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Nov 17 at 12:14 history edited Thomas Fritsch CC BY-SA 4.0
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Nov 17 at 11:36 comment added Thomas Fritsch @dutchrunner No, by using $$p=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}$$ you always get a velocity $v$ smaller than $c$., regardless how big $p$ may be,
Nov 17 at 11:31 comment added dutchrunner But for instance, if I use 10^-14 m, I get a momentum that leads to a velocity that is almost equal to c, if I use 10^-15m, I get a momentum that leads to a speed that is more then the speed of light. So something goes wrong here
Nov 17 at 11:22 history answered Thomas Fritsch CC BY-SA 4.0