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Homework Help: Help with metric equations (and other questions)

  1. May 17, 2010 #1
    1. They commonly begin with ds^2=...
    Long story short, I'm teaching myself advanced physics because while I attend a university, I am unable to compete with other students of quantitative sciences because my number skills are piss poor: I would bomb on exams. I'm more of an Arts and Social Sciences guy.

    Now I have a basic knowledge of calculus: differentiation, integrals, vectors, gradients, curl, divergence etc, and also a TI-89 Titanium handy (this calculator is also one of the reasons I can't do quantitative courses in university- it's a crutch (but I never saw why they should be banned anyway- if you become a respected theoretical physicist is there still a ban on those for you? I think not)), so if anyone feels like helping this physics noob out, that is what I already know.

    I know that metrics are often used to describe the topology of space time, but how? For that matter, how do matrices like the Minkowski metric [-1, 1, 1, 1] (for short) describe space time? Me no understand, hurrdurr.

    Also, what do physicists mean by "timelike curves"? It seems kind of an oxymoron to me.

    2. For example, the metric of a traversable wormhole is ds^2=-c^2 dt^2+dl^2+(k^2+l^2)(dTHETA^2+sin^2THETA dPHI^2) (sorry, I have no idea how to use... latex, is it?)

    3. The Wikipedia article on metrics is too technical for me to understand. I require a theoretical understanding along side quantitative know-how.

    Thank you very much for your knowledge, should you decide to help.
  2. jcsd
  3. May 17, 2010 #2


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    hmm, well, you do realize that this is fairly technical stuff, right? We can try to explain it using analogies etc. but they're never perfect, and you can't really understand it properly without knowing the underlying math.
    It all starts with the concept of a coordinate mapping. Remember that a coordinate mapping is simply a way of assigning a set of numbers to every point in space such that no two points correspond to the same set of numbers. A good coordinate system is continuous, meaning that nearby points have nearby numbers. More precisely: consider two points A and B, and say A is identified by the set of numbers [itex]\vec{x}_A[/itex] and B is identified by the set of numbers [itex]\vec{x}_B[/itex]. A continuous coordinate mapping satisfies
    [tex]\lim_{AB\to 0}\vec{x}_A - \vec{x}_B = 0[/tex]
    where AB is the measured distance between A and B - that is, think of it as a physical measurement taken with an arbitrarily tiny ruler, not as something you calculate.

    Now, when I said you should think of AB as a physical measurement, that may seem a little weird because in the way you're used to thinking, there's probably no distinction between the concept of "distance between two points" and a formula like the Pythagorean Theorem [tex]\sqrt{(x_A - x_B)^2 + (y_A - y_B)^2 + (z_A - z_B)^2}[/tex]. But in fact, they are distinctly different concepts. If you have any two points in space, you can certainly measure the distance between them, but there's really no reason to assume that the distance will be given by the Pythagorean theorem. In fact, there are plenty of counterexamples. Think about latitude and longitude, which set up a coordinate mapping on the Earth's surface. If you have two points at the same latitude but different longitudes, say 25° and 30°, you could compute the difference of coordinates by subtracting one from the other to get 5°. But that doesn't tell you anything about the distance between the points. It could be a few hundred miles, if the points are at the equator, or barely an inch if the points are near one of the poles.

    So I've tried to explain that distance is a physical quantity that you can't get just from subtracting coordinates. But it makes sense that, if you're given the coordinates of two points, you should be able to calculate the distance between them somehow, right? The metric is what allows you to do that. Think of it this way: given two points, A and B, you can definitely compute the difference between the two points' coordinates, [itex]\Delta \vec{x} = \vec{x}_B - \vec{x}_A[/itex], because you have the coordinates. And you can also go out into space and locate each of those points, and then measure the distance AB between them. Then you can put those numbers together into an equation
    [tex](AB)^2 = \sum_{i,j}(\Delta x)_i g_{ij} (\Delta x )_j[/tex]
    where [itex]g_{ij}[/itex] is a matrix of whatever numbers are needed to make the equation work for all possible pairs of points. That matrix is the metric.

    Well, actually not all points, just infinitesimally separated points, because the metric can be position-dependent. To compute a finite distance between two points, you split the path between the two points up into differential segments, calculate the distance (which we write as [itex]\mathrm{d}s[/itex]) for each segment, and add them all up.
    [tex]s = \int \mathrm{d}s[/tex]

    Anyway, to do a simple example, the Minkowski metric is
    [tex]\eta = \begin{pmatrix}-1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix}[/tex]
    That corresponds to an infinitesimal distance (a.k.a. interval) of
    [tex]\mathrm{d}s^2 = -c^2\mathrm{d}t^2 + \mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2[/tex]
    You can see that part of that is the plain old Pythagorean theorem. So, in a sense, the fact that the Pythagorean theorem works tells us that we live in a space whose metric is pretty darn close to the Minkowski metric. The negative sign on the time term comes about because of the way time and space distort when you change velocity. (It's special relativity, basically)

    Imagine a particle moving through space. For ease of visualization, let's say that the space is flat and 2-dimensional (although the same argument works for 3D curved space or any other kind of space). Now, imagine taking "snapshots" of the particle's position every so often and stacking them up on top of each other. You would basically be creating a 3D space out of repeated snapshots of the 2D space, and the extra third dimension (along which the snapshots are stacked) corresponds to time. You could connect the dots, i.e. connect the particle's positions from one snapshot to the next, and you'd wind up with a curve, called a world line.

    Now, at any point on this curve, you can draw a tangent vector to the curve. There are three possibilities:
    -Its space component can be greater than its time component. This is called a "spacelike vector."
    -Its space component can be equal to its time component. This is called a "null vector."
    -Its space component can be less than its time component. This is called a "timelike vector."
    It turns out (and in fact, you could do the math to show this using the contents of this post) that if the tangent vector is spacelike, that would correspond to the particle moving faster than light. If the tangent vector is null, that corresponds to the particle moving at the speed of light, and of course if the tangent vector is timelike, that corresponds to the particle moving slower than light. A timelike curve is a world line for which the tangent vector at any point along it is timelike, which means that a timelike curve is the only kind of world line that can represent a physical particle. (Unless tachyons are real, but that's just speculation)
  4. May 17, 2010 #3
    That was the awesomest and clearest reply I could have expected. Thank you for making things much clearer to me.
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