[geocentrism] Re: Angular momentum
- From: Mike <mboyd@xxxxxxxxxxxx>
- To: geocentrism@xxxxxxxxxxxxx
- Date: Wed, 27 Oct 2004 10:25:42 +0100
Dear Neville,
> Mike <mboyd@xxxxxxxxxxxx> wrote:
>
>> "This is incorrect. Fot (u1-v1) to be -ve all that is required is
>> that u1 < v1 which will be true for the particle that gained velocity
>> in the collision."
>
> The particles, 1 and 2, are as labelled in the diagram (Fig. 1 on
> "The laws of physics," on www.midclyth.supanet.com ). Particle 1
> cannot go through Particle 2, and hence v1 cannot be greater than u1.
Well in that case you have made your abritrary choice and then get the
inequality you derived, and there is nothing preposterous about it. If
you *swapped the labels over* in your diagram then *u1 would be less
than v1*. Your statement above completely contradicts your claim in the
conclusion that the choice of labelling was arbitrary.
> The rest of the posting is irrelevant (and also keeps referring to
> Newton's experimental "law," though not by name, which I have already
> denied).
This "law" you keep trying to bring in is a strawman. Either you make
your hidden assumption explicit, or you change the '>' to a '<>' in your
result, or you have an incorrect mathematical step in your proof. In
all three cases your conclusion doesn't hold. I'll try again with a
slightly different approach, I hope you won't just claim this is
irrelevant this time - all I am doing here is algebraically manipulating
your proof!
Let:
a = (u1 - v1)
b = (u2 - v2)
c = (u1 + v1)
d = (u2 + v2)
then in your proof
(u1 - v1) + (u2 - v2) = 0
(u1 + v1)(u1 - v1) + (u2 + v2)(u2 - v2) > 0
=> (u1 + v1) - (u2 + v2) > 0
ie (u1 + v1) > (u2 + v2)
becomes
a + b = 0
ca + db > 0
=> c - d > 0
ie c > d
so:
a = -b
ca + db > 0
does it follow that
c - d > 0 (i.e. c > d)
?
No! Either a or b must be negative (if they are not zero). The above
step assumed b was negative. If we assume instead that a was negative
then we would have got
c - d < 0 (i.e. c < d)
If a and b could have been labelled either way round and we don't know
which way round it is (which we don't in general) then the correct
result is:
c - d <> 0 (i.e. c <> d)
You cannot maintain that the choice of labelling was arbitrary for the
purpose of your conclusion and then appeal to the choice of labelling in
your diagram to support the result the conclusion was drawn from.
Now lets plug some numbers in to this algebraically simplified version
of the crucial step in your proof.
Lets assume that all the velocities are positive. If your case holds in
general then it must also hold for a collision where this is assumption
is true.
Then, due to their definitions above, c and d will be positive and
either a or b will be negative. We have to chose values that satisfy
a = -b
ca + db > 0
otherwise we don't even start with true statements.
So lets just chuck some numbers in that satisfy that, any will do to
illustrate the point. Remember though that when we swap a for b we also
have to swap c for d due to the way they are defined in terms of
u1,u2,v1,v1, or if you prefer, do satisfy the two equations above.
let:
a = 3, b = -3, c=5, d=4
Then:
a = -b
3 = -3
ca + db > 0
5.3 + 4.-3 > 0 (i.e. 15 - 12 > 0)
c - d > 0 (i.e. c > d)
5 - 4 > 0 (i.e. 5 > 4)
now if we had made the other choice of labelling:
let:
a = -3, b = 3, c=4, d=5
then:
a = -b
-3 = -(3)
ca + db > 0
4.-3 + 5.3) > 0 (i.e. -12 + 15 > 0)
c - d < 0 (i.e. c < d) !!!!!!!
4 - 5 < 0 (i.e. 4 < 5)
Do you see how the '>' flipped to a '<' purely because of our choice of
labelling? According to the logic of your proof we would have had the
following preposterous result:
3=-(-3)
4.-3 + 5.3 > 0
=> 4 > 5 !!!!!!!!!!
Note that in the above, the signs of c and d did not change so this
change in choice of labelling was not a case of changing the convention
as to which direction we assign positive.
What I have shown you here is that symbolically you can derive c > d or
c < d depending on your choice of labelling and that it does work in
either case if you substitute some numbers in there that satisfy the two
equations in question and does not work if you assume that step in your
proof is correct. If the labelling is arbitrary then you can only
conclude that:
(u1 + v1) <> (u2 + v2)
Again, I am not seeking to ridicule you or get the upper hand as it
were. I make mistakes like this (and some far worse) all the time. It
took me a while to spot the error in your proof, had I seen your proof
in a standard text book I would have probably just believed it assuming
someone else had already checked it thoroughly. Please be man enough to
accept that you were in error (or show me exactly where I am in error in
above - I am not infalible either - but please stick to the point).
Regards,
Mike.
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