Re: .Re: Why all the fascination with E = mc^2 ??
From: Old Man (nomail_at_nomail.net)
Date: 06/06/04
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Date: Sun, 6 Jun 2004 00:59:16 -0500
"Leonard Pardin" <leoppard@MailAndNews.com> wrote in message
news:d746a243.0406051614.4024f88c@posting.google.com...
> D.McAnally@i'm_a_gnu.uq.net.au (David McAnally) wrote in message
news:<c9stf3$l9e$1@bunyip.cc.uq.edu.au>...
> > leoppard@MailAndNews.com (Leonard Pardin) writes:
> >
> >
> > Einstein's First Postulate (the Principle of Relativity) states that the
> > physical laws relative to all inertial frames of reference are
identical.
> > There is no physical experiment that can distinguish between them. This
> > means that no inertial frame of reference is more special than any
other.
>
> If that's true, then the inertial frame that is stationary
> relative to the radiating body must be indistinguishable from any
> other frame. Yet, in that frame, no loss of mass is demonstrated.
>
> And that is the crux of our disagreement. Us common folk think of
> a reference frame as a place, but we can think of it as simply a
> coordinate system. As a practical matter,though, the coordinate
> system that follows the object to be analyzed may be the most
> important frame depending on what we are seeking.
>
> If I am on a ship in the ocean studying another distant ship, I may
> want to know his speed relative to my ship. In that case, the
> reference frame centered on my ship is the most important. If I want
> to know how fast the other ship is traveling relative to the water,
> the sea would be the most important frame.
>
> But if I want to know how fast one of the crew is running along
> the deck of the other ship relative to the other ship, the reference
> frame surrounding that other ship is the most important frame. In
> order to calculate that information, adjustments to my own frame of
> reference must be made. I must triangulate the other ship to get its
> size, and distance. Then I have compute the distance the crewman runs
> and time it. In other words, the information I am trying to calculate
> is what is actually happening in the other ship's frame of reference.
> But the ultimate answer, the true answer, that which is really
> happening on the other ship, is best measured by using the reference
> frame of that other ship.
>
> If our ships were space ships traveling some hefty percentage of
> the speed of light, the data collected on my own ship would be skewed,
> and adjustments would have to be made using the Lorentz equations. If
> all physical laws must be the same for all frames, then the final
> result of my calculations should be the same as the final result of
> the calculations made aboard the other ship. That's where the action
> is. If it isn't, then either my calculations are wrong or the laws of
> physics don't apply equally to all reference frames.
Idiot. The laws are invariant, but a velocity vector
isn't a law and isn't Lorentz invariant.
ds^2 = - c^2 dt^2 + dx^2 + dy*2 + dz^2
ds is invariant, but nothing on the RHS is. Of relevance
here, v_x = dx / dt isn't invariant. Go figure.
[Old Man]
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