Jump to content

SkySurveyBanner.jpg.21855908fce40597655603b6c9af720d.jpg

Stationary stars?


discovered799

Recommended Posts

The long read in Wiki. But some stuff I didn't know (wasn't known) before! ;)

https://en.wikipedia.org/wiki/List_of_nearest_stars_and_brown_dwarfs

Stars still move at significant speeds (80km/s!) compared to that we experience
on earth. Compared to the lifetime of our SPECIES (units of 10000 years) some
nearby stars may move significantly. Their closest approach may be (has been)
near to the Oort cloud and may send asteroids / comets on OUR direction... :eek:

https://en.wikipedia.org/wiki/Scholz's_star

But no point worrying (mostly)! :p

Link to comment
Share on other sites

  • Replies 32
  • Created
  • Last Reply
4 hours ago, vlaiv said:

One particle universe is in superposition of all imaginable states - all motion and spin vectors at the same time, and only once we measure it we will find it in particular state - moving or not, spinning or not ... :D

 

I agree , I think :) 

My understanding is that measurement is an independent act to the state of an object.  What I mean is, it doesn't need to be measured in order that in may spin, it can have spin surely without ever being measured - only thing is we would not know its condition.  I know that's a bit further down the rabbit hole and I try to avoid that stuff  :) 

I guess what I'm trying to do is get back to an earlier proposition in the thread that all motion is relative.  I'm not sure it is.  What I am asking is , does it need to be?  A spinning object surely does not need the presence of other objects  (the relative reference) in order to spin.  Sure if we want to measure the spin then I can see that we may need a relative position from which to make the measurement. However, the act of measurement is surely separate from the condition of being (state) of the object's motion. So I guess I'm throwing my hat in to say that the one particle universe can indeed have motion, we just can't measure it.  I'll also add that I am well out of my depth here - I'm an engineer (the younger slower brother of physics):) 

Jim 

Link to comment
Share on other sites

24 minutes ago, Macavity said:

Their closest approach may be (has been)
near to the Oort cloud and may send asteroids / comets on OUR direction... :eek:

https://en.wikipedia.org/wiki/Scholz's_star

But no point worrying (mostly)! :p

I love your quote at the end Chris - no point worrying (mostly). Sounds like something you would find in The Hitchhikers Guide To The Galaxy :) 

 

Link to comment
Share on other sites

Back to the OP's question, I always understood that the parallax effect we observe owed more to the local movement of Earth around the sun than it did to any movement of the distant stars. I thought the effect of the local movement to be the more dominant of the two. So even if a distant star was somehow truly static we would still observe a parallax angle of a star between us and the background (static star) when viewing between 6 months apart. 

Jim

Link to comment
Share on other sites

Given that Hubble is F/24 2.4m telescope, and with current common sensor pixel size, and 1/16 of pixel centroid accuracy, I would estimate Hubble parallax precision

to be of order of 1mas, giving distances of up to 1000pc or ~3261 Ly.

This is what I would say is theoretical max. Actual ability (given installed sensors) is probably 1 order of magnitude less than that, so somewhere around 300 Ly - 500 Ly.

Now that I've written that above, I've actually found following quote from Wiki that is quite different:

"The Hubble telescope WFC3 now has a precision of 20 to 40 microarcseconds, enabling reliable distance measurements up to 3,066 parsecs (10,000 ly) for a small number of stars.[11] This gives more accuracy to the Cosmic distance ladder and improves the knowledge of distances in the Universe, based on the dimensions of the Earth's orbit."

Found on this page:

https://en.wikipedia.org/wiki/Stellar_parallax

That differs in that it is 3 times more than my theoretical maximum (???) I guess that using Hubble and equipment on Hubble one should be able to do centroid with much higher precision (order of 1/100 instead of 1/16 assumed in my case).

Just checked WFC3 operates on 0.04"/pixel resolution - giving correct 40 mas precision stated above.

Link to comment
Share on other sites

1 hour ago, lightning8675309 said:

Could you show your work regarding the calculation in detail? Also, what telescope, resolution and equation are you using? Very patiently waiting!

It is really simple reasoning.

I took specs for Hubble telescope - being 2.4m and F/24 - that gives 57.6m focal length. Most modern sensors have pixel size of around 4um (although it turns out that they are using more sensible ~16um pixel camera for WFC3 on Hubble), but let's do calculations based on 4um pixel size.

To obtain resolution in arc seconds per pixel, we use following formula: 206.3 x pixel size (in um) / focal length (in mm)

This gives 0.014"/pixel or 14mas (milli arc seconds per pixel). Current algorithms for centroid (finding center of a star in pixelated image) are at least 1/16 pixel size accurate.

Therefore 14mas / 16 ~ 1mas accuracy of centroid. 1mas parallax precision is enough for 1000pc distances (1 parsec is parallax at 1 arc second).

So my estimate was off on two counts. First WFC3 has larger pixel size than modern sensors (in line with largest pixel sizes in CCD cameras, of 12-24um instead of 4). Also estimate of 1/16 centroid accuracy is probably more suited for short exposure with small SNR. Long exposure by Hubble and the fact that fitting of Airy PSF is done rather than Gaussian possibly yields higher precision needed for distances of ~3000pc

Link to comment
Share on other sites

Archived

This topic is now archived and is closed to further replies.

  • Recently Browsing   0 members

    • No registered users viewing this page.
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue. By using this site, you agree to our Terms of Use.