If you are a physicist or at least know a bit about the theory of relativity, your answer to this question might be a confident "no". Sure, if nothing can exceed the speed of light, one would need an infinite amount of time to travel an infinite distance. But let's put the relativistic laws of physics aside and imagine a pure Newtonian universe. In such a universe, an object can travel arbitrarily fast. But would it be possible for someone to travel "to infinity" in a finite amount of time? The answer is yes.

Let's prove this with an example. Imagine an object at $x = 0$ which is at rest at $t = 0$. Between $t = 0$ and some $t = T \gt 0$, a force $F(t)$ pointing along the $x$ axis is applied to the object. This force is given by: $$ F(t) = \displaystyle\frac{\alpha}{(T - t)^3} $$ for some constant $\alpha \gt 0$. For $0 \leq t \lt T$ the value of $F(t)$ is positive and finite, so it points along the positive $x$ axis. Its value increases as $t \rightarrow T$ (here and in what follows, $t \rightarrow T$ means $t$ approaches $T$ from the left side, so $t \lt T$) and diverges at $t = T$.

If the mass of the object is $m$, Newton's second law of motion states that (below the notation $\dot{q}$ is used to represent $dq/dt$ and $\ddot{q}$ to represent $d^2q/dt^2$): $$ F(t) = ma(t) = m\ddot{x}(t) = \displaystyle\frac{\alpha}{(T - t)^3} $$ where $a(t)$ is the acceleration experienced by the object at time $t$. This implies that: $$ \ddot{x}(t) = \displaystyle\displaystyle\frac{\alpha}{m}\frac{1}{(T - t)^3} \Longrightarrow \int_{t' = 0}^{t' = t} \ddot{x}(t')dt' = \int_{t' = 0}^{t' = t} \frac{\alpha}{m}\displaystyle\frac{1}{(T - t')^3}dt' $$ The integral of the term on the right-hand side can be computed by letting $w := T - t'$, $dw = -dt'$: $$ \dot{x}(t) - \dot{x}(0) = \frac{\alpha}{m}\int_{w = T}^{w = T - t} \displaystyle\frac{(-1)}{w^3}dw = \frac{\alpha}{m}\displaystyle\left(\frac{1}{2w^2}\right) \bigg|_{w = T}^{w = T - t} $$ Since the object is initially at rest ($\dot{x}(0) = 0$), we get: $$ \boxed{ \dot{x}(t) = \displaystyle\frac{\alpha}{2m}\left[ \displaystyle\frac{1}{(T - t)^2} - \frac{1}{T^2} \right] } \label{%INDEX_vel} $$ This expression gives us the velocity $v(t) = \dot{x}(t)$ of the object at time $t$ (for $0 \leq t \lt T$). For all $0 \lt t \lt T$ we have $\dot{x}(t) \gt 0$ since in this interval $(T - t) \lt T$. The velocity diverges at $t = T$.

Integrating equation \eqref{%INDEX_vel} with respect to time yields: $$ \int_{t' = 0}^{t' = t}\dot{x}(t')dt' = \frac{\alpha}{2m}\left[ \int_{t' = 0}^{t' = t}\displaystyle\frac{1}{(T - t')^2}dt' -\int_{t' = 0}^{t' = t}\frac{1}{T^2}dt' \right] $$ Computing the second integral on the right-hand side is a trivial task since the integrand $1/T^2$ is a constant. To integrate the first term on the right-hand side, let $w := T - t'$, $dw = -dt'$. We then obtain: $$ \begin{eqnarray} x(t) - x(0) &=& \frac{\alpha}{2m}\left[ \int_{w = T}^{w = T - t}\displaystyle\frac{(-1)}{w^2}dw - \displaystyle\frac{t}{T^2} \right] \nonumber\\[5pt] &=& \frac{\alpha}{2m}\left[ \displaystyle\left(\frac{1}{w}\right)\bigg|_{w = T}^{w = T - t} - \displaystyle\frac{t}{T^2}\right] \nonumber\\[5pt] &=& \displaystyle\frac{\alpha}{2m}\left[ \displaystyle\frac{1}{T - t} - \displaystyle\frac{1}{T} - \displaystyle\frac{t}{T^2}\right] \nonumber\\[5pt] &=& \displaystyle\frac{\alpha}{2m}\left[ \displaystyle\frac{1}{T - t} - \displaystyle\frac{(T + t)}{T^2} \right] \label{post_1bb7826792e3d5f8d4af531e1fca6075_pos_meq} \end{eqnarray} $$ The object is initially at $x = 0$, so $x(0) = 0$. A bit more algebraic work on the right-hand side of equation \eqref{post_1bb7826792e3d5f8d4af531e1fca6075_pos_meq} yields: $$ \boxed{ x(t) = \displaystyle\frac{\alpha}{2mT^2} \displaystyle\frac{t^2}{(T - t)} } $$ Since $\dot{x}(t) \gt 0$ for $0 \lt t \lt T$, $x(t)$ increases monotonically during this time period. As $t \rightarrow T$, $x(t)$ diverges. In other words: $$ x(t) \rightarrow \infty \quad \textrm{when} \quad t \rightarrow T $$ so the object reaches "the infinity" in a finite amount of time (namely, in time $T$).

Although theoretically possible, one would have to come up with a way to produce such a force on an object. But that task, dear reader, I will leave to you! ;-)

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