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The Necessity Alternate Theories

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  The Necessity Alternate Theories kefischer <emoneyjoe@iglou.com> - 2015-10-08 13:01 -0400
    Re: The Necessity Alternate Theories Natalina Nazvarova <natana42@hotmail.com> - 2015-10-08 17:09 +0000
    Re: The Necessity Alternate Theories Odd Bodkin <bodkinodd@gmail.com> - 2015-10-08 13:30 -0500
      Re: The Necessity Alternate Theories Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2015-10-08 22:45 +0200
        Re: The Necessity Alternate Theories Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2015-10-08 22:58 +0200
          Re: The Necessity Alternate Theories Natalina Nazvarova <natana42@hotmail.com> - 2015-10-08 21:56 +0000
            Re: The Necessity Alternate Theories Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2015-10-09 06:39 +0200
              Re: The Necessity Alternate Theories Natalina Nazvarova <natana42@hotmail.com> - 2015-10-09 10:59 +0000
                Re: The Necessity Alternate Theories Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2015-10-09 20:29 +0200
                  Re: The Necessity Alternate Theories Natalina Nazvarova <natana42@hotmail.com> - 2015-10-09 18:44 +0000
                    Re: The Necessity Alternate Theories Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2015-10-10 09:40 +0200
                      Re: The Necessity Alternate Theories Natalina Nazvarova <natana42@hotmail.com> - 2015-10-10 12:08 +0000
                        Re: The Necessity Alternate Theories paparios <paparios@gmail.com> - 2015-10-10 05:25 -0700
                          Re: The Necessity Alternate Theories Natalina Nazvarova <natana42@hotmail.com> - 2015-10-10 12:37 +0000
                          Re: The Necessity Alternate Theories Maciej Woźniak <mlwozniak@wp.pl> - 2015-10-10 14:54 +0200
                            Re: The Necessity Alternate Theories paparios <paparios@gmail.com> - 2015-10-10 06:02 -0700
                        Re: The Necessity Alternate Theories Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2015-10-10 17:49 +0200
                          Re: The Necessity Alternate Theories Natalina Nazvarova <natana42@hotmail.com> - 2015-10-10 16:08 +0000
                            Re: The Necessity Alternate Theories Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2015-10-10 18:13 +0200
                          Re: The Necessity Alternate Theories Natalina Nazvarova <natana42@hotmail.com> - 2015-10-10 16:16 +0000
                            Re: The Necessity Alternate Theories Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2015-10-10 19:08 +0200
                              Re: The Necessity Alternate Theories Natalina Nazvarova <natana42@hotmail.com> - 2015-10-10 18:02 +0000
                                Re: The Necessity Alternate Theories Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2015-10-10 20:17 +0200
                                  Re: The Necessity Alternate Theories Natalina Nazvarova <natana42@hotmail.com> - 2015-10-10 18:37 +0000
                                    Re: The Necessity Alternate Theories Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2015-10-10 20:54 +0200
                                      Re: The Necessity Alternate Theories Natalina Nazvarova <natana42@hotmail.com> - 2015-10-10 19:00 +0000
                                        Re: The Necessity Alternate Theories benj <nobody@gmail.com> - 2015-10-10 16:40 -0400
                  Re: The Necessity Alternate Theories underante <underante@yahoo.com> - 2015-10-09 15:31 -0700
    Re: The Necessity Alternate Theories kefischer <emoneyjoe@iglou.com> - 2015-10-09 02:42 -0400
      Re: The Necessity Alternate Theories Natalina Nazvarova <natana42@hotmail.com> - 2015-10-09 11:44 +0000
        Re: The Necessity Alternate Theories kefischer <emoneyjoe@iglou.com> - 2015-10-09 08:00 -0400

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#366490 — The Necessity Alternate Theories

The Necessity Alternate Theories


          Essentially all gravitational research now involves 
the same type of theory having the same anticipated type 
of effect.

         The type of theory involves action of some type
to cause objects to appear to change motion due to
gravity.

         And the type of effect involves an apparent
change in motion caused by another entity of some
kind.


       The problem that exists is the question,
"what if the apparent change of motion is NOT
a change of motion at all?".
       While General Relativity may consider all 
motion in space to be inertial, and all motion
in spacetime to be straight lines, there may
be the possibility that all motion is inertial AND
in straight lines.
       Should gravity research consider such a
vacuous possibility?     Only if an alternate
theory models it in a rational way.

