Re: I am Trying To Learn Relativity

On Jul 11, 4:41 pm, Mike <elea...@xxxxxxxx> wrote:
On Jul 11, 1:28 pm, Bruce Richmond <bsr3...@xxxxxxxxxxx> wrote:
On Jul 11, 4:07 am, Mike <elea...@xxxxxxxx> wrote:

On Jul 10, 8:26 pm, Bruce Richmond <bsr3...@xxxxxxxxxxx> wrote:

On Jul 10, 5:08 pm, Mike <elea...@xxxxxxxx> wrote:

On Jul 10, 2:49 pm, PD <TheDraperFam...@xxxxxxxxx> wrote:

[snip]

There are no paradoxes in relativity. Relativity is completely self-
consistent. There are puzzles that reveal superficial understanding of
relativity, and they are designed as teaching tools, but there are no
logical contradictions in relativity.

If there are no paradoxes in relativity, call these people and cry
again once more to remove their webpage:

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/bugrivet.html#c1

The bug gets squashed either way so there is no paradox.  These things
have been explained elswhere so I'm not wasting my time doing so here.

Reference please. I stated a referen ce, whether you loike it or not,
for eah of my statements.

What part of "I'm not wasting my time" did you not understand?  Google
"paradox".

Google paradox? [snip]

Yeah, do your own leg work. I owe you nothing. If you wish to stay
stupid it's your choice.

Even more importantly you present trivial
examples of Kinematics not relevant to the problem at hand.

Let's see, it showed the correct choice of rest frame makes
calculations easier, that the results we get are not intuitive, and
provided an example of frame jumping which you showed earlier that you
had problems with. That's three things making it relevant, and I
notice you didn't explain why the kinetic energy was different for the
two directions or where the frame jumping took place.

Don't worry, I wont take the time to write anything like that again
knowing it will probably get snipped.

The tone of your voice is of someone who is policing or thinks of
itslef as an authority.  I GOT BAD NEWS FOR YOU. You have no idea of
the problem and you miss all the point, probably both on purpose but
laos from an  inability to understand them

I'm not the one that came here asking questions. Why were you asking
questions if you have all the answers?

It simple, much simplers than you think and Petkov in the reference I
gave puts it in terms of "The vicious circle -- to determine whether
two events are simultaneous we need to know the one-way velocity of
light between them, but to determine the one-way velocity of light we
need to know that the two events are simultaneous"

"The vicious circle" is vicious because it looks like there is no good
way to break it.

Lorentz broke it by saying the speed of light in the ether frame was
c. He then showed how it could be measured c in moving frames,
provided everything changed size in the direction of motion.

Einstein broke the circle by saying the speed of light is c in all
frames by definition. But in doing so he had to give up absolute time
and space.

It sounds like you want to keep the speed of light c in all frames but
not give up absolute time. I'm not going to write out the setup with
the cars again, but explain how you think light can travel at the same
speed relative to the two cars traveling at different speeds. If you
can't your theory is dead in the water.

I will say it even in a more simple way. Einstein commited the formal
fallacy of denying the antecedent when he concluded that simultaneity
is relative. He first showed that if

tB - tA = tA' - tB then the clocks are synched.

He then shows a case where the anteceeant, tB - tA = tA' - tB, is
false. He thne conludes that the clocks are not synched. This is the
formal fallacy of denying the antecedant, or aht is notoriously known
as the inverse error.

However there are more possibilities as to why tB - tA = tA' - tB  is
false. A simple one is that the rod is moving, which was actually the
case. That has nothing to do with whether the clcoks are synched. If
the motion is taking into account, one can know why the antecedent is
false.

"If the motion is taking into account" The observer considers himself
to be at rest. There is no motion to take into account. You have
also written the equation incorrectly three times, so don't try to
claim it was a typo. The correct form is

tB - tA = t'A - tB

http://www.fourmilab.ch/etexts/einstein/specrel/www/

If you put two identical clocks at A and B then you can record the
time of events happening at A using the clock there, and events
happening at B using the clock there. That gives you an A time and a
B time, but it gives you no way to relate when things happen at A to
when things happen at B without first synchronizing the clocks.

