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

Thomas Smid says...

>Daryl McCullough wrote:

>> The point is that x' and t' are *functions* of x and t.
>>....
>>....
>>....
>>
>> Let's put in all the functional dependence explicitly. The
>> transformation equations are these
>>
>> x'(x,t) + c t'(x,t) = (A+B) (x + ct)
>> x'(x,t) - c t'(x,t) = (A-B) (x - ct)
>
>You *assumed* they are functions of x and t when writing down your
>transformation
>
>x'=Ax+Bct
>ct'=Bx+Act
>
>but this actually contradicts the use of t and t' as independent
>variables in x(t)=ct and x'(t')=ct'.

t' is not independent of x. t' is a *function* of x and t. You
cannot choose x, t, x' and t' independently. You can choose *two*
independently, but then the other two are determined.

>As 'independent' means just this,

t' is not independent of x and t.

>you are actually not entitled to use t or t' with an argument like
>t'(x,t)

Yes, you certainly are.

>But OK, let's just for the time being hypothetically assume that t' is
>not an independent variable.
>
>>Since x2(t)
>> is unequal to x1(t), the corresponding value of t' in (4b)
>> is *not* equal to the corresponding value of t' in (3c).
>>
>> Let's put in all the functional dependence explicitly. The
>> transformation equations are these
>>
>> x'(x,t) + c t'(x,t) = (A+B) (x + ct)
>> x'(x,t) - c t'(x,t) = (A-B) (x - ct)
>>
>> Now, you want to specialize to the case x1(t) = ct in
>> the first equation, and specialize to the case x2(t) = -ct
>> in the second equation. Fine. That gives us:
>>
>> x'(ct,t) + c t'(ct,t) = (A+B) (2ct)
>> x'(-ct,t) - c t'(-ct,t) = (A-B) (-2ct)
>>
>> Now, we use the fact that x' = ct' in the top
>> equation, and x' = -ct' in the bottome equation to get:
>>
>>
>> 2 c t'(ct,t) = (A+B) (2ct)
>> -2 c t'(-ct,t) = (A-B) (-2ct)
>>
>> Your mistake was assuming that t'(ct,t) = t'(-ct,t).
>
>Why should it be a mistake?

You are trying to show that Einstein's derivation led
to an inconsistency. If you make an additional assumption
that Einstein *didn't* make, then any inconsistency you
derive is an inconsistency in the combination of *your*
assumptions and *Einstein's* assumptions. It doesn't
prove anything about Einstein's derivation.

>There is nothing in our previous equations
>that would have prevented me from making this assumption

On the contrary, it leads to a contradiction.

>which, as mentioned above merely implies that t and t'
>are independent variables.

They aren't independent variables. Given x and t, t' is
determined.

> (1) x1'(x1,t) + c t'(x1,t) = (A+B) (x1 + ct)
> (2) x2'(x2,t) - c t'(x2,t) = (A-B) (x2 - ct)
>
>and additionally we have
>
>(3) x1'(x1,t) - c t'(x1,t) = 0
>(4) x2'(x2,t) + c t'(x2,t) = 0
>(5) x1=ct
>(6) x2=-ct
>
>Now by virtue of (3)-(6), (1) and (2) become
>
>(7) 2ct'(x1,t)=(A+B)2x1
>(8) 2ct'(-x1,t)=(A-B)2x1
>
>The question is how ct'(-x1,t) relates to ct'(x1,t). For this we
>consider the original transformation for ct'
>
>(9) ct'(x1,t) = Bx1+Act
>
>which by virtue of Eq.(5) becomes
>
>(10) ct'(x1,t) = (B+A)x1
>
>which obviously means that
>
>(11) ct'(-x1,t) = -(B+A)x1 =-ct'(x1,t)

Thomas, you forget that x1 in equation 10 is *not*
an independent variable, it is a *function* of t.
In particular, x1 = ct, so (10) should be written
as follows:

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

Now, if you replace t by -t, you get the correct
version of (11)

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

[Rest deleted, since it relies on a false step]

--
Daryl McCullough
Ithaca, NY

.

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