The proof of mass vector.
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From: Ka-In Yen on 28 Jan 2007 19:31 On Jan 26, 4:05 pm, "Pmb" <peter102560_nos...(a)comcast.net> wrote: > "yen, ka-in" <yenk...(a)yahoo.com.tw> wrote in messagenews:1169773194.320952.105590(a)s48g2000cws.googlegroups.com... > > > > > On Jan 17, 10:07 am, "Pmb" <peter102560_nos...(a)comcast.net> wrote: > >> Can someone tell me please how this thread got started and what was meany > >> by > >> proving something was a "mass vector". At best inertial mass is a tensor. > >> See MTW page 159. There is Exercise 5.4 "Inertial Mass Per Unit Volume". > >> This is one of the things I've been studying over the years. > > > Please refer to: > >http://www.geocities.com/redlorikee/mdb2.html > You're assigning vector values where scalars belong. E.g. you have the > "length" of a string, a scalar quantity, a vector value by assigning it a > direction. However you chose the + direction for no reason and > the -direction works just as well. You stated this > "k/<a,b,c>=[k<a,b,c>]/[<a,b,c>^2] > " as an equality whereas it is not. The right side is a well defined vector > whereas the left side is not a vector at all. Dear Pete, Thank you for your comment. Inverse of a vector is widely accepted by mathmaticians and physicists; Clifford proved 1/A = A/A^2 (where A is a vector). According to the above equation, we have k/<a,b,c>=[k<a,b,c>]/[<a,b,c>^2]
From: Pmb on 28 Jan 2007 20:01 "Ka-In Yen" <yenkain(a)yahoo.com.tw> wrote in message news:1170030684.922322.140380(a)a75g2000cwd.googlegroups.com... > > > On Jan 26, 4:05 pm, "Pmb" <peter102560_nos...(a)comcast.net> wrote: >> "yen, ka-in" <yenk...(a)yahoo.com.tw> wrote in >> messagenews:1169773194.320952.105590(a)s48g2000cws.googlegroups.com... >> >> >> >> > On Jan 17, 10:07 am, "Pmb" <peter102560_nos...(a)comcast.net> wrote: >> >> Can someone tell me please how this thread got started and what was >> >> meany >> >> by >> >> proving something was a "mass vector". At best inertial mass is a >> >> tensor. >> >> See MTW page 159. There is Exercise 5.4 "Inertial Mass Per Unit >> >> Volume". >> >> This is one of the things I've been studying over the years. >> >> > Please refer to: >> >http://www.geocities.com/redlorikee/mdb2.html >> You're assigning vector values where scalars belong. E.g. you have the >> "length" of a string, a scalar quantity, a vector value by assigning it a >> direction. However you chose the + direction for no reason and >> the -direction works just as well. You stated this >> "k/<a,b,c>=[k<a,b,c>]/[<a,b,c>^2] >> " as an equality whereas it is not. The right side is a well defined >> vector >> whereas the left side is not a vector at all. > > Dear Pete, > > Thank you for your comment. Inverse of a vector is widely > accepted by mathmaticians and physicists; Clifford proved > 1/A = A/A^2 (where A is a vector). > According to the above equation, we have > k/<a,b,c>=[k<a,b,c>]/[<a,b,c>^2] I didn't say that the inverse of a vector didn't exist. I said it wasn't a vector. Pete
From: Ka-In Yen on 29 Jan 2007 19:43 On Jan 29, 9:01 am, "Pmb" <som...(a)somewhere.net> wrote: > "Ka-In Yen" <yenk...(a)yahoo.com.tw> wrote in messagenews:1170030684.922322.140380(a)a75g2000cwd.googlegroups.com... > > Thank you for your comment. Inverse of a vector is widely > > accepted by mathmaticians and physicists; Clifford proved > > 1/A = A/A^2 (where A is a vector). > > According to the above equation, we have > > k/<a,b,c>=[k<a,b,c>]/[<a,b,c>^2] > I didn't say that the inverse of a vector didn't exist. I said it wasn't a > vector. Dear Pete, Everytime you find 1/A in a equation, you can replace it with A / A^2. Assuming A = a<u>, where <u> is a unit vector, we have 1/<u> = <u> / <u>^2 = <u>. This equation tells that the direction of a vector, unit vector, is not changed by "division". In 3D vector algebra, the direction of a vector is changed by -1 only. 1/A = 1/ (a<u>) = (1/a) <u> 1/A includes two informations: 1) magnitude 2) direction; so it's a vector.
From: Pmb on 29 Jan 2007 20:00 "Ka-In Yen" <yenkain(a)yahoo.com.tw> wrote in message news:1170117823.060160.318280(a)h3g2000cwc.googlegroups.com... > > On Jan 29, 9:01 am, "Pmb" <som...(a)somewhere.net> wrote: >> "Ka-In Yen" <yenk...(a)yahoo.com.tw> wrote in >> messagenews:1170030684.922322.140380(a)a75g2000cwd.googlegroups.com... >> > Thank you for your comment. Inverse of a vector is widely >> > accepted by mathmaticians and physicists; Clifford proved >> > 1/A = A/A^2 (where A is a vector). >> > According to the above equation, we have >> > k/<a,b,c>=[k<a,b,c>]/[<a,b,c>^2] >> I didn't say that the inverse of a vector didn't exist. I said it wasn't >> a >> vector. > > Dear Pete, > Everytime you find 1/A in a equation, ..... I'd have to look up its meaning. What does this1/i mean? > .....you can replace it with > A / A^2. Assuming A = a<u>, where <u> is a unit vector, we > have 1/<u> = <u> / <u>^2 = <u>. This equation tells that the > direction of a vector, unit vector, is not changed by "division". > In 3D vector algebra, the direction of a vector is changed by -1 > only. > 1/A = 1/ (a<u>) = (1/a) <u> > 1/A includes two informations: 1) magnitude 2) direction; so > it's a vector. More later Pete
From: Ka-In Yen on 4 Feb 2007 19:50
On Jan 30, 9:00 am, "Pmb" <som...(a)somewhere.net> wrote: > "Ka-In Yen" <yenk...(a)yahoo.com.tw> wrote in message > > news:1170117823.060160.318280(a)h3g2000cwc.googlegroups.com... > > > > > On Jan 29, 9:01 am, "Pmb" <som...(a)somewhere.net> wrote: > >> "Ka-In Yen" <yenk...(a)yahoo.com.tw> wrote in > >> messagenews:1170030684.922322.140380(a)a75g2000cwd.googlegroups.com... > >> > Thank you for your comment. Inverse of a vector is widely > >> > accepted by mathmaticians and physicists; Clifford proved > >> > 1/A = A/A^2 (where A is a vector). > >> > According to the above equation, we have > >> > k/<a,b,c>=[k<a,b,c>]/[<a,b,c>^2] > >> I didn't say that the inverse of a vector didn't exist. I said it wasn't > >> a > >> vector. > > > Dear Pete, > > Everytime you find 1/A in a equation, ..... > > I'd have to look up its meaning. What does this1/i mean? > > > .....you can replace it with > > A / A^2. Assuming A = a<u>, where <u> is a unit vector, we > > have 1/<u> = <u> / <u>^2 = <u>. This equation tells that the > > direction of a vector, unit vector, is not changed by "division". > > In 3D vector algebra, the direction of a vector is changed by -1 > > only. > > 1/A = 1/ (a<u>) = (1/a) <u> > > 1/A includes two informations: 1) magnitude 2) direction; so > > it's a vector. > > More later Dear Pete, Do you have any further questions? |