I How to Measure the Energy of an Electromagnetic Wave?

[DavAeroEng]

We know a way of measuring energy of a electromagnetic wave is the Poynting vector, which is independent of the frequency. But let's say we want to make two different electromagnetic waves, with different wavelengths and so different frequency, but with the same amplitude (so same Poynting vector).In my opinion the guy in the image is spending more energy for the wave with higher frequency. But the Poynting vector, doesn't care. So is there another property of Electromagnetic Waves that accounts for the energy that the guy used to make the wave?

Also think of it in terms of kinetic energy, let's say you have a rope attached to a rotating wheel. The rotational kinetic energy of the wheel is: K= 1/2 * I * w . I = 1/2 *m*r^2. w is the angular speed. We know w=2*pi*frequency. While r we could say is the same as the amplitude of the wave of the rope. Well it is pretty simple to see that the energy to make a higher frequency wave, is higher. Naturally also the amplitude has influence over the energy of the associated rope wave.

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[mfb]

Electromagnetic waves and mechanical waves in a rope are not the same. The difference matters here. In a rope, the same displacement amplitude has a higher energy if the wavelength is shorter, but for electromagnetic waves that is not the case.

 

[DavAeroEng]

Electromagnetic waves and mechanical waves in a rope are not the same. The difference matters here. In a rope, the same displacement amplitude has a higher energy if the wavelength is shorter, but for electromagnetic waves that is not the case.

Ok I may get this point but so Can you give an explanation to what happens if to an apple's temperature is gamma ray and a radio is shooted at it? Given the fact that the amplitude of both waves are the same.

The temperature of the apple raise more with gamma rays or will be the same by using radio waves, at the same final time T?

 

[osilmag]

Gamma rays have more energy than radio waves, however they might pass right through the apple because their wavelength is so small. I don't know if the situation you give is valid.

The energy of an EM photon is found from E=hf.

 

[tech99]

We make an EM wave by accelerating a charge back and forth. The amplitude of movement will be less at higher frequencies, as with sound, but the power radiated will be the the same, because it depends on both velocity and amplitude. For similar reasons, the antenna for high frequencies is smaller. The radiated power is the work done on the charge and when we say that higher frequencies have more power, we mean that the photons have more energy at the higher frequency.
When an EM waves strike or pass an object the power abstracted from the wave depends on physical factors about the object, such as its size and material. In the case of radio waves and an apple, if the wavelength is long compared with the diameter of thr apple, the energy absorbed will be small.
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