Thomas Smid says...

Daryl McCullough wrote:

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

imply that A=0 or B=0. Obviously, that's incorrect, because
one solution to these 4 equations is this: A = 5/3, B = -4/3

x' = 5/3 x - 4/3 ct
ct' = -4/3 x + 5/3 ct

You haven't actually solved your equations (1)-(4) as they contain D
and E which you replaced by A and B (and it is here where the
contradiction occurs (as I mentioned before))

I'm sorry. I thought it was clear what D and E were, but to be
explicit: The 4 equations above have a solution

A=5/3

B=-4/3

D=-4/3

E=5/3

So it is incorrect to say that the 4 equations imply A=0 or B=0.
There is no contradiction.

(5) ct'(x1,t)=(B+A)x1

But that equation is *not* true for all x and t. It is only
true when x = ct.

(6) ct'(x2,t)=(B-A)x2

But that equation is *not* true for all x and t. It is only
true when x = -ct.

These are conditions you interprete into the validity of the variables
which aren't fixed mathematically anywhere.

I don't know what you are talking about.

As I said, these are
the four conditions we have talked about so far:

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

Equations 1 and 2 say "forall e". Equations 3 and 4
have conditions. You are ignoring these conditions.

There is only one function
t'(x,t) defined here

If you don't like events, then the same information can
be conveyed in terms of x and t:

1. forall x, forall t, x'(x,t) = A x + B c t
2. forall x, forall t, t'(x,t) = D x + E c 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)

i.e. if you are serious about this issue you
should also split t into t1 and t2, but I wonder how far you will be
getting this way in the derivation of the Lorentz transformation.

Also you still haven't answered my question why you are not changing
the sign convention and let x1=ct and x2=ct.

You were the one who said that x2 was -ct. Now you want to say
that x2 = +ct. That doesn't make any sense.

--
Daryl McCullough
Ithaca, NY

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

Daryl McCullough wrote:

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

imply that A=0 or B=0. Obviously, that's incorrect, because
one solution to these 4 equations is this: A = 5/3, B = -4/3

x' = 5/3 x - 4/3 ct
ct' = -4/3 x + 5/3 ct

You haven't actually solved your equations (1)-(4) as they contain D
and E which you replaced by A and B (and it is here where the
contradiction occurs (as I mentioned before))

I'm sorry. I thought it was clear what D and E were, but to be
explicit: The 4 equations above have a solution

A=5/3

B=-4/3

D=-4/3

E=5/3

So it is incorrect to say that the 4 equations imply A=0 or B=0.
There is no contradiction.

(5) ct'(x1,t)=(B+A)x1

But that equation is *not* true for all x and t. It is only
true when x = ct.

(6) ct'(x2,t)=(B-A)x2

But that equation is *not* true for all x and t. It is only
true when x = -ct.

These are conditions you interprete into the validity of the variables
which aren't fixed mathematically anywhere.

I don't know what you are talking about.

As I said, these are
the four conditions we have talked about so far:

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

Equations 1 and 2 say "forall e". Equations 3 and 4
have conditions. You are ignoring these conditions.

There is only one function
t'(x,t) defined here

If you don't like events, then the same information can
be conveyed in terms of x and t:

1. forall x, forall t, x'(x,t) = A x + B c t
2. forall x, forall t, t'(x,t) = D x + E c 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)

i.e. if you are serious about this issue you
should also split t into t1 and t2, but I wonder how far you will be
getting this way in the derivation of the Lorentz transformation.

Also you still haven't answered my question why you are not changing
the sign convention and let x1=ct and x2=ct.

You were the one who said that x2 was -ct. Now you want to say
that x2 = +ct. That doesn't make any sense.

Daryl, even sub-retards like Marcel Luttgens and Androcles
would laugh their bottoms off over this Smid character.
Have you considered asking him to solve an equation like
4x - 3 = 5
?
Are you still not convinced that he is playing a game?

Dirk Vdm

Daryl McCullough wrote:
Thomas Smid says...

