Understanding the Maxwell Stress Tensor

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SUMMARY

The discussion focuses on the Maxwell Stress Tensor, which is crucial for expressing electromagnetic forces. The equation presented, $$\vec F = \oint_{s} \vec T \cdot d \vec a - \epsilon \mu \frac{\partial }{\partial t} \oint_{V} \vec S d\tau$$, highlights the relationship between stress tensor $$\vec T$$ and electromagnetic momentum. Participants clarify that in static cases, the second term vanishes due to a zero magnetic field. The conversation also delves into the application of the Generalized Stokes Theorem in understanding how the tensor captures internal field dynamics.

PREREQUISITES
  • Understanding of Maxwell's equations
  • Familiarity with tensor calculus
  • Knowledge of the Generalized Stokes Theorem
  • Basic concepts of electromagnetic field theory
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  • Study the derivation and applications of the Maxwell Stress Tensor
  • Explore the implications of the Generalized Stokes Theorem in physics
  • Investigate the relationship between electromagnetic momentum and energy density
  • Review examples of calculating forces using the Maxwell Stress Tensor in various geometries
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Physicists, electrical engineers, and students studying electromagnetism who seek to deepen their understanding of the Maxwell Stress Tensor and its applications in electromagnetic theory.

JD_PM
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Homework Statement
Maxwell Stress Tensor
Relevant Equations
##\vec F = \oint_{s} \vec T \cdot d \vec a - \epsilon \mu \frac{\partial }{\partial t} \oint_{V} \vec S d\tau ##
The elecromagnetic force can be expressed using the Maxwell Stress Tensor as:

$$\vec F = \oint_{s} \vec T \cdot d \vec a - \epsilon \mu \frac{\partial }{\partial t} \oint_{V} \vec S d\tau $$

(How can I make the double arrow for the stress tensor ##T##?)

In the static case, the second term drops out because the magnetic field is zero.

I have read (from Griffiths) that ##T## represents the force per unit area acting on the surface. But then he states:

Screenshot (394).png


Let's see if understand what's going on:

Screenshot (396).png


I'd say that the stress tensor on the triangle projected on xy plane would be : ##T_{ik}##. Is this correct?

Besides, I am seeking for understanding it. After solving an example exercise

Screenshot (397).png


(The exercise is not solved above of course)

But I want to focus on what he states at the end:

Screenshot (401).png


How can the tensor 'sniff out what is going inside'?

I don't understand why.
 

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This is the same thing as the
fundamental theorem of calculus ##\int^b_a dx\, \frac{df}{dx} = f\left(b\right)-f\left(a\right)##
Stokes theorem: ##\int_S d^2 r\, \mathbf{\hat{n}}.\boldsymbol{\nabla}\times\mathbf{F}=\oint_{\partial S} dl \mathbf{\hat{l}}.\mathbf{F}##
Gauss Theorem: ##\int_V d^3 r\, \boldsymbol{\nabla}.\mathbf{F}=\oint_{\partial V} d^2 r\, \mathbf{\hat{n}}.\mathbf{F}##

etc. Generally this is known as Generalized Stokes Theorem. The point is that integrating a form (##f## or ##\mathbf{F}##) on the boundary of a manifold (##\partial S## or ##\partial V##) is equal to the integral of the exterior derivative (##\frac{df\left(x\right)}{dx},\, \boldsymbol{\nabla}\times\mathbf{F},\, \boldsymbol{\nabla}.\mathbf{F}##) over that manifold.

In your case ##\mathbf{f}=\boldsymbol{\nabla}.\mathbf{T}+\dots## so

##\mathbf{F}=\int_V d^3 r\, \mathbf{f} = \int_V d^3 r\, \boldsymbol{\nabla}.\mathbf{T}+\dots = \oint_{\partial V} d^2r\, \mathbf{\hat{n}}.\mathbf{T}+\dots##
 
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Thanks for your reply.

I understand that ##\epsilon \mu \oint_{V} \vec S d\tau## represents the momentum stored in the fields and
##\oint_{s} \vec T \cdot d \vec a## is the momentum per unit time flowing in through the surface.
 
I think you are correct, but check your units (just in case)
 

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