To break "The vicious circle" you mentioned above Einstein wrote, "We
have not defined a common ``time'' for A and B, for the latter cannot
be defined at all unless we establish by definition that the ``time''
required by light to travel from A to B equals the ``time'' it
requires to travel from B to A."

The above tells you that by agreement between A and B light travels at
the same speed in both directions. If we want to know how long it
took light to get from A to B we subtract the time it left A from the
time it arived at B and get tB - tA. To get the time for the trip
back we subtract the time it left B from the time it arrived back at
A. Since the time it left A and the time it arrived back at A are two
different times he used t' to represent the arrival time back at A.
So that leg of the round trip is t'A - tB. Since the two legs take
the same time by definition we can write

tB - tA = t'A - tB

As for your claim, "He then shows a case where the anteceeant, tB - tA
= tA' - tB, is false. He thne conludes that the clocks are not
synched." We set up the clocks on the rod so that they tick in
unison with the clocks in the stationary system, as it sounds you
would like to do. With the rod moving at v we send a beam from A to
B. We are measuring things in the "stationary system" now. The light
travels away from the point where it was emitted as observed by a
stationary observer in the stationary system. The rod is moving at v
relative to the stationary observer. So the stationary observer
calculates that the light is moving down the rod at c-v. It can't be
moving down the rod at c in the stationary system because it can't
move at c relative to both the stationary observer and A any more than
an object could move at the same speed relative to the two cars that
were moving relative to each other like I described before. This does
not contradict the second postulate. The statonary observer never
measured light to travel at any speed other than c in his frame. The
c-v was the calculated closing speed of light with the moving observer
B. The stationary observer was not in on the agreement between A and
B that the light travels at c between them. So it takes L/(c-v) for
the light to travel from A to B as measured in the stationary system.

The light reflects off B and travels back moving at c while A
continues moving toward it at v for a closing speed of c+v. So the
return trip takes the shorter time of L/(c+v). The trip from A to B
took longer than the trip from B to A. Which is fine since the beam
had to travel further to catch up to B than it had to travel back to
A, as measured in the stationary system.

But on the rod, where the observers consider themselves to be at rest,
the light travels the same distance in both directions. And since we
have synced the clocks on the rod to be in agreement with the
stationary clocks they must agree that the trip from A to B was
measured to take longer than the trip from B to A. So by the
definition we provided earlier A and B must consider their clocks to
be out of sync. By adjusting the clock at B to read a bit later they
can measure the times for both legs of the trip to be the same. They
would also have to adjust the tick rate slightly to measure light to
travel at c. (The tick rate had been adjusted away from its standard
setting to get the clock to stay in sync with the clocks in the
"stationary system".) The clocks will then be in sync according to
the definition provided earlier.

Next, the experimental predictions of SR are identical to those that
assume absolute simultaneity, like an absolute spacem aether, or light
propagation medium. You seem not to know that and you profess yourself
as an expert. You are obviously not one. TOm Roberts who is close to
an expert, has admitetd this several times.

LOL, you mean "an absolute spacem aether" like LET. Yes, I agree the
predictions are the same, and I know Tom agrees also. Of course the
clocks in LET don't show absolute time. The clocks moving relative to
the ether slow down and go out of sync with the ether frame clocks.
And moving matter contracts, and you can't even tell which frame *is*
the ether frame, so there is no way you can say the clocks read in
absolute time. If by "an absolute spacem aether" you mean one like
what was assumed prior to the MMX you are of course just plain wrong.

If you mean something else by "an absolute spacem aether" please
explain what it is.

As for "the experimental predictions of SR are identical to those that
assume absolute simultaneity" that is wrong also. SR predicted that
the two legs of the round trip for light along the rod would take the
same time. When absolute simultanity was forced on the moving clocks
the two legs took different times.


Relativity of simultaneity is a bogus concept that is a re4sult of a
trivial error by Einstein. It does not affect the physical
significance of SR. It just makes it indistinguishable from a larger
class of theories where relativity of simultaneity is not an issue and
that can be the result, for example, of a speed od source dependence
of one way speed of light.

You are trying to talk the talk but it is coming out gibberish.

Good luck to you too,

Bruce
.

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