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relativistic equation

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Richard Hachel

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Jan 21, 2022, 12:40:12 PM1/21/22
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Can anyone solve?

dt'= ∫ (1/dt + a)^(-1/2)

R.H.


mitchr...@gmail.com

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Jan 21, 2022, 2:47:32 PM1/21/22
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How is addition of velocity going to work if
it comes from what is not relative first?

Mitchell Raemsch

Thomas 'PointedEars' Lahn

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Jan 21, 2022, 6:31:06 PM1/21/22
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Richard Hachel wrote:

> Can anyone solve?
>
> dt'= ∫ (1/dt + a)^(-1/2)

A lot is missing here, for example: What is the integration variable?
Is “a” a constant with respect to that variable?

And how do you think this equation came about?


PointedEars
--
“Science is empirical: knowing the answer means nothing;
testing your knowledge means everything.”
—Dr. Lawrence M. Krauss, theoretical physicist,
in “A Universe from Nothing” (2009)

sergio

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Jan 21, 2022, 10:55:41 PM1/21/22
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what does dt' mean ? you have 2 symbols for derivative stuck onto a t...

t' usually is the first derivative with respect to time....

what is the integrand with respect to ?

do you mean ∫ (1/dt + a)^(-1/2) dt ? or ∫ (1/dt + a)^(-1/2) da ?




Richard Hachel

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Jan 22, 2022, 7:41:09 AM1/22/22
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∫ (1/dt + a)^(-1/2) dt

a is a constante.

R.H.

sergio

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Jan 22, 2022, 9:13:29 AM1/22/22
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the other item to clear up is dt'

normally d(t)/dt = t',

still dont know what dt' is

Python

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Jan 22, 2022, 9:51:02 AM1/22/22
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Lengrand (aka Hachel) has the habit to never define anything just
producing usually meaningless equation. Given that his "hobby" is
a demented attempt to rebuild Relativity from ill-defined concepts
you may assume that t' is a time coordinate in a frame while t is
a time coordinate in another one, but you'll never know for sure.

There is far worse than that in the original "equation"... 1/dt
makes absolutely no sense in the integrated expression.


sergio

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Jan 22, 2022, 10:07:06 AM1/22/22
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whoa! three mistakes in one equation
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