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Can light made of photons in theory topple lets say a car? I know basic photon energy is hv where v is frequency. So according to conservation of linear momentum, if high enough and large amount of photon falls, would not it topple objects in particle nature of light.

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  • $\begingroup$ Related: Solar sails, Wikipedia. $\endgroup$
    – lemon
    Mar 23, 2016 at 12:11
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    $\begingroup$ Yes in fact new york weighs more on a sunny day than on a cloudy one. youtube.com/watch?v=Do1lm9IevYE $\endgroup$
    – Jaywalker
    Mar 23, 2016 at 12:34
  • $\begingroup$ Solar sail propulsion is now being developed by NASA scientists and their partners in industry ! $\endgroup$
    – drvrm
    Mar 23, 2016 at 12:40
  • $\begingroup$ Remember: if you use a highly reflective car, you get twice the momentum transfer because the photons "reverse" direction rather than being absorbed. $\endgroup$ Mar 23, 2016 at 14:02
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    $\begingroup$ There are far more efficient ways of toppling a car, but yes, it is perfectly possible. With the kind of energy required, there may not be much of a car left afterwards though. $\endgroup$
    – Neil
    Mar 23, 2016 at 15:05

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This depends on what the car is made of, but the likely answer is no.

For light, the relation $$E = pc $$ holds, so $$\frac{dE}{dt} = c\frac{dp}{dt} $$ $\frac{dE}{dt}$ is power, and $\frac{dp}{dt} $ is force, so for light $$F = \frac{P}{c} $$ for absorption, and $$F = \frac{2P}{c} $$ for reflection.

So, how much force do you need to apply to a car to tip it over? As a complete SWAG, let's say 1000 kgF, or about 10,000 newtons. Plugging this in, we get $$P = Fc = 10^4\times 3\times10^8 = 3\times10^{12} = 3 \text{ TW}$$ And yes, the T stands for terawatts, as in 3 million million watts. To put it in perspective, the entire annual US electrical power production is 2 TW-hr. To put it another way, that's the equivalent of burning about 70,000 tons of oil per second, so the car is going to get pretty hot. I suspect such a light beam would either vaporize the car or punch a hole in it before it tipped over, although this would be a matter to address using rather more sophisticated analysis than I'm willing to do.

You could, of course, specify that the car is perfectly reflective, in which case it would only take one half the power. Make sure to wear your safety goggles.

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    $\begingroup$ +1 for "Make sure to wear your safety goggles." That and your 1 meter thick lead plating hazmat suit. $\endgroup$
    – Neil
    Mar 23, 2016 at 15:31
  • $\begingroup$ @Neil - You are too kind. Actually, you don't necessarily need lead. If the light is visible, and nothing in the neighborhood (except potentially you) is getting hit by reflections, there is no ionizing radiation involved. $\endgroup$ Mar 23, 2016 at 15:41
  • $\begingroup$ It's not all that hard to estimate the tipping force - "assume" a rectangular block of uniform density, total mass 800 kg, 1.5 m high, so CM is 75 cm; calculate the torque about the CM, etc. $\endgroup$ Mar 24, 2016 at 14:45
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    $\begingroup$ @CarlWitthoft - Yeah, I could do that. Or you could, if you think it adds to the discussion. Feel free to contribute a more precise answer. I just did it as a Fermi experiment, and I'm pretty sure my SWAG is good within a factor of 10 either way, so I didn't bother with unnecessary precision and just got on with it. $\endgroup$ Mar 24, 2016 at 18:43

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