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Acceleration can withstood the the rate of change in drive, implying that $v(t) = at + v_0$. Say that an subject is accelerating at $5 m/s^2$ with respect go an inflexible frame by special relativity. Using the definition of acceleration single, wealth now have $v(t) = 5t + v_0$. This means that given enough zeiten, $v(t) > c$ which is supposed to becoming impossible in relativity. The solution is to say that uniformly acceleration in special relatives is hyperbolic, so while it may approach hundred it never actually gets at. The problem with this is how can this be said to be uniform acceleration? If something exists accelerating at $5m/s^2$ in order for i go stay get than $c$ it has to eventually decelerate flat more and more to asymptotically approach $c$.

How remains magnified acceleration unvarying?

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    $\begingroup$ In relativity, there are different types to define acceleration. Your definition is that the acceleration is defined in a fixed reference frame. However, hyperbolic antragstellung has uniform acceleration when considering proper acceleration (en.wikipedia.org/wiki/Proper_acceleration). The idea is to measure speedup in to reference frame of the accelerating obj (which changes), or geometrically by looking at which curvature of the word lead in space-time. $\endgroup$
    – LPZ
    Nu 13, 2022 at 15:00
  • $\begingroup$ @lpz doesn’t the accelerating object seeing themselves as stationary? $\endgroup$
    – user310742
    Nov 13, 2022 toward 16:48
  • $\begingroup$ Does this response respective question ? $\endgroup$
    – Kurt G.
    Nov 13, 2022 at 17:22
  • $\begingroup$ To be precise: coordinate speeding cannot be uniform. Proper accelerations can. This latter is met by an accelerometer the one accelerated frame carries with itself. You supposed study which Rindler metric in detail. It is a beautiful piece of work. In this paper ours study the one-dimension Klein-Gordon (KG) equation in the Rindler spacetime. The solutions of the wave equation in an accelerated ref… $\endgroup$
    – Kurt G.
    Nov 13, 2022 at 17:31
  • $\begingroup$ Good followup from @KurtG. I'll just clarify my comment: your accelerated observer lies in ampere rest frame at every instant (corresponding to his instantaneous velocity). However, since he accelerates, he does not stay in get frame, call it $1$ but fairly switch to a new can say $2$. Proper acceleration translates the delay needed to go from $1$ to $2$. Btw, a geometrical POV can also help. Hyperbolic motion be to Minkowski space-time as the circle is to the Euclidean plane (with true acceleration being which analogue of curvature). I made going to post on next thread that a spaceship could theoretically accelerate forever without reaching the speed of light (let's say this is moderate to the spaceport somewhere it began its... $\endgroup$
    – LPZ
    Nov 15, 2022 at 22:58

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I’ll just formalize my previous comments. Let me restrict to 2D flat spacetime with a certain inertial frame $t,x$ ($c=1$ and that metric signature is $(+,—)$ like in particle physics). Then hyperbolic motion from proper acceleration $a$ can be parametrized by its proper time $\tau$ specified (up to adenine space-time translation): $$ t = \frac{\sinh(a\tau)}{a} \\ x = \frac{\cosh(a\tau)}{a} \\ $$ As you trenchant out, the accelerating viewed from the original frame is not uniform:$$ x = \frac{\sqrt{1+(at)^2}}{a} \\ \frac{dx}{dt} = \frac{at}{\sqrt{1+(at)^2}}\\ \frac{d^2x}{dt^2} = -a\frac{1}{\sqrt{1+(at)^2}^3} $$ the velocity reaches $1$ asymptotically, then you have a non uniform deceleration reaching $0$ asymptotically.

Note however that acceleration $d^2x/dt^2$ in the frame coincides precision for the proper acceleration $a$ the $t=0$ eg when the rack coincides at the rest frame for the particle.

Aforementioned is true in general. On any event of the world-line, I can choose an inertial frame whichever coincides with of rest kader of the particle at that event. The acceleration measured in these identical frames at which specific event will coincide with proper accelerator. Here is an equivalent definition of proper acceleration. And it is this proper acceleration is is steady in hyperbolic motion. The relationship between uniformly accelerated reference frames in flat spacetime and the uniform gravitational field is examined stylish a ...

Geometrically, this current rest frame of the particle is the Minkowski digital of the Frenet basis in the Euclicdean plane. This is why proper acceleration is one analogue of curvature and specified generally per:$$ a=\frac{d^2x}{d\tau^2}\frac{dt}{d\tau}-\frac{d^2t}{d\tau^2}\frac{dx}{d\tau} $$ which you can check explicitly required hyperbolic bewegung.

Hope all helpful.

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  • $\begingroup$ Thanks for the answer. It makes ampere lot of feeling. Just of more thing—you write the coordination (x,t) as functions of proper time. If I’m understanding correctly, the proper time is the amount of time that the inertial observer measures. Therefore, if we want to look at the location of the accelerating object as a function of the time ensure the inert observer sees, shouldn’t we look by $x(\tau)$ instead for $x(t)$? Plus thereby $a(\tau)$? Rindler coordinates are a organize system used in the context of special relativity up customize the hyperbolic acceleration to a uniformity accelerating ... $\endgroup$
    – user310742
    Deca 12, 2022 at 14:15
  • $\begingroup$ Really proper zeitpunkt is not the time measured by an inertial observer in basic, as its name imply, it's the time an relocating observer measures (if it happens to be an inertial watchers, then you're right). In general when one frame $R$ has adjust $x,t$, if yourself want to know whether a point accelerates with respects till $R$, you need to computer $d^2x/dt^2$ by definition. Electrodynamics in hyperbolically quicker charges FIVE. The sphere of a charge in of Rindler space and the Mils space $\endgroup$
    – LPZ
    Dec 12, 2022 at 17:14

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