      And the type of effect in that alternate theory
must be different enough to warrant consideration.


      The Divergent Matter model provides a
rational physics where objects moving in
space do not deviate from inertial motion
or straight lines.
      
      And they type of effect is totally different,
even though it still "appears" that gravity
works as Newton and Einstein described.


      The big differences are in astrophysics
and cosmology.      Many researchers still
"talk" of gravity being an attraction, and seem
to think that two objects move together under
the effect of gravity.
      In the Divergent Matter model this does
not happen, all objects simply continue their
motion, as inertial, and in straight lines, the
apparent effect of gravity is just "apparent".


     Even though some of these apparent
effects of Divergent Matter have been 
described, all responses have avoided
detailed discussion of the processes
involved.

      Any serious in-depth rebuttal will be 
appreciated, if it is the detailed physics
of the described physical processes and
their effects.

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#366492

kefischer wrote:

> The Necessity Alternate Theories 
>           Essentially all gravitational research now involves
> the same type of theory having the same anticipated type of effect.

Incomputable.

>          The type of theory involves action of some type
> to cause objects to appear to change motion due to gravity. 
>          And the type of effect involves an apparent
> change in motion caused by another entity of some kind.

It depends on the type of theory. In thermodynamics the theories are not 
predicting motion, other than that of the heat/density etc.

>        The problem that exists is the question,
> "what if the apparent change of motion is NOT a change of motion at
> all?".

You just broke my other bullshit-meter. Good bye.

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#366500

On 10/8/2015 12:01 PM, kefischer wrote:
>   While General Relativity may consider all
> motion in space to be inertial, and all motion
> in spacetime to be straight lines, there may
> be the possibility that all motion is inertial AND
> in straight lines.

Gobbledygook.

In general relativity, motion is inertial and worldlines are locally 
straight.

"Motion in spacetime" is an oxymoron.

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

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#366507

Odd Bodkin wrote:

> In general relativity, motion is inertial and worldlines are locally
> straight.

Hmmm.  I think it is only safe to say that in GR _spacetime_ is locally 
_flat_.

> "Motion in spacetime" is an oxymoron.

In light of current events, one cannot resist to exclaim:

“You’re just not thinking fourth-dimensionally!”

;-)

Motion is generally characterized by a change of coordinates in a suitable 
frame of reference.

The trivial motion in spacetime, where there are four coordinates, three 
spatial and one temporal, is to do nothing, thereby following along a time-
like world line of maximum proper time, a geodesic:

Then your 3 spatial coordinates in spacetime, x₁, x₂, and x₃ – your 
(3-)position – in a suitable frame of reference (e.g., the surface of Terra 
on which you are standing/sitting/lying), do not change, but your one other, 
temporal coordinate t changes continuously (which you can observe by looking 
at a working, co-moving watch/clock, and if you wait a really long time, by 
observing your hair and nails growing, and yourself aging).  (The proper 
time interval for uniform motion in spacetime is

   ∆τ = √((∆t²) – (∆x_i)²∕c²).

As a result of not moving spatially,

  ∆x_i = 0,

so that the proper time interval between arbitrary events on that curve is 
maximal.  [Using Landau/Lifshitz signature convention, and Einstein 
summation convention.])

See also:

<https://en.wikipedia.org/wiki/Four-velocity> (describes the rate of change 
of four-position, i.e. motion, in spacetime);

<https://en.wikipedia.org/wiki/World_line#Usage_in_physics> pp. (describes 
the [two] possibilities for motion in spacetime).

Also, STFW for peer-reviewed papers and physics books on motion in different 
kinds of space(-)time, such as 
<http://scitation.aip.org/content/aip/journal/jmp/56/3/10.1063/1.4913882> 
and 
<https://books.google.com/books?id=DCLUBwAAQBAJ&pg=PA14&dq=motion+%22in+spacetime%22&hl=de&sa=X&ved=0CDQQ6wEwA2oVChMInrXiss2zyAIViMcUCh0jHwbQ#v=onepage&q=motion%20%22in%20spacetime%22&f=false>


PointedEars
-- 
Q: What did the female magnet say to the male magnet?  
A: From the back, I found you repulsive, but from the front
   I find myself very attracted to you.
(from: WolframAlpha)

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#366510

Thomas 'PointedEars' Lahn wrote:

> […]  (The proper time interval for uniform motion in spacetime is
> 
>    ∆τ = √((∆t²) – (∆x_i)²∕c²).

