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Watsisname
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04 Aug 2017 11:47

Marko S. wrote:
Source of the post And also: Does the Moon looks like it travels from west to east during the eclipse because it travels slower than Sun on our sky? Because a lot of people have trouble answering this question. i think I got that right. Well, that's what I would answer if someone asked me.

That's correct. :)  Both the Moon and Sun move across the sky east to west, but when you add in the orbital motion of the Moon around the Earth (which is counterclockwise viewed from above the north pole, or west to east), it crosses the sky more slowly.  West to east relative to the Sun and the stars.

Starlight Glimmer wrote:
Source of the post Lets say we had a piece of wood, about the width of a 2x4, a lightyear long somehow. Lets say it was in space, and you were in a spacesuit with some form of rocketpack, you turn it on and push the piece of would. Would the entire thing move as you push 1 end? Thatd break the speed of light though. 

FFT has is right. The block does not move all at once.  The far end of the block is at rest until the information that you pushed it can reach it, which in a solid or 'rigid' body is always slower than the speed of light in vacuum.

FastFourierTransform wrote:
Source of the post But I think there is an interesting issue here appart from this. Would you say that the rod has been deformed? or is acting as a deformed body? or is totally rigid but in a very special geometry (space-time)?. With this I mean for example that if you consider the fact that simultaneity can be broken could you consider from the frame of reference of the traveling perturbation that the rod is moving as a rigid body? Wouldn't from that frame of reference all the parts of the rod moved at the same time?

Yes, the rod has been very slightly deformed, and the force of the push propagates down its length as a sound wave (a distortion of the structure.)  So the concept of 'rigidity' is always slightly wrong -- there is no such thing as a perfectly rigid body in nature.  

You could consider the frame of reference of the wave propagating down the rod.  If the wave propagates at the speed of light, then it is correct that the whole rod moves all at once according to you.  But then the rod's length is also infinitely short, because of the relativistic length contraction!  You will also find that clocks fixed to the ends of the rod read very different times -- a synchronized network of clocks in the rod's frame are not in sync according to you!
 
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04 Aug 2017 12:41

Watsisname wrote:
Marko S. wrote:
Source of the post And also: Does the Moon looks like it travels from west to east during the eclipse because it travels slower than Sun on our sky? Because a lot of people have trouble answering this question. i think I got that right. Well, that's what I would answer if someone asked me.

That's correct. :)  Both the Moon and Sun move across the sky east to west, but when you add in the orbital motion of the Moon around the Earth (which is counterclockwise viewed from above the north pole, or west to east), it crosses the sky more slowly.  West to east relative to the Sun and the stars.

Starlight Glimmer wrote:
Source of the post Lets say we had a piece of wood, about the width of a 2x4, a lightyear long somehow. Lets say it was in space, and you were in a spacesuit with some form of rocketpack, you turn it on and push the piece of would. Would the entire thing move as you push 1 end? Thatd break the speed of light though. 

FFT has is right. The block does not move all at once.  The far end of the block is at rest until the information that you pushed it can reach it, which in a solid or 'rigid' body is always slower than the speed of light in vacuum.

FastFourierTransform wrote:
Source of the post But I think there is an interesting issue here appart from this. Would you say that the rod has been deformed? or is acting as a deformed body? or is totally rigid but in a very special geometry (space-time)?. With this I mean for example that if you consider the fact that simultaneity can be broken could you consider from the frame of reference of the traveling perturbation that the rod is moving as a rigid body? Wouldn't from that frame of reference all the parts of the rod moved at the same time?

Yes, the rod has been very slightly deformed, and the force of the push propagates down its length as a sound wave (a distortion of the structure.)  So the concept of 'rigidity' is always slightly wrong -- there is no such thing as a perfectly rigid body in nature.  

You could consider the frame of reference of the wave propagating down the rod.  If the wave propagates at the speed of light, then it is correct that the whole rod moves all at once according to you.  But then the rod's length is also infinitely short, because of the relativistic length contraction!  You will also find that clocks fixed to the ends of the rod read very different times -- a synchronized network of clocks in the rod's frame are not in sync according to you!

When you account for all this, you'll discover the pulse actually is travelling through the rod at the same speed in all frames of reference.

