# Homework Help: Geodesics in a Given, Arbitrary Metric, dt Coefficient Only

1. Mar 15, 2016

### MattRob

Not a formal course - just a question I decided to try to tackle with what I've gleaned from Stanford's lectures on Youtube, but still putting this here on account of this.

So, I've been watching the Stanford GR series, and I have two motivations for messing around with this type of metric;
1. Given gravitational time dilation, I've been curious for quite awhile if a region with nothing but time flowing at a different rate as a function of position would have any sort of gravitational effect.
2. A bit more whimsically, in fantasy settings there's sometimes some sort of magical gateway in-between worlds with differing flow rates of time (for example, Narnia). I couldn't help but wonder if this, similar to 1, wouldn't result in a cataclysmically powerful gravitational effect.

1. The problem statement, all variables and given/known data

We somehow have a region of spacetime where time is flowing differently in one part of the universe than another. A gradient of the flow of time, if you will.

Here's our metric tensor for a simple linear function for this, arbitrarily chosen to make the $dt^2$ coefficient a function of position:

$g = \begin{pmatrix} -kx^1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}$

Where $x^0 = t, x^1 = x, x^2 = y, x^3 = z$

2. Relevant equations

Giving the metric

$dτ^2 = kxdt^2 - dx^2 - dy^2 - dz^2$

The Geodesic equation,

$\frac{d^2x^μ}{ds^2} = - {\Gamma^{μ}_{αβ}} \frac{dx^α}{ds} \frac{dx^β}{ds}$

And the Christoffel Symbol:

${\Gamma^μ_{αβ}} = \frac{1}{2} g^{μm} (\frac{∂g_{mα}}{∂x^β} + \frac{∂g_{mβ}}{∂x^α} - \frac{∂g_{αβ}}{∂x^m})$

3. The attempt at a solution

Fortunately, though there are 40 Christoffels, we can rule out most of these...

Every partial derivative that is not the partial derivative of $g^{00}$ with respect to $x^1$ is going to equal zero.

Only the diagonal terms of $g^{μm}$ where $μ = m$ will be non-zero.

$α = 0$, $m = 0$ OR $β = 0$ must be true for that Christoffel symbol to not be zero. Since $m = μ$ must also be true or the $g^{μm}$ will be zero, this implies that $α = 0$, $μ = 0$ OR $β = 0$ must be true.

However, one of those three, $α, β, μ$, must also equal 1, or else we will not have any partial derivatives with respect to $x^1$, thus they will all be zero.

ALSO, if $μ ≠ 0$, then $α = β = 0$ must be true, since neither $g_{mα}$ nor $g_{mβ}$ will be $g_{00}$, since we require that $μ = m$. Thus the only possible $g_{μν}$ term that could be $g_{00}$ is $g_{αβ}$ if $μ ≠ 0$, and that will only be $g_{00}$ if $α = β = 0$.

So for our Christoffel Symbol ${\Gamma^{μ}_{αβ}} ≠ 0$ , for $μ = 0$, $α = 0$ or $β = 0$, and since ${\Gamma^{μ}_{βα}} = {\Gamma^{μ}_{αβ}}$, this gives us four Christoffels:

${\Gamma^{0}_{00}}$
${\Gamma^{0}_{01}}$
${\Gamma^{0}_{02}}$
${\Gamma^{0}_{03}}$

And for $μ ≠ 0$, $α = β = 0$ must be true, as stated above, giving us three more:

${\Gamma^{1}_{00}}$
${\Gamma^{2}_{00}}$
${\Gamma^{3}_{00}}$

Finally, if none of the terms $μ$, $α$, or $β$ are 1, then there is no partial derivative with respect to $x^1$, so it will be zero. This only leaves us with:

${\Gamma^{0}_{01}}$
${\Gamma^{1}_{00}}$

These equations are nice and short since only diagonal terms of $g$ are non-zero. It also pays to keep in mind that only $\frac{∂g_{00}}{∂x^1}$ terms will be non-zero;
${\Gamma^μ_{αβ}} = \frac{1}{2} g^{μm} (\frac{∂g_{mα}}{∂x^β} + \frac{∂g_{mβ}}{∂x^α} - \frac{∂g_{αβ}}{∂x^m})$

${\Gamma^0_{01}} = \frac{1}{2} g^{00} (\frac{∂g_{00}}{∂x^1} + \frac{∂g_{01}}{∂x^0} - \frac{∂g_{01}}{∂x^0}) = \frac{1}{2} (-kx^1) (\frac{∂}{∂x^1}(-kx^1)) = \frac{1}{2} (-kx^1) (-k) = \frac{1}{2} (k^2x^1)$
${\Gamma^1_{00}} = \frac{1}{2} g^{11} (\frac{∂g_{10}}{∂x^0} + \frac{∂g_{10}}{∂x^0} - \frac{∂g_{00}}{∂x^1}) = \frac{1}{2} (-(\frac{∂}{∂x^1}(-kx^1))) = \frac{k}{2}$

