# Tag Info

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

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

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

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".

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 ...

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.

4

Every star or galaxy contains some elements, and each element emits a particular frequency. Here are the lines of the Sun (https://en.wikipedia.org/wiki/Fraunhofer_lines) In particular, Hydrogen is present almost everywhere and Hydrogen lines are visible in most galaxy spectra. The Hydrogen-alpha line is particularly strong in many galaxies. This ...

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 ...

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: $$h\nu_e - ... 3 In addition to the special-relativistic version of the Doppler effect, there are other sources of redshift due to general relativity, in particular gravitational redshift (a consequence of gravitational time dilation which can also manifest as a blueshift) and cosmological redshift (a consequence of an expanding spacetime). Hubble's law is due to the ... 3 If I understand you correctly, you want to desribe sound waves from the viewpoint of a moving observer. To do this, you just have to take the standard wave equation and perform a variable substitution \mathbf{x}\mapsto \mathbf{y}:=\mathbf{x}-\mathbf{v}\cdot t. You start with$$ \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

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

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 ...

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 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 ...

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 $\nu^'$ ...

2

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 ...

2

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 ...

2

Yes, though it isn't quite as you think. You've probably heard of the Doppler effect. Even without the help of relativity we can see ultraviolet light if the source is moving away from us fast enough because the Doppler effect reduces the frequency of the light. We still get this effect in special relativity and it's called the relativistic Doppler effect. ...

2

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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