The proof of mass vector.
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From: PD on 18 Nov 2005 11:49 Ka-In Yen wrote: > PD wrote: > > Ka-In Yen wrote: > > > The proof of mass vector. > > > > > > Ka-In Yen > > > yenkain(a)yahoo.com.tw > > > http://www.geocities.com/redlorikee > > > > > > > > > Introduction: > > > In this paper, 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 > > > > > > <a,b,c> m, or <a:3> m > > > > > > or > > > > > > l m<i,j,k> = <a,b,c> m > > > > > > where l=abs(sqrt(a^2+b^2+c^2)) is the magnitude of the vector, > > > and <i,j,k> 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<i:3>. Where l m is the magnitude of the length, and > > > <i:3> is a 3-D unit vector which gives the direction of the > > > string. Then the linear mass density of the string is: > > > > > > M/(l<i:3>)=(M/l) (kg/m)<i:3> > > > > > > The direction, <i:3>, is not changed by "division", so we > > > can move <i:3> from denominator to numerator. A direction > > > is changed by -1 only. A proof is found in Clifford algebras: > > > > > > [Proof] > > > k/<a,b,c>=[k<a,b,c>]/[<a,b,c>^2] > > > =(k/l) <i,j,k> > > > where l is the magnitude of <a,b,c>, and <i,j,k> is the > > > unit vector of <a,b,c>. > > > [Proof] > > > > > > > > > 3. Surface mass density is a vector. > > > > > > A parallelogram has two vectors: l m<i:3> and h m<j:3>. <i:3> > > > and <j:3> are unit vectors. The area vector of the parallelogram > > > is the cross product of these two vectors. > > > > > > l m<i:3> X h m<j:3>= lh (m^2 )<i:3>X<j:3> > > > = lh abs(sin(theta)) (m^2)<k:3> > > > > > > Where theta is the angle between <i:3> and <j:3>. <k:3> is > > > a unit vector which is perpendicular to <i:3> and <j:3>. > > > For AXB=-BXA, an area has two directions. > > > > > > We can divide the area vector by the length vector. > > > > > > lh*abs(sin(theta))<k:3>/[l<i:3>] > > > =h<i:3>X<j:3>/<i:3> > > > =h(<i:3>X<j:3>)X<i:3> > > > (The direction, <i:3>, is not changed by "division", and > > > the division is replaced by a cross product.) > > > =-h<i:3>X(<i:3>X<j:3>) > > > =-h[<i:3>(<i:3>o<j:3>)-<j:3>(<i:3>o<i:3>)] > > > (where o is dot product.) > > > =-h(cos(theta)<i:3>-<j:3>) > > > =h(<j:3>-cos(theta)<i:3>) m > > > > > > The result is a rectangle, not the original parallelogram. We > > > can test the result. > > > > > > h(<j:3>-cos(theta)<i:3>)Xl<i:3>=lh m^2<j:3>X<i:3> > > > > > > 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<i:3>; then the surface mass density is > > > > > > M kg/(A m^2<i:3>)=M/A (kg/m^2)<i:3> > > > > > > > > > 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<i:3> of a string to Bernoulli's > > > equation(where <i:3> is a unit vector), > > > > > > P*A<i:3> + k*A<i:3>*v^2/2=C*A<i:3> > > > F<i:3> + L<i:3>*v^2/2=C*A<i:3> > > > (where F is the magnitude of force, and L is the magnitude > > > of linear mass density.) > > > > > > These two equations are well used in the theory "Magnetic force: > > > Combining Drag force and Bernoulli force of ether dynamics." > > > For detail, please refer to my site: > > > http://www.geocities.com/redlorikee > > > > > > I've already explained to you, KY, that your mathematical steps 1 by 1 > > are ok. The physical meaning that you associate with those steps is > > what is faulty. > > Dear PD, > > Thank you for your comment. The vector of linear density and surface > density have been used for a long time. > > Current density(J) is a surface density and a vector; its unit is > A/m^2. > > Electric field(E) is linear density and a vector; Its unit is V/m. > > Displacement(D) is surface density and a vector; Its unit is coul/m^2. > > Reference: Classical Electrodynamics(J.D. Jackson) p.820 > First of all, I find it interesting that you define anything that has a power of length in the denominator as a density. That's not completely unreasonable. But to assume that because *one* quantity has density-like units and is a vector does not mean that *all* quantities that have comparable units are also vectors. I'll give you an example of this. Torque and energy have identical units: N-m. However, one is explicity defined as a scalar quantity and the other is explicity defined as a vector quantity. No comparison of the units can gloss over the fact that the two quantities are inherently different in nature. They *behave* differently under coordinate transformations. PD