Oh so its basically just a pulse? Weird. 
 
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Watsisname
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04 Aug 2017 12:59

Ja.  It's basically a mechanical vibration or sound wave, so its speed is determined by the properties of the material.  So if you try shoving a solid beam, you can't make the other end move sooner than it takes for the sound to reach it.  In most materials the speed is quite fast though (hundreds or thousands of meters per second) so this isn't very obvious.  List of sound speeds in some common materials.
 
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04 Aug 2017 13:27

Watsisname wrote:
Source of the post That's correct.  Both the Moon and Sun move across the sky east to west, but when you add in the orbital motion of the Moon around the Earth (which is counterclockwise viewed from above the north pole, or west to east), it crosses the sky more slowly.  West to east relative to the Sun and the stars.
Thanks, science officer! :) I was hoping I was correct!
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Watsisname
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04 Aug 2017 23:39

spaceguy wrote:
Source of the post What are those whitish 'cracks' you see around the surface of the sun, notably around sunspots? They're visible optically but only near the solar limb. Spicules perhaps?

Those are called 'plages', which are bright features in the Sun's chromosphere, and most apparent with an H-alpha filter. Spicules are smaller, and dark against the background.

I knew about spicules previously, but not the plages, so that was cool to learn. :)  Solar observing is very fascinating.
 
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05 Aug 2017 07:08

Watsisname wrote:
spaceguy wrote:
Source of the post What are those whitish 'cracks' you see around the surface of the sun, notably around sunspots? They're visible optically but only near the solar limb. Spicules perhaps?

Those are called 'plages', which are bright features in the Sun's chromosphere, and most apparent with an H-alpha filter. Spicules are smaller, and dark against the background.

I knew about spicules previously, but not the plages, so that was cool to learn. :)  Solar observing is very fascinating.

Ah, thank you. In my image, I don't think there was any H-filtering on the left image. Or is that not how it works?
 
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Mosfet
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05 Aug 2017 09:14

spaceguy wrote:
Source of the post Ah, thank you. In my image, I don't think there was any H-filtering on the left image. Or is that not how it works?

The image on the left is taken with a common WL filter (white light), the other one with a CaK filter. It shows the emission of the Calcium K band, slightly different than Hidrogen-alpha.
I didn't knew that one, cool.
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07 Aug 2017 13:43

Watsisname, thank you for your answer regarding planetary density :)

I have one new question. Some planets in SE orbit very close to their parent stars. So close that they are classified as scorched. Some of these have large moons which are so bright on the "night" sky of their planets, that their apparent magnitude rivals the Sun. Assuming I could somehow survive on the surface of such planet, could I sunbath in the moonlight, or even get lunar sunburns?
 
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07 Aug 2017 14:13

An'shur wrote:
Source of the post Assuming I could somehow survive on the surface of such planet, could I sunbath in the moonlight, or even get lunar sunburns?

The the moon would either have to reflect or emit sufficiently UV light.  An icy/snowy surface reflects UV light well (which is why there is snowblindness).  But in your scenario, icy moons can be ruled out.  Our earthly moon has poor reflectivity and even poorer UV reflectivity.  I suspect that the UV reflectivity of molten rock is very low also.  Molten rock will emit some UV, though.  But compared to the IR it's very little.  In reality, if we disregard any heat on such planets, I think the heat from the moon would be a far greater concern than UV from it.  So, in practise, no, you wont get a lunar sunburn before you get burnt otherwise.  For lunar sunburns, I think you need to look for stars that are UV bright and a cool, icy moon filling a large area of the sky.
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09 Aug 2017 10:51

Watsisname wrote:
The challenges would actually work out to be about the same in that case.   The idea is that you're trying to tease a particular spectrum (in this case a reflection spectrum of the planet) out of a signal that is almost completely due to something you're not interested in (the star).  So you have to remove the star somehow.

This is doable, and the main technique is to block out the star with an occulting disk.  But it works best if the planet is large and far from the star (same idea as for direct imaging).  When this does work it is a very powerful tool that can reveal a lot about the composition of the atmosphere and of clouds and haze layers.  The glints themselves would be a very small change in this spectrum though, and most likely still lost in the noise.

