# What is the amplitude of the electric field in a laser?

I'm looking for reliable informations about the amplitude (not the intensity), in volt/meter, of the electric field in a typical laser.

Or in other words : what are the typical amplitudes of the monochromatic plane waves in an intense electromagnetic wave ?

EDIT : I'm interested in the highest amplitudes values we could find in an intense electromagnetic wave. Maybe not "typical waves", after all.

I know that the maximum theoretical value of an electric field (from quantum mechanics) is about $10^{18} \text{volt/meter}$. But this is for static fields, not waves. What about intense waves ?

I hope the question is clearer now.

• Here is a link to a field strength calculator. This is for radio frequency applications, but it's completely frequency independent, so it's just as valid for light: giangrandi.ch/electronics/anttool/tx-field.shtml – CuriousOne Apr 1 '16 at 14:43
• What about the case of a laser ? – Cham Apr 1 '16 at 14:58
• Electromagnetic wave is electromagnetic wave. As long as we are talking about the propagation in vacuum, the values are independent of frequency. One would have to use a correction factor for non-monochromatic light, but I assume by laser you do mean monochromatic waves. – CuriousOne Apr 1 '16 at 15:00
• Cham, you touched a sore spot. – HolgerFiedler Apr 1 '16 at 19:13
• @Cham It is a weakness of all theories about EM radiation that there isn't prediction about the amplitude of EM waves. As long as the field of a point charge is defined as to be infinite and as long as the constituent of electric field, magnetic field and photons are not explored, the wave-particle-duality will not be shifted by a deeper theory. – HolgerFiedler Apr 2 '16 at 15:43

The electric field strength is related to the power of the laser by the Poynting vector. This is given by:

$$\mathbf{S} = \mathbf{E} \times \mathbf{H}$$

and the magnitude of $\mathbf{S}$ is the power. Assuming we can treat your laser as a plane wave (which seems reasonable) then $\mathbf{E}$ and $\mathbf{H}$ are at right angles so the power is simply:

$$P = EH$$

and $H = E/\eta$ so we end up with:

$$P = \frac{E^2}{\eta}$$

In this expression $P$ is the peak power but what we really want is the average power, because that's what your laser spec will give. As it happens this just introduces a factor of a half:

$$P_\text{av} = \frac{E^2}{2\eta}$$

Remember that $P_\text{av}$ is the power per unit area so you need to take the power of your laser and divide by the beam area. Then substitute in the equation above and solve for $E$.

• Well, I don't know much about real laser power (in watts/m^2). So what amplitudes would you give, in volt/meter ? – Cham Apr 1 '16 at 15:16
• @Cham: I think you should research laser powers. It won't take much Googling. – John Rennie Apr 1 '16 at 15:21
• I don't think this would be so simple. In your calculation above, you should take the distance to the source into consideration. Also, what is $\eta$ ? My question is just about the typical amplitude value in an experiment with lasers. – Cham Apr 1 '16 at 15:41
• @ThePhoton, it is implied that the laser is in empty space. No air. So this is irrelevant. If you prefer, you could remove the word "intense", and consider the highest amplitude values we could find in any laser. It is certainly NOT above $10^{18} \text{volt/meter}$ ! – Cham Apr 1 '16 at 16:24
• I see you have a very loose interpretation of the term "laser pointer". – The Photon Apr 1 '16 at 16:39

When you say things like

My question is just about the typical amplitude value in an experiment with lasers.

you run into trouble, because "typical experiments" use laser powers which cover many orders of magnitude.

In general, you find the electric field from the intensity $I$ of the light, which is equal to the amount of power per unit area transmitted by the light, using the formula $$I=\frac{c\varepsilon_0}{2}E^2,$$ which you can invert as $$E=\sqrt{\frac{2I}{c\varepsilon_0}}.$$

Now, to get the intensity, you need the laser power and the spot size, and here is where you run into trouble, because those can vary enormously depending on what you want to achieve with the experiment. However:

• The spot size is usually not much of a problem. For a regular laser pointer you can take it to be of the order of $1\:\mathrm{mm^2}$, but if you focus tightly then you can get down to focal spots just bigger than the wavelength, so for visible light on the order of $1\:\mathrm{\mu m}^2$.

• On the low end, many experiments run on the single-photon per shot regime, with maybe one shot every microsecond, which is overall a very small laser power.

• For a regular laser pointer, the power will be limited to about $1\:\mathrm{mW}$. (Any higher than that and it becomes dangerous.)

• On the high end, the most intense lasers currently in use produce electric fields so strong that when an electron oscillates driven by the laser electric field, (i) it reaches the relativistic regime, and (ii) its cycle-averaged energy of oscillation becomes several times greater than the electron rest energy $m_ec^2$. That means that the electron's motion will spontaneously produce electron-positron pairs, with the positron later on recombining and emitting gamma rays. These are not nice experiments to stand around.

In terms of intensity, though, this happens somewhere along the $10^{19}\:\mathrm{W/cm^2}$ mark, but the record is probably rather higher than $10^{22}\:\mathrm{W/cm^2}$ these days.

That's enough to calculate the quantities you need; I have purposefully left some boxes empty so you can flex your muscles a bit.

• With respect to the final bullet point, we wish we could make laser beams intense enough to rip electron-positron pairs from the vacuum, but sadly we are a long way from that yet. This is known as the Schwinger limit, about 10^18 V/m, or a focused intensity of about 10^24 W/cm^2. It is interesting because the pairs would immediately interact with the electric field of the beam, breaking the linearity of electromagnetism. It is some way from being realized. – Calchas Jul 18 '17 at 0:02
• +1 I really appreciate this answer, as it gives actual experimental numbers to work with and think about. Equations are good, but getting some numbers for developing a mental scale is very helpful. – KF Gauss Feb 4 at 23:42

A simple rule: Electric field (in V/m) = 2745*sqrt(intensity), whereas the intensity is given in W/cm²