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A professor asked me about the (c=1) equation: $$x^2 - t^2 = 1/g^2$$ which I used in a paper. Or with $c$: $$x^2 - (ct)^2 = c^4/g^2.$$

I told him that it was the exact equation of motion for a relativistic particle undergoing a constant acceleration $g$. He had some doubts about the equation.

So I went back to my office and derived the equation over again, but it was quite messy. It's such a simple equation. Basically it says that to convert particle motion from Newtonian to relativistic, you have to replace parabolas with hyperbolas. Instead of speeding up indefinitely, the particle speed approaches $\pm c$. Is there a simple or short derivation?

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Show your professor chapter 6 of Gravitation by Misner, Thorne and Wheeler. The derivation there is short, though only because they skip all the messy algebra! –  John Rennie May 17 '12 at 7:46
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4 Answers 4

up vote 12 down vote accepted

Yes--- this is directly analogous to the following statement from geometry: The plane curve with everywhere constant curvature is the circle.

You could prove this by integrating the condition of constant curvature (which is just as messy as in relativity), or by doing an arclength parametrization (which is clean, see below). But easiest of all is to note that rotations around the center take the circle to itself, and every point on the circle transitively to every other point. Since curvature is rotationally invariant, the curvature at all points is the same.

Similarly in relativity, the hyperbola

$$ x^2-t^2 = R^2 $$

is invariant under boosts, and the boosts act transitively. So you conclude that the curvature is constant. The curvature is the rest-frame acceleration, and the result is that the hyperbola has a constant acceleration.

The magnitude of the acceleration can be seen by looking at small t's (since at t=0, the velocity is 0, so relativity reduces to Newton/Galileo)

$$ x = \sqrt{ R^2 +t^2} = R + {t^2\over 2R} + ... $$

So that the acceleration a is given by

$$ a={1\over R}$$

It's the inverse radius of the Hyperbola, just like the curvature is the inverse radius of a circle in geometry.

Arclength parametrization

There are many problems in geometry/relativity that are simplified by arclength parametrization. In this parametrization, x(s) and y(s) obey

$$\dot{x}^2 + \dot{y}^2 = 1$$

so that, differentiating

$$\dot{x}\ddot{x} = - \dot{y}\ddot{y}$$

The curvature is the magnitude of the second derivative

$$ \ddot{x}^2 + \ddot{y}^2 = {1\over R^2}$$

and substituting for $\ddot{y}$ using the relation gives

$$ (1+ {\dot{x}^2\over\dot{y}^2}) \ddot{x}^2 = {1\over R^2}$$

and simplifying

$$ {\ddot{x} \over \sqrt{1-\dot{x}^2}} = {1\over R }$$

This is an ordinary differential equation in $\dot{x}$ which integrates simply to give that x(s) is a sinusoid, and then y(s) is also a sinusoid.

$$ {d\over ds}(\sin^{-1}(\dot{x})) = {1\over R} $$

$$ \dot{x}= \sin ({s-s_0\over R}) $$

$$ x - x_0 = R \cos({(s-s_0)\over R}) $$

In relativity, the same manipulations give a differntial equation for x with flipped sign under the square-root, and give the hyperbolic trigonometric functions.

Here is an answer where I make use of arclength parametrization for a nontrivial problem, which can't be solved by symmetry: Is there an intuitive reason the brachistochrone and the tautochrone are the same curve?

There are probably many fascinating curve-geometry problems in relativity which are analogs of the Brachistochrone and Isochrone, but nobody has ever formulated these. It's not hard to do, but they might not have an immediate physical interpretation.

Two definitions of constant acceleration

There are two different definitions possible for "constant acceleration". One is that you have a constant acceleration in your rest frame, and this obviously gives the invariant circle-like curve, the hyperbola. The other definition is a particle's motion in a constant E field.

$$ {d\over dt} {dx\over d\tau} = E $$

The two definitions look different superficially, but they are the same. This is shown here: Knowing the mass and force acting on a particle, how do we derive the relativistic function for velocity with respect to time?

