Emission of light from a surface

In summary: The sun should similarly be a point source.In summary, the conversation is about a person in the lighting industry who believes that according to Gauss's divergence theorem and Maxwell's Laws, light is only emitted orthogonal to the surface it is emitted from. However, the speaker has tried to present real world examples that contradict this premise, but the person continues to stick to their mathematical argument. The speaker is seeking a good mathematical argument to counter this belief, as the person presents themselves as an expert and may influence others with their mention of Gauss and Maxwell. The speaker also points out the flaw that following this argument to its logical conclusion would mean a heated sphere appears
  • #1
Eric Bretschneider
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I came across someone in the lighting industry who insists that because of Gauss's divergence theorem and Maxwell's Laws that when light is emitted from a surface that it is only emitted orthogonal to the surface. I have tried to point out numerous real world examples that contradict the premise, but this person keeps going back to the equation and insists they are right.

Does anyone have a good mathematical argument I can present? The concern I have is this person presents themselves as an expert and there are people who will be awed by the mention of Gauss and Maxwell and won't follow the argument to logical (and impossible) conclusions.
 
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  • #2
Eric Bretschneider said:
I came across someone in the lighting industry who insists that because of Gauss's divergence theorem and Maxwell's Laws that when light is emitted from a surface that it is only emitted orthogonal to the surface. I have tried to point out numerous real world examples that contradict the premise, but this person keeps going back to the equation and insists they are right.

Does anyone have a good mathematical argument I can present? The concern I have is this person presents themselves as an expert and there are people who will be awed by the mention of Gauss and Maxwell and won't follow the argument to logical (and impossible) conclusions.

But what exactly does it mean to say that it is "... only emitted orthogonal to the surface.. "? Does it mean that the wavevector is normal to the surface? Or does he mean that the Poynting vector is normal to the surface? While those two are often in the same direction, this is not true for all cases, especially when the medium of propagation is involved.

Zz.
 
  • #3
ZapperZ said:
But what exactly does it mean to say that it is "... only emitted orthogonal to the surface.. "? Does it mean that the wavevector is normal to the surface? Or does he mean that the Poynting vector is normal to the surface? While those two are often in the same direction, this is not true for all cases, especially when the medium of propagation is involved.

Zz.
The Poynting vector is normal to the surface. They equate emission from the surface of an LED chip to emission from a semiconductor laser. It is all related to the intensity distribution of light from LEDs vs other light sources.

They have dismissed obvious counter examples and continue to focus on their mathematical derivations.
 
  • #4
Mentors' note: An unhelpful and off-topic argument has been removed from this thread. All members are reminded that we're hear to advance people's understanding of physics.
 
  • #5
Nugatory said:
Mentors' note: An unhelpful and off-topic argument has been removed from this thread. All members are reminded that we're hear to advance people's understanding of physics.

What I am hoping for is a good argument as to why Gauss's divergence theorem combined with Maxwell's laws don't apply to generation/emission of photons. There seems to be a leap of faith between electric field lines and the path a photon would take.
 
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  • #6
Eric Bretschneider said:
I came across someone in the lighting industry who insists that because of Gauss's divergence theorem and Maxwell's Laws that when light is emitted from a surface that it is only emitted orthogonal to the surface. I have tried to point out numerous real world examples that contradict the premise, but this person keeps going back to the equation and insists they are right.

Does anyone have a good mathematical argument I can present? The concern I have is this person presents themselves as an expert and there are people who will be awed by the mention of Gauss and Maxwell and won't follow the argument to logical (and impossible) conclusions.
If light is reflected from a surface it can radiate in any direction. So if the electrons of the surface are excited somehow with some particular phase relation, the surface will presumably radiate a beam in any direction. And if they are excited with random phase, as perhaps by heating, then presumably the light will go in all directions.
 
  • #7
tech99 said:
If light is reflected from a surface it can radiate in any direction. So if the electrons of the surface are excited somehow with some particular phase relation, the surface will presumably radiate a beam in any direction. And if they are excited with random phase, as perhaps by heating, then presumably the light will go in all directions.

What frustrates me the most is that the following this person's argument to its logical conclusion means that a sphere heated to incandescence would appear as a point source. The sun should similarly be a point source.

That is why I am looking for the flaw in the logic of combining Gauss and Maxwell.
 
  • #8
Eric Bretschneider said:
What frustrates me the most is that the following this person's argument to its logical conclusion means that a sphere heated to incandescence would appear as a point source. The sun should similarly be a point source.
Seems to me that that's a very good qualitative argument right there. Have you asked him what he thinks of that?
 
  • #9
phinds said:
Seems to me that that's a very good qualitative argument right there. Have you asked him what he thinks of that?
They ignore that and argue that I don't understand Gauss and Maxwell.

I am struggling to see how they could be combined to explain emission of photons from a surface. Gauss describes the direction of electric field lines. Maxwell links electrical and magnetic fields and does describe photons.

I am guessing that someone made a huge leap when they combined Gauss and Maxwell and derived a result that requires photons to be emitted in the direction of the electric field. If there is nothing else I'm missing then I will continue on.
 
