49

Electromagnetic radiation will continue to travel until it is absorbed. Some of your wifi signal is escaping to space where it may continue traveling for a very long time. However, the strength of your wifi signal will degrade with distance according to the inverse square law. So if you double the distance between your device and the wifi transmitter, your ...


16

There are some technological issues to solve with putting any large telescope into space - and a space telescope is required at UV wavelengths. It is not possible to optimise such an instrument to work at both UV and IR wavelengths because of issues like cooling, mirror coatings and such-like. The simple angular resolution limit of a telescope goes $\lambda/...


15

You're correct in that the sharp dropoff is simply because there are very few planned major telescopes operating in the UV range, whereas there are a substantial number planned in the infrared range. As I mentioned in my answer you linked to, CHARA and the EELT, two of the top planned infrared/visible projects, will use new adaptive optics technology, making ...


13

Signal-to-noise ratio In addition to what others have said, it is very important to understand the difference between just detecting something and decoding a useful signal from it. The CMB is essentially random noise – in fact, that's how it was discovered in the first place! Still, in normal conditions it is easily drowned by other noise sources and was ...


10

There's higher quantity of atoms in your 20cm wall than there is in the 13.8 billion light-years travelling to the CMB, so the wifi waves hit atoms on their travel. Space has an average density of 5.9 protons per cubic meter, that's 10^-25 g/m3, and there are only 1.22*10^26m to travel to the CMB source. The CMB is an omni directional transmitter source, it'...


9

The possible planet 9 is thought to be about 10 Earth masses and is unlikely to be a gas giant (it may be the core of an "interrupted" gas giant). As such, it will not be generating significant luminosity itself and would be rocky, or more likely, icy in character. It would thus only be seen by reflected light. The considerations for what wavelength to ...


7

tl;dr: Yes. Feynman's beads on a string argument The other answers skirt what I think is the issue that the OP is asking about. In a lossless medium a spherical wave packet itself, caused by a disturbance, will not "loose energy" itself. If you integrate over a large volume you get a constant energy versus time. Of course the flux per unit area will ...


7

Direct reflection of sunlight is the most likely scenario for a ninth planet discovery, however that does not hold if the object has a very low albedo. I assume you are interested in what wavelengths the planet would radiate. For the surface temperature, the rotation of the planet is important. If it is locked with one side facing the sun, or rotates very ...


6

There are two basic ways to detect such an object. First is to detect it through reflected sunlight. Second is from the heat that it produces. We already know that the reflected light of such an object likely would be around a 16.5 magnitude. To determine the infrared, we have to estimate the temperature The temperature very much depends on the composition. ...


5

Standard cosmological models predicts that the cosmological redshift and the speed of light are wavelength-independent. This result is confirmed observationally e.g. by Ferreras & Trujillo (2016), who used $\sim500\,000$ SDSS galaxy spectra down to a precision of $\Delta z \sim 10^{-6}$ and $\Delta z \sim 10^{-5}$ for galaxies at $z<0.1$ and $z>0.1$...


5

From Genzel et al. (2010), here's part of Fig. 7.7.1: This is part of the spectral energy distribution of Sagittarius A*, a flot of $\nu$ (frequency) vs. $\nu L_{\nu}$ (frequency times luminosity). For comparison, visible light is in wavelengths from $\sim4\times10^{14}\text{ Hz}$ to $\sim8\times10^{14}\text{ Hz}$, which happens to be around the bottom of ...


4

"Was it as loud as we think" is difficult to answer, since it's opinion-based. But since sound is nothing but longitudal oscillations in a gaseous medium, Big Bang was not at all silent. If the Universe were completely homogeneous, it would stay like that. But primordial quantum fluctuations ensured that space was a tiny bit more dense in some places, and a ...


4

Well, I don't think this question is entirely answerable. The true answer, I think, is that there really are no limits, as Rob Jeffries commented. However, using Wikipedia as my only source, the crab pulsar holds the current record for most energetic gamma ray emissions at 80 TeV (wavelength of about $1.5\times10^{-11}$nm). Whereas the longest detected ...


4

Regarding visible light, the nominal range is usually given as 400 nm (violet) to 700 nm (red). For light of the former wavelength to get Doppler-shifted to the latter means a z value of (700–400)/400 = 0.75, corresponding to a recession of 152,000 km/s, just over half the speed of light.


4

It's just as simple as taking the flux at some wavelength (just a number) and using this number to represent a visible intensity. If you only have one wavelength then you can only get a monochrome picture. However, if you have flux information at more than one wavelength, let's say three, you can use the flux at the longest wavelength to represent red (r), ...


