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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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#479211

Thomas Heger wrote:
> Am 06.12.2018 um 19:49 schrieb RichD:
>> Everybody knows E = mc², and figures it's well understood.
>> I disagree.  It all depends on what '=' means.
>
> The equation is meant as quantitative description of the nuclear decay:

Not only.

> A radioactive sample looses mass as it sends out energy in the form of 

_loses_ (same pronunciation, different spelling, different meaning)

> rays and particles.
> 
> The numerical values of the measured quantities follow that equation.

Yes, that was Einstein’s original idea in 1905 (“Does the Inertia of a Body
Depend Upon Its Energy Content?”).  But that is not the end of it.

For example, the Sun produces energy (to sustain herself against her own
gravity, as a side effect spending heat to her surrounding planets) by
fusing nuclei because the resulting nucleus has less mass than the sum of
the masses of the progenitor nuclei (mass defect).  This energy is so-called
binding energy; it would be required to tear these nuclei apart again.

So, IOW, the Sun converts mass to energy.  Because they really two aspects
of the same thing, they are *equivalent* (hence the *equals* sign).

Likewise, when in nuclear *fission* reactors we *split nuclei (earlier than
they would do that naturally), we gain energy because (put simply, due to
the number of nucleons as compared to the fusion case) the mass of the
resulting nuclei is less than that of the progenitor nucleus.

> So: certain NUMBERS are equal.

_Quantities_, not numbers.  Were it only numbers, physics would be much
easier (and much more boring).

The quantity “energy” (E) literally has *units* of mass times speed squared
(among other ways).

    [You can understand this using Newtonian mechanics if you consider that
     one way to say it is that the energy of a physical system is a measure
     of the capability of that system to do work.

     Work W is force F applied over a distance s:

       W = F s.

     Force F is mass m moved with acceleration “a”:

       F = m a.

     Acceleration “a” is speed (change) v per time (change) t:

       a = v/t

     Speed v is distance s traveled per time t:

       v = s/t

     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.]

> 'Equal' means, that the number on the left side of the equation sign is 
> the same as on the right side, if all the algebraic calculations are 
> carried out on bot sides (and differences in sign are leveled out).

No, differences in sign are NOT "leveled out".

-- 
PointedEars

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

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#479228

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.

Energy is a scalar unit, hence cannot become negative and has no direction.

Work is a vectorial unit and has a direction and can eventually become 
negative
> 
>> 'Equal' means, that the number on the left side of the equation sign is
>> the same as on the right side, if all the algebraic calculations are
>> carried out on bot sides (and differences in sign are leveled out).
> 
> No, differences in sign are NOT "leveled out".
> 

Sure they are.

Since energy is positive and the LOSS of mass is negative, we need to 
add a negative sign somewhere.

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.


TH

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#479239

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.

> hence cannot become negative

But it can.

> and has no direction.

A matter of interpretation.  There are processes that transport energy in a
particular direction.

> Work is a vectorial unit

Nonsense.

> and has a direction and can eventually become negative

No, that the value of work can become negative is indicative instead of work
performed *on* a system as compared to work performed *by* a system.

This is only a *convention*.

For example, if you push a piston to reduce the volume of the gas in a
cylinder, then you perform work *on* that system, and the convention is
that then this work has a positive sign.

If instead the piston is pushed outwards, increasing the volume of the gas
in a cylinder, then work is performed *by* the system on the environment,
and it follows that this work must have a negative sign.

Thus, the first law of thermodynamics is

  dU = δQ + δW = δQ − p dV

with

  dU – infinitesimal change in inner energy
  δQ − infinitesimal change in heat
  δW – infinitesimal change in work/infinitesimal work performed
  p  – pressure
  dV – infinitesimal change in volume

    [One writes δX instead of dX to indicate that X is not a state variable
     but depends on the process performed; to get from δX to X you have to
     integrate over a path.]

>>> 'Equal' means, that the number on the left side of the equation sign is
>>> the same as on the right side, if all the algebraic calculations are
>>> carried out on bot sides (and differences in sign are leveled out).
>> No, differences in sign are NOT "leveled out".
> 
> Sure they are.

No, they are not.

> Since energy is positive and the LOSS of mass is negative, we need to 
> add a negative sign somewhere.

Milchmädchenrechnung.

> 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.

