The proof of mass vector.
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From: Bilge on 12 Apr 2006 01:29 Ka-In Yen: > >Bill Hobba wrote: >> "Ka-In Yen" <yenkain(a)yahoo.com.tw> wrote in message >> news:1144112932.833871.196670(a)i39g2000cwa.googlegroups.com... >> > 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. >> >> > >> >> > M/(l<i:3>)=(M/l) (kg/m)<i:3> >> >> You can not divide by vectors. >> > Why? >> >> http://www.mcasco.com/qa_vdq.html > >Dear Bill, > >Thank you for the information you provide. You were misled by >mathematician. Mathematicians play vectors without unit(meter >for example); that's not for physicists. Well, I'm a physicist and as far as I can tell, you haven't yet said anything that makes an physical sense, regardless of what you want to claim about mathematics. > Area = Length * Height > Height = Area / Length > >I learned the above equations when I was a pupil in elementary >school. Dividing an area by a length, we always get the height >of a rectangle(although infinite number of parallelograms have >the same area and length). What does that have to do with dividing by vectors? All you wrote were magnitudes. >Physicists have been doing vector-by-vector-division for a >hundred years. The equation of magnetic force is "vector division >by vector". No, it is not. >F=iLXB (X is corss product). Note that L is a vector and B is a pseudovector and those are not dividing anything.
From: Ka-In Yen on 13 Apr 2006 21:48 Dear Bilge, Thank you for your comment. Bilge wrote: > Ka-In Yen: > > > >With Clifford's method, we can get the same result. > > No, you don't. Feel free to write down the expression in terms of > the clifford algebra. > Clifford proves k / <a,b,c> = k<a,b,c> / <a,b,c>^2 [Proof] k/<a,b,c>=[k<a,b,c>]/[<a,b,c>^2] =(k/l) <i,j,k> where l=sqrt(a^2+b^2+c^2) is the magnitude of <a,b,c>, and <i,j,k>=<a,b,c>/l is the unit vector of <a,b,c>. [End of proof] > >Do you have any strong reason to reject Clifford's method? > > >In 3D VECTOR algebra, we have to divide a mass by a length > >VECTOR, and linear mass density is a VECTOR. > > Wrong. Given a linear mass density lying along -a < x < a, what > direction does it point? We are talking about 3D vector algebra, your question is 1D. two points : (-a, y, z) m and (a, y, z) m (m is meter). length vector: (a,y,z)m - (-a,y,z)m = <2a,0,0>m mass of a straight wire between the above two points is M kg. linear mass density = M kg / <2a,0,0>m = (M/2a) <1,0,0> kg/m
From: Bilge on 14 Apr 2006 00:40 Ka-In Yen: >Dear Bilge, >Thank you for your comment. > >Bilge wrote: >> Ka-In Yen: >> > >> >With Clifford's method, we can get the same result. >> >> No, you don't. Feel free to write down the expression in terms of >> the clifford algebra. >> > >Clifford proves k / <a,b,c> = k<a,b,c> / <a,b,c>^2 Define your notation. >[Proof] > k/<a,b,c>=[k<a,b,c>]/[<a,b,c>^2] > =(k/l) <i,j,k> > where l=sqrt(a^2+b^2+c^2) is the magnitude of <a,b,c>, >and <i,j,k>=<a,b,c>/l is the unit vector of <a,b,c>. >[End of proof] > >> >Do you have any strong reason to reject Clifford's method? >> >> >In 3D VECTOR algebra, we have to divide a mass by a length >> >VECTOR, and linear mass density is a VECTOR. >> >> Wrong. Given a linear mass density lying along -a < x < a, what >> direction does it point? > >We are talking about 3D vector algebra, your question is 1D. Oh, in other words, wires don't exist in 3-d? > >two points : (-a, y, z) m and (a, y, z) m (m is meter). >length vector: (a,y,z)m - (-a,y,z)m = <2a,0,0>m >mass of a straight wire between the above two points is M kg. > >linear mass density = M kg / <2a,0,0>m > = (M/2a) <1,0,0> kg/m >
From: Ka-In Yen on 16 Apr 2006 23:14 Dear Bilge, Thank you for your comment. Bilge wrote: > Ka-In Yen: > >Bill Hobba wrote: > >> http://www.mcasco.com/qa_vdq.html > >Thank you for the information you provide. You were misled by > >mathematician. Mathematicians play vectors without unit(meter > >for example); that's not for physicists. > Well, I'm a physicist and as far as I can tell, you haven't yet > said anything that makes an physical sense, regardless of what you > want to claim about mathematics. > > Area = Length * Height > > Height = Area / Length > > > >I learned the above equations when I was a pupil in elementary > >school. Dividing an area by a length, we always get the height > >of a rectangle(although infinite number of parallelograms have > >the same area and length). > > What does that have to do with dividing by vectors? All you wrote > were magnitudes. According to mathematician's opinion: "Again there are two unknowns, |V| and u, in the equation so there are infinitely many answers. Therefore cross division is also undefined." ---- http://www.mcasco.com/qa_vdq.html That's not true. Stupid mathematicians hinder the development of 3D vector algebra. An area vector is A<i:3>m^2, and its length is l<j:3>m. where <i:3> and <j:3> are unit vectors and m is meter. We can divide the area vector by the length vector, and we get the height(vector) of rectangle. A<i:3>m^2 / l<j:3>m =(A/l) <i:3>x<j:3> m (x is cross product) =(A*sin(theta)/l) <k:3> m (<k:3>=(<i:3>x<j:3>)/sin(theta)) where theta is the angle between <i:3> and <j:3>. <k:3> is a unit vector and perpendicular to <i:3> and <j:3>. Or A<i:3>m^2 / l<j:3>m =A/(l <i:3>x<j:3>) m =A/(l*sin(theta)) <k:3> m > > >Physicists have been doing vector-by-vector-division for a > >hundred years. The equation of magnetic force is "vector division > >by vector". > > No, it is not. > > >F=iLXB (X is corss product). > > Note that L is a vector and B is a pseudovector and those are > not dividing anything. As soon as you accept Clifford's method, you will realize that LXB is vector by vector division.
From: Eric Gisse on 17 Apr 2006 02:25
Ka-In Yen wrote: > Dear Bilge, > Thank you for your comment. > > Bilge wrote: > > Ka-In Yen: > > >Bill Hobba wrote: > > >> http://www.mcasco.com/qa_vdq.html > > >Thank you for the information you provide. You were misled by > > >mathematician. Mathematicians play vectors without unit(meter > > >for example); that's not for physicists. > > Well, I'm a physicist and as far as I can tell, you haven't yet > > said anything that makes an physical sense, regardless of what you > > want to claim about mathematics. > > > Area = Length * Height > > > Height = Area / Length > > > > > >I learned the above equations when I was a pupil in elementary > > >school. Dividing an area by a length, we always get the height > > >of a rectangle(although infinite number of parallelograms have > > >the same area and length). > > > > What does that have to do with dividing by vectors? All you wrote > > were magnitudes. > > According to mathematician's opinion: > "Again there are two unknowns, |V| and u, in the equation so there are > infinitely many answers. Therefore cross division is also undefined." > ---- http://www.mcasco.com/qa_vdq.html > > That's not true. Stupid mathematicians hinder the development > of 3D vector algebra. Only a crank complains about "stupid mathematicians". [snip] Yet for all your idiotic notation, you are unable to demonstrate that your vector "division" has an inverse. |