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the famous formula

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  the famous formula RichD <r_delaney2001@yahoo.com> - 2018-12-06 10:49 -0800
    Re: the famous formula pnalsing@gmail.com - 2018-12-06 11:30 -0800
    Re: the famous formula dlzc <dlzc1@cox.net> - 2018-12-06 11:43 -0800
      Re: the famous formula RichD <r_delaney2001@yahoo.com> - 2018-12-07 10:15 -0800
        Re: the famous formula dlzc <dlzc1@cox.net> - 2018-12-07 11:13 -0800
          Re: the famous formula RichD <r_delaney2001@yahoo.com> - 2018-12-09 00:02 -0800
            Re: the famous formula dlzc <dlzc1@cox.net> - 2018-12-09 15:23 -0800
              Re: the famous formula Libor Striz <poutnik@privacy.net> - 2018-12-10 09:54 +0100
                Re: the famous formula dlzc <dlzc1@cox.net> - 2018-12-10 05:43 -0800
                  Re: the famous formula Tom Roberts <tjroberts137@sbcglobal.net> - 2018-12-12 11:08 -0600
                    Re: the famous formula dlzc <dlzc1@cox.net> - 2018-12-12 09:35 -0800
                    Re: the famous formula Rich Carpenter <lcug@kigcohc.pa> - 2018-12-12 19:20 +0000
                      Re: the famous formula Tom Roberts <tjroberts137@sbcglobal.net> - 2018-12-12 19:28 -0600
                        Re: the famous formula Scott Richman <sunn@idhnonow.iv> - 2018-12-13 16:13 +0000
                    Re: the famous formula "David (Kronos Prime) Fuller" <fuller.david@hotmail.com> - 2018-12-14 04:56 -0800
                      Re: the famous formula "David (Kronos Prime) Fuller" <fuller.david@hotmail.com> - 2018-12-14 05:23 -0800
          Re: the famous formula Tom Roberts <tjroberts137@sbcglobal.net> - 2018-12-12 10:28 -0600
            Re: the famous formula dlzc <dlzc1@cox.net> - 2018-12-12 09:12 -0800
              Re: the famous formula Tom Roberts <tjroberts137@sbcglobal.net> - 2018-12-14 12:21 -0600
    Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-06 20:50 +0100
      Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-06 22:47 +0100
        Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-07 12:06 +0100
          Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-07 16:15 +0100
            Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-08 14:18 +0100
              Re: the famous formula Odd Bodkin <bodkinodd@gmail.com> - 2018-12-08 13:43 +0000
                Re: the famous formula "David (Kronos Prime) Fuller" <fuller.david@hotmail.com> - 2018-12-08 06:01 -0800
                  Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-08 18:18 +0100
                    Re: the famous formula Libor Striz <poutnik4REMOVEnntp@CAPITALSgmail.com.INVALID> - 2018-12-08 19:26 +0100
                      Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-09 19:35 +0100
                        Re: the famous formula Libor Striz <poutnik4REMOVEnntp@CAPITALSgmail.com.INVALID> - 2018-12-09 20:24 +0100
                          Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-10 18:48 +0100
                            Re: the famous formula Rich Carpenter <lcug@kigcohc.pa> - 2018-12-10 19:37 +0000
                            Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-10 22:27 +0100
                              Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-10 22:35 +0100
                              Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-12 16:45 +0100
                                Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-13 17:21 +0100
                        Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-09 19:03 +0100
                        Re: the famous formula Libor Striz <poutnik4REMOVEnntp@CAPITALSgmail.com.INVALID> - 2018-12-09 20:37 +0100
                        Re: the famous formula Odd Bodkin <bodkinodd@gmail.com> - 2018-12-09 22:20 +0000
                          Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-10 18:40 +0100
                            Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-10 21:59 +0100
                            Re: the famous formula Odd Bodkin <bodkinodd@gmail.com> - 2018-12-10 21:41 +0000
                              Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-11 16:31 +0100
                                Re: the famous formula Odd Bodkin <bodkinodd@gmail.com> - 2018-12-11 18:24 +0000
                                  Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-12 16:32 +0100
                                    Re: the famous formula Odd Bodkin <bodkinodd@gmail.com> - 2018-12-12 16:27 +0000
                                      Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-14 15:13 +0100
                                        Re: the famous formula Odd Bodkin <bodkinodd@gmail.com> - 2018-12-14 14:28 +0000
                                          Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-15 13:35 +0100
                                            Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-15 13:52 +0100
                                              Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-15 19:50 +0100
                                                Re: the famous formula Odd Bodkin <bodkinodd@gmail.com> - 2018-12-15 20:33 +0000
                                                Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-15 22:02 +0100
                                            Re: the famous formula Odd Bodkin <bodkinodd@gmail.com> - 2018-12-15 12:55 +0000
                                    Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-13 23:14 +0100
                                [OT] Work (was: the famous formula) Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-11 21:56 +0100
                    Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-08 20:14 +0100
              Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-09 19:22 +0100
          Re: the famous formula Libor Striz <poutnik4REMOVEnntp@CAPITALSgmail.com.INVALID> - 2018-12-07 17:57 +0100
            Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-08 14:03 +0100
              Re: the famous formula Gary Harnagel <hitlong@yahoo.com> - 2018-12-08 05:29 -0800
                Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-08 18:43 +0100
                  [OT] Work (was: the famous formula) Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-08 21:53 +0100
                    Re: [OT] Work Thomas Heger <ttt_heg@web.de> - 2018-12-09 19:08 +0100
                      Re: [OT] Work Libor Striz <poutnik4REMOVEnntp@CAPITALSgmail.com.INVALID> - 2018-12-09 19:31 +0100
                      Re: [OT] Work Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-09 18:58 +0100
                        Re: [OT] Work Thomas Heger <ttt_heg@web.de> - 2018-12-10 18:52 +0100
                          Re: [OT] Work Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-10 22:28 +0100
                            Re: [OT] Work Thomas Heger <ttt_heg@web.de> - 2018-12-11 16:36 +0100
                              Re: [OT] Work Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-11 20:24 +0100
                  Re: the famous formula Gary Harnagel <hitlong@yahoo.com> - 2018-12-08 14:02 -0800
              Re: the famous formula Libor Striz <poutnik4REMOVEnntp@CAPITALSgmail.com.INVALID> - 2018-12-08 16:56 +0100
                Re: the famous formula Ejí Utorka <edew@aewedoc.cl> - 2018-12-08 16:05 +0000
                  Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-08 18:35 +0100
                    Re: the famous formula Ejí Utorka <edew@aewedoc.cl> - 2018-12-08 17:58 +0000
                      Re: the famous formula Libor Striz <poutnik4REMOVEnntp@CAPITALSgmail.com.INVALID> - 2018-12-08 19:34 +0100
                        Re: the famous formula Ejí Utorka <edew@aewedoc.cl> - 2018-12-08 18:54 +0000
            Re: the famous formula Ejí Utorka <edew@aewedoc.cl> - 2018-12-08 16:09 +0000
    Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-06 22:16 +0100
    Re: the famous formula Odd Bodkin <bodkinodd@gmail.com> - 2018-12-07 16:07 +0000
    Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-07 22:47 +0100
    Re: the famous formula "Ross A. Finlayson" <ross.finlayson@gmail.com> - 2018-12-09 15:06 -0800
      Re: the famous formula benj <benj.fake@gmail.com> - 2018-12-12 13:44 -0500

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#479561

Thomas Heger wrote:
> Am 09.12.2018 um 23:20 schrieb Odd Bodkin:
>> Thomas Heger <ttt_heg@web.de> wrote:
>>> work is NOT 'forth times path'
>>>
>>> but ' force times distance'.
>>
>> No, it is not that either. It is the integral of the scalar product between
>> two vectors, the force and the differential displacement. 

