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Discussion: Photon Yoga

An object has the shape of a square and has side length \( a \). Light beams are shone on the object from a big machine. If \( m \) is the mass of the object, \( P \) is the power per unit area of the photons, \( c \) is the speed of light, and \( g \) is the acceleration of gravity, prove that the minimum value of \( P \) such that the bar levitates due to the light beams is \[ P = \dfrac {4cmg}{5a^2}. \]

Note: This problem was originally proposed by Trung Phan for the IPhOO.

Note by Ahaan Rungta
2 years, 9 months ago

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Are the photons striking perpendicularly? Are they rebounding? (Please clarify.)

If both of above assumptions are valid, using classical mechanics, I get \(P = \frac{cmg}{2a^2}\) Jatin Yadav · 2 years, 9 months ago

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@Jatin Yadav This is the issue. We did not mention any assumptions! Ahaan Rungta · 2 years, 9 months ago

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each photon (assuming it strikes and rebounds perpendicularly) grants a momentum of 2E/c E=2Pta^2/c so we need mgt=2E/c=2Pta^2/c so P=cmg/2a^2 Mvs Saketh · 2 years, 9 months ago

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Ahaan, can you suggest something Jatin and I might be missing? Josh Silverman Staff · 2 years, 9 months ago

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@Josh Silverman We gave a statement to prove and said nothing about how the light is reflected. Some investigation should suggest that it's reflected in the way that the probability for the ray to scatter in any direction is equal. Ahaan Rungta · 2 years, 9 months ago

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@Ahaan Rungta Hey, ahaan, even if we assume that the incidence and scattering is random, we get \(P = \dfrac{\pi cmg}{4 a^2}\), not \(\dfrac{4cmg}{5a^2}\)

Initial vertical momentum/sec = \(\dfrac{Pa^2}{c} \displaystyle \int_{- \frac{\pi}{2}}^{\frac{\pi}{2}} \cos \theta \frac{d \theta}{\pi} = \dfrac{2 P a^2}{\pi c}\).

Final vertical momentum/sec = \(- \dfrac{2P a^2}{\pi c}\)

Hence, Force = \(\dfrac{4P a^2}{\pi c}\)

Hence, \(P_{min} = \dfrac{\pi cmg }{4 a^2}\), which is quite close to the answer you show, but not exactly equal! Jatin Yadav · 2 years, 9 months ago

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@Jatin Yadav @jatin yadav and all others, sorry to bother you here. How much are you expecting in mains Jatin? Vijay Raghavan · 2 years, 9 months ago

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@Jatin Yadav You did this problem in two dimensions. It is a three-dimensional problem. You need to take into account a trigonometric factor, resulting in a different integral. If you want, I can tell you the correct integral but I'd prefer not to as I want everybody to discover the correct solution themselves! Ahaan Rungta · 2 years, 9 months ago

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@Ahaan Rungta

we ... said nothing about how the light is reflected

Right, so, given that it's unspecified we are free to think of whatever arrangement for the geometry of the light shining onto the surface, right?

Light shining perpendicularly onto the square results in the most efficient momentum transfer to the square.

Like Jatin said, under that assumption (that the light shines perpendicularly), it is possible the levitate the bar with \(P = \frac{cmg}{2a^2}\) which is less than \(\frac{4cmg}{5a^2}\), contradicting the claim in the problem that \(P = \frac{cmg}{2a^2}\) is the minimum power that can achieve levitation.

The only possibilities are that Jatin and I are reading the problem wrong, or there is something wrong with our argument, or possibly that the claim is wrong. Can you perhaps point out where we might be going wrong in our reading of the problem? Josh Silverman Staff · 2 years, 9 months ago

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@Josh Silverman

Right, so, given that it's unspecified we are free to think of whatever arrangement for the geometry of the light shining onto the surface, right?

Which is why we scored everybody who got a \( \frac {1}{2} \)-coefficient 5/7. I will come up with a better reasoning soon. Ahaan Rungta · 2 years, 9 months ago

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