So, Nothing can go faster than speed of light.
How come the Energy be real if C² isn't real?
Is it even allowed to square C?
How can we use C² as a constant as it has not representations in nature?
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- 1$\begingroup$ You should try to come out from the hypotheses and preconditions created for yourself. $\endgroup$peterh– peterh2015-09-27 12:23:53 +00:00Commented Sep 27, 2015 at 12:23
- 4$\begingroup$ Real in what sense? $\endgroup$Qmechanic– Qmechanic ♦2015-09-27 12:39:44 +00:00Commented Sep 27, 2015 at 12:39
- 2$\begingroup$ 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. $\endgroup$Apoorv– Apoorv2015-09-27 12:45:36 +00:00Commented Sep 27, 2015 at 12:45
- $\begingroup$ 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. $\endgroup$HDE 226868– HDE 2268682015-09-27 13:25:44 +00:00Commented Sep 27, 2015 at 13:25
- 1$\begingroup$ Related: physics.stackexchange.com/q/18250 $\endgroup$dmckee --- ex-moderator kitten– dmckee --- ex-moderator kitten2015-09-27 16:29:23 +00:00Commented Sep 27, 2015 at 16:29
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2 Answers
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"$c$" is here a constant in the formula. The formula is not about about the speed of an object but energy, so there is no problem. Actually this formula is correct for an object who is at rest in his frame of reference, so with no speed at all.
Why "$c$" appear like that? It's like asking why "$\pi$" appear in some equations who have nothing to do with circle. You have to study special relativity to understand the demonstration who lead to that result. But one of the reason of this "$c$" is the gamma factor in the Lorentz transformation :
$$ \gamma = \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} $$
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8 It's just a derivation to find the energy . Overall mc^2 makes sense, which is the energy. For example - Kinetic energy = (1/2)mv^2. Here v^2 does not make any sense.
- 3$\begingroup$ $v^2$ does make sense because it is the logical progression of the math and it contributes to the correct units. Integrate $\frac{dp}{dt}dx$. The $v^2$ comes from that. $\endgroup$Bill N– Bill N2015-09-27 13:44:01 +00:00Commented Sep 27, 2015 at 13:44
- $\begingroup$ That's what i'm saying- c^2 must be coming from similar logical progression $\endgroup$Syed Jaffri– Syed Jaffri2015-09-27 13:45:45 +00:00Commented Sep 27, 2015 at 13:45
- $\begingroup$ but v^2 does not have any physical meaning, it's not real. Just like c^2 $\endgroup$Syed Jaffri– Syed Jaffri2015-09-27 13:47:25 +00:00Commented Sep 27, 2015 at 13:47
- $\begingroup$ It absolutely does. It is the square of the magnitude of the velocity. Now, if you want to say it's abstract, that's different. Abstract things can be real, just as concrete things are real. Do you think momentum is real? I do. But it's abstract. $\endgroup$Bill N– Bill N2015-09-27 13:51:10 +00:00Commented Sep 27, 2015 at 13:51
- $\begingroup$ Then why one does not give any name to v^2 $\endgroup$Syed Jaffri– Syed Jaffri2015-09-27 13:58:29 +00:00Commented Sep 27, 2015 at 13:58