Science questions not worth a thread I: I'm a moron!

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Can you guys suggest any good YouTube channels for science content?
 
The air is still polarized, if only slightly. Obviously the effect is much weaker than in condensed matter, but I see no fundamental difference. A photon in air could be called a polariton, even if its properties are very close to that of a photon in vacuum.

Imagine light travelling from A to B in a very dilute gas consisting of a noble gas. The gas can be treated as solid spheres, an atom not interacting with things outside a radius R. The gas is so dilute that the chance for a light ray to hit a sphere between A and B is significantly lower than 1. (A and B must be quite close!) We measure the time of flight between A and B. What will the speed of light be? Will some of the photons be able to travel from A to B with c, while others take longer, or will they all take the same time to make the journey?
 
Imagine light travelling from A to B in a very dilute gas consisting of a noble gas. The gas can be treated as solid spheres, an atom not interacting with things outside a radius R. The gas is so dilute that the chance for a light ray to hit a sphere between A and B is significantly lower than 1. (A and B must be quite close!) We measure the time of flight between A and B. What will the speed of light be? Will some of the photons be able to travel from A to B with c, while others take longer, or will they all take the same time to make the journey?

You have to consider that the minimum size of a photon is on the order of the wavelength cubed. So if the atoms are not supposed to be interacting with the photons at all, the gas has to be extremely dilute to the point where we are talking about a high vacuum. And in that case the approach of light in a medium breaks down and you have to consider the light in a vacuum interacting with single photons.

Anyway, photons are very, very strange and measuring a speed for them is very tricky. Photons are delocalized (we can produce single photons hundreds of meters long) and it is hard to say how long a photon takes from A to B, when the photon is in both places at once, especially when you cannot map out the envelope of a photon in a single shot. What you can do is to make many measurements and take averages. Then you can take the differences of the mean arrival time at point A and B and get a speed. But that will be the speed of the pulse and not the speed of a single photon.

In your example, the distribution will be slightly shifted (and distorted) towards later arrival times at point B. But if the atom interacts with the photon, it will become entangled with the photon, so you could select on a state change of the atom and thus on those photons that are delayed. There you would indeed be able to detect a difference in the mean arrival time with respect to those where the state of the atom did not change.

In a real medium this would quickly become impossible, as the space of states you would have to watch grows very quickly.
 
Pluto actually does have a thin nitogren atmosphere at times.


Yes, you could see all the stars because the sun is so small and dim at that distance. It'd be just like nighttime here, mostly. You could also probably see almost all of them except those close to the sun in the sky.
Apparent magnitude of the sun at Pluto perihelion is -19.38, apparent magnitude at aphelion is -18.25, full moon on Earth is -12.74, sun on earth is -26.78.

The sun's apparent magnitude on an overcast day is -19, IIRC. :p
 
How do I find the galactic anti-podal point of a star other than Sol, as seen from Earth?

Meaning, the anti-pode of, say, Deneb for example, with Sgr A as the reference frame center, but the observer is puny little Earth.

For an analogy of what I'm trying to do, take Sgr A as seen from Earth, double the distance between us and Sgr A, and boom, you found Sol's anti-pode in this galaxy. Now, try calculating Sol's anti-pode if you were at a distant star. :p
 
Yeah, bite-sized videos wouldn't cut it for me. Sixtysymbols looked good, and I'm on the fence about Numberphile. Wonder what other prestigious university channels I can subscribe to other than Stanford and the University of California.
 
Questions:

1. What would an apparent magnitude of -24.73 look like to the human eye?

2. How bout when pointing a typical video camera at said light source?

3. Any difference with anything with said light source positioned a few degrees of arc from the sun?
 
This seems like a good place to ask this question. I'll post the video and ask the question.

Why does the boiling water turn to fog instead of freezing on the way down the building? I would think at -41 (I'm assuming celcius) it would be so cold, the water would freeze on the way down.


Link to video.
 
The water's still at a high vapour pressure (since it's boiling), so the tendency is to condense into vapour instead of freezing solid.
 
Questions:

1. What would an apparent magnitude of -24.73 look like to the human eye?

2. How bout when pointing a typical video camera at said light source?

3. Any difference with anything with said light source positioned a few degrees of arc from the sun?

As near as makes no difference, the same as looking directly at the sun. There's a measurable difference between the apparent magnitude of the sun and your figure, but your eye would be so overwhelmed that you'd not personally be able to tell.

