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Dirk Van de moortel · Guest

strich.9993@gmail.com <strich.9993@gmail.com> wrote in message
b78d0757-8597-4332-9e5a-28ae3074b5ce@h60g2000hsg.googlegroups.com
[quote]On Oct 9, 11:15 am, Uncle Ben <b...@greenba.com> wrote:

You never defined E1 and M1. But I can guess your meaning. When the
two frames had no relative velocity, I agree that you could
synchronize the clocks, but when frame M is moving w.r.t. E you
have lost the synchronization. (All of this being what SR says, not
necessarily what is true.)

It>s good you agree that E1=M1.

We proved in RS101 that M2=M1, and in the special case of E where it
did not move at all, E2=E1. The rest is simple math.
[/quote]
The "Simple Math" of representing 8 quantities with 4 variables:

As was explained to you in
http://groups.google.com/group/sci.physics.relativity/msg/e6df3b52be1e7860 ,
as a reply to
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/DisprovenForGood.html ,
1) before M is set in motion:
Tick/tock on clock at rest in E:
rate w.r.t E-clock: dte/dte = 1
rate w.r.t M-clock: dte/dtm = 1
Tick/tock on clock at rest in M:
rate w.r.t E-clock: dtm/dte = 1
rate w.r.t M-clock: dtm/dtm = 1

2) after M is set in motion:
Tick/tock on clock at rest in E:
rate w.r.t E-clock: dte/dte = 1
rate w.r.t M-clock: dte/dtm = 1/gamma (was 1 before)
Tick/tock on clock at rest in M:
rate w.r.t E-clock: dtm/dte = 1/gamma (was 1 before)
rate w.r.t M-clock: dtm/dtm = 1

Condensing these 8 quantities into 4 variables is called fumbling,
or cheating, and in "Allyou!r" case, probably both.

Dirk Vdm

Dirk Van de moortel · Guest

Strich.9 <strich.9992@gmail.com> wrote in message
5f1434fb-775f-407d-b739-9baf07d57dd3@t41g2000hsc.googlegroups.com
[quote]On Oct 9, 11:29 am, "Dirk Van de moortel"
dirkvandemoor...@nospAm.hotmail.com> wrote:

Condensing these 8 quantities into 4 variables is called fumbling,

Creating new variables where there is no need, is called cheating, or
maybe just stupidity.

But let me humor this idiot. Justify those extra variables, without
prematurely invoking the same conclusion you want to prove.

Be careful of the premature part. That has always been your waterloo.
[/quote]
Brilliant, this "Creating new variables where there is no need"
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/YourWaterloo.html

And specially, that phrase "But let me humor this idiot", is
100% typical of "Allyou!". Check it out at
http://www.google.com/search?&q=Allyou%21+site:users%2etelenet%2ebe

Dirk Vdm

doug · Guest

strich.9993@gmail.com wrote:

[quote]On Oct 9, 10:17 am, papa_r...@hotmail.com wrote:

On 9 oct, 09:50, strich.9...@gmail.com wrote:

On Oct 9, 9:26 am, PD <TheDraperFam...@gmail.com> wrote:

On Oct 9, 8:19 am, strich.9...@gmail.com wrote:

On Oct 8, 8:59 pm, Uncle Ben <b...@greenba.com> wrote:

On Oct 8, 3:49 pm, strich.9...@gmail.com wrote:

On Oct 8, 3:29 pm, fishfry <BLOCKSPAMfish...@your-mailbox.com> wrote:

In article
1dd26d6f-6d3e-4ca3-b299-0320b3fc6...@f40g2000pri.googlegroups.com>,

"Strich.9" <strich.9...@gmail.com> wrote:

"Relativity = Stupidity"

apparently the physics trolls are worked up about relativity the same
way the math trolls are worked up about set theory.

Please see:http://groups.google.com/group/sci.physics/browse_frm/thread/d6a4f635...

I>m halfway to proving relativity wrong. Let me finish this
relativity thing, then perhaps I can help resolve the Cantor
'dilemma'.

Strich you are not even near showing that you know what SR says, much
less proving it wrong.

Uncle Ben- Hide quoted text -

- Show quoted text -

I>m sorry but your reply has no argumental weight (no logic).

