# Ds notation for the metric

1. Aug 11, 2008

### DrGreg

When giving an explicit formula for a metric in terms of a coordinate chart, e.g. the Schwarzschild metric, it is conventional to use the notation $ds^2 = g_{\alpha\beta} dx^\alpha dx^\beta$ (with the right hand side explicitly written out).

This notation has an unfortunate side-effect. If you take it literally, then s, evaluated along a spacetime curve, can be either real or imaginary, depending on whether the curve is spacelike or timelike and what your metric signature is. This could be potentially confusing to students of the subject.

Are you aware of any alternative notation that authors have used to get around this notational ugliness?

You might be tempted to write something like dQ instead of ds2, and then consider separately $ds = \sqrt{dQ}$ and $d\tau = \sqrt{-dQ/c^2}$. However that notation is not "infinitesimally correct" as you can't equate a squared infinitesimal with a single infinitesimal.

So are there any neater solutions?

This question arose because of this comment in another thread:

2. Aug 11, 2008

### MeJennifer

I fail to see the problem. What do you think the confusion is?

3. Aug 11, 2008

### DrGreg

Well, kev seems a little confused. And imaginary numbers are unnecessary; why make something more complicated than it really is?

4. Aug 11, 2008

### MeJennifer

It is not more complicated than it really is, as you wrote:

That is eaxtly true and simply a mathematical feature!

If you want to avoid any possible confusion then use complexified spacetime (see for instance Penrose) which no longer depends on the signature of the metric.

5. Aug 11, 2008

### DrGreg

I'm not disputing you can use complex numbers if you want to. I'm saying you don't need to, especially not when introducing the concept of the metric for the first time.

6. Aug 11, 2008

### MeJennifer

I disagree, it must be made clear to students that either timelike or spacelike distances must be imaginary and that this fully depends on the signature of the metric. This is a basic property of Minkowski and Lorentzian metrics.

7. Aug 11, 2008

### yuiop

In the other thread https://www.physicsforums.com/showthread.php?t=248015 you did obtain dtau from the square root of ds^2, when you said:

"if ds2 < 0, you have a timelike curve (by definition) and $d\tau = \sqrt{-ds^2/c^2}$ represents proper time"

I might be wrong, but I think in this case it is acceptable to take the square root of ds^2 because when the metric is written out with explicit brackets it is:

$$(ds)^2 = (dx)^2 - (c dt)^2$$ rather than

$$d(s^2) = d(x^2) - c^2 d(t^2)$$ that might be assumed when the brackets are left out.

For example the expression for acceleration:

$$\frac{d^2 x}{d t^2}$$ is not the same thing as $$\frac{(dx)^2}{(dt)^2}$$

and while it is acceptable to take the square root of the latter it is not acceptable to take the square root of the former and assume the answer is dx/dt. On that basis your suggestion of using something like dQ may be acceptable.

Earlier in https://www.physicsforums.com/showpost.php?p=1825950&postcount=15" in the same thread you said:

"...It's true that, for radial motion,
$$\frac {ds}{dr} = \frac {1} {\sqrt{1 - r_s/r}} \rightarrow \infty$$"

a student might not be clear whether you are using ds to refer to proper distance or proper time because you have not defined the metric signature for ds in that post. This is sort of the crux of the point I am getting at. Different sources use different conventions for the signature and that is surely going to be a source of confusion for students referring to different texts and has the additional overhead of always having to explicitly state what signature you are using. (Note that you have again taken the square root of the squared infinitesimal ds^2 to obtain the above equation)

Would it not to be better to have a convention that everyone sticks to?

Perhaps, something like always using ds to mean proper distance where $ds = \sqrt{c^2 (dt)^2 - (dx)^2}$ and always using d(tau) to mean a proper time interval where $d\tau = \sqrt{(dt)^2 - (dx)^2 / c^2}$.

I agree with Jennifer here. It is a nice mathematical feature and to me it has a physical interpretation. For example when the result of the expression $d\tau = \sqrt{dt^2 - dx^2 / c^2}$ is imaginary, it implies that no real particle can travel the distance dx in a time interval of less than dt, which in turn implies no real particle can travel faster than the speed of light which is not a bad thing.

