Bender Posted April 25, 2017 Posted April 25, 2017 (edited) I think it would matter when you measure mass and what the efficiency of the processes involved are. Immediately after excitation (and ignoring that there are several capturing events) one would assume a mass increase of the photosystem as outlined by others above. However, that energy is used to e.g. split water and initiate electron transfer and excess energy is lost as heat and/or fluorescence (i.e. radiation). So measuring the total mass of the system later in the process should yield a different mass as one would expect from the initial absorbed energy alone. For such an experiment, you would obviously have to measure all inputs and outputs, including heat transfer through radiation or conduction. The main point was that the additional mass of sugar, or any other substance produced by plants, comes from radiation. This process does not have to be perfectly efficient. Hypothetically, if you could measure everything except for the heat lost, you could calculate those losses from the energy input and the mass difference. Still, you can approach it as a black box problem where you know nothing about the process. Edited April 25, 2017 by Bender
DrKrettin Posted April 25, 2017 Posted April 25, 2017 So from what has been said, is it the case that the mass of a spinning Earth is greater than that of a non-spinning one by virtue of the increase in kinetic energy Iw2? (that's an I omega squared)
Bender Posted April 25, 2017 Posted April 25, 2017 That is my understanding, yes. The rotational kinetic energy of earth is 2.138×1029 J. So converting that to mass is about 2×1012 kg, which is pretty negligible compared to the total mass of 6×1024 kg. According to Wikipedia, the Earth loses about 16 tons of this kinetic energy because the moon slows it down. The kinetic energy of the Earth moving around the sun at a speed of 30 km/s is 6e24.(30000)²/2=3×1033 J, or about 3×1016 kg, which is still negligible. What I don't know is how to deal with gravitational potential energy, because I don't know what reference point to pick. 1
swansont Posted April 25, 2017 Posted April 25, 2017 So from what has been said, is it the case that the mass of a spinning Earth is greater than that of a non-spinning one by virtue of the increase in kinetic energy Iw2? (that's an I omega squared) Rotational energy yes (as Bender has calculated) but translational KE does not, as this is accounted for separately. E2 = p2c2 + m2c4 Any translational energy is accounted for with the momentum term. That's why the mass term is the rest mass. 1
Bender Posted April 25, 2017 Posted April 25, 2017 Rotational energy yes (as Bender has calculated) but translational KE does not, as this is accounted for separately. E2 = p2c2 + m2c4 Any translational energy is accounted for with the momentum term. That's why the mass term is the rest mass. Right. But for small speeds (which the orbital speed of the Earth is): [math]E \approx \frac{mv^2}{2}+mc^2[/math] Does that [math]\frac{mv^2}{2}[/math] not contribute to the mass of the Earth?
humility Posted April 25, 2017 Author Posted April 25, 2017 Noooo! Stupid long radiation. It feels so wrong tor there to be so much of an energy that cant be used for anything even though it can be interacted with. But I guess even black holes release long radiation.
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