1
$\begingroup$

So I'm researching if the sun is transparent or not and have found that it is made up of gases. From what I know the gases the sun is made from are transparent. So why can we see the sun if it's made of transparent things.

$\endgroup$
6
  • 1
    $\begingroup$ Why can you see a candle flame if it's made of transparent things? Because it's hot and glowing. $\endgroup$ – user93237 Oct 3 '17 at 19:31
  • $\begingroup$ @SamuelWeir It is the soot particles in a candle flame that are glowing. And the flame is still transparent - a laser will shine through it. $\endgroup$ – Pieter Oct 3 '17 at 22:21
  • $\begingroup$ @Pieter The flame is essentially plasma - ionised molecules and atoms. There may be particulates (soot) in the flame, but that is not what's making it glow. As for transparency - the ions and free electrons react to the electromagnetic waves so there is absorption and attenuation. It's not very transparent. $\endgroup$ – Oscar Bravo Oct 6 '17 at 7:46
  • $\begingroup$ @OscarBravo Have you tried to shine a laser through a candle flame or not? And it really is the soot that glows. Compare the candle flame with the much hotter bluish flame of a Bunsen burner. $\endgroup$ – Pieter Oct 7 '17 at 23:14
  • $\begingroup$ @Pieter The OP was asking about the Sun, not candles. The Sun is largely plasma. Plasma is electrically conducting and so attenuates EM radiation. $\endgroup$ – Oscar Bravo Oct 13 '17 at 7:36
3
$\begingroup$

This has an answer on several levels.

As others have indicated, the physics reason is because the Sun itself is opaque to light; it isn't transparent.

There are many sources of opacity in the gases that make up the Sun. These arise in processes that can be classified as: bound-bound (radiative transitions between bound states in ions and atoms that occur at discrete energies); bound-free (transitions of electrons from bound to free, or ionised, states and vice-versa - e.g. photoelectric effect); or free-free (involving the acceleration of free electrons - e.g. inverse bremsstrahlung).

Bound-free and free-free processes produce absorption over a continuum and if you have enough gas, then it becomes "optically thick" (has an optical depth much greater than unity) to radiation. That is, a photon has no chance of passing unabsorbed through the Sun.

Now the next part is determining what the Sun looks like. Because it is roughly in thermal equilibrium and opaque, the sun emits (roughly) blackbody radiation. The temperature of the blackbody is the temperature of the upper layer of the Sun's gas where photons are able to escape (the Sun gets hotter as you move inwards). This is somewhat wavelength dependent, which is why the spectrum of the Sun is not a perfect Planck function. The opacity source that is mostly important here is continuum opacity due to bound-free transitions with H$^{-}$ ions (Wildt 1939). In the Sun this leads to a relationship between temperature and optical depth such that photons escape the "surface" from gas that is at around 5800K. The escaping blackbody radiation peaks (by Wien's law) in the visible part of the spectrum. This layer is what we refer to as the Sun's photosphere or visible "surface".

Why can we see this radiation? Well that's down to the development of the eye and evolution.

$\endgroup$
5
  • $\begingroup$ Is it really important that the sun is opaque? For example, is glowing heated glass opaque? I would think that opaque or not, we see the glow. (As an aside it's quite good that (at least the outer layer of ) the sun is opaque so that we are not exposed to the radiation produced in the core.) As another aside, almost everything besides the vacuum is opaque when it's hundreds of thousands of kilometers thick. $\endgroup$ – Peter - Reinstate Monica May 9 '19 at 21:44
  • $\begingroup$ @PeterA.Schneider It is a truism that you cannot see things if they are transparent. An "optically thin" gas is almost transparent. The spectrum is entirely different and nothing like a blackbody. $\endgroup$ – ProfRob May 10 '19 at 9:32
  • $\begingroup$ With all due respect -- I thought "transparent" means "does not absorb nor reflect", so it won't heat up from absorption and it won't be visible by means of reflecting ambient light. It may well emit light and hence become visible, may it not? Take a laser, gas or solid: The material must be pretty transparent (at least at the given wavelength) because the light must be able to travel between the mirrors many times without too much absorption, but the laser is still pretty well visible. (Though it's not exactly blackbody radiation, granted.) $\endgroup$ – Peter - Reinstate Monica May 10 '19 at 9:48
  • $\begingroup$ @PeterA.Schneider we are talking about things (stars) that are in thermodynamic equilibrium. What you propose is not feasible for such an object and indeed the example of a laser is as far from a thermal equilibrium/blackbody situation as it could be. $\endgroup$ – ProfRob May 10 '19 at 12:44
  • $\begingroup$ I see; because of reciprocity only matter which absorbs at a given wavelength emits radiation from heat at that wavelength... that should indeed mean that glass becomes intransparent (or glows only faintly) when heated to, say, 1300K. Interesting. I would still say that opacity is not a reason but a consequence of being in a state which leads to heat radiation emission, but one may argue that the two are inextricably linked. $\endgroup$ – Peter - Reinstate Monica May 10 '19 at 13:41
1
$\begingroup$

when electrically neutral, gases are transparent. the gases that form the sun are so hot that their outermost electrons have been torn loose (this is called ionization), rendering the gas electrically conductive. light cannot travel through anything that is electrically conductive, so you cannot "see" through the sun even though it is made of gas.

to clarify: I meant transparent at wavelengths your eye uses to observe the sun.

$\endgroup$
1
  • 1
    $\begingroup$ Neutral gases (I assume you mean un-ionised) are not necessarily transparent. Certainly not at all wavelengths. Pedantically, it is also not possible to arrange for a gas to have exactly zero ionisation unless it has zero density and zero temperature. So all gases have some opacity and then it is simply a question of how much of the gas there is. The Earth's atmosphere is mostly "neutral" but we can see that. $\endgroup$ – ProfRob Oct 6 '17 at 9:21
0
$\begingroup$

It is actually made up of plasma. Plasma means ionized gas. The Sun is a big ball of hot ionized gas and we can see it because it emits light. As Samuel said, we can see a candle flame because it is hot and it is glowing.

The main process that happens in the Sun is proton-proton fusion. There is a chain of processes in which the protons inside the Sun collide and fuse into Helium. In these processes there is a release of energy in the form of charged particles, neutrinos and light.

$\endgroup$
3
  • 5
    $\begingroup$ And significantly plasmas are opaque to light! $\endgroup$ – tfb Oct 3 '17 at 21:16
  • 1
    $\begingroup$ Well, it becomes irrelevant whether or not the plasma is opaque to light if itself emits light. $\endgroup$ – pathintegral Oct 3 '17 at 22:07
  • 1
    $\begingroup$ @pathintegral The Sun emits a black-body spectrum because it is opaque. Kirchoff's law. $\endgroup$ – Pieter Oct 3 '17 at 22:23

Not the answer you're looking for? Browse other questions tagged or ask your own question.