Groups > sci.physics.relativity > #479361

Re: the famous formula

From	Libor Striz <poutnik4REMOVEnntp@CAPITALSgmail.com.INVALID>
Newsgroups	sci.physics.relativity
Subject	Re: the famous formula
Date	2018-12-08 19:34 +0100
Organization	news.eternal-september.org
Message-ID	<puh2ro$hr2$1@dont-email.me> (permalink)
References	(5 earlier) <g71to8Fu610U1@mid.individual.net> <pugpjb$fbf$1@dont-email.me> <pugq4s$1dms$2@gioia.aioe.org> <g72dmjF34g5U1@mid.individual.net> <puh0oa$1qiv$1@gioia.aioe.org>

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Ejí Utorka <edew@aewedoc.cl> Wrote in message:
> Thomas Heger wrote:
> 
>> Am 08.12.2018 um 17:05 schrieb Ejí Utorka:
>>> Libor Striz wrote:
>>> 
>>>> Thomas Heger <ttt_heg@web.de> Wrote in message:
>>>>> Am 07.12.2018 um 17:57 schrieb Libor Striz:
>>>>>> Work  is a scalar, not a vector.
>>>>>> dW = vec_F . d(vec_L)
>>>>>> W = path integral of dW along path L.
>>>>>> dW as a dot product of 2 vectors is a scalar.
>>>>>>
>>>>> I do not regard a distance as a vector.
>>>>
>>>> Neither do I. But the infinitesimal path is, especially on context of
>>>> the definition of the work  diferential as scalar product of 2 vectors
>>>> dW = F . dL = |F| * |dL| * cos(fi)
>>>> Where fi is the angle between vector of force and infinitesimal
>>>>   vector of path.
>>> 
>>> dL is the vector, not the L,
>> 
>> Well, no.
>> A length is usually a scalar, not a vector.
> 
> But that's not a length. Reread what I wrote. Basic physics. When you 
> differentiate something. What's wrong with you people.
> 

Do not confuse a distance and a length of a path.

The closed path length around a circle is 2.pi.r.
The distance between its endpoints is zero.
The elements dL along this circular path are vectors  and F . dL
 gives infinitesimal scalar of work dW.

-- 
Libor Striz aka Poutnik ( a pilgrim/wanderer/wayfarer)

"Humour is the only effective weapon against stupidity."
Miloš Forman


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                Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-15 19:50 +0100
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