The upshot of the linked argument is that while the $d/dt$ is not the same as $d/d\tau$, there is a geometric projection, the projection is the same hyperbolic trigonometric factor as the one involved in taking the x-component, rather than the component perpendicular to the motion, to define acceleration, so the end result is the same.

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My goal was to write down an answer that would be much shorter than yours ;-) but the content would be so similar that I decided to plus-one yours instead... –  Luboš Motl May 17 '12 at 10:56
    
@LubošMotl: Thanks Lubos, I added a little to make it more worth your upvote. –  Ron Maimon May 17 '12 at 17:28
    
You want $x=\sqrt{R^2+t^2}$ not $x^2$. Also +1. –  kηives May 20 '12 at 21:16
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Your expression is a solution to the integration of the transformation, $\vec g = \gamma^3\vec a$, for acceleration along $\vec V$ between the lab and proper frames with initial conditions $\vec u = 0$, $\vec x = c^2/g$ at $t = 0$. I'll give a derivation for this standard transformation.

Let frames $S'$ and $S$ be in standard configuration with $S'$ moving with velocity $\vec V$ along $\vec x$. Differentiating the velocity transformation for $\vec u$ wrt t gives the transformation of acceleration along $\vec V$ $$\begin{align*}\vec a_\parallel &= \frac d {dt}\left(\frac{\vec u'_\parallel + \vec V}{1 + \frac{\vec u'\cdot\vec V}{c^2}}\right)\\ &= \frac {d\vec u'_\parallel}{d\tau} \frac{d\tau}{dt}\frac {d}{d\vec u_\parallel} \left(\frac{\vec u_\parallel + \vec V}{1 + \frac{\vec u'\cdot\vec V}{c^2}}\right)\\ &= \vec a'_\parallel\frac{1}{\gamma(1 + \frac{\vec u'\cdot\vec V}{c^2})}\frac{1}{\gamma^2(1 + \frac{\vec u'\cdot\vec V}{c^2})^2}\\ &= \vec a'_\parallel\frac{1}{\gamma^3(1 + \frac{\vec u'\cdot\vec V}{c^2})^3}\end{align*}$$ For completion, the same can be done for the acceleration perpendicular to $\vec V$, $\vec a_\perp$ which Jackson asks to prove in problem 11.5 in Classical electrodynamics, 3rd edition

$$ {{\vec a}_\parallel } = \frac{1}{{{\gamma ^3}{{\left( {1 + \frac{{\vec v\cdot\vec u'}}{{{c^2}}}} \right)}^3}}}{{\vec a'}_\parallel }{{\vec a}_ \bot } = \frac{1}{{{\gamma ^2}{{\left( {1 + \frac{{\vec v\cdot\vec u'}}{{{c^2}}}} \right)}^3}}}\left( {{{\vec a'}_ \bot } + \frac{{\vec v}}{{{c^2}}} \times (\vec a' \times \vec u')} \right)$$ which reduce to the important standard expressions for a proper frame $$\gamma^3\vec a_\parallel = \vec a'_\parallel\qquad\gamma^2\vec a_\perp = \vec a'_\perp$$

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@RonMaimon we could sit here all day with you criticizing its lack of geometry, me defending its precise simplicity in answering the op's question. Personally, I think it's good that we have three answers that range from my simple approach based upon the use of the Lorentz transformation of 3-vectors and time as introduced by Einstein, to your more powerful geometrical approach introduced by Minkowski two years later. Without destroying its current simplicity and audience, how do you think my answer could be improved? –  John McVirgo May 21 '12 at 15:03
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You are right, it is good to have this answer here, I deleted my comment, and will make no more. It is useful for someone who has not learned to think covariantly yet. –  Ron Maimon May 21 '12 at 19:24
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I think that it is enough to use Dimensional analysis:

$$x^2 - (ct)^2 = c^4/g^2.$$
LHS = $[L]^2$ ( the subtraction of areas is an area )
RHS = $[L]^{4}[T]^{-4}\cdot [T]^{4}[L]^{-2}=[L]^{2}$ you got an area.

