Gaussian beam

1. Jun 16, 2010

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

2. Jun 16, 2010

Redbelly98

Staff Emeritus
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 εo Eo2
where Eo is the amplitude of the electric field.

3. Jun 17, 2010

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

Last edited: Jun 17, 2010
4. Jun 17, 2010

Antiphon

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

5. Jun 18, 2010

Redbelly98

Staff Emeritus
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.

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?

6. Jun 18, 2010

Bob S

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

I(x,y) = [I0/(σxσy2π)] exp[-x2/(2σx2) - y2/(2σy2)]

where ∫I(x,y) dx dy = I0, and σx and σy 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

7. Jun 18, 2010

Redbelly98

Staff Emeritus
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.

8. Jun 19, 2010

Integral

Staff Emeritus
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.

9. Jun 19, 2010

Bob S

We consider a normalized bi-gaussian electromagnetic radiation beam travelling 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/(σxσy2π)] exp[-x2/(2σx2) - y2/(2σy2)] watts/m2

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

Bob S

Last edited: Jun 19, 2010