Re: Mathematical Inconsistencies in Einstein's Derivation of the Lorentz Transformation

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

.

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