this way (ie, that this hypothesis involves a
logical contradiction?) Or are you just using "illogical" to mean
"implausible"?
I mean illogical.
The PoR
stood the test of time until Einstein who corrupted it in favour of
his own
insistence concerning the speed of light, which he stated in 1905 was
"only apparently irreconcilable" and in 1920 recognized was
irreconcilable. He
rejected the PoR in favour of c = (c+v)/(1+v/c). Trouble is, he used
c+v to derive the composition of velocities.
In relativity it is true that if I see you moving in one direction
with velocity v, and I see a light beam moving in the opposite
direction with velocity c, then the distance between you will grow at
the rate (c+v), in my reference frame. But this is not a problem,
since you will *not* measure the light beam to be moving at velocity
(c+v) relative to yourself.
That is your assertion. Assertions carry no weight. If I approach a
source
of sound I'll certainly measure 731.4 m/s +v. Doppler shift will be
evident also.
If I approach a source of light, likewise. Are you saying I'd see no
shift?
No it isn't. The burden of proof is upon the claimant, and Einstein
failed to prove
his case. All I have to show is the error in his math/logic.
But it is easy to prove that if all the fundamental laws of physics
are Lorentz-invariant, then if observers use rulers and clocks at rest
relative to themselves and synchronize the clocks using Einstein's
procedure, then the Lorentz transformations will give the correct rule
for transforming between different observer's measuring system.
It is also an objective fact that the laws of electromagnetism exhibit
Lorentz-invariance, as do all our current fundamental laws (like
quantum field theory). So unless someone like you can come up with
some new fundamental laws which don't exhibit Lorentz-invariance, this
is sufficient to prove that the Lorentz transformation will give the
correct rule for transforming between the measurements of different
networks of rulers and clocks.
If you think my understanding of his meaning is incorrect, could
you explain why?
I do not know what you understanding is.
I just explained it in the post you responded to--here it is again:
"I assume he's not talking about how fast the light is moving in a
given frame, but rather how fast the light is moving away from some
other object, as seen not in the object's own frame but in a frame
where the object itself is moving at velocity v.
You admit you are making assumption. Im not in the assumption game.
Einstein makes many assumptions, many invalid.
This is the worst one.
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))
Where does that ½ come from?
Consider McCullough's silly little puzzle, which I'll embellish
slightly.
Sam and Joe are walking along carrying a 32 ft ladder between
them, at 3 fps. A mosquito flies from Sam to Joe and back to
Sam again at 5 fps, ground frame. How long does it take the
mosquito to make the round trip?
x' is the 32 ft ladder, c the speed of the mosquito and v the speed
of Sam and Joe.
Wait, first you said the mosquito flew at 5 fps, then you said at
c...are you assuming the speed of light is 5 fps in this
thought-experiment, or do you no mean c to be the speed of light?
I've clearly stated "c the speed of the mosquito", which is 5 fps.
In either case, I'd solve it like this. In the ground frame, the end
of the ladder starts out at position x=32 feet, moving at 3 fps, and
the mosquito starts out at x=0 feet, and moves at 5 fps, so in this
frame you'd calculate the time t for them to meet using the equation:
0 + 5t = 32 + 3t
which gives t=16 seconds, meeting at position x=80. Then the mosquito
flies backwards from that position at velocity -5 fps, and meanwhile
the other end of the ladder starts at x=48 feet and moves forward at 3
fps, so to find where they meet you could use this equation:
80 - 5t = 48 + 3t
which gives t = 4 seconds. So, the total time is 20 seconds. Notice
that I calculated everything from the ground frame, without ever
switching to the mosquito's frame or the ladder's frame.
Well, carry on, then. This is only algebra, prove that time for the
mosquito
is less, the ladder is shorter and that the speed of the mosquito is 5
fps in the moving frame. You can do (c+v)/(1+v/c) when you come to § 5
"with the help of the equations of transformation developed in § 3 "-
Einstein.
Introduction.
Section 1
Section 2
Section 3 Use c+v, c - v. Derive LT.
Section 4
Section 5 c= (c+v)/(1+v/c) from section 3.
The speeds of the mosquito in the moving frame are 8 and 3.
This makes the speed of the mosquito 5 in the moving frame
(Lorentz invariant)
I don't call that logical or consistent, but you do. We cannot agree
on what is logical.
The problem as I see it is that the time for the moquito to travel
from 0 to 80 and back to 60 at 5 fps = 100/5 = 20 is the same
time as in the moving frame, a distance of 32 at 8 fps + 32 at 2 fps,
and Einstein has used ½ of 20 seconds in the stationary frame
for the one way trip taking 16 seconds in the moving frame.
I call that assinine. The light never gets back to the origin of the K
frame, there is no half to consider.
Answer. 16 seconds to reach Joe and 4 seconds to return.
So (16+4)/2 = 16 ?
I don't think so. Einstein tries to justify it by saying Sam cannot
know when the mosquito reaches Joe, so he'll simply use the ½.
That is assumption, not mathematics.
What are you talking about?
I'm talking about the statement
"From the origin of system k let a ray be emitted at the time tau0 along
the X-axis to x', and at the time tau1 be reflected thence to the origin
of the co-ordinates,
arriving there at the time tau2; we then must have ½(tau0+tau2) = tau1.
Why MUST we have that? I would only be true if the speed of the mosquito
was 5 in the moving frame, and we have already stated it to be 2 and 8.
It doesn't become 5 until section 5, and a priori, we haven't reached
that section.
I say (16/20) * (0+20) = 16, not ½(tau(0,0) +tau(0, 20)) = tau(32, 16).
Einstein would agree with my analysis above,
even if the speed of light was 5 feet per second; notice that I always
assumed the mosquito was travelling at 5 fps in the ground frame, I
never switched to a different frame.
I expect Einstein would. I don't.
As far as I'm concerned the light never gets back to the origin in
the stationary frame, so there is no half to consider.
So if we look at light from a given star at two points in the earth's
orbit, the first when the earth is moving away from the star and the
second when it's moving towards it, why isn't any difference in the
speed of light measured?
It is. You'll see as doppler-shift.
Androcles.
A doppler-shift is just a change in the wavelength, not in the
velocity.
Doppler's equation is (for one axis)
(c+v)
f' = f. ---------
(c+u)
where u is the velocity of the source and v the velocity of the
observer.
Where there is no medium, this reduces to
f' = f. (c+v)/c.
The wavelength does not change.
Androcles.