Intensity & Amplitude of Gaussian Beam

[russel.arnold]

Can tell me the exact relationship between the intensity and the amplitude for a gaussian beam
?
I know that I is proportional to |A|² .. but i want the value of this proportianality constant

 

[Redbelly98]

It's more properly called the irradiance (assuming you want the power per unit area), and for any electromagnetic wave (not just Gaussian beams) the relation is

I = ½ c ε_o E_o²​

where E_o is the amplitude of the electric field.

 

[Bob S]

I(x) = [I₀/σ(2π)^0.5] exp[-x²/(2σ²)]

where ∫I(x) dx = I₀ and σ is the rms width.

Bob S

 

[Antiphon]

In engineering units you drop the "c" In the above intensity equation and you'll have Watts.

 

[Redbelly98]

I(x) = [I₀/σ(2π)^0.5] exp[-x²/(2σ²)]

where ∫I(x) dx = I₀ and σ is the rms width.

Bob S

But that isn't in terms of the amplitude, as requested by the OP. You're also assuming the beam is Gaussian in just 1 transverse dimension. If the beam has a circularly symmetric intensity profile, things are different.

In engineering units you drop the "c" In the above intensity equation and you'll have Watts.

Actually, dropping the c gives the energy density in J/m³.
I'm using SI units. I'm not aware of any official "engineering units", unless you also mean the SI system?

 

[Bob S]

But that isn't in terms of the amplitude, as requested by the OP. You're also assuming the beam is Gaussian in just 1 transverse dimension. If the beam has a circularly symmetric intensity profile, things are different.

I believe this is the correct normalized Gaussian beam distribution in two dimensions:

I(x,y) = [I₀/(σ_xσ_y2π)] exp[-x²/(2σ_x²) - y²/(2σ_y²)]

where ∫I(x,y) dx dy = I₀, and σ_x and σ_y are the rms widths in the x and y dimensions.

At any point x and y, the intensity is equal to the square of the amplitude.

Bob S

 

[Redbelly98]

I believe this is the correct normalized Gaussian beam distribution in two dimensions:

I(x,y) = [I₀/(σ_xσ_y2π)] exp[-x²/(2σ_x²) - y²/(2σ_y²)]

where ∫I(x,y) dx dy = I₀, and σ_x and σ_y are the rms widths in the x and y dimensions.

I₀ is perhaps a poor choice of variable name, in that it has different units than I(x,y). I.e., I₀ is in Watts and I(x,y) is W/m².

At any point x and y, the intensity is equal to the square of the amplitude.

Bob S

Not in SI units, or Gaussian-cgs units either. Assuming "the amplitude" refers to the electric field (And I can't imagine what else it would mean.)

(Electric field)² has units of energy/m³ in Gaussian units, or [energy/(charge*distance)]² in SI units. Neither is consistent with intensity = energy/(time*distance²), so I maintain that the relationship is proportional, but not equal.

 

[Integral]

I had this image on hand, thought I would share it. This is a power cross section of a gaussian laser beam. Each color represents a power level. Other beam parameters are given to the left.

 

[Bob S]

I₀ is perhaps a poor choice of variable name, in that it has different units than I(x,y). I.e., I₀ is in Watts and I(x,y) is W/m².Not in SI units, or Gaussian-cgs units either. Assuming "the amplitude" refers to the electric field (And I can't imagine what else it would mean.)

(Electric field)² has units of energy/m³ in Gaussian units, or [energy/(charge*distance)]² in SI units. Neither is consistent with intensity = energy/(time*distance²), so I maintain that the relationship is proportional, but not equal.

We consider a normalized bi-gaussian electromagnetic radiation beam traveling in the z direction.

The Poynting vector is P(x,y) = E(x,y) x H(x,y) watts/m² at every point in space, where E(x,y) is in volts per meter, and H(x,y) is in amps per meter.

E(x,y)/H(x,y) = Z₀ = 377 ohms in free space

So the power density is w(x,y) = E²(x,y)/(2·Z₀) watts/m²

or E(x,y) = sqrt[2·Z₀·w(x,y)] volts/m

where the bi-gaussian distribution is w(x,y) = [W₀/(σ_xσ_y2π)] exp[-x²/(2σ_x²) - y²/(2σ_y²)] watts/m²

where the total beam power is W₀ = ∫w(x,y) dx dy watts

Bob S