Radiation dose in Joules/Kilogram, so is Gravitational Potential


Not sure if this will add to your discussion, but Volt^2 might be easy to remember if you think of dosage in Photons per second, an Ampere of unit charge as (1/e = 6.2415091E18 per second) and energy in Watt seconds. Watts = Volts*Amperes = Volts^2/Ohm. The total absorbed dose in Joules/Kilogram is Watts*Seconds/Kilogram.

The photons are not considered to have charge, but if the end result is energy deposited and you use electron volts, then how it gets there is partly irrelevant. There are so many tens of thousands of rules of thumb on the Internet in the many technologies. It is hard to memorize them all perfectly. I am looking into replacing all the Internet servers and browsers with AI assisted methods, so you just ask, “Computer! and it will likely know what you are doing and what you are likely to ask. These groups who store their knowledge on paper (PDF is worse than paper since it is often locked and hard to get to) waste vast amount of human time. It is almost criminal, perhaps it ought to be.

Joules/Kilogram is the unit of the gravitational potential. At the Earth’s surface it is about GM/R = 3.986004418E14/6371000 = 62.6 MegaJoules/Kilogram. But we walk through it, or it flows through us as we walk, with little resistance or drag. Only at “relativistic” speeds is it noticeable. X-rays are so inefficient. I wonder if there is a better way.

Richard K Collins

About: Richard K Collins

Director, The Internet Foundation Studying formation and optimized collaboration of global communities. Applying the Internet to solve global problems and build sustainable communities. Internet policies, standards and best practices.

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