Daryl McCullough wrote:

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

imply that A=0 or B=0. Obviously, that's incorrect, because
one solution to these 4 equations is this: A = 5/3, B = -4/3

x' = 5/3 x - 4/3 ct
ct' = -4/3 x + 5/3 ct

You haven't actually solved your equations (1)-(4) as they contain D
and E which you replaced by A and B (and it is here where the
contradiction occurs (as I mentioned before))

I'm sorry. I thought it was clear what D and E were, but to be
explicit: The 4 equations above have a solution

A=5/3

B=-4/3

D=-4/3

E=5/3

So it is incorrect to say that the 4 equations imply A=0 or B=0.
There is no contradiction.

No there isn't a contradiction as you restricted again the validity of
the equations:

if you have two arbitrary events e1 and e2 , the transformation
equations are

x'(e1) = A x(e1) + Bc t(e1)
ct'(e1)= B x(e1) + Ac t(e1)

x'(e2) = A x(e2) + Bc t(e2)
ct'(e2)= B x(e2) + Ac t(e2)

These equations must hold for all events e1 and e2 simultaneously,
whether x(e1) and x(e2) are positive or negative, and thus also for
further events at -x(e1) and -x(e2), so we can actually double up the
equations to

(1a) x'(e1) = A x(e1) + Bc t(e1)
(1b) ct'(e1)= B x(e1) + Ac t(e1)

(2a) -x'(e1) = -A x(e1) + Bc t(e1)
(2b) ct'(e1)= -B x(e1) + Ac t(e1)

(3a) x'(e2) = A x(e2) + Bc t(e2)
(3b) ct'(e2)= B x(e2) + Ac t(e2)

(4a) -x'(e2) = -A x(e2) + Bc t(e2)
(4b) ct'(e2)= -B x(e2) + Ac t(e2)

Note that at this point e1 and e2 are arbitray and we have not defined
a relationship x(e2)=-x(e1) yet. Still it is obvious from (1b) and (2b)
for instance that B=0.



(5) ct'(x1,t)=(B+A)x1

But that equation is *not* true for all x and t. It is only
true when x = ct.

(6) ct'(x2,t)=(B-A)x2

But that equation is *not* true for all x and t. It is only
true when x = -ct.

These are conditions you interprete into the validity of the variables
which aren't fixed mathematically anywhere.

I don't know what you are talking about.

As I said, these are
the four conditions we have talked about so far:

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

Equations 1 and 2 say "forall e". Equations 3 and 4
have conditions. You are ignoring these conditions.

There is only one function
t'(x,t) defined here

If you don't like events, then the same information can
be conveyed in terms of x and t:

1. forall x, forall t, x'(x,t) = A x + B c t
2. forall x, forall t, t'(x,t) = D x + E c 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)

i.e. if you are serious about this issue you
should also split t into t1 and t2, but I wonder how far you will be
getting this way in the derivation of the Lorentz transformation.

Also you still haven't answered my question why you are not changing
the sign convention and let x1=ct and x2=ct.

You were the one who said that x2 was -ct. Now you want to say
that x2 = +ct. That doesn't make any sense.

On the contrary, it makes perfect sense if you have separate variables
for the two directions. You only need negative numbers if you have one
variable. Our calendar is for instance the best example for this: you
have one scale named AD and another named BC, both measured positive.
Obviously you could do the same for a spatial scale as well. If you
want, you can define the location of the light signal by a spherical
coordinate e.g. by r(phi,t) (where phi is a corresponding angle and
thus r always positive).
I am sure you are not suggesting that the derivation of the Lorentz
transformation depends on the use of a Cartesian coordinate system.

Thomas

"Thomas Smid" wrote in message ups.com...

[snip]

I am sure you are not suggesting that the derivation of the Lorentz
transformation depends on the use of a Cartesian coordinate system.

It depends on linear algebra, which is something you seem to
pretend never to have heard about. So either you are lying
about this, or you are lying about having a masters in physics.
Or you have severe brain rot. What will it be?

Dirk Vdm

Daryl McCullough schrieb:

Thomas Smid says...

