Thomas Smid says...

Thomas, this is pretty silly. We have the following
constraints:

1. forall x, forall t, x'(x,t) = A x + Bc t
2. forall x, forall t, ct'(x,t) = D x + Ec t
3. forall x, forall t, if x=ct, then x'(x,t) = c t'(x,t)
4. forall x, forall t, if x=-ct, then x'(x,t) = -c t'(x,t)

Obviously, these constraints are not inconsistent, since
they have at least one solution,

A = 5/3,
B = -4/3
D = -4/3
E = 5/3

If the equations are inconsistent, then there is no solution.
Also, you claim that B=0. That is obviously *not* implied by
equations 1-4, since we have a solution with B nonzero. So
there is something wrong with your reasoning.

If we have a solution with B nonzero, and you prove that
the only solution is B=0, then obviously you made a mistake.
B=0 is *not* the only solution.

--
Daryl McCullough
Ithaca, NY

Daryl McCullough wrote:
Thomas Smid says...

Thomas, this is pretty silly. We have the following
constraints:

1. forall x, forall t, x'(x,t) = A x + Bc t
2. forall x, forall t, ct'(x,t) = D x + Ec t
3. forall x, forall t, if x=ct, then x'(x,t) = c t'(x,t)
4. forall x, forall t, if x=-ct, then x'(x,t) = -c t'(x,t)

Obviously, these constraints are not inconsistent, since
they have at least one solution,

A = 5/3,
B = -4/3
D = -4/3
E = 5/3

If the equations are inconsistent, then there is no solution.
Also, you claim that B=0. That is obviously *not* implied by
equations 1-4, since we have a solution with B nonzero. So
there is something wrong with your reasoning.

If we have a solution with B nonzero, and you prove that
the only solution is B=0, then obviously you made a mistake.
B=0 is *not* the only solution.

Then insert A=5/3 and B=-4/3 into

x1'=Ax1+Bct
x2'=Ax2+Bct=-Ax1+Bct

which gives

x1'=5/3*x1-4/3*ct
x2'=-5/3*x1-4/3*ct

and thus

x1'+x2'=-8/3*ct

whereas it should be zero.

Other combinations of the transformation equations may be consistent,
but if a certain subset of them is not consistent, the whole set is
obviously not consistent either.

Thomas

Thomas Smid says...

Daryl McCullough wrote:

1. forall x, forall t, x'(x,t) = A x + Bc t
2. forall x, forall t, ct'(x,t) = D x + Ec t
3. forall x, forall t, if x=ct, then x'(x,t) = c t'(x,t)
4. forall x, forall t, if x=-ct, then x'(x,t) = -c t'(x,t)

Obviously, these constraints are not inconsistent, since
they have at least one solution,

A = 5/3,
B = -4/3
D = -4/3
E = 5/3

Can you please acknowledge that you understand
that equations 1-4 have the solution A=E=5/3, B=D=-4/3?

Plug it in. Check your work.

Okay, now if you agree that B=-4/3 is a correct solution,
then it follows that your proof that B=0 is an *incorrect*
proof. B is *not* necessarily 0.

Then insert A=5/3 and B=-4/3 into

x1'=Ax1+Bct
x2'=Ax2+Bct=-Ax1+Bct

which gives
x1'=5/3*x1-4/3*ct
x2'=-5/3*x1-4/3*ct

and thus

x1'+x2'=-8/3*ct

whereas it should be zero.

No, it shouldn't be zero. Show how x1' + x2' = 0 follows
from equations 1-4 above. It doesn't.

You are making the same mistakes over and over and over.
Look, we are considering two functions x'(x,t) and t'(x,t).
You keep introducing new functions and then forgetting that
they are functions. You want to introduce a new function
x1(t) = ct. Fine. You want to introduce a new
function x2(t) = -ct. Fine. Then you can define functions
x2'(t) = x'(x1(t),t) = x'(ct,t) and
x1'(t) = x'(x2(t),t) = x'(-ct,t). Why in the
world do you think you are justified in saying

x2'(t) = - x1'(t)?

Show me which of my equations 1-4 imply that.

Where you are getting confused is this: If e1 and e2 are
two events such that

x(e1) = ct(e1)
x(e2) = -ct(e2)
t'(e1) = t'(e2)

then in that circumstance

x'(e1) = -x'(e2)

What this implies for our functions x2' and x1' is the following:

x2'(t(e1)) = -x1'(t(e2))

But t(e1) is not equal to t(e2).

You are getting confused because you are failing to keep track
of the functional dependencies.

You keep thinking that if

t(e1) = t(e2)

then

t'(e1) = t'(e2)

and vice-versa. That's not true.

--
Daryl McCullough
Ithaca, NY

"Daryl McCullough" wrote in message ...
Thomas Smid says...

