Talk:Maxwell stress tensor

message: As a maths graduate (many years ago) who skipped the courses that would have taught me this, but did a fair bit of mathematical analysis - a request:
If my inference is correct, then can a small note, or else a link, please be inserted to make plain the notation that
:  E^2=\mathbf{E}\cdot \mathbf{E}
-in other words that the unbold, unsubscripted E is just the norm of the bold E vector?
And similarly for B. Else, if I have this wrong, then can words be inserted that would prevent me from making this assumption.
Thanks.  — Preceding unsigned comment added by 83.217.170.175 (talk) 14:51, 30 July 2012 (UTC)
replies: text: |\mathbf{V}|^2 = \mathbf{V} \cdot \mathbf{V} = V^2  for any vector \mathbf{V} in \mathbb{R}^3, it is just a notation to drop the bold font, or the arrow, meaning the norm of the vector. Heitorpb (talk) 09:10, 12 October 2015 (UTC)

message: The article currently says:
"For cylindrical objects, such as the rotor of a motor, this is further simplified to: :\sigma_{r t} = \frac{1}{\mu_0} B_r B_t - \frac{1}{2 \mu_0} B^2 \delta_{r t} \,."
Wouldn't it be clearer to simply say :\sigma_{r t} = \frac{1}{\mu_0} B_r B_t \,.?
After all, the Kroenecker delta is zero, because r is not t. I've been staring at this equation wondering why anyone would want to leave that term in the expression. I will wait a while to see if there are any objections, and then change it.--JB Gnome (talk) 02:06, 2 February 2014 (UTC)
thread_title: Cylindrical Objects

message: If one had a positive point charge at r=0 surrounded by a negatively-charged spherical shell at r=1 with equal, but opposite charge, then clearly there would be an electric field extending from r=0 to r=1. Also, an electric field of a positive charge positioned just outside the spherical shell at r=1+ε (ε>0) would have an electric field extending to infinity, and this field would clearly overlap the field interior of the spherical shell. The Lorentz force computation considered by the net E-field from the sphere at r=1+ε and the charge positioned at r=1+ε returns zero force, and yet, the energy computation (or the Maxwell stress tensor rather) does not return a zero result and predicts a force. I'm not the only one who is concerned about a similar problem. See Advances in Applied Science Research, 2011, 2 (2): 99-102 for an example of two positron fields whose fields are limited to some radius r from the respective centers of each:
http://connection.ebscohost.com/c/articles/64287573/nuclear-force-electrostatics
A copy of the article can be found at Pelagia Research Library (http://pelagiaresearchlibrary.com/advances-in-applied-science/vol2-iss2/AdASR-2011-2-2-99-102.pdf).siNkarma86—Expert Sectioneer of Wikipedia86 = 19+9+14 + karma = 19+9+14 + talk 14:21, 31 July 2013 (UTC)
replies: text: I don't think the electric field won't be equal to zero unless all the charges on and within the sphere aren't allowed to move. Like, on a conductor surface charges can also move and polarize. If they can't move, then they all tug and push on the outside charge so then there's zero Lorentz force and E field. Coulomb's Law is a special case of the Jefimenko equations. In the general case where charge distributions can move, along with the Coulomb q/r^2, you also get a (dq/dt)/(rc) term and a (dJ/dt)/(rc^2) term. 2600:1702:3C50:74A0:6D74:219:A7A3:EB73 (talk) 21:38, 26 September 2023 (UTC)
thread_title: Lorentz force vs. Maxwell stress tensor