Newton's second law of motion states that $f = ma$. However, in this equation, theoretically there could be a value of $f$ and $m$ that results in an acceleration that is enough to push an object past the speed of light. In my case I have a value of newtons that is large enough (I think) to accelerate an object past the speed of light if $f=ma$ held. Is there any way, knowing that value of newtons, and the mass of the object it is acting on to find the acceleration in an Einsteinian universe?
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1Are you asking this in the context of special or general relativity? – Danu Jul 26 '14 at 01:44
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1Why is the size of the force relevant? If it were possible to accelerate past c with some force F, then it would also be possible with any smaller force F'; it would just take longer. – Jul 26 '14 at 02:32
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Newton's second law was stated as $\mathbf F=\mathrm d\mathbf p\ /\ \mathrm dt$. Later it was derived in the form $\mathbf F=m\mathbf a$ since mass does not change, does it? And much later some german patent officer stuck his nose in Maxwell equations and people started thinking that Newton's second law was wrong. The second law is valid in both "einsteinian" and "newtonian" universe. – Crowley Jun 28 '16 at 14:51
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In special relativity, you are better off thinking of things as four-vectors, rather than three-vectors. In that case, you generalize momentum to 4-momentum ${}^{1}$(and you take time derivatives with respect to the clock of the spaceship).
Then, you have $\bf F = \frac{d{\bf p}}{dt}$
Since momentum is given by $p = \frac{mv}{\sqrt{1-v^{2}/c^{2}}}$, and energy is given by $mc^{2}/\sqrt{1-v^{2}/c^{2}}$, it becomes clear that no matter how large of a force that we apply, the velocity of the object never goes beyond $c$.
${}^{1}$What is the fourth component, you say? Why, it turns out to be the energy! And it turns out that the way we have defined things fixes $E^{2} - p^{2}c^{2} = m^{2}c^{4}$
Zo the Relativist
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In the sentence "Since momentum ... it becomes clear that...," I don't see the logical connection. – Jul 26 '14 at 02:29
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@BenCrowell: force increases the momentum, but the asymptote in the formula for momentum allows the momentum to increase without bound, even while $v < c$ – Zo the Relativist Jul 26 '14 at 02:32
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What I'm saying is that the answer doesn't present a complete argument. You haven't connected the dots. – Jul 26 '14 at 02:34
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@BenCrowell: and with elementary questions like this, I'd rather provide hints and places to look and things to think about than I would provide a comprehensive answer. – Zo the Relativist Jul 26 '14 at 02:42
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The question is phrased in terms of dynamical concepts like force and mass, but there's a more fundamental kinematical answer that trumps these issues.
If an object is moving with speed $u$, and you then apply a boost $v$, the object's new speed is not $u+v$ but rather $(u+v)/(1+uv/c^2)$. This is always less than $c$. Therefore it's not possible to accelerate an object past $c$ by any continuous process.
Dynamics does give a nice way of showing that this is also impossible through any discontinuous process (imagine the transporters on Star Trek or something). The definition of mass in relativity is $m^2c^4=E^2-p^2c^2$. If you use the correct relativistic generalization of $E$ and $p$ here, then you find that an object with $v<c$ always has a mass that's a real number, while an object with $v>c$ has a mass that's an imaginary number. Since mass is a permanent characteristic of a particle, there is nothing we can do to, say, an electron to give it $v>c$.
Appealing to the finiteness of force is not the strongest dynamical argument against accelerating a particle past $c$. There is no principle in physics that says that forces must be finite. If an electron and a positron collide head-on, they're going to exert infinite forces on one another at the instant when they touch. We can even have singularities in the motion of a particle without a singularity in the force [Xia 1992].
Z. Xia, “The Existence of Noncollision Singularities in Newtonian Systems,” Annals Math. 135, 411-468, 1992.