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Assuming the spacetime principle, if the space is modified the time does too. So if the velocity in the space is increase, does the time slow down? What happens if the speed is the speed of light, does the time stops?

Qmechanic
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Alejandro
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3 Answers3

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No, for several reasons.

First, the idea of time "slowing down" is a little bit of a misnomer. If you were traveling at relativistic speeds, you would not perceive the passage of time any differently than you do right now. It's only when you compare your clocks to an observer in another reference frame (let's me, sitting in my living room, at rest with respect to the ground) that you would notice that your clock shows a smaller passage of time than mine does. So within your own frame of reference, it makes no sense to talk about time ticking slow or fast, it's only when you compare your clock to a clock in another reference frame that you can compare passages in time.

Also, and this is probably the larger hang-up to your question, is that a particle with mass can never reach the speed of light. With truly astounding amounts of energy, you could get arbitrarily close to the speed of light, but never reach it. The reason is because at relativistic speeds, the equation for kinetic energy is different than the classical equation for kinetic energy that you're used to.

Classically, the kinetic energy of a moving particle is given by $$KE=\frac{1}{2}mv^2$$ where we can reach any finite velocity that we want, so long as we do enough work on the object to increase its kinetic energy to the proper amount. However, it turns out that this is equation is only an approximation for small velocities. Relativistically, the equation for kinetic energy is $$KE=\frac{mc^2}{\sqrt{1-\frac{v}{c}^2}}-mc^2$$. What this equation shows you is that for any finite amount of kinetic energy on an object $v<c$. If you have trouble seeing this, attack the equation from a different angle. Let's say I'm crazy, and it is possible to travel at the speed of light. So let's take $v=c$ and substitute that into our relativistic kinetic energy equation. Oh no! You'll see that we get a divide by zero error, because our entire denominator goes to zero. This means we would need an infinite amount of kinetic energy for our object and that's just not possible.

So I hate to burst your bubble, but no light-speed travel for you anytime soon. (Or ever).

Sean
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Its not possible to stop time but using relativity it can be thought of to be slowed down .
Nothing can be faster than the speed of light so its not possible .
Even when we near it , energy tends to become infinity .

stud
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Things will be bigger/smaller/slower with speed as $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$.

Thus, as $v \rightarrow c$,

  1. Mass (or energy) goes to infinite,
  2. Time goes to zero.

But it is only a limit, which is unreachable (at least in the special relativity), because of (1).

peterh
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    This is reachable for photons however, since those are already massless. – DK2AX Mar 05 '15 at 11:03
  • Ok, but this is only relative to something not moving, which is already a challenge to understand. meaning a gets accelerated relative to b and then moves close to speed of light. However, there is not change in the sensation of time for a; it does not slow down and-as mentioned in the answer-definitively does not stop. – mikuszefski Mar 05 '15 at 11:29
  • @mikuszefski To me is it unclear what do you understand on "not moving". The speed of light is an absolute thing in SR, but "moving" (in the sense of $v=0$) is not. And with $v \rightarrow c$, the mass goes to infinity in every reference frame. – peterh Mar 05 '15 at 12:17
  • @andynitrox Yes, but they are going always with $c$, thus the formula above can't be applied on them. And the question is definitely about slower-than-c things. – peterh Mar 05 '15 at 12:18
  • Ah yes I missed that part of the question. – DK2AX Mar 05 '15 at 12:37
  • @peterh I basically agreed with your answer, despite some details; but on the comment I must say no. 1: the speed v must be relative to something. 2: if I am inside a rocket starting from earth accelerating to 0.999 c I still have 90 kg in my reference frame. In any case I believe that attaching gamma to the mass is rather unfortunate. This is a property of the momentum 4-vector transformation. Decomposing it afterwards and saying it's the mass that changes, is an arbitrary and non necessary choice. I can go further and decompose mass into volume and density and say density changes... – mikuszefski Mar 05 '15 at 13:15
  • @mikuszefski Afaik I never stated anything against your (1) or (2). I answered simply to a simple question, but not against your (1), and also not against your (2). – peterh Mar 05 '15 at 13:23
  • @peterh OK, then I just misinterpreted your "the mass goes to infinity in every reference frame". No problem, cheers. – mikuszefski Mar 05 '15 at 13:48
  • @mikuszefski Well, I had to write: "in every nonaccelerating reference frame"? – peterh Mar 05 '15 at 13:58
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    This is a very incomplete answer. For example: What is $1/\sqrt{1-v^2/c^2}$? Where did it come from? What "things" do you mean? How does mass/energy go to $\infty$? How does time go to 0? – Kyle Kanos Mar 05 '15 at 15:21
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    @KyleKanos $v$, $c$, $\infty$ and such have their common meanings. "Things" means here mass points moving slower than light. Thank you. – peterh Mar 05 '15 at 23:46
  • Lovely response. I try helping and all I get is a non sequitor. Do you ever actually respond in a coherent manner, or is it always this way? – Kyle Kanos Mar 06 '15 at 03:38
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    @KyleKanos It was a simple question, which deserved a simple answer. I couldn't include a 200 page long book with the axiomatical description of the special relativity. I don't really understand, what is not clear to you. You know also SR, probably better as me. But this was exactly (the Lorentz-transformation of the spacetime coordinates) which couldn't been written there, because the question required clearly the possible simplest answer. – peterh Mar 06 '15 at 04:04
  • I state precisely what is unclear with your post in my first comment. Please parse that and, if you feel up to it, fix your post to actually answer the question. – Kyle Kanos Mar 06 '15 at 04:10
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    I'm also not suggesting that I don't understand your post. I know what you mean to say, but you are doing a terrible job at actually saying it. What I am saying is that John Q Public who doesn't have any physics knowledge would look at your post would have no clue what you were talking about. – Kyle Kanos Mar 06 '15 at 04:12
  • @KyleKanos Yes, it is unfortunately true. In similar cases I vote up the more exact, but not so simple (and not so fast) answers. – peterh Mar 06 '15 at 04:42