Thus the propositions remain undisputed. In summary:

Clocks E and M in inertial frames E and M are at rest with respect to
one another with rate E1 and M1 respectively. Obviously E1 = M1. Let
frame M with clock M now move with respect to E, with relative
constant
velocity v. Let E2 and M2 be the rates of the clocks after the move
with respect to frame E and M.

From RS 101 we learned that M2 = M1.
Commonsense tells us E2 = E1.

Since E1 = M1 in the premise, then E2 = M2.

Nope. This assumes that E2 is the rate of the E clock in some absolute
sense, independent of frame.
E2=E1 in the E frame, but E2 =/= E1 in the M frame.
You cannot evaluate a statement in one frame and say that this is a
true statement absolutely or irrespective of frame.

Commonsense tells you that the M clock is stationary in the M frame,
both in snapshots 1 and 2. By your logic, we should just insist that
the M clock has the property of being stationary, period. But this
would be obviously idiotic, as you>ve just declared the difference
between the snapshots 1 and 2 is that the M clock is set in motion in
between.

It should be apparent to ANYONE not an idiot that the property of
"stationary" is a frame-dependent statement. It is also obvious to
ANYONE not an idiot that the clock rate remaining constant is a frame-
dependent statement.

Since time does not change with motion, then obviously, Einstein>s LTE
is rendered imaginary and unnecessary.- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

What can you not understand?

Answer this simple question: When clock M moves away, does the rate of
clock E in frame E changes? Yes or no? (clue: the clock still ticks
and tocks)

You idiotically insist in that comparing synchronicity of two clocks
at one point in spacetime, implies those same clocks will maintain
synchronicity at a different point in spacetime, while one of the
clocks is moving with respect to the other. That is wrong and has been
shown to be wrong both theoretically and experimentally.

Again, try to read the following paragraph from Landau and Lifshitz
book:

"Suppose some clocks are moving in uniform rectilinear motion relative
to an inertial system K. A reference frame K' linked to the latter is
also inertial. Then from the point of view of an observer in the K
system the clocks in the K' system fall behind. And conversely, from
the point of view of the K' system, the clocks in K lag. To convince
ourselves that there is no contradiction, let us note the following.
In order to establish that the clocks in the K' system lag behind
those in the K system, we must proceed in the following fashion.
Suppose that at a certain moment the clock in K' passes by the clock
in K, and at that moment the readings of the two clocks coincide. To
compare the rates of the two clocks in K and K' we must once more
compare the readings of the same moving clock in K' with the clocks in
K. But now we compare this clock with different clocks in K with those
past which the clock in K' goes at this new time. Then we find that
the clock in K' lags behind the clocks in K with which it is being
compared. We see that to compare the rates of clocks in two reference
frames we require several clocks in one frame and one in the other,
and that therefore this process is not symmetric with respect to the
two systems. The clock that appears to lag is always the one which is
being compared with different clocks in the other system. If we have
two clocks, one of which describes a closed path returning to the
starting point (the position of the clock which remained at rest),
then clearly the moving clock appears to lag relative to the one at
rest. The converse reasoning, in which the moving clock would be
considered to be at rest (and vice versa) is now impossible, since the
clock describing a closed trajectory does not carry out a uniform
rectilinear motion, so that a coordinate system linked to it will not
be inertial."

Miguel Rios- Hide quoted text -

- Show quoted text -

Do not change the topic. The calculations are simple.

Let E1=M1.

If E1=E2 and M1=M2

Then E2=M2.

Where does it become confusing for you?
[/quote]
It is simple to me. It does seem very confusing to you.
E2 does not equal M2. This is an experimental fact.
You are wrong.

Guest

On 9 oct, 16:40, strich.9...@gmail.com wrote:
[quote]On Oct 9, 4:19 pm, PD <TheDraperFam...@gmail.com> wrote:

What part of equality do you not understand?

In the beginning, E1=M1. Meaning the E and M clocks were ticking in
synch. clock E was the same rate whether it was measured from E or
from M, and clock M was the same rate whether it was measured from E
or from M. At the end, E2=M2. Meaning the clocks are still ticking
in synch.

