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tech / sci.math / relativistic equation

relativistic equation

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From: r.hac...@tiscali.fr (Richard Hachel)

by: Richard Hachel - Fri, 21 Jan 2022 17:40 UTC

Re: relativistic equation

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by: mitchr...@gmail.com - Fri, 21 Jan 2022 19:47 UTC

Re: relativistic equation

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From: PointedE...@web.de (Thomas 'PointedEars' Lahn)
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Subject: Re: relativistic equation
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by: Thomas 'Pointed - Fri, 21 Jan 2022 23:30 UTC

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)

Re: relativistic equation

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Newsgroups: sci.math
Subject: Re: relativistic equation
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by: sergio - Sat, 22 Jan 2022 03:55 UTC

On 1/21/2022 11:40 AM, Richard Hachel wrote:
>
>
> Can anyone solve?
>
> dt'= ∫ (1/dt + a)^(-1/2)
>
> R.H.
>

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 ?

Re: relativistic equation

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From: r.hac...@tiscali.fr (Richard Hachel)

by: Richard Hachel - Sat, 22 Jan 2022 12:40 UTC

Le 22/01/2022 à 04:55, sergio a écrit :
> On 1/21/2022 11:40 AM, Richard Hachel wrote:
>>
>>
>> Can anyone solve?
>>
>> dt'= ∫ (1/dt + a)^(-1/2)
>>
>> R.H.
>>
>
> 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 ?

∫ (1/dt + a)^(-1/2) dt

a is a constante.

R.H.

Re: relativistic equation

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From: inva...@invalid.com (sergio)
Newsgroups: sci.math
Subject: Re: relativistic equation
Date: Sat, 22 Jan 2022 08:13:18 -0600
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by: sergio - Sat, 22 Jan 2022 14:13 UTC

On 1/22/2022 6:40 AM, Richard Hachel wrote:
> Le 22/01/2022 à 04:55, sergio a écrit :
>> On 1/21/2022 11:40 AM, Richard Hachel wrote:
>>>
>>>
>>> Can anyone solve?
>>>
>>> dt'= ∫ (1/dt + a)^(-1/2)
>>>
>>> R.H.
>>>
>>
>> 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 ?
>
> ∫ (1/dt + a)^(-1/2) dt
> a is a constante.
> R.H.

the other item to clear up is dt'

normally d(t)/dt = t',

still dont know what dt' is

Re: relativistic equation

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Subject: Re: relativistic equation
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by: Python - Sat, 22 Jan 2022 14:50 UTC

sergio wrote:
> On 1/22/2022 6:40 AM, Richard Hachel wrote:
>> Le 22/01/2022 à 04:55, sergio a écrit :
>>> On 1/21/2022 11:40 AM, Richard Hachel wrote:
>>>>
>>>>
>>>> Can anyone solve?
>>>>
>>>> dt'= ∫ (1/dt + a)^(-1/2)
>>>>
>>>> R.H.
>>>>
>>>
>>> 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 ?
>>
>> ∫ (1/dt + a)^(-1/2) dt
>> a is a constante.
>> R.H.
>
> the other item to clear up is dt'
>
> normally d(t)/dt = t',
>
> still dont know what dt' is

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.

Re: relativistic equation

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Newsgroups: sci.math
Subject: Re: relativistic equation
Date: Sat, 22 Jan 2022 09:06:54 -0600
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by: sergio - Sat, 22 Jan 2022 15:06 UTC

On 1/22/2022 8:50 AM, Python wrote:
> sergio wrote:
>> On 1/22/2022 6:40 AM, Richard Hachel wrote:
>>> Le 22/01/2022 à 04:55, sergio a écrit :
>>>> On 1/21/2022 11:40 AM, Richard Hachel wrote:
>>>>>
>>>>>
>>>>> Can anyone solve?
>>>>>
>>>>> dt'= ∫ (1/dt + a)^(-1/2)
>>>>>
>>>>> R.H.
>>>>>
>>>>
>>>> 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 ?
>>>
>>> ∫ (1/dt + a)^(-1/2) dt
>>> a is a constante.
>>> R.H.
>>
>> the other item to clear up is dt'
>>
>> normally d(t)/dt = t',
>>
>> still dont know what dt' is
>
> 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.
>
>

whoa! three mistakes in one equation

Subject	Author
relativistic equation	Richard Hachel
Re: relativistic equation	mitchr...@gmail.com
Re: relativistic equation	Thomas 'PointedEars' Lahn
Re: relativistic equation	sergio
Re: relativistic equation	Richard Hachel
Re: relativistic equation	sergio
Re: relativistic equation	Python
Re: relativistic equation	sergio