4

Before I start, let me just say that this topic is vastly more complicated than you've presented and what I will be showing. The trouble here is that ultimately, everything you've done and I will do (to a lesser extent) uses approximations and assumptions. Because of this, it can sometimes be hard to understand the true underlying physics when looking at ...


4

So the two observed lines must be two of the three suggested lines, red-shifted by the same amount. That means the ratio of their wavelengths will be unchanged, so we need two of the suggested lines whose wavelength is close to the ratio $404.7/317.6$. It's easy to check that only the first two are close to that ratio, so they must be those two. So the ...


3

If the redshift is $z$, then the wavelength of light you observe is at a wavelength that is $1+z$ times longer than the wavelength emitted in the frame of reference of the galaxy. In this case $z=0.373$, so when you observe light centered on say the 3.4 micron band, then the light was emitted at a wavelength of 3.4/1.373 microns.


3

If you were in deep space, i,e., a spot somewhere between galaxies, then the night sky would look a lot like the sky seen by astronauts in the ISS, when they look away from both the ecliptic plane and the galactic plane, minus the individual Galactic stars. That is, it is like our deep sky photos in those directions that are dominated by galaxies, but no ...


3

For more fun, let's posit an experiment 10 000 years from now. I found some forum which provided this data for a transmitter: The Taldom transmitter is a large facility for longwave and shortwave broadcasting located near Taldom, Russia. It transmits on two longwave frequencies, on 153 kHz with 300 kW and on 261 kHz with a power of 2500 kW, the latter is ...


3

As the above commentator says, our technology isn't sufficiently sensitive to pick up everyday radio broadcasts from so far away, and even if we could it wouldn't be worth sending them a message. The earliest we could hope to receive a reply would be in 20,000 years time, by which time there might be no humans left on Earth to receive it, as a result of ...


3

If you have bins of equal log wavelength increment then a velocity shift corresponds directly to a pixel shift. That is the reason for binning in equal log wavelength increments. See this question for a short proof. All you need to do is work out what each pixel corresponds to in terms of a velocity. $$\Delta V = c\, \Delta \log \lambda$$


2

Virtually all KBO searches use mosaic CCD cameras operating in the visible part of the spectrum, usually with a red filter like SDSS-r' (centered at approx. 622nm SDSS filters) on big (~4m telescopes). This is what we did with the OSSOS survey I was (lightly) involved with (http://www.ossos-survey.org/about.html). Although cold KBOs are brightest in the ...


2

Someone mentioned redshift. Note that the cosmic background radiation limits redshift we can detect now but there is also a much older neutrino background radiation even though we can't do science on it (at least not yet). 10^20 electron volts (approximately) is the most powerful cosmic ray ever detected. By an order of magnitude or two I think. I once saw ...


2

In empty space, just like a light wave, they spread out, becoming less intense as they get further from their source, but never vanishing completely. At some stage the waves from a distant event might become undetectable in local noise, or so weak that quantum effects might become relevant, but essentially they never die out. As @uhoh points out, they do ...


2

It would be easier if you could include an example of the kind of image you mean, but I can guess that you're thinking of those drawings where there's a sine wave drawn in space and suggesting that that's what a wave "looks like". This plot from EMS 2- Wave Properties of Light shows "crest", "trough" and "rest position" when explaining light waves, and ...


2

$$v_r = c\left( \frac{\lambda - \lambda_0}{\lambda_0}\right),$$ where $\lambda$ is the observed wavelength and $\lambda_0$ is the wavelength at rest. This gives an equivalent velocity with respect to the central wavelength of the line. In your plot, what is shown is a line profile where the wavelength separation from $\lambda_0$ for that line (or from the ...


2

Steve Linton answers correctly that the line ratios must be (nearly) identical and hence identifies the two lines as Lyman $\alpha$ and C IV, but it's not true that the two obtained redshifts are suspiciously alike. In fact, when you do the calculation, you get (from $z \equiv \lambda_\mathrm{obs}/\lambda_\mathrm{rest} - 1$), $$ \begin{array}{rcl} z_{\mathrm{...


2

The WISE craft (Widefield Infrared Survey Explorer) surveys the sky in 4 wavelength bands; 3.4, 4.6, 12, and 22 $\mu$m. They are only using 3.4 and 4.6 since the coolant ran out. Source: FAQ "The Near-Earth Object Wide-field Infrared Survey Explorer at IPAC". When it's observing an object with an estimated redshift of, say, 0.373 (the lowest ...


1

It's a wave, and like any other wave, it loses flux (power over area) with the square of the distance. That is, if you double the distance it travels, your wave will have a quarter of the flux it had. If you triple the distance, your wave will have a ninth of the flux it had. This is a property of waves and can be observed in light intensity, sound, and any ...


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