Let me show you in painful detail. Note that M2=E2. Observer M in M
looks at his clock M. It is ticking at rate M2. Observer M in M then
looks at clock E. It is ticking at rate E2, which is equal to rate
M2. Also, observer E in E looks at his clock E. It is ticking at
rate E2. Observer E in E then looks at clock M. It is ticking at M2,
which is equal to E2.

I feel bad that you have schizophrenized your logic so bad that you
have lost the ability to see an identity relation anymore.
[/quote]
Your calculations are incomplete, since when you say M2=E2 you are not
considering that those two measures are in different frames, one of
them moving at a high speed with respect to the other.The relations
should be as follows:

Let the rate of clock E, E1, measured in frame E defined by
coordinates (x,y,z,t), be equal to the rate of clock M, M1, measured
in frame M defined by coordinates (x',y',z',t'), at initial point
(x,y,z,t) = (x',y',z',t') = (0,0,0,0)

Clocks E and M are very accurate. The rate of clock E, E1, at
(0,0,0,0) will be equal to the rate of clock E, E2, at (0,0,0,t2),
both rates measured in frame E.
Similarly, the rate of clock M, M1, at (0,0,0,0) will be equal to the
rate of clock M, M2 at (x2',0,0,t2').
So both clocks maintain their rates as measured in their own frames.

Then it can be shown that the rate E2 at (0,0,0,t2) IS NOT EQUAL to
the rate M2 at (x2',0,0,t2')

Miguel Rios

Eric Gisse · Guest

On Oct 9, 1:32 pm, papa_r...@hotmail.com wrote:
[snip]

What the f ck is wrong with you?

Is it not yet obvious that he is not interested in learning and wants
to soak up your time in shit like this?

Spaceman · Guest

dkelvey@hotmail.com wrote:
[quote]On Oct 8, 10:33 am, "Strich.9" <strich.9...@gmail.com> wrote:
"Relativity = Stupidity"

That is a very difficult equation to teach.

We are only using one example to teach the subject. Past experience
of Professor Androcles on teaching this subject has shown that pupils
learn extremely slowly, if at all. This series of classes hopes to
teach this difficult subject on a step by step basis.

Example 1: Clocks E and M in inertial frames E and M are at rest with
respect to one another. Clock rates are of course equal. Let frame M
with clock M now move with respect to E, with relative constant
velocity v.

In "R=S 101", it was proven that the rate of clock M, in frame M did
not change. Many pupils reached this conclusion with much pain, but
eventually everybody learned this now accepted fact.

We skip "R=S 102" since the course material teaches that the rate of
clock E in frame E did not change after the external clock M moved.
It is common sense that since no change occurred in frame E, then no
change must occur in the rate of clock E. Of course, the course "R=S
102" may be opened for those who are having difficulty in the
succeeding course.

Here in "R=S 103" we use still the same example above, and we put
together the fact that since the rate of clock M did not change after
it moved (see course RS 101) and the rate of clock E did not change
after the M clock moved (course RS 102), then the conclusion must be
that clocks E and M are in synch before and after the M clock moved.

Let me repeat the argument:

Rates of clocks E and M did not change after clock M moved. Since
the rates were equal before the move, then the rates must be equal
after the move. In short, the clocks stay in synch.

The floor is now open for questions...

Hi
The problem is that experiments have shown that the clocks ARE no
longer in sync when brought back together in the same location.
[/quote]
Clock malfuctions are older than relativity.
It is nice how you ignore the history of clocks to support
a theory that is also ignorant of clock malfunctions

[quote]So far, relativity has predicted exactly how much those clocks should
have shifted, with precission.
[/quote]
Relativity fails to correctly predict ALL clocks rate changes.
Relativity fails to understand history of clock and the history of
more than one type of clock.

