I Does all infrared get detected as heat?

[LightningInAJar]

TL;DR Summary

Non heat IR?

There are different ranges of IR. Are they all represented as heat? If not maybe Webb telescope can read those only so our own sun isn't as much of an obstacle? Maybe xray instead?

 

[FactChecker]

The Sun is very bright in the entire electromagnetic spectrum. (see https://astronomy.com/magazine/ask-...part-of-the-spectrum-does-the-sun-emit-energy )

 

[russ_watters]

There are different ranges of IR. Are they all represented as heat? If not maybe Webb telescope can read those only so our own sun isn't as much of an obstacle? Maybe xray instead?

I think you are confused about what both "heat" is and maybe light/EM radiation. The definition of heat is simply "thermal energy transfer". And EM radiation transfers thermal energy (or at least heats up the recipient). Thus, any electromagnetic radiation can be considered heat.

https://en.wikipedia.org/wiki/Elect...d_electromagnetic_radiation_as_a_form_of_heat

Similarly, all warm objects emit EM radiation, thus there is another term for it, called "thermal radiation":
https://en.wikipedia.org/wiki/Thermal_radiation

It's my understanding that all/hot objects emit in all frequencies as a function of temperature, but for higher frequencies (x-ray, gamma) the temperature required to emit them efficiently are unachievably high. As a result they are typically generated through other means.

 

[Orodruin]

There are not different ranges of IR. IR refers to the part of the electromagnetic radiation spectrum with just slightly longer wavelength than the visible spectrum.

IR is not heat. It is electromagnetic radiation. Heat is an energy transfer between thermodynamic systems that is not in the form of work or mass transfer.

X-rays are definitely not IR. They are an entirely different part of the electromagnetic spectrum. They are also used in astronomy.

 

[russ_watters]

IR is not heat. It is electromagnetic radiation. Heat is an energy transfer between thermodynamic systems that is not in the form of work or mass transfer.

IR (and other EM radiation) is energy transfer not in the form of work or mass transfer...

Or did you mean that IR isn't exclusively, heat?

 

[Orodruin]

IR (and other EM radiation) is energy transfer not in the form of work or mass transfer...

Not always. It can be part of a thermodynamic system such as a photon gas. It can be heat but must not be so.

 

[russ_watters]

Not always. It can be part of a thermodynamic system such as a photon gas. It can be heat but must not be so.

It sounds like you are saying that if you draw a system boundary such that the energy transfer is internal it doesn't count.

 

[Orodruin]

It sounds like you are saying that if you draw a system boundary such that the energy transfer is internal it doesn't count.

Just like internal forces don’t count in a free body diagram.

The radiation escaping the system is heat. Radiation inside the system is part of the system.a situation where radiation is part of your system with no way of escaping is the early Universe. On the contrary, entropy from other sectors is dumped into the photon gas heating it up. This is why there are funny factors in the ratio between the CMB and CNB temperatures - neutrinos decoupled before the other particles were done dumping entropy into the remaining relativistic species and were therefore not heated in the same way as the photons.

 

[russ_watters]

Just like internal forces don’t count in a free body diagram.

Ok...but that just means that if you draw a different system boundary/diagram, maybe they do. So in my opinion this line of reasoning creates an unnecessary distinction-without-a-difference.

 

[Orodruin]

Ok...but that just means that if you draw a different system boundary/diagram, maybe they do. So in my opinion this line of reasoning creates an unnecessary distinction-without-a-difference.

Are you arguing that the distinction between internal and external forces is unnecessary?

I think it is highly relevant and many times, such as in the example I gave, there really is no reason to draw your system in such a way that radiation is heat.

 

[russ_watters]

Are you arguing that the distinction between internal and external forces is unnecessary?

No, I'm arguing that the distinction of whether EM is "heat" depending on the system boundary is unnecessarily narrow and leads to contradiction (the same EM being "heat" and "not heat" at the same time). Also, that isn't what you said to begin with. You said "IR is not heat" when I think it should have been "IR is not always heat".

 

[Devin-M]

So when photons hit the JWST NIRCam sensor, some of them are absorbed by kicking electrons from valence to conduction in the photodiodes and others of the same wavelength are absorbed without kicking the electrons to a different energy level (the percentage of photons of a certain wavelength that do elevate electrons is known as the quantum efficiency).

Are both forms of absorption considered heat transfer considering the former case leads to a charge building up in an in pixel capacitor in a typical CMOS imaging sensor?

“An MOS capacitor is integrated in a pixel and accumulated charges in a photodiode are transferred to the in-pixel capacitor multiple times depending on the maximum incident light intensity”
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6479534/

PS This shows the quantum efficiency for NIRCam factoring in various filters:

 

[snorkack]

There are many different ranges of IR. 850 nm is infrared (the limit of eye is blueward of 800 nm) but so is 85 000 nm (where is the limit between IR and microwave?).