Daryl McCullough wrote:

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

imply that A=0 or B=0. Obviously, that's incorrect, because
one solution to these 4 equations is this: A = 5/3, B = -4/3

x' = 5/3 x - 4/3 ct
ct' = -4/3 x + 5/3 ct

You haven't actually solved your equations (1)-(4) as they contain D
and E which you replaced by A and B (and it is here where the
contradiction occurs (as I mentioned before))

I'm sorry. I thought it was clear what D and E were, but to be
explicit: The 4 equations above have a solution

A=5/3

B=-4/3

D=-4/3

E=5/3

So it is incorrect to say that the 4 equations imply A=0 or B=0.
There is no contradiction.

(5) ct'(x1,t)=(B+A)x1

But that equation is *not* true for all x and t. It is only
true when x = ct.

(6) ct'(x2,t)=(B-A)x2

But that equation is *not* true for all x and t. It is only
true when x = -ct.

These are conditions you interprete into the validity of the variables
which aren't fixed mathematically anywhere.

I don't know what you are talking about.

As I said, these are
the four conditions we have talked about so far:

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

Equations 1 and 2 say "forall e". Equations 3 and 4
have conditions. You are ignoring these conditions.

There is only one function
t'(x,t) defined here

If you don't like events, then the same information can
be conveyed in terms of x and t:

1. forall x, forall t, x'(x,t) = A x + B c t
2. forall x, forall t, t'(x,t) = D x + E c 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)

i.e. if you are serious about this issue you
should also split t into t1 and t2, but I wonder how far you will be
getting this way in the derivation of the Lorentz transformation.

Also you still haven't answered my question why you are not changing
the sign convention and let x1=ct and x2=ct.

You were the one who said that x2 was -ct. Now you want to say
that x2 = +ct. That doesn't make any sense.
On the contrary, it makes perfect sense if you have separate variables
for the two directions. You only need negative numbers if you have one
variable. Our calendar is for instance the best example for this: you
have one scale named AD and another named BC, both measured positive.
Obviously you could do the same for a spatial scale as well. If you
want, you can define the location of the light signal by a spherical
coordinate e.g. by r(phi,t) (where phi is a corresponding angle and
thus r always positive).
I am sure you are not suggesting that the derivation of the Lorentz
transformation depends on the use of a Cartesian coordinate system.

Thomas

"Thomas Smid" wrote in message oups.com...

Any particular reason why you always post your crap
twice?

Dirk Vdm

Thomas Smid says...

Daryl McCullough wrote:

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

The 4 equations above have a solution

A=5/3

B=-4/3

D=-4/3

E=5/3

No there isn't a contradiction as you restricted again the validity of
the equations:

if you have two arbitrary events e1 and e2 , the transformation
equations are

x'(e1) = A x(e1) + Bc t(e1)
ct'(e1)= B x(e1) + Ac t(e1)

x'(e2) = A x(e2) + Bc t(e2)
ct'(e2)= B x(e2) + Ac t(e2)

These equations must hold for all events e1 and e2 simultaneously,
whether x(e1) and x(e2) are positive or negative, and thus also for
further events at -x(e1) and -x(e2),

What does "further events at -x(e1) and -x(e2)" mean?

I think that maybe what you mean is this: let e3 and e4
be events such that

x(e3) = -x(e1)
x(e4) = -x(e2)

I'll assume that's what you mean.

so we can actually double up the equations to

(1a) x'(e1) = A x(e1) + Bc t(e1)
(1b) ct'(e1)= B x(e1) + Ac t(e1)

(2a) -x'(e1) = -A x(e1) + Bc t(e1)

That doesn't make any sense. If you want to consider
some event e3 such that x(e3) = - x(e1),
and t(e3) = t(e1), then you have

(2a) x'(e3) = A x(e3) + Bc t(e3)
= - A x(e1) + B c t(e1)

Note: the left side is in terms of e3, not e1. You seem
to be assuming that

x'(e3) = - x'(e1)

but there is no reason to assume that.

You were the one who said that x2 was -ct. Now you want to say
that x2 = +ct. That doesn't make any sense.