Correction:

  ∆τ = √((∆t)² – (∆x_i)²∕c²).


PointedEars
-- 
Q: What did the female magnet say to the male magnet?  
A: From the back, I found you repulsive, but from the front
   I find myself very attracted to you.
(from: WolframAlpha)

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#366520

Thomas 'PointedEars' Lahn wrote:

> Thomas 'PointedEars' Lahn wrote:
>> […]  (The proper time interval for uniform motion in spacetime is
>> 
>>    ∆τ = √((∆t²) – (∆x_i)²∕c²).
> 
> Correction:
> 
>   ∆τ = √((∆t)² – (∆x_i)²∕c²).

Why should it be like that? I bet you dont even know what that /i/ stands 
for. Real life you MUST always specify that /i/ (plus all the others)

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#366539

The ’nym-shifting troll wrote as "Natalina Nazvarova":

> Thomas 'PointedEars' Lahn wrote:
>> Thomas 'PointedEars' Lahn wrote:
>>> […]  (The proper time interval for uniform motion in spacetime is
>>> 
>>>    ∆τ = √((∆t²) – (∆x_i)²∕c²).
>> 
>> Correction:
>> 
>>   ∆τ = √((∆t)² – (∆x_i)²∕c²).
> 
> Why should it be like that?

Because the metric of a space is in general

   ds² = g_µν dx^µ dx^ν

and in a flat (Minkowski) spacetime (3+1-space) the metric tensor can be 
written as (the matrix)

         (c²  0   0   0)
     g = (0  −1   0   0)
         (0   0  −1   0)
         (0   0   0  −1)

where one can define the indexes starting from 0, so that

  dx^0 = dt

from which follows

   ds² =  c² dt² − (dx^i)² =   c² dt² −  (dx^1)² − (dx^2)² − (dx^3)²

                           =   c² dt² − ((dx^1)² + (dx^2)² + (dx^3)²)

   ds                      = √(c² dt² − ((dx^1)² + (dx^2)² + (dx^3)²)))

And

  ∆τ = ∫ dτ = ∫ ds/c
       W
            = ∫ 1∕c ds

            = ∫ 1∕c √(      c² dt² − ((dx^1)²    + (dx^2)² + (dx^3)²)).

            = ∫     √(1∕c² (c² dt² − ((dx^1)²    + (dx^2)² + (dx^3)²))).

            = ∫     √(         dt² − ((dx^1)²∕c² + (dx^2)²∕c²
                                                 + (dx^3)²∕c²)).

along the world line W (“the proper time interval ∆τ between two events on a 
world line W equals the sum of infinitesimal proper time intervals dτ along 
W”).  And if the motion is uniform, then

         ∆τ = ∫  √(dt² − ((dx^1)²∕c² + (dx^2)²∕c² + (dx^3)²∕c²))

            =    √(∆t² − ((∆x_1)²∕c² + (∆x_2)²∕c² + (∆x_3)²∕c²))

            =    √(∆t² − (∆x_i)²∕c²). ∎

See also:

<http://mathworld.wolfram.com/MetricTensor.html>

and

<https://en.wikipedia.org/wiki/Proper_time#In_special_relativity> pp.

> I bet you dont even know what that /i/ stands for.
            ^^^^ YSCIB

In this notation, “i” is the 1-based index of the spatial coordinate.

You lose.

> Real life you MUST always specify that /i/ (plus all the others)

No, real life includes Einstein notation, also known as Einstein summation 
convention, where repeated indexes (here: i) are summed over:

  a_i b^i := ∑ a_i b^i := a₁ b^1 + a₂ b^2 + …
             i

(b^2 here does not mean „b-squared“ – that would be and is above 
b² accordingly –, but the second component of the contravariant tensor b, 
which in this example is a coordinate vector).  And so,

  (∆x_i)² := ∆x_i ∆x_i

          := ∆x₁ ∆x₁ + ∆x₂ ∆x₂ + ∆x₃ ∆x₃

           = (∆x₁)²  + (∆x₂)²  + (∆x₃)²

for 3 spatial dimensions.