The same is true for the Lorentz time transform:

$$t = t'\sqrt(1-v^2/c^2)}$$

where t is the proper time, then if we ask what is the proper time of a particle traveling with relative velocity v=2c, the answer is imaginary, again implying no real particle can travel faster than the speed of light.

In the Newtonian context the classic equation for constant acceleration is:

$$s = ut +at^2/2$$

where s is distance, u is initial velocity, a is acceleration and t is time.

Expressed in terms of time the equation becomes:

$$t = (\sqrt{(u^2+2as)}-u)/a$$

Now if a ball is thrown directly upwards with initial velocity u = 8 m/s and acceleration a = -10m/s^2 at what time does it get to a height s = 6m? The imaginary result is telling us something useful about the physical result in the real world, which in this case is that the ball never gets to height of 6 meters above the ground.

So rather than being an unfortunate side effect imaginary numbers can be useful.

If we adopt a convention for proper distance of $ds^2 = c^2 (dt)^2 - (dx)^2$ then your summary can adapted to:

if ds is imaginary, you have a spacelike curve and two events separated by the proper distance ds can not be causally connected.
if ds is real, you have a timelike curve and two events separated by the proper distance ds can be causally connected.
if ds= 0, you have a lightlike (a.k.a. null) curve (by definition), the worldline of a photon.

Similarly, using a convention of $d\tau^2 = (dt)^2 - (dx)^2/c^2$ for the proper time interval, then your summary can adapted to:

if $d\tau$ is imaginary, you have a spacelike curve and two events separated by an imaginary proper time interval can not be causally connected.
if $d\tau$ is real, you have a timelike curve and two events separated by a real proper time interval can be causally connected.
if $d\tau$ = 0, you have a lightlike (a.k.a. null) curve (by definition), the worldline of a photon.

By the way, I have finally decided to become a paid up member of PF in some sort of recognition of all the detailed feedback from members like DrGreg, George, Mentz and mysearch to name a few and the very concise but usually very accurate comments from Jeniffer. I have learned something from you all (sometimes very slowly I know :P) so thanks again ;)

Last edited by a moderator: Apr 23, 2017
8. Aug 12, 2008

### DrGreg

There is no doubt that symbol ds2 means (ds)2 and does not mean d(s2) (and that wasn't my question, in case there's any misunderstanding).

I agree wholeheartedly, it would, but the world don't have one. If you consult various text books etc you will see both conventions are used, although to be fair most authors pick one convention and stick to it; it was remiss of me to switch conventions mid-thread in my earlier reply in the other thread.

The comment I would make here is that you are not really making use of the imaginary value as such; you are just noting that there is no real solution to the equation which is saying that the equation is invalid in that situation.

There are other branches of science and engineering where complex numbers play an invaluable part, e.g. electronic engineering and quantum theory, but my impression is that in relativity you can usually avoid complex numbers altogether provided you accept that some equations have no (real) solution.

(Yes, there are approaches such as using the Euclidean metric with $(ict, \textbf{x})$ or even using quaternions, but these topics are off the beaten track and seem to be dismissed as out-of-date, as far as I can tell.)

9. Aug 12, 2008

### Fredrik

Staff Emeritus
You have to point this out to them in some way, but I would choose not to mention complex numbers at all. I would instead explain how the length of a curve is defined in Riemannian geometry (or just in $\mathbb R^3$) and then explain why that definition doesn't work for arbitrary curves in spacetime. We can only define the length of the curve if it's either timelike or spacelike. In the former case we define it as the integral of $\sqrt{-ds^2}$ and call it "proper time". In the latter case we define it as the integral of $\sqrt{ds^2}$ and call it "proper length". (I'm using the -+++ convention).

10. Aug 12, 2008

### MeJennifer

Note thought that mathematically this does not work for Lorentzian manifolds and is the reason why $\Delta s$ or $\Delta \tau$ is measured with affine parameters instead.

11. Aug 12, 2008

### K.J.Healey

I always felt like you lost information writing it that way. Why not just always give "g" as a matrix, with a paired coordinate vector. Once you do the product you lose the structure.

12. Aug 12, 2008

### cesiumfrog

What structure do you think you lose?

13. Aug 12, 2008

### Fredrik

Staff Emeritus
Not really. You know that g is symmetric, so if you e.g. have $ds^2=-dt^2+2dtdx+4dx^2$, you know that $g_{01}=g_{10}=1$. So what information do you lose?