Correct.  (AISB)

> 'Work' is the name of a quantity in physics.
> 
> It relates to the real world, hence is not a product of either one or 
> two vectors.

I am getting tired of telling you this: You have not the slightest clue what
you are talking about.

Work is *precisely* one of those physical quantities that are *calculated*,
i. e. that can NOT be observed, and can NOT be measured directly.

> 'Displacement' is actually not the same thing as 'distance'.
> 
> Displacement is a vector, since you have a position before and after 
> that movement.
> 
> But if you have - say - the distance between Berlin and New York, this 
> does not include any movement.

  “I find your argument strewn with gaping defects in logic.”

    –Spock

> What the hack is so difficult about that?

This is what you should be asking yourself, about yourself.

Why is it so hard for you to understand these basic concepts of physics?

Why is it so hard for you to take up a textbook and actually *read* about
them *before* you are talking about them?

In general: Why do you keep talking about things as if you knew about them
*although* you do NOT (and probably, deep down, you *know* that you do NOT
know)?

-- 
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.

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#479567

Thomas Heger <ttt_heg@web.de> wrote:
> Am 09.12.2018 um 23:20 schrieb Odd Bodkin:
>> Thomas Heger <ttt_heg@web.de> wrote:
>>> Am 08.12.2018 um 19:26 schrieb Libor Striz:
>>>> Thomas Heger <ttt_heg@web.de> Wrote in message:
>>>> 
>>>>> 
>>>>> Vectors have a direction.
>>>>> 
>>>>> A Joule for instance or a kilo-Watt-hour has no direction.
>>>> 
>>>> Why do you then say work is a vector ?
>>>> The unit of work  is 1 Joule.
>>> 
>>> 
>>> We have a similar problem with 'speed' and 'velocity'.
>>> 
>>> Both have units like m/s , but one is a vector quantity (velocity) and
>>> one is not (speed).
>>> 
>>> Speed is the magnitude of velocity.
>>> 
>>> https://examples.yourdictionary.com/examples-vector-scalar-quantity-physics.html
>>> 
>>>> The acting force has direction.
>>>> The path along which the force is acting has direction as well
>>>> But the work as integral of scalar product of these 2 vectors is a
>>>> scalar, without direction.
>>> 
>>> work is NOT 'forth times path'
>>> 
>>> but ' force times distance'.
>> 
>> No, it is not that either. It is the integral of the scalar product between
>> two vectors, the force and the differential displacement. 
> 
> 
> 'Work' is the name of a quantity in physics.

As is velocity. 

> 
> It relates to the real world, hence is not a product of either one or 
> two vectors.

I disagree. There are LOTS of real world quantities that are related in
some way to other real world quantities. Velocity (a vector) is related to
differential displacement and differential time. Momentum is related to
mass and velocity. 

> 
> 'Displacement' is actually not the same thing as 'distance'.

That is correct. And work is characterized by its relationship to
DISPLACEMENT in general, not distance. The relation to distance only
applies in a simple sub case. 

> 
> Displacement is a vector, since you have a position before and after 
> that movement.
> 
> But if you have - say - the distance between Berlin and New York, this 
> does not include any movement.
> 
> What the hack is so difficult about that?
> 
> ...
> 
> 
> TH
> 



-- 
Odd Bodkin — Maker of fine toys, tools, tables

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#479610

Am 10.12.2018 um 22:41 schrieb Odd Bodkin:
> Thomas Heger <ttt_heg@web.de> wrote:
>> Am 09.12.2018 um 23:20 schrieb Odd Bodkin:
>>> Thomas Heger <ttt_heg@web.de> wrote:
>>>> Am 08.12.2018 um 19:26 schrieb Libor Striz:
>>>>> Thomas Heger <ttt_heg@web.de> Wrote in message:
>>>>>
>>>>>>
>>>>>> Vectors have a direction.
>>>>>>
>>>>>> A Joule for instance or a kilo-Watt-hour has no direction.
>>>>>
>>>>> Why do you then say work is a vector ?
>>>>> The unit of work  is 1 Joule.
>>>>
>>>>
>>>> We have a similar problem with 'speed' and 'velocity'.
>>>>
>>>> Both have units like m/s , but one is a vector quantity (velocity) and
>>>> one is not (speed).
>>>>
>>>> Speed is the magnitude of velocity.
>>>>
>>>> https://examples.yourdictionary.com/examples-vector-scalar-quantity-physics.html
>>>>
>>>>> The acting force has direction.
>>>>> The path along which the force is acting has direction as well
>>>>> But the work as integral of scalar product of these 2 vectors is a
>>>>> scalar, without direction.
>>>>
>>>> work is NOT 'forth times path'
>>>>
>>>> but ' force times distance'.
>>>
>>> No, it is not that either. It is the integral of the scalar product between
>>> two vectors, the force and the differential displacement.
>>
>>
>> 'Work' is the name of a quantity in physics.
> 
> As is velocity.
> 
>>
>> It relates to the real world, hence is not a product of either one or
>> two vectors.
> 
> I disagree. There are LOTS of real world quantities that are related in
> some way to other real world quantities. Velocity (a vector) is related to
> differential displacement and differential time. Momentum is related to
> mass and velocity.

This is correct. But I had something against this statement:

".. It is the integral of the scalar product between two vectors..."

I said: no, it is not, since a vector is a mathematical construct.

A number is also a mathematical construct. And you cannot multiply two 
brainchilds and hope to get something real.


First statement: a quantity in physics (work) is defined as being the 
product of one quantity (force)and some sort of length.

We could take the integral of tiny displacements and regard this as 
vector (for instance).

This would be ok.

But mainly the vector is a representation of something real, not real 
itself.

I wanted to address the question, wether the second component of this 
product is a vector or a scalar.

A vector (vectorial quantity) has a direction and a value, while a 
scalar has no direction and only a value.

In my opinion work is a quantity, which essential has a direction.

This is so because it is a product of force (which has a direction) and 
something with length (distance or path).

I suggested to take distance and not path, because then work become a 
vectorial quantity and energy the associated scalar.

This would have benefits, since it is more like our understanding of the 
terms.

As example I take this:

you lift a heavy box by 1m.

This is work (done by you) and the work applied had a direction (up) and 
an amount (in Nm - depending on the weight).

If work is a scalar, then the information of direction is discarded for 
no obvious real.

But more serious is the interpretation of work as form of energy. This 
would then create work as 'kind of invisible stuff'.

And this would raise some illogic questions, like: where did the work go?

Sure: it went into the box.

But that is not true, since the work is not done by the box, but by the 
person that lifted it.

To make the box contain that work, work had to become kind of 'stuff'.


TH

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#479618

Thomas Heger <ttt_heg@web.de> wrote:
> Am 10.12.2018 um 22:41 schrieb Odd Bodkin:
>> Thomas Heger <ttt_heg@web.de> wrote:
>>> Am 09.12.2018 um 23:20 schrieb Odd Bodkin:
>>>> Thomas Heger <ttt_heg@web.de> wrote:
>>>>> Am 08.12.2018 um 19:26 schrieb Libor Striz:
>>>>>> Thomas Heger <ttt_heg@web.de> Wrote in message:
>>>>>> 
>>>>>>> 
>>>>>>> Vectors have a direction.
>>>>>>> 
>>>>>>> A Joule for instance or a kilo-Watt-hour has no direction.
>>>>>> 
>>>>>> Why do you then say work is a vector ?
>>>>>> The unit of work  is 1 Joule.
>>>>> 
>>>>> 
>>>>> We have a similar problem with 'speed' and 'velocity'.
>>>>> 
>>>>> Both have units like m/s , but one is a vector quantity (velocity) and
>>>>> one is not (speed).
>>>>> 
>>>>> Speed is the magnitude of velocity.
>>>>> 
>>>>> https://examples.yourdictionary.com/examples-vector-scalar-quantity-physics.html
>>>>> 
>>>>>> The acting force has direction.
>>>>>> The path along which the force is acting has direction as well
>>>>>> But the work as integral of scalar product of these 2 vectors is a
>>>>>> scalar, without direction.
>>>>> 
>>>>> work is NOT 'forth times path'
>>>>> 
>>>>> but ' force times distance'.
>>>> 
>>>> No, it is not that either. It is the integral of the scalar product between
>>>> two vectors, the force and the differential displacement.
>>> 
>>> 
>>> 'Work' is the name of a quantity in physics.
>> 
>> As is velocity.
>> 
>>> 
>>> It relates to the real world, hence is not a product of either one or
>>> two vectors.
>> 
>> I disagree. There are LOTS of real world quantities that are related in
>> some way to other real world quantities. Velocity (a vector) is related to
>> differential displacement and differential time. Momentum is related to
>> mass and velocity.
> 
> This is correct. But I had something against this statement:
> 
> ".. It is the integral of the scalar product between two vectors..."
> 
> I said: no, it is not, since a vector is a mathematical construct.

