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Why do the exhaust flames from cryogenic stage engines appear to be separated from the nozzle?

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    $\begingroup$ In the top photo, the very bright area is where there's an internal shock wave in the plume, which very quickly changes the properties of the gas (e.g. density), which makes the plume visible to the eye at a very clear (shock) boundary. The answers below explain why the plume would be visible or not for given fluid properties. $\endgroup$ – Adam Wuerl Sep 18 '13 at 23:38
  • $\begingroup$ It's a good question - can you show a link to where the images are from? At least the top one? Thanks! $\endgroup$ – uhoh Jun 24 '16 at 2:31
  • $\begingroup$ The intensity of the blue radiation appears to scale non-linearly with pressure. The abrupt increase in pressure at a static shock wave raises the pressure sufficiently for the blue radiation to appear. See this answer and the linked thesis for more information. Ya, it's actually called "blue radiation"! $\endgroup$ – uhoh Jul 10 '16 at 8:16

They might appear as the flame is detached from the nozzle, but that's in essence merely an illusion and the burn is there, all over the plume of the nozzle exhaust. It is however nearly translucent due to high purity of cryogenic propellants and by the chemical reaction producing molecules having high translucency. Visibility of the plume, unless you're launching at night, has actually little to do with its burn. The burn itself, as the propellant components react (fuel with the oxidizer) will be mostly in the non-visible spectrum with propellants that don't use additives / fuel impurities to shift the heat radiation to the visible spectrum. You have answered yourself why the flame appears invisible to the naked eye, but you didn't mention why it appears to be separated from the engine's nozzle, as per your question.

Both images you attached to your question, judging by the color of the exhaust plume and light refraction, use LOX/LH2 or liquid diatomic oxygen O2 and liquid diatomic hydrogen H2 in 1:2 ratio as their oxidizer + fuel propellant components, producing superheated but pure water H2O at the nozzle exhaust, and unless these plumes combine with atmospheric or propellant impurities, they will appear nearly translucent, especially given the pressure they're ejected from the nozzle injectors into the expansion chamber, where they combine. If you look closely at the images you attached, you'll however still notice a somewhat typical light refraction of water at the test facility railings behind them, more pronounced on the top photograph.

OK, so you're probably asking yourself now, what about the more white core of the exhaust plume? That is simply an area of negative pressure that forms different shapes depending on the nozzle size, shape, and of course the fuel injection rate. With some nozzles, much of this can be adjusted depending on the current performance requirements of the engine, so its position might vary, or it might even repeat itself more frequently, as for example on this image of the alcohol fueled Armadillo Aerospace's quad vehicle showing visible banding, or shock diamonds, in the exhaust plume:

          enter image description here

In short though, the areas of negative (back-) pressure will cause this banding and additional walls forming within the plume. With molecules high refraction index, the light will scatter at these edges in a zigzag pattern (more likely much more complex with vortices forming actually) and block any direct view of the background. To rather leave this explanation to Wikipedia on Plume physics:

Rocket plume varies depending on the rocket engine, design altitude, altitude, thrust and other factors.

Carbon rich exhausts from kerosene fuels are often orange in color due to the black body radiation of the unburned particles, in addition to the blue Swan bands. Peroxide oxidiser based rockets and hydrogen rocket plumes contain largely steam and are nearly invisible to the naked eye but shine brightly in the ultraviolet and infrared. Plumes from solid rockets can be highly visible as the propellant frequently contains metals such as elemental aluminium which burns with an orange-white flame and adds energy to the combustion process.

Some exhausts, notably alcohol fueled rockets, can show visible shock diamonds. These are due to cyclic variations in the plume pressure relative to ambient creating shock waves that form 'Mach disks'.

The shape of the plume varies from the design altitude, at high altitude all rockets are grossly under-expanded, and a quite small percentage of exhaust gases actually end up expanding forwards.

In short, the answer to the question is not exactly in the burn itself, since these bands, walls, and any other shapes that form within the plume as a result of a backpressure would do that just as well, if there wasn't any burn to start with. It is due to light refraction and high pressure, high velocity ejecta forming walls travelling at the opposite direction each side of the wall (detached burn), or because of the jets intersecting (diamonds).

  • $\begingroup$ You said the brighter region is an area of negative pressure (understood as lower pressure). This is different from what I read elsewhere, for example, @Thomas answer to the same question. Those other sources all said that the brighter region is the first Mach disk region with higher pressure and density. $\endgroup$ – verdelite Apr 16 at 2:28
  • $\begingroup$ If it is really a lower pressure region (To me it looks a little bit higher than the position of the first Mach disk), could water vaper there be condensed into water droplets? If so, the droplets can be illuminated by bright light from the combustion chamber to give off visible light. $\endgroup$ – verdelite Apr 16 at 2:35
  • $\begingroup$ As to water vapor condensation, we may compare to the air plane vapor cone phenomenon, bbc.com/future/article/… $\endgroup$ – verdelite Apr 16 at 3:13

What you see is the first Mach disk (a standing shock wave), which causes a sudden increase in temperature, pressure, and density.

