# Tag Info

115

It does! However it doesn't change the frequency enough to matter. An FM transmission is not a precise frequency. Instead it spans a range of about 100 or 200 kHz depending on which country you are in. So your FM radio actually accepts a range of frequencies either side of the central frequency. Let's suppose you're travelling at the maximum speed ...

36

The instantaneous change occurs when you consider the Doppler shift in only one dimension. In three dimensions you can consider the correction when the velocity vector and the separation vector are not parallel. Usually such corrections go like $\cos\theta$, where $\theta$ is the angle between the two vectors, but more complicated things are possible. ...

33

Let us consider that you are at rest and the car, which emits at frequency $f_0$, approaches you with speed $v$. The frequency you receive increases to $$f_1=f_0\frac{c}{c-v},$$ where $c$ is the speed of sound. When the car get passed you the perceived frequency is reduced to $$f_2=f_0\frac{c}{c+v}.$$ The ratio is $$\frac{f_1}{f_2}=\frac{c+v}{c-v}.$$ Now ...

32

The asymmetry comes from the medium. In the first case the source is at rest with respect to the medium and in the second case the receiver is at rest with respect to the medium. These two cases are not physically equivalent. In the case of the source at rest with respect to the medium the wavelength of the wave is isotropic, but not when it is moving with ...

30

To further add to John Rennie's answer: you don't even need autotuning for a frequency drift of the magnitude John calculates (10Hz): all FM receivers I've ever dealt with (I qualified as an electrical engineer in 1985 and worked a few short years in communications before returning to study) demodulated with a phase locked loop detector, whose job it is to ...

19

John and Rod already pointed out that the expected frequency shift from driving a car is "small"; I would like to expand a little more on the way FM works. FM = Frequency Modulation. The carrier (nominal center frequency) is being modulated - that is, in order to convey the audio content, the frequency is actually moving around deliberately in order to get ...

18

You can derive the relativistic Doppler shift from the Lorentz transformations. Let's start in the frame of the moving rocket, and let's take two events corresponding to nodes in the emitted wave (i.e. 1/$f$). Then in the rocket's frame the two events are (0, 0) and ($\tau$, 0), where $\tau$ is the period of the radiated wave. To see what the period of the ...

18

It has to do with something called the Doppler effect. Looking at it from the cosmic ray's point of view, the light it hits head on has a really high energy, and the light that hits it from behind is even colder/lower energy than what we see ($2.7$ Kelvin). If you want to stick to our point of view, then yes a photon hitting it from behind would boost it, ...

18

The object cannot occupy your same place as he passes you, so let us assume that the trajectory is a straight line that passes next to you. As the object approaches, the component of the velocity in your direction diminishes, to the point of being zero when the object is next to you. Thus the doppler effect will change continuously, from blue to zero to red ...

16

You do see the Doppler shift of the Earth's motion with respect to the CMB. It imprints a dipole component on the CMB temperature map. The motion of the Earth around the Sun is a modest 30 km/s, but even this doppler shifts inferred temperatures by a factor $\simeq v/c$ and needs to be taken out of the CMB analysis. A larger effect is the motion of the Sun ...