I disagree. A vector is a way to characterize a physical quantity. Velocity
is a vector, momentum is a vector, force is a vector. All of those are
real. 

> 
> A number is also a mathematical construct. And you cannot multiply two 
> brainchilds and hope to get something real.
> 
> 
> First statement: a quantity in physics (work) is defined as being the 
> product of one quantity (force)and some sort of length.

“the product” is too vague. Given that the two things you are multiplying
are vectors, you have to be precise about what you mean by “product”. In
this case “product” means scalar- or dot-product. 

> 
> We could take the integral of tiny displacements and regard this as 
> vector (for instance).
> 
> This would be ok.
> 
> But mainly the vector is a representation of something real, not real 
> itself.
> 
> I wanted to address the question, wether the second component of this 
> product is a vector or a scalar.
> 
> A vector (vectorial quantity) has a direction and a value, while a 
> scalar has no direction and only a value.
> 
> In my opinion work is a quantity, which essential has a direction.

Your opinion is wrong here. 

> 
> This is so because it is a product of force (which has a direction) and 
> something with length (distance or path).
> 
> I suggested to take distance and not path, because then work become a 
> vectorial quantity and energy the associated scalar.

And you are trying to rig things to defend your misconception. That is not
the way work is defined or calculated. How it is properly defined and
calculated is well documented in beginner textbooks. That is not  as you
described. 

> 
> This would have benefits, since it is more like our understanding of the 
> terms.
> 
> As example I take this:
> 
> you lift a heavy box by 1m.
> 
> This is work (done by you) and the work applied had a direction (up) and 
> an amount (in Nm - depending on the weight).

No. The work does not have a direction. The force has a direction (up). The
displacement has a direction (up). In this case, the force is constant and
always collinear with the displacement. This is the simplest of cases but
is by no means the general one. Here, though the force and the displacement
have directions, the work itself does not by DEFINITION. You have simply
incorrectly ascribed that direction (up) to the work. 

This is a beginner error. 


> 
> If work is a scalar, then the information of direction is discarded for 
> no obvious real.
> 
> But more serious is the interpretation of work as form of energy. This 
> would then create work as 'kind of invisible stuff'.
> 
> And this would raise some illogic questions, like: where did the work go?
> 
> Sure: it went into the box.
> 
> But that is not true, since the work is not done by the box, but by the 
> person that lifted it.
> 
> To make the box contain that work, work had to become kind of 'stuff'.
> 
> 
> TH
> 
> 
> 
> 



-- 
Odd Bodkin — Maker of fine toys, tools, tables

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#479684

Am 11.12.2018 um 19:24 schrieb Odd Bodkin:

>>>>>>
>>>>>> We have a similar problem with 'speed' and 'velocity'.
>>>>>>
>>>>>> Both have units like m/s , but one is a vector quantity (velocity) and
>>>>>> one is not (speed).
>>>>>>
>>>>>> Speed is the magnitude of velocity.
>>>>>>
>>>>>> https://examples.yourdictionary.com/examples-vector-scalar-quantity-physics.html
>>>>>>
>>>>>>> The acting force has direction.
>>>>>>> The path along which the force is acting has direction as well
>>>>>>> But the work as integral of scalar product of these 2 vectors is a
>>>>>>> scalar, without direction.
>>>>>>
>>>>>> work is NOT 'forth times path'
>>>>>>
>>>>>> but ' force times distance'.
>>>>>
>>>>> No, it is not that either. It is the integral of the scalar product between
>>>>> two vectors, the force and the differential displacement.
>>>>
>>>>
>>>> 'Work' is the name of a quantity in physics.
>>>
>>> As is velocity.
>>>
>>>>
>>>> It relates to the real world, hence is not a product of either one or
>>>> two vectors.
>>>
>>> I disagree. There are LOTS of real world quantities that are related in
>>> some way to other real world quantities. Velocity (a vector) is related to
>>> differential displacement and differential time. Momentum is related to
>>> mass and velocity.
>>
>> This is correct. But I had something against this statement:
>>
>> ".. It is the integral of the scalar product between two vectors..."
>>
>> I said: no, it is not, since a vector is a mathematical construct.
> 
> I disagree. A vector is a way to characterize a physical quantity. Velocity
> is a vector, momentum is a vector, force is a vector. All of those are
> real.

There is still a difference between 'vector' and 'physical quantity 
characterized by a vector'.

A vector is a mathematical construct, while quantities have units and a 
physical meaning.

You may characterize that quantity by an appropriate vector, but this 
does not make a vector a quantity.


>>
>> A number is also a mathematical construct. And you cannot multiply two
>> brainchilds and hope to get something real.
>>
>>
>> First statement: a quantity in physics (work) is defined as being the
>> product of one quantity (force)and some sort of length.
> 
> “the product” is too vague. Given that the two things you are multiplying
> are vectors, you have to be precise about what you mean by “product”. In
> this case “product” means scalar- or dot-product.


I was trying to show, that the product of a force and a path is not the 
right idea.

You are correct, since it is always done that way.

But it is wrong, in my opinion. So let me tell you, why work should be a 
vectorial quantity:


Force is a quantity (represented by a vector) with a direction (in space).

This vector contains enough information about the direction.

A path is actually a construct composed out of points in Euclidean 
space, which the object in question reaches at certain points in time.

First problem: the points are unimportant and do not belong into the 
calculation.

IOW: if you lift a heavy stone by one meter, the work should be equal, 
if you lift the same stone a little bit later at a different position 
(same gravity assumed).

So the location does not belong into the equation and also time 
(absolute time) does not belong there neither.

If we kick that out of the equation, we have got left 'distance' (in the 
direction of the movement), which is something we need.

This distance is a scalar.

If we multiply the scalar value of distance with a vector for force, 
then we get a good and satisfying result.


The benefit is, that the quantity would match to the usual understanding 
of the term 'work'.

It is also matched by a counterpart, similar to Newton's actio equals 
reactio.

The work is then done by the active part (in case of the stone: by you) 
and balanced by the passive part (the stone).


If work is a scalar quantity, you cannot easily do this,

Instead it would require to invent something like 'working stuff'.

But work is not a substance, but an activity (done by e.g. material 
objects).


...


