The proof of mass vector.
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From: Ka-In Yen on 10 Apr 2006 01:54 Eric Gisse wrote: > > > Ka-In Yen wrote: > > > > Bill Hobba wrote: > > > > > "Ka-In Yen" <yenkain(a)yahoo.com.tw> wrote in message > > > > > news:1144028073.121452.279020(a)j33g2000cwa.googlegroups.com... > > > > > > 2. Linear mass density is a vector. > > > > > > M/(l<i:3>)=(M/l) (kg/m)<i:3> > > > > > > > > > > You can not divide by vectors. > > > > > > > > Dear Bill, > > > > Can you tell me how you get the following three kind of vectors? > > > > > > > > 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 > > The quantities are defined to be vectors and are operated on by tools > from vector analysis and calculus. You are inventing your own tools > because you have no clue what you are doing. With Clifford's method, we can get the same result. 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.
From: Eric Gisse on 10 Apr 2006 02:08 Ka-In Yen wrote: > Eric Gisse wrote: > > > > Ka-In Yen wrote: > > > > > Bill Hobba wrote: > > > > > > "Ka-In Yen" <yenkain(a)yahoo.com.tw> wrote in message > > > > > > news:1144028073.121452.279020(a)j33g2000cwa.googlegroups.com... > > > > > > > 2. Linear mass density is a vector. > > > > > > > M/(l<i:3>)=(M/l) (kg/m)<i:3> > > > > > > > > > > > > You can not divide by vectors. > > > > > > > > > > Dear Bill, > > > > > Can you tell me how you get the following three kind of vectors? > > > > > > > > > > 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 > > > > The quantities are defined to be vectors and are operated on by tools > > from vector analysis and calculus. You are inventing your own tools > > because you have no clue what you are doing. > > With Clifford's method, we can get the same result. 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. Idiot.
From: Ka-In Yen on 11 Apr 2006 21:22 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. 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). Physicists have been doing vector-by-vector-division for a hundred years. 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.
From: Eric Gisse on 11 Apr 2006 21:41 Ka-In Yen wrote: > 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. A meter is a length, not a direction. You have no idea what you are talking about. [snip idiocy]
From: Bilge on 11 Apr 2006 23:47
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. >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? |