# 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