TH

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#479689

Thomas Heger <ttt_heg@web.de> wrote:
> Am 11.12.2018 um 19:24 schrieb Odd Bodkin:
> 
>>>>>>> 
>>>>>>> We have a similar problem with 'speed' and 'velocity'.
>>>>>>> 
>>>>>>> Both have units like m/s , but one is a vector quantity (velocity) and
>>>>>>> one is not (speed).
>>>>>>> 
>>>>>>> Speed is the magnitude of velocity.
>>>>>>> 
>>>>>>> https://examples.yourdictionary.com/examples-vector-scalar-quantity-physics.html
>>>>>>> 
>>>>>>>> The acting force has direction.
>>>>>>>> The path along which the force is acting has direction as well
>>>>>>>> But the work as integral of scalar product of these 2 vectors is a
>>>>>>>> scalar, without direction.
>>>>>>> 
>>>>>>> work is NOT 'forth times path'
>>>>>>> 
>>>>>>> but ' force times distance'.
>>>>>> 
>>>>>> No, it is not that either. It is the integral of the scalar product between
>>>>>> two vectors, the force and the differential displacement.
>>>>> 
>>>>> 
>>>>> 'Work' is the name of a quantity in physics.
>>>> 
>>>> As is velocity.
>>>> 
>>>>> 
>>>>> It relates to the real world, hence is not a product of either one or
>>>>> two vectors.
>>>> 
>>>> I disagree. There are LOTS of real world quantities that are related in
>>>> some way to other real world quantities. Velocity (a vector) is related to
>>>> differential displacement and differential time. Momentum is related to
>>>> mass and velocity.
>>> 
>>> This is correct. But I had something against this statement:
>>> 
>>> ".. It is the integral of the scalar product between two vectors..."
>>> 
>>> I said: no, it is not, since a vector is a mathematical construct.
>> 
>> I disagree. A vector is a way to characterize a physical quantity. Velocity
>> is a vector, momentum is a vector, force is a vector. All of those are
>> real.
> 
> There is still a difference between 'vector' and 'physical quantity 
> characterized by a vector'.

Not in a physics discussion, normally. 
When I talk about velocity, I can refer to it as a vector and that means
the physical quantity characterized as a vector. The two are synonymous and
understood. Maybe not by you. 

> 
> A vector is a mathematical construct, while quantities have units and a 
> physical meaning.
> 
> You may characterize that quantity by an appropriate vector, but this 
> does not make a vector a quantity.
> 
> 
>>> 
>>> A number is also a mathematical construct. And you cannot multiply two
>>> brainchilds and hope to get something real.
>>> 
>>> 
>>> First statement: a quantity in physics (work) is defined as being the
>>> product of one quantity (force)and some sort of length.
>> 
>> “the product” is too vague. Given that the two things you are multiplying
>> are vectors, you have to be precise about what you mean by “product”. In
>> this case “product” means scalar- or dot-product.
> 
> 
> I was trying to show, that the product of a force and a path is not the 
> right idea.

If you want to redefine work, using the same term that has been used with a
particular understanding for centuries, good luck with that. 

So many amateur hacks come into this group with a lack of background
reading and understanding of basic concepts, and then they make mistakes
because the concepts do not mean what they think those mean. And then they
complain, “Well, it SHOULD mean what I think it means.” 

No, it should not. What SHOULD happen is that amateur hacks should educate
themselves on the meaning of basic concepts before discussing them. 

> 
> You are correct, since it is always done that way.
> 
> But it is wrong, in my opinion. So let me tell you, why work should be a 
> vectorial quantity:
> 
> 
> Force is a quantity (represented by a vector) with a direction (in space).
> 
> This vector contains enough information about the direction.
> 
> A path is actually a construct composed out of points in Euclidean 
> space, which the object in question reaches at certain points in time.
> 
> First problem: the points are unimportant and do not belong into the 
> calculation.
> 
> IOW: if you lift a heavy stone by one meter, the work should be equal, 
> if you lift the same stone a little bit later at a different position 
> (same gravity assumed).
> 
> So the location does not belong into the equation and also time 
> (absolute time) does not belong there neither.
> 
> If we kick that out of the equation, we have got left 'distance' (in the 
> direction of the movement), which is something we need.
> 
> This distance is a scalar.
> 
> If we multiply the scalar value of distance with a vector for force, 
> then we get a good and satisfying result.
> 
> 
> The benefit is, that the quantity would match to the usual understanding 
> of the term 'work'.
> 
> It is also matched by a counterpart, similar to Newton's actio equals 
> reactio.
> 
> The work is then done by the active part (in case of the stone: by you) 
> and balanced by the passive part (the stone).
> 
> 
> If work is a scalar quantity, you cannot easily do this,
> 
> Instead it would require to invent something like 'working stuff'.
> 
> But work is not a substance, but an activity (done by e.g. material 
> objects).
> 
> 
> ...
> 
> 
> TH
> 



-- 
Odd Bodkin — Maker of fine toys, tools, tables

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#479807

Am 12.12.2018 um 17:27 schrieb Odd Bodkin:

>>>>>>>> https://examples.yourdictionary.com/examples-vector-scalar-quantity-physics.html
>>>>>>>>
>>>>>>>>> The acting force has direction.
>>>>>>>>> The path along which the force is acting has direction as well
>>>>>>>>> But the work as integral of scalar product of these 2 vectors is a
>>>>>>>>> scalar, without direction.
>>>>>>>>
>>>>>>>> work is NOT 'forth times path'
>>>>>>>>
>>>>>>>> but ' force times distance'.
>>>>>>>
>>>>>>> No, it is not that either. It is the integral of the scalar product between
>>>>>>> two vectors, the force and the differential displacement.
>>>>>>
>>>>>>
>>>>>> 'Work' is the name of a quantity in physics.
>>>>>
>>>>> As is velocity.
>>>>>
>>>>>>
>>>>>> It relates to the real world, hence is not a product of either one or
>>>>>> two vectors.
>>>>>
>>>>> I disagree. There are LOTS of real world quantities that are related in
>>>>> some way to other real world quantities. Velocity (a vector) is related to
>>>>> differential displacement and differential time. Momentum is related to
>>>>> mass and velocity.
>>>>
>>>> This is correct. But I had something against this statement:
>>>>
>>>> ".. It is the integral of the scalar product between two vectors..."
>>>>
>>>> I said: no, it is not, since a vector is a mathematical construct.
>>>
>>> I disagree. A vector is a way to characterize a physical quantity. Velocity
>>> is a vector, momentum is a vector, force is a vector. All of those are
>>> real.
>>
>> There is still a difference between 'vector' and 'physical quantity
>> characterized by a vector'.
> 
> Not in a physics discussion, normally.

What you are apparently trying to say is, that if physicists chat at 
lunch brake or in a bar with a bear, they take the numbers for the thing 
and the vector for the quantity.

Well, maybe..

In my view there is a difference between a model and the system it tries 
to model. There is also a difference between the mathematical 
representation of some type of behavior and the mechanisms acting in 
that system.

The statement, that I disagreed with was, that work is a scalar, because 
it is a product of two vectors.

This is wrong, since it is not a product of two vectors, but a product 
of a physical quantity called 'force', which behaves like a vector, and 
something we are still debating.

You said, this second term should be a vector, too, while I wanted to 
use a scalar.

> When I talk about velocity, I can refer to it as a vector and that means
> the physical quantity characterized as a vector. The two are synonymous and
> understood. Maybe not by you.

Sure.

But where is the connection to the problem in question?

>>
>> A vector is a mathematical construct, while quantities have units and a
>> physical meaning.
>>
>> You may characterize that quantity by an appropriate vector, but this
>> does not make a vector a quantity.
>>
>>
>>>>
>>>> A number is also a mathematical construct. And you cannot multiply two
>>>> brainchilds and hope to get something real.
>>>>
>>>>
>>>> First statement: a quantity in physics (work) is defined as being the
>>>> product of one quantity (force)and some sort of length.
>>>
>>> “the product” is too vague. Given that the two things you are multiplying
>>> are vectors, you have to be precise about what you mean by “product”. In
>>> this case “product” means scalar- or dot-product.
>>
>>
>> I was trying to show, that the product of a force and a path is not the
>> right idea.
> 
> If you want to redefine work, using the same term that has been used with a
> particular understanding for centuries, good luck with that.

