Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Minkowski Metric Questions

  1. Jul 4, 2014 #1
    I have recently been studying the tensors on the left side of the Einstein field equations, but I have been studying and deriving them in 3-D. I would now like to move on to adding time into the mixture.

    I have some questions regarding the Minkowski metric [itex]\eta[/itex][itex]\mu[/itex][itex]\nu[/itex].

    First, I know that the elements of this metric are: x00= -1. The other elements on the diagonal are 1. Every other element is 0. My question about this is: Where does the negative come from? What makes the temporal component opposite to that of the spatial elements?

    2nd: I know that ds2= g[itex]\mu[/itex][itex]\nu[/itex]dx[itex]\mu[/itex]dx[itex]\nu[/itex].

    In normal Euclidean space, ds2 is the square of the distance between two points in space. It is essentially like the c2 in Pythagorean's theorem a2 + b2 = c2

    However, what would ds2 represent if I throw in the temporal component and do something like:

    [itex]\eta[/itex][itex]\mu[/itex][itex]\nu[/itex]dx[itex]\mu[/itex]dx[itex]\nu[/itex]

    I heard that it might be the square of proper time, but I am not sure about this. I know that in Minkowski Cartesian coordinates, x0 = ct (where c is usually set to equal 1). x1= x. x2= y. x3 = z.

    When I used this information to calculate [itex]\eta[/itex][itex]\mu[/itex][itex]\nu[/itex]dx[itex]\mu[/itex]dx[itex]\nu[/itex] , I ended up getting this as my result:

    ds2 = -c2t2 + x2 + y2+ z2= -t2+ x2 + y2+ z2 (since c = 1).

    If it were just spatial components then the meaning of ds2 would be apparent to me, but that temporal component makes it not so as apparent because it looks to me like you are either subtracting a measure of time from the square of a distance in space, or you are just subtracting a measure of space from what would be the normal 3D distance in space (because c is a velocity and speed times time equals distance). Can someone please confirm for me what ds2 is in the case of 4 dimensions?


    Finally, I need to know how I should go about deriving 4 dimensional metric tensors in coordinate systems other than Cartesian. Take spherical coordinates for instance.

    I can easily derive the 3D metric tensor for spherical coordinates because I know the transformation properties from Cartesian to spherical. Therefore I know the vector that I would differentiate to get the tangential vectors:

    R= <rsin(θ)cos(ø), rsin(θ)sin(ø), rcos(θ)> and
    x1=r
    x2
    x3

    However, if I try to switch to 4 dimensions and transform between Cartesian and spherical in 4D, then how would the temporal component of Cartesian coordinates (x0= ct) transform into a temporal component for spherical components?

    Would the ct of Cartesian coordinates just remain the same ct in spherical? Does this apply to all types of coordinate systems? Is the temporal component just always ct (as opposed to some spatial components changing to stuff like r or θ or ø)?

    Also, would this mean that to derive the tangential vectors for a 4D metric tensor that I would have to differentiate the vector with respect to t in addition to all of the other components?

    That is all. Thank you.
     
  2. jcsd
  3. Jul 5, 2014 #2
    If anyone wants clarity of what I am asking on any of my questions, I will give some examples of what I mean.

    Example:

    We know that ds2 = -(ct)2 + x2 + y2 + z2 in Cartesian 4D coordinates.

    Now assuming that c = 1, let's say that x= 3 units of space , y=4 units of space, z= 5 units of space and then t= 6 seconds.

    This equation would turn into:

    ds2= -36 + 9 + 16 +25 = 14

    ds2 = 14 and therefore ds = [itex]\sqrt{14}[/itex]

    Now what exactly would this result mean? What quantity would [itex]\sqrt{14}[/itex] actually represent?

    Also, what if ds2 turns out to be negative? That would make ds an imaginary number. What would this kind of result mean?
     