Daryl McCullough wrote:

1. forall x, forall t, x'(x,t) = A x + Bc t
2. forall x, forall t, ct'(x,t) = D x + Ec t
3. forall x, forall t, if x=ct, then x'(x,t) = c t'(x,t)
4. forall x, forall t, if x=-ct, then x'(x,t) = -c t'(x,t)

Obviously, these constraints are not inconsistent, since
they have at least one solution,

A = 5/3,
B = -4/3
D = -4/3
E = 5/3

Can you please acknowledge that you understand
that equations 1-4 have the solution A=E=5/3, B=D=-4/3?

Plug it in. Check your work.

Okay, now if you agree that B=-4/3 is a correct solution,
then it follows that your proof that B=0 is an *incorrect*
proof. B is *not* necessarily 0.

Then insert A=5/3 and B=-4/3 into

x1'=Ax1+Bct
x2'=Ax2+Bct=-Ax1+Bct

which gives
x1'=5/3*x1-4/3*ct
x2'=-5/3*x1-4/3*ct

and thus

x1'+x2'=-8/3*ct

whereas it should be zero.

No, it shouldn't be zero. Show how x1' + x2' = 0 follows
from equations 1-4 above. It doesn't.

You are making the same mistakes over and over and over.

Deliberately.
Until *you* give it up. Bet?

Remember that I won this kind of bet *many* times before ;-)

Dirk Vdm

Dirk Van de moortel says...

You are making the same mistakes over and over and over.

Deliberately.
Until *you* give it up. Bet?

Remember that I won this kind of bet *many* times before ;-)

No, you haven't. Androcles gave up first. He plonked *me*. So
I win.

--
Daryl McCullough
Ithaca, NY

"Daryl McCullough" wrote in message ...
Dirk Van de moortel says...

You are making the same mistakes over and over and over.

Deliberately.
Until *you* give it up. Bet?

Remember that I won this kind of bet *many* times before ;-)

No, you haven't. Androcles gave up first. He plonked *me*. So
I win.

If you look at it as like to a game, no argument on that one :-)
But I was talking more generally. There's more retards than
Androcles. And there's more people who try to help (or play
with) them. We're not alone...

Anyway, enjoy :-)
Dirk Vdm

Daryl McCullough wrote:
Thomas Smid says...

Daryl McCullough wrote:

1. forall x, forall t, x'(x,t) = A x + Bc t
2. forall x, forall t, ct'(x,t) = D x + Ec t
3. forall x, forall t, if x=ct, then x'(x,t) = c t'(x,t)
4. forall x, forall t, if x=-ct, then x'(x,t) = -c t'(x,t)

Obviously, these constraints are not inconsistent, since
they have at least one solution,

A = 5/3,
B = -4/3
D = -4/3
E = 5/3

Can you please acknowledge that you understand
that equations 1-4 have the solution A=E=5/3, B=D=-4/3?

Plug it in. Check your work.

Okay, now if you agree that B=-4/3 is a correct solution,
then it follows that your proof that B=0 is an *incorrect*
proof. B is *not* necessarily 0.

Then insert A=5/3 and B=-4/3 into

x1'=Ax1+Bct
x2'=Ax2+Bct=-Ax1+Bct

which gives
x1'=5/3*x1-4/3*ct
x2'=-5/3*x1-4/3*ct

and thus

x1'+x2'=-8/3*ct

whereas it should be zero.

No, it shouldn't be zero. Show how x1' + x2' = 0 follows
from equations 1-4 above. It doesn't.

You are making the same mistakes over and over and over.
Look, we are considering two functions x'(x,t) and t'(x,t).
You keep introducing new functions and then forgetting that
they are functions. You want to introduce a new function
x1(t) = ct. Fine. You want to introduce a new
function x2(t) = -ct. Fine. Then you can define functions
x2'(t) = x'(x1(t),t) = x'(ct,t) and
x1'(t) = x'(x2(t),t) = x'(-ct,t). Why in the
world do you think you are justified in saying

x2'(t) = - x1'(t)?

Show me which of my equations 1-4 imply that.

Where you are getting confused is this: If e1 and e2 are
two events such that

x(e1) = ct(e1)
x(e2) = -ct(e2)
t'(e1) = t'(e2)

then in that circumstance

x'(e1) = -x'(e2)

What this implies for our functions x2' and x1' is the following:

x2'(t(e1)) = -x1'(t(e2))

But t(e1) is not equal to t(e2).

You are getting confused because you are failing to keep track
of the functional dependencies.

You keep thinking that if

t(e1) = t(e2)

then

t'(e1) = t'(e2)

and vice-versa. That's not true.

I have algebraically proved above that it is true and you haven't given
any valid objection against it (as I pointed out already, your
'functional dependences' argument is algebraically completely
irrelevant).