From: Ka-In Yen on 19 Nov 2005 20:12 PD wrote: > Ka-In Yen wrote: > > PD wrote: > > > > > > I've already explained to you, KY, that your mathematical steps 1 by 1 > > > are ok. The physical meaning that you associate with those steps is > > > what is faulty. > > > > Dear PD, > > > > Thank you for your comment. The vector of linear density and surface > > density have been used for a long time. > > > > Current density(J) is a surface density and a vector; its unit is > > A/m^2. > > > > Electric field(E) is linear density and a vector; Its unit is V/m. > > > > Displacement(D) is surface density and a vector; Its unit is coul/m^2. > > > > Reference: Classical Electrodynamics(J.D. Jackson) p.820 > > > > First of all, I find it interesting that you define anything that has a > power of length in the denominator as a density. That's not completely > unreasonable. > > But to assume that because *one* quantity has density-like units and is > a vector does not mean that *all* quantities that have comparable units > are also vectors. > > I'll give you an example of this. Torque and energy have identical > units: N-m. However, one is explicity defined as a scalar quantity and > the other is explicity defined as a vector quantity. No comparison of > the units can gloss over the fact that the two quantities are > inherently different in nature. They *behave* differently under > coordinate transformations. Dear PD, Thank you for your question. I suggest different notations for them: N X m (X is cross product) for torque and Nm for energy. Ka-In Yen Magnetic force: Drag force and Bernoulli force of ether dynamics. http://www.geocities.com/redlorikee
From: PD on 20 Nov 2005 14:30 Ka-In Yen wrote: > PD wrote: > > Ka-In Yen wrote: > > > PD wrote: > > > > > > > > I've already explained to you, KY, that your mathematical steps 1 by 1 > > > > are ok. The physical meaning that you associate with those steps is > > > > what is faulty. > > > > > > Dear PD, > > > > > > Thank you for your comment. The vector of linear density and surface > > > density have been used for a long time. > > > > > > Current density(J) is a surface density and a vector; its unit is > > > A/m^2. > > > > > > Electric field(E) is linear density and a vector; Its unit is V/m. > > > > > > Displacement(D) is surface density and a vector; Its unit is coul/m^2. > > > > > > Reference: Classical Electrodynamics(J.D. Jackson) p.820 > > > > > > > First of all, I find it interesting that you define anything that has a > > power of length in the denominator as a density. That's not completely > > unreasonable. > > > > But to assume that because *one* quantity has density-like units and is > > a vector does not mean that *all* quantities that have comparable units > > are also vectors. > > > > I'll give you an example of this. Torque and energy have identical > > units: N-m. However, one is explicity defined as a scalar quantity and > > the other is explicity defined as a vector quantity. No comparison of > > the units can gloss over the fact that the two quantities are > > inherently different in nature. They *behave* differently under > > coordinate transformations. > > Dear PD, > > Thank you for your question. I suggest different notations > for them: N X m (X is cross product) for torque and Nm for energy. > An interesting suggestion but not generally required. There is also the subtlety when one is referring to a vector quantity or the (scalar) magnitude of the same vector quantity, as you've noticed. PD
