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7

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: ...


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

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 ...


6

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 ...


5

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 ...


5

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 $\nu \lambda=c$. ...


5

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.


5

What you hear in this experiment is the combination of the Doppler effect and the beat. As John Rennie points out, the frequency change due to the Doppler effect would be hardly audible. However, the frequency between the two tuning forks will now be slightly different, which results in a intensity modulation, called the "beat".


4

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" ...


4

(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, ...


4

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 ...


4

Your thoughts are basically right; the essential point is that sound waves travel through a medium at a certain speed, $c_s$, and as a result, there is an asymmetry between the effects due to the velocity $v_o$ of the observer relative to the medium and the velocity $v_s$ of the source relative to the medium, but no such asymmetry exists in the Doppler ...


4

The doppler shift causes a shift in wavelength at the origin of the wave (the frequency of the source never changes). This results in a shift in frequency for the observer. In the link below you can see the emission of the wave for a moving source causes the wavelength to be shorter in front and longer behind. The actual source isn't changing in ...


4

Yes it does have an effect as you correctly reasoned and is even used in cold atoms technology where it is known as the Doppler cooling technique.


3

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 ...


3

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) ...


3

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 ...


3

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) = ...


3

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 ...


3

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 ...


3

I don't have Carroll's book, but I don't recognise the description you give of the derivation of the red shift, and in particular I don't see why the relativistic Doppler shift is relevant. The derivation I'm familiar with is to say that the change in potential energy is $mgd$, where $m$ is the effective mass given by $E = h\nu = mc^2$. So: $$ h\nu_e - ...


3

That's right, running away from a gamma source fast enough would shift them into the visible portion of the spectrum. It goes without saying that he'd have to run quite fast: $$\frac{\lambda_{\rm obs}}{\lambda_{\rm emit}} = \sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}$$ Picking rough round numbers for gamma radiation at $\lambda=10\;\rm pm$ and visible at ...


2

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}} ...


2

One neat trick for middle ranges requires a dynamically bound system whose components have measurable proper motions. There are a reasonable number of globular clusters that qualify. If you project those motions across the sky, they will appear to come together (to some approximation) in two places (one forward and one backward), and the directions of those ...


2

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 ...


2

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 ...


2

There is a list on Wikipedia. Radar guns use an optical Doppler effect to measure speed. Their acoustic equivalent is used in medicine, where it's called an ultrasound and used to measure blood flow or other sorts of motion in the body. Animals that use echolocation can use the Doppler shift to gain information about the motion of their surroundings. A ...


2

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 $c = 1$): ...


2

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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