Momentum per unit area per unit time in EM

In summary, the conversation discusses problem 8.5 in Griffiths 3rd edition, where an infinite parallel plate capacitor is considered with different charge densities on the upper and lower plates. The conversation then addresses finding all nine elements of the stress tensor in the region between the plates, the force per unit area on the top plate, and the momentum per unit area, per unit time crossing the xy plane. The conversation also includes a discussion on finding the recoil force per unit area on the top plate. The relevant equations used are Maxwell's Stress Tensor, the equation for momentum density, and the equation for the change in momentum over time. The final solution for part (c) is that ##\int_S \overleftrightarrow{T} \cdot
  • #1
athrun200
277
0

Homework Statement


In fact it is problem 8.5 in Griffiths 3rd ed p357

Consider an infinite parallel plate capacitor with the lower plate carrying the charger density [itex] - \sigma [/itex], and the upper plate carrying the charge density [itex] + \sigma [/itex].
(a) Determine all nine elements of the stress tensor, in the region between the plates. Display your answer as a 3x3 matrix
(b)Use Eq 8.22,[itex]\overrightarrow F = \oint\limits_S {T \cdot d\overrightarrow a } [/itex] for [itex]\overrightarrow S = 0[/itex], to determine the force per unit area on the top plate.
(c) What is the momentum per unit area, per unit time, crossing the xy plane?
(d) Find the recoil force per unit area on the top plate.


Homework Equations


Maxwell's Stress Tensor
[itex]\overrightarrow F = \oint\limits_S {T \cdot d\overrightarrow a } - {\varepsilon _0}{{\bar \mu }_0}\frac{d}{{dt}}\int {Sd\tau } [/itex]

Momentum density
[itex]{\wp _{em}} = {\varepsilon _0}{\mu _0}\overrightarrow S [/itex]

[itex]\frac{d}{{dt}}({\wp _{em}} + {\wp _{mech}}) = \nabla \cdot T[/itex]



The Attempt at a Solution


Part (a) and (b) are easy for me.

(a) [itex]{T_{xy}} = {T_{xz}} = {T_{yz}} = ... = 0[/itex] and by using some equations we can find [itex]{T_{xx}},{T_{yy}}{\rm{ and }}{T_{zz}}[/itex]
So the answer is [itex]T = \frac{{{\sigma ^2}}}{{2{\varepsilon _0}}}\left( {\begin{array}{*{20}{c}}
{ - 1}&0&0\\
0&{ - 1}&0\\
0&0&1
\end{array}} \right)[/itex]
(b) is also easy, use that equation provided we can find the answer [itex]\overrightarrow f = \frac{{{\sigma ^2}}}{{2{\varepsilon _0}}}\widehat z[/itex]

The problem is (c)
I think the equation [itex]\frac{d}{{dt}}({\wp _{em}} + {\wp _{mech}}) = \nabla \cdot T[/itex] is useful to solve it with [itex]{\wp _{mech}}=0[/itex]
However on the right hand side we have [itex]\nabla \cdot T[/itex]. The div of a tensor would be? I learned only the div over a vector, but not a tensor.
 
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  • #2
athrun200 said:
The problem is (c)
I think the equation [itex]\frac{d}{{dt}}({\wp _{em}} + {\wp _{mech}}) = \nabla \cdot T[/itex] is useful to solve it with [itex]{\wp _{mech}}=0[/itex]

I think a more relevant equation for part (c) is equation 8.28. Note how Griffiths interprets the last term of this equation (just below equation 8.29). So, how can you interpret ##\int_S \overleftrightarrow{T} \cdot d\vec{a}## over a surface S in terms of momentum?
 
  • #3
TSny said:
I think a more relevant equation for part (c) is equation 8.28. Note how Griffiths interprets the last term of this equation (just below equation 8.29). So, how can you interpret ##\int_S \overleftrightarrow{T} \cdot d\vec{a}## over a surface S in terms of momentum?

##\int_S \overleftrightarrow{T} \cdot d\vec{a}## is the momentum transferred to the surface per unit time

So [itex]\overset\leftrightarrow T[/itex] is the momentum transferred to the surface per unit time, per unit area, thus it is the answer?

The momentum per unit area, per unit time perpendicular to xy is [itex]\overset\leftrightarrow {{T_{zz}}}[/itex] which is [itex]\frac{{{\sigma ^2}}}{{2{\varepsilon _0}}}[/itex]
While [itex]\overset\leftrightarrow {{T_{xx}}} = \overset\leftrightarrow {{T_{yy}}} = - \frac{{{\sigma ^2}}}{{2{\varepsilon _0}}}[/itex] is the momentum per area, per time parallel to the surface?
 
  • #4
athrun200 said:
##\int_S \overleftrightarrow{T} \cdot d\vec{a}## is the momentum transferred to the surface per unit time

So [itex]\overset\leftrightarrow T[/itex] is the momentum transferred to the surface per unit time, per unit area, thus it is the answer?

The momentum per unit area, per unit time perpendicular to xy is [itex]\overset\leftrightarrow {{T_{zz}}}[/itex] which is [itex]\frac{{{\sigma ^2}}}{{2{\varepsilon _0}}}[/itex]
Yes.

While [itex]\overset\leftrightarrow {{T_{xx}}} = \overset\leftrightarrow {{T_{yy}}} = - \frac{{{\sigma ^2}}}{{2{\varepsilon _0}}}[/itex] is the momentum per area, per time parallel to the surface?

Well, ##T_{xx}## is the x-component of momentum per unit area per unit time crossing a surface oriented perpendicular to the x axis. Similarly for ##T_{yy}##.
 
  • #5
Thanks a lot!
 

1. What is momentum per unit area per unit time in electromagnetic (EM) radiation?

Momentum per unit area per unit time in EM radiation is a measure of the amount of momentum that is transferred through a specific area over a specific time period. It is also known as radiation pressure or radiation force.

2. How is momentum per unit area per unit time in EM radiation calculated?

The formula for calculating momentum per unit area per unit time in EM radiation is force per unit area (pressure) multiplied by the speed of light. This can also be written as the energy per unit volume divided by the speed of light.

3. What is the significance of momentum per unit area per unit time in EM radiation?

Momentum per unit area per unit time in EM radiation is important because it helps us understand the transfer of energy and momentum through electromagnetic waves. It is also relevant in fields such as astrophysics and laser technology.

4. Can momentum per unit area per unit time in EM radiation be measured?

Yes, momentum per unit area per unit time in EM radiation can be measured using various techniques such as optical tweezers, radiation pressure experiments, and laser doppler velocimetry. These methods allow for the detection and measurement of the force exerted by EM radiation on a surface.

5. How does momentum per unit area per unit time in EM radiation relate to the wave-particle duality of light?

Momentum per unit area per unit time in EM radiation is a concept that is used to describe the wave-like nature of light. It helps to explain how photons, which are particles of light, can also behave like waves and transfer momentum through electromagnetic fields. This phenomenon is known as the wave-particle duality of light.

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