1. Jun 7, 2017

Pushoam

1. The problem statement, all variables and given/known data

2. Relevant equations
Radiation pressure i.e.R.P is the instantaneous pressure exerted by radiation on a perfectly absorbing surface perpendicular to the propagation vector of the radiation.
Maxwell has shown that R.P = u , (1) where u is energy density
R.P = F/A, (2)
F $\hat n$ = dp/dt $\hat n$ , (3)
S = εc2E×B = εc2 EB $\hat n$ = cu$\hat n$, (4)

3. The attempt at a solution
Using eqns. 1,2,3,4
R.P.= u = EB εc = F/A = (dp/dt)/A =lim (Δp/Δt)/A: Δt→0

ε EB = $\lim_{Δt \rightarrow 0}\frac{ Δp}{c A ~Δt}$:
Since change in the momentum is in the direction of poynting vector, we can write,

ε E×B = $\lim_{Δt \rightarrow 0}\frac{ Δp}{c A ~Δt}\hat n$ i.e. change in the momentum per unit volume , (5)

Now, according to the book,
change in the momentum i.e. Δp $\hat n$= pv( c A Δt),
where pv is the momentum density

So, ε E×B = pv, (6)

Now, my question is : momentum density means momentum per unit volume or change in momentum per unit volume?

I think momentum density is the momentum transported to the surface by the radiation , when the energy contained in one unit volume of the radiation gets absorbed by the surface.
So, here momentum density is change in the momentum per unit volume, not momentum per unit volume.
Is this correct?

2. Jun 7, 2017

zenmaster99

Momentum density is momentum per unit volume. But how you calculate it might involve how much that momentum would change as the momentum-carrying material leaves a certain volume. You calculated the momentum density by calculating how much pressure was induced by a certain amount of momentum in a certain amount of volume being absorbed by your surface in a certain amount of time.

Saying it another way, you used the idea of radiation pressure to compute momentum density, but pressure is a change in momentum at the surface of interest. That momentum came from the volume immediately on one side of the surface. The amount of momentum density times the volume that impinges on your surface in some amount of time will be absorbed by your surface, thus giving you the radiation pressure.

Does that help?

3. Jun 7, 2017

Pushoam

Thank you, zenmaster99.
What I have got till now is :
Momentum density is the momentum stored in one unit volume of radiation.
When one unit volume of radiation gets absorbed by an ideal absorbing surface , then the change in the momentum of this surface i.e. Δp $\hat n$ = pv × one unit volume
where $\hat n$ is the direction of pv .

My mistake was : I applied both the concepts Δp $\hat n$ and pv to the radiation.

While the concept of Δp $\hat n$ applies to the surface and the concept of pv applies to the radiation.

Is it correct?

4. Jun 8, 2017

zenmaster99

Pushoam,

Momentum density is just the momentum of any fluid (radiation, dust, liquids, etc.) per unit volume.

But, yes, when the unit volume strikes the surface, it will transfer its momentum, which is normal to the surface, to the surface (assuming it is absorbed and not reflected, in which case it will transfer double that momentum to the surface). So I think you are saying the same thing.

ZM