  4. Jul 5, 2014 #3

    Dale

    Staff: Mentor

    You have asked a lot of questions and I don't think that I can answer all of them. But basically the difference between the signs corresponds to a measurement made with a clock out with a rod. If you take a measurement with a rod then you will square it and add it, and if you take the measurement with a clock then you will square it and subtract. If the total is positive then it is called spacelike and if the total is negative then it is called timelike.
     
  5. Jul 5, 2014 #4

    ChrisVer

    User Avatar
    Gold Member

    So let's try to clarify some things. I prefer using the unit convention of c=1.
    [itex]ds^{2}[/itex] is the invariant length, proper time, proper length, 1st fundamental form ... these are just its names. It's inveriant because it remains the same under a general coordinate transformation (Poincare Group) when you have a general metric [itex]g_{\mu \nu}(x)[/itex] or under Lorentz transformations when you have the Minkowski metric (Lorentz Group).
    That means that the transformation :
    [itex] x^{\mu} \rightarrow \bar{x}^{\mu}= L^{\mu}_{\nu} x^{\nu} [/itex]
    is supposed to transform the metric as well such that [itex]ds^{2}[/itex]. This means that the metric defines an isometry on the Minkowski spacetime.
    The Minkowskian spacetime is not the same as the one in Euclidean space, because they have different signatures (signs of the metric's eigenvalues). However you can always go from a Minkowskian spacetime to an Euclidean one, by doing a transformation of time: [itex]t \rightarrow it [/itex] (then the - in the dt component will become a + because of the [itex]i^{2}[/itex] ).
    That's the reason why the trajectories of least length on Minkowski spacetime differ to those of the Euclidean one.... In case for example you'd find a solution that goes as [itex]cos(t)[/itex] in Euclidean, this would have to be identical to a solution in Minkowski that goes as [itex] cos(it) = cosh(t)[/itex].
    Nevermind... though...
    In the Minkowski spacetime, you have 3 possibilities for [itex]ds^{2}[/itex] indeed...
    [itex]ds^{2} > 0 [/itex] your vector is called space-like
    [itex]ds^{2} =0 [/itex] your vector is called light-like or null
    [itex]ds^{2} <0 [/itex] your vector is called time-like
    http://en.wikipedia.org/wiki/Spacetime#Spacetime_intervals

    They don't change the ability of measuring lengths (although they have to be redefined). They are just dividing your space in 3 distinct spaces which cannot commute by any Lorentz Transformation...

    Now for the 4 dim Minkowski metric nothing will change when you go to some other coordinate system. In that case you will have to make the proper transformations of [itex]x^{\mu} \rightarrow \bar{x}^{\mu}(x)[/itex] and thus transform the metric accordingly as a rank 2 tensor...
    For example in spherical coords:
    [itex]x^{0}=t \rightarrow \bar{x}^{0}=t [/itex]
    [itex]x^{1}=x \rightarrow \bar{x}^{1}= r [/itex]
    [itex]x^{2}=y \rightarrow \bar{x}^{2}= \theta [/itex]
    [itex]x^{3}=z \rightarrow \bar{x}^{3}= \phi [/itex]
    and so your metric will change:

    [itex] \bar{n}_{\mu \nu} = \frac{ \partial x^{\rho}}{\partial \bar{x}^{\mu}}\frac{ \partial x^{\sigma}}{\partial \bar{x}^{\nu}} n_{\rho \sigma}[/itex]
    For the Minkowski metric this is easy to evaluate, because you have that [itex]\rho=\sigma[/itex]
    and you know how [itex]x^{\rho}[/itex] depends on [itex]\bar{x}^{\mu}[/itex] in spherical coords:
    [itex] t=t[/itex]
    [itex] x= r \sin (\theta) \cos (\phi) [/itex]
    [itex] y= r \sin (\theta) \sin (\phi) [/itex]
    [itex] z= r \cos(\theta) [/itex]
     
