14 replies to this topic

[Lucullus]

I recently wondered how you can calculate the temperature at a telescope's focus depending on the telescope aperture and the observed star. E.g. what is the temperature increase in Kelvin at the telescope focus if observing Rigel with a 12" Cassegrain?

[rob1986]

I would expect that its small enough to be negligable

[bokemon]

magnitude of a star tells you it's photon / energy flux.

Multiply by aperature of telescope

It all gets concentrated into (approximately) a spot the size of the airy disk (depends on the scope's f number)

[Redbetter]

The theoretical answer is that in perfect seeing it could be focused to a tiny disk (because even the largest close stars are quite distant) that could reach the photosphere temperature of the star in question.  Of course seeing would mess with that to a huge degree in Earth's atmosphere spreading the light over a much greater area than just the focused disk should occupy, and thereby reducing the maximum temperature.  Additionally there would be some losses due to the atmospheric absorption and scattering.  For a truly tiny apparent size, losses to the surrounding atmosphere, conduction, etc. would also significantly reduce the maximum. 

 

The Sun, being much larger in apparent size at its distance, can be focused to several thousand degrees C by a very large mirror.  Link

 

This is a matter of black body radiation emission (which goes in both directions...hence subcooling and dewing under the clear night sky.).  You can't get a temperature higher than the hotter source or colder than the colder source, but various aspects will reduce the maximum achievable temperature, or increase the minimum achievable temperature.

Edited by Redbetter, 27 March 2025 - 07:45 AM.

[TOMDEY]

No one has explained it correctly yet --- I'll revisit later to see if there is progress. Hints: 1) It's really very simple, but no one here has yet hit on it. Start with the simplest idealized case first and apply Emmy Noether's Theorem 2) fold in whether the star is truly resolved or not 3) review the meanings of black body, gray body, and temperature and fold those into your declaration. Historical context: Early scientists took measurements of the temperature of celestial bodies (earliest was the moon) using bolometers at the image. I'd suggest that you try to solve the ~thought experiment~ "closed book" before looking it up. Radiometry is a fascinating topic and very fundamental to physics in the context of conservative fields --- in this case --- the electromagnetic field present in all empty space (what we call the vacuum). This is where Emmy's comes to the fore.     Tom

[moefuzz]

No one has explained it correctly yet --- I'll revisit later to see if there is progress. Hints: 1) It's really very simple, but no one here has yet hit on it. Start with the simplest idealized case first and apply Emmy Noether's Theorem 2) fold in whether the star is truly resolved or not 3) review the meanings of black body, gray body, and temperature and fold those into your declaration. Historical context: Early scientists took measurements of the temperature of celestial bodies (earliest was the moon) using bolometers at the image. I'd suggest that you try to solve the ~thought experiment~ "closed book" before looking it up. Radiometry is a fascinating topic and very fundamental to physics in the context of conservative fields --- in this case --- the electromagnetic field present in all empty space (what we call the vacuum). This is where Emmy's comes to the fore.     Tom

 

 

 

 

 

To put it simply, I suspect that you'd have to measure the increase in heat at specific infrared temperatures based on the radius of the "star" being measured and then -thru a transducer- (an electrical circuit) convert this to watts.

 

 

It would then be necessary to measure and subtract the atmospheric interference (a variable based on the thermal radiation from the atmosphere) as well as subtract the  cosmic microwave background (CMB) radiation which is a known and more or less consistent at .4 picowatts.

 

 

In the old days this was done using a prism to separate the light into it's spectrum and then utilize and measure the component of the infrared spectrum.

 

As an analogy and of more recent years, an easy way of measuring temperatures (in the advent of electricity) is very often done using a bi-metal tang or bi-metal gauge - not unlike the old bi-metal gauges on your dash that indicated the temperature of your vehicles engine.

The bimetal tang was a thin strip laid with 2 different coatings, one on each side of the metal strip. Given that different elements react differently to current (or in our case infrared heat), coating the strip on each side with a different element would cause the strip to deflect to one side or the other at which point you could measure the deflection as a difference in resistance and quantize this in volts or watts which can then be read by the gauges needle wherein the difference in resistance/electrical/fields would be visually marked and displayed on your temperature gauge as a measure of hot/cold.

 

In essence the principle would be the same or similar to measure the infrared spectrum of light,

just measure the difference in resistance of 2 dissimilar elements and convert this difference into watts while again allowing for both thermal radiation from the atmosphere as well as the cosmic microwave background.

 

 

 

 

p.s. edit

 

To me that seems fairly simple as laid out by your implication that it's "simplest idealized case" or 'keep it simple stupid' (kiss).

 

Not sure how much more simple of a method you could make to convert the radius of a star and it's infrared emissions than via a comparator of "normal versus hot" thru easily measured interactions of two different reactive coatings and known/normalized math and equations.

