# GR equations of motion

1. Feb 5, 2008

### Mentz114

These are the EOMs for a 4D space-time that has a plane singularity at x=0. They are simpler than most, but as I've confessed before, I need help with even the simplest DE's. This is not a home-work question.

$$x\frac{d^2\tau}{dt^2} - \frac{d\tau}{dt}\frac{dx}{dt} = 0$$

$$x^3\frac{d^2x}{dt^2} + \frac{m^2}{2}(\frac{d\tau}{dt})^2 - \frac{x^2}{2}(\frac{dx}{dt})^2 = 0$$

Any solutions gratefully received and will be acknowledged.

M

2. Feb 6, 2008

### Rainbow Child

If $x(t)=c_1,\,\tau(t)=c_2$ then the equations are satisfied.
If $x(t)\neq 0,\,\tau(t) \neq c_2$ then from the 1st equation we have

$$\frac{\tau''}{\tau'}=\frac{x'}{x}\Rightarrow x=C\,\tau'$$

Plugging this result to the 2nd equation we have

$$\tau'\,(2\,C^4\,\tau'\,\tau'''-C^4\,(\tau'')^2+m^2)=0\Rightarrow \tau'\,\tau'''-\frac{1}{2}\,(\tau'')^2+\frac{m^2}{2\,C^4}=0 \quad \text{since} \quad \tau'\neq 0$$

Now set $\tau'=\frac{m}{C^2}\,y$, thus

$$y\,y''-\frac{1}{2}\,y'^2+\frac{1}{2}=0 \quad (\ast)$$

With the substitution
$$y(t) \rightarrow \theta,\,t\rightarrow u(\theta),\,y'(t) \rightarrow \frac{1}{u'(\theta)},\, y''(t) \rightarrow -\frac{u''(\theta)}{u'(\theta)^3}$$
makes the last equation

$$\frac{u''(\theta)}{u'(\theta)\,(u'(\theta)^2-1)}=\frac{1}{2\,\theta}$$

which is a 1st order separable ODE.
Thus from the last one we can find $u(\theta)$ and from $(\ast), y(t)$ and finally $\tau(t)$

Last edited: Feb 7, 2008
3. Feb 6, 2008

### Mentz114

Thanks again, RainbowChild. Unfortunately what you've left looks just as difficult as the original to me and I wouldn't know where to begin to solve it.

I'll go and look up 'separable'.

[later]
So, I need to write the last equation the form

$$\frac{dy}{dx} = h(x)g(y)$$

but I don't understand your last equation, are the primes still differentiation wrt t ?

the nearest I can get is this

$$\frac{dU}{dt} = \frac{1}{\theta}U(U^2-1)$$

I'm sorry, I'm at sea here.

Last edited: Feb 6, 2008
4. Feb 7, 2008

### Rainbow Child

Ok, let me throw you a life jacket!

For

$$\frac{u''(\theta)}{u'(\theta)\,(u'(\theta)^2-1)}=\frac{1}{2\,\theta}$$

let $u'(\theta)=z(\theta)$ thus

$$\frac{z'(\theta)}{z(\theta)\,(z(\theta)^2-1)}=\frac{1}{2\,\theta}\Rightarrow \int \frac{d\,z}{z\,(z^2-1)}=\int \frac{d\,\theta}{2\,\theta} \Rightarrow \ln\frac{z^2-1}{z^2}=\ln\theta+C_1 \Rightarrow$$

$$z(\theta)=\pm \frac{1}{\sqrt{1-C_1\,\theta}}\Rightarrow u'(\theta)=\pm \frac{1}{\sqrt{1-C_1\,\theta}}\Rightarrow u(\theta)=\pm \frac{2}{C_1}\,\sqrt{1-C_1\,\theta}-C_2$$

which yields to

$$t=\pm \frac{2}{C_1}\,\sqrt{1-C_1\,y(t)}-C_2\Rightarrow y(t)=-\frac{1}{4\,C_1}\,\left((t+C_2)^2\,C_1^2-4\right)$$

Now use $\tau'=\frac{m}{C^2}\,y$, in order to calculate $\tau(t)$

I hope that clears up things!

5. Feb 7, 2008

### Mentz114

That helps a lot, thanks. My head is above water.

M