"Max Keon" wrote in message
...

"George Dishman" wrote in message
ups.com...
Max, I'm going trim almost all your reply so we can
focus on the question of signed velocities. We can
come back to the other bits after that.

Max Keon wrote:
"George Dishman" wrote in message
oups.com...

.. Take Mercury's maximum
radial velocity relative to the Sun on its journey to and from
its perihelion, for example. v of course cannot be negative,
regardless of direction.
...
Now try your second equation which applies for inward
motion and hence v is negative. Let's take a similar toy
value of -0.01c:

What on earth is a negative velocity? A car moving toward me at
60km/h is not moving at -60km/h toward me. Nor can it be moving
at -60km/h away from me if it's going the other way. Wherever
it's going it is traveling at +60km/h.

That is the speed, not the velocity.

Why shouldn't speed and velocity be expected to abide by the same
rules? The dictionary doesn't specify any unique properties for
velocity that can't be applied to speed.

Your dictionary should have told you that velocity
includes direction. I explained the difference but
you snipped it:

Velocity is a vector and includes the direction so if
we think in terms of a car then we have a 2D situation
and we can define the velocity as north-south and
east-west components. For a car moving at 80km/h
going east, the velocity would be (0, 80). If it was going
west, it would be (0, -80). North is (80,0) while west is
(-80, 0). The velocity of a car moving with a speed of
80km/h in a north-easterly direction is (56.6, 56.6).

Speed is a scalar and is the magnitude of the velocity
is if an object is moving at (vn, ve) then its speed is
sqrt(vn^2 + vs^2).

This is going to be critical for resolving our disagreements
so please make sure you understand it. Note the velocities
are all shown as number pairs. Try examples in all four
quadrants to see where the components are negative and
positive:

http://www.phy.ntnu.edu.tw/ntnujava/viewtopic.php?t=68

If you don't have Java installed you can get it he

http://www.java.com/en/

This doesn't use Java:

http://www.glenbrook.k12.il.us/GBSSC...ors/u3l1b.html

If I'm following along at
a constant distance behind the car, the car's velocity is zero
relative to me. If I increase my speed by 10km/h, is the car
traveling at -10km/h ?

Relative to you yes. The distance between you is
reducing so to get a negative distance change in
a positive time you must have a negative relative
velocity. Of course both cars are still moving forward
so the first might be moving at 80km/h while the
second is doing 90km/h and the relative velocity
is 80-90 = -10.

I'll try to abide by your new set of rules and use a + - switch
to identify velocity direction. I should still get the same
answer though.

You will if you stick with just one of your equations,
the changed sign inherent in using the standard
convention means you don't need to swap between c+v
and c-v.

How can anyone possibly have any confidence in
mathematics if it's going to give the conflicting result that
you're proposing?

This time I'll adopt your "toy value" system.
G = 1: M = 1: r = 1: c = 1.
# = (c+v) * -(G*M/r^2)+(G*M/r^2)
Note that -(G*M/r^2) = -1 and (G*M/r^2) = 1.

For v = .01 (c+v=1.01)
# = 1.01 * -1 + 1 = -.01

That is negative so indicates the anisotropic force is
towards the Sun.

For v = -.01 (c+v=.99)
# = .99 * -1 + 1 = .01

That is positive so indicates the anisotropic force is
away from the Sun. I believe those answers are both
what you expected.

Since you didn't include the + (G*M/r^2) in your analysis, while
the difference between the two results is still the same, they
are both negative. # = 1.01 * -1 = -1.01
# = .99 * -1 = -.99

Yes, we tend to lay aside the normal gravity when
discussing your slight change. Overall for a positive
velocity indicating motion away from the Sun at 0.01c
the total pull towards the Sun is increased to 101% of
the usual Newtonian value while for an object moving
towards the Sun at 0.01c the pull is reduced to 99%.
Correct me if I'm wrong but I thought that was what
you expected.

If you don't get the same answer when you start shifting numbers
about, you've done something wrong. It's hardly likely that the
math will be at fault.

Indeed but my 'shifting about' always gave the same
answers as your equations.

((c+v)^2/c^2)^.5*G*M/r^2-(G*M/r^2)

always gives the same answers as

-(G*M/r^2) * (-v/c)

while

((c-v)^2/c^2)^.5*G*M/r^2-(G*M/r^2)

always gives the same answers as

-(G*M/r^2) * (+v/c)

The rules are mostly he

http://www.mathsisfun.com/associativ...tributive.html

but also (x^a)^b = x^(a*b) so (x^2)^0.5 = x^1 = x or simply
the square root of the square of x is just x.

The problem with the direction was something quite different,
you didn't realise physics uses signed velocities and that
change of sign means you don't need to use two different
equations, a single equation does the job.

George