16

The ordinary Doppler effect is independent of relativity; it's basically just a fact of kinematics. It's not even really a wave phenomenon; it also applies to particles. For example, the Doppler effect explains why your car windshield gets wetter faster when you're driving than when you're parked. The formula for the Doppler effect is $$f_o = \frac{v - v_o}{... 14 You can do such estimations! It even turns out that you don't need perfect pitch. As a sanity check, picture a train going past you, blowing its horn. Its horn consists of three notes forming a chord (incidentally, the chord was chosen because it was annoying). Now let the train go by you. It's still the same chord, just with a lower root. If the ... 14 As Dale has said, the asymmetry is due to the medium in which waves propagate. The propagation is tied to the medium (the propagation speed is only v_s with respect to this medium), so it is not simply the relative velocities of the source and observer that matters. In the relativistic version of the Doppler effect, however, the situation is entirely ... 13 Yes, the sound can be reversed. Thanks to JiK, we have this animation (Python source code) of a supersonic jet moving forwards that can illuminate what is going on: The red circle represents the first sound produced by the object, the blue circle the second sound produced by the object and the remaining (black) circles representing the sounds produced ... 13 The question is, in antenna's frame after acceleration, (which is indeed inertia) how can the frequency of the light still be shifted? The light frequency was always shifted in this frame. (i.e. the frequency was always different in this frame than in the frame where the receiver is intially not moving) You have two inertial frames, we can call them "A" ... 13 Yes, the de Broglie wavelength of a particle depends on the relative velocity between the particle and an observer. I find it easier to think about the de Broglie frequency instead of the wavelength. They are related by v=f\lambda or f=v/\lambda, where v=p/m. If the particle is moving towards the observer, the frequency appears higher, and if the ... 13 When thinking about the Doppler shift, I think it's important to decouple the wave from its source, that is: the source is not a property of the wave itself. So, in the relativistic case, there isn't any rest frame. All motion is relative, so that velocity symmetry is mandatory. A photon on its own does not have a rest frame, nor does it have an intrinsic ... 12 Temperature is related to kinetic energy in the rest frame of the fluid/gas. In non-relatvistic kinetic theory the distribution function is$$ f(p) \sim \exp\left(-\frac{(\vec{p}-m\vec{u})^2}{2mT}\right) $$where \vec{u} is the local fluid velocity. The velocity can be found by demanding that the mean momentum in the local rest frame is zero. Then \vec{u}... 12 Sean E. Lake's answer is excellent, and should be the accepted answer. I just wanted to provide an alternative way of seeing the same thing. Thermal equilibrium When particles interact, they exchange energy. This tends to bring an ensemble of interacting particles in thermal equilibrium, where each particle has the same distribution of energies, no matter ... 11 While not explicitly an answer, there is a nice historical connection to your question. In effect, the first public experiment that decisively illustrated the Doppler effect was almost exactly what you're describing. In 1845, Christrophe Ballot placed one group of trumpet players on a moving train and another group at a station. Having tuned everyone up ... 11 This asymmetry is best understood analyzing the case when the source moves towards the receiver approximately at wave propagation speed c. Then the Doppler ratio will be$$ \frac ff_0 = \frac {c}{c-c} = \infty  It means that receiver will not register any waves until the source will arrive fully at the receiver, and then the receiver will experience an ...

10

According to Bohr model, the absorption and emission lines should be infinitely narrow, because there is only one discrete value for the energy. There are few mechanism on broadening the line width - natural line width, Lorentz pressure broadening, Doppler broadening, Stark and Zeeman broadening etc. Only the first one isn't described in Bohr theory - it's ...

10

Spectroscopy is done on the starlight. Say we look for hydrogen lines. We know where they’ll appear in the spectrum of laboratory hydrogen. If the starlight is redshifted then all the hydrogen lines will be shifted by the same amount. As can be seen in an example spectrum below. Once the shift is quantified, we can work back what the unshifted spectrum of ...

9

The word "apparent" means "as observed at a particular point X". Different observers will observe different frequencies depending on their relative velocity to the source. This doesn't change the frequency of the sound that is generated; just the frequency of the sound that arrives at the ear of the observer.

9

The answer is yes, the atom does absorb radiation that does not exactly match the transistion frequency. This is due to the Doppler effect that everyone knows from an ambulance with siren driving by. The frequency you hear is higher if the ambulance moves towards you and lower if it drives away from you. It's the same with the atom. If the atom moves (and ...

9

[June 19,2016: thoroughly revised, giving a more detailed, comparative presentation and better references] General case. In relativistic thermodynamics, inverse temperature $\beta^\mu$ is a vector field, namely the multipliers of the 4-momentum density in the exponent of the density operator specifying the system in terms of statistical mechanics, using the ...

9

Update 1: 1) Note added in proof: The photon stress-energy densities obtained below more or less heuristically are identical to those obtained in more rigorous approaches from the electromagnetic stress-energy density tensor. 2) The physical reason why the stress-energy argument retrieves the detailed balance result in the OP, but is inequivalent to simply ...

8

The linewidths come out very naturally from Maxwell's Equations by treating the atom as a tiny classical antenna. I do the calculations for the 2p-1s transition in hydrogen on my blogsite here: The Semi-Classical Calculation The idea is that from the Schroedinger equation, the superposition of the s and p states gives you get a tiny oscillating dipole about ...

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