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Uncle Ben
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 Posted: Thu Oct 09, 2008 1:17 pm Post subject: Re: Relativity = Stupidity 103 |
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On Oct 8, 3:20 pm, strich.9...@gmail.com wrote:
[quote]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 -
Pardon the wording. The reference in both cases is frame M. The rate
of clock M in frame M remains the same, so that its rate at frame M
prior to its move, and its rate at frame M, after it moved, are
equal. This was debated in RS 101 and eventually agreed upon. Thanks.- Hide quoted text -
- Show quoted text -
[/quote]
(I suppose you mean "rate IN frame M," not "rate AT frame M".)
Not so fast there, Strich!
But in order to stay synchronized, the two clocks have to keep the
same rate w.r.t. the same frame. Now you say that M>s rate is
constant with respect to frame M, not frame E, so there is no problem
in having the rates differ w.r.t. a single frame. Relativity survives
your criticism. You have lost the argument.
Uncle Ben |
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| Posted: Thu Oct 09, 2008 1:19 pm Post subject: Time is shown to be a constant |
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On Oct 8, 8:59 pm, Uncle Ben <b...@greenba.com> wrote:
[quote]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 -
[/quote]
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.
Since time does not change with motion, then obviously, Einstein>s LTE
is rendered imaginary and unnecessary. |
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PD
Guest
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| Posted: Thu Oct 09, 2008 1:26 pm Post subject: Re: Time is shown to be a constant |
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On Oct 9, 8:19 am, strich.9...@gmail.com wrote:
[quote]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.
[/quote]
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.
[quote]
Since time does not change with motion, then obviously, Einstein>s LTE
is rendered imaginary and unnecessary.[/quote] |
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Guest
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| Posted: Thu Oct 09, 2008 1:50 pm Post subject: Re: Time is shown to be a constant |
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On Oct 9, 9:26 am, PD <TheDraperFam...@gmail.com> wrote:
[quote]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 -
[/quote]
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) |
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Guest
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| Posted: Thu Oct 09, 2008 1:52 pm Post subject: Re: Relativity = Stupidity 103 |
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On Oct 9, 9:17 am, Uncle Ben <b...@greenba.com> wrote:
[quote]On Oct 8, 3:20 pm, strich.9...@gmail.com 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 -
Pardon the wording. The reference in both cases is frame M. The rate
of clock M in frame M remains the same, so that its rate at frame M
prior to its move, and its rate at frame M, after it moved, are
equal. This was debated in RS 101 and eventually agreed upon. Thanks.- Hide quoted text -
- Show quoted text -
(I suppose you mean "rate IN frame M," not "rate AT frame M".)
Not so fast there, Strich!
But in order to stay synchronized, the two clocks have to keep the
same rate w.r.t. the same frame. Now you say that M>s rate is
constant with respect to frame M, not frame E, so there is no problem
in having the rates differ w.r.t. a single frame. Relativity survives
your criticism. You have lost the argument.
Uncle Ben- Hide quoted text -
- Show quoted text -
[/quote]
You forget that we calibrated the clocks beforehand when they were at
rest to each other, so we know E1 = M1. Do you deny that E1=M1? |
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Guest
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| Posted: Thu Oct 09, 2008 2:17 pm Post subject: Re: Time is shown to be a constant |
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On 9 oct, 09:50, strich.9...@gmail.com wrote:
[quote]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)
[/quote]
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 |
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Guest
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| Posted: Thu Oct 09, 2008 2:40 pm Post subject: Re: Relativity = Stupidity 103 |
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On Oct 9, 10:33 am, david.bostw...@chemistry.gatech.edu (David
Bostwick) wrote:
[quote]In article <7da2d2d9-d177-4c33-b924-bcb7d003b...@r66g2000hsg.googlegroups..com>, strich.9...@gmail.com wrote:
[...]
Too bad you remain stupid. KE requires a reference frame. Time does
not. Did Enstein ever say time stops in a single reference frame? I
thought so.
Ah, the favorite retort of those who can>t handle criticism - "Everyone>s
wrong but me, and you>re a stupid doo-doo head."
OK, you>ve convinced me that you>re obviously smarter than Albert, Richard,
and the rest of those silly so-called "experts."
[/quote]
Ah, another cheerleader in the making. May I remind you, cheering is
not an argument. |
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Guest
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| Posted: Thu Oct 09, 2008 2:43 pm Post subject: Re: Time is shown to be a constant |
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On Oct 9, 10:17 am, papa_r...@hotmail.com wrote:
[quote]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 -
[/quote]
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? |
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Guest
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| Posted: Thu Oct 09, 2008 3:00 pm Post subject: Re: Relativity = Stupidity 103 |
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On Oct 9, 10:48 am, david.bostw...@chemistry.gatech.edu (David
Bostwick) wrote:
[quote]In article <8f871697-5164-47d9-8041-972e6fe54...@o4g2000pra.googlegroups.com>, strich.9...@gmail.com wrote:
On Oct 9, 10:33=A0am, david.bostw...@chemistry.gatech.edu (David
Bostwick) wrote:
In article <7da2d2d9-d177-4c33-b924-bcb7d003b...@r66g2000hsg.googlegroups> >..com>, strich.9...@gmail.com wrote:
[...]