Granted, this is not the trivial real *troll* life that you are living, 
where you are blathering about tensors without having the slightest clue 
about them; but this is *sci*.physics.relativity, so you will have to get 
used to it.

[1] cf. <https://youtu.be/h96SW0PfQcg?list=PLQrxduI9Pds1fm91Dmn8x1lo-O_kpZGk8&t=1016>


PointedEars
-- 
Q: Who's on the case when the electricity goes out?  
A: Sherlock Ohms.

(from: WolframAlpha)

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#366564

Thomas 'PointedEars' Lahn wrote:

> The ’nym-shifting troll wrote as "Natalina Nazvarova":
> 
>> Thomas 'PointedEars' Lahn wrote:
>>> Thomas 'PointedEars' Lahn wrote:
>>>> […]  (The proper time interval for uniform motion in spacetime is
>>>> 
>>>>    ∆τ = √((∆t²) – (∆x_i)²∕c²).
>>> 
>>> Correction:
>>> 
>>>   ∆τ = √((∆t)² – (∆x_i)²∕c²).
>> 
>> Why should it be like that?
> 
> Because the metric of a space is in general
> 
>    ds² = g_µν dx^µ dx^ν

What's this crap?? No tensors content inside the above equality I asked 
about, cretin.

> and in a flat (Minkowski) spacetime (3+1-space) the metric tensor can be
> written as (the matrix)

!! 

[snip loads of crap, misunderstood stuff stolen from whom knows where]

> See also:  <http://mathworld.wolfram.com/MetricTensor.html>

Ohh, I see... LOL

>> Real life you MUST always specify that /i/ (plus all the others)
> 
> No, real life includes Einstein notation, also known as Einstein
> summation convention, where repeated indexes (here: i) are summed over:

Has nothing to do with Einstein, pussycat.

[snip more crap]

> Granted, this is not the trivial real *troll* life that you are living,
> where you are blathering about tensors without having the slightest clue

Has nothing to do with tensors, idiot, you dont even know what you wrote. 
Here we go once again

    ∆τ = √((∆t)² – (∆x_i)²∕c²).

No tensors. Which seems to be consistent. As you, an almost half of an 
engineer, are discombobulated when it comes to the domain of tensors.

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#366616

The ’nym-shifting troll wrote as "Natalina Nazvarova":

> Thomas 'PointedEars' Lahn wrote:
>> The ’nym-shifting troll wrote as "Natalina Nazvarova":
>>> Thomas 'PointedEars' Lahn wrote:
>>>> Thomas 'PointedEars' Lahn wrote:
>>>>> […]  (The proper time interval for uniform motion in spacetime is
>>>>> 
>>>>>    ∆τ = √((∆t²) – (∆x_i)²∕c²).
>>>> 
>>>> Correction:
>>>> 
>>>>   ∆τ = √((∆t)² – (∆x_i)²∕c²).
>>> 
>>> Why should it be like that?
>> 
>> Because the metric of a space is in general
>> 
>>    ds² = g_µν dx^µ dx^ν
> 
> What's this crap??

It is the metric …

> No tensors content inside the above equality I asked about, cretin.

… that leads to the equality that I stated, if only you cared to read.

But you do not; you are not interested in the truth, not interested in 
learning, you are just a ‘nym-shifting troll, right?

>> Granted, this is not the trivial real *troll* life that you are living,
>> where you are blathering about tensors without having the slightest clue
> 
> Has nothing to do with tensors,

Yes, it has.  The equation is that way because of the metric of spacetime, 
and because of the signature of the metric tensor of spacetime.

> idiot,

Not pleased to meet you.

> you dont even know what you wrote.

I know.  I also know that either you have not read it, or not understood it, 
or you have both read and understood it and pretend that you did not because 
that would take away your twisted rationale for trolling.  In either case 
further reading your postings, let alone replying to you, appears to be a 
waste of time.  Please prove me wrong there.

> Here we go once again
> 
>     ∆τ = √((∆t)² – (∆x_i)²∕c²).
> 
> No tensors.

Actually, the x_i are the components of a tensor.  This is obvious to anyone 
who has the slightest idea what a tensor is.

> Which seems to be consistent.

Of course it is consistent.  You can read how one must arrive at this by 
reading what I wrote much more carefully than you did, if that.