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all this does is demonstrate that the dimensions are consistent. –  tmac May 18 '12 at 6:00
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Here we elaborate on @tmacs comment. Define two dimensionful constants $t_0:=c/g$ and $x_0:=c^2/g$. Then $t/t_0$ and $x/x_0$ are dimensionless. Moreover, any relation of the form $f(x/x_0,t/t_0)=0$, where $f$ is an arbitrary function of two dimensionless variables, would be consistent with dimensional analysis. –  Qmechanic May 18 '12 at 19:07
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Consider the particle's view from its own (accelerating) reference frame. Since $g$ is a proper acceleration, the particle always measures a constant acceleration for itself. Combined with the fact that by definition, its velocity is zero in that reference frame, you know that the particle's trajectory has to be invariant under a combined time translation and Lorentz boost. (In other words, the trajectory has to locally look the same after some time has elapsed and the particle's coordinate velocity has changed accordingly.) The only trajectory that satisfies that condition is a hyperbola.

Mathematically, you can come up with this by repeatedly applying a Lorentz transformation. The transformation can be represented as

$$\begin{pmatrix}ct \\ x\end{pmatrix} \to \begin{pmatrix}\gamma & \beta\gamma \\ \beta\gamma & \gamma\end{pmatrix}\begin{pmatrix}ct \\ x\end{pmatrix}$$

(if you have any questions about negative signs, ask me when I've gotten some sleep :-P) Suppose the particle starts with $x'(0) = 0$, which corresponds to setting up a coordinate system in which it starts out at rest. Using regular old kinematics, you can figure out that in the vicinity of $t = 0$, its path should follow $x(t) = \frac{1}{2}gt^2 + \mathcal{O}(t^3)$ (after all, this should match the nonrelativistic result over short time scales). So an infinitesimal time $\delta t$ after $t = 0$, its speed is $\beta c = g\delta t$, and the corresponding Lorentz transformation is

$$\begin{pmatrix}1 & g\delta t/c \\ g\delta t/c & 1\end{pmatrix}$$

Now, if the particle's path is invariant under the time translation with corresponding boost, then if you want to advance $\delta t$ further forward in time, you just do the same thing again, and again, and again. The only catch is that this increment $\delta t$ is proper time, because you're always measuring it in the particle's own rest frame.

In general, to figure out the particle's path after a finite proper time $\tau$, you just apply the transformation $N$ times, using the limit $N\to\infty$ with $N\delta t = \tau$:

$$\begin{align}\begin{pmatrix}1 & g\delta t/c \\ g\delta t/c & 1\end{pmatrix}^N &= \left[\mathbb{1}_2 + \begin{pmatrix}0 & g/c \\ g/c & 0\end{pmatrix}\delta t\right]^{(\tau/\delta t)} \\ &= \exp\left[\begin{pmatrix}0 & g/c \\ g/c & 0\end{pmatrix}\tau\right] \\ &= \begin{pmatrix}\cosh\frac{g\tau}{c} & \sinh\frac{g\tau}{c} \\ \sinh\frac{g\tau}{c} & \cosh\frac{g\tau}{c}\end{pmatrix}\end{align}$$

Apply this to the starting position four-vector $\begin{pmatrix}0 \\ x(0)\end{pmatrix}$ and you have your coordinate time and coordinate position,

$$\begin{pmatrix}ct(\tau) \\ x(\tau)\end{pmatrix} = \begin{pmatrix}x(0)\sinh\frac{g\tau}{c} \\ x(0)\cosh\frac{g\tau}{c}\end{pmatrix}$$

and if you solve this to actually put $x$ in terms of $t$, you get

$$\begin{align}x(0)^2\cosh^2\frac{g\tau}{c} - x(0)^2\sinh^2\frac{g\tau}{c} &= x(0)^2 \\ x(t)^2 - t^2 &= x(0)^2\end{align}$$

To figure out what $x(0)$ is, you can take the second derivative with respect to proper time,

$$\frac{\mathrm{d}^2}{\mathrm{d}\tau^2}x(\tau) = x(0)\frac{g^2}{c^2}\cosh\frac{g\tau}{c} = g \quad\to\quad x(0) = \frac{c^2}{g}$$

I don't know if that's the simple proof you're looking for. It's physically intuitive, but as is often the case, when you jump into the math, it's easy to get bogged down in details.

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