Another spectroscopic technique is to take the absorption spectrum.  In that case you're looking at the light which was passed through the atmosphere of the planet en route to Earth, rather than being reflected off of it (no glints in this case).  It's like seeing the light of all the sunsets and sunrises on the planet simultaneously.  Then the idea is to take the spectrum when the planet is transiting in front of its star, and then substract the spectrum immediately before or after the transit.  You'll be left with just the absorption spectrum of the planet's atmosphere, which again reveals a lot of information.  The benefit of this technique is that it is more applicable to planets that are very close to their stars, since they are more likely to be transiting and transit more often.

But haven't we visually detected exoplanets? I remember seeing a gif showing 5 giant exoplanets orbiting a single star and the view was from above- the kind of thing I wish we could do with our own solar system!

By the way, I have another question about the program.  I am trying to get an accurate view of the stars and constellations I see in my night sky, and proper sunrise and sunset times, moon phases, etc.  Is there any way to do this?  I was able to adjust the longitude to mine- 74W- but no matter what I do, I can't get the latitude to change :(  How do I get it to 41N?  Orbiting the earth and then stopping the clock when the longitude reaches 74W works, but the latitude is still stuck at 1S and Orion looks upside down......  Is there a direct way to change the latitude?
Last edited by A-L-E-X on 09 Aug 2017 10:54, edited 2 times in total.
 
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09 Aug 2017 10:58

JackDole wrote:
A-L-E-X wrote:
Source of the post Or in general, how do I change the angle of view?  The default is 45 degrees I believe.

You can also use the console. (Open it with the tilde '~')
For example, type 'fov 30'

I like this!  Maybe I can use something as simple and elegant as this to choose the lat and long of where I land on the planet? I want to use it for Earth to see the stars as I see them and the proper times for sunrise and sunset and the moon phases!  I got the long right in a roundabout way (by orbiting Earth until the long is -74 or 74W but I can't get the lat to change from -1 or 1S to +41 or 41N).
 
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09 Aug 2017 11:09

Watsisname wrote:
In this case I'm not sure of a satisfying explanation for why, other than this is the way the universe works.  Or we can say that if this wasn't the way it worked, then it would violate deep principles of physics and lead to problems.  Like, the location of a black hole's event horizon depending on how fast you're moving, which in turn would make it such that a particle does or does not get swallowed by a black hole depending on your frame of reference, which is equivalent to saying an event both does and does not occur, and that's bad and leads to paradoxes.

Probably the best answer I can give is that it is because of the relationship between space and time -- the Lorentz transformation.  From there you can show that the speed of light can only be approached asymptotically with relativistic velocity addition formulas, without requiring any reference to mass or energy whatsoever!  It is simply a property of the geometry of space-time.

In particular, if you consider an object moving away from you (let's say in the +X direction) at half the speed of light, and a second object moving at half the speed of light relative to the first, also in the +X direction, then the second object is not moving at the speed of light relative to you.

Maybe I'll derive the velocity addition formula later. :)

But I really like paradoxes and I hope we become advanced enough as a species one day to create a few ourselves :)
About black holes, I remember reading about this particular paradox and it was resolved using the cloning theorem, that the particle both falls into the black hole and doesn't- there is a duplicate created :)

I don't mind breaking physics, I find the universe/multiverse/omniverse a far more interesting place if our science(s) are proven wrong on a regular basis- it keeps us humble and inquisitive.
 
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09 Aug 2017 12:40

A-L-E-X wrote:
Source of the post Maybe I can use something as simple and elegant as this to choose the lat and long of where I land on the planet?

You can select the Earth and enter it in the console 'LandTo <Lat> <Lon> <Height>'.
 
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09 Aug 2017 13:35

JackDole wrote:
Source of the post You can select the Earth and enter it in the console 'LandTo <Lat> <Lon> <Height>'.

Are yousure? The console tells me "unknown comand"
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09 Aug 2017 14:10

XBrain130 wrote:
Source of the post Are yousure? The console tells me "unknown comand"

'LandTo' only works up to SE 0.980. In SE 0.981, the commands are very different.
In SE 0.981 you have to enter something like this: 'Goto {Time 2.0 Height 200 Lat 63.069167 Lon -151.007778}'

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