Re: A little lesson for sqrt(144) year olds.

From: Jesper Pedersen (jesper_at_befunk.com)
Date: 07/03/04 

Date: Sat, 3 Jul 2004 17:05:10 +0200

"Androcles" <androc1es@nospamblueyonder.co.uk> wrote in message
news:TJyFc.2789$WY.24394841@news-text.cableinet.net...
>
<snip>
> | And I am familiar with Einstein's work on special
> | relativity.
>
> If that had any ring of truth to it, you might have recognised the passage
> I quoted.
>
>
> | Seems like you have a bit of learning to do though.
>
> Seems like you have a bit of lying to do.
>
>
> | "Einstein's
> | system k" for one makes no sense at all.
>
> For someone that cannot figure out which newsgroup he is posting to, I
doubt
> very much anything makes sense to you.
>
> | In which direction is it moving, IS
> | it moving at all, is it accelerated, etc? These are quite vital
specifics.
>
> Read Einstein, since you claim you are "familiar with Einstein's work on
> special relativity", but are clearly not.

For someone who continuously acuses me of arrogance and bad manners, you
sure do mount a lot of personal attacks.

> |
> | > So we have v > 0
> | > | and k moving with constant velocity v along the x-axis.
> | >
> | > Very good.
> |
> | Thank you.
> |
> | > | <snip intermediates>
> | > | > See that "IF" at the beginning?
> | > | > Does that mean
> | > | > xi really is equal to (x-vt)*gamma,
> | > | > or does it depend on the "If"?
> | > | > What IF we place x' = x+vt?
> | > | > Androcles
> | > |
> | > | If you choose x'=x+vt, then a point at rest in system k no longer
has
> | > values
> | > | x', y, z independent of time.
> | >
> | > What does it have values of, then?
> |
> | Well, x' is no longer independent of time, which should be painfully
> obvious
> | to anyone who has ever studied physics.
>
> You really are arrogant, aren't you?
>
> | Look at time t. The system has now
> | moved v*t along the positive x-axis.
>
> Well, well. So it has. How painfully obvious.
>
> | Thus a point at rest in k which was
> | originally at (x, y, z)=(x1, y1, z1) is now at (x, y, z)=(x1+v*t, y1,
z1).
> | The coordinate system of your choice,
>
> Einstein's choice.

No, Einstein's choice is x-v*t. You are the one going on about choosing
x+v*t instead.

> (x', y, z), then places this same
> | point in (x', y, z)=(x1+2v*t, y1, z1) which is by no means independent
of
> | time.
>
> Really?

Yes, really.

> | This is the reasoning behind choosing x'=x-vt, since this effectively
> | eliminates the time dependence of the x' coordinate.
>
> Explain "effectively". Here, I'll give you a diagram. One that should be
> painfully obvious to anyone who has ever studied physics.
> t0 x'------------x---x''-------
> t1 ----x'--------x-------x''---
> t2 ---------x'---x------------x''
> > vt >

How is it that you think that the translation of the x' and the x''
coordinate is in the same direction, even though they are given as x+/- v*t
respectively?

> Which is also
> t0 ---------x'------------x---x''
> t1 ---------x'--------x-------x''
> t2 ---------x'---x------------x''
> > vt >
> Now tell me how choosing x' = x-vt makes it "effectively" independent of
> time and chosing x'' = x+vt does not.

Because x' = x(t)-vt gives, since x(t)=x(0)+vt, that x' = x(0)+vt-vt = x(0),
which is time-independent. Obviously, choosing x'' = x+vt does not yield the
same result.

> | > | For this to be true you would have to invert
> | > | the direction of movement of k along the x-axis, meaning that you
new
> v'
> | > | = -v, in which case you end up with the same expression that you get
> | > | choosing x'=x-vt.
> | >
> | > Are you telling me that the RHS of the equation which you carefuly
> snipped
> | > to deliberately induce confusion doesn't change?
> |
> | I did not snip anything to induce confusion. If anything, I am trying
very
> | hard to avoid confusing you. But yes, I am saying that the entire
equation
> | stays the same.
>
> Really. Proof, please.
> ½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))"
> is the same as
> ½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c+v))"
> because
> tau(x',0,0,t+x'/(c+v)) = tau(x',0,0,t+x'/(c-v))
> Yes, I suppose it is if v = 0, which is trivial.
> Conclusion: special relativity is valid for all values of v that are equal
> to zero.

You misunderstood what I said. If you go through the deriviation, choosing
v<0 rather than v>0 and then x' = x+vt instead of x' = x-vt, then it is
quite obvious that you arrive at the same result.

> | > | Sounds like you have not understood the fundamental
> | > | reasoning behind the choice of x'=x-vt. It is not arbitrary, given
the
> | > | initial movement of k.
> | >
> | > Is that what is "sounds like"?
> | > It sounds to me as if you are guesser and a snipper, ignoring what you
> | have
> | > no answer for, and certainly you have not understood the fundamental
> | > reasoning behind (2/3 + 1/3)/2 = 1/3, or that cars may be driven
> forwards
> | as
> | > well as backwards, in which case (2/3 + 1/3)/2 = 2/3.
> |
> | I have throughout this "debate" tried to maintain a sober tone of
writing.
>
>
<snip personal stuff>
Let's keep this to a minimum.

> | to make it sure that
> | noone believed that you jumped even faster to your false conclusion.
>
> Well done. You've made it very clear that you consider Einstein's false
> conclusion was jumped to. Make sure you tell that to "noone" so they'll
> believe you when you say are "familiar with Einstein's work on special
> relativity".

No, I made it very clear that I believe that you jumped to a wrong
conclusion regarding the choice of x'.

<snip more personal stuff>
> | The notion of a non-time-dependent coordinate
> | system (x', y', z') = (x' - vt, y', z') is not something related solely
to
> | special relativity.
>
> It isn't related to anything at all.
> Since when did x' = x'-vt? When t = 0 or v = 0, I suppose. Heck, if you
> can't even get your notation right, it's little wonder nothing makes sense
> to you. Spelling mistakes I can tolerate, but that (X, y', z') = (X - vt,
> y', z') is a monumental gaff.

Yes. That was an obvious typo.

<snip remainding personal stuff>

I am not gonna take part in this discussion anymore, since it is clear that
it will get neither of us anywhere.

/ Jesper P

Relevant Pages

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