  6. Jul 5, 2014 #5

    ChrisVer

    User Avatar
    Gold Member

    Also another easy way to see what [itex]ds^{2}[/itex] is forgetting for now a Minkowski space, and try to evaluate how on a surface you can measure/define invariant distances.
    One possibility is to take your surface defined by the vector [itex]X=F(x^{a})[/itex] (parametrized the surface with [itex]x^{a}[/itex], so imagine x^{i}=const as grids covering the surface )
    Then changes of this vector along the [itex]x^{a}[/itex] parameter are going to be vectors in the tangent space of each of the point on your surface.
    [itex] \frac{\partial X}{\partial x^{a}} [/itex]
    In order to create an invariant out of THE WHOLE SPACE it, you need to take its square.
    [itex] \frac{\partial \vec{X}}{\partial x^{a}}\frac{\partial \vec{X}}{\partial x^{b}} [/itex]

    This is the embedded metric, defining distances on the submanifold you parametrized... it doesn't depend on your general manifold (because we took the dot product to create an invariant in that space) but only on the intrinsic space of your surface in consideration, which is of at least 1 dimension less (eg a sphere is a two dimensional manifold, which you can embed into the [itex]R^{3}[/itex] three dimensional Euclidean space by using the parametrization ([itex]\theta,\phi [/itex]) and using the unit vector.
    In four dimensions the vector product should be taken accordingly in order to get invariants.
     
  7. Jul 5, 2014 #6

    Nugatory

    User Avatar

    Staff: Mentor

    It doesn't really "come from" anywhere; although that metric does interact nicely with the Lorentz transformations, it's more the case that experiments have confirmed that the spacetime we live in is described by the Minkowski metric rather than a Euclidean one.

    Intuitively, the negative sign on the 00 component reflects the (fairly self-evident) fact that the time dimension is different from the three spatial dimensions.

    Be aware that the -+++ signs on the four diagonal coordinates is just a convention; you'll find some texts (generally older) that use a +--- convention and others map x, y, and z to the 0, 1, and 2 coordinate and use x3 for t so the metric is +++- or -+++. No matter which you use, the t coordinate is "special" in that it it has a different sign from the three spatial coordinates.

    If we use the most common -+++ convention, then:
    - ##ds^2## is always the square of the interval (metric distance) between two nearby points.
    - If ##ds^2## is negative, then it is the negative of the square of the proper time between the two events. If an inertial observer passes through both events he will interpret them as two events happening at the same place (where he is) and with one of them happening ##\sqrt{-ds^2}## seconds after the other. Other observers may measure a different amount of time between the two events, but no observer will consider them to be simultaneous, and all observers ill agree about whoich happened first. These are the "timelike" intervals and we say that the two events are timelike-separated"
    - If ##ds^2## is positive, then there can be no observer who can pass through both events; no observer can ever interpret them as two events happening at the same place. Different observers will disagree about whether the two events are simultaneous, and if not, which happened first. These are the "space-like" intervals.
    - If ##ds^2## is zero, then the path of a beam of light would pass through both of them. This implies the invariant speed of light; you can see this for yourself by considering that ##ds^2=0## implies that ##(cdt)^2=\sqrt{dx^2+dy^2+dz^2}=(dr)^2## where r is the spatial distance between the two events, so we have ##dR/dt=c##.

    Whether you use spherical or cartesian coordinates, you're still using the same clock, so the transformation of the t coordinate is trivial: t'=t. It's only if you're using different clocks in the two coordinate systems that t' will be a non-trivial function of t. And yes, you do have to do the differentiation with respect to t, but as long as ##dt'/dt=1## and ##dr/dt=d\theta/dt=d\phi/dt=0## these derivatives don't change the result.
     
    Last edited: Jul 5, 2014
  8. Jul 5, 2014 #7
    Thanks. This helped a lot. Let me see if I can summarize what you are saying just so that I know that I fully get what you are saying. If you could, I would appreciate if you check my 5 summary points below for accuracy (especially 3 and 4: I had just a few lingering questions about those).

