A black body absorbs all energy. It doesn't reflect or transmit energy. It also absorbs all light and doesn't reflect any light. Why, then, can we see it? For instance, burnt platinum is 98-99% black body and yet is visible.

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    $\begingroup$ stars are near perfect black bodies because they reflect an insignificant amount of energy. how does that fit into your argument? $\endgroup$ Feb 26, 2018 at 10:13
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    $\begingroup$ @Cursed not "because they reflect an insignificant amount of energy". They are near enough to black body radiation but not due to reflection. They are not in equilibrium with the vacuum around them, so they just emit the black body of their temperature, while they absorb whatever small radiation hits them. The +1 votes are for the "stars are near perfect black bodies" $\endgroup$
    – anna v
    Feb 27, 2018 at 4:42

5 Answers 5


There are some misconceptions here :

Blackbody absorbs all energy.

So far so good.

It doesn't reflect or transmit energy.

No. Any object that did that would get hotter and hotter as it absorbed more and more energy and just retained it. A blackbody, like any object, will radiate energy away as photons - light at many different wavelengths.

In respect of light energy, it also absorbs all light and doesn't reflect any light ray.

See the previous paragraph.

then why can we see it?.....for instance burnt Pt which is 98-99% blackbody

Because a blackbody does not exist in total isolation and it will try to reach thermal equilibrium with the world around it. To do that it either has to radiate energy away (if it's hotter than the world around it) or absorb energy (if it's cooler).

Even when it absorbs it still radiates some energy, and it simply radiates less than it absorbs.

But when you talk about being "visible" in a human sense, this is a different thing. If you walk into a room with every object and surface emitting light you can see, and then I pop something into it that actually did not emit light at a visible wavelength you would still detect that visually - it would seem like a "hole" in the scene - a silhouette.

For something to be invisible to a human, i.e. not be detectable visually, it would need to be perfectly transparent (at least at visible wavelengths) - all light would have to pass through it without distortion. Alternatively, all light would need to go around it.

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    $\begingroup$ "I pop something into it that actually did not emit light at a visible wavelength you would still detect that visually - it would seem like a "hole" in the scene - a silhouette." - Just like this basketball coated in Vantablack. $\endgroup$
    – Samuel
    Feb 26, 2018 at 20:49
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    $\begingroup$ Your rejection of the statement "It doesn't reflect or transmit energy" is troubling me. Blackbodies indeed do not reflect or transmit EM waves, by definition of their being perfect absorbers. They do, however, emit -- as you have correctly gone on to say $\endgroup$
    – binaryfunt
    Feb 26, 2018 at 23:54

One must know the definition of a black body and its radiation:

"Blackbody radiation" or "cavity radiation" refers to an object or system which absorbs all radiation incident upon it and re-radiates energy which is characteristic of this radiating system only, not dependent upon the type of radiation which is incident upon it

Black, like white, is not a frequency in the electromagnetic spectrum. white is a combination of the frequencies of the spectrum that the retina decides as "white ", and black is the absence of detecting anything in the visible. It is called color perception.

Your example of platinum is interesting, because I am sure you are seeing it with light reflected from it , even if reflection is only 2% as you say, does it look black?

enter image description here

Normalized response spectra of human cones, to monochromatic spectral stimuli, with wavelength given in nanometers.

Our eyes see the frequencies within the visible spectrum, the other frequencies register as black, if they exist. At night an infrared camera sees, our eyes see black.

A perfect black body in equilibrium with the surrounding temperature will be absorbing incident radiation and emitting black body radiation according to the temperature scales of this radiation. At temperatures where our eyes can exist we perceive it as black, because these are infrared frequencies. We only see visible light reflections on bodies, not their black body radiation.

Outside equilibrium, if the emission has parts in the visible, our eyes and brain will see it as visible we will see the visible part of its black body radiation, as with the sun, or red hot iron at the blacksmith.

  • $\begingroup$ Soyou are saying," our human eyes detect the absence of visible frequencies as Black? " $\endgroup$ Feb 26, 2018 at 13:34
  • $\begingroup$ @annav to twist that around to make it more precise... If there are no frequencies that the receptors we have can detect, then we only see black. (I don't have receptors capable of seeing x-rays, but stick me (and a lamp) inside a lead box and I will still be able to see, despite the lack of X-rays) (this is a response to your comment, your answer has this correct) $\endgroup$
    – Baldrickk
    Feb 26, 2018 at 16:45
  • $\begingroup$ Corrected after comment :Yes, exactly. To sense light our eye cones have to interact with it and then a signal goes to the brain. If there are no receptors for all the frequencies coming from a certain space we see that space as black. the picture linked in a comment to an answer is indicative. nerdist.com/wp-content/uploads/2017/03/… $\endgroup$
    – anna v
    Feb 26, 2018 at 17:21
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    $\begingroup$ @MohammadMizanurRahaman Yes. Although we can also see something as black simply by contrast with something way brighter. Imagine a projector screen. It's white, smooth and very reflective. Then you turn the projector on and display some black text on a white page. The black text isn't any darker than the white screen you were looking at just before! It's just that with the projector making the area around it even brighter white the contrast makes the unlit areas appear as black. $\endgroup$
    – Joren
    Feb 27, 2018 at 11:03

A black body absorbs all electro-magnetic energy it receives and does indeed emit radiation to stay in thermal equilibrium. If it is very hot, it will be visible. As you say, a black body does not reflect energy: It absorbs and emits.