> As you, an almost half of an engineer, are discombobulated when it comes
> to the domain of tensors.

I had not thought that the gibberish you present here as English could get 
more hilarious, but I was wrong.  So thanks for the laugh and the unlikely 
addition of “discombobulated” to my vocabulary.


PointedEars
-- 
Q: How many theoretical physicists specializing in general relativity
   does it take to change a light bulb?  
A: Two: one to hold the bulb and one to rotate the universe.
(from: WolframAlpha)

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#366618

Thomas 'PointedEars' Lahn wrote:

>>> Because the metric of a space is in general
>>>    ds² = g_µν dx^µ dx^ν
>> 
>> What's this crap??
> 
> It is the metric … 
> … that leads to the equality that I stated, if only you cared to read.

Hmm, let's watch how you come from the one to another, in small steps, so 
I/we can learn. Thanks, you are great. Don't bother, when you get stuck, 
we will ask some others. Just do it as far as you can.

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#366699

Natalina Nazvarova wrote:

> Thomas 'PointedEars' Lahn wrote:
>>>> Because the metric of a space is in general
>>>>    ds² = g_µν dx^µ dx^ν
>>> 
>>> What's this crap??
>> 
>> It is the metric …
>>> […]
>> … that leads to the equality that I stated, if only you cared to read.
> 
> Hmm, let's watch how you come from the one to another, in small steps, so
> I/we can learn.

Metric of a manifold:

  ds² = g_µν dx^µ dx^ν                                      (1)

    g_µν – (components of the) metric tensor

Metric tensor of Minkowski spacetime for events (x^0, x^1, x^2, x^3) using 
signature (+,−,−,−):

         (g₀₀  g₀₁  g₀₂  g₀₃)   (c²  0   0   0)
  g_µν = (g₁₀  g₁₁  g₁₂  g₁₃) = (0  −1   0   0)             (2)
         (g₂₀  g₁₁  g₂₂  g₂₃)   (0   0  −1   0)
         (g₃₀  g₃₁  g₃₂  g₃₃)   (0   0   0  −1)

Applying (2) in (1):

  ds² =   c²   dx^0 dx^0
        + (−1) dx^1 dx^1
        + (−1) dx^2 dx^2
        + (−1) dx^3 dx^3

  ds² = c² dx^0 dx^0 − dx^1 dx^1 − dx^2 dx^2 − dx^3 dx^3    (3)

Defining the time coordinate:

  dt := dx^0                                                (4)

Using (4) in (3):

  ds² = c² dt dt −  dx^1 dx^1 − dx^2 dx^2 − dx^3 dx^3

      = c² dt²   −  dx^1 dx^1 − dx^2 dx^2 − dx^3 dx^3

Writing the spatial part as a sum:

      = c² dt²   − (dx^1 dx^1 + dx^2 dx^2 + dx^3 dx^3)

                    3
      = c² dt²   −  ∑  dx^i dx^i
                   i=1

Using Einstein summation convention:

      =   c² dt² − dx^i dx^i

  ds² =   c² dt² − (dx^i)²

  ds  = √(c² dt² − (dx^i)²)                                 (5)

Defining proper time using an instantaneous rest frame:

      ds² = c² dτ² − (dx_τ^i)²

  dx_τ^i := 0

      ds² =  c² dτ²

          = (c  dτ)²

      ds  =  c  dτ

       dτ = ds∕c                                            (6)

Proper time interval between two events on world line W:

  ∆τ  = ∫ dτ
        W

Using (6):

      = ∫ ds∕c
        W

      = ∫ 1∕c ds
        W

Using (5):

      = ∫ 1∕c √(      c² dt² − (dx^i)²)
        W

      = ∫     √(1∕c² (c² dt² − (dx^i)²))
        W
 
  ∆τ  = ∫     √(         dt² − (dx^i)²∕c²)                  (7)
        W

Uniform motion:

   W : [a, b] ↦ ℝ^(1,3)

   W(a) = (t₁, x^1₁, x^2₁, x^3₁)
   W(b) = (t₂, x^1₂, x^2₂, x^3₂)

            b
   ∫ dt   = ∫  dt  = (t₂ − t₁)     =: ∆t                    (8)
   W        a

            b
   ∫ dx^i = ∫ dx^i = (x^i₂ − x^i₁) =: ∆x^i                  (9)
   W        a

Using (8) and (9) in (7):

  ∆τ  =       √(       (∆t)² − (∆x^i)²∕c²)         

Considering irrelevance of how tensors transform:

  ∆τ  = √((∆t)² − (∆x_i)²∕c²). ∎


PointedEars
-- 
A neutron walks into a bar and inquires how much a drink costs.
The bartender replies, "For you? No charge."