All forms of electromagnetic radiation can turn into heat. This includes visible light and UV. But the ways to sense them, and sensitivity to them, are different.
For example, direct sunlight perceptibly heats up skin. It can be perceived as heat, even though it is also visible light.
Moonlight is also absorbed by skin and turned into heat. But the amount of moonlight turned into heat in skin is negligible and imperceptible compared to internal heat of the skin.
Yet while skin cannot sense moonlight, eyes can, and so can film. The tiny amount of high energy photons in moonlight can excite retinol in eyes, or break down silver bromide, the way the large numbers of low energy photons emitted by eyes and film themselves, or the underlying thermal motion of eyes and film, cannot.

How does a telescope distinguish IR radiation of external and weak sources from its own internal heat?

 

[Drakkith]

How does a telescope distinguish IR radiation of external and weak sources from its own internal heat?

It doesn't. That's why IR telescopes are usually cooled to a temperature well below whatever objects they will be observing and why the Spitzer telescope had to stop long wavelength observations once it ran out of liquid helium coolant. The noise from the telescope's internal IR swamped these long wavelength observations with too much noise to get useful images.

 

[Orodruin]

Also, that isn't what you said to begin with. You said "IR is not heat" when I think it should have been "IR is not always heat".

That’s just word wrangling and ”IR is not heat” is technically just saying that there is not a equivalence.

No, I'm arguing that the distinction of whether EM is "heat" depending on the system boundary is unnecessarily narrow and leads to contradiction (the same EM being "heat" and "not heat" at the same time)

You are missing the point. The same thing can be said about internal vs external forces.

 

[LightningInAJar]

What limitations does Webb have that Hubble doesn't besides higher resolution? I assume Hubble doesn't need to hide in Earth's shadow? Can Webb photograph earth?

 

[russ_watters]

Posts regarding imaging sensors and the 2nd law of thermo split/moved...again.

https://www.physicsforums.com/threads/jwst-sensor-thread-split.1016876/

 

[russ_watters]

That’s just word wrangling and ”IR is not heat” is technically just saying that there is not a equivalence.

Fair enough. I'm generally a fan of concise wording, but I submit that if I misunderstood your intent the OP may have as well. So thanks for clarifying.

 

[Drakkith]

What limitations does Webb have that Hubble doesn't besides higher resolution? I assume Hubble doesn't need to hide in Earth's shadow? Can Webb photograph earth?

Hubble doesn't need extreme cooling of its sensors like JWST does. Hubble also has a smaller mirror, reducing its resolution and light gathering compared to JWST, and observes targets in a different part of the EM spectrum. Hubble is designed to image from UV down through visible light to the near-IR, the part of the IR band that's closest to visible light.

 

[collinsmark]

What limitations does Webb have that Hubble doesn't besides higher resolution? I assume Hubble doesn't need to hide in Earth's shadow? Can Webb photograph earth?

First question:

Neither telescope permanently hides in Earth's shadow (neither Hubble Space Telescope [HST] nor James Web Space Telescope [JWST]).

HST:

HST orbits Earth in a rather conventional low-Earth-orbit. Its orbit isn't as low as the International Space Station (ISS), but it's still pretty low: only about 540 km (330 miles) above Earth's surface. That means it's in direct sunlight a little more than half the time. It has a comparatively small sunshield sort of thing that keeps direct sunlight from entering the optical tube, but that's about it. Maybe some insulation.

By "conventional" orbit I mean that HST orbits around Earth in a elliptical orbit (along with precession and such), as you might expect from most any other orbiting satellite.

JWST:

The sun-shield side of JWST is in direct sunlight all of the time. So from that point of view, JWST is in direct sunlight way more often than HST.

JWST completely relies on its own sunshield to keep it in the shade (from the Sun). There is no hiding in Earth's shadow.

If JWST was directly at the Lagrange 2 (L2) point (which it isn't), the Earth would annular eclipse the Sun (i.e., not a total eclipse).

JWST instead orbits around the L2 point in such a way that it's always in direct sunlight; it never passes through the Earth's shadow (and again, even if it did, there is no umbra of Earth's shadow: penumbra only at that distance).

Second question:

No, JWST cannot image Earth. JWST must keep its sunshield side on the side where the Sun is. Because JWST always roughly in the vicinity of the L2 point, it means that both Earth and the Sun are roughly in the same direction, more-or-less.

If JWST were to rock itself over to point its optics toward Earth, some of the delicate equipment would end up in the sunlight and bad things would happen.

 

[LightningInAJar]

That's a little bit of a bummer. The pictures of Jupiter were quite sharp. Would of been curious to see Earth in higher resolution.

 

[glappkaeft]

That's a little bit of a bummer. The pictures of Jupiter were quite sharp. Would of been curious to see Earth in higher resolution.

Images of the Earth with much higher resolution than what JWST could deliver are not hard to find.

 

[hutchphd]

In fact I can take one out of my back window as I write this. We are fascinated by selfies, aren't we...