On the contrary, it makes perfect sense if you have separate variables
for the two directions.

That doesn't make any sense. I have no idea what you mean.

--
Daryl McCullough
Ithaca, NY

Daryl McCullough wrote:
Thomas Smid says...

Daryl McCullough wrote:

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

The 4 equations above have a solution

A=5/3

B=-4/3

D=-4/3

E=5/3

No there isn't a contradiction as you restricted again the validity of
the equations:

if you have two arbitrary events e1 and e2 , the transformation
equations are

x'(e1) = A x(e1) + Bc t(e1)
ct'(e1)= B x(e1) + Ac t(e1)

x'(e2) = A x(e2) + Bc t(e2)
ct'(e2)= B x(e2) + Ac t(e2)

These equations must hold for all events e1 and e2 simultaneously,
whether x(e1) and x(e2) are positive or negative, and thus also for
further events at -x(e1) and -x(e2),

What does "further events at -x(e1) and -x(e2)" mean?

I think that maybe what you mean is this: let e3 and e4
be events such that

x(e3) = -x(e1)
x(e4) = -x(e2)

I'll assume that's what you mean.

so we can actually double up the equations to

(1a) x'(e1) = A x(e1) + Bc t(e1)
(1b) ct'(e1)= B x(e1) + Ac t(e1)

(2a) -x'(e1) = -A x(e1) + Bc t(e1)

That doesn't make any sense. If you want to consider
some event e3 such that x(e3) = - x(e1),
and t(e3) = t(e1), then you have

(2a) x'(e3) = A x(e3) + Bc t(e3)
= - A x(e1) + B c t(e1)

Note: the left side is in terms of e3, not e1. You seem
to be assuming that

x'(e3) = - x'(e1)

but there is no reason to assume that.


Sorry, but I deleted the section you commented on before you posted
your reply. I noticed that it might not improve my argument but that
the introduction of further events may actually confuse things further
rather than clarify them.

So I want to get back to the issue of not writing the arguments of
variables here (which you used in order to prevent me from making
certain substitutions in the equations).
As I indicated before, the arguments of variables are irrelevant for
the algebra. The value of a variable on the left hand side of an
equation is only determined by the expression on the right hand side
and is not in any way affected by writing any arguments or not.
In this sense, if I have the identies

(1) x2=-x1; x2'=-x1'

and the further relationships

(2) x1'=Ax1+Bct
(3) x2'=Ax2+Bct

then there is nothing to prevent me from substituting the mathematical
identies (1) into (3) which thus yields

(4) -x1'=-Ax1+Bct

and thus

(5) x1'=Ax1-Bct

and hence by a comparison with (2)

(6) B=0.

Every other conclusion is in my opinion mathematically incorrect and
would involve making further assumptions or restrictions that are not
inherent in the equations.

Thomas

Daryl McCullough wrote:
Thomas Smid says...

Daryl McCullough wrote:

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

The 4 equations above have a solution

A=5/3

B=-4/3

D=-4/3

E=5/3

No there isn't a contradiction as you restricted again the validity of
the equations:

if you have two arbitrary events e1 and e2 , the transformation
equations are

x'(e1) = A x(e1) + Bc t(e1)
ct'(e1)= B x(e1) + Ac t(e1)

x'(e2) = A x(e2) + Bc t(e2)
ct'(e2)= B x(e2) + Ac t(e2)

These equations must hold for all events e1 and e2 simultaneously,
whether x(e1) and x(e2) are positive or negative, and thus also for
further events at -x(e1) and -x(e2),

What does "further events at -x(e1) and -x(e2)" mean?

I think that maybe what you mean is this: let e3 and e4
be events such that

x(e3) = -x(e1)
x(e4) = -x(e2)

I'll assume that's what you mean.

so we can actually double up the equations to

(1a) x'(e1) = A x(e1) + Bc t(e1)
(1b) ct'(e1)= B x(e1) + Ac t(e1)

(2a) -x'(e1) = -A x(e1) + Bc t(e1)

That doesn't make any sense. If you want to consider
some event e3 such that x(e3) = - x(e1),
and t(e3) = t(e1), then you have

(2a) x'(e3) = A x(e3) + Bc t(e3)
= - A x(e1) + B c t(e1)

Note: the left side is in terms of e3, not e1. You seem
to be assuming that

x'(e3) = - x'(e1)

but there is no reason to assume that.