(from: WolframAlpha)

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#366717

Thomas 'PointedEars' Lahn wrote:

> Metric of a manifold:  ds² = g_µν dx^µ dx^ν        (1)

How come, explain in details, so morons like use can understand.

>     g_µν – (components of the) metric tensor

This must be the tensor, not its components. I am not the only moron 
around here.

>          (g₀₀   g₀₁   g₀₂  g₀₃)    (c²  0   0   0)
>   g_µν = (g₁₀  g₁₁  g₁₂  g₁₃) = (0  −1   0   0)             (2)
>          (g₂₀  g₁₁  g₂₂  g₂₃)   (0   0  −1   0)
>          (g₃₀  g₃₁  g₃₂  g₃₃)   (0   0   0  −1)

These must are the components. But does not imply that so many have to be 
zeroed. You could just use a diagonal matrix, in stead of the so many 
insignificant zeroes, and represent it simpler by a "vector" notation.

> Applying (2) in (1):
> ds² = c² dx^0 dx^0 + (−1) dx^1 dx^1 + (−1) dx^2 dx^2 + (−1) dx^3 dx^3
> ds² = c² dx^0 dx^0 − dx^1 dx^1 − dx^2 dx^2 − dx^3 dx^3    (3)

This must give a vector, not what you have here. You have (multiply) two 
vectors and a Matrice, which you call "metric".

> Defining the time coordinate:

We stop here till you clarify yourself adequately.

> (from: WolframAlpha)

I see.

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#366718

El sábado, 10 de octubre de 2015, 9:08:29 (UTC-3), Natalina Nazvarova  escribió:
> Thomas 'PointedEars' Lahn wrote:

> 
> I see.

No, you do not!

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#366719

paparios wrote:

> El sábado, 10 de octubre de 2015, 9:08:29 (UTC-3), Natalina Nazvarova 
> escribió:
>> Thomas 'PointedEars' Lahn wrote:
> 
> 
>> I see.
> 
> No, you do not!

Now I dont, right. Can you?

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#366720

Użytkownik "paparios"  napisał w wiadomości grup 
dyskusyjnych:16fd781a-ec63-47b3-a5e5-b79c26db57ac@googlegroups.com...

El sábado, 10 de octubre de 2015, 9:08:29 (UTC-3), Natalina Nazvarova 
escribió:
> Thomas 'PointedEars' Lahn wrote:

>
> I see.

|No, you do not!

No, she doesn't. Only relativistic morons - see!!! All others
are blind!!!! 

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#366721

El sábado, 10 de octubre de 2015, 9:54:34 (UTC-3), Maciej Woźniak  escribió:
> Użytkownik "paparios"  napisał w wiadomości grup 
> dyskusyjnych:16fd781a-ec63-47b3-a5e5-b79c26db57ac@googlegroups.com...
> 
> El sábado, 10 de octubre de 2015, 9:08:29 (UTC-3), Natalina Nazvarova 
> escribió:
> > Thomas 'PointedEars' Lahn wrote:
> 
> >
> > I see.
> 
> |No, you do not!
> 
> No, she doesn't. Only relativistic morons - see!!! All others
> are blind!!!!

For sure you are both blind AND dumb!

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#366728

Natalina Nazvarova wrote:

> Thomas 'PointedEars' Lahn wrote:
>> Metric of a manifold:  ds² = g_µν dx^µ dx^ν        (1)
> 
> How come, explain in details, so morons like use can understand.

Follow the references I give (not only in this thread), and STFW; read 
carefully; ask *specific* questions *where they are on-topic*.  I am neither 
able nor willing to give you a course in multi-dimensional differential 
geometry here.

Just a hint: “ds” is the line element of the manifold; it determines how 
distances in the manifold are measured.

>>     g_µν – (components of the) metric tensor
> 
> This must be the tensor, not its components.

