Daryl McCullough wrote:


1' forall events e, (x'(e)-ct'(e)) = lambda (x(e)-ct(e))

OK let's choose then the 'event' x(e)=-ct(e)
This yields
-2ct'(e)=-lambda*2ct(e)

2' forall events e, (x'(e)+ct'(e)) = mu (x(e)+ct(e))

OK let's choose then the 'event' x(e)=ct(e)
This yields
2ct'(e)=mu*2ct(e)

and hence lambda=mu.


Also, if all the equations apply in any case, I should be entitled to
make the transformation ct'(x2,t)=ct'(-x1,t)=-(B+A)x1. Not allowing me
to do this means that you are applying double standards here.

But anyway, why don't you make life easier for yourself and avoid
negative x-coordinates in the first place (as I suggested above
already). If you split x into x1 and x2 there is no point anyway having
x2 negative. As the sign of the measuring unit is only a convention,
you can just take x1(t)=ct and x2(t)=ct (imagine you have two rulers
going into opposite directions, one with a scale entitled x1 and one
with x2).

Thomas

"Thomas Smid" wrote in message oups.com...
Daryl McCullough wrote:


1' forall events e, (x'(e)-ct'(e)) = lambda (x(e)-ct(e))

OK let's choose then the 'event' x(e)=-ct(e)
This yields
-2ct'(e)=-lambda*2ct(e)

Yes, but only for events that also satisfy
x(e) = -c t(e)


2' forall events e, (x'(e)+ct'(e)) = mu (x(e)+ct(e))

OK let's choose then the 'event' x(e)=ct(e)
This yields
2ct'(e)=mu*2ct(e)

Yes, but only for events that also satisfy
x(e) = c t(e)


and hence lambda=mu.

No, stupid.
If you want to combine the system of two equations
{ -2ct'(e)=-lambda*2ct(e)
{ 2ct'(e)=mu*2ct(e)
you can do that only for events that satisfy
{ x(e) = -c t(e)
{ x(e) = +c t(e)
in other words, for events that satisfy
{ x(e) = 0
{ t(e) = 0
so the system of equations is
{ -2*0 = - lambda*2*0
{ 2c*0 = mu*2c*0
or in other words
{ 0 = 0
{ 0 = 0
from which you cannot deduce that
lambda = mu

Are you Marcel Luttgens in disguise?

Dirk Vdm

"Thomas Smid" wrote in message
oups.com...
Daryl McCullough wrote:


1' forall events e, (x'(e)-ct'(e)) = lambda (x(e)-ct(e))

OK let's choose then the 'event' x(e)=-ct(e)
This yields
-2ct'(e)=-lambda*2ct(e)

2' forall events e, (x'(e)+ct'(e)) = mu (x(e)+ct(e))

OK let's choose then the 'event' x(e)=ct(e)
This yields
2ct'(e)=mu*2ct(e)

and hence lambda=mu.


No, you are talking about two *different* events here. So, you should let
e1 be the event where x(e1)=-ct(e1) and let e2 be the different event where
x(e2)=ct(e2). You then have

2ct'(e1)=lambda*2ct(e1)

and

2ct'(e2)=mu*2ct(e2)

Since event e1 is not the same event as e2, you cannot conclude that lambda
= mu.

Todd

"Todd" wrote in message news:4n1Ve.324465$_o.140936@attbi_s71...

"Thomas Smid" wrote in message
oups.com...
Daryl McCullough wrote:


1' forall events e, (x'(e)-ct'(e)) = lambda (x(e)-ct(e))

OK let's choose then the 'event' x(e)=-ct(e)
This yields
-2ct'(e)=-lambda*2ct(e)

2' forall events e, (x'(e)+ct'(e)) = mu (x(e)+ct(e))

OK let's choose then the 'event' x(e)=ct(e)
This yields
2ct'(e)=mu*2ct(e)

and hence lambda=mu.


No, you are talking about two *different* events here.

So, you should let
e1 be the event where x(e1)=-ct(e1) and let e2 be the different event where
x(e2)=ct(e2). You then have

2ct'(e1)=lambda*2ct(e1)

and

2ct'(e2)=mu*2ct(e2)

Since event e1 is not the same event as e2, you cannot conclude that lambda
= mu.

It can be the same event if, as I have shown, it is the
(x,t) = (x',t') = (0,0) event. But then the equations
reduce to 0 = 0 and lambda and mu disappear from
the equations.
This is a common mistake made by those who never
had elementary linear algebra.
Smid seems to have decided to turn this mistake
into a Fine Art :-)

Dirk Vdm

Thomas Smid says...