You must distinguish between the 'metaphysical' definition and the 
technical.

If a certain term is used, we need to talk about its meaning.

In case of 'work' we have something in mind. This would be somehow like 
"work is the quantity we have, if we lift a stone for a certain hight".

That is a little rough, but gives an idea about what we are actually 
trying to define.

A more technical definition would be like this:

work for lifting a stone in Earth gravitational field is equal to weight 
times additional hight (by what the stone is lifted).


This is still not very good, but defines, what we are trying to define: 
the physical quantity 'work'.


Now we should look, whether or not a certain definition would actually 
fit to what we had in mind.

'additional height' is not a vector and it is also not a path.

It is not a vector, because 'height' is meant as radius of the Earth 
surface at a certain point (where the thing to lift is right now).

The larger radius with additional height is actually the former radius 
multiplied by a scalar.

'additional height' is also not a path, because 'path' is meant as set 
of points, which are reached one after the other at certain times.

These points and the times, when they are passed are not interesting in 
this context and do not belong to the metaphysical construct 'work'.




...


TH

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#479808

Thomas Heger <ttt_heg@web.de> wrote:
> Am 12.12.2018 um 17:27 schrieb Odd Bodkin:
> 
>>>>>>>>> https://examples.yourdictionary.com/examples-vector-scalar-quantity-physics.html
>>>>>>>>> 
>>>>>>>>>> The acting force has direction.
>>>>>>>>>> The path along which the force is acting has direction as well
>>>>>>>>>> But the work as integral of scalar product of these 2 vectors is a
>>>>>>>>>> scalar, without direction.
>>>>>>>>> 
>>>>>>>>> work is NOT 'forth times path'
>>>>>>>>> 
>>>>>>>>> but ' force times distance'.
>>>>>>>> 
>>>>>>>> No, it is not that either. It is the integral of the scalar product between
>>>>>>>> two vectors, the force and the differential displacement.
>>>>>>> 
>>>>>>> 
>>>>>>> 'Work' is the name of a quantity in physics.
>>>>>> 
>>>>>> As is velocity.
>>>>>> 
>>>>>>> 
>>>>>>> It relates to the real world, hence is not a product of either one or
>>>>>>> two vectors.
>>>>>> 
>>>>>> I disagree. There are LOTS of real world quantities that are related in
>>>>>> some way to other real world quantities. Velocity (a vector) is related to
>>>>>> differential displacement and differential time. Momentum is related to
>>>>>> mass and velocity.
>>>>> 
>>>>> This is correct. But I had something against this statement:
>>>>> 
>>>>> ".. It is the integral of the scalar product between two vectors..."
>>>>> 
>>>>> I said: no, it is not, since a vector is a mathematical construct.
>>>> 
>>>> I disagree. A vector is a way to characterize a physical quantity. Velocity
>>>> is a vector, momentum is a vector, force is a vector. All of those are
>>>> real.
>>> 
>>> There is still a difference between 'vector' and 'physical quantity
>>> characterized by a vector'.
>> 
>> Not in a physics discussion, normally.
> 
> What you are apparently trying to say is, that if physicists chat at 
> lunch brake or in a bar with a bear, they take the numbers for the thing 
> and the vector for the quantity.
> 
> Well, maybe..
> 
> In my view there is a difference between a model and the system it tries 
> to model. There is also a difference between the mathematical 
> representation of some type of behavior and the mechanisms acting in 
> that system.
> 
> The statement, that I disagreed with was, that work is a scalar, because 
> it is a product of two vectors.
> 
> This is wrong, since it is not a product of two vectors, but a product 
> of a physical quantity called 'force', which behaves like a vector, and 
> something we are still debating.

The correct assessment is that you are debating it. This is in effect the
same thing as you saying that it is open to debate what “penguin” means and
that to you, a penguin is a large striped cat found in Asia. 

It isn’t really up to debate. The term “work” is defined clearly, has been
defined clearly for a couple of centuries, and that definition contrasts
with a meaning that you are saying makes sense to you. Surely you
understand that what makes sense to you is largely irrelevant in the
context of the definitions of basic concepts. Here, you are uneducated in
the meanings of basics concepts, and you are attempting to make a case
that, in the absence of education, what you come up just by “thinking about
it” in a vacuum of ignorance should nevertheless be considered seriously. I
disagree. Basic concepts are to be learned, not reinvented by “free
thinkers” such as yourself. 

> 
> You said, this second term should be a vector, too, while I wanted to 
> use a scalar.
> 
>> When I talk about velocity, I can refer to it as a vector and that means
>> the physical quantity characterized as a vector. The two are synonymous and
>> understood. Maybe not by you.
> 
> Sure.
> 
> But where is the connection to the problem in question?
> 
>>> 
>>> A vector is a mathematical construct, while quantities have units and a
>>> physical meaning.
>>> 
>>> You may characterize that quantity by an appropriate vector, but this
>>> does not make a vector a quantity.
>>> 
>>> 
>>>>> 
>>>>> A number is also a mathematical construct. And you cannot multiply two
>>>>> brainchilds and hope to get something real.
>>>>> 
>>>>> 
>>>>> First statement: a quantity in physics (work) is defined as being the
>>>>> product of one quantity (force)and some sort of length.
>>>> 
>>>> “the product” is too vague. Given that the two things you are multiplying
>>>> are vectors, you have to be precise about what you mean by “product”. In
>>>> this case “product” means scalar- or dot-product.
>>> 
>>> 
>>> I was trying to show, that the product of a force and a path is not the
>>> right idea.
>> 
>> If you want to redefine work, using the same term that has been used with a
>> particular understanding for centuries, good luck with that.
> 
> You must distinguish between the 'metaphysical' definition and the 
> technical.
> 
> If a certain term is used, we need to talk about its meaning.
> 
> In case of 'work' we have something in mind. This would be somehow like 
> "work is the quantity we have, if we lift a stone for a certain hight".
> 
> That is a little rough, but gives an idea about what we are actually 
> trying to define.
> 
> A more technical definition would be like this:
> 
> work for lifting a stone in Earth gravitational field is equal to weight 
> times additional hight (by what the stone is lifted).
> 
> 
> This is still not very good, but defines, what we are trying to define: 
> the physical quantity 'work'.
> 
> 
> Now we should look, whether or not a certain definition would actually 
> fit to what we had in mind.
> 
> 'additional height' is not a vector and it is also not a path.
> 
> It is not a vector, because 'height' is meant as radius of the Earth 
> surface at a certain point (where the thing to lift is right now).
> 
> The larger radius with additional height is actually the former radius 
> multiplied by a scalar.
> 
> 'additional height' is also not a path, because 'path' is meant as set 
> of points, which are reached one after the other at certain times.
> 
> These points and the times, when they are passed are not interesting in 
> this context and do not belong to the metaphysical construct 'work'.
> 
> 
> 
> 
> ...
> 
> 
> TH
> 



-- 
Odd Bodkin -- maker of fine toys, tools, tables

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#479859

Am 14.12.2018 um 15:28 schrieb Odd Bodkin:
...
>> The statement, that I disagreed with was, that work is a scalar, because
>> it is a product of two vectors.
>>
>> This is wrong, since it is not a product of two vectors, but a product
>> of a physical quantity called 'force', which behaves like a vector, and
>> something we are still debating.
> 
> The correct assessment is that you are debating it. This is in effect the
> same thing as you saying that it is open to debate what “penguin” means and
> that to you, a penguin is a large striped cat found in Asia.
> 
> It isn’t really up to debate. The term “work” is defined clearly, has been
> defined clearly for a couple of centuries, and that definition contrasts
> with a meaning that you are saying makes sense to you.

The concept of vectors is much younger than the concept of the quantity 
'work'.

