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What Could That Be??


t_ilan_s

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I took in the Geminid meteor shower some widefield pictures.

After I put them on my computer I notice there is something strange in one photo.

white dot appeared in the center and then disappeared:

The first picture with the dot:

Taken on: 2012/12/14 01:47:57

The second picture without the dot:

Taken on: 2012/12/14 01:48:14

post-26995-0-61791300-1356029307_thumb.j

post-26995-0-49735300-1356029324_thumb.j

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It's the precursor to the Mayan apocalypse; you have spotted the Earth's evil twin planet returning from its 7,800 year long orbit to obliterate us all. It is currently moving so fast you would only have captured it for a fleeting moment.

Or more boringly, it is most likely a head on hit from a cosmic ray (or more properly from one of the particles created by a cosmic ray hitting the upper atmosphere). Entirely likely it has come from another galaxy and been travelling for millions or even billions of years to get here.

Useful paper on using DSLRs to detect cosmic rays:

http://scitation.aip.org/getpdf/servlet/GetPDFServlet?filetype=pdf&id=AERSCZ000009000001010111000001&idtype=cvips&doi=10.3847/AER2010009&prog=normal

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Unlikely to be a cosmic ray capture, as the exposure time is too short. According to the paper in the link, we could all experience this phenomenon when we take our very long exposure dark frames.

And we all thought it was just hot pixels..

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Unlikely to be a cosmic ray capture, as the exposure time is too short. According to the paper in the link, we could all experience this phenomenon when we take our very long exposure dark frames.

And we all thought it was just hot pixels..

The exposure time has no bearing on whether a particle from cosmic ray will register on your image or not; it is (to all intents and purposes) an instantaneous process. If the particle hits the sensor element, or tracks through several of them depending on the angle of the impact, it will dislodge many electrons in to the sensor well(s). The higher energy the particle, the more electrons and the brighter the dot (single sensor element) or trail (multiple elements) will appear.

The likelihood of capturing one or more hits is dependent on the mean frequency of particles per unit of area at your location. This will vary as described in the paper by latitude, altitude, time of day, current solar conditions, etc. As stated, the mean rate of muons (most likely to create these effects) is 1 per square centimetre per minute at sea level. So if you have a 2 square centimetre chip you should expect to see somewhere in the region of two strikes per minute of exposure. The camera/chip's sensitivity probably enters in to it as well, but CMOS (DSLR) sensors and CCDs are both susceptible to this effect.

The OP doesn't state the length of the exposures, but from the timestamps they can only be 17 seconds at most. Assuming a 2cm square chip and the mean rate of 1 hit per square cm per minute, that equates to one hit per 30 seconds, and you would have to expose 1.76 x 17 second exposures to expect a hit, or to make it simple there should be a hit on every second exposure.

Clearly YMMV depending on all sorts of factors as described, but it is erroneous to assume that you will not get a cosmic ray hit unless you undertake long exposures. I can assure you that many of your images, long and short exposures will have these artefacts in them. Distinguishing them from dark current/noise and hot pixels can be done using the normal process of subtracting a well matched master dark frame, and using a defect mapping/elimination tool if your dark frames don't fully deal with hot pixels.

Of course if you stack multiple exposures as we tend to do, then the stacking process will reject the cosmic ray artefacts as they will only appear in a given place in one image from the stack (which is what you want it do of course!) Your daylight exposures will have cosmic ray impacts in them too, but they're not going to be easy to pick out!

If you don't believe what I am saying, I really would urge you to try the experiments proposed in the paper for yourself. You can do it in the middle of the day easily enough and you probably have much better processing software, techniques and skills that the paper's author to do so.

The theory that this is a head-on view of a meteor travelling straight at the observer seems unlikely to me. A meteor trail will be in the region of tens to hundreds of miles long and the distance from the observer will be in the same range. The probability of being in the spot where the start and end of the trail are precisely aligned is going to be very low indeed. And even if one did happen to be in that location, all of the light from the entire length trail would be concentrated on one spot in the image, so for a typical Geminid you would expect the spot to be very bright. I will concede that you might capture a very faint meteor end on, making it visible where a side on view would not register. It's still very unlikely that you would be in the right spot at the right time though.

