Re: Are *observed* SR effects real?



On Aug 4, 2:15 pm, mluttg...@xxxxxxxxxx wrote:

Before this string gets too befouled by Wilson and Potter, I want to
ask you if you have further questions before we move on to units.

Let's move on!

Wonderful.

So let's go back to those surveyors and the discovery that there is an
invariant
x'^2 + y'^2 = x^2 + y^2.
Now (again borrowing from Taylor and Wheeler), suppose that the
surveyors had had this long-standing rule from antiquity that east-
west distances are measured in miles and north-sound distances are
measured in kilometers. (Yes, I know it's stupid to do so, but they
had certain things they had used for a long time, like "northish rate"
which was a measure of how far you were going north for every so much
that you went east. It was a good analog for what we would call a
bearing angle or a heading today.)
Then the formula above still works, but you have to either convert
miles to kilometers to miles before you can add the x and y pieces.
So, including the conversion factor b from miles to kilometers, the
expression above becomes
x'^2 + (y'/b)^2 = x^2 + (y/b)^2.
Or of course you could change the x value to kilometers rather than
the y value to miles. Either will work:
(bx')^2 + y'^2 = (xb)^2 + y^2.

"But," you might ask, "Why do this at all? Why not just settle on
measuring all x and y distances in the same units?" It seems like a
reasonable question, and the surveyors splutter that this is the way
they've always measured east-west and north-south distances, and there
are even separate east-west and north-south *standards* in some
standards bureau someplace, and besides, if you did that, then the
"northish rate" would become unitless and who can make sense of a
unitless "northish rate"?

Whatever. In any case, the b doesn't add any physics. It's just a
*conversion factor* to change kilometers to miles or back again. It's
purely an artifact of the units chosen to measure x and y.

Now, if you're with me at all, it probably isn't lost on you that the
same identical situation is present in the expression for the
spacetime interval.
t'2 - (x'/c)^2 = t^2 - (x/c)^2
The presence of the factor of c is *purely* due to the choice of
measuring t and x with different units, and accordingly, c is just a
conversion factor that takes seconds to meters or vice versa.

If we chose more sensible units, we would decide on one or the other
-- say, meters for time t and meters for distance x; or alternatively,
seconds for time t and seconds for distance x. It's not hard to
convert 2316 m to seconds. Just use the conversion factor c = 3.00E8 m/
s. If we use these sensible units to begin with, then the expression
for the invariant becomes much more sensible:
t'^2 - x'^2 = t^2 - x^2.

This is why I hedged when you said that the spacetime interval is a
time. I wasn't sure how you were guessing that, but took a shot that
you were looking at the units. But in sensible units, time and space
have the *same* units. So it doesn't make much sense to say it is one
or the other.

In this system of units (called "natural units" if you want to look it
up), the value of c is 1 and it is unitless, and all material things
travel with a speed 0<=v<1.

Questions? We're about ready to wrap up the basics lesson.

But just as a note about the exercise, I did for fun an extract of
just the long text messages where I explained this and that, leaving
out all the questions and answers and side-trips and quoted passages,
and it still comes up with over 20 pages of material, and almost a
month of conversation. In contrast, in a 45-contact-hour university
course on relativity, all of this would have been done in 2 or 3 1-
hour class sessions, certainly less than a week. Moreover, it would
have been one evening's reading in a textbook. The point is plain:
While it is possible to learn this stuff for free on the internet, it
is *highly* inefficient in comparison with other venues, and on top of
that you have to slog through piles and piles of noise.

PD

PD
.



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