So you are incorrect in your assumption, that work is regarded as 
product of two vectors since many centuries.


> .. Here, you are uneducated in
> the meanings of basics concepts, and you are attempting to make a case
> that, in the absence of education, what you come up just by “thinking about
> it” in a vacuum of ignorance should nevertheless be considered seriously. I
> disagree. Basic concepts are to be learned, not reinvented by “free


Of course I am 'educated' in this field.

But I wanted to express, that I do not agree to mainstream concepts 
about this term.

I have written, why I think so, but you have not written any reply to my 
arguments and are spitting out statements of unjustified arrogance.


TH

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#479860

Thomas Heger wrote:
> Am 14.12.2018 um 15:28 schrieb Odd Bodkin:
>> It isn’t really up to debate. The term “work” is defined clearly, has been
>> defined clearly for a couple of centuries, and that definition contrasts
>> with a meaning that you are saying makes sense to you.
> 
> The concept of vectors is much younger than the concept of the quantity 
> 'work'.

The *term* is younger, but that does not mean anything.

> So you are incorrect in your assumption, that work is regarded as 
> product of two vectors since many centuries.

The abstraction of a complex number as a pair of two numbers was introduced
by William Rowan Hamilton (1805–1865).  That was about two (“a couple of”)
centuries ago now.

The geometric concept of putting arrows atop each other is so basic that it
was known earlier already; it is in the “Mechanics” of Heron of Alexandria
(first century CE):

<http://www.math.mcgill.ca/labute/courses/133f03/VectorHistory.html>

(first Google hit for “vector history”)

>> .. Here, you are uneducated in the meanings of basics concepts, and you
>> are attempting to make a case that, in the absence of education, what
>> you come up just by “thinking about it” in a vacuum of ignorance should
>> nevertheless be considered seriously. I disagree. Basic concepts are to
>> be learned, not reinvented […]
>
> Of course I am 'educated' in this field.

Once again you are using your silly straight-apostrophe quotes correctly.

You are "educated" in that you do not have the slightest clue what you are
talking about.

-- 
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.

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#479887

Am 15.12.2018 um 13:52 schrieb Thomas 'PointedEars' Lahn:

>>> It isn’t really up to debate. The term “work” is defined clearly, has been
>>> defined clearly for a couple of centuries, and that definition contrasts
>>> with a meaning that you are saying makes sense to you.
>>
>> The concept of vectors is much younger than the concept of the quantity
>> 'work'.
> 
> The *term* is younger, but that does not mean anything.

Newton wrote in the 17th century and vectors came into fashion in the 
19th century.

https://www.researchgate.net/publication/244957729_A_History_of_Vector_Analysis


> 
>> So you are incorrect in your assumption, that work is regarded as
>> product of two vectors since many centuries.
> 
> The abstraction of a complex number as a pair of two numbers was introduced
> by William Rowan Hamilton (1805–1865).  That was about two (“a couple of”)
> centuries ago now.
> 

not quite. As far as I remember this was in the 1850th.

> The geometric concept of putting arrows atop each other is so basic that it
> was known earlier already; it is in the “Mechanics” of Heron of Alexandria
> (first century CE):
> 
> <http://www.math.mcgill.ca/labute/courses/133f03/VectorHistory.html>
> 
> (first Google hit for “vector history”)
> 
>>> .. Here, you are uneducated in the meanings of basics concepts, and you
>>> are attempting to make a case that, in the absence of education, what
>>> you come up just by “thinking about it” in a vacuum of ignorance should
>>> nevertheless be considered seriously. I disagree. Basic concepts are to
>>> be learned, not reinvented […]
>>
>> Of course I am 'educated' in this field.
> 
> Once again you are using your silly straight-apostrophe quotes correctly.

No.

In German we use an active term for learning ('lernen'), while 'being 
educated' is a passive phrase, which has no counterpart in German, 
except maybe in prisons ('belehrt werden').

The entire concept of 'education' is actually British and adopted by 
institutions in the USA.

TH

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#479896

Thomas Heger <ttt_heg@web.de> wrote:
> Am 15.12.2018 um 13:52 schrieb Thomas 'PointedEars' Lahn:
> 
>>>> It isn’t really up to debate. The term “work” is defined clearly, has been
>>>> defined clearly for a couple of centuries, and that definition contrasts
>>>> with a meaning that you are saying makes sense to you.
>>> 
>>> The concept of vectors is much younger than the concept of the quantity
>>> 'work'.
>> 
>> The *term* is younger, but that does not mean anything.
> 
> Newton wrote in the 17th century and vectors came into fashion in the 
> 19th century.

Newton offered no concept of work. Work was introduced in the 19th century.



> 
> https://www.researchgate.net/publication/244957729_A_History_of_Vector_Analysis
> 
> 
>> 
>>> So you are incorrect in your assumption, that work is regarded as
>>> product of two vectors since many centuries.
>> 
>> The abstraction of a complex number as a pair of two numbers was introduced
>> by William Rowan Hamilton (1805–1865).  That was about two (“a couple of”)
>> centuries ago now.
>> 
> 
> not quite. As far as I remember this was in the 1850th.
> 
>> The geometric concept of putting arrows atop each other is so basic that it
>> was known earlier already; it is in the “Mechanics” of Heron of Alexandria
>> (first century CE):
>> 
>> <http://www.math.mcgill.ca/labute/courses/133f03/VectorHistory.html>
>> 
>> (first Google hit for “vector history”)
>> 
>>>> .. Here, you are uneducated in the meanings of basics concepts, and you
>>>> are attempting to make a case that, in the absence of education, what
>>>> you come up just by “thinking about it” in a vacuum of ignorance should
>>>> nevertheless be considered seriously. I disagree. Basic concepts are to
>>>> be learned, not reinvented […]
>>> 
>>> Of course I am 'educated' in this field.
>> 
>> Once again you are using your silly straight-apostrophe quotes correctly.
> 
> No.
> 
> In German we use an active term for learning ('lernen'), while 'being 
> educated' is a passive phrase, which has no counterpart in German, 
> except maybe in prisons ('belehrt werden').
> 
> The entire concept of 'education' is actually British and adopted by 
> institutions in the USA.

Excuse me?

> 
> TH
> 



-- 
Odd Bodkin — Maker of fine toys, tools, tables

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#479897

Thomas Heger wrote:
> Am 15.12.2018 um 13:52 schrieb Thomas 'PointedEars' Lahn:
>>>> It isn’t really up to debate. The term “work” is defined clearly, has been
>>>> defined clearly for a couple of centuries, and that definition contrasts
>>>> with a meaning that you are saying makes sense to you.
>>>
>>> The concept of vectors is much younger than the concept of the quantity
>>> 'work'.
>> The *term* is younger, but that does not mean anything.
> 
> Newton wrote in the 17th century

He was prolific in the 17th *and* 18th century.

> and vectors came into fashion in the 19th century.

Which, AISB, *was* about 200 years ago (it is the 21st century now),
contrary to your claims.

But once again, when the *term* was first used is completely irrelevant.
The *concept* was known before.

In particular, as the referenced source says, Newton *used* the concept in
the «Principia», where he proved geometrically that Kepler’s Second Law of
Planetary Motion (put simply, “a planet moves in such a way that a line in
the *direction* *from* the Sun *to* the planet covers equal areas in equal
times”) holds if one assumes that there is a force *directed* *towards* the Sun.

This is conceptually no different from saying that the rate of change of the
area A covered by the radial vector r⃗ (pointing from the Sun to the planet)
is constant if the radial vector and the force vector F⃗ are (anti)parallel,
so that their vector product is zero:

    dA/dt = |r⃗ × dr⃗/dt|

  d²A/dt² = |dr⃗/dt × dr⃗/dt +  r⃗ × d²r⃗/dt²|

    [product rule of differentiation: (u v)' = u' v + u v']

          = |0⃗             +  r⃗ × a⃗  |
          = |                 r⃗ × F⃗∕m|
          = |(r⃗ × F⃗)∕m|
          = |0⃗/m|
          = |0⃗|
  d²A/dt² = 0.