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Jeez!! remind me not to get in an argument with you Ian. I wouldn't stand a chance.

I have experienced these odd artefacts in normal images but could never explain them. I just clone them out. Maybe cosmic ray hits are the cause.

t_ilan_s I still can't get the links to work. They just take me to a googlemail page. could you just post them like you did the others.

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I was left feeling slightly unsatisfied by my 'seems unlikely to me' point above regarding an end-on meteor being imaged. Of course I wanted to know 'how unlikely?', so here it is. (Note, this is a fairly unsatisfactory but reasonable method based on areas only, I suspect someone better at maths would do it using cones and solid angles projected on to the Earth's surface, any takers?)

- Let's assume that the near end of the meteor trail is 2 arcseconds in apparent diameter. This doesn't seem like an unreasonable assumption. Most meteors burn out or break up before they reach an altitude of 50km or so, and a fair chunk of the atmospheric disturbance that causes seeing occurs below this altitude. Given the size of a typical meteor (sand grain or similar), it can be treated as a point source at these distances, and thus be limited by seeing, for which (in the UK at least) a limit of about 2 arcseconds of resolution is a good rule of thumb. It may be better in Israel where the OP is located, but if so that actually makes it even less likely to get an end-on image.

- Hopefully it is obvious that due to the seeing limitation, the apparent diameter far end of the meteor trail must also be 2 arcseconds (conveniently avoiding me having to spend the rest of the afternoon reading up on calculating solid angles, cones, etc.)

- Let's also assume we are imaging at 2 arcseconds per pixel, or somewhere in that region.

- Let's also assume that we can see the whole sky (flat horizon).

- Finally let's ignore atmospheric extinction near the horizon, so that we can see any meteor above the horizon. (Not true in practice but doesn't affect the probability I am going to calculate).

To get an "end-on" image, we must be located so that the far end (top) of the meteor trail is hidden behind the near end (bottom) of the trail. As per our assumptions, both ends of the meteor trail have an apparent diameter of 2 arcseconds. So for a meteor to appear 'end-on', i.e. as a dot in the image, the start and end of the trail must be within the same 2 arcsecond diameter circle in the sky.

Thus the question we have to answer is, "How many 2 arcsecond diameter areas are visible from a given spot on the earth?" The answer is as follows:

The radius of the sky sphere is:

Radius = Circumference / (2 x Pi)

Radius = (360 degrees x 3,600 arcseconds) / (2 x Pi)

Radius = 206,264.81 arcseconds

The area of the sky sphere is:

Area = 4 x Pi x Radius2

AreaSky = 4 x Pi x 206,264.81 arcseconds2

AreaSky = 534,638,377,792.47 square arcseconds

Since only half the sky is visible (assuming a flat horizon), we can see:

AreaVisible = 267,319,188,896.24 square arcseconds

The area of a 2 arcsecond diameter circle is:

AreaMeteor = Pi x Radius2

AreaMeteor = Pi x (2/1)2

AreaMeteor = Pi (conveniently enough!)

Number of 2 arcsecond diameter circles in half the sky area is:

NumberCircles = AreaVisible / AreaMeteor

NumberCircles = 267,319,188,896.24 / Pi

NumberCircles = 85,090,340,592.30 circles

Thus for any given meteor, the probability of you being in the right place to image it perfectly end on is 1 in 85,090,340,592

For comparison, the probability of winning the EuroMillions lottery is 1 in 116,531,800. Therefore you are 730 times more likely to win the big jackpot tonight than you are to image a meteor end-on.

Compare that with the 1 in 2 probability of catching a cosmic ray in a 17 second exposure calculated in my previous post. I hope you will see which is the more likely explanation for the mysterious dot!

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Well I have looked at the frames before and after the dot appeared and confess that I can find no other explanation. So it could be a cosmic ray hit ater all.

Its nothing other than a Satalite if your looking early in the morning a geo stationary type, just a blip from the solar panels.....nothing special at all...... :rolleyes:

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