PD · Guest

On Oct 9, 3:40 pm, strich.9...@gmail.com wrote:
[quote]On Oct 9, 4:19 pm, PD <TheDraperFam...@gmail.com> wrote:

On Oct 9, 3:17 pm, strich.9...@gmail.com wrote:

On Oct 9, 3:33 pm, PD <TheDraperFam...@gmail.com> wrote:

On Oct 9, 1:16 pm, strich.9...@gmail.com wrote:

On Oct 9, 2:03 pm, PD <TheDraperFam...@gmail.com> wrote:

On Oct 9, 12:59 pm, strich.9...@gmail.com wrote:

On Oct 9, 1:45 pm, PD <TheDraperFam...@gmail.com> wrote:

On Oct 9, 12:38 pm, strich.9...@gmail.com wrote:

On Oct 9, 1:30 pm, PD <TheDraperFam...@gmail.com> wrote:

On Oct 9, 8:50 am, strich.9...@gmail.com wrote:

On Oct 9, 9:26 am, PD <TheDraperFam...@gmail.com> wrote:

On Oct 9, 8:19 am, strich.9...@gmail.com wrote:

On Oct 8, 8:59 pm, Uncle Ben <b...@greenba.com> wrote:

On Oct 8, 3:49 pm, strich.9...@gmail.com wrote:

On Oct 8, 3:29 pm, fishfry <BLOCKSPAMfish....@your-mailbox.com> wrote:

In article
1dd26d6f-6d3e-4ca3-b299-0320b3fc6...@f40g2000pri.googlegroups.com>,

"Strich.9" <strich.9...@gmail.com> wrote:
"Relativity = Stupidity"

apparently the physics trolls are worked up about relativity the same
way the math trolls are worked up about set theory.

Please see:http://groups.google.com/group/sci..physics/browse_frm/thread/d6a4f635...

I>m halfway to proving relativity wrong. Let me finish this
relativity thing, then perhaps I can help resolve the Cantor
'dilemma'.

Strich you are not even near showing that you know what SR says, much
less proving it wrong.

Uncle Ben- Hide quoted text -

- Show quoted text -

I>m sorry but your reply has no argumental weight (no logic).

Thus the propositions remain undisputed. In summary:

Clocks E and M in inertial frames E and M are at rest with respect to
one another with rate E1 and M1 respectively. Obviously E1 = M1. Let
frame M with clock M now move with respect to E, with relative
constant
velocity v. Let E2 and M2 be the rates of the clocks after the move
with respect to frame E and M.

From RS 101 we learned that M2 = M1.
Commonsense tells us E2 = E1.

Since E1 = M1 in the premise, then E2 = M2.

Nope. This assumes that E2 is the rate of the E clock in some absolute
sense, independent of frame.
E2=E1 in the E frame, but E2 =/= E1 in the M frame.
You cannot evaluate a statement in one frame and say that this is a
true statement absolutely or irrespective of frame.

Commonsense tells you that the M clock is stationary in the M frame,
both in snapshots 1 and 2. By your logic, we should just insist that
the M clock has the property of being stationary, period. But this
would be obviously idiotic, as you>ve just declared the difference
between the snapshots 1 and 2 is that the M clock is set in motion in
between.

It should be apparent to ANYONE not an idiot that the property of
"stationary" is a frame-dependent statement. It is also obvious to
ANYONE not an idiot that the clock rate remaining constant is a frame-
dependent statement.

Since time does not change with motion, then obviously, Einstein>s LTE
is rendered imaginary and unnecessary.- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

What can you not understand?

Answer this simple question: When clock M moves away, does the rate of
clock E in frame E changes? Yes or no? (clue: the clock still ticks
and tocks)

No. This has been answered before. This seems to be very confusing to
you.

It will help if, rather than using just E1 to denote the rate of clock
1, you use a notation that denotes the clock rate in a particular
reference frame.

For example, you could use E1(in E) to denote the rate of clock E in
state 1 (before the onset of M>s motion) in frame E.

Please rewrite your equations, then.

Note also that you have a choice:
- Your frame M can be one that is moving relative to frame E at all
times, which is what an inertial reference frame would have to do
- Your frame M can be tied to clock M while it changes its motion, in
which case M is not an inertial reference frame and you>ll be able to
say less about clock rates at times 1 and 2.

PD- Hide quoted text -

- Show quoted text -

By E1, E2 we mean the rates of clock E in frame E (before and after).
By M1, M2 we mean the rates of clock M in frame M (before and after).

E1=M1 was defined at the outset as the frames are at rest to each.

Ah, but here we have a problem. You have a choice, you see.
- You can choose FRAMES E and M to be inertial, in which case they
will always have to be in motion relative to each other, and clock M
will change from being stationary in frame E to stationary in frame M.
- You can choose frame M to be noninertial, in which case you cannot
make the statement that M1=M2.

