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Yet another way. Consider a wave of a form $f(kx-\omega t)$ in the emitting frame, being observed as some other wave $f^{\prime} (k^{\prime} x^{\prime}-\omega^{\prime} t^{\prime})$ in a frame moving at $v$ along the x-axis. We actually don't care what $f^{\prime}$ is, only the argument of the function matters. So use the Lorentz transforms to change $x$ and ...

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Neither origins nor emitters nor receivers are required: consider a photon with momentum $\vec{p}$: it has 4-veclocity $p_{\mu}=(||p||/c, \vec{p})$. To compute what another observer moving at $\vec{v}$ sees, do a Lorentz transform: $p'_{\mu} = \Lambda_{\mu}^{\ \nu}p_{\nu}$. From this you'll get the relativistic (time-dilated) Doppler effect and stellar ...

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The motion does affect the observed continuous spectrum. The flux at each wavelength is doppler shifted to a new wavelength. The result is of course a new continuum, so a doppler shifted continuous spectrum is rather hard to distinguish from one that is not. In particular, unless the continuum has some feature or break in its slope, then it can be ...

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That's correct. Imagine a sound emitter and receiver traveling at twice the speed of sound in the same direction with the emitter behind the receiver. Even though the relative velocity is zero, the sound will never reach the receiver because sound doesn't travel fast enough. At this point, the frequency calculated from Doppler shift formulas would have ...

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As the particle is moving in counterclockwise direction, the velocity of the source at point A will be along the line joining the source and detector and hence the apparent frequency will be maximum at that point. Therefore apply the equation for Doppler and maximum apparent frequency detected by detector. Moreover calculate the time taken by source to reach ...

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Lets take a simpler case: instead of a song, the fighter pilot emits two pulses separated by some time. After the first pulse is emitted, he will indeed outrun it - he is after all going faster than the sound speed. He will then emit a second pulse in front of the first (and outrun that as well). As a distant observer, you will indeed then hear the second ...

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In case of a travelling observer , there is change in wavelength,and the magnitude and sign of change in wavelength depends on the velocity of the observer. Let's say an observer moves towards a stationary source emitting pulses with a frequency of 'f' with a velocity vo . A pulse reaches the observer and by the next time a pulse reaches the observer, the ...

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The rate emission of photons is a measure of total power: 100 photons per second vs 1,000,000,000 photons per second. The individual photon emitters are, for example, electrons being driven back and forth in a radio circuit: the photons are quanta of radio waves, with the frequency of the oscillator circuit setting the energy of each photon by Planck's ...

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I believe I found the solution for my question, it was actually fairly simple (no calculus or explicit use of the Doppler effect!). In the received signal $r(t)$, the value at any time $t$ is equal to the sum of the perturbations that would reach the observer at time $t$. Assuming that the speed of the source is significantly lower than the speed of a wave ...

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I would assume only the distance between the source and the observer, $x(t)$, has any effect on the frequency scaling. Wave Dispersion There are multiple effects that matter here if the medium is dispersive (i.e., the wave frequency depends upon the wave number or another way of stating this is that the phase speed depends upon the wave frequency ...

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