Transverse Wave Incident on Absorbing Surface: Pressure & Energy Density

Click For Summary
SUMMARY

This discussion focuses on the analysis of a transverse wave incident on a perfectly absorbing surface, specifically examining the relationship between pressure and energy density. The participants clarify that the electric field (E) is oriented in the x-direction and the magnetic field (B) in the y-direction, with both fields being independent of their respective spatial coordinates. The goal is to demonstrate that the pressure exerted on the absorbing surface is equivalent to the energy density per unit volume, utilizing the conservation of linear momentum and the properties of electromagnetic waves in a vacuum.

PREREQUISITES
  • Understanding of electromagnetic wave properties, specifically transverse waves
  • Familiarity with the concepts of electric field (E) and magnetic field (B)
  • Knowledge of conservation of linear momentum in physics
  • Basic calculus, particularly partial derivatives and vector calculus (nabla operator)
NEXT STEPS
  • Study the derivation of pressure from energy density in electromagnetic waves
  • Learn about the application of the conservation of momentum in electromagnetic theory
  • Explore the mathematical implications of the nabla operator in vector calculus
  • Investigate the characteristics of perfectly absorbing surfaces in wave physics
USEFUL FOR

Physicists, electrical engineers, and students studying electromagnetic theory who are interested in the dynamics of wave interactions with materials.

ptabor
Messages
14
Reaction score
0
Suppose a light wave is traveling in the z-direction.
Then the Electric field is in the x direction, and the B field is in the y direction.

My understanding is that these E and B fields are independent of the x and y directions, respectively. This is to say, that if I take the partial derivative of E with respect to x, I get zero - Likewise for B and y.

I'm trying to show that for a transverse plane wave incident normally on a perfectly abosrbing surface, starting from the conservation of linear momentum, that the pressure exerted on the screen is equal to the energy density per unit volume.

Since we're in vacuum there is no charge or current density, and I only have to consider the momentum density of the wave itself (no mechanical momentum density for the charges). I reduce the expression to something that is the energy density plus the scalar product of B and nabla (or del, whichever you prefer) acting on B - likewise for E. I need to show these are zero.
 
Physics news on Phys.org
ptabor said:
Suppose a light wave is traveling in the z-direction.
Then the Electric field is in the x direction, and the B field is in the y direction.

My understanding is that these E and B fields are independent of the x and y directions, respectively. This is to say, that if I take the partial derivative of E with respect to x, I get zero - Likewise for B and y.

I'm trying to show that for a transverse plane wave incident normally on a perfectly abosrbing surface, starting from the conservation of linear momentum, that the pressure exerted on the screen is equal to the energy density per unit volume.

Since we're in vacuum there is no charge or current density, and I only have to consider the momentum density of the wave itself (no mechanical momentum density for the charges). I reduce the expression to something that is the energy density plus the scalar product of B and nabla (or del, whichever you prefer) acting on B - likewise for E. I need to show these are zero.

I am probably asking a stupid question...but what do you mean by nabla of B? You are not talking about [itex]{\vector \nabla} \cdot {\vec B}[/itex] since this is a scalar. So you mean what exactly? ([itex]{\partial B_x \over \partial x } + \ldots[/itex]? But that's not nabla applied to B)

:-( The tex editor does not accept my symbol \nabla
 
Last edited:
[itex]{\vector \nabla} \cdot {\vec B}[/itex]
Hmm...
[tex]\nabla[/tex]
[itex]\nabla[/itex] [itex]{\nabla}[/itex]
[itex]\vec\nabla[/itex]
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 15 ·
Replies
15
Views
3K
Replies
2
Views
3K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K