We have shown already that M2=M1, and E2=E1. The conclusion is
straightforward: M2=M1, M1=E1, E1=E2, or M2=E2.

Note that the conclusion relates the 'new' times in the now moving M
frame to the E frame.

What is it that you cannot follow here?- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

The last resort. Same way Einstein resolved the Twin paradox in SR by
cheating and using GR.

Said nothing about GR. But it does help to know what an inertial
reference frame is.

Not valid. But go ahead, give it a try.

I noticed you have not answered my simple question about the equality
of M2 and E2.

Because you have not denoted which frames M2 and E2 are specific to.

PD- Hide quoted text -

- Show quoted text -

I guess you were sleeping. Let me repeat:

By E1, E2 we mean the rates of clock E in frame E (before and after).
By M1, M2 we mean the rates of clock M in frame M (before and after).

E1=M1 was defined at the outset as the frames are at rest to each.

We have shown already that M2=M1, and E2=E1. The conclusion is
straightforward: M2=M1, M1=E1, E1=E2, or M2=E2.

Yes, this is fine. The final clock rate of clock M in frame M is the
same as the final clock rate of clock E in frame E. However, this
doesn>t say at all whether the final clock rate of clock E in frame E
is the same as the final clock rate of clock M in frame E. Nor does it
say that the final clock rate of clock E in frame M is the same as the
final clock rate of clock M in frame M. Relativity says that the final
clock rates of the two clocks, when viewed from the *same frame*, are
different. What is it about this that you don>t understand?

What is it that you cannot follow again? (Take some coffee so you can
keep up)- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

Scroll back to the E2=M2 that you agreed with. That means the rates
of the moving clock M in M and the rate of 'clock E while M is
moving', in E, are the same, meaning still synchronized.

No it doesn>t mean that at all. For them to be synchronized, it means
the rates are the same when viewed from the *same frame*. You aren>t
doing that.

Perhaps you missed something very basic.

Where can
you insert an inequality here that Relativity demands?- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

What part of equality do you not understand?

In the beginning, E1=M1. Meaning the E and M clocks were ticking in
synch.
[/quote]
No, it does not mean that. For the clocks to be in synch you would
need to compare the clock rate of clock E in frame E with the clock
rate of clock M in frame E, or alternatively you could compare the
clock rate of clock E in frame M with the clock rate of clock M in
frame M. You haven>t done that, in either states 1 or 2.

[quote] clock E was the same rate whether it was measured from E or
from M, and clock M was the same rate whether it was measured from E
or from M.
[/quote]
That>s true if you are going to have a frame M that is initially
coincident with frame E and then accelerates to have motion relative
to frame E. However, now you can>t say M1=M2. The way you determine
whether clock rates are the same is if you take a SINGLE pair of
physical events and see if the time elapsed between those events, as
measured by a clock at rest in that frame, is the same. You have TWO
pairs of events, a tick and a tock before the acceleration, and a tick
and a tock after the acceleration.

[quote] At the end, E2=M2. Meaning the clocks are still ticking
in synch.
[/quote]
Nope.

[quote]
Let me show you in painful detail. Note that M2=E2. Observer M in M
looks at his clock M. It is ticking at rate M2.
Observer M in M then
looks at clock E. It is ticking at rate E2,
[/quote]
No sir. Observer in M cannot see the clock E ticking at rate E2. E2 is
the rate observed in the *E* frame, by definition.
You cannot possibly say that *M* looks at the clock E and sees rate
E2.

What happens in observation is this:
The observer in E looks at the clock E and writes down on a piece of
paper: "E clock is ticking at 0.75 Hz."
The observer in M looks at the clock M and writes down on a piece of
paper: "M clock is ticking at 0.75 Hz."
Now, they can pass these notes to each other via pneumatic tube and
this completely satisfies the equation M2 = E2, because that>s
precisely what that equation means.

But in observation it is ALSO true that:
The observer in E looks at the clock M and writes down on a piece of
paper: "M clock is ticking at 0.60 Hz."
The observer in M looks at the clock E and writes down on a piece of
paper: "E clock is ticking at 0.60 Hz."
Note that neither of these rates is E2 or M2, according to their
definitions, so the equation M2=E2 has no bearing whatsoever on these
observations.

