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Minkowski metric

  1. Mar 9, 2009 #1
    I have just been learning in some of my maths courses that a metric space is a set which has an operation mapping 2 members of the set to the reals, called "distance", which respects certain axioms, one of which is that the distance between two members of the set is greater than or equal to zero.

    In what sense then is the Minkowski metric a metric, since it allows two points to have a negative interval separating them, thus violating this axioms of a metric?
     
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  3. Mar 9, 2009 #2

    Fredrik

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    It's a "metric tensor", which is something different than the kind of "metric" that's a part of the definition of a metric space.
     
  4. Mar 9, 2009 #3

    Ben Niehoff

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    No, it's the same sort of metric (in that it defines the notion of distance). The catch is that the geometry of relativity is pseudo-Riemannian, which means the metric is no longer required to be positive definite.

    In classical differential geometry, the metric tensor is defined to be strictly positive definite. In this case, the manifold is a "metric space" in the usual sense of the word, with a notion of distance that obeys our intuitions.
     
  5. Mar 9, 2009 #4

    Fredrik

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    It's not the same at all. For example, a "metric" [itex]d:X\times X\rightarrow X[/itex] doesn't have to be bilinear. There doesn't even have to be an addition defined on X. Both concepts have something to do with distances, but a metric tells you the distance between two points, and a metric tensor can be used to define the length of a curve between two points. That length may be negative, as Ben pointed out, but that's not the main difference. The main difference is what sort of objects they're acting on.

    I think it's probably true that given a Riemannian metric tensor on a smooth manifold, you can use it to define a metric on the manifold, which turns it into a metric space. The distance between two points would be defined as the length of the shortest curve between those points. (The reason I say "probably" is that I'm not sure if e.g. the sphere with opposite points identified has a manifold structure, or if the axiom d(x,y)≥d(x,z)+d(z.y) will hold in this case. I could probably look that up, but I'm guessing someone will explain it to me if I don't :wink:).

    It should also be mentioned that when someone uses the word "metric" in the context of general relativity, they always mean "metric tensor".
     
    Last edited: Mar 9, 2009
  6. Mar 10, 2009 #5
    How can the length of a curve between two points be negative? Sorry I dont know much about any geometry except the good old Euclidean kind
     
  7. Mar 10, 2009 #6

    Fredrik

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    It can't. That was a mistake by me. If we simply replace the scalar product in the standard definition of "length of a curve" with a Lorentzian metric, then the "length" could sometimes be imaginary, but never negative. However, that's not how these things are usually dealt with. I'll try to explain.

    The length (the good old Euclidean kind) of a curve [itex]\vec C:(a,b)\rightarrow\mathbb R^n[/itex] can be defined as

    [tex]\int_a^b \sqrt{\vec C'(t)\cdot\vec C'(t)}dt[/tex]

    The motivation for this is that the length should be approximately equal to a sum of contributions of the form [itex]\sqrt{\Delta x^2+\Delta y^2+\Delta z^2}[/itex] along the curve, and if we write [itex]\Delta x=(\Delta x/\Delta t)\Delta t[/itex] and similarly for y and z, that square root can be written as

    [tex]\sqrt{\Big(\frac{\Delta x}{\Delta t}\Big)^2+\big(\frac{\Delta y}{\Delta t}\Big)^2+\Big(\frac{\Delta z}{\Delta t}\Big)^2}\Delta t[/tex]

    which is approximately equal to

    [tex]\sqrt{v_x^2+v_y^2+v_z^2}\Delta t=\sqrt{\vec v\cdot\vec v}\Delta t[/tex]

    Note that the definition contains a scalar product of two tangent vectors of the curve. In GR, we don't have a scalar product. Instead, we have the metric, which when it acts on two arbitrary tangent vectors at a point in the manifold satisfies all the conditions that a scalar product is required to satisfy, except the condition [itex]\langle v,v\rangle\geq 0[/itex] for all vectors v. We can have g(v,v)<0 for some vectors v (for example, any vector in the time direction).

    Since the metric is the closest thing to a scalar product we have, the closest thing to a "length" of a curve [itex]C:(a,b)\rightarrow M[/itex] that we can define is this:

    [tex]\int_a^b\sqrt{\pm g_{C(t)}(\dot C_t,\dot C_t)}dt[/tex]

    where [tex]\dot C_t[/tex] is the tangent vector of the curve C at the point C(t). (Note that there's a dot above Ct. It's barely visible on my screen).

    It wouldn't make much sense to use this definition on arbitrary curves. (The "length" defined this way wouldn't be a useful concept). Therefore, we only define the "length" for curves such that [tex]g_{C(t)}(\dot C_t,\dot C_t)>0[/tex] everywhere on the curve (in which case we use the + sign and call the result of the integration "proper length"), and curves such that [tex]g_{C(t)}(\dot C_t,\dot C_t)<0[/tex] everywhere on the curve (in which case we use the - sign and call the result of the integration "proper time"). The former type of curve is said to be "spacelike" and the latter type is said to be "timelike".
     
  8. Mar 11, 2009 #7
    Ok I think I understand a little better now. Ive just started a course on differential geometry so hopefully this will clear things up properly.

    Thank you Fredrik!
     
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