Unless you meant, if the star we were orbiting looked like -24.73, what daytime look like? I wanna say like earth viewed during a cloudless day with some mild sunglasses on.
 
As near as makes no difference, the same as looking directly at the sun. There's a measurable difference between the apparent magnitude of the sun and your figure, but your eye would be so overwhelmed that you'd not personally be able to tell.

Unless you meant, if the star we were orbiting looked like -24.73, what daytime look like? I wanna say like earth viewed during a cloudless day with some mild sunglasses on.
I know when I enter apparent magnitude values into wolfram alpha, sometimes (I dunno how often), it says what it looks like if you were x distance from a 100 watt incandescent light, but dunno how accurate that is.

EDIT: What I meant was basically your first paragraph; a two-suns-in-the-sky scenario. Would my eye (or even a video camera for that matter) still be unable to tell a difference between the two, even though the second sun would technically be only about 15%-20% as luminous if our sun and said second one had an angular separation of about 6°?
 
I know when I enter apparent magnitude values into wolfram alpha, sometimes (I dunno how often), it says what it looks like if you were x distance from a 100 watt incandescent light, but dunno how accurate that is.

It could only be accurate if apparent magnitude was defined in a unit that could be accurately converted into lux. But apparently it isn't, though you can only get approximations that way. If the star has a spectrum that closely resembles a blackbody radiation spectrum, the approximation should be very good (if you apply the right conversion factors). For an arbitrary light source, there is missing information, so the comparison could be far off.


EDIT: What I meant was basically your first paragraph; a two-suns-in-the-sky scenario. Would my eye (or even a video camera for that matter) still be unable to tell a difference between the two, even though the second sun would technically be only about 15%-20% as luminous if our sun and said second one had an angular separation of about 6°?

If the two stars are close together you could probably tell which one is which by determining which one hurts your eyes more. I am not sure you could tell with a quick glance, as both would be saturating your eyes.

If you were looking at one of them, I don't think you would be able to tell which one it was (baring any obvious differences in color or size, of course). Human vision works on a logarithmic scale, so a fivefold increase does not make a large difference and it gets worse when the eyes are close to saturation (It is an interesting experiment to try to tell the power of a laser by looking at the brightness of the spot it makes).

Video cameras work on a linear scale, so in principle you have an easier time distinguishing them. However the stars would have to be in the range of the camera and not just saturate the pixels. That would not be the case with standard settings, so the question would be how far the settings of the camera could be adjusted beyond the standard settings (That is if the chip is not damaged by the light). If you had a set of neutral density filters, you could put filters on the camera until the light was in the range of the sensor and then you could easily tell the difference. This would also be safe for the camera. If the stars had the same spectrum, the filters would not have to be strictly neutral density, so you could use makeshift filters by holding sheets of paper in front of the lens, for example.
 
It could only be accurate if apparent magnitude was defined in a unit that could be accurately converted into lux. But apparently it isn't, though you can only get approximations that way. If the star has a spectrum that closely resembles a blackbody radiation spectrum, the approximation should be very good (if you apply the right conversion factors). For an arbitrary light source, there is missing information, so the comparison could be far off.




If the two stars are close together you could probably tell which one is which by determining which one hurts your eyes more. I am not sure you could tell with a quick glance, as both would be saturating your eyes.

If you were looking at one of them, I don't think you would be able to tell which one it was (baring any obvious differences in color or size, of course). Human vision works on a logarithmic scale, so a fivefold increase does not make a large difference and it gets worse when the eyes are close to saturation (It is an interesting experiment to try to tell the power of a laser by looking at the brightness of the spot it makes).

Video cameras work on a linear scale, so in principle you have an easier time distinguishing them. However the stars would have to be in the range of the camera and not just saturate the pixels. That would not be the case with standard settings, so the question would be how far the settings of the camera could be adjusted beyond the standard settings (That is if the chip is not damaged by the light). If you had a set of neutral density filters, you could put filters on the camera until the light was in the range of the sensor and then you could easily tell the difference. This would also be safe for the camera. If the stars had the same spectrum, the filters would not have to be strictly neutral density, so you could use makeshift filters by holding sheets of paper in front of the lens, for example.
Cool; thanks! :)

When you say "spectrum", you're referring to the spectral type of the stars, correct? In that case, the second star I'm thinking of is a K dwarf. :)
 
At -40 degrees, that temperature could be either Celsius or Fahrenheit. :)
 
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