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Guest
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 Posted: Thu Oct 09, 2008 5:17 pm Post subject: Re: Time is shown to be a constant |
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On Oct 9, 12:58 pm, Dirk Bruere at NeoPax <dirk.bru...@gmail.com>
wrote:
[quote]strich.9...@gmail.com wrote:
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.
Yes, they just last longer when moving fast.
Mere coincidence.
[/quote]
I bet you got that from Wikipedia. Let me humor this idiot as well.
What would be the reason the muon lasts longer when it is moving
fast? Go ahead, look it up again. Make sure you understand the
answer. |
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Guest
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| Posted: Thu Oct 09, 2008 5:39 pm Post subject: Re: Time is shown to be a constant |
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On Oct 9, 1:30 pm, PD <TheDraperFam...@gmail.com> wrote:
[quote]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 -
[/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? |
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Guest
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| Posted: Thu Oct 09, 2008 5:44 pm Post subject: Re: Relativity = Stupidity 103 |
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On Oct 9, 2:41 pm, doug <x...@xx.com> wrote:
[quote]
Except, of course, your expression does not agree with reality.
You cannot change reality with a poorly thought out example.
[/quote]
Let me humor another idiot. What reality are you talking about? |
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Guest
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| Posted: Thu Oct 09, 2008 7:23 pm Post subject: Re: Relativity = Stupidity 103 |
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On Oct 9, 3:17 pm, Dirk Bruere at NeoPax <dirk.bru...@gmail.com>
wrote:
[quote]
Esp cuckoo clocks
[/quote]
Ah, a Swiss troll. |
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Lloyd
Guest
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| Posted: Thu Oct 09, 2008 7:30 pm Post subject: Re: Relativity = Stupidity 103 |
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On Oct 8, 3:49 pm, strich.9...@gmail.com wrote:
[quote]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'.
[/quote]
Uh-huh. What>s next, proving atoms don>t exist? |
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doug
Guest
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| Posted: Thu Oct 09, 2008 7:31 pm Post subject: Re: Time is shown to be a constant |
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strich.9991@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]
Except that experimentally this is not true. Your "logic" is wrong.
You are assuming that which you want to prove. That may work in
your world but not in the real world.
[quote]
Since time does not change with motion, then obviously, Einstein>s LTE
is rendered imaginary and unnecessary.
[/quote]
Wrong assumptions, wrong conculsion. Try again. |
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Lloyd
Guest
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| Posted: Thu Oct 09, 2008 7:33 pm Post subject: Re: Relativity = Stupidity 103 |
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On Oct 9, 1:04 pm, "Spaceman" <space...@yourclockmalfunctioned.duh>
wrote:
[quote]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.
[/quote]
Specify which ones.
[quote]Relativity fails to understand history of clock and the history of
more than one type of clock.
[/quote]
Theories do not understand anything. They are not alive. |
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David Bostwick
Guest
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| Posted: Thu Oct 09, 2008 7:33 pm Post subject: Re: Relativity = Stupidity 103 |
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In article <7da2d2d9-d177-4c33-b924-bcb7d003bba8@r66g2000hsg.googlegroups.com>, strich.9991@gmail.com wrote:
[...]
[quote]
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.
[/quote]
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." |
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David Bostwick
Guest
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| Posted: Thu Oct 09, 2008 7:48 pm Post subject: Re: Relativity = Stupidity 103 |
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In article <8f871697-5164-47d9-8041-972e6fe542b3@o4g2000pra.googlegroups.com>, strich.9993@gmail.com wrote:
[quote]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.
[/quote]
Hey, it>s at least as good as repeating nonsense. Better, in my not-so-humble
opinion. |
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Guest
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| Posted: Thu Oct 09, 2008 8:17 pm Post subject: Re: Time is shown to be a constant |
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On Oct 9, 3:33 pm, PD <TheDraperFam...@gmail.com> wrote:
[quote]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 -
[/quote]
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. Where can
you insert an inequality here that Relativity demands? |
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PD
Guest
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| Posted: Thu Oct 09, 2008 8:19 pm Post subject: Re: Time is shown to be a constant |
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On Oct 9, 3:17 pm, strich.9...@gmail.com wrote:
[quote]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.
[/quote]
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.
[quote] Where can
you insert an inequality here that Relativity demands?[/quote] |
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Dirk Van de moortel
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| Posted: Thu Oct 09, 2008 8:29 pm Post subject: Re: Time is shown to be a constant |
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strich.9993@gmail.com <strich.9993@gmail.com> wrote in message
c2b8b434-b4af-4473-b17c-dba2e6892fe0@p49g2000hsd.googlegroups.com
[snip to essence]
[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?
[/quote]
You remind me of another logic-idiot, calling itself "Allyou!"
Oh yes, if your kind of logic allows you to use 4 variables
E1, M1, E2 and M2 for 8 carefully defined quantities, then
the theory is certainly illogical.
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 |
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Guest
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| Posted: Thu Oct 09, 2008 8:40 pm Post subject: Re: Time is shown to be a constant |
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On Oct 9, 4:19 pm, PD <TheDraperFam...@gmail.com> wrote:
[quote]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 -
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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 -
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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 -
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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 -
[/quote]
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. |
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David Bostwick
Guest
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| Posted: Thu Oct 09, 2008 8:57 pm Post subject: Re: Relativity = Stupidity 103 |
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In article <ebac37f6-fba5-49b1-88dd-6f88f4424d32@p58g2000hsb.googlegroups.com>, strich.9993@gmail.com wrote:
[...]
[quote]You do repeat nonsense. What should we expect from the poor cousin of
Emory?
[/quote]
Yeah, Coke loves them a lot more than us.
I continue to be overwhelmed by your logic. Tech and Emory are different, and
that relationship obviously means that your opinions are correct. I bow to
your mental powers. |
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Spaceman
Guest
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| Posted: Thu Oct 09, 2008 8:58 pm Post subject: Re: Relativity = Stupidity 103 |
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PD wrote:
[quote]On Oct 9, 8:04 am, strich.9...@gmail.com wrote:
Did Enstein ever say time stops in a single reference frame?
*Stopping* does is not a requirement for frame-dependence.
[/quote]
If the clock stopped, how much time did it stop for?
:) |
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