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| From | moroney@world.std.spaamtrap.com (Michael Moroney) |
|---|---|
| Newsgroups | sci.physics |
| Subject | Re: people that do not understand that if you have a Magnetic Field which is always a vector, that multiples thereof are also vectors |
| Date | 2016-09-19 22:26 +0000 |
| Organization | The World : www.TheWorld.com : Since 1989 |
| Message-ID | <nrpolu$pu8$3@pcls7.std.com> (permalink) |
| References | <34f08e43-b8ca-46f8-8673-828eb384691a@googlegroups.com> <nrnf87$rc0$1@gioia.aioe.org> <207e1f0a-f452-484b-b2e0-982711b337e3@googlegroups.com> <nrp1i3$vv9$3@pcls7.std.com> <19906c3f-010e-4a2b-b30c-41e847e19aae@googlegroups.com> |
Archimedes Plutonium <plutonium.archimedes@gmail.com> writes: >On Monday, September 19, 2016 at 10:52:42 AM UTC-5, Michael Moroney wrote: >> Archimedes Plutonium <plutonium.archimedes@gmail.com> writes: >> >All of the above are multiples of the vector space Magnetic Field, hence, >> >all are vectors. >> No, whether a derived unit is a vector or a scalar depends on >> how it is derived. If the dot product is taken between two >> vectors, you get a scalar. The cross product of two vectors >> is another vector. For example, voltage is the dot product >> of the distance vector and the electric field vector, thus it >> is a scalar. A (3D) vector has a direction, while a scalar is >> just a value. >>=20 >> >You got bamboozled by Old Physics >>=20 >> >AP >The Magnetic Field is always a vector, regardless of any operation placed >upon it. That is the meaning of dipole. Something like a parrot of >understanding cannot grasp for themselves until told. Dipole means >Magnetism requires a Field to mediate the two poles, and hence a vector >is involved. The magnetic field is a vector, yes, but not for the reasons you give. I suspect you really don't understand what a vector is, or what the dot products and cross products of vectors are. >And no matter what operations you place on magnetism you never get rid of >the vector field unless you get rid of the magnetism itself. For the cross product, yes, not the dot product. Again, do you even know what they are? >You still cling to the stupidity of a Electric Field which would be in your > imagination (m/s) times kg /A*s^2. Unable to fathom that your electric-field >is just a Magnetic Field times speed. A vector times speed. You can't just "times" two vectors. You can take the dot product, or the cross product, or create a 2d matrix/tensor from them. The electric field is a vector, the electric potential is a scalar. They are different. Electrodynamics requires and uses both. >All these terms are vectors because they all are multiples of the Magnetic >Field Vector Some are scalars, some vectors, depending on how they are derived. The base units are an oversimplification, they don't reveal how the unit was derived. In the list of base units, both dot products and cross products are represented by multiplication, so information is lost. For example, torque and energy both have the same units, (kg*m^2/s^2) and both can be derived from force and distance, but one is a vector and the other a scalar. And they are very different.
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How to gather the missing terms in Faraday Law, and yes, Maxwell goofed here Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-09-18 15:07 -0700
How to gather the missing terms in Faraday Law, and yes, Maxwell goofed here Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-09-18 17:25 -0700
Re: How to gather the missing terms in Faraday Law, and yes, Maxwell goofed here Serigo <invalid@invalid.com> - 2016-09-18 20:32 -0500
Magnetic Field in most-everything Re: How to gather the missing terms in Faraday Law, and yes, Maxwell goofed here Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-09-18 19:38 -0700
Re: Magnetic Field in most-everything Re: How to gather the missing terms in Faraday Law, and yes, Maxwell goofed here moroney@world.std.spaamtrap.com (Michael Moroney) - 2016-09-19 15:51 +0000
people that do not understand that if you have a Magnetic Field which is always a vector, that multiples thereof are also vectors Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-09-19 13:07 -0700
Re: people that do not understand that if you have a Magnetic Field which is always a vector, that multiples thereof are also vectors Serigo <invalid@invalid.com> - 2016-09-19 16:15 -0500
Re: people that do not understand that if you have a Magnetic Field which is always a vector, that multiples thereof are also vectors Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-09-19 14:29 -0700
Re: people that do not understand that if you have a Magnetic Field which is always a vector, that multiples thereof are also vectors Serigo <invalid@invalid.com> - 2016-09-19 19:51 -0500
Faraday law looks like (V*R^-1)' = A*R + B + constant; Ampere law looks like (A*R)' = B + V*A^-1 + constant Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-09-19 19:41 -0700
Re: Faraday law looks like (V*R^-1)' = A*R + B + constant; Ampere law looks like (A*R)' = B + V*A^-1 + constant Serigo <invalid@invalid.com> - 2016-09-20 09:55 -0500
Re: Faraday law looks like (V*R^-1)' = A*R + B + constant; Ampere law looks like (A*R)' = B + V*A^-1 + constant "hanson" <hanson@quick.net> - 2016-09-20 09:03 -0700
Re: people that do not understand that if you have a Magnetic Field which is always a vector, that multiples thereof are also vectors moroney@world.std.spaamtrap.com (Michael Moroney) - 2016-09-19 22:26 +0000
This moron thought a magnetic field is only a vector dependent on how it was derived Re: people that do not understand Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-09-19 19:23 -0700
Re: This moron thought a magnetic field is only a vector dependent on how it was derived Re: people that do not understand Serigo <invalid@invalid.com> - 2016-09-19 21:41 -0500
(A*R)' = A'*R + R'*A and derivative (V*R^-1)' = V'*R^-1 + V*(R^-1)' Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-09-19 22:12 -0700
derivative of magnetic field Re: (A*R)' = A'*R + R'*A and derivative (V*R^-1)' = V'*R^-1 + V*(R^-1)' Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-09-19 23:45 -0700
Re: (A*R)' = A'*R + R'*A and derivative (V*R^-1)' = V'*R^-1 + V*(R^-1)' Serigo <invalid@invalid.com> - 2016-09-20 10:09 -0500
Re: This moron thought a magnetic field is only a vector dependent on how it was derived Re: people that do not understand moroney@world.std.spaamtrap.com (Michael Moroney) - 2016-09-20 04:08 +0000
Re: This moron thought a magnetic field is only a vector dependent on how it was derived Re: people that do not understand Serigo <invalid@invalid.com> - 2016-09-20 10:28 -0500
Re: This moron thought a magnetic field is only a vector dependent on how it was derived Re: people that do not understand moroney@world.std.spaamtrap.com (Michael Moroney) - 2016-09-20 16:40 +0000
Someone in Old Physics trying to flesh out Faraday's law Re: How to gather the missing terms in Faraday Law, and yes, Maxwell goofed here Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-09-18 20:03 -0700
Re: Someone in Old Physics trying to flesh out Faraday's law Re: How to gather the missing terms in Faraday Law, and yes, Maxwell goofed here Serigo <invalid@invalid.com> - 2016-09-18 22:31 -0500
Old Physics Faraday law math does not match the experiment Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-09-18 20:51 -0700
math that matches is the Product Rule in Differentiation that gives a Lenz law inside of Faraday law Re: Old Physics Faraday law math does not match the experiment Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-09-19 02:47 -0700
trying to teach a brainwashed that velocity on Magnetic Field is just another Magnetic Field Re: How to gather the missing terms in Faraday Law, and yes, Maxwell goofed here Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-09-18 20:20 -0700
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