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In the physics section there is doubt, that a mathematician would object to deriving something by dividing an equation by a variable ##m##, and then applying that derived result to all values of ##m## including ##m = 0##:
To context is this proof that Newtonian gravity affects particles with ##m = 0##:
to a situation where ##m = 0##. In my opinion no, because the above step implicitly assumes that ##m ≠ 0##, otherwise it would not be a valid operation. So whatever you derive by doing this step cannot be applied when ##m = 0##.
DaleSpam said:Even if some mathematician would object (which I doubt), it is clear that in physics canceling variables on both sides of an equation is a common and well-established technique. It is part of nearly every non-trivial proof that I have ever seen. So even if a mathematician would object to ##mx=my## for all m implying ##x=y## I don't think that any physicist would object to ##ma=mg## implying ##a=g##.
To context is this proof that Newtonian gravity affects particles with ##m = 0##:
My question is not about physics, which can postulate whatever it wants. I would like to hear a mathematicians opinion, on whether you can apply something, that you derived by doing this:DaleSpam said:CKH said:It seems to me that Newtonian gravity as described by Newton could not predict an effect on something mass-less.
Newtonian gravity is a force that exists between masses ##(F=G m_1 m_2 / r^2)##. In his equations there would be zero gravitational force acting on zero mass. Since ##F=ma##, ##F/m=a##, but in this case that is ##0/0=a## which is undefined.
##F=ma##
##GMm/r^2=ma##
##gm=ma##
##g=a##The acceleration is independent of mass, and is well defined.
DaleSpam said:##gm=ma##
##g=a##
to a situation where ##m = 0##. In my opinion no, because the above step implicitly assumes that ##m ≠ 0##, otherwise it would not be a valid operation. So whatever you derive by doing this step cannot be applied when ##m = 0##.
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