Re: Division by zero

Den 02.02.2025 10:40, skrev Mikko:

On 2025-02-02 06:58:32 +0000, Thomas Heger said:
 

Am Sonntag000002, 02.02.2025 um 03:19 schrieb Ross Finlayson:

On 02/01/2025 01:36 AM, Mikko wrote:

On 2025-02-01 08:14:08 +0000, Thomas Heger said:
>

Hi NG
>
I'm actually not really certain, but found an error in Einstein's 'On
the electrodynamics of moving bodies' which is quite serious.
>
>
See page six, roughly in the middle:
>
There we find an equation, which says this:
>
∂τ/∂y= 0

>
Do you mean on page 899 (9th page of the article) in §3?
The operation is not division but a partial derivative.
>

Now, 'tau' is a time belonging to the moving system k.

>
Yes, but it is also a number that is computed from coordinates of K.
>

This system k moves along the x-axis of system K with velocity v,
while x- and xsi-axis coincide and etha- and y axis remain parallel.
>
In other words v_y is permanently zero,

>
Yes,
>

 or: ∂y=0.

>
No. ∂y is not a number but a part of an operator. There are points with
different values of y and ∂/∂y refers to a line where t, x, and z (but not
y) have the same value at every point.
>
See https://en.wikipedia.org/wiki/Partial_derivative

 Did you read https://en.wikipedia.org/wiki/Partial_derivative ?
 

Thomas Heger wouldn't understand it if he tried to read it.
Back in 2020 I tried to explain this equation:
(from the same page as the above)
  1/2*[ tau(0,0,0,t) + tau(0,0,0,t+ x'/(c-v)+x'/(c+v)) ]
   = tau ( x',0,0,t+x'/(c-v))
Thomas idea was that:  1/2*tau(0,0,0,t) = tau(0,0,0,t/2)
So I defined a simpler function and wrote
a simpler equation:
|Den 23.03.2020 17:41, skrev Thomas Heger:
|> Am 23.03.2020 um 10:10 schrieb Paul B. Andersen:
|>>
|>> Given the linear function f(x',t) = x'+2t
|>>
|>>  0.5*[f(0,1)+f(0,2)] = f(1,1)    (3 = 3)
|>>
|>>  0.5*[f(0,k)+f(0,2k)] = f(k,k)   (3k = 3k)
|>>
|>>  0.5*[2+4] = 3      [1+2] = 3
|>>  0.5*[2k+4k] = 3k   [1k+2k] = 3k
|> no
|>
|> 1/2 * f(0,1) = f(0*x', 1/2*1*t) = f(0,1/2*t)= 1/2*t
|> + 1/2 * f(0,2)= f(0, t)=t
|> ------------------------
|> = 0.5*[f(0,1)+f(0,2)] = f(0, 1.5 *t)=1.5*t
|>
|>
|> TH
Thomas Heger seems incapable to learn, so he probably still don't
know what a function is.
--
Paul
https://paulba.no/