    [Proof:

                (x₁, y₁, z₁) × (x₂, y₂, z₂)
             := (y₁ z₂ − y₂ z₁, z₁ x₂ − z₂ x₁, x₁ y₂ − x₂ y₁).

     If x₂ = x₁, y₂ = y₁, and z₂ = z₁, then

                (y₁ z₁ − y₁ z₁, z₁ x₁ − z₁ x₁, x₁ y₁ − x₁ y₁)
              = (0, 0, 0).

     Also, if x₁ and x₂ are arbitrary, and y₁ = z₁ = y₂ = z₂ = 0, then

                (x₁, y₁, z₁) × (x₂, y₂, z₂)
              = (0 · 0 − 0 · 0, 0 · x₂ − 0 · x₁, x₁ · 0 − x₂ · 0)
              = (0, 0, 0).

     Likewise for x₁ = x₂ = z₁ = z₂ = 0, and x₁ = y₁ = x₂ = y₂ = 0,
     respectively.

     NB: The magnitude of the vector product of two vectors a⃗ and b⃗ is equal
         to the area of the parallelogram produced by them: A = |a⃗ × b⃗|.

                 a⃗
            ---------->
           ^\ \ \ \ \ ^
        b⃗ /\ \ A \ \ / b⃗
         /\ \ \ \ \ /    __
        '---------->    |PE
             a⃗

     It should be obvious that if both vectors have the same direction
     (parallel) or opposite direction (antiparallel), the area of that
     parallelogram must be zero.  And only the null vector has
     magnitude zero: |0⃗| = |(0, 0, 0)| = √(0² + 0² + 0²) = 0.]

(If you do not understand this, then you should shut up about vectors NOW.)

In English:

[en]
<http://www.cornell.edu/video/richard-feynman-messenger-lecture-2-relation-mathematics-physics<

In German:

[de] <https://www.youtube.com/watch?v=SeORz7UV2wI>

> [pointless, clueless crackpot quibbling and blathering]

-- 
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.

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#479861

Thomas Heger <ttt_heg@web.de> wrote:
> Am 14.12.2018 um 15:28 schrieb Odd Bodkin:
> ...
>>> The statement, that I disagreed with was, that work is a scalar, because
>>> it is a product of two vectors.
>>> 
>>> This is wrong, since it is not a product of two vectors, but a product
>>> of a physical quantity called 'force', which behaves like a vector, and
>>> something we are still debating.
>> 
>> The correct assessment is that you are debating it. This is in effect the
>> same thing as you saying that it is open to debate what “penguin” means and
>> that to you, a penguin is a large striped cat found in Asia.
>> 
>> It isn’t really up to debate. The term “work” is defined clearly, has been
>> defined clearly for a couple of centuries, and that definition contrasts
>> with a meaning that you are saying makes sense to you.
> 
> The concept of vectors is much younger than the concept of the quantity 
> 'work'.

Both were introduced in the 19th century. I stand by my statement.

> 
> So you are incorrect in your assumption, that work is regarded as 
> product of two vectors since many centuries.
> 
> 
>> .. Here, you are uneducated in
>> the meanings of basics concepts, and you are attempting to make a case
>> that, in the absence of education, what you come up just by “thinking about
>> it” in a vacuum of ignorance should nevertheless be considered seriously. I
>> disagree. Basic concepts are to be learned, not reinvented by “free
> 
> 
> Of course I am 'educated' in this field.
> 
> But I wanted to express, that I do not agree to mainstream concepts 
> about this term.

Which is as irrelevant as disagreeing on the mainstream concept of mammal.
It is a concept established by DEFINITION, not by argument by appeal. 

> 
> I have written, why I think so, but you have not written any reply to my 
> arguments and are spitting out statements of unjustified arrogance.
> 
> 
> TH
> 



-- 
Odd Bodkin -- maker of fine toys, tools, tables

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#479776

Thomas Heger wrote:
> Am 11.12.2018 um 19:24 schrieb Odd Bodkin:
>>> ".. It is the integral of the scalar product between two vectors..."
>>>
>>> I said: no, it is not, since a vector is a mathematical construct.
>> I disagree. A vector is a way to characterize a physical quantity. Velocity
>> is a vector, momentum is a vector, force is a vector. All of those are
>> real.
> 
> There is still a difference between 'vector' and 'physical quantity 
> characterized by a vector'.

No, there is not.  All we can have are models about the world.

> A vector is a mathematical construct, while quantities have units and a 
> physical meaning.

False dichotomy.

If there is an object that is moving in a frame of reference at the same
time with 3 m/s in the x-direction, 4 m/s in the y-direction, and 0 m/s
in the z-direction, then it has the (relative) velocity

      ⎛3 m/s⎞   ⎛3⎞
  v⃗ = ⎜4 m/s⎟ = ⎜4⎟ m/s,
      ⎝0 m/s⎠   ⎝0⎠

and it has the (relative) speed

  v = |v⃗| = √((3² + 4² + 0²) m²/s²) = 5 m/s.

  ∂y/∂t
  ^
  :
  +       ^^
  :      / :
  +     /  :
  :  v⃗ /   :
  +   /    :
  :  /     :
  + /      :
  :/       :
  '--+--+->+-------> ∂x/∂t

> You may characterize that quantity by an appropriate vector, but this 
> does not make a vector a quantity.

Blithering fool, shut up until you have a minimum clue.

F’up2 where you belong

-- 
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.

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#479645 — [OT] Work (was: the famous formula)

Thomas Heger wrote:
> Am 10.12.2018 um 22:41 schrieb Odd Bodkin:
>> Thomas Heger <ttt_heg@web.de> wrote:
>>> It relates to the real world, hence is not a product of either one or
>>> two vectors.
>>
>> I disagree. There are LOTS of real world quantities that are related in
>> some way to other real world quantities. Velocity (a vector) is related to
>> differential displacement and differential time. Momentum is related to
>> mass and velocity.
> 
> This is correct. But I had something against this statement:

In English, that is “I objected to”.

> ".. It is the integral of the scalar product between two vectors..."
> 
> I said: no, it is not,

But it is.  Your ignorant claims do not change reality; they do not change
the fact that calculating the quantity of work this way actually works (no
pun intended).

In fact, for work to be a scalar quantity (which is the case, despite your
claims to the contrary and your contradicting yourself in the next posting),
it MUST be so that the factors of the product are vectors.

Because we do not calculate a vector (cross) product (×) here; only then
would the result be a vector.

If only one of the factors would NOT be a vector, then the ("dot") product
of them *would* be a vector, because multiplying a vector with a scalar
results in a vector that has all its components multiplied with the scalar:

              ⎛v_x⎞   ⎛a · v_x⎞
  a · v⃗ = a · ⎜v_y⎟ = ⎜a · v_y⎟.
              ⎝v_z⎠   ⎝a · v_z⎠

The same equation for work also works with scalars *only*, and *then* the
common product operation as well because, as I have already showed, the
scalar product in one dimension is equivalent to the common product, and
the magnitude of a vector with one component is simply the component itself:

                 ⎛Fₓ'⎞   ⎛drₓ'⎞
  F⃗'(r⃗) · dr⃗' =  ⎜ 0 ⎟ · ⎜ 0  ⎟ = Fₓ · drₓ + 0 · 0 + 0 · 0.
                 ⎝ 0 ⎠   ⎝ 0  ⎠

  |F⃗'(r⃗)|  = √(Fₓ'²  + 0² + 0²) = √(Fₓ'²)  = Fₓ'
  |dr⃗'|    = √(drₓ'² + 0² + 0²) = √(drₓ'²) = drₓ'.