[quote]which is equal to rate
M2. Also, observer E in E looks at his clock E. It is ticking at
rate E2. Observer E in E then looks at clock M. It is ticking at M2,
which is equal to E2.

I feel bad that you have schizophrenized your logic so bad that you
have lost the ability to see an identity relation anymore.[/quote]

Dirk Van de moortel · Guest

strich.9991@gmail.com <strich.9991@gmail.com> wrote in message
48e0f942-981b-42ef-bf38-311b272f5c0d@t41g2000hsc.googlegroups.com

[snip]

[quote]By E1, E2 we mean the rates of clock E in frame E (before and after).
By M1, M2 we mean the rates of clock M in frame M (before and after).

E1=M1 was defined at the outset as the frames are at rest to each.

We have shown already that M2=M1, and E2=E1. The conclusion is
straightforward: M2=M1, M1=E1, E1=E2, or M2=E2.

Note that the conclusion relates the 'new' times in the now moving M
frame to the E frame.

What is it that you cannot follow here?
[/quote]
Here, imbecile:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/YourWaterloo.html

Dirk Vdm

doug · Guest

strich.9993@gmail.com wrote:

[quote]On Oct 9, 11:15 am, Uncle Ben <b...@greenba.com> wrote:

You never defined E1 and M1. But I can guess your meaning. When the
two frames had no relative velocity, I agree that you could
synchronize the clocks, but when frame M is moving w.r.t. E you
have lost the synchronization. (All of this being what SR says, not
necessarily what is true.)

It>s good you agree that E1=M1.

We proved in RS101 that M2=M1, and in the special case of E where it
did not move at all, E2=E1. The rest is simple math.
[/quote]
Except, of course, your expression does not agree with reality.
You cannot change reality with a poorly thought out example.

doug · Guest

strich.9993@gmail.com wrote:

[quote]On Oct 9, 12:26 pm, doug <x...@xx.com> wrote:

E2 does not equal M2. This is an experimental fact.

By experiment, you mean the muon experiment. The interpretation was
erroneous, which is what I am showing now.
[/quote]
You are showing no such thing. The experiment gives the correct
result. What this means is that you do not understand or like
the result of the experiment. That is your problem, not the
world>s problem.

[quote]By fact, you mean, what?
[/quote]
See the experiments of the last century.

xxein · Guest

On Oct 9, 9:10 am, Uncle Ben <b...@greenba.com> wrote:
[quote]On Oct 8, 9:17 pm,xxein<xxe...@bellsouth.net> wrote:

On Oct 8, 2:32 pm, Uncle Ben <b...@greenba.com> wrote:

On Oct 8, 1:33 pm, "Strich.9" <strich.9...@gmail.com> wrote:

"Relativity = Stupidity"

That is a very difficult equation to teach.

We are only using one example to teach the subject. Past experience
of Professor Androcles on teaching this subject has shown that pupils
learn extremely slowly, if at all. This series of classes hopes to
teach this difficult subject on a step by step basis.

Example 1: Clocks E and M in inertial frames E and M are at rest with
respect to one another. Clock rates are of course equal. Let frame M
with clock M now move with respect to E, with relative constant
velocity v.

In "R=S 101", it was proven that the rate of clock M, in frame M did
not change. Many pupils reached this conclusion with much pain, but
eventually everybody learned this now accepted fact.

We skip "R=S 102" since the course material teaches that the rate of
clock E in frame E did not change after the external clock M moved.
It is common sense that since no change occurred in frame E, then no
change must occur in the rate of clock E. Of course, the course "R=S
102" may be opened for those who are having difficulty in the
succeeding course.

Here in "R=S 103" we use still the same example above, and we put
together the fact that since the rate of clock M did not change after
it moved (see course RS 101) and the rate of clock E did not change
after the M clock moved (course RS 102), then the conclusion must be
that clocks E and M are in synch before and after the M clock moved..

Let me repeat the argument:

Rates of clocks E and M did not change after clock M moved. Since the
rates were equal before the move, then the rates must be equal after
the move. In short, the clocks stay in synch.

The floor is now open for questions...

OK, I>ll be straight man in this little show.