If you knew what you are talking about, you would know that Einstein already
basically suggested

  ∆m = ∆E∕c²

in the 1905 paper that you cited (he used words instead of writing an equation).

-- 
PointedEars

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

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#479322

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.

> 
>> hence cannot become negative
> 
> But it can.

There is a difference between mathematics and the real world.

If we want to regard physics as science, we must always prefer the 'real 
world perspective' (over the mathematical).

So a quantity in nature is called 'energy' and this cannot become 
negative (by definition of the term 'energy').

> 
>> and has no direction.
> 
> A matter of interpretation.  There are processes that transport energy in a
> particular direction.

The '*direction of transport' seems to correspond to 'arrow of time'.

So a positive direction of time should always be regarded as positive, 
hence the path determines the direction, where signs are positive.


Since you cannot create (or transport) something, that absorbs energy 
without a trace, you must regard the path of
something as positive and its energetic features also as positive.
> 
>> Work is a vectorial unit
> 
> Nonsense.
> 
>> and has a direction and can eventually become negative
> 
> No, that the value of work can become negative is indicative instead of work
> performed *on* a system as compared to work performed *by* a system.
> 
> This is only a *convention*.
> 
> For example, if you push a piston to reduce the volume of the gas in a
> cylinder, then you perform work *on* that system, and the convention is
> that then this work has a positive sign.
> 
> If instead the piston is pushed outwards, increasing the volume of the gas
> in a cylinder, then work is performed *by* the system on the environment,
> and it follows that this work must have a negative sign.

You address the problem of pulling a piston out and creating a lower 
pressure.

But this is - of course - also positive work.

If the piston pushes outwards against some kind of resistance, then we 
have also positive work, done by that piston.

Work is therefor positive, if seen from one side and negative, if seen 
from the other.

This is why work should be a vector, while energy is not.

..

TH

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#479325

Thomas Heger <ttt_heg@web.de> 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.
> 
>> 
>>> hence cannot become negative
>> 
>> But it can.
> 
> There is a difference between mathematics and the real world.
> 
> If we want to regard physics as science, we must always prefer the 'real 
> world perspective' (over the mathematical).
> 
> So a quantity in nature is called 'energy' and this cannot become 
> negative (by definition of the term 'energy').
> 
>> 
>>> and has no direction.
>> 
>> A matter of interpretation.  There are processes that transport energy in a
>> particular direction.
> 
> The '*direction of transport' seems to correspond to 'arrow of time'.
> 
> So a positive direction of time should always be regarded as positive, 
> hence the path determines the direction, where signs are positive.
> 
> 
> Since you cannot create (or transport) something, that absorbs energy 
> without a trace, you must regard the path of
> something as positive and its energetic features also as positive.
>> 
>>> Work is a vectorial unit
>> 
>> Nonsense.
>> 
>>> and has a direction and can eventually become negative
>> 
>> No, that the value of work can become negative is indicative instead of work
>> performed *on* a system as compared to work performed *by* a system.
>> 
>> This is only a *convention*.
>> 
>> For example, if you push a piston to reduce the volume of the gas in a
>> cylinder, then you perform work *on* that system, and the convention is
>> that then this work has a positive sign.
>> 
>> If instead the piston is pushed outwards, increasing the volume of the gas
>> in a cylinder, then work is performed *by* the system on the environment,
>> and it follows that this work must have a negative sign.
> 
> You address the problem of pulling a piston out and creating a lower 
> pressure.
> 
> But this is - of course - also positive work.
> 
> If the piston pushes outwards against some kind of resistance, then we 
> have also positive work, done by that piston.
> 
> Work is therefor positive, if seen from one side and negative, if seen 
> from the other.
> 
> This is why work should be a vector, while energy is not.
> 
> ..
> 
> TH
> 

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. 

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

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#479326

Odd Bodkin escrito 

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. 

Using “current mainstream Dimensionally accurate laws of physics”, The Entire universe is a giant violation of the law of conservation of Energy & all the laws of physics.

That’s why the mainstream is furiously groping in the dark for (dark energy & matter) to balance their error. 

To imply “Negative Energy as valid” is mysticism & witch doctoring 

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#479345

Am 08.12.2018 um 15:01 schrieb David (Kronos Prime) Fuller:
> Odd Bodkin escrito
> 
> 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.