At the nozzle exit, the exhaust gases have a comparably low temperature due to the high expansion ratio. In case of your SSME, the temperature is about 1200 K. The hydrogen-oxygen flame plume does not soot, and the water vapor or the other combustion products are not hot enough to radiate in the visible.

Over the shock, the temperature jumps above 3000 K, which causes thermal excitation of the water vapor. You can sometimes see this as a faint reddish glow. The origin of the dominant broad-band blue radiation from hydrogen flames is actually unclear. Most likely, it is a chemiluminescence recombination of radicals present in the plume. This is suddenly enhanced over the shock because of the high density and temperature, causing a great number of molecular collisions.

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    $\begingroup$ Your answer actually packs a lot of scientific detail into eight sentences, and makes a lot of sense! Do you think you could add a link or reference for the 1200K temperature? Or any place I could read more about the chemiluminescence being suddenly enhanced by pressure? $\endgroup$ – uhoh Jun 22 '16 at 2:07
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    $\begingroup$ I computed the temperature of 1200K myself using basic gas dynamics and the official data for chamber pressure, ROF, etc.. I know that I checked the overall results against other sources, but it's too long ago for me to remember which. $\endgroup$ – Thomas Jun 23 '16 at 20:58
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    $\begingroup$ The impact on chemiluminescence originates primarily from the higher density, not pressure (which is naturally coupled, though). If you're interested, there is a nice PhD thesis on it;) nbn-resolving.de/urn/… $\endgroup$ – Thomas Jun 23 '16 at 21:05
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    $\begingroup$ re: 1200K, OK great, thanks! $\endgroup$ – uhoh Jun 24 '16 at 0:10
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    $\begingroup$ This is fascinating! I've asked more about the blue light in this question. $\endgroup$ – uhoh Jun 24 '16 at 3:13

It's because of black body radiation; as the temperature increases the maximum light wavelength gets shifted to ultraviolet (near blue) regions which humans cannot see.

This article says:

Hydrogen flame, like alcohol flame emits little visible radiation but yet emits UV radiation.

According to Wikipedia:

With increasing oxygen supply, less black body-radiating soot is produced due to a more complete combustion and the reaction creates enough energy to excite and ionize gas molecules in the flame, leading to a blue appearance. The spectrum of a premixed (complete combustion) butane flame on the right shows that the blue color arises specifically due to emission of excited molecular radicals in the flame, which emit most of their light well below ~565 nanometers in the blue and green regions of the visible spectrum.

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    $\begingroup$ It's not just rockets. Firefighters have to be careful not to walk into a hydrogen flame if they are in a facility that stores hydrogen. $\endgroup$ – Loren Pechtel Sep 20 '13 at 3:54
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    $\begingroup$ For some reason this link http://www.en.wikipedia.org/wiki/Flame isn't working for me. But if I delete the www. that does work for me http://en.wikipedia.org/wiki/Flame. $\endgroup$ – uhoh Jun 22 '16 at 2:33
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    $\begingroup$ The first sentence can't be the reason that the flame is at first not emitting significant visible light. Things that are hot enough to emit strong thermal UV light will be extremely bright in the visible. As something gets hotter, the intensity at longer wavelength part of the spectrum just continues to increase, since the overall radiated power increases so rapidly with temperature: $\sim T^4 $ $\endgroup$ – uhoh Jun 22 '16 at 2:51
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    $\begingroup$ The first sentence of this isn't correct physics. It's true that higher temperature emitters shift to shorter wavelengths (i.e. UV) for the peak emission, but the emission at any given wavelength only goes up: As you raise the temperature, you still get more and more and more red light, it just that the shorter wavelengths start to go up even more. For example, see the 1000m, line on this figure: sunearthday.gsfc.nasa.gov/2009/images/ttt65_image1.jpg $\endgroup$ – Bob Jacobsen Jul 30 '18 at 15:34
  • $\begingroup$ This should not be due to blackbody radiation. See Russell Borogove's answer to the question, space.stackexchange.com/questions/44475/… , and the paper he referred to therein. In a common yellow-bright flame, the blackbody radiation comes from hot soot (carbon particles) in the flame. In the hydrogen-oxygen flame it is totally different. $\endgroup$ – verdelite Apr 16 at 2:22

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