No, unfortunately you still do not understand Einstein summation convention.
Actually, “g” is the metric tensor, the “g_µν” are its components, and “µ” 
and “ν” are variables for the indices of its components.

Connecting the dots for you again (now using “_00” instead of Unicode 
subscript “₀₀” etc.), for ν from 0 to 3, and for µ from 0 to 3 (since 
spacetime is a four-dimensional manifold):

  ds² =   g_µν dx^µ dx^ν

      =   g_00 dx^0 dx^0 + g_01 dx^0 dx^1 + g_02 dx^0 dx^2 + g_03 dx^0 dx^3
        + g_10 dx^1 dx^0 + g_11 dx^1 dx^1 + g_12 dx^1 dx^2 + g_13 dx^1 dx^3
        + g_20 dx^2 dx^0 + g_21 dx^2 dx^1 + g_22 dx^2 dx^2 + g_23 dx^2 dx^3
        + g_30 dx^3 dx^0 + g_31 dx^3 dx^1 + g_32 dx^3 dx^2 + g_33 dx^3 dx^3

> I am not the only moron around here.

That might be true, but different than you imply.

>>          (g₀₀   g₀₁   g₀₂  g₀₃)    (c²  0   0   0)
>>   g_µν = (g₁₀  g₁₁  g₁₂  g₁₃) = (0  −1   0   0)             (2)
>>          (g₂₀  g₁₁  g₂₂  g₂₃)   (0   0  −1   0)
                  ^^^
Typo; has to be g₂₁.

>>          (g₃₀  g₃₁  g₃₂  g₃₃)   (0   0   0  −1)
> 
> These must are the components.

On the near right-hand side of the equation you can see the tensor with its 
components and on the far right-hand side their corresponding values.  The 
far left-hand side is therefore maybe better written just “g” here to avoid 
confusion, but you can find both variants in the literature.  (The variable 
indexes of “g” indicate that it is a tensor, and how it transforms; they are 
required for expressing operations with the tensor exactly.)

> But does not imply that so many have to be zeroed.

But it does; that is how you write the metric tensor for Minkowski 
spacetime.  Otherwise the spacetime would not be a Minkowski one, it would 
not be flat (measuring distances would depend on the four-position).

And if you replace the components with their values in the expansion above, 
you can see that only the summands where µ = ν remain as the coefficients of 
those where µ ≠ ν are all 0.  (That is what multiplying with a diagonal 
matrix left-hand side does.)

Leaving only

  ds² =   g_µν dx^µ dx^ν

      =   g_00 dx^0 dx^0
                         + g_11 dx^1 dx^1
                                          + g_22 dx^2 dx^2
                                                           + g_33 dx^3 dx^3

Effectively, *exactly* as I have written in my previous follow-up.

> You could just use a diagonal matrix,

Obviously I already have.  The metric tensor of Minkowski spacetime *is* 
that matrix.

> in stead of the so many insignificant zeroes,

[I learned just now: The plural of the English numeral “zero” is _zeros_.  
“zeroes” is the third person singular of the verb “(to) zero” (to make sth. 
zero) instead:

<http://www.oxforddictionaries.com/definition/english/zero>
<http://www.oxforddictionaries.com/definition/american_english/zero>]

A diagonal matrix *has* zeros except in the main diagonal; therefore, too, 
the zeros are _not_ insignificant.  Which you could have seen if you had 
quoted me properly so that the columns of the matrices were still aligned.

<http://mathworld.wolfram.com/DiagonalMatrix.html>
<https://en.wikipedia.org/wiki/Diagonal_matrix>

> and represent it simpler by a "vector" notation.

Utter nonsense.

First of all, vectors are tensors of rank 1 (one index suffices to address a 
component), and matrices are tensors of rank 2 (two indexes are required for 
addressing a component).

<http://mathworld.wolfram.com/Tensor.html>
<https://en.wikipedia.org/wiki/tensor>

Second, tensor notation and Einstein summation notation are ways – so far 
the best ways, therefore *the* ways in relativity – of *avoiding* the 
cumbersome/tedious/impossible task of having to write vectors and matrices, 
and tensors of higher rank, with their components, either in rows and 
columns, or as sums (see below), explicitly.  As you can see above (count 
the lines or characters if you do not trust your eyes).

