# Why is there more energy in a gamma wave vs. lower frequencies?

If energy is a function of frequency and amplitude, why are microwaves and lower frequency waves considered “less energetic” than gamma waves only as a function of their frequency and wavelength?

One always refers to the energy per quanta (photons). However, you can, of course, transport a given amount of energy by a lot of photons each having a very small energy i.e. large wavelength. But you can also pack all energy in only one single photon of extreme small wavelength and transmit all energy at once.

• Case study: a handheld UV light emits photons with more energy per photon... but the flames of a fire emit so many "lower energy" photons that they can warm your body when its cold outside. Jul 12 at 20:01

One reason is Wien's Displacement Law, which states the maximum emission in the blackbody spectrum is proportional to temperature:

$$\lambda^{-1}_{max}=bT$$

where

$$b\propto \frac{k}{\hbar c}$$

so the now cold CMB radiates at 2.7K, the Earth's atmosphere (250K) emits shorter microwaves, red hot coals start around 800K, and the pits of recently detonated nuclear weapons radiate thermal X-rays in the tens of millions of Kelvin.

Note: the total emission goes as $$\sigma T^4$$, so that argument may not clear up frequency vs. amplitude.

Another way to look at it is to realize a lone EM wave packet doesn't actually have a frequency, nor wavelength, nor energy. Suppose you're in deep space, and you have a wave packet of $$L=1000 \lambda$$, where $$\lambda$$ is the wavelength. A peak electric field $$\mathcal{E}$$ over a cross-sectional areas of $$A$$ defines the amplitude and volume, respectively. It's basically a "box" of propagating EM.

What is the energy in the box? For simplicity, ignore the magnetic field, then the energy density is

$$E = A\times L\times(\epsilon_0\bar{\mathcal{E}}^2) = 1000\lambda A \epsilon_0\bar{\mathcal{E}}^2$$

The total number of photons in the box is:

$$N = \frac{E}{\frac{h}{\lambda c}} = 1000\lambda^2 A \frac{c\epsilon_0}h\bar{\mathcal{E}}^2$$

Those results describe the wave packet from your reference frame. The energy and wavelength are specific too you . Another observer moving against the direction of propagation at $$+v$$ will see:

1. Length contracted box: $$L\rightarrow L'=L/\gamma$$
2. Length contracted wavelength: $$\lambda\rightarrow \lambda'=\lambda/\gamma$$
3. A boosted transverse electric field: $$\mathcal{E}\rightarrow \mathcal{E}'=\gamma\mathcal{E}$$

(with $$\gamma=1/\sqrt{1+v^2/c^2}$$) For a total energy of:

$$E' = 1000\lambda' A \epsilon_0\bar{\mathcal{E'}}^2 = 1000\frac{\lambda}{\gamma} A \epsilon_0(\gamma\bar{\mathcal{E}})^2 = \gamma E$$

and number of photons:

$$N'= 1000\lambda'^2 A \frac{c\epsilon_0}h\bar{\mathcal{E}'}^2 =1000\big(\frac{\lambda}{\gamma}\big)^2 A \frac{c\epsilon_0}h(\gamma\bar{\mathcal{E}})^2 = N$$

So the wavelength shortens and the energy increases, while the total number of photons is unchanged.

Thus: shorted wavelength is associated with higher energy, regardless of amplitude.

• Ha! Amazing answer, thanks… but can you dial it back a couple notches on the math scale…? “ Another way to look at it is to realize a lone EM wave packet doesn't actually have a frequency, nor wavelength, nor energy. ” Then how is energy “emerging?” Jul 14 at 14:45