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Can light waves be canceled by merging them with their inverted waves? Seems like it would violate conservation of energy but waves are added together when they overlap, right? Where is the flaw in this logic? I'm thinking polarized laser light, added to its opposite, might becomes dark again. -- Appreciating your collective wisdom.

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Yes - light waves can destructively interfere. This is the principle behind interferometers. There is no violation of energy conservation because the energy of two waves doesn't add. The energy is proportional to the square of the amplitude, and the amplitudes add so $E \sim \left( A_1 + A_2 \right)^2 \sim A_1^2 + A_2^2 + 2 A_1 \cdot A_2$. The third term is an interference term between the two waves which can be negative and cancel the others. –  Michael Brown Jan 18 '13 at 2:21
    
@MichaelBrown that would make a pretty good answer. –  David Z Jan 18 '13 at 2:47
    
Thank you Michael. Wow, physics is fun. Thanks for the term interferometer. Thus, we can make a wave with no amplitude by having two waves merge - and then use it to detect minor shifts in gravity, rotation etc because of the resulting distortions to the overlapped waves. Coool. –  David Jan 18 '13 at 2:51
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Possible duplicate: physics.stackexchange.com/questions/23930/… –  Steve B Jan 18 '13 at 3:19
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1 Answer 1

Yes - light waves can destructively interfere. This is the principle behind interferometers. There is no violation of energy conservation because the energy of two waves doesn't add. The energy is proportional to the square of the amplitude, and the amplitudes add. So $E\sim\left(A_1+A_2\right)^2\sim A^2_1+A^2_2+2A_1\cdot A_2$. The third term is an interference term between the two waves which can be negative and cancel the other two terms.

EDIT: Eduardo Guerras (validly) brings up a point about global cancellation versus local cancellation of waves. A local cancellation is simply the interference phenomenon I mentioned and there is absolutely no problem with it. Global cancellation is a different beast. You cannot, using any real combinations of sources, arrange for the global cancellation of waves. In fact, a global cancellation of just about anything is a sign that you have used an inappropriate mathematical abstraction, such as infinite plane waves, which simply don't exist in reality.

There is a sense in which global cancellation is fine, but pointless. If you have one wave (take a scalar wave $\phi\left(x,y,x,t\right)$ instead of an electromagnetic wave for simplicity) $$\phi_1\left(x,y,x,t\right)$$ and another $$\phi_2\left(x,y,x,t\right)$$ which just happens to be $\phi_2 = -\phi_1$ everywhere at all times, then the sum of the two is clearly zero and you have global cancellation. But in a very trivial this is the same as there being no wave at all to start with. What you definitely cannot do is produce such a pair of waves from a source - a simple application of the wave equation in this instance shows that the source must vanish. So the whole idea of a global cancellation is trivial and pointless in practice. That is why I assumed you were referring to a local cancellation, but I should have been clear from the start.

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Note that when $A_1$ and $A_2$ are in phase (both positive max or both negative max), you get extra energy. What ends up happening is that this doesn't cancel everywhere; it makes an interference pattern where it's higher energy some places but lower energy others. And if it interferes negatively when it strikes a surface, you get zero absorption on that surface - it all gets either reflected or it passes through. –  Will Cross Jan 18 '13 at 4:16
    
I don't think that this is a good answer (-1) since it leads to the (wrong) idea that you can make light "vanish" by simply making it interfere with itself. That is wrong. Interferences happen only in a certain region of space, whatever the experimental set up you invent, but there is always non-interfering light elsewhere. If you don't agree with that, then please show a gedankenexperiment in which light destructively interferes with itself all over the place! –  Eduardo Guerras Valera Jan 18 '13 at 8:07
    
If that "infinitely extended interference region" where possible, then there would be indeed energy conservation violation (the OP is right about that!) but this simply doesn't happen. Among other facts, it is geometrically impossible. –  Eduardo Guerras Valera Jan 18 '13 at 8:15
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The addition of amplitudes is a mathematical model. But if you managed to be pumping electricity to a light source but there were no detectable light absolutely anywhere, then that would be an energy conservation violation and your passport to the next Nobel Prize. –  Eduardo Guerras Valera Jan 18 '13 at 8:30
    
Eduardo you are so right. –  Marty Green Jan 18 '13 at 10:15
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