# Relativity and differential equation

I have a question regarding Misner, Charles W.; Thorne, Kip S.; Wheeler, John Archibald (1973), Gravitation ISBN 978-0-7167-0344-0. It is a book about Einstein's theory of gravitation.

In page 166 of chapter 6.2 about Hyperbolic Motion, the authors present a person feeling constant acceleration $$g$$ along the direction $$x^1$$. The authors get the following equations:

$$a^0 = \frac{du^0}{d \tau} = gu^1$$ $$a^1 = \frac{du^1}{d \tau} = gu^0$$

And the system is solved by the authors to get:

$$t=g^{-1} \sinh{g\tau}$$ $$x=g^{-1} \cosh{g\tau}$$

So obviously $$u^0 = t$$ and $$u^1 = x$$, but I am not sure about the change of variable between $$t$$ and $$\tau$$ to make the appearance of the Lorentz factor.

How do the authors solve the differential equation?

• $u$ is the four velocity. $u^0 = dx^0/d\tau$, i.e. $dt/d\tau$, and $u^1 = dx^1/d\tau$ Jun 26 '19 at 16:13
• Take another derivative to get uncoupled second-order ODEs for $u^0$ and $u^1$. They look like harmonic oscillators but with opposite sign. Solve these, then integrate to get $x^0$ and $x^1$. Jun 26 '19 at 16:19
• equation 1 ${\frac {d}{d\tau}}t \left( \tau \right) =gx \left( \tau \right)$ equation 2 ${\frac {d}{d\tau}}x \left( \tau \right) -gt \left( \tau \right)$ take $\frac{d}{dt}$ of equation (1) with equation (2) you get: ${\frac {d^{2}}{d{\tau}^{2}}}t \left( \tau \right) -{g}^{2}t \left( \tau \right)$ The solution is $t(\tau)=\frac{1}{g}\,\sinh(g\,\tau)$
– Eli
Jun 26 '19 at 17:31
• @Eli No, see John Rennie’s comment. Also, your solution doesn’t satisfy your equation. Jun 26 '19 at 19:33
• @G. Smith I see: equation (1) is correct but equation (2) must be equal to zero. The differential equation is : ${\frac {d^{2}}{d{\tau}^{2}}}t \left( \tau \right) -{g}^{2}t \left( \tau \right) =0$ and the solution is what i wrote bevor? check it please.
– Eli
Jun 26 '19 at 20:20

solution:

$${\frac {d}{d\tau}}u_{{0}} \left( \tau \right) -gu_{{1}} \left( \tau \right) =0\tag 1$$

$${\frac {d}{d\tau}}u_{{1}} \left( \tau \right) -gu_{{0}} \left( \tau \right) =0\tag 2$$

and the constraint condition that

$$dsq=\left( {\frac {d}{d\tau}}u_{{0}} \left( \tau \right) \right) ^{2}- \left( {\frac {d}{d\tau}}u_{{1}} \left( \tau \right) \right) ^{2}= \epsilon\tag 3$$

where $$\epsilon=0$$ or $$1$$

with $$\frac{d}{d\tau}eq(1)$$ and equation (2) we obtain:

$${\frac {d^{2}}{d{\tau}^{2}}}u_{{0}} \left( \tau \right) -{g}^{2}u_{{0} } \left( \tau \right) =0 \tag 4$$ $$\Rightarrow$$ $$u_0(\tau)=1/2\,{\frac { \left( gA+B \right) {{\rm e}^{g\tau}}}{g}}+1/2\,{\frac { \left( -B+gA \right) {{\rm e}^{-g\tau}}}{g}}\tag 5$$ where $$A=u_0(0)$$ and $$B=D(u_0)(0)$$ are arbitrary initial conditions

with equation (3) and (2) we get for dsq

$$dsq(\tau)=\left( {\frac {d}{d\tau}}u_{{0}} \left( \tau \right) \right) ^{2}- (g\,u_0(\tau))^2= \epsilon$$

thus: for $$dsq(0)=\epsilon$$ we can obtain the initial condition $$B=B(A)$$ and get:

$$B=\sqrt {{g}^{2}{A}^{2}+\epsilon}$$

with $$A=0$$ and $$\epsilon=1$$ we get:

$$u_0(\tau)=1/2\,{\frac {{{\rm e}^{g\tau}}}{g}}-1/2\,{\frac {{{\rm e}^{-g\tau}}}{g }}=\frac{1}{g}\sinh(g\tau)$$

$$u_1(\tau)={\frac {1/2\,{{\rm e}^{g\tau}}+1/2\,{{\rm e}^{-g\tau}}}{g}}=\frac{1}{g}\cosh(g\tau)$$ and for $$A=1 \,,\epsilon=0$$ we get

$$u_0(\tau)=e^{g\,\tau}$$ $$u_1(\tau)=e^{g\,\tau}$$

• How did you get the constraint condition? And why should we pick A=0 and epsilon = 1? Jun 30 '19 at 16:04
• @PackSciences This is because the 4 velocities $u_\mu\,u^\mu=c^2=1=\epsilon$ , and for example with $A=0$ you can fulfil this requirement.
– Eli
Jul 1 '19 at 6:58
• Why is epsilon = 0 possible? Why the condition A=0 imposed by the question? What in the question tells A=0 or epsilon=1? Jul 1 '19 at 16:13
• @PackSciences for particle with mass $(\frac{ds}{d\tau})^2=c^2=\epsilon=1$ for a massless particle (light) $\epsilon=0$ so you get the light cone this mean that $u_0(\tau)=u_1(\tau)$ or if you want $t=x$.
– Eli
Jul 1 '19 at 16:30