-5

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?

lopata
  • 139
  • You should try to come out from the hypotheses and preconditions created for yourself. – peterh Sep 27 '15 at 12:23
  • 3
    Real in what sense? – Qmechanic Sep 27 '15 at 12:39
  • 2
    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. – Apoorv Sep 27 '15 at 12:45
  • 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. – HDE 226868 Sep 27 '15 at 13:25
  • $\vec v=c^2\vec p /\sqrt{(mc^2)^2+|c\vec p|^2}$ 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 $c^2.$ – Timaeus Sep 27 '15 at 14:57
  • 1
    Related: http://physics.stackexchange.com/q/18250/ – dmckee --- ex-moderator kitten Sep 27 '15 at 16:29

2 Answers2

3

"$c$" 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 :

$$ \gamma = \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} $$

rdhox
  • 146
-1

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.

Syed Jaffri
  • 71
  • 3
    $v^2$ 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. – Bill N Sep 27 '15 at 13:44
  • That's what i'm saying- c^2 must be coming from similar logical progression – Syed Jaffri Sep 27 '15 at 13:45
  • but v^2 does not have any physical meaning, it's not real. Just like c^2 – Syed Jaffri Sep 27 '15 at 13:47
  • 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. – Bill N Sep 27 '15 at 13:51
  • Then why one does not give any name to v^2 – Syed Jaffri Sep 27 '15 at 13:58
  • No need to give a name to everything. Do you have a name for every function of velocity (e.g. $e^{-v/v_0}$) or every possible Lagrangian? Most certainly not, but it doesn't mean they aren't real. – Soba noodles Sep 27 '15 at 14:35
  • Those things are real, but not v^2 – Syed Jaffri Sep 27 '15 at 14:57
  • Again it boils down to what you consider "real". Bill N's comment gives a nice insight into the matter. – Soba noodles Sep 27 '15 at 20:27