The proof of mass vector.
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From: Ka-In Yen on 6 Mar 2006 00:47 Am I right? The proof of mass vector. Ka-In Yen yenk...(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: Ka-In Yen on 2 Apr 2006 21:34 Is it useful? The proof of mass vector. Ka-In Yen yenk...(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: Bill Hobba on 2 Apr 2006 23:32 "Ka-In Yen" <yenkain(a)yahoo.com.tw> wrote in message news:1144028073.121452.279020(a)j33g2000cwa.googlegroups.com... > > Is it useful? > > > The proof of mass vector. > > Ka-In Yen > yenk...(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> You can not divide by vectors. Rest of rubbish snipped. Bill
From: Eric Gisse on 3 Apr 2006 00:12 Bill Hobba wrote: > "Ka-In Yen" <yenkain(a)yahoo.com.tw> wrote in message > news:1144028073.121452.279020(a)j33g2000cwa.googlegroups.com... > > > > Is it useful? > > > > > > The proof of mass vector. > > > > Ka-In Yen > > yenk...(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> > > You can not divide by vectors. > > Rest of rubbish snipped. > > Bill http://groups.google.com/group/sci.physics/msg/f2cd8c816d2706d0?dmode=source
From: Ka-In Yen on 3 Apr 2006 21:08
Dear Bill Hobba, Thank you for your comment. Bill Hobba wrote: > "Ka-In Yen" <yenkain(a)yahoo.com.tw> wrote in message > news:1144028073.121452.279020(a)j33g2000cwa.googlegroups.com... > > Is it useful? > > 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> > / > You can not divide by vectors. Why? |