Advanced Guesstimation 1: The Temperature of Lightning

The Question

The temperature of lightning is hotter than the surface of the Sun, averaging 20000 K, and ranging from 15000 – 60000 degrees Fahrenheit (8600 – 33000 K), whereas the surface of the Sun is averages 5800 K. The temperature range of lightning is measured using spectroscopy, but is this is mathematically demonstrable? Here we will calculate the temperature of lightning using basic electromagnetic theory and the Stefan-Boltzmann Law. Hopefully the presented model will be accurate enough to produce accurate results.


Altitude of storm clouds

2000 – 16000 m (average 3000 m)

Base area of storm clouds

108 m2


107 – 108 V

Resistivity of air

1.3 – 3.3 x 1016 Ω m


Figure 2. Plasma storm on Earth

Proposition 1

Treating the Earth and cloud as a capacitor, the current of lightning can be estimated.

Treating the Earth and cloud as a parallel plate capacitor, the stored energy is equal to the product of its capacitance and voltage.

Q = CV

Q = \frac{{\epsilon}_{0} kA}{h} V

Here, h is the base altitude of the cloud in metres, and A is the area of the base of the cloud in square metres. The dielectric constant k of air is approximately one, so the stored energy is approximately

Q = \frac{(8.854E-12)(1)(10E8)}{3000} (10E8) = 30 Coulombs

The average current is defined as passing charge density divided change of time.

i = \frac{\Delta Q}{\Delta t} = \frac{30}{0.2} = 150 Amps

Proposition 2

Treating the flash of lightning as radiation from heat, the temperature of lightning can be estimated.

Suppose the energy stored in an Earth-cloud capacitor is suddenly released. The energy is transferred between the Earth and cloud through the atmospheric medium, which if assumed to be the path of least work, would discharge minimum quantities of stored energy. This being said, the particles of a path traced from the Earth to the storm cloud would ionize the atmospheric gas particles, producing plasma that will radiate light, which can be measured by spectroscopic instruments.

The power dissipated is {i}^{2} \rho \frac{h}{A}, where i is the average current of the Earth-cloud capacitor, \rho  is the resistivity of air, h is the base altitude of the storm cloud, and A is the base area of the storm cloud. Equating the power dissipated to Stefan-Boltzmann law, which relates power to temperature, we arrive at

{i}^{2} \rho \frac{h}{A} = e\sigma {A}_{S} {T}^{4}

where {A}_{S} is the surface area of the body of ionized air. The emissivity of the atmosphere is approximately one. I will assume the body of ionized air to be contained in a cylindrical volume.

{i}^{2} \rho \frac{h}{A} = \sigma 2\pi rh{T}^{4}

\frac{{i}^{2} \rho}{2\pi rA \sigma} = {T}^{4}

Interestingly, the temperature of lightning is independent on altitude. The main factor for storing all that energy is in the size of the cloud. Also, if the lightning spreads out over a larger cylindrical space, the temperature decreases.


Q = \frac{(8.854E-12)(1)(10E8)}{3000} (10E8) = 30 Coulombs

i = \frac{\Delta Q}{\Delta t} = \frac{30}{0.2} = 150 Amps

T = \sqrt[4]{\frac{{150}^{2} (3.3E16)}{2\pi (100)(10E8)(5.67E-8)}} = 21366Kelvins

This figure is within the 8600K-33000K temperature range!


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