Re: Are *observed* SR effects real?

Uncle Ben wrote:
On Aug 4, 6:44 pm, "Spaceman" <space...@xxxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:
PD wrote:
On Aug 4, 5:02 pm, "Spaceman" <space...@xxxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:
PD wrote:
On Aug 4, 3:40 pm, "Spaceman"
<space...@xxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
PD wrote:
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

I gather you actually mean
t'^2 and not t'2

Yes.

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.

But here is the problem PD,
c is not a conversion factor for such and a meter is a distance
that should not be "converted" with time to another distance.
It is a speed.
A conversion factor does not include the unit being converted
inside the factoring itself.

Of course it does. To convert 8 quarts to gallons, you use the
conversion factor 1/4 gallon/quart.

Devide the gallon by (4) or multiple the quart times (4)
the (4) is unitless.
The (4) contains nothing but a number that has nothing in it
about the quart, nor the gallon.

I see you do not remember your 6th grade math. And you just had it
last spring! And I even just pointed you to 6th grade math sites to
refresh your memory, and your head is as wiped clean as a brand new
bedpan!

If I divide (8 quarts) by (4 no units), I get (2 quarts), which is
obviously not right.

Poor PD,
He does not know how to convert to a different measurement
now. he thinks deviding quarts by 4 would give you quarts again
instead of the gallons you were trying to "convert" to by using
the unitless 4.
Sheesh
You use a unitless 4 incorrectly above and you are too ignorant and
too stupid
to understand that c is not unitless like that 4 is.

So,
You can keep thinking the c does not have the distance or the time
built in and look like the fool you are presenting yourself to be,
and keep the silly 5th grade etc remarks going, or you could
actually think about what the unitless 4 has for quarts or gallons
in it?

Yup. And the 6th grade math teacher should think about it too,
right?

The 6th grade math teacher should smack you in the head for
thinking that c (a speed) is unitless at all you stupid ass.

--
James M Driscoll Jr
Creator of the Clock Malfunction Theory
Spaceman- Hide quoted text -

- Show quoted text -

When you program your computer in some language like Fortran, and x
represents some volume of gasoline, what type of number is x in your
program? Probably a floating point number, and not a new kind of
number called a "gallon number."

In other words, dimensional anaysis is a optional tool, not a
necessary procedure. (I once heard of a physics student so entranced
with dimensional analysis that he denied the reality of x^4, because
x^3 represents a cube in space and there is no fourth spatial
dimension.)

Ben,
c is not unitless,
c is not dimensionless.
c is also not even a constant to all frames
just as "negative" is not negative to all frames.
I think you might want to stop backing up the nonsense
that ignores relativity just to support it.
Which way is negative to all frames of reference Ben?
No hypothesis can be correct to all frames so don't even
try to bull that bull*** either unless you can prove
nothing is "relative".
:)

--
James M Driscoll Jr
Creator of the Clock Malfunction Theory
Spaceman


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