Re: origin of inertia

From: Bjoern Feuerbacher (feuerbac_at_thphys.uni-heidelberg.de)
Date: 03/16/05 

Date: Wed, 16 Mar 2005 12:49:28 +0100

Ken S. Tucker wrote:
> Bjoern Feuerbacher wrote:
>
>>Ken S. Tucker wrote:
>>
>>>Dirk Van de moortel wrote:
>>>
>>>
>>>><aleksandar.vukelja@gmail.com> wrote in message
>>>
>>>news:1110831894.432040.57570@z14g2000cwz.googlegroups.com...
>>>
>>>
>>>>>>But there is nothing to agree or disagree.
>>>>>>All you have been saying was wrong. I have given you
>>>>>>two formal proofs of your errors.
>>>>>>I am telling you how it works, because you clearly
>>>>>>haven't understood the very basics of what you think
>>>>>>you are talking about.
>>>>>>I am trying to *help* you to avoid embarrassing yourself.
>>>>>>Of course, it is up to you to accept the help or reject it.
>>>>>>If/when you change your mind, you can go back to my
>>>>>>previous replies. The best attitude, is to wonder and
>>>>>>ask questions where you don't understand something.
>>>>>
>>>>>Now this comment of yours goes beyond scientific discussion, and
>>>>>towards more usual dismissive bull on this group, which actually
>>>>>deserves no reply.
>>>>
>>>>But we can't reach a point of scientific discussion before
>>>>you have some idea what you are talking about.
>>>>I was trying to help you get some idea.
>>>>
>>>>Dirk Vdm
>>>
>>>
>>>I'm having a problem following the approach
>>>to this apparently simple problem.
>>>
>>>Would it be easier to agree the invariant
>>>
>>>ds^2 = g_uv dx^u dx^v
>>>
>>>is true and then in SR simplify it to,
>>>
>>>ds^2 = (cdt)^2 - dx^2
>>
>>Right.
>>
>>
>>
>>>When dx/dt = constant I think that will
>>>integrate to (c=1),
>>>
>>>s^2 = t^2 - x^2 = t'^2 - x'^2.
>>
>>Why on earth do you think one can get this from
>>integrating ds^2 = dt^2 - dx^2???
>
>
> It is true when ds=0,

Wrong in general.

ds=0 implies dt^2 = dx^2, which in turn implies after
taking the square root (paying attention to the two different
possible signs) dt = \pm dx. Integrating gives t = \pm x + k, with an
arbitrary constant k. So we get in general
   t^2 - (\pm x+k)^2 = 0
Your t^2 - x^2 = 0 is just a *very* special case and not the
general solution!

> and forces a sphere of light to appear that way in all CS's.
>
> Sorry, that integration was over simplified,
> Ken get's one ding-bat point, it's true only
> when s=0, which I should have specified.

s=0 does not help you in any way here.

[snip]

>>>Suppose one were to set
>>>
>>>dt' = dt/g where g is a constant then
>>>
>>>dx' = dx/g .
>>
>>And what has that to do with the question at hand?
>
>
> Integrate and find out.

I did. It indeed has nothing to do with the question at hand.

> These poor SOB's (Dirk) are still yakking in 1905 language.

1) They don't.
2) You haven't even understood that language, but nevertheless
try using more sophiscated language.

> BTW, don't forget to add some arbituary constants of
> integration.

Oh, I obviously would never have noticed that if you had not
told me!

Say, what do *you* think why I asked you why on earth you
thought that the integration works as you said originally?

[snip]

>>>Anyway that's how it's done in GR, I have
>>>refs if anyone wants.
>>
>>Please provide them. It will be interesting to see
>>what you have misunderstood this time.
>
>
> Of course, look up GR, hopefully that comic book
> you use will be available at U of Hide-a-beer,
> unless you guy's ran short of toilet paper again.

No reference? No surprise.

Stop trolling.

Bye,
Bjoern

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