    1. The negative sign really just comes from the fact that the temporal dimension is simply different from the spatial ones and the fact that experimentation simply seems to agree with this metric.

    2. If ds2 is negative, then this number is the negative of the square of the proper time between two events and both events happened at the same place according to the observer.
    In other words, if ds2 = -144 then the proper time interval between the two events is 12 seconds (or whatever unit of time you use) according to the observer (although another observer may see a different time interval). All observers will agree about which event happened first. This is called time-like.


    3. If ds2 is positive, then the two events happened in different places and different observers will disagree on which event took place first or whether they were simultaneous or not. You essentially can not deduce the proper time between the two events. Therefore, if ds2= 25 then ds= 5 and you can't really tell when the two events happened in relation to each other. This is space like. I have just one more question about this part. In the example that I just gave, what significance would the number 5 (as opposed to any other positive number) hold besides telling me that the interval is space like? in other words, would it mean any thing different or give me any different information if ds was to equal 6 instead of 5, or does it just not matter what the number is as long as it is positive?


    4. If ds2= 0, then the two events were either simultaneous or they were just the right distance and time interval apart for a beam of light to get to both of them in the nick of time. I have just one question about this part. Would any observer be able to determine which of these events happened first or if they were simultaneous?


    5. Regardless of the coordinate system, as long as you are using the same clock, then t is just t. Therefore, in Cartesian, cylindrical, spherical, curvilinear or any other coordinate system, the t axis is just the t axis and you don't have to do any transformations like you have to do with spatial coordinates (such as how you have to transform x into rsin(θ)cos(ø)).
     
  9. Jul 5, 2014 #8

    Nugatory

    User Avatar

    Staff: Mentor

    It matters. When ##ds^2## is positive, we call ##ds## the "proper distance" (as opposed to "proper time"), and it is the spatial distance between the two events according to an observer for whom the two events are simultaneous. The physical significance of this is clear if you consider the two events "At time T the left-hand end of a rod is here" and "At time T the right-hand end of the rod is there" - the proper is the length of the rod in a frame in which the rod is at rest. It's easy to see this if you consider that if the t coordinate of both events is the same, then ##dt## is zero and the expression for ##ds^2## just reduces to the three-dimensional Euclidean distance.

    There's a symmetry here: Proper distance is equal to the coordinate distance in a frame in which the two events happen at the same time; proper time is equal to the coordinate time in a frame in which the two events happen at the same place.


    If ##ds^2=0## and the two events did not happen at the same place, then they are not simultaneous for any observer. If they did happen at the same place, then they're the same event (##ds=dx=dy=dz=dt=0## so we know that all four coordinates are the same for the two events - and an event is identified by its four coordinates).

    Pretty much, yes; "using the same clock" is just another way of saying that we'll assign the same time coordinate to events in both coordinate systems. I did insert the word "spatial" to stress that in in these transformations, the primed spatial coordinates are functions of the unprimed spatial coordinates but not of t.
     
  10. Jul 6, 2014 #9
    I see. So proper distance is essentially the Euclidean distance between two simultaneous events. Proper time is the time interval measured by an observer who sees two events happening in the same place.

    I thought about something however. In the case of proper distance, in order for the events to be simultaneous, dt would have to equal 0. However, I noticed that ds2 can still be positive even when dt does not equal 0. For example:

    If we set c=1, dt=2 , dx =3, dy=4 and dz = 5 then the formula would turn out to be:

    ds2= -4+ 9+16+ 25= 46 and ds = [itex]\sqrt{46}[/itex]

    As you can see, the result of this is a positive number despite dt not being 0. Yet, this is also not the Euclidean distance (because the Euclidean distance would have been [itex]\sqrt{50}[/itex] since 9 + 16 + 25 = 50).

    How would I interpret a result like this where the two events are not simultaneous, but ds is still a positive number? What would ds mean in this case?