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

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


> Using “current mainstream Dimensionally accurate laws of physics”, The Entire universe is a giant violation of the law of conservation of Energy & all the laws of physics.
> 
> That’s why the mainstream is furiously groping in the dark for (dark energy & matter) to balance their error.
> 
> To imply “Negative Energy as valid” is mysticism & witch doctoring
> 

Energy must be a scalar quantity and always positive.

You can - nevertheless - subtract energy, which is somehow similar to 
negative energy.

But in the real word there should be a thing or something similar, which 
has an attribute, that you could eventually call 'negative energy'.

Such an 'energy absorber' should absorb energy and add that to its own 
'negative energy' to nothing (zero).

But I have never heard of such a thing and actually cannot imagine one.

Sure, things can absorb heat, for instance. But then (positive) thermal 
energy is transferred from a donator to an absorber (or: from source to 
sink, if you prefer these terms).

There is still no negative energy involved.

TH

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#479358

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.

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.

What is a vector and does have direction is e.g an energy flow,
e.g power per square meter.


-- 
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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#479472

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'.

Since force is a vector quantity and distance a scalar quantity, work 
must be a vector quantity.
> 
> What is a vector and does have direction is e.g an energy flow,
> e.g power per square meter.


Energy flow has a direction and energy has not.

So work and energy are not the same thing.

They are related, but not equal.

TH

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#479482

Thomas Heger <ttt_heg@web.de> Wrote in message:
> 

> 
> We have a similar problem with 'speed' and 'velocity'.
> 
There is no problem with them,
but with people confusing both.
> 
> 
> work is NOT 'forth times path'
> 
> but ' force times distance'.

Well, that is definition of work from physics lessons
of the 8th grade of the basic school.

Students of a high school are taught
work is force times distance' times cosine of their angle.

Students at a collage or a university already know that magnitude
 and direction of force can be variable
and the path the force is acting along need not to be straight.

Therefore the integral definition of work along a generally curved
 path - the endpoint distance can be even zero 
- is the path integral of force
W = integral (path L)[ vec_F . vec_dL]
Where .  means scalar aka dot product of vector F and vector dL.

Imagine you are walking across a forest along a curved path.

You have always some direction 
of your motion ( vector dL)
and the length of your path
is  bigger
than distance of endpoints of your trip.



> 
> Since force is a vector quantity and distance a scalar quantity, work 
> must be a vector quantity.
>> 
>> What is a vector and does have direction is e.g an energy flow,
>> e.g power per square meter.
> 
> 
> Energy flow has a direction and energy has not.
> 
> So work and energy are not the same thing.
> 
> They are related, but not equal.
> 
> TH
> 
> 


-- 
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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#479541

Am 09.12.2018 um 20:24 schrieb Libor Striz:

>>
>> work is NOT 'forth times path'
>>
>> but ' force times distance'.
> 
> Well, that is definition of work from physics lessons
> of the 8th grade of the basic school.
> 
> Students of a high school are taught
> work is force times distance' times cosine of their angle.

Sorry, no.

The angle is meant as the angle between the applied force and the actual 
movement. Even if this is correct, the force in the equation

work =force times distance

is meant - of course - as the component of the force in the actual 
direction of movement.

> 
> Students at a collage or a university already know that magnitude
>   and direction of force can be variable
> and the path the force is acting along need not to be straight.
> 
> Therefore the integral definition of work along a generally curved
>   path - the endpoint distance can be even zero
> - is the path integral of force
> W = integral (path L)[ vec_F . vec_dL]
> Where .  means scalar aka dot product of vector F and vector dL.
> 
Now you provided a mathematical proof, that collages produce brainwashed 
zombies.


> Imagine you are walking across a forest along a curved path.

'Scary'...

So they scare the poor students, to believe in any nonsense they are 
supposed to swallow.



TH

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#479556

Thomas Heger wrote:

>> Well, that is definition of work from physics lessons of the 8th grade
>> of the basic school.
>> Students of a high school are taught work is force times distance'
>> times cosine of their angle.
> 
> Sorry, no.
> The angle is meant as the angle between the applied force and the actual
> movement. Even if this is correct, the force in the equation

Not the actual movement, but rather the next. As numerically it takes 
place in a loop piecewise. Thus it depends on the scheme used. Piece of 
cake, for the experienced. You guys are not even new beginners.