  • #10
Eric Bretschneider said:
They ignore that and argue that I don't understand Gauss and Maxwell.
Well you could always quote Feynman. "If a theory doesn't fit the experiments, it's wrong", and his theory doesn't fit experiments.
 
  • #11
They are making a mistake in their derivation. Gauss divergence theorem and Maxwell’s law do not imply that all emission is normal to the surface.

If they were correct then you could not get a penumbra from a light source with a smooth surface.
 
  • #12
Eric Bretschneider said:
They ignore that and argue that I don't understand Gauss and Maxwell.

I am struggling to see how they could be combined to explain emission of photons from a surface. Gauss describes the direction of electric field lines. Maxwell links electrical and magnetic fields and does describe photons.

I am guessing that someone made a huge leap when they combined Gauss and Maxwell and derived a result that requires photons to be emitted in the direction of the electric field. If there is nothing else I'm missing then I will continue on.

While I'm waiting for the turkey to get to room temperature before I stick it into the oven (don't worry, I've washed my hands and there's no risk of salmonella contamination), I'll address a few of the things that I'm puzzled here.

1. Gauss's Divergence Theorem, as stated in the first post, is a mathematical operation. It gives us the ability to take the volume integral of a divergence, and relate it to an area integral over a closed surface. When the divergence theorem is applied to the differential form of Gauss's Law, we get the integral form of Gauss's law. So

Divergence theorem ≠ Gauss's Law

But in your posts, it appears that you have used the two terms to mean the same thing.

2. Remember I asked earlier on what exactly is going out orthogonally from the surface. You clarified that it is the Poynting vector. If this is true, and if they are also insisting that the E-field is orthogonal to the surface, then these two are contradicting each other. Since the Poynting vector is E x H, it is impossible for both the Poynting vector and E to be normal to the surface.

The problem here is that we are seeing all of this only through your lens. All we have is your interpretation of it, based on your understanding. This can easily be inaccurate, and in fact, unfair to these folks. So unless you have references to their actual derivation, I do not think it is worthwhile to tackle this any further, because we can all easily be in a tizzy over a misunderstanding.

Zz.
 
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  • #13
ZapperZ said:
While I'm waiting for the turkey to get to room temperature before I stick it into the oven (don't worry, I've washed my hands and there's no risk of salmonella contamination), I'll address a few of the things that I'm puzzled here.

1. Gauss's Divergence Theorem, as stated in the first post, is a mathematical operation. It gives us the ability to take the volume integral of a divergence, and relate it to an area integral over a closed surface. When the divergence theorem is applied to the differential form of Gauss's Law, we get the integral form of Gauss's law. So

Divergence theorem ≠ Gauss's Law

But in your posts, it appears that you have used the two terms to mean the same thing.

2. Remember I asked earlier on what exactly is going out orthogonally from the surface. You clarified that it is the Poynting vector. If this is true, and if they are also insisting that the E-field is orthogonal to the surface, then these two are contradicting each other. Since the Poynting vector is E x H, it is impossible for both the Poynting vector and E to be normal to the surface.

The problem here is that we are seeing all of this only through your lens. All we have is your interpretation of it, based on your understanding. This can easily be inaccurate, and in fact, unfair to these folks. So unless you have references to their actual derivation, I do not think it is worthwhile to tackle this any further, because we can all easily be in a tizzy over a misunderstanding.

Zz.

I am really trying to make this as simple as possible. Is there a way to combine Gauss and Maxwell to come up with photons only being emitted in the direction perpendicular to the surface? This is equivalent to saying that light emission from a flat surface is virtually the same as light emission from a laser (well collimated).

It is disturbing that this is being taught to college level students.

https://www.crcpress.com/Understanding-LED-Illumination/Khan/p/book/9781466507722
 

Related to Emission of light from a surface

1. What is the emission of light from a surface?

The emission of light from a surface refers to the process by which an object emits visible light due to the release of energy from within its atoms or molecules. This can occur through various mechanisms such as thermal radiation, fluorescence, phosphorescence, or luminescence.

2. What factors affect the emission of light from a surface?

The emission of light from a surface can be affected by several factors, including the temperature of the object, the composition of its surface material, and the type of energy source that is causing the emission. Other factors such as external light sources and the object's surface properties can also impact the emission of light.

3. What is the difference between emission and reflection of light from a surface?

Emission and reflection of light from a surface are two different processes. While emission refers to the release of light from an object, reflection is the bouncing back of light from a surface without being absorbed. Reflection occurs when light hits a smooth surface, while emission can occur from any type of surface, regardless of its texture.

4. What are some real-world applications of emission of light from a surface?

The emission of light from a surface has many practical applications, such as in lighting systems, displays, and optical communication technologies. It is also utilized in medical imaging techniques, such as fluorescence microscopy, and in various industrial processes, such as welding and heat treatment.

5. Can the emission of light from a surface be controlled or manipulated?

Yes, the emission of light from a surface can be controlled and manipulated through various methods. For example, the intensity and wavelength of the emitted light can be altered by changing the temperature or chemical composition of the surface. Additionally, external factors such as electric and magnetic fields can also influence the emission of light from a surface.

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