Daryl McCullough wrote:


1' forall events e, (x'(e)-ct'(e)) = lambda (x(e)-ct(e))

OK let's choose then the 'event' x(e)=-ct(e)

Okay, but you need to distinguish between variables
and constants. Let e1 represent some particular event
such that

x(e1) = -ct(e1)

This yields
-2ct'(e)=-lambda*2ct(e)

Right. For that particular event. So for that event, we have

t'(e1) = lambda t(e1)

2' forall events e, (x'(e)+ct'(e)) = mu (x(e)+ct(e))

OK let's choose then the 'event' x(e)=ct(e)

Okay, that's some new event, so don't use the same label.
So let e2 be some event such that x(e2)=ct(e2).

This yields
2ct'(e)=mu*2ct(e)

For that particular event e2, we have

2ct'(e2) = mu * 2 c t(e2)

or

t'(e2) = mu t(e2)

and hence lambda=mu.

No, there is one event e1 satisfying x(e1)=-ct(e1). There is a
*different* event e2 satisfying x(e2) = +ct(e2). They aren't the
same event (unless t(e1) = 0).

Thomas, you just keep making algebraic mistake after algebraic
mistake.

Also, if all the equations apply in any case, I should be entitled to
make the transformation ct'(x2,t)=ct'(-x1,t)=-(B+A)x1.

As I have told you several times, x2 is not a variable. It is
(by assumption) equal to -ct. x1 is not a variable. It is equal
to +ct. So you can't go from

ct'(x1,t) = (B+A) x1
to
ct'(-x1,t) = -(B+A)x1

That amounts to going from

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

to

ct'(-ct,t) = - (B+A) ct

That's clearly incorrect, and you keep doing it.

--
Daryl McCullough
Ithaca, NY

"Dirk Van de moortel" wrote
in message ...

"Todd" wrote in message
news:4n1Ve.324465$_o.140936@attbi_s71...

"Thomas Smid" wrote in message
oups.com...
Daryl McCullough wrote:


1' forall events e, (x'(e)-ct'(e)) = lambda (x(e)-ct(e))

OK let's choose then the 'event' x(e)=-ct(e)
This yields
-2ct'(e)=-lambda*2ct(e)

2' forall events e, (x'(e)+ct'(e)) = mu (x(e)+ct(e))

OK let's choose then the 'event' x(e)=ct(e)
This yields
2ct'(e)=mu*2ct(e)

and hence lambda=mu.


No, you are talking about two *different* events here.

So, you should let
e1 be the event where x(e1)=-ct(e1) and let e2 be the different event
where
x(e2)=ct(e2). You then have

2ct'(e1)=lambda*2ct(e1)

and

2ct'(e2)=mu*2ct(e2)

Since event e1 is not the same event as e2, you cannot conclude that
lambda
= mu.

It can be the same event if, as I have shown, it is the
(x,t) = (x',t') = (0,0) event. But then the equations
reduce to 0 = 0 and lambda and mu disappear from
the equations.
This is a common mistake made by those who never
had elementary linear algebra.
Smid seems to have decided to turn this mistake
into a Fine Art :-)

Dirk Vdm


Yes, I overlooked the trivial case (x,t) = (x',t') = (0,0) in which event e1
is the same as event e2. But then, as you show, this case leads nowhere.
Thanks.

Todd

Daryl McCullough wrote:
Thomas Smid says...

Daryl McCullough wrote:


1' forall events e, (x'(e)-ct'(e)) = lambda (x(e)-ct(e))

OK let's choose then the 'event' x(e)=-ct(e)

Okay, but you need to distinguish between variables
and constants. Let e1 represent some particular event
such that

x(e1) = -ct(e1)

This yields
-2ct'(e)=-lambda*2ct(e)

Right. For that particular event. So for that event, we have

t'(e1) = lambda t(e1)

2' forall events e, (x'(e)+ct'(e)) = mu (x(e)+ct(e))

OK let's choose then the 'event' x(e)=ct(e)

Okay, that's some new event, so don't use the same label.
So let e2 be some event such that x(e2)=ct(e2).

This yields
2ct'(e)=mu*2ct(e)

For that particular event e2, we have

2ct'(e2) = mu * 2 c t(e2)

or

t'(e2) = mu t(e2)

and hence lambda=mu.

No, there is one event e1 satisfying x(e1)=-ct(e1). There is a
*different* event e2 satisfying x(e2) = +ct(e2). They aren't the
same event (unless t(e1) = 0).

