Hot answers tagged doppler-effect
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why do textbooks never mention this?
Because in order to travel at supersonic speeds, human beings must be enclosed in a rigid metal tube of some sort. Also, these metal tubes they ride in at those speeds generally tend to be insulated against noise from the outside.
As for trying to place some sort of microphone outside said metal tube, the ...
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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 ...
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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 ...
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You can use a reflector with gaps. Then the light from a car will alternate between reflecting and not reflecting at a rate dependent on their velocity towards the reflector. Please excuse my crude diagram:
As the car moves right to left, gaps in the reflector will cause it to appear to flash on an off.
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There are numerous distance indicators used for within the galaxy. The most common way is by using intrinsic magnitude. By knowing how bright an object would be if we were close, we can determine how far away it is by how dim it is. There are many types of stars where we have a rough idea of how bright they should be due to characteristics of the star:
...
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I would have to see the words in context, but the description "apparent frequency" seems strange to me. The frequency your ear detects is exactly what the most sophisticated scientific instrument would measure. The Doppler shift is real in that the frequency your ear detects is really the sound frequency in your frame.
I would guess "apparent frequency" ...
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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 $\nu \lambda=c$. ...
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I tracked down a spectrum of the sky at an altitude somewhere below 51° and overlaid it on the colors of the spectrum:
From this diagram, it appears that the intensity of the light admitted through the atmosphere diminishes significantly before reaching violet. Unless the perceiving retina was overpoweringly tuned to ...
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This is a great question, as it is both centrally important to modern astrophysics and cosmology, and it is misunderstood by very many people, including scientists themselves. Now the full, rigorous treatment requires general relativity, which I won't discuss in detail here. However, this is a topic that can be explained somewhat intuitively, so I'll give ...
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In addition to the special-relativistic version of the Doppler effect, there are other sources of redshift due to general relativity, in particular gravitational redshift (a consequence of gravitational time dilation which can also manifest as a blueshift) and cosmological redshift (a consequence of an expanding spacetime).
Hubble's law is due to the ...
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There is an important difference.
With sound waves, it makes a difference whether the source or the receiver is moving, because what really matters is their speed relative to the medium (air in this case).
With EM waves, on the other hand, it only matters what the relative speed between emitter and receiver is, i.e. it doesn't matter whether the source ...
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(I'm considering the speakers are emitting some kind of music or something nonperiodic, the situation gets a bit boring if you consider a uniform source)
It basically means Alice hears nothing. Atleast, not until Bob crosses (at which time your equation is no longer valid, the $-$ in the denominator becomes a $+$). She hears a sonic boom as Bob crosses her, ...
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Lemme make sure I have the assumptions correct:
1) The buoy is at rest in a known inertial frame in a known plane; the boats are in the same plane.
2) Each boat may make a "single measurement" of the frequency. I will assume this means that the time of measurement is long enough that the boat can resolve the frequency to arbitrary precision, but the ...
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If I understand you correctly, you want to desribe sound waves from the viewpoint of a moving observer. To do this, you just have to take the standard wave equation and perform a variable substitution $\mathbf{x}\mapsto \mathbf{y}:=\mathbf{x}-\mathbf{v}\cdot t$.
You start with
$$
\partial_t^2 p(\mathbf{x},t) = c^2\nabla^2 p(\mathbf{x},t)
= ...
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This is something that originally motivated quantum mechanics, and gave Planck's quantization law as natural. If you have a moving light source, and the stationary light source emits a radio-pulse of frequency $\omega$, if you boost the thing so that it is moving in the direction of the outgoing pulse, the frequency and energy have the same transformation ...
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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 ...
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The (non-relativistic) Doppler effect is the result of a Galilean transformation, and non-relativistic quantum mechanics is invarient under Galilean transformations so systems described with QM automatically show Doppler effects. There isn't any sense in which QM has to "explain" the Doppler effect.
The same applies to relativistic (Lorentz) ...
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The apparent paradox can be analyzed more clearly if we simplify it a bit. Let's assume we have one object with mass $M$ emitting two photons in opposite directions, each of them with momentum $p$.
In the rest frame of the object, choosing the X-axis to align with the emitted photons, we will have the following four-momenta (using units where $c = 1$):
...
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The argument is Einstein is a little annoying to read, here's a simpler version (I put it on Wikipedia under mass-energy equivalence a long time ago. It also appears somewhere on Terry Tao's blog, and it's the right way to make the argument).
Consider a mass M which is stationary. It emits two identical pulses of light in opposite directions, each of equal ...
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These thought experiments should always be made as simple as possible. Emitting in all directions is a needless complication. Consider Einstein's original derivation of E=mc^2 with only two light rays pointing in opposite directions (as Ron said above, this is a necessary assumption- it means the momentum of the light rays won't kick the bulb in one ...
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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 ...
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Actually No. Here's a Flash simulation for Doppler effect which I found useful while googling. It really does exist and not some random appearance..! I don't know whether the term Apparent frequency is good to use or not...
We know that Sound is a form of mechanical vibration. When the source moves away from the stable observer at some noticeable speeds ...
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The equation is correct, though whether it's $(c + u)/(c - u)$ or $(c - u)/(c + u)$ depends on your sign convention. As it stands your equation is correct if $u$ is positive when the source is approaching you and negative when the source is moving away from you. The Wikipedia article you cite actually uses the opposite convention.
Anyhow, provided $\nu^'$ ...
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The short answer to the (original) title question of "Does redshifts/blueshifts occur on Earth?" is yes, of course it does.
So why don't you notice, right?
The thing to note is that redshift from relative motion is proportional to the relative velocity divided by the speed of light.
The speed of light is 300,000 kilometers per second.
The fastest ...
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In the classical Doppler effect, you should think of $v_s$ and $v_r$ to be the velocities of the source and receiver relative to the medium, at the time when the signal was sent and received. In an accelerated reference frame, it is essentially the medium that is moving. Remember that the Doppler shift concerns the frequency sent by the source compared to ...
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The end of the equation, where it's written $\dots = 1+gh$, shouldn't be there, right? This just means that the denominator was erased. The last $=$ only holds within an approximation, $gh\ll 1$, up to the leading term.
At any rate, due to the difference-of-squares identity,
$$\frac{1+gh}{\sqrt{1-g^2h^2}} = \sqrt{\frac{1+gh}{1-gh}} = \sqrt{\frac{1+v}{1-v}} ...
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Maybe it is better to ask the other way around. Why is the frequency the only property that is independent? Although, mentioning the Doppler effect, you already found an example, showing that this is not exactly true. Hence the frequency depends on the source-velocity relative to me, and my velocity relative to the source. So "independent" is tricky and the ...
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Yes, you won't see the purple light because it's in the UV to Xray, depending on the actual speed and the angle you are looking at (sideways there'd be no change, if you go straight, and you better go straight).
Behind you there would be red-shift.
If you go fast enough, the 4 K background radiation (residue of the big bang, normally micro-wave ...
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The source puts out the same frequency; the observer only perceives a different frequency because of the relative motion between them i.e. the actual frequency of the source remains unchanged and that this apparent change in frequency is due to the relative velocity between the source and the observer.
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If I understand your question correctly, you're asking if there is an argument using light for some kind of mass - energy equivalence in the context of non-relativistic mechanics.
I think you might be forgetting that the classical theory of light, Maxwell's equations, are relativistically covariant and it was this tension between Maxwell's theory and ...
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