The proof of mass vector.
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From: Ka-In Yen on 13 Nov 2005 20:26 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
From: Eric Gisse on 13 Nov 2005 21:29 Ka-In Yen wrote: > The proof of mass vector. > > Ka-In Yen > yenkain(a)yahoo.com.tw > http://www.geocities.com/redlorikee Geocities's place as a shithole continues to be secured by those who inhabit it. > > > 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 Your terminology will confuse those who are actually educated because <a,b,c> is actually the triple scalar product of a,b,c multiplied by a scalar quantity "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. Taking the absolute value of a manifestly positive object is pointless. Your nonstandard way of expressing vectors should be corrected. > > 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. Physics uses it correctly. The difference is in your imagination. [snip - LENGTH IS NOT A VECTOR] > > 4. Mass vector in physics. > > Mass vector has been found in two equations: 1) the velocity > equation of string. 2) Bernoulli's equation. It was explained to you last time that you are mistaken. > > 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. Whats the square root of a vector? Your misconceptions were pointed out to you the last time you mentioned your "examples". > > 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.) These are all scalar quantities. > > 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 There is zero useful content on your site.
From: PD on 14 Nov 2005 10:39 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. PD
From: Ka-In Yen on 16 Nov 2005 20:29 Eric Gisse wrote: > Ka-In Yen wrote: > > The proof of mass vector. > > > > Ka-In Yen > > yenkain(a)yahoo.com.tw > > http://www.geocities.com/redlorikee > > Geocities's place as a shithole continues to be secured by those who > inhabit it. > > > > > > > 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 > > Your terminology will confuse those who are actually educated because > <a,b,c> is actually the triple scalar product of a,b,c multiplied by a > scalar quantity "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. > > Taking the absolute value of a manifestly positive object is pointless. > Your nonstandard way of expressing vectors should be corrected. Dear Eric, Thank you for your comment. I will correct it. The above notation will be used in this thread only. > > > > > 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. > > Physics uses it correctly. The difference is in your imagination. > > [snip - LENGTH IS NOT A VECTOR] > The equation of magnetic force: F=iLXB L is a length vector. Please refer to http://www.physics.wsu.edu/academics/labs/102Labs/Current_Balance(8-12-02).htm > > > > 4. Mass vector in physics. > > > > Mass vector has been found in two equations: 1) the velocity > > equation of string. 2) Bernoulli's equation. > > It was explained to you last time that you are mistaken. > > > > > 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. > > Whats the square root of a vector? > Good question!!! An area vector is A m^2<i:3>. <i:3> is a unit vector. What's the result of sqrt( A m^2<i:3> )? Or what's a square giving the area vector A m^2<i:3>? We can find a pair of vectors of a square: l m<j:3> and l m<k:3>. <j:3> and <k:3> are unit vector. These two vectors meet: i) l^2=A. ii) <j:3>X<k:3>=<i:3>. These three vectors are perpendicular to each other. In fact, we can find infinite pairs of vectors on a plane which meet i) and ii). Ka-In Yen Magnetic force: Drag and Bernoulli force of ether dynamics. http://www.geocities.com/redlorikee
From: Ka-In Yen on 17 Nov 2005 21:14
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 Ka-In Yen Magnetic force: Drag and Bernoulli force of ether dynamics. http://www.geocities.com/redlorikee |