117

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 ...


34

Yes, gravitational waves will undergo the same red-shift as any wave that propagates at $c$. There were probably very violent gravitational waves in the very early universe. If those waves hadn't been red-shifted, they'd be ripping us apart right now. If so, could observations of them be used like red-shifted electromagnetic waves from distant sources are ...


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 ...


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

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, ...


17

So does this mean that the wavelength of a particle depends upon the relative motion between the particle and the observer? Yes. The second statement is more or less equivalent to the first. Side note: The strange thing about this relationship is that it means that, e.g., wave train seen by one observer as consisting of one wavelength can be seen by ...


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}{...


15

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 ...


14

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 ...


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

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

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" ...


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

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}...


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 ...


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 ...


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

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 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.


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 ...


7

The first image shows an object traveling at Mach 1 ($v=c$). The second one shows the object traveling at some supersonic velocity ($v>c$). For both the cases, the longitudinal pressure waves pile up. Say the observer is standing in the ground and the object is traveling at $c$. The observer can't hear the pitch of sound because, the waves reach him all ...


7

[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 ...


6

I can't claim any experimental experience in this area (fortunately :-) but I thought it was interesting enough to be worth a bit of Googling. The results suggest there is a difference between shells and bombs. There is an extensive collection of eye witness accounts of WW2 at http://www.bbc.co.uk/history/ww2peopleswar/categories/, and searching this ...


6

The Doppler cooling limit is due to the fact that as the atoms absorb photons and spontaneously emit them in random directions they will not only have momentum no smaller than that of a laser photon, but they must also scatter one photon momentum in a random direction every natural lifetime of the excited state. If this lifetime is short then the random walk ...


6

The blackbody spectrum of the sun is the following, given $T=5778 K$. I admit I'm just copying from Wikipedia. $$I(\nu,T) =\frac{ 2 h\nu^{3}}{c^2}\frac{1}{ e^{\frac{h\nu}{kT}}-1}$$ The comic suggests that the reflection from scattering transforms the above spectrum by $1/\lambda^4$ (as in, it is multiplied by this). Light is a wave, so $\nu \lambda=c$. ...


6

I believe CuriousOne is correct, however it does not make any sense to define the threshold frequency for the photoelectric effect in anything but the rest frame of the metal. From the rest frame of the spaceship, the metal plate is rushing towards it and the apparent threshold frequency is lowered, but an occupant of the spaceship should realise that this ...


6

Since there seem to be no other attempts so far, I'll take a final stab, in the context of the particle annihilation problem. Short answer: As already suggested in several comments and one other answer, the paradox resolves itself when we take into account the restoring forces against the gas, either from the resonator walls or, if we dismiss the ...


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