Between

In "R=S 101", it was proven that the rate of clock M, in frame M did
not change. Many pupils reached this conclusion with much pain, but
eventually everybody learned this now accepted fact.

and

Here in "R=S 103" we use still the same example above, and we put
together the fact that since the rate of clock M did not change after
it moved (see course RS 101) and the rate of clock E did not change
after the M clock moved (course RS 102), then the conclusion must be
that clocks E and M are in synch before and after the M clock moved..

you made a significant change in wording. You went from "the rate of
clock
M, in frame M did not change" to "the rate of clock M did not
change".

The first statement is true, but the second one is incomplete.
It is stated as if time were absolute.

The rate of clock M did not change RELATIVE to what frame?
If frame M, it is true.
If frame E, it is not, according to the theory of RELATIVITY.

The theory of RELATIVITY says that the rate of a clock is RELATIVE to
its velocity.
Clock M still has zero velocity "in frame M", but velocity v "in frame
E."

[I have always resisted the language "in a frame of reference." It is
much clearer to say "with respect to" a frame of reference. That
avoids the silly fight over which frame something is "in". It makes
no sense to argue over which frame the object is "with respect to,"
since the object can be observed with respect to any frame of
reference.)

So, I didn>t pose a question, but an argument. The question, dear
teacher, is,
"Am I not right?" Note that I don>t even claim that SR is true
(although I believe that it has not been proved false). I just claim
that it is not self-contradictory.

Uncle Ben- Hide quoted text -

- Show quoted text -

xxein: But it is self-contradictory. You just haven>t gotten to the
objective/logical level yet.

I still like SRT-GRT for what it allows us to do with a confidence.
It is a physics though: not the physic. As such, it is a mere
engineering tool.- Hide quoted text -

- Show quoted text -

There is no contradiction between a clock having one rate w.r.t. one
frame of reference and another rate w.r.t. a different frame of
reference. You may think it false, and experiement will judge whether
it is or not, but there is no contradiction.

If you think otherwise, tell us what the contradiction is.

Uncle Ben- Hide quoted text -

- Show quoted text -
[/quote]
xxein: Semantics, but I know your point.

A clockrate... "another rate w.r.t. a different frame of reference"
should be "without respect to another FOR".

When you accelerate to another inertial velocity frame, your clock
still has the same observable rate. It is not the same as it was, but
"wrt'" your current velocity frame it is not observed at "another
rate".

Just nitpicking to aid the novices. OK?

doug · Guest

strich.9991@gmail.com wrote:

[quote]On Oct 9, 2:41 pm, doug <x...@xx.com> wrote:

Except, of course, your expression does not agree with reality.
You cannot change reality with a poorly thought out example.

Let me humor another idiot. What reality are you talking about?
[/quote]
The world we live. It may not be true in the world of your delusions
but that is your worry.

Dirk Bruere at NeoPax · Guest

Spaceman wrote:
[quote]dkelvey@hotmail.com wrote:
On Oct 8, 10:33 am, "Strich.9" <strich.9...@gmail.com> wrote:
"Relativity = Stupidity"

That is a very difficult equation to teach.

We are only using one example to teach the subject. Past experience
of Professor Androcles on teaching this subject has shown that pupils
learn extremely slowly, if at all. This series of classes hopes to
teach this difficult subject on a step by step basis.

Example 1: Clocks E and M in inertial frames E and M are at rest with
respect to one another. Clock rates are of course equal. Let frame M
with clock M now move with respect to E, with relative constant
velocity v.

In "R=S 101", it was proven that the rate of clock M, in frame M did
not change. Many pupils reached this conclusion with much pain, but
eventually everybody learned this now accepted fact.

We skip "R=S 102" since the course material teaches that the rate of
clock E in frame E did not change after the external clock M moved.
It is common sense that since no change occurred in frame E, then no
change must occur in the rate of clock E. Of course, the course "R=S
102" may be opened for those who are having difficulty in the
succeeding course.

Here in "R=S 103" we use still the same example above, and we put
together the fact that since the rate of clock M did not change after
it moved (see course RS 101) and the rate of clock E did not change
after the M clock moved (course RS 102), then the conclusion must be
that clocks E and M are in synch before and after the M clock moved.

