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The proof of mass vector
By: The Math Guy - 12/8/05 (7:12 PM)
Introduction:
In this forum, we will prove that linear mass density and
surface mass density are vector, and the application of mass
vector is presented.
1. The unit of vector.
In physics, The unit of three-dimensional cartesian coordinate
systems is meter. In this paper, a point of 3-D coordinate
system is written as
(p1,p2,p3) m, or (p:3) m
and a vector is written as
m, or m
or
l m = m
where l=abs(sqrt(a^2+b^2+c^2)) is the magnitude of the vector,
and is a unit vector which gives the direction of
the vector.
For three reasons, a magnitude of a vector can not add to a
scalar:
i) The magnitude belongs to the set of vector; it's a
portion of a vector. Scalar belongs to a field.
ii) The magnitude is real non-negative number, but scalar
is real number.
iii) The unit of magnitude is meter, but scalar has no unit.
This is a major difference between physics and mathematics.
5m+3 is meaningless.
2. Linear mass density is a vector.
The mass of a string is M kg, and the length of the string
is l m. Where l m is the magnitude of the length, and
is a 3-D unit vector which gives the direction of the
string. Then the linear mass density of the string is:
M/(l)=(M/l) (kg/m)
The direction, , is not changed by "division", so we
can move from denominator to numerator. A direction
is changed by -1 only. A proof is found in Clifford algebras:
[Proof]
k/=[k]/[^2]
=(k/l)
where l is the magnitude of , and is the
unit vector of .
[Proof]
3. Surface mass density is a vector.
A parallelogram has two vectors: l m and h m.
and are unit vectors. The area vector of the parallelogram
is the cross product of these two vectors.
l m X h m= lh (m^2 )X
= lh abs(sin(theta)) (m^2)
Where theta is the angle between and . is
a unit vector which is perpendicular to and .
For AXB=-BXA, an area has two directions.
We can divide the area vector by the length vector.
lh*abs(sin(theta))/[l]
=hX/
=h(X)X
(The direction, , is not changed by "division", and
the division is replaced by a cross product.)
=-hX(X)
=-h[(o)-(o)]
(where o is dot product.)
=-h(cos(theta)-)
=h(-cos(theta)) m
The result is a rectangle, not the original parallelogram. We
can test the result.
h(-cos(theta))Xl=lh m^2X
The magnitude of the area vector is conserved, but the direction
is opposite.
The mass of a round plate is M kg, and the area vector is
A m^2; then the surface mass density is
M kg/(A m^2)=M/A (kg/m^2)
4. Mass vector in physics.
Mass vector has been found in two equations: 1) the velocity
equation of string. 2) Bernoulli's equation.
i) For waves on a string, we have the velocity equation:
v=sqrt(tau/mu). v is velocity of wave, tau is tension
applying to string, and mu is linear mass density of
string. We can rewrite the equation:
mu=tau/v^2.
In the above equation, the mu is parallel to tau, and both
of them are vector.
ii) Bernoulli's equation is:
P + k*v^2/2=C (P is pressure, k is volume density, and v is
velocity. Here we neglect the gravitational term.)
Multiplying cross area vector A m^2 of a string to Bernoulli's
equation(where is a unit vector),
P*A + k*A*v^2/2=C*A
F + L*v^2/2=C*A
(where F is the magnitude of force, and L is the magnitude
of linear mass density.)
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RE: The proof of mass vector
By: FlyOnTheWall - 12/9/05 (9:18 AM)
About Damn time. I have been racking my brain trying to figure out the final portion. Thanks!
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