Plugging these Christoffel Symbols into their respective Geodesic equations, and switching back to more familiar notation after doing so:

Form:

$\frac{d^2x^μ}{ds^2} = - {\Gamma^{μ}_{αβ}} \frac{dx^α}{ds} \frac{dx^β}{ds}$

One:

$\frac{d^2x^0}{ds^2} = - {\Gamma^{0}_{01}} \frac{dx^0}{ds} \frac{dx^1}{ds} = - \frac{1}{2} (k^2x^1) \frac{dx^0}{ds} \frac{dx^1}{ds}$

$\frac{d^2t}{ds^2} = - \frac{1}{2} (k^2x) \frac{dt}{ds} \frac{dx}{ds}$

And the other:

$\frac{d^2x^1}{ds^2} = - {\Gamma^{1}_{00}} \frac{dx^0}{ds} \frac{dx^0}{ds} = - \frac{k}{2} \frac{dx^0}{ds} \frac{dx^0}{ds}$
$\frac{d^2x}{ds^2} = - \frac{k}{2} \frac{dt}{ds} \frac{dt}{ds}$

Now here's the first issue I was really worried about; switching from one notation to the other. I'm somewhat familiar with coordinate time dt and proper time dτ, but does ds in the geodesic equation correspond to dτ?

From what I saw I think so, which would mean the two geodesic equations are:

$\frac{d^2t}{dτ^2} = - \frac{1}{2} (k^2x) \frac{dt}{dτ} \frac{dx}{dτ}$

\begin{align} \frac{d^2x}{dτ^2} &= - \frac{k}{2} \frac{dt}{dτ} \frac{dt}{dτ} \\ \frac{d^2x}{dt^2}&= - \frac{k}{2} \end{align}

And, going back to SI units, in order to make the units balance, I think this is the correct way to convert them?

\begin{align} \frac{d^2x}{dt^2}&= - \frac{kc^2}{2} \end{align}

Since we have a squared time term in the denominator on the left side, to go back from geometric units, we put a $c^2$ next to the $dt^2$ to set the time units back to seconds, and then multiply both sides by that $c^2$.

Now, here's something I find odd:

The time dilation is a simple, linear function of position, for dx = dy = dz = 0, it's simply $dτ^2 = kxdt^2$.

It's immediately apparent that dt goes to zero at x = 0, so there must be a horizon there. I'm not sure exactly how to interpret the dτ and dt, but I recall from the Schwarchild solution that the smaller the dt coefficient, then the "slower" that object's clocks will appear to tick - that is to say, closer to the horizon, things will appear to experience time more slowly to a distant observer.

Applying that same logic, it would seem that a space-coordinate stationary observer at x = 1 would experience only 1 second for every two seconds a space-coordinate stationary observer at x = 2 experiences. That is, an observer at x = 2 would see a clock at x = 1 tick half as fast, and an observer at x = 1 would see a clock at x = 2 tick twice as fast.

Okay, well, just as I'm typing this it starts to make sense - my original objection/confusion was that these geodesics have the same coordinate acceleration at x << 1, where 1 unit of x corresponds to an arbitrarily large fractional change in dt, and at x >> 1, where 1 unit of x corresponds to an arbitrarily small change in dt.

That seems counter-intuitive at first, since if the time dilation gradient is the driving phenomenon, shouldn't the geodesic coordinate acceleration be greater where the gradient is "steeper"? But, after thinking about it, I realize that this is analogous to a uniform electric field in-between two infinitely large plates: because it's symmetric, there's no space for the "flux" to spread out in, thus it doesn't dissipate by any power law, so it's equally strong everywhere.

So, going back to the very top of this post, I'm somewhat thrilled to see an answer to #1, though it's completely failed to serve as any kind of approximation for #2, haha. At the very least, I think it does tell me that if I'm trying to solve for #2, I could probably treat the "portal" (fully recognizing this is not a possible spacetime construct, but a fun "what if" scenario, if not merely to probe the theory's workings) as something like a Schwarzchild solution, but introduce a boundary condition where dt/dτ is at a particular value, then see what the gravitational field down to that point would be like.

In that case, it does somewhat answer the question in #2 that it would, in fact, be a rather cataclysmic event. I don't imagine creating a wardrobe-sized surface with a gravitational effect a billion billion times stronger than Earth's surface gravity would do much good for the surrounding atmosphere or countryside.

More than the result, though, I'm just thrilled to answer anything with the actual equations of GR. After some years of impatiently devouring whatever I could understand about the field, it's rather thrilling to finally, actually mess around with some of the harder maths of it, even if it is a good deal beyond my formal education...

So, did I do it right?

2. Mar 20, 2016