Thomas, you just keep making algebraic mistake after algebraic
mistake.

Also, if all the equations apply in any case, I should be entitled to
make the transformation ct'(x2,t)=ct'(-x1,t)=-(B+A)x1.

As I have told you several times, x2 is not a variable. It is
(by assumption) equal to -ct. x1 is not a variable. It is equal
to +ct. So you can't go from

ct'(x1,t) = (B+A) x1
to
ct'(-x1,t) = -(B+A)x1

That amounts to going from

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

to

ct'(-ct,t) = - (B+A) ct

That's clearly incorrect, and you keep doing it.

We have gone through all this already. Each time the algebra is not
working in your sense, you are trying to change the formalism by adding
arguments to the variables. This does not change anything about the
algebraic conclusions (as we have seen before when we did the split
into x1 and x2). If you have ever written a computer program to solve
mathematical equations, then you know that variables are characterized
by simple names that have no arguments (unless they are arrays).
Arguments are only introduced in analytical maths in order to better
keep track of things when doing operations like differentiation and
integration etc. Again, they do not change anything about the algebra
and the solution of the equations (all the algebraic connections are
given through the equations between the variables, whether you denote
the latter with arguments or not).

Assume you have a function f(x) which is defined for all numbers x
(positive as well as negative) and you have the two equations

(1) f(x)=(A+B)x
(2) f(-x)=(A-B)x
then you can *logically* conclude from (1) that
(3) f(-x)=-(A+B)x
and thus by comparison with (2)
(4) A=0.

In the same sense, if your function t'(x,t) is defined on all numbers x
(positive and negative) and you have the relationships

(5) ct'(x1,t)=(B+A)x1
(6) ct'(x2,t)=(B-A)x2
(7) x2=-x1

then you can *logically* conclude from (5)

(8) ct'(x2,t)=ct'(-x1,t)=-(B+A)x1

and from (6)

(9) ct'(x2,t)=(B-A)x2=(A-B)x1

and thus by comparing (8) and (9)

(10) A=0.

This is straightforward algebra and I can not see how you can arrive at
any other conclusion.

Thomas

Thomas Smid says...

Daryl McCullough wrote:

As I have told you several times, x2 is not a variable. It is
(by assumption) equal to -ct. x1 is not a variable. It is equal
to +ct. So you can't go from

ct'(x1,t) = (B+A) x1
to
ct'(-x1,t) = -(B+A)x1

That amounts to going from

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

to

ct'(-ct,t) = - (B+A) ct

That's clearly incorrect, and you keep doing it.

We have gone through all this already. Each time the algebra is not
working in your sense,

Each time, you fail to keep track of dependencies between
variables, and it gets you into trouble. As I have said
before, if you want to show that *I'm* making a mistake,
then you need to be *more* careful than I am. But you are
consistently *less* careful.

you are trying to change the formalism by adding
arguments to the variables.

I'm saying that that is the *correct* way to reason
about these things. If you want to show *my* reasoning
leads to a contradiction, then you need to use *my*
reasoning. If you want to show that *your* reasoning
leads to a contradiction, then by all means, use your
own reasoning.

You are the one who introduced x1 and said that it was
equal to ct. You are the one who introduced x2 and said
that it was equal to -ct. Thus it makes no sense to go
from

ct'(x1,t) = (B+A) x1
to
ct'(-x1,t) = -(B+A)x1

How can it make any sense, when x1 is just another
name for ct?

This does not change anything about the
algebraic conclusions

The correct algebraic conclusion is that you are having
trouble with algebra. Have you ever heard of *checking*
your results? You say that the equations

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

Let's check: Suppose x=ct. Then x' = 5/3 ct - 4/3 ct = 1/3 ct
Then t' = -4/3 ct + 5/3 ct = 1/3 ct. So x' = ct'. Check.

Suppose x = -ct. Then x' = -5/3 ct - 4/3 ct = -3 ct.
Then t' = +4/3 ct + 5/3 ct = +3 ct. So x' = -ct'. Check.

So, obviously my 4 equations above are consistent with both
A and B being nonzero. So if you come up with A=0 or B=0,
then you are adding something new. If that something new
leads to a contradiction, then that's *your* contradiction,
not mine.

Assume you have a function f(x) which is defined for all numbers x
(positive as well as negative) and you have the two equations

(1) f(x)=(A+B)x
(2) f(-x)=(A-B)x
then you can *logically* conclude from (1) that
(3) f(-x)=-(A+B)x
and thus by comparison with (2)
(4) A=0.