  • $\begingroup$ you are saying if it is very hot, it will be visible.But so for I know in room temperature burnt Pt is visible $\endgroup$ Feb 26, 2018 at 9:56
  • $\begingroup$ Then it is not a perfect black body. Some light is reflected and this is what you see. A black body at room temperature is invisible to the naked eye. $\endgroup$
    – Aziraphale
    Feb 26, 2018 at 10:02
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    $\begingroup$ In a lit room, our sense of vision has no problem noticing a black object. People may run into glass doors, but they won't run into a wall of platina-black. $\endgroup$
    – user137289
    Feb 26, 2018 at 10:08
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    $\begingroup$ @MohammadMizanurRahaman This is what a nearly-perfect blackbody at room temperature looks like. Eerie, isn't it? $\endgroup$
    – Chris
    Feb 26, 2018 at 10:37
  • $\begingroup$ @Pieter, I think it would still be easy to run into. With nothing to see, there's no way to judge depth. You could see a line on the carpet where the wall was, but you could hold your hand out and not know where the wall was until you poked it. $\endgroup$
    – JPhi1618
    Feb 26, 2018 at 18:38

You are combining several different issues into one question.

A black body absorbs all energy. It doesn't reflect or transmit energy.

You are treating "reflecting" and "transmitting" as the same. They are not. A perfect black body does not reflect any light, but it does transmit light, in the sense that it radiates light (that is, it glows from the heat of the energy it absorbs).

Why, then, can we see it?

In a steady-state situation, a black body would not be visible. That is, if the amount of light were constant across time and space, then the black body would look like every other part of space, and therefore not be distinguishable. But then, if all parts of space looked the same, then nothing would be distinguishable. If light varies across time and space, then a black body would be detectable as it "averages out" the variation; if a black body is surrounded by some objects that are hot and some that are cold, then the black body will radiate at an intermediate temperature, and therefore look different from the hot and cold objects. If the temperature of the black body's surroundings changes over time, then the temperature of the black body will take time to adjust. If the surrounding are heating up, then the black body will be radiating at a lower temperature than the surrounding, and if the surroundings are cooling down, the black body will be at a higher temperature than its surroundings. (There's a similar concept behind videos showing that areas exposed to moonlight are colder than areas in the shade. While this makes it seem like moonlight makes things colder, what's really going on is that whatever is giving shade is absorbing energy during the day and giving it off at night, and so areas in the shade are cooler than their surrounding during the day, but warmer than their surroundings during the night.)

And that, really, is what it means for something to be "seen": light coming from it is different from light coming from elsewhere. A uniformly yellow object surrounded by objects the exact same shade and intensity of yellow will be just as invisible as a black object surrounded by blackness.

For instance, burnt platinum is 98-99% black body and yet is visible.

And here you are conflating several further ideas. A perfect black body is 100% black, not 99%. The concept is more a theoretical construct for thinking through thermodynamic processes than an actual phenomenon that one encounters in everyday life. The human eye can detect objects at something like a millionth of the intensity of sunlight. So even something that is 99% black will, in direct sunlight, be reflecting enough light for it to be visible. Now, if there are other objects, they will be reflecting even more light, drowning out the light from a "black" object, but then the black object will be detectable by noting that there is a region of space with less light.

Another issue you seem a bit unclear on is that the term "black body" is generally used in the context of objects interacting through radiation. If you put a block of platinum on a table, the amount of heat exchange between the platinum and the table due to conduction will be orders of magnitude larger than the amount of heat that they are exchanging through radiation. Thus, standard black body analysis does not apply to that situation.

  • $\begingroup$ Nitpick: OP clearly treats "reflection" and "transmission" as separate, or he wouldn't be mentioning both. "Emmission" is yet another thing, which OP missed and you're grouping under transmission. For reference "absorption" = receive, "emmission" = send, "transmission" = pass through and "reflection" = return to sender. $\endgroup$
    – timuzhti
    Feb 27, 2018 at 0:37

Different visible wave lengths make sense of different colors in our brain.If there are no visible wavelengths, it makes sense of black. In respect of black body, it doesn’t reflect any visible elecro-magnetic wave as you mention.So it’s black and visible

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    $\begingroup$ Are you answering the original question? If so, it at best confusing because the burnt platinum is no black body, as the above answers already pointed out. $\endgroup$
    – Aziraphale
    Feb 26, 2018 at 18:56

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