<http://mathworld.wolfram.com/EinsteinSummation.html>
<https://en.wikipedia.org/wiki/Einstein_notation>
 
>> Applying (2) in (1):
>> ds² = c² dx^0 dx^0 + (−1) dx^1 dx^1 + (−1) dx^2 dx^2 + (−1) dx^3 dx^3
>> ds² = c² dx^0 dx^0 − dx^1 dx^1 − dx^2 dx^2 − dx^3 dx^3    (3)
> 
> This must give a vector,

How good then that it does.

> not what you have shere.

You can write each vector as the sum of the products of its components and 
the corresponding orthonormal basis vectors of the coordinate system.

Homework assignment: Determine the orthonormal basis vectors of Minkowski 
spacetime for events given by the coordinates (t, x₁, x₂, x₃).

> You have (multiply) two vectors and a Matrice, which you call "metric".

The proper term is _matrix_.  Its plural is “matrices” or “matrixes”.

<http://mathworld.wolfram.com/Matrix.html>
<http://en.wikipedia.org/wiki/Matrix>
<http://en.wiktionary.org/wiki/matrix>
<http://www.oxforddictionaries.com/definition/english/matrix>

BTW, “metric” is not a term that I coined or introduced, and it is sensible 
one: it means “measurement” (of the distance between two points in a space); 
from Ancient Greek μέτρον ‎(métron) “measure”).

<http://mathworld.wolfram.com/Metric.html>
<https://en.wikipedia.org/wiki/Metric_tensor_(general_relativity)>

(Where there are differences, consider the first source the correct one, 
respectively.  Wikipedia articles may have been written/edited by interested 
laymen only.)

>> (from: WolframAlpha)
> 
> I see.

No, you don’t.


PointedEars
-- 
Q: What did the nuclear physicist order for lunch?  
A: Fission chips.

(from: WolframAlpha)

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#366731

Thomas 'PointedEars' Lahn wrote:

> BTW, “metric” is not a term that I coined or introduced, and it is
> sensible one: it means “measurement” (of the distance between two points
> in a space); from Ancient Greek μέτρον ‎(métron) “measure”).

You cant. To do that you need another metric, which you dont have yet. 
Generally, you CANNOT measure a distance without having a metric already 
in your hand.

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#366733

Natalina Nazvarova wrote:

> Thomas 'PointedEars' Lahn wrote:
>> BTW, “metric” is not a term that I coined or introduced, and it is
>> sensible one: it means “measurement” (of the distance between two points
>> in a space); from Ancient Greek μέτρον ‎(métron) “measure”).
> 
> You cant.

You _can’t_ spell.

> To do that you need another metric, which you dont have yet.

Utter nonsense.  A manifold has one metric only.

> Generally, you CANNOT measure a distance without having a metric already
> in your hand.

I have “a metric already in my hand”.  Minkowski "gave it to me".

You are not reverting to troll mode, are you?  That would be a shame.


PointedEars
-- 
A neutron walks into a bar and inquires how much a drink costs.
The bartender replies, "For you? No charge."

(from: WolframAlpha)

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#366734

Thomas 'PointedEars' Lahn wrote:

> Thomas 'PointedEars' Lahn wrote:
> 
>> Natalina Nazvarova wrote:
>>> But does not imply that so many have to be zeroed.
>> 
>> But it does; that is how you write the metric tensor for Minkowski
>> spacetime.  Otherwise the spacetime would not be a Minkowski one, it
>> would not be flat (measuring distances would depend on the
>> four-position).
>> 
>> And if you replace the components with their values in the expansion
>> above, you can see that only the summands where µ = ν remain as the
>> coefficients of those where µ ≠ ν are all 0.  (That is what multiplying
>> with a diagonal matrix left-hand side does.)
>> 
>> Leaving only
>> ds² = g_µν dx^µ dx^ν = g_00 dx^0 dx^0 + g_11 dx^1 dx^1
>> + g_22 dx^2 dx^2 + g_33 dx^3 dx^3

What is this??

> Sorry for the confusion, of course it is
> = g_00 dx^0 dx^0 − g_11 dx^1 dx^1 − g_22 dx^2 dx^2 − g_33 dx^3 dx^3.
> instead.

Which is still incorrect of course. Nevermind, a you postulated already, 
as that being "a vector" (without having the demanded UNIT vectors 
attached at it). Then your another blunder, that those terms should stand 
for "values". LOL.

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