Right. But we are not in that situation. We have a function
of *two* variables x and t.

In the same sense, if your function t'(x,t) is defined on all numbers x
(positive and negative) and you have the relationships

(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.

(7) x2=-x1

then you can *logically* conclude from (5)

(5) only applies when x=ct. (6) only applies when x=-ct.
So for what values of x and t do *both* 5 and 6 hold?
That's a simple question, Thomas, and it has a simple
answer:

If x=ct, and x=-ct, then x=0 and ct=0.

That's the *only* case in which both 5 and 6 hold.

(8) ct'(x2,t)=ct'(-x1,t)=-(B+A)x1

Right. In the special case x=0 and t=0, it follows that

ct'(x2,t) = ct'(-x1,t) = -(B+A)x1

because in that special case, x1 = 0, x2 = 0, t = 0, t' = 0.

and from (6)

(9) ct'(x2,t)=(B-A)x2=(A-B)x1

Only in the special case x1 = x2 = t = t' = 0. Plug in these
values for x1,x2,t,t', and you get

0 = (B-A) * 0 = (A-B) 0

You cannot conclude that B-A = A-B except by dividing by zero.

and thus by comparing (8) and (9)

(10) A=0.

This is straightforward algebra

Yes, it is a straight-forward example of division by zero.

and I can not see how you can arrive at
any other conclusion.

By actually taking care that I not divide by zero.

--
Daryl McCullough
Ithaca, NY

Daryl McCullough wrote:
Thomas Smid says...

Daryl McCullough wrote:

As I have told you several times, x2 is not a variable. It is
(by assumption) equal to -ct. x1 is not a variable. It is equal
to +ct. So you can't go from

ct'(x1,t) = (B+A) x1
to
ct'(-x1,t) = -(B+A)x1

That amounts to going from

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

to

ct'(-ct,t) = - (B+A) ct

That's clearly incorrect, and you keep doing it.

We have gone through all this already. Each time the algebra is not
working in your sense,

Each time, you fail to keep track of dependencies between
variables, and it gets you into trouble. As I have said
before, if you want to show that *I'm* making a mistake,
then you need to be *more* careful than I am. But you are
consistently *less* careful.

you are trying to change the formalism by adding
arguments to the variables.

I'm saying that that is the *correct* way to reason
about these things. If you want to show *my* reasoning
leads to a contradiction, then you need to use *my*
reasoning. If you want to show that *your* reasoning
leads to a contradiction, then by all means, use your
own reasoning.

You are the one who introduced x1 and said that it was
equal to ct. You are the one who introduced x2 and said
that it was equal to -ct. Thus it makes no sense to go
from

ct'(x1,t) = (B+A) x1
to
ct'(-x1,t) = -(B+A)x1

How can it make any sense, when x1 is just another
name for ct?

This does not change anything about the
algebraic conclusions

The correct algebraic conclusion is that you are having
trouble with algebra. Have you ever heard of *checking*
your results? You say that the equations

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

Let's check: Suppose x=ct. Then x' = 5/3 ct - 4/3 ct = 1/3 ct
Then t' = -4/3 ct + 5/3 ct = 1/3 ct. So x' = ct'. Check.

Suppose x = -ct. Then x' = -5/3 ct - 4/3 ct = -3 ct.
Then t' = +4/3 ct + 5/3 ct = +3 ct. So x' = -ct'. Check.

So, obviously my 4 equations above are consistent with both
A and B being nonzero. So if you come up with A=0 or B=0,
then you are adding something new. If that something new
leads to a contradiction, then that's *your* contradiction,
not mine.

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))



Assume you have a function f(x) which is defined for all numbers x
(positive as well as negative) and you have the two equations

(1) f(x)=(A+B)x
(2) f(-x)=(A-B)x
then you can *logically* conclude from (1) that
(3) f(-x)=-(A+B)x
and thus by comparison with (2)
(4) A=0.

Right. But we are not in that situation. We have a function
of *two* variables x and t.

In the same sense, if your function t'(x,t) is defined on all numbers x
(positive and negative) and you have the relationships

(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.

(7) x2=-x1

then you can *logically* conclude from (5)

(5) only applies when x=ct. (6) only applies when x=-ct.
So for what values of x and t do *both* 5 and 6 hold?
That's a simple question, Thomas, and it has a simple
answer:

If x=ct, and x=-ct, then x=0 and ct=0.

That's the *only* case in which both 5 and 6 hold.