But this – F⃗'(r⃗), dr⃗' − is *a special case* where the force and the
displacement vector are (anti-)parallel (e.g. they have *the same*
direction; and we rotate our coordinate system so that both are
[anti-]parallel to one of the coordinate axes).

The *general* solution is indeed

  W = ∫  F⃗(r⃗) · dr⃗
       C

for the total work done by a force field F⃗(r⃗) on a particle moving along a
curve C that is described by the tangent vector dr⃗ on it:

                            __
        C      F(r⃗) _   dr⃗  .':
         \         :\     .'
          \ __       \  .' `..__. r⃗₂
         .-'  '-.--._..'    _.-':
        r⃗₁          .'  _.-'
       ^            _.-'
      /         _.-'
  z  / y    _.-'
  ^ /.^ _.-'                      __
  :/'.-'                         |PE
  +---> x

As nice result this explains why the orbital speed of an object (e.g. a
planet,  “P”) in one point r⃗ of an orbit C (ignoring any friction,
precession, and gravitational radiation) is *exactly the same* when it
returns to the same point, even though the gravitational field of the
central body (e.g. Sol, “S”) has done work on it on the way: *in total* that
gravitational field did NOT do *any* work on it (this was a question in an
exercise in freshman physics):

          ___   <.
     .-'''   '''-. '.
   .               . '
  :           S---->P '
   '             r⃗ '
     '-...___...-'   __
                    |PE

    [Drawing proper ellipses in ASCII art is *hard*.]

                      r⃗
  W = ∫  F⃗(r⃗) · dr⃗  = ∫  F⃗(r⃗) · dr⃗ = 0.
       C              r⃗

    [You do not even have to do the integration or – you *really* do NOT
     want to do this – parametrize the ellipse.  Just remember the rule of
     thumb that the definite integral of a function yields the area under
     the curve of that function from the lower bounds of integration to
     the upper one?  *Of course* that area is zero if the bounds are
     *the same*, and so is the value of this integral:

       F⃗(r⃗)                __
        ^                 |PE
        :             W
        :    __     .'
        :..':/:`._.'
        :   :/:/.'`..--.-''
        :   :/://////:
        :   :/://////:
        '---+-+------+-----> r⃗
            r⃗₁       r⃗₂
      dr⃗ -->: :<--

    ]

> since a vector is a mathematical construct.
Like a curve/path/trajectory.  Like (your) Minkowski spacetime.  Like…

It is also a physical construct, because some (not all) physical quantities
have a magnitude *and* a direction.  This includes the infinitesimal segment
of the path of motion (dr⃗) and the force (F⃗), which are scalar-multiplied
here (also known as the “dot product” in English).

It is completely hopeless to try to understand physics without at least a
basic understanding of mathematics.  That includes linear algebra, so
vectors and the dot/scalar product.  Had you studied what you are talking
about, you would know this.

Good grief, even Wikipedia confirms it.

F’up2 sci.physics

-- 
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.

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#479376

Thomas Heger wrote:

> Am 08.12.2018 um 15:01 schrieb David (Kronos Prime) Fuller:
>> You make a couple of mistakes.
>>
>> First of all, energy is not positive-definite. I don’t know why you think it is.
>>
>> Secondly, a scalar can have either a positive or negative sign.
>> Being signed doesn’t make it a vector.
> 
> Vectors have a direction.

Yes.  Also they follow certain transformation rules, which scalars do not.

> A Joule for instance or a kilo-Watt-hour has no direction.

That is what I told you.

> Numbers often have signs, what does not turn them into vectors.

Then why have *you* claimed the opposite before?  Why have *you* claimed
that work (which can have the unit kWh) would be a vector?

-- 
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.

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#479486

Thomas Heger wrote:
> Am 07.12.2018 um 16:15 schrieb Thomas 'PointedEars' Lahn:
>> Thomas Heger wrote:
>>> Am 06.12.2018 um 22:47 schrieb Thomas 'PointedEars' Lahn:
>>>>        Then energy/work is
>>>>
>>>>          E = W = F s = m a s = m v/t s = m s/t² s = m s²/t² = m (s/t)² = m v².
>>>>
>>>>            The far right-hand side should ring a bell if you think of kinetic
>>>>            energy, the energy of motion: T = E_kin = 1/2 m v² (Newtonian).
>>>>
>>>>        c is the *speed* of light, so E₍₀₎ = m c² makes sense dimensionally.]
>>>
>>> Energy and work are not the same thing.
>>
>> Dimensionally they are exactly the same because of how energy is defined
>> (AISB in the part that you did not quote).
>>
>>> Energy is a scalar unit,
>>
>> Energy is a _quantity_, not a unit.
> 
> Well, yes. English is a foreign language for me and occasionally I make 
> mistakes.

English is not my native language either.  You make *ridiculous*, *beginner*
mistakes in virtually *all* scientific matters because you have not the
slightest clue what you are talking about.

The main reason for that is that you are either lazy or stupid.  You do not
want to understand or are unable to understand the underlying concepts.  So
you think that just using throwing around scientific terminology without
regard to its meaning would be enough.  Well, it is not.

>>> hence cannot become negative
>> But it can.
> 
> There is a difference between mathematics and the real world.

Says the person who cluelessly claims that a “skewed time axis” and “a
transformation of matter to radiation when objects are moving” would be
real.  Hilarious.

In the real world, Minkowski spacetime applies locally, and distances in it
can be negative.

You have repeatedly stated that Minkowski spacetime is the real and only
truth for you, so you cannot wind around the fact now that even according to
your own distorted idea of reality distances *can* become negative.

Of course, as you have no clue what you are talking about when talking about
Minkowski spacetime, too, you were completely unaware of that fact.

Hopeless case.

> [not even wrong]

-- 
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.

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#479241

Thomas Heger <ttt_heg@web.de> Wrote in message:
> 
> Energy is a scalar unit, hence cannot become negative and has no direction.

Being a scalar does not say it cannot be negative. 

Total energy is positive, indeed.

But e.g. the kinematic energy of a bound object is negative,
as it's kinetic energy is smaller than the energy needed
to break the bound.

> 
> Work is a vectorial unit and has a direction and can eventually become 
> negative

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.

> 
> But I personally think, the equation (E=m*c²) needs also 'deltas', since 
> 'mass lost' addresses a difference, for what you should use a delta.
> 
> 
dE/dm is not function of m,
therefore dE/dm=E/m=c^2

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

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


----Android NewsGroup Reader----
http://usenet.sinaapp.com/

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#479321

Am 07.12.2018 um 17:57 schrieb Libor Striz:
> Thomas Heger <ttt_heg@web.de> Wrote in message:
>>
>> Energy is a scalar unit, hence cannot become negative and has no direction.
> 
> Being a scalar does not say it cannot be negative.
> 
> Total energy is positive, indeed.
> 
> But e.g. the kinematic energy of a bound object is negative,
> as it's kinetic energy is smaller than the energy needed
> to break the bound.
> 
>>
>> Work is a vectorial unit and has a direction and can eventually become
>> negative
> 
> 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.

Sure. It is often done, but I do not agree.

The path along a line is not a distance, since a path is a function of time.

The reason why a distance is not a vector is this:
a vector has a direction, but a distance between two spot on the same 
line has no direction, since it is the same distance from A to B as from 
B to A.

Since force is directed, work should be a vector and the amount is then 
energy.

TH
>>
>> But I personally think, the equation (E=m*c²) needs also 'deltas', since
>> 'mass lost' addresses a difference, for what you should use a delta.
>>
>>
> dE/dm is not function of m,
> therefore dE/dm=E/m=c^2
> 

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