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Sep 30, 2015 at 23:29 review Reopen votes
Sep 30, 2015 at 23:49
Sep 30, 2015 at 23:10 history edited lopata CC BY-SA 3.0
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Sep 27, 2015 at 16:29 comment added dmckee --- ex-moderator kitten Related: physics.stackexchange.com/q/18250
Sep 27, 2015 at 14:57 comment added Timaeus $\vec v=c^2\vec p /\sqrt{(mc^2)^2+|c\vec p|^2}$ which is less than (or equal to) $c$ since $c\vec p /\sqrt{(mc^2)^2+|c\vec p|^2}\leq 1.$ And $|\vec v|=c$ when $m=0.$ Finally, $c^2$ and $c$ have different units so it is wrong to compare them because to which one is larger depends in your unit system. If you use mks then $c^2$ is numerically bigger than $c$ but if you use natural units they are numerically equal and if you use lightyear per hour then numerically $c$ is bigger than $c^2.$
Sep 27, 2015 at 13:58 history closed Bill N
David Z
Needs details or clarity
Sep 27, 2015 at 13:47 review Low quality answers
Sep 27, 2015 at 14:03
Sep 27, 2015 at 13:25 comment added HDE 226868 If you're asking why $c^2$ gives a greater value than $c$, and that means that there is a quantity greater that the speed of light, then the error is interpreting $c^2$ as a speed - which it isn't, if you look at the units.
Sep 27, 2015 at 12:45 answer added Syed Jaffri timeline score: -1
Sep 27, 2015 at 12:45 comment added Apoorv It is unclear what you are asking. c is a real number and a universal constant and so is c^2. It is always permitted to square real numbers.
Sep 27, 2015 at 12:39 comment added Qmechanic Real in what sense?
Sep 27, 2015 at 12:34 answer added rdhox timeline score: 3
Sep 27, 2015 at 12:23 comment added peterh You should try to come out from the hypotheses and preconditions created for yourself.
Sep 27, 2015 at 12:20 history asked lopata CC BY-SA 3.0