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Well i know that it needs infinite energy..but the term infinite is itself not physical so i am not satisfied with this answer ...but what would happen if we provide a constant acceleration to an object for hundreds of year ...isn't its velocity will go on increasing with time ?? some told me as velocity increases ...so time dilation occurs and time slows down ...so acceleration will not work ...but i didn't get it ..its all messed up ...please help ..

Can i get an answer only in term of forces and time i mean without using infinite energy concept ...i hope you will get what i wanna ask about

Qmechanic
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Abhay
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    This really just requires an understanding of asymptotes – mmainville Sep 02 '17 at 20:29
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    Possible duplicate of Accelerating particles to speeds infinitesimally close to the speed of light? And: https://physics.stackexchange.com/q/10252/75633 ; https://physics.stackexchange.com/q/267852/75633 ; https://physics.stackexchange.com/q/80365/75633 ; and many others. – stafusa Sep 02 '17 at 20:40
  • This may sound a bit unconventional, but you first need to understand what the rest mass is. "Rest" is the key here. In other words, the rest mass is a localized energy, the energy that does not fly away with the speed of light. So its nature of being localized, as in not flying away with the speed of light, is incompatible with it flying with the speed of light. If you remove the energy localization, e.g. annihilate an alectron and positron and produce two photons instead, then your energy flies away with the speed of light, but in two opposite directions. – safesphere Sep 02 '17 at 22:13
  • An excellent answer see https://physics.stackexchange.com/a/234929/46708 – HolgerFiedler Sep 06 '17 at 06:53

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Well the kinetic energy of an object can be written as $$E=(\gamma-1)m_0c^2$$where $\gamma$ is the realativistic factor $$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ $m_0$ is the rest mass and $c$ is the speed of light. From this you can see that $$\lim_{v\to c}\gamma=\infty$$ This means that the energy tends towards infinity as you get faster. Obviously we have no concept of infinite energy - and neither do we have to, as you can't reach the energy required to get to $c$.

Hope this helps :)

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You have the answer in your own question. You only have to get rid of classical prejudices and understand that time is not universal anymore. Different observers do measure different time rates. If your velocity is higher, your time runs slower. For usual velocities this is never noted, but if you travel at really high $v$'s (e.g. 10% of c) you will start noting that. If you try to reach $c$, our time just freezes, so there's no next instant in which your velocity has increased over $c$.

FGSUZ
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  • Thanks for your answwer it helped alot ... however Stafusa's answer was also quite satisfying but yours is pure practical – Abhay Sep 03 '17 at 07:07
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if we provide a constant acceleration [...] answer only in term of forces

The problem is this "if": it can't be realized, you can't keep the acceleration constant.

You can keep the force constant, perhaps, but what you are going to observe then is that, even though the force is constant, the acceleration starts decreasing noticeably as you get close to light speed, tending to zero as the speed tends to $c$. So you never reach it.

The explanation of why that happens could be given in the lines of:

The (relativistic) mass $m$ of an object of (rest) mass $m_0$ (at $v=0$) grows with the speed $v$ according to $m = m_0/\sqrt{1-v^2/c^2}$, which clearly diverges for $v\to c$.

Notice, though, that that would be probably misleading, as it's nowadays pretty much agreed on, that "In reality, the increase of energy with velocity originates not in the object but in the geometric properties of spacetime itself." Nonetheless, it gives still probably a helpful intuition if someone wants to reason in terms of forces.

stafusa
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By acceleration, I assume you mean the 3-acceleration, the one that was learnt before relativity.

If the 3-acceleration is constant, the 3-velocity increases linearly, and the mass $m=m_0/\sqrt{1-v^2/c^2}$ increases, too. So the 3-force $F=ma$ is increasing linearly with the mass. So as you can see, as 3-velocity approaches $c$, the mass $m$ approaches infinity, so does the 3-force.

I hope this answers your question. It is difficult not to use the concept of "infinity".

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The answer is as trivial as it could only be. To accelerate a body to a velocity greater the velocity of light $c$ you need a body or at least a system which is faster interacting with your body as $c$. As long as an interaction faster $c$ is not observed you won't have the "leverage" to accelerate something behind $c$.

HolgerFiedler
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    This is not how acceleration works. Mind explaining a bit further? – Javier Sep 02 '17 at 20:45
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    Wrong answer. The speed of light is not a technical limitation, but the limitation of the hyperbolic geometry of spacetime. I this geometry, speeds faster than light simply do not exist topologically. – safesphere Sep 02 '17 at 21:38
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    Note that a rocket with a mass ratio larger than $e$ can exceed the exhaust speed of it's own exhaust, providing an explicit counter example to the principle suggested here. – dmckee --- ex-moderator kitten Sep 03 '17 at 04:05
  • @safesphere Did you say that the speed of light is a theoretical construct and not derived from observation? – HolgerFiedler Sep 03 '17 at 15:41
  • @dmckee Please share your knowledge or give a source how a rocket get a velocity greater the velocity of the exhaust. – HolgerFiedler Sep 03 '17 at 15:43
  • @HolgerFiedler: Yes, the speed of light is not a measured value. In natural units, it equals one (e.g. one light second per second). These units are not up to scale in real life, so we use the metric system instead, in which this unity is multiplied by a whole number for convenience. So c=299792458 meters per second. It is not a measured value, but a postulated whole number with no fractions. You cannot measure the speed of light, it is theoretically given as the unity that for every day convenience is multiplied by this whole number. – safesphere Sep 03 '17 at 16:28
  • @safesphere Hope this is helpful https://physics.stackexchange.com/a/355296/46708 – HolgerFiedler Sep 03 '17 at 19:58
  • @HolgerFiedler: Yes, everyone confirmed what I told you and then you selected the worst answer. – safesphere Sep 03 '17 at 20:21
  • @safesphere You are confused by the possible base units. They are indeed arbitrarily selectable. Furthermore derive the speed of light without measuring it. c is a natural unit and can't be derived without measurement. – HolgerFiedler Sep 03 '17 at 20:38
  • @HolgerFiedler: Not sure why you are duplicating the same comment in different threads. You had no understanding on the subject and wanted to learn. This isn't going well, because you are getting wrong ideas and setting your mind on them. This makes you much worse off than before, because before you didn't know the subject, but you knew that you didn't know it. Now you still don't know the subject, but wrongly believe that you know it. This is much worse, because no one can help hardminded beliefs. Definitely not me. So I wish you best of luck while learning or not, it's all up to you. – safesphere Sep 03 '17 at 21:27
  • @safesphere The best answer I have seen is this: https://physics.stackexchange.com/a/234929/46708. – HolgerFiedler Sep 06 '17 at 06:52