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For an object to create different EM waves, it needs to increase the temperature, so what if we or some material could be so hot, that it would emit ultraviolet light, and thanks to that be invisible for the human eye.

I have a lot of questions about this and I would like you to answer me.

  1. At which temperature does an object emit UV light?
  2. If an object emits UV light, we wouldn't see anything or we would see some type of violet light?
  3. Is there any material that can get to that temperature without melting?
  4. Is there any powerful insulator?

Thanks for reading this, and for breaking my dream of making someone invisible, I invite you to day-dream and imagine stupid questions.

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  • $\begingroup$ If you are just searching for ways to make you invisible, as the answer of Brionius shows, you primarily need to prevent reflection, which means you have to send light around the body. This can be done to a certain way via Metamaterial cloaking: en.wikipedia.org/wiki/Metamaterial_cloaking $\endgroup$ – Martin Jun 11 '15 at 9:06
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    $\begingroup$ Made an account here just to up vote for "Thanks for reading this, and for breaking my dream of making someone invisible, I invite you to day-dream and imagine stupid questions." that enthusiasm should be rewarded :) $\endgroup$ – Vincent Jun 11 '15 at 10:14
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    $\begingroup$ The sun emits UV light. $\endgroup$ – Random832 Jun 11 '15 at 13:12
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    $\begingroup$ When you heat a metal it becomes red hot,means it actually emitting most of the photons of colour red,but that doesnt mean that they are not emitting photons of other colour.(wavelenght). $\endgroup$ – Paul Jun 12 '15 at 1:59
  • $\begingroup$ Even if an object doesn't give off any visible light that doesn't make it invisible, that just makes it black. Put it in front of something less black and you will be able to see it in silhouette. $\endgroup$ – bdsl Jun 12 '15 at 13:44
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You're right that as the temperature increases, shorter wavelengths receive a higher proportion of thermally radiated power, and longer wavelengths a smaller proportion, because of the shifting Boltzmann distribution of your molecules' kinetic energy, and therefore the shifting power spectrum of the light they emit.

However, most of the objects you see around you are not visible because they're thermally radiating, they're visible because visible light from the sun or a lightbulb are reflecting off their surface.

For example, right now you are "glowing" mainly in the infrared spectrum. You are emitting almost no visible light due to thermal motion of your molecules, because you are too cold to emit an appreciable amount of visible light by that process. But even though you aren't thermally radiating visible light, you are still reflecting light from external sources, and thus are perfectly visible.

It's also worth pointing out that while as an object gets hotter, it does radiate a higher percentage of its thermal radiation in higher frequencies, and a lower percentage in lower frequencies, as an absolute measure, the amount of power in any given frequency band actually increases with temperature. You can see this visually in the power spectrum curves in this graph. If you pick any wavelength, the amount of power is greater in each successive temperature curve. That means that when you heat an object hot enough to radiate primarily in the UV, it will actually be brighter in the visible spectrum than at cooler temperatures.

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  • $\begingroup$ excellent point about the visibility at normal temperatures. $\endgroup$ – Floris Jun 11 '15 at 1:28
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    $\begingroup$ So, would this mean that at the temperatures necessary to generate UV light that you would be radiating so much light in the visible spectrum that you would burn out people's retina's and hence be invisible? $\endgroup$ – Klors Jun 11 '15 at 14:19
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    $\begingroup$ Yeah probably. On the other hand you would have been vaporized, and then ionized, at that temperature, so you'd be no better off. You'd be invisible because you'd very shortly be in a gaseous state. $\endgroup$ – Brionius Jun 11 '15 at 14:28
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    $\begingroup$ @Brionius - Hot gases and plasma are visible. That's how astronomy started. $\endgroup$ – Jirka Hanika Jun 12 '15 at 8:43
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    $\begingroup$ @Jirka Yes, of course, but if you vaporized and ionized a human being, they would very quickly dissipate into the atmosphere and become invisible. $\endgroup$ – Brionius Jun 12 '15 at 9:31
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Planck's Law gives us the intensity of black body radiation as a function of temperature:

$$B(\lambda,T)=\frac{2hc^2}{\lambda^5}\cdot \frac{1}{e^{\frac{h c}{\lambda k_B T}}-1}$$

If we plot a normalized plot of this curve for different temperatures, you see the following:

enter image description here

As you can see, it does look like the higher temperatures make the relative intensity in the visible part of the spectrum lower. However, if we plot the absolute intensity (I can't figure out how to show the visible colors on this plot, sorry) you need a log scale to see what is going on:

enter image description here

(vertical axis: Watt / steradian / square meter / meter - units of spectral radiance).

As you can see, at the higher temperatures the entire curve is shifted upwards - the intensity in the visible part of the spectrum gets higher.

Of course, with that much UV you will probably make everyone who looks at you blind in an instant, after which you would indeed be "invisible". But I don't think that is what you had in mind with your question...

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  • $\begingroup$ Good point about the amount of power in each band increasing with temperature. $\endgroup$ – Brionius Jun 11 '15 at 2:16
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Unfortunately it does not work that way. If a body is hot enough to emit in UV it will still emit in visible light, and will actually emit more EM radiation at optical wavelengths that when it was colder and the peak was in visible light. So, no invisibility here. It is due to Plank's law, a good graph can be found here.

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1) It's true that the peak wavelength for a black body decreases with temperature. But let's say you want to know what temperature has what peak wavelength. Well, you can Google on "peak wavelength temperature calculator" and try for yourself. But I'll give you the short form. Since visible light is in the range of 400 to 700 nm, your body would have to be at about 8000 degrees Kelvin to have a peak in the UV. You might notice a whole lot of visible light being emitted, not to mention your surroundings bursting in to flame (it's hard to be invisible at a time like that). Just for comparison, iron melts at 1800 degrees K.

2) But let's say you could somehow reflect all light that hits you, but it is somehow transformed into ultraviolet. Would you be invisible? No, you would appear dead black. That is, you would emit no light that your eyes can detect, and that is the definition of black. Plus, of course, you would not be transparent, so you would be easy to see.

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To summarize: hotter => bluer, but more radiant => brighter. Something which is very hot, but invisible, would need to be small and have an energy output proportionate to the lower surface area.

An energy source of a given number of watts is indeed easier to "hide" if it emits mainly at higher frequencies. This is what makes cobalt-60 dangerous when carelessly scrapped, although that is not at a black-body spectrum. (Also, I'm not sure which part of its radiation is worse, the gamma rays or the electrons.)

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To make an object "invisible", one important thing you have to do is to stop it from blocking the light behind it from the observer's perspective.

If you have an opaque object that emits/reflects no visible light, then what you will see is a black silhouette of the object. The only way it will be invisible then is if it's against a completely black background, which means it's not preventing any light reaching the observer.

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  • $\begingroup$ You're absolutely wrong. To make an object invisible the light must pass through that object without being reflected or bent. If it stops absorbing light it will be a perfect mirror. If it absorbs all the light - totally dark. $\endgroup$ – Creative Magic Jun 12 '15 at 6:06
  • $\begingroup$ Perhaps I should have said explicitly that it should not absorb, reflect, or bend the light coming from behind it. It should behave like a transparent object - with the same refractive index as the air around it. I was just trying to convey that not emitting/reflecting visible light wasn't the main issue. $\endgroup$ – mwfearnley Jun 12 '15 at 23:17

protected by Qmechanic Jun 12 '15 at 6:00

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