Re: It is not always c!

On Aug 17, 6:14 pm, "Gerald L. O'Barr" <glob...@xxxxxxxxx> wrote:
Subject: Re: It is not always c!

PD <TheDraperFam...@xxxxxxxxx> wrote:
Gerald L. O'Barr <globarr...@xxxxxxxxx> wrote:
Subject: Re: It is not always c!
globarr

<deletes by O'Barr>

O'Barr wrote:
If you say it is only an approximation, how about
showing it the exact way. What is the true and
exact way, versus my way, what ever you think that
is. Surely you can give a simple example.
PD wrote:
Sure. That's been given to you before.
The exact and correct way is (u+v)/(1+uv/c^2).
If you put in 30 mph and 60 mph, you'll see why the
approximation has been used for a long time.
If you put in 1.5E8 m/s and 0.8E8 m/s, you'll see
why the approximation isn't to be taken to be good
in all cases.

O'Barr comments:
Well, at least you have said something. But you
did not really say what you were doing, and what u
and v stood for, etc. And you did not explain why
you want to do this, and what requires you to do
this, and why it is wrong if you do not do this.

u and v are two velocities as measured in that frame.
I want to do this because it agrees with measurement.
You will see it works for 30 mph and 60 mph.
It also works for 1.5E8 m/s and 0.8E8 m/s.

All that I see you are doing is this:
I see you trying to jump frames.

No. It is the relative velocity as measured in that frame.

And the
signs do not make sense, if we were taking the
differences between two velocities.

Alright, then if you can't absorb the signs into the values of u and
v,
then use (u-v)/(1-uv/c^2).

And you really
did not define which is which, etc. It was really
not anything you have really applied to the specific
problem being discussed. But you did do something.

Now to repeat: If anyone did what you did (but
did it correctly, not as you just pushed it out) all
that would be done would be to change the measurement
to what a different frame would measure it to be.

No, that was not my intent. I'm using that to describe the relative
velocity between two objects as measured in a single frame. You'll
note that it agrees superbly with experiment in every single case.

Once if you change frames, then it will no longer be
the relative velocity, as measured in the frame that
I said was being used. So what you are doing is
violating the directions that had been given, you are
not finding the answer for the frame that was
specified. Why do you not give the answer that was
requested?
And why would you want to use a frame that was
not specified? Why do this, if your request was to
know what was measured in frame X, how could you not
use the data that was measured in frame X? Surely
you do know that the answer you are implying is not
the data that is actually measured in the frame
specified.

O'Barr wrote:
PD, before we go too far, I am going to say some
good things about Jeckyl. He is my friend. I
have no enemies. The ones who give me the most
trouble on
this net are my friends if they are being correct,
being correct at least in their own minds. Now
vector summations are perfect in SR,
PD wrote:
I have no idea what you mean by "perfect". They
don't correspond to any physically meaningful
quantity in the case of velocity, if both of the
vectors are nonzero, even in only one frame.

You can stay in that one frame, and you can perform
a vector subtraction of two velocities, and the
units are the same so you're not doing anything
stupid, and you'll come up with a number that is in
m/s or [L]/[T] or whatever so that it looks
something like a velocity, and you can even give a
name if you like by calling it "closing velocity" or
"departing velocity". That doesn't mean that it
corresponds to any measurable velocity in that frame
or for that matter any measurable velocity in any
other frame. Your definition of what constitutes
"perfect" is a little odd.

O'Barr comments:
Boy, this is getting weird. Are you trying to say
that in Einstein's train example, that c-v was not
the exact rate of speed that light was moving in the
train, as measured in the track frame?

Yes, exactly.

If it is not
exactly c - v, then relativity is not exact.

I don't think so. You need to reread what Einstein wrote.

It was
c - v that was used, and it had to be exactly c - v
if we are to get what Einstein got.

Nonsense.

QED!

PD wrote:
I can take the force applied tranversely to one end
of the bar and multiply it by the dimension of the
bar that is perpendicular to both the direction of
the force and the direction of the moment arm, and
it will have units of torque (even though it is not
the torque as commonly defined) and I can even give
it a name "transverse torque" which is perfectly
defined by the prescription I just gave of the
mathematical operation, and there won't be a thing
wrong with it. It just doesn't correspond to
anything physically useful.

O'Barr comments:
Well, I guess we all have our problems, but for
now, let us keep our problems to what is more simple,
just simple SR kinematics.

O'Barr wrote:
when they are done
in the same frame. Let us repeat again what a
measurement system is in SR. In SR, you put into
place a simple 3-D grid! This 3-D grid is
theoretically perfect. It consists of simple
rulers, which do not change. And in such a simple
3-D system, additions, subtractions,
multiplications, you name it, everything works.
PD wrote:
Except when you try to combine two velocities by
vector addition subtraction. Then it only works
approximately, and better when the velocities are
small.

O'Barr comments:
PD, I am not here to teach you SR. I should not
have to say this again: When you stay in the same
frame, vector additions/subtractions are exact. Only
when you are going from one frame to another do you
need to use your fancy equation.

I disagree. It works superbly inside a single frame.

[rest snipped]

PD

.

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