# Tag Info

7

For the first question, yes. Because the surface of the sun is close to a blackbody emitter, it radiates at all wavelengths below the peak. So radio waves are included. However, the longer the wavelength, the less the power that is put into that portion of the spectrum. Radio is so far from visible light on the EM spectrum that the solar radiation in ...

1

The gray model results from starting with the assumption that we have a plane-parallel slab: The light ray from the source (i.e., the star's atmosphere) travels at some angle, $\theta$, from normal, $z=0$. Since the light is coming from an angle, we need to account for that by modifying the radiative transfer equation to have a vertical optical depth, ...

3

It has nothing to do with the modulation, AM or FM; it is because the wavelength difference. The so-called AM band is between 540kHz and 1600kHz, so its wavelength is about 300m, or so. The FM radio operates in the 88MHz to 108MHz band, or around 3m. The longer 300m EM wave reflects from the gap between the metal in the bridge and the ground (the latter is ...

0

This same question confused me a lot "when I was just a lad". Your eye responds to the frequency of light, not the wavelength. Whatever happens to the wavelength between source and retina is immaterial to the color. The frequency does not change, as you point out in (4).

1

Spectral Power – Image B (top right). The nature of images C and D is explained by the spectral power image, B. You can see from spectral power image that most energy is concentrated at low frequencies (the brighter middle part of the image). The higher frequencies (moving away from the centre) have less energy in them because there is less brightness in ...

2

This still doesn't explain to what the importance of the factor of i is Euler's formula $$e^{i\omega} = \cos \omega + i\sin \omega$$ Thus, the real part of $e^{i\omega}$ is $\cos \omega$: $$\cos \omega = \frac{e^{i\omega} + e^{-i\omega}}{2}$$ and the imaginary part of $e^{i\omega}$ is $\sin \omega$: $$\sin \omega = \frac{e^{i\omega} - ... 1 This the continuation of my comment: Every (real) quantity in time domain can be written like$$ A(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty}\mathrm{d}\omega\, a(\omega)\exp(i\omega t) $$In oder to guarantee the reality property of A(t), a(\omega) which in general takes on complex values has to fulfill a^{\star}(\omega) =a(\omega^{\star}) where ... 1 The use of complex frequencies in physics rather common, so I will explain it in a general context. Probably it will be sufficient, otherwise somebody else can put it in the context of general-relativity. If you consider a mode of frequency \omega = \omega_R + i\omega_I then the oscillation has time-dependence like \exp(-i\omega t) = \exp(-i\omega_R t) ... 2 Fourier frequencies, and particularly complex ones, are best thought about in terms of oscillating exponentials rather than sines and cosines. That is, you express the function of interest f(t) as some sort of superposition (sum, series, or integral transform) of complex exponentials e^{-i\omega t}:$$ f(t)=\sum_\omega\!\!\!\!\!\!\!\!\int \,a_\omega ...

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