    Similarly, in the case of proper time, in order for two events to take place in the same place, dx dy and dz would have to all equal 0. However, if dt= 8, dx=3 , dy=4 and dz=5, then

    ds2= -64 + 9 + 16 + 25 = -14

    What would this result mean in the case where ds2 is negative, but the spatial distances do not equal 0?
     
  11. Jul 6, 2014 #10

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    I think someone already said the answer: If [itex]ds^2[/itex] is spacelike ([itex]\delta x > c \delta t)[/itex], then that means that in some other frame, the two events are simulaneous. Using the Lorentz transformations, in another frame:

    [itex]\delta t' = \gamma (\delta t - v/c^2 \delta x)[/itex]

    So if you choose [itex]v[/itex] such that [itex]v = c^2 \delta t/\delta x[/itex], then in that frame, [itex]\delta t' = 0[/itex]. (Note that this is only possible if [itex]\delta x > c \delta t[/itex], otherwise, you'll get [itex]v > c[/itex], which doesn't correspond to any frame.)

    In contrast, if [itex]\delta x < c \delta t[/itex], and we transform to a new frame, we have:
    [itex]\delta x' = \gamma (\delta x - v \delta t)[/itex]

    If we choose [itex]v = \delta x/\delta t[/itex], then in this new frame, [itex]\delta x' = 0[/itex]. (Note this is only possible if [itex]\delta x < c \delta t[/itex], otherwise, you'll get [itex]v > c[/itex]).
     
    Last edited: Jul 6, 2014
  12. Jul 6, 2014 #11

    Ibix

    User Avatar
    Science Advisor

    You can always consider this in the 1d case - where the spatial separation between the events is along the x-axis. Then
    [tex]ds^2=dx^2-c^2dt^2[/tex]
    It's then relatively easy to see what the three possible cases mean:

    [itex]ds^2=0[/itex] means that [itex]dx^2=c^2dt^2[/itex] - in other words, that it is only possible to make it from one event to the other at the speed of light.

    [itex]ds^2>0[/itex] means that [itex]dx^2>c^2dt^2[/itex] - in other words, the distance between the two events is too great for even light to travel between them. There is no frame in which these events are co-located, although there is one in which they are simultaneous. In some frames, one happens before the other; in other frames, the other happens before the one.

    [itex]ds^2<0[/itex] means that [itex]dx^2<c^2dt^2[/itex] - in other words, things slower than light can make it from one event to the other. There is one frame in which these events are co-located, but no frames in which they are simultaneous. Observers all agree on which one happened first.

    This has consequences. For example, events that are so far apart that not even light can cross the gap in the time available cannot be causally connected (i.e., event A cannot cause event B because nothing can travel fast enough for A to have an effect so far away). That is true in all frames, and it is why it doesn't matter that one observer says that A happened before B while another says B happened before A. Neither could affect the other.
     
    Last edited: Jul 6, 2014
  13. Jul 7, 2014 #12

    Nugatory

    User Avatar

    Staff: Mentor

    It means the two events are space-like separated, so there will be some observer for whom they're simultaneous - but you're not that observer. It's still very important, because if the two events are spacelike separated, there can be no cause and effect relationship between them.


    Pretty much the same thing, but the other way around: there will be some observer for whom the time between these two events is equal to the proper time - but you're not that observer. It's still very important because if the two events are timelike separated, then all observers will agree about which happened first, and it is at least possible for there to be a cause and effect relationship between them.
     
  14. Jul 7, 2014 #13

    ChrisVer

    User Avatar
    Gold Member

    Also in that case the invariant separation, in order to be interpreted as distance, needs to be redefined as:
    [itex] d τ^{2} =- ds^{2}[/itex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Minkowski Metric Questions
  1. Minkowski metric (Replies: 6)

  2. The minkowski metric (Replies: 3)

  3. Minkowski metric (Replies: 8)

  4. Minkowski Metric (Replies: 36)

Loading...