 

But the analogy of a simple bi-metal heat gauge popped to mind upon reading this as even the earliest of automobiles comparative dissimilar coated bi-metal gauges can be made for pennies using very few parts which by that definition garners all the elements of kiss in it's ~100 year old design.

 

moe

Edited by moefuzz, 27 March 2025 - 08:08 PM.

[Dino]

You measure the colour of the light and that tells you the temperature of the surface.

[moefuzz]

You measure the colour of the light and that tells you the temperature of the surface.

This is a given but perhaps re read the original OPs question as he is being a little more specific rather than just asking  of the standard or basic knowledge of colour temperatures.

Edited by moefuzz, 27 March 2025 - 03:52 PM.

[Bubbagumps]

The question is a bit ambiguous.  Absorption of photons can lead to an increase in internal heat energy within a material which in turn leads to an increase in temperature. Electromagnetic energy can heat an object via absorption. But light itself does not have a temperature.

 

As far as absorption of starlight heating the actual optical components of the telescope, the effect is miniscule and completely negligible. The addition of heat energy to a material via EM absorption primarily occurs as a result of the absorption of photons within the infrared energy range. It's why a fire or a heat source feels hot from a distance. Absorption of infrared photons are jostling the molecules in your skin. The energy of the photon is translated into the total heat energy of the absorbing material and this manifests itself physically as an increase in temperature. 

[ziggeman]

Its very interesting. And there are many clever people here with great knowledge in physics and related. So what was I thinking when I first thought: 'Cant you calculate this when you image this with a camera-sensor..the sensor efficiency . .
'Hiding'..

[garret]

1) At what wavelength will the measured temperature be highest?

2) How much energy does a particular star emit at that wavelength?

3) What is the transmission of the telescope at that wavelength?

 

Can you burn paper with the moonlight?

Answer: https://www.cloudyni...with-moonlight/

 

[moefuzz]

 

 

Can you burn paper with the moonlight?

Answer: https://www.cloudyni...with-moonlight/

 

 

Give me a few weeks and I'll give it a try! (next full moon)

 

I was playing around with my 25 inch mirror a few weeks ago and posted this video to YouTube..

 

 

 

Lighting fires with a WWll US Army surplus mirror

 

(youtube embedded, opens in a new tab)

 

LIGHTING INSTANT FIRES WITH A WW-ll ERA ARMY SURPLUS MIRROR

 

 

 

 

I posted a thread here which goes into a little more detail about the mirror

https://www.cloudyni...xperimentation/

 

Once you get it aimed precisely at the sun (culminated?) you can hear an instant

and fairly loud 'Pop' as it physically knocks a hole (depression) in the

wood igniting it instantly on fire.

 

And it will melt steel.

 

more to come at the thread,

moe

.

[Pickwick]

I do not really understand your question, you want to calculate the temperature increase at your pixel with the telescope pointed to Rigel, but that's going to be a supertiny fraction of degree, absolutely not detectable.

You shall anyway:

 

- Convert from magnitude to power flux.

- Multiply it by the area of your scope pi*r^2 where r is half diameter. This is the incident power

- Estimate the absorption rate of your pixel, let's assume 100%.

- Put the incident power equal to sigma*Star_Area*(Tpixel^4-t_amb^4) where you basically assume no heat transfer through convection and/or conduction from/to your pixel. It would be better in this case run a simulation and keep everything into account.

 

Which according to me is meaningless.

 

If instead your question was about measuring the temperature of the star: take the spectrum and compare it with the black-body radiation, the maximum shall satisfy the law of Wien.

[moefuzz]

I think TOMDEYS post is to the point and fair relevant but he has not blessed us with an update..

Sounds like we are all missing the mark

[deSitter]

I recently wondered how you can calculate the temperature at a telescope's focus depending on the telescope aperture and the observed star. E.g. what is the temperature increase in Kelvin at the telescope focus if observing Rigel with a 12" Cassegrain?

Not really a well-formed question, but I think you are really asking what is the ratio in say watts per square meter of the focused disk for a star at N astronomical units from us, and it is 1/N^2 the value for the Sun, times the relative luminosity of the star compared to the Sun.

 

A fun fact is that if you put the Sun very far away, but no so far that it cannot be focused to a perceptible disk, then the surface brightness would be the same as for the actual Sun - that is, HIGH. That's a consequence of the above. But the watts per square meter would go down in proportion to the relative areas of the focused disks. So say from Pluto, the Sun will be only the size of Jupiter as seen from Earth, but the surface brightness will be the same as here. So if you turned your telescope on it, and magnified it so that it was 1/2 degree in diameter - about 43x - it would be no different than staring at the actual Sun from your front yard.

 

-drl