Too bad you remain stupid. =A0KE requires a reference frame. =A0Time doe> >s
not. =A0Did Enstein ever say time stops in a single reference frame? =A0> >I
thought so.
Ah, the favorite retort of those who can>t handle criticism - "Everyone>s
wrong but me, and you>re a stupid doo-doo head."
OK, you>ve convinced me that you>re obviously smarter than Albert, Richar> >d,
and the rest of those silly so-called "experts."
Ah, another cheerleader in the making. May I remind you, cheering is
not an argument.
Hey, it>s at least as good as repeating nonsense. Better, in my not-so-humble
opinion.- Hide quoted text -
- Show quoted text -
[/quote]
You do repeat nonsense. What should we expect from the poor cousin of
Emory? |
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Uncle Ben
Guest
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| Posted: Thu Oct 09, 2008 3:15 pm Post subject: Re: Relativity = Stupidity 103 |
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On Oct 9, 9:52 am, strich.9...@gmail.com wrote:
[quote]On Oct 9, 9:17 am, Uncle Ben <b...@greenba.com> wrote:
On Oct 8, 3:20 pm, strich.9...@gmail.com 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 -
Pardon the wording. The reference in both cases is frame M. The rate
of clock M in frame M remains the same, so that its rate at frame M
prior to its move, and its rate at frame M, after it moved, are
equal. This was debated in RS 101 and eventually agreed upon. Thanks.- Hide quoted text -
- Show quoted text -
(I suppose you mean "rate IN frame M," not "rate AT frame M".)
Not so fast there, Strich!
But in order to stay synchronized, the two clocks have to keep the
same rate w.r.t. the same frame. Now you say that M>s rate is
constant with respect to frame M, not frame E, so there is no problem
in having the rates differ w.r.t. a single frame. Relativity survives
your criticism. You have lost the argument.
Uncle Ben- Hide quoted text -
- Show quoted text -
You forget that we calibrated the clocks beforehand when they were at
rest to each other, so we know E1 = M1. Do you deny that E1=M1?- Hide quoted text -
- Show quoted text -
[/quote]
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.) |
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Guest
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| Posted: Thu Oct 09, 2008 3:46 pm Post subject: Re: Relativity = Stupidity 103 |
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On Oct 9, 11:15 am, Uncle Ben <b...@greenba.com> wrote:
[quote]
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.)
[/quote]
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. |
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Guest
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| Posted: Thu Oct 09, 2008 3:49 pm Post subject: Re: Time is shown to be a constant |
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On Oct 9, 12:26 pm, doug <x...@xx.com> wrote:
[quote]E2 does not equal M2. This is an experimental fact.
[/quote]
By experiment, you mean the muon experiment. The interpretation was
erroneous, which is what I am showing now.
By fact, you mean, what? |
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Strich.9
Guest
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| Posted: Thu Oct 09, 2008 4:07 pm Post subject: Re: Time is shown to be a constant |
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On Oct 9, 11:29 am, "Dirk Van de moortel"
<dirkvandemoor...@nospAm.hotmail.com> wrote:
[quote]Condensing these 8 quantities into 4 variables is called fumbling,
[/quote]
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. |
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Strich.9
Guest
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| Posted: Thu Oct 09, 2008 4:20 pm Post subject: Re: Time is shown to be a constant |
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On Oct 9, 12:14 pm, "Dirk Van de moortel"
<dirkvandemoor...@nospAm.hotmail.com> wrote:
[quote]Strich.9 <strich.9...@gmail.com> wrote in message
5f1434fb-775f-407d-b739-9baf07d57...@t41g2000hsc.googlegroups.com
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..
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
[/quote]
I have humored the idiot and he did not deliver. Idiot is hereby
dismissed. |
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dkelvey@hotmail.com
Guest
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| Posted: Thu Oct 09, 2008 4:58 pm Post subject: Re: Relativity = Stupidity 103 |
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On Oct 8, 10:33 am, "Strich.9" <strich.9...@gmail.com> wrote:
[quote]"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...
[/quote]
Hi
The problem is that experiments have shown that the clocks ARE no
longer in sync when brought back together in the same location.
So far, relativity has predicted exactly how much those clocks should
have shifted, with precission.
It may be that relativity isn>t the best absolute way of determining
what is happening but we have not found a better one.
Can you explain why the clocks are no longer in sync, even though
your thought experiment contradicts what really does happen?
At least Spaceman realizes that the clocks will no longer be in sync,
in real life. The only thing he seems to miss is that everything that
experiences time, clocks, people, atoms, plants and what ever,
all experience the same malfuction. Since they all experience the
same malfuction, they can not, within their own frame of reference,
know that there is a malfunction.
Dwight |
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