Sorry, but I deleted the section you commented on before you posted
your reply. I noticed that it might not improve my argument but that
the introduction of further events may actually confuse things further
rather than clarify them.

So I want to get back to the issue of not writing the arguments of
variables here (which you used in order to prevent me from making
certain substitutions in the equations).
As I indicated before, the arguments of variables are irrelevant for
the algebra. The value of a variable on the left hand side of an
equation is only determined by the expression on the right hand side
and is not in any way affected by writing any arguments or not.
In this sense, if I have the identities

(1) x2=-x1; x2'=-x1'

and the further relationships

(2) x1'=Ax1+Bct
(3) x2'=Ax2+Bct

then there is nothing to prevent me from substituting the mathematical
identities (1) into (3) which thus yields

(4) -x1'=-Ax1+Bct

and thus

(5) x1'=Ax1-Bct

and hence by a comparison with (2)

(6) B=0.

Every other conclusion is in my opinion mathematically incorrect and
would involve making further assumptions or restrictions that are not
inherent in the equations.

Thomas

Daryl McCullough wrote:
Thomas Smid says...

Daryl McCullough wrote:

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

The 4 equations above have a solution

A=5/3

B=-4/3

D=-4/3

E=5/3

No there isn't a contradiction as you restricted again the validity of
the equations:

if you have two arbitrary events e1 and e2 , the transformation
equations are

x'(e1) = A x(e1) + Bc t(e1)
ct'(e1)= B x(e1) + Ac t(e1)

x'(e2) = A x(e2) + Bc t(e2)
ct'(e2)= B x(e2) + Ac t(e2)

These equations must hold for all events e1 and e2 simultaneously,
whether x(e1) and x(e2) are positive or negative, and thus also for
further events at -x(e1) and -x(e2),

What does "further events at -x(e1) and -x(e2)" mean?

I think that maybe what you mean is this: let e3 and e4
be events such that

x(e3) = -x(e1)
x(e4) = -x(e2)

I'll assume that's what you mean.

so we can actually double up the equations to

(1a) x'(e1) = A x(e1) + Bc t(e1)
(1b) ct'(e1)= B x(e1) + Ac t(e1)

(2a) -x'(e1) = -A x(e1) + Bc t(e1)

That doesn't make any sense. If you want to consider
some event e3 such that x(e3) = - x(e1),
and t(e3) = t(e1), then you have

(2a) x'(e3) = A x(e3) + Bc t(e3)
= - A x(e1) + B c t(e1)

Note: the left side is in terms of e3, not e1. You seem
to be assuming that

x'(e3) = - x'(e1)

but there is no reason to assume that.


Sorry, but I deleted the section you commented on before you posted
your reply. I noticed that it might not improve my argument but that
the introduction of further events may actually confuse things further
rather than clarify them.

So I want to get back to the issue of not writing the arguments of
variables here (which you used in order to prevent me from making
certain substitutions in the equations).
As I indicated before, the arguments of variables are irrelevant for
the algebra. The value of a variable on the left hand side of an
equation is only determined by the expression on the right hand side
and is not in any way affected by writing any arguments or not.
In this sense, if I have the identities

(1) x2=-x1; x2'=-x1'

and the further relationships

(2) x1'=Ax1+Bct
(3) x2'=Ax2+Bct

then there is nothing to prevent me from substituting the mathematical
identities (1) into (3) which thus yields

(4) -x1'=-Ax1+Bct

and thus

(5) x1'=Ax1-Bct

and hence by a comparison with (2)

(6) B=0.

Any other conclusion from the transformation equations would only
indicate that they are in fact algebraically inconsistent. So either
way it shows that they don't make any sense as they would only be
consistent if all variables are identically zero.


Thomas