Question

Although it deals with inertial reference frames, the special theory of relativity describes accelerating objects without difficulty. Of course, uniform acceleration no longer means $dv/dt=g,$ where $g$ is a constant, since that would have $v$ exceeding $c$ in a finite time. Rather, it means that the acceleration experienced by the moving body is constant: In each increment of the body's own proper time $d\tau ,$ the body acquires velocity increment $dv=gd\tau$ as measured in the inertial frame in which the body is momentarily at rest. (As it accelerates, the body encounters a sequence of such frames, each moving with respect to the others.) Given this interpretation: a) Write a differential equation for the velocity $v$ of the body, moving in one spatial dimension, as measured in the inertial frame in which the body was initially at rest (the "ground frame"). You can simplify your equation, remembering that squares and higher powers of differentials can be neglected. b) Solve this equation for $v(t),$ where both $v$ and $t$ are measured in the ground frame. c) Verify that your solution behaves appropriately for small and large values of $t$. d) Calculate the position of the body $x(t),$ as measured in the ground frame. For convenience, assume that the body is at rest at ground-frame time $t=0,$ at ground-frame position $x=c^2/g$ e) Identify the trajectory of the body on a space-time diagram (Minkowski diagram, for Hermann Minkowski) with coordinates $x$ and $ct,$ as measured in the ground frame. f) For $g=9.81\mathrm{m}/{\mathrm{s}}^2,$ calculate how much time it takes the body to accelerate from rest to $70.7\mathrm{\%}$ of $c,$ measured in the ground frame, and how much ground-frame distance the body covers in this time.

Short answer

Question: A particle undergoes uniform proper acceleration g. The equation for its velocity is derived as v(t)=c*sinh(gt). Calculate the time it takes for the particle to reach 70.7% of the speed of light and the distance traveled up to that point. (Hint: Use g=9.81 m/s²) Answer: To calculate the time it takes for the particle to reach 70.7% of the speed of light, set v(t)=0.707c in the velocity equation, v(t)=c*sinh(gt). Solving for t, you get t=arcsinh(0.707)/g. Substitute g=9.81 m/s² and solve for the time in seconds. To find the distance traveled up to that point, substitute this time into the position equation x(t)= (c²/g) * (cosh(gt)−1) and calculate the distance.

Step-by-step solution

Integral of dv/dt=g over dt

The problem statement gives us that dv=g*dτ. We thus need to multiply both sides by dt/dτ. Use the well-known result from special relativity that dt/dτ=γ, where γ=1/√(1−v²/c²) is the Lorentz factor. We thus obtain dv= γgdt which can be written as dv/dt= γg.

Solve differential equation dv/dt= γg

To solve this differential equation, separate the variables: dv/γ= g dt. Substitute γ=1/√(1−v²/c²): √(1−v²/c²) dv= g dt. Integrate both sides with boundaries on the left from 0 to v and on the right from 0 to t The integral will become: arcsinh(v/c)=gt

Solve for v(t)

While arcsinh(v/c)=gt; we can simply bring down gt to the other side by taking sinh on both sides: sinh(gt)= v/c. So the solution to the differential equation is v(t)=c*sinh(gt)

Small and Large t Behavior of v(t)

For small values of t, sinh(gt)≈gt. So, the velocity v(t) behaves as v(t)≈c*gt which is the non-relativistic result when velocities are much less than the speed of light. For large values of t, sinh(x) approaches e^x/2 for large x, so the speed v(t) approaches c as t→∞ which is consistent with the fact that no object can exceed the speed of light.

Solving for x(t)

To calculate x(t), we need to integrate v(t) over time: x(t)=∫_0^tv(t')dt'. Substituting v(t') from step 3 and integrating we obtain x(t)= (c²/g) * (cosh(gt)−1)

Identifying the Trajectory on a Minkowski Diagram

On a Minkowski diagram the path followed by a particle is given by it's "world line". In this case, the world line of the accelerating particle is given by x(t) and ct. In terms of ct this is ct(t)= (c²/g) * sinh(gt). The world line is a hyperbola, characteristic of uniformly accelerating particles in special relativity.