From: Lynx Xiong on 20 Nov 2005 15:20 PD wrote: > Ka-In Yen wrote: > > PD wrote: > > > Ka-In Yen wrote: > > > > PD wrote: > > > > > > > > > > I've already explained to you, KY, that your mathematical steps 1 by 1 > > > > > are ok. The physical meaning that you associate with those steps is > > > > > what is faulty. > > > > > > > > Dear PD, > > > > > > > > Thank you for your comment. The vector of linear density and surface > > > > density have been used for a long time. > > > > > > > > Current density(J) is a surface density and a vector; its unit is > > > > A/m^2. > > > > > > > > Electric field(E) is linear density and a vector; Its unit is V/m. > > > > > > > > Displacement(D) is surface density and a vector; Its unit is coul/m^2. > > > > > > > > Reference: Classical Electrodynamics(J.D. Jackson) p.820 > > > > > > > > > > First of all, I find it interesting that you define anything that has a > > > power of length in the denominator as a density. That's not completely > > > unreasonable. > > > > > > But to assume that because *one* quantity has density-like units and is > > > a vector does not mean that *all* quantities that have comparable units > > > are also vectors. > > > > > > I'll give you an example of this. Torque and energy have identical > > > units: N-m. However, one is explicity defined as a scalar quantity and > > > the other is explicity defined as a vector quantity. No comparison of > > > the units can gloss over the fact that the two quantities are > > > inherently different in nature. They *behave* differently under > > > coordinate transformations. > > > > Dear PD, > > > > Thank you for your question. I suggest different notations > > for them: N X m (X is cross product) for torque and Nm for energy. > > > > An interesting suggestion but not generally required. There is also the > subtlety when one is referring to a vector quantity or the (scalar) > magnitude of the same vector quantity, as you've noticed. > > PD why wont you eat your sandwitch first
From: Ka-In Yen on 21 Nov 2005 21:28
Ka-In Yen wrote: > The proof of mass vector. > Introduction: > In this paper, we will prove that linear mass density and > surface mass density are vector, and the application of mass > vector is presented. > 3. Surface mass density is a vector. > A parallelogram has two vectors: l m<i:3> and h m<j:3>. <i:3> > and <j:3> are unit vectors. The area vector of the parallelogram > is the cross product of these two vectors. > l m<i:3> X h m<j:3>= lh (m^2 )<i:3>X<j:3> > = lh abs(sin(theta)) (m^2)<k:3> > Where theta is the angle between <i:3> and <j:3>. <k:3> is > a unit vector which is perpendicular to <i:3> and <j:3>. > For AXB=-BXA, an area has two directions. .... > We can divide the area vector by the length vector. > lh*abs(sin(theta))<k:3>/[l<i:3>] > =h<i:3>X<j:3>/<i:3> > =h(<i:3>X<j:3>)X<i:3> > (The direction, <i:3>, is not changed by "division", and > the division is replaced by a cross product.) > =-h<i:3>X(<i:3>X<j:3>) > =-h[<i:3>(<i:3>o<j:3>)-<j:3>(<i:3>o<i:3>)] > (where o is dot product.) > =-h(cos(theta)<i:3>-<j:3>) > =h(<j:3>-cos(theta)<i:3>) m > The result is a rectangle, not the original parallelogram. We > can test the result. > h(<j:3>-cos(theta)<i:3>)Xl<i:3>=lh m^2<j:3>X<i:3> > The magnitude of the area vector is conserved, but the direction > is opposite. The equation of magnetic force is "vector division by vector". F=iLXB (X is corss product). L is a length vector; assuming L=l m<i:3>, <i:3> is a unit vector. B is magnetic flux density. Its unit is tesla, or Wb/m^2. Wb, Weber, is the unit of magnetic flux. Assuming B= b (Wb/m^2)<j:3>, <j:3> is a unit vector. Since B is a vector of surface density, we can rewrite it: B= b Wb/(m^2<j:3>), <j:3> is moved to denominator. LXB= l m<i:3> X b (Wb/m^2)<j:3> = l m<i:3> * b Wb /(m^2<j:3>) = lb Wb m<i:3>/(m^2<j:3>) It's VDV. Ka-In Yen Magnetic force: Drag and Bernoulli force of ether dynamics. http://www.geocities.com/redlorikee |