geordief 86 Posted May 6 I have it on authority **that tensors enable one to model the electromagnetic interaction between a " moving "charge and an electric conductor in such a way as to dispense with frames of reference . I wonder if anyone might have anything to say on the subject and whether it can be shown in more detail how this is the case. **https://www.markushanke.net/tensors-for-laypeople/ 0 Share this post Link to post Share on other sites

MigL 875 Posted May 7 Where is Markus when you need him... This is way above my pay grade, but, since Electomagnetism is a vector force, the explanation that Markus gives for a tensor as a black box which takes a vector input, processes it, and produces a vector output completely independent of co-ordinate system, seems to make sense. Thanks a lot, now I have to do some research. Or maybe someone more knowledgeable will step up. 0 Share this post Link to post Share on other sites

studiot 1690 Posted May 7 (edited) 2 hours ago, MigL said: This is way above my pay grade, but, since Electomagnetism is a vector force, the explanation that Markus gives for a tensor as a black box which takes a vector input, processes it, and produces a vector output completely independent of co-ordinate system, seems to make sense. Yeah, that's true it works rather like this:- In elementary electromagnetism we learn that the flux density vector B is proportional to the field strength vector H by the magnetic permeability μ, which is taken as a constant. So B = μH This is simply the multiplication of a vector by a scalar so the vector B points in the same direction as H. However the medium may not be isotropic so the permeability may vary with orientation or direction. So now we have three quantities which possess directional characterisitcs. The effect of this is that μ has to be regarded as an object which maintains the above equation. This cannot be a vector since the defined vector product of two vectors produces a vector perpendicular to their plane. The object that is needed is a rank 2 tensor and so the permeability becomes a rank 2 tensor. We can now regard μ aas a tensor of rank zero when taken as just a constant. Now you get a vector into the black multiplication box and the correct vector out. Edited May 7 by studiot 1 Share this post Link to post Share on other sites

geordief 86 Posted May 8 14 hours ago, studiot said: Yeah, that's true it works rather like this:- In elementary electromagnetism we learn that the flux density vector B is proportional to the field strength vector H by the magnetic permeability μ, which is taken as a constant. So B = μH This is simply the multiplication of a vector by a scalar so the vector B points in the same direction as H. However the medium may not be isotropic so the permeability may vary with orientation or direction. So now we have three quantities which possess directional characterisitcs. The effect of this is that μ has to be regarded as an object which maintains the above equation. This cannot be a vector since the defined vector product of two vectors produces a vector perpendicular to their plane. The object that is needed is a rank 2 tensor and so the permeability becomes a rank 2 tensor. We can now regard μ aas a tensor of rank zero when taken as just a constant. Now you get a vector into the black multiplication box and the correct vector out. Is it possible (ie not too complicated for me) to ask where the covector space ** fits into the relationship of those 3 vectors? (if only to perhaps light my path a little into this territory I am only now learning a bit about) ** I presume it must be there somewhere... 0 Share this post Link to post Share on other sites

studiot 1690 Posted May 8 (edited) 39 minutes ago, geordief said: Is it possible (ie not too complicated for me) to ask where the covector space ** fits into the relationship of those 3 vectors? (if only to perhaps light my path a little into this territory I am only now learning a bit about) ** I presume it must be there somewhere... First read this extract about covectors. http://www.rpi.edu/dept/phys/Courses/PHYS4210/S10/NotesOnVectors.pdf Now look at pages 7 and 8 of this document https://www.grc.nasa.gov/www/k-12/Numbers/Math/documents/Tensors_TM2002211716.pdf These explains the connection between the dot product of two vectors, one a covector, and tensors in general, including the permeability tensor. See what you can get out of these two documents they are free to download. Edited May 8 by studiot 1 Share this post Link to post Share on other sites

MigL 875 Posted May 9 Thanks for 'stepping up', Studiot. While I've used ( OK, played with ) tensors before, I've never really had the "Ahaa" moment, where they suddenly make sense. 0 Share this post Link to post Share on other sites