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#479562

Thomas Heger wrote:
> Am 09.12.2018 um 20:24 schrieb Libor Striz:
>>> work is NOT 'forth times path'
>>>
>>> but ' force times distance'.
>>
>> Well, that is definition of work from physics lessons
>> of the 8th grade of the basic school.
>>
>> Students of a high school are taught
>> work is force times distance' times cosine of their angle.
> 
> Sorry, no.

Yes they are, for all schools comparable to the U.S. American “high school”.

Probably you never went to one.

> The angle is meant as the angle between the applied force and the actual 
> movement.

“Movement” is not a physical quantity, you blithering fool.

The angle between the force vector (F⃗) and the distance vector (ds⃗, and
s⃗) is for motion on a straight line *equal* to the angle between the force
vector and the *velocity* (vector) v⃗, because both the distance vector and
the velocity vector are *tangent* vectors on the trajectory then.

Therefore (AISB), we *can* substitute

  W = ∫ F⃗ · ds⃗ = ∫ F⃗ · ds⃗/dt dt = ∫ F⃗ · v⃗ dt.

And for a constant force F⃗, we can eliminate the constant from the integral
(because adding products with the same factor is the same as adding the
other factors and then multiplying the sum with the common factor –
basically *Grundschul*-arithmetics):

  W = ∫ F · v⃗ dt = F ∫ ds⃗/dt dt = F r⃗.

For motion parallel to one axis:

  W = ∫ F · v dt = F ∫ ds/dt dt = F ∫ ds = F s.

    [See below for the proof.
     I have used “s” now instead of ”r” to make it more obvious.]

> Even if this is correct,

It is.

> the force in the equation
> 
> work =force times distance
> 
> is meant - of course - as the component of the force in the actual 
> direction of movement.

Which is why it is a vector, in general.

Only in the case where the motion is parallel to one of the coordinate axes
one can simplify and consider both it and the infinitesimal displacement a
scalar, because *then*

  F⃗ · ds⃗ = (F_x, 0, 0) · (ds_x, 0, 0) = F_x · ds_x + 0 · 0 + 0 · 0
         = F_x · ds_x = F ds

    [(e.g. for motion parallel to the x axis)]

Because the scalar product (·) is defined in 3 spatial Euclidean dimensions as

  a⃗ · b⃗ = (a_x, a_y, a_z) · (b_x, b_y, b_z) = a_x b_x + a_y b_y + a_z b_z

and

  a⃗ · b⃗ = |a⃗| · |b⃗| · cos(∡(a⃗, b⃗)) = |a⃗| · |b⃗| · cos φ,

with φ being the angle between a⃗ and b⃗ (*highschool* geometry).

Which is *precisely* what Libor Striz wrote above.

>> Students at a collage or a university already know that magnitude
>>   and direction of force can be variable
>> and the path the force is acting along need not to be straight.
>>
>> Therefore the integral definition of work along a generally curved
>>   path - the endpoint distance can be even zero
>> - is the path integral of force
>> W = integral (path L)[ vec_F . vec_dL]
>> Where .  means scalar aka dot product of vector F and vector dL.
>>
> Now you provided a mathematical proof, that collages produce brainwashed 
> zombies.

Blithering fool, I’ll remind you of that the next time you are cluelessly
babbling about the purely mathematical concept of Minkowski spacetime as
formulated by Minkowski in 1908.

-- 
PointedEars

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

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#479565

Thomas 'PointedEars' Lahn wrote:
> Only in the case where the motion is parallel to one of the coordinate axes
> one can simplify and consider both it and the infinitesimal displacement a
> scalar, because *then*
> 
>   F⃗ · ds⃗ = (F_x, 0, 0) · (ds_x, 0, 0) = F_x · ds_x + 0 · 0 + 0 · 0
>          = F_x · ds_x = F ds
> 
>     [(e.g. for motion parallel to the x axis)]

I should say: “… where the motion is in the direction of the force, and we
choose the orientation of the coordinate axes so that the direction of one
coincides with the direction of either quantity.”

  [One can see that this is equally true if only either
   the y- or the z-components of the quantities are non-zero.]

-- 
PointedEars

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

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#479685

Am 10.12.2018 um 22:27 schrieb Thomas 'PointedEars' Lahn:
>> Now you provided a mathematical proof, that collages produce brainwashed
>> zombies.
> Blithering fool, I’ll remind you of that the next time you are cluelessly
> babbling about the purely mathematical concept of Minkowski spacetime as
> formulated by Minkowski in 1908.