Calculation of Time to reach 70.7% of the speed of light

Set v(t)=0.707c in the velocity equation derived in step 3 and solve for t. You will find that t=arcsinh(0.707)/g. Substitute the value of g=9.81 m/s² to get the time (in seconds) needed to reach this speed. The distance travelled is then found by substituting this time into the position equation derived in step 3.

Key concepts

Inertial Reference Frames

Imagine you're sitting in a smoothly gliding train, not feeling any jerks or bumps. Everything inside the train appears to be at rest relative to you. This is essentially what physicists refer to as an inertial reference frame—a viewpoint or a framework where an observer is not experiencing any acceleration. According to the special theory of relativity, the laws of physics are the same in all inertial reference frames, whether in motion or at rest, as long as they move with a constant velocity relative to each other.

In the context of the given exercise, the 'ground frame'—the inertial frame in which the body was initially at rest—is the reference point from which we measure the velocity and acceleration of the moving body. But as the body accelerates, it passes through a sequence of momentarily inertial frames, because from the perspective of each of those frames, the body seems not to be accelerating at that instant in time.

This concept is pivotal in understanding how special relativity treats acceleration. Even though acceleration involves changing velocity, it can still be analyzed by 'jumping' between infinitely many infinitesimally different inertial frames, each applicable for just an instant. This approach enables the use of special relativity's equations for scenarios that, at first glance, may appear to involve non-inertial (accelerating) frames.

Lorentz Factor

At the heart of many special relativity equations lies the Lorentz factor, symbolized by $\gamma$. The Lorentz factor comes into play when velocities approach the speed of light, denoted as $c$. It's a measure of time dilation, length contraction, and the relativity of simultaneity — all hallmark effects in the special theory of relativity. Its formula is $\gamma =\frac{1}{\sqrt{1-(v^2/c^2)}}$, where $v$ is the relative velocity.

In the exercise, the Lorentz factor appears when we convert the proper time increment $d\tau$ to the ground frame time increment $dt$, effectively scaling the acceleration felt in the body's momentarily inertial frame to the ground frame. As the body’s velocity $v$ increases, the Lorentz factor $\gamma$ also increases, approaching infinity as $v$ approaches the speed of light, $c$. This behavior has profound implications: it ensures that, according to the equations of special relativity, no object can exceed the speed of light, and it provides the mathematical framework for correctly predicting how velocities add in relativistic contexts.

By incorporating the Lorentz factor into the differential equation for velocity as in Step 1 of the solution, we see that the body's acceleration in the ground frame diminishes as its velocity nears $c$—an intuitive result once you're familiar with the warping effects on space and time predicted by relativity.

Minkowski Diagram

To visualize events in spacetime as posited by the special theory of relativity, physicists employ a Minkowski diagram. Named after Hermann Minkowski, who extended Einstein's work on special relativity, this two-dimensional diagram simply graphs space and time coordinates. On such a diagram, time typically runs vertically, and space horizontally. Lines of simultaneity—where all events along the line happen at the same time—are parallel to the space axis, and lines of equivalent location—events happening at one place—are parallel to the time axis.

In our exercise, a Minkowski diagram helps illustrate the world line of a body under uniform acceleration. The trajectory, rather than being a straight line as it would in non-relativistic physics, is a hyperbolic curve. This curve reflects the relativistic effect of time dilation as the body accelerates—time stretches out, making it impossible for the object to reach the speed of light within a finite amount of ground frame time.

Using such a diagram not only helps in visualizing and understanding the special theory of relativity but also in solving problems intertwined with relativistic speeds. The trajectory on the Minkowski diagram provides a graphical solution to the exercise, tracing out the path the body follows through spacetime and enforcing the concept that, no matter how much the object accelerates, it will never cross the light-speed barrier.