Let me repeat the argument:

Rates of clocks E and M did not change after clock M moved. Since
the rates were equal before the move, then the rates must be equal
after the move. In short, the clocks stay in synch.

The floor is now open for questions...

Hi
The problem is that experiments have shown that the clocks ARE no
longer in sync when brought back together in the same location.

Clock malfuctions are older than relativity.
It is nice how you ignore the history of clocks to support
a theory that is also ignorant of clock malfunctions

So far, relativity has predicted exactly how much those clocks should
have shifted, with precission.

Relativity fails to correctly predict ALL clocks rate changes.
Relativity fails to understand history of clock and the history of
more than one type of clock.
[/quote]
Esp cuckoo clocks

--
Dirk

http://www.transcendence.me.uk/ - Transcendence UK
http://www.theconsensus.org/ - A UK political party
http://www.onetribe.me.uk/wordpress/?cat=5 - Our podcasts on weird stuff

Michael Press · Guest

In article
<BLOCKSPAMfishfry-471AD3.12294708102008@comcast.dca.giganews.com>,
fishfry <BLOCKSPAMfishfry@your-mailbox.com> wrote:

[quote]In article
1dd26d6f-6d3e-4ca3-b299-0320b3fc6e5a@f40g2000pri.googlegroups.com>,
"Strich.9" <strich.9992@gmail.com> wrote:

"Relativity = Stupidity"

apparently the physics trolls are worked up about relativity the same
way the math trolls are worked up about set theory.
[/quote]
Fit subject matter for a troll is that which can be expressed
in everyday language or has a consequence expressible in everyday
language. Nobody will troll with the Riemann-Roch theorem,
action-angle variables, or Le Chatelier>s principal. Everyone
understands infinity, relativity, and entropy.

--
Michael Press

doug · Guest

Lloyd wrote:

[quote]On Oct 9, 1:04 pm, "Spaceman" <space...@yourclockmalfunctioned.duh
wrote:

dkel...@hotmail.com wrote:

On Oct 8, 10:33 am, "Strich.9" <strich.9...@gmail.com> wrote:

"Relativity = Stupidity"

That is a very difficult equation to teach.

We are only using one example to teach the subject. Past experience
of Professor Androcles on teaching this subject has shown that pupils
learn extremely slowly, if at all. This series of classes hopes to
teach this difficult subject on a step by step basis.

Example 1: Clocks E and M in inertial frames E and M are at rest with
respect to one another. Clock rates are of course equal. Let frame M
with clock M now move with respect to E, with relative constant
velocity v.

In "R=S 101", it was proven that the rate of clock M, in frame M did
not change. Many pupils reached this conclusion with much pain, but
eventually everybody learned this now accepted fact.

We skip "R=S 102" since the course material teaches that the rate of
clock E in frame E did not change after the external clock M moved.
It is common sense that since no change occurred in frame E, then no
change must occur in the rate of clock E. Of course, the course "R=S
102" may be opened for those who are having difficulty in the
succeeding course.

Here in "R=S 103" we use still the same example above, and we put
together the fact that since the rate of clock M did not change after
it moved (see course RS 101) and the rate of clock E did not change
after the M clock moved (course RS 102), then the conclusion must be
that clocks E and M are in synch before and after the M clock moved.

Let me repeat the argument:

Rates of clocks E and M did not change after clock M moved. Since
the rates were equal before the move, then the rates must be equal
after the move. In short, the clocks stay in synch.

The floor is now open for questions...

Hi
The problem is that experiments have shown that the clocks ARE no
longer in sync when brought back together in the same location.

Clock malfuctions are older than relativity.
It is nice how you ignore the history of clocks to support
a theory that is also ignorant of clock malfunctions

So far, relativity has predicted exactly how much those clocks should
have shifted, with precission.

Relativity fails to correctly predict ALL clocks rate changes.

Specify which ones.
[/quote]
Spaceman thinks that all clocks except the one in Greenwich
malfunction and change rates. He says that is the reason that
GMT is used as a time reference. (Seriously, he says this).

[quote]

Relativity fails to understand history of clock and the history of
more than one type of clock.

Theories do not understand anything. They are not alive.
[/quote]
Sounds a lot like spaceman.


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