You don't seem to realize that the original equations

(1) x'=Ax+Bct
(2) ct'=Bx+Act

are inconsistent upon changing the signs of x and x' (i.e. when
applying the equations to a light signal travelling in the opposite
direction) because from (1) and (2) we have

(3) x'+ct'=(A+B)(x+ct) ,

but by changing the signs of x and x' we have

(4) -x'=-Ax+Bct
(5) ct'=-Bx+Act

and by multiplying (4) by -1

(6) x'=Ax-Bct
(7)ct'=-Bx+Act ,

so from (6) and(7)

(8) x'+ct'=(A-B)(x+ct) ,

which is inconsistent with (3) unless B=0 (or x=x'=0 and t=t'=0)

Thomas

Thomas Smid says...

I have algebraically proved above that it is true

Every single one of your proofs starts off with an
error. You can *see* that it is an error by checking
your work. You claim to have proved that B=0 is the
only solution to the equations

1. forall x, forall t, x'(x,t) = A x + Bc t
2. forall x, forall t, ct'(x,t) = D x + Ec t
3. forall x, forall t, if x=ct, then x'(x,t) = c t'(x,t)
4. forall x, forall t, if x=-ct, then x'(x,t) = -c t'(x,t)

You can *see* that that is incorrect by trying
the solution A=5/3 B=-4/3 D=-4/3 E=5/3. If you
prove something that is false, then that is
evidence that you made a mistake. I explicitly
pointed out what your mistake was, but you
don't need me to point it out in order for you
to admit that you made a mistake:

You proved that B=0 was the only solution.
B=0 is *not* the only solution.
Therefore, your proof is wrong.

That is the simplest, and most airtight rule
for mathematics. If you prove something that
turns out to be false, then you've made a mistake.
You proved something that turned out to be false.
Therefore, you made a mistake.

--
Daryl McCullough
Ithaca, NY

"Daryl McCullough" wrote in message ...
Thomas Smid says...

I have algebraically proved above that it is true

Every single one of your proofs starts off with an
error. You can *see* that it is an error by checking
your work. You claim to have proved that B=0 is the
only solution to the equations

1. forall x, forall t, x'(x,t) = A x + Bc t
2. forall x, forall t, ct'(x,t) = D x + Ec t
3. forall x, forall t, if x=ct, then x'(x,t) = c t'(x,t)
4. forall x, forall t, if x=-ct, then x'(x,t) = -c t'(x,t)

You can *see* that that is incorrect by trying
the solution A=5/3 B=-4/3 D=-4/3 E=5/3. If you
prove something that is false, then that is
evidence that you made a mistake. I explicitly
pointed out what your mistake was, but you
don't need me to point it out in order for you
to admit that you made a mistake:

You proved that B=0 was the only solution.
B=0 is *not* the only solution.
Therefore, your proof is wrong.

That is the simplest, and most airtight rule
for mathematics. If you prove something that
turns out to be false, then you've made a mistake.
You proved something that turned out to be false.
Therefore, you made a mistake.

Actually that is debatable.
Deliberately writing something completely wrong in order
to annoy someone, isn't really making a mistake. It's more
like lying and deceiving, don't you agree?
But I do agree that the very act of thinking that he can get
away with this kind of deceptive behaviour, can be called
a mistake ;-)

Dirk Vdm

Dirk Van de moortel says...

Deliberately writing something completely wrong in order
to annoy someone, isn't really making a mistake. It's more
like lying and deceiving, don't you agree?

I don't know about that explanation. It seems to give
Thomas too much credit to assume that he *understands*
his errors, but is making them intentionally in order
to win some argument. I don't see any evidence of such
understanding. But maybe that's part of his
deviousness---that he carefully hides his own understanding?

The question is: what possible motivation could someone
have to lie about the Lorentz transformations? Are you
thinking that some people (Thomas and maybe Androcles)
actually know that Einstein's derivations were correct,
but are trying to cast doubt on them anyway?

Okay, I guess I have a hypothesis as to what may ge
going on, that's almost the same as your dishonesty
hypothesis, but doesn't credit them with understanding.

Thomas and Androcles and the other relativity-bashers
*sincerely* believe that relativity is wrong, and that
there is something wrong with Einstein's derivation.
However, they also know that they don't have the energy
or mathematical ability to figure out exactly where the
mistake is. But they reason: It doesn't matter exactly
what the mistake is---if it's mistaken, then people
shouldn't be using relativity.

So, in the spirit of "The ends justify the means" they
are using methods that they know are incorrect to try
to convince people of a *correct* (to them, anyway)
conclusion: that relativity is nonsense. If the conclusion
is correct, who cares about the picky details?

--
Daryl McCullough
Ithaca, NY