(8) ct'(x2,t)=ct'(-x1,t)=-(B+A)x1

Right. In the special case x=0 and t=0, it follows that

ct'(x2,t) = ct'(-x1,t) = -(B+A)x1

because in that special case, x1 = 0, x2 = 0, t = 0, t' = 0.

and from (6)

(9) ct'(x2,t)=(B-A)x2=(A-B)x1

Only in the special case x1 = x2 = t = t' = 0. Plug in these
values for x1,x2,t,t', and you get

Let's make things clearer by considering again your equations involving
lambda and mu for the two directions x1 and x2

(1)(x1'-ct')=lambda*(x1-ct)
(2)(x1'+ct')=mu*(x1+ct)

(3)(x2'-ct')=lambda*(x2-ct)
(4)(x2'+ct')=mu*(x2+ct)

Now since both signals are supposed to travel in opposite directions,
we have in addition the global condition

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

Note that these are mathematical identities i.e. we can replace x2 and
x2' by -x1 and -x1' respectively in (3) or (4) as we please without
changing anything at all about the equations (whether we replace the
variables themselves or the (not written) arguments)
So (3) therefore becomes

(6) (-x1'-ct')=lambda*(-x1-ct)
or
(7) (x1'+ct')=lambda*(x1+ct)

Now we know that equations(1)-(4) must hold simultaneously (after all
they are all functions of the same variable t,t'), so from (7) and (2)
we have therefore

(8) lambda=mu

or

(9) A-B=A+B

and hence

(10) B=0.

Previously, I had A=0, but I think this is due to the fact that I
changed the sign of one equation for the ct-transformation, whereas it
should have been

ct'(x1,t)=(A+B)x1
ct'(x2,t)=ct'(-x1,t)=-(B-A)x1=(A-B)x1

i.e. also B=0 from a comparison of both equations.I reckon this is the
correct result, but it does not really make a difference anyway as in
both cases the transformation is only consistent if x=x'=t=t'=0.


Thomas

Daryl McCullough wrote:
Thomas Smid says...

Daryl McCullough wrote:

As I have told you several times, x2 is not a variable. It is
(by assumption) equal to -ct. x1 is not a variable. It is equal
to +ct. So you can't go from

ct'(x1,t) = (B+A) x1
to
ct'(-x1,t) = -(B+A)x1

That amounts to going from

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

to

ct'(-ct,t) = - (B+A) ct

That's clearly incorrect, and you keep doing it.

We have gone through all this already. Each time the algebra is not
working in your sense,

Each time, you fail to keep track of dependencies between
variables, and it gets you into trouble. As I have said
before, if you want to show that *I'm* making a mistake,
then you need to be *more* careful than I am. But you are
consistently *less* careful.

you are trying to change the formalism by adding
arguments to the variables.

I'm saying that that is the *correct* way to reason
about these things. If you want to show *my* reasoning
leads to a contradiction, then you need to use *my*
reasoning. If you want to show that *your* reasoning
leads to a contradiction, then by all means, use your
own reasoning.

You are the one who introduced x1 and said that it was
equal to ct. You are the one who introduced x2 and said
that it was equal to -ct. Thus it makes no sense to go
from

ct'(x1,t) = (B+A) x1
to
ct'(-x1,t) = -(B+A)x1

How can it make any sense, when x1 is just another
name for ct?

This does not change anything about the
algebraic conclusions

The correct algebraic conclusion is that you are having
trouble with algebra. Have you ever heard of *checking*
your results? You say that the equations

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

Let's check: Suppose x=ct. Then x' = 5/3 ct - 4/3 ct = 1/3 ct
Then t' = -4/3 ct + 5/3 ct = 1/3 ct. So x' = ct'. Check.

Suppose x = -ct. Then x' = -5/3 ct - 4/3 ct = -3 ct.
Then t' = +4/3 ct + 5/3 ct = +3 ct. So x' = -ct'. Check.

So, obviously my 4 equations above are consistent with both
A and B being nonzero. So if you come up with A=0 or B=0,
then you are adding something new. If that something new
leads to a contradiction, then that's *your* contradiction,
not mine.

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))


Assume you have a function f(x) which is defined for all numbers x
(positive as well as negative) and you have the two equations

(1) f(x)=(A+B)x
(2) f(-x)=(A-B)x
then you can *logically* conclude from (1) that
(3) f(-x)=-(A+B)x
and thus by comparison with (2)
(4) A=0.

Right. But we are not in that situation. We have a function
of *two* variables x and t.

In the same sense, if your function t'(x,t) is defined on all numbers x
(positive and negative) and you have the relationships

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

Thomas