My assumption is, that Minkowski's spacetime is real.

The world we observe is internal structure of this real physical entity.


I call that concept 'structured spacetime'.


I have written kind of book about this idea, which you find here:

https://docs.google.com/present/view?id=dd8jz2tx_3gfzvqgd6


I had to change a few fundamental assumptions about our world, but the 
concept works quite well.

E.g. I try to promote a certain mathematical construct as good 
representation. That is a certain algebra, based on bi-quaternions.

(Those are also called 'complex four vectors'.)

My assumption is, that nature uses at a certain fundamental level 
methods similar to mathematics and uses that specific algebra.

This would build a field I call 'spacetime'.

By being somewhere an observer sees the internal structure from that 
point. This is then like a 'cut', that generates the world we see.

But this is only the world seen from that point, while infinitely many 
other points exist, from which other worlds could be seen.

TH

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#479751

Thomas Heger wrote:
> Am 10.12.2018 um 22:27 schrieb Thomas 'PointedEars' Lahn:
>>> Now you provided a mathematical proof, that collages produce brainwashed
>>> zombies.
>> Blithering fool, I’ll remind you of that the next time you are cluelessly
>> babbling about the purely mathematical concept of Minkowski spacetime as
>> formulated by Minkowski in 1908.
> 
> My assumption is, that Minkowski's spacetime is real.

Which is why your argument is inconsistent.

Minkowski spacetime is originally a *purely* mathematical construct, yet you
claim it is real.  (Moreover, you claim that it is real while special
relativity is not, ignoring that special relativity and Minkowski spacetime
are *one and the same thing*.)

Now, all observation indicates that Minkowski spacetime *is* real *locally*.

But *so is the definition of work* as it has been given to you.

*That* is the problem with your argument: You are not accepting *all*
observable reality, only the *part* of observable reality that you *like*
(because you are still able to understand it).

That is NOT how science, NOT how a *scientist* *to be taken seriously*,
are working.  Scientists do NOT pick and choose, they formulate their
conclusions based on *all* available information.

It is only crackpots, *wannabe* scientists, who pick and choose.

-- 
PointedEars

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

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#479484

Thomas Heger 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'.

We do not have a problem.  You have one.

Your problem is that you have not the slightest clue what you are talking
about, and that either you have an attention deficiency that lets you forget
what you said before – or you are simply a pathological liar:

,-<g6v2h6Fb0lhU1@mid.individual.net>
|
| […]
|
| Work is a vectorial unit and has a direction and can eventually become
| negative

[sic!]

-- 
PointedEars

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

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#479485

Thomas Heger <ttt_heg@web.de> Wrote in message:
> 
> 
> Energy flow has a direction and energy has not.

Neither the work has.

Read any textbook from a high school or better
how the term work  is defined on context of physics.
You will realize it is a scalar.

> So work and energy are not the same thing.

There is no so, 
as the latter cannot be implied from the former,
no matter if the latter is true or not.

> 
> They are related, but not equal.
> 
You confuse equal and identical.
Non identical quantities can still  be equal.
And the are not identical,
as work has 2 meanings:
1/ a phenomena of a process of transformation of energy to a
 mechanical energy (in narrow sense) or non-thermal energy ( wider
 sense, e.g electrochemical work )
2/ the amount of this transformed energy.


> 
> 


-- 
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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#479502

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. If the force
varies over location, such as with a spring, then the force cannot be
pulled out of the integral as a constant. If the displacement does not
follow a straight line, then the cosine of the angle between the two
vectors in the integrals is not constant, and again the answer may be not
simple. It is only in the case where the force is independent of the
position in the integral, and the displacement follows a straight line,
then the work is a simple scalar product between the force vector and the
displacement vector. And it is only in an even simpler case, when the force
vector is in the same direction as the displacement, that the work is force
times distance. 

You have incorrectly taken the extremely simple case to be the general
definition of work, and that’s a bad mistake to make. 

> 
> Since force is a vector quantity and distance a scalar quantity, work 
> must be a vector quantity.
>> 
>> What is a vector and does have direction is e.g an energy flow,
>> e.g power per square meter.
> 
> 
> Energy flow has a direction and energy has not.
> 
> So work and energy are not the same thing.
> 
> They are related, but not equal.
> 
> TH
> 
> 



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

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#479540

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.

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

'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.

What the hack is so difficult about that?

...


TH

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