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GR metric meaning

  1. Nov 29, 2004 #1
    I've been trying to work out the meaning of the metric in General relativity. I have a few ideas, but nothing's really come together.
    These are what I think is right, from SR: the space-time distance is a quantity which is agreed upon by all observers, a fundamental property of the interval. The proper time is the time between two events measured by an observer who also measures the spatial interval to be zero. It is [tex]\frac{ds}{c^2}[/tex].
    Now the book I'm reading says about the Schwarzschild metric that [tex]dt[/tex] in the expression for the spacetime distance is the time measured by an observer who is in a region of spacetime so far away from the object that spacetime there can be considered flat. This is different from what I understood about the metric, namely that [tex]dt, dr, d\theta, d\phi [/tex] were the coordinate differences as measured by anyone. Furthermore, from my understanding it seem that [tex]d\tau[/tex] is not [tex]\frac{ds}{c^2}[/tex] anymore but [tex]ds c^{-2} \left( 1 - \frac{r_0}{r} \right)^{-1}[/tex]. This does seem wrong. Please could some kind person shed light on my confusion, in particular the meaning of the coordinates in the metric equation. Thanks in advance.
    Last edited: Nov 29, 2004
  2. jcsd
  3. Nov 29, 2004 #2
    You really should purchase "Exploring Black Holes" by J.A. Wheeler and E.F. Taylor (ISBN 0-201-38423-X). This is the most incredibly understandable book on GR I have ever seen -- in a few lessons you will be computing the effects of a black hole using Schwartzchild all by yourself.

    BTW - Wheeler and Taylor agree with your author.
    Last edited: Nov 29, 2004
  4. Nov 29, 2004 #3


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    Close - people usually write

    ds^2 = c^2 dt^2 - (dx^2 + dy^2 + dz^2)

    for the metric of flat space-time. Well, actually there are a number of different sign conventions, but this is a common sign convention which is the closest to your post.

    It's easy to see that dt = ds /c when dx,dy,dz=0, not ds/c^2. I'm not sure how you got ds/c^2, I don't think it's quite right.

    The book is talking about the coordinate time t, which is different from proper time tau. The coordinate time is indeed what the book describes.

    A local observer can and does measure his own local times and distances which are different from the Schwarzschild coordinates.

    The coordinate times are used just to have a "standard" coordinate system. There are actually many possible coordinate systems, the Schwarzschild coordinate system is just one of the most common. Given the path that a particular observer is taking through space-time, the more physical observed proper times can be derived from the coordinate times. But one must specify the path to be able to perform this calculation.
  5. Dec 1, 2004 #4
    So what is signficant about the observer's position - is this considered a transformation. This is different from SR, where the metric was an invariant under transformations?
    When you transform to a frame in free fall, what happens to the matric? Will it become minkowskian from the equivalence principle, even though the proper time is that measured in a frame not following a geodesic? Or it may stay the same, which I suspect, but will the position of the observer in the mass' potential well matter?

    donjennix: I'll be on the look out for that book.

    pervect: yes, sorry, it of course ds/c.

    Thanks for the replies.
    Last edited: Dec 1, 2004
  6. Dec 1, 2004 #5
    Let me copy/paste a section of Wheeler and Taylor to see if it helps --"wristwatch time" is the observed time of the person in the GR space. I use "tau", "phi" etc. to keep from using weird fonts. phi is the angle around the plane of rotation about the black hole. I see that my superscript "2's" indicating squares have ceased being superscripts so please interpret. The "reduced circumference", r, is a radius of course -- it is just not the radius you would measure as you got closer to the black hole. It is the circumference (which is not affected by GR dilations) divided by 2pi.

    "The Schwarzschild metric describes the separation between two neighboring events in the vicinity of a spherically symmetric, nonrotating center of gravitational attraction.
    d(tau)2 = {1-2M/r}dt2 – dr2/{1-2M/r} – r2d(phi)2 [A]
    • dtau is the wristwatch time between the two events as measured on a wristwatch that moves directly from one event to the other.
    • dt is the time between the events measured on a clock far from the center (page 2-27)
    • r is the reduced circumference: circumference divided by 2pi (page 2-7).
    • M is the mass of the center of attraction measured in units of meters (page 2-13).
    Equation [A] is called the timelike version of the Schwarzschild metric, useful when a clock can be carried between the two events at less than the speed of light. If this is not possible, then we use the spacelike version, equation [11] on page 2-19.

    d(sigma)2 = -{1-2M/r}dt2 + dr2/{1-2M/r} – r2d(phi)2

    Here d(sigma) is the proper distance between the two events: the distance between them recorded by a measuring rod moving in such a way that the two events occur at the same time in its rest frame."
    Last edited: Dec 1, 2004
  7. Dec 2, 2004 #6


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    In a non-rotating free fall frame, the metric will always be diagonal. With the correct choice of units, the diagonal metric will be Minkowskian. The correct unit choice represents measuring distances with respect to your local rulers, and time with respect to your local clocks.

    So the answer is basically that a free-fall frame will look Minkowskian - if it doesn't rotate, and if you also scale the coordiantes correctly.

    The metric transforms like any other tensor with respect to a change in coordinates. The explicit equations for this are

    ga'b' = gab Laa'Lbb'

    It may look intimidating, but like all tensor equations you just sum over the repeated indices as far as doing the calculation.

    Here the primed coordinates are the transformed coordinates, and the tensor L describes the coordinate transformation. The same matrix L would describe how a "vector" xa transformed, for instance, like this

    xa' = Laa' xa

    So if you know how to transform a vector, transforming the metric isn't much different in principle, except that you have to apply the transformation matrix twice.

    With the convention that subscripts on the matrix L alwas run from north-west to south-easth, it's easy to remember this equation too, if you think of the superscripts as "cancelling out" a repeated subscript.

    You might compare this tensor notation with the usual rule for how a matrix A transforms in standard linear algebera via the matrix L if you're ambitious, I should point out that gab would transform just like a matrix would. Unfortunately, matrix notation is a bit of a dead end, it can handle mixed tensors of rank two (such as the previously mentioned gab but not the more general tensors of rank two such as gab or gab, and higher order tensors are Right Out. Since many important tensors in GR are of rank greater than two, matrix notation can't handle deal with them.
  8. Dec 2, 2004 #7
    Note: That assertion assumes one is using Minkowski coordinates [i.e. (ct, x, y, z) and not somerthing like (ct, r, theta, phi)] and in the case of curved spacetime g_uv = diag(1, -1, -1, -1) only applies locally.
    L does not a represent a tensor.
    Note: In SR [itex]x^{\alpha}[/itex] is the prototype for the Lorentz 4-vector if L represents the Lorentz transformation. In such a case [itex]x^{\alpha}[/itex] is a displacement vector and represents a spacetime displacement from an event which is chosen to be the origin of coordinates.

  9. Dec 2, 2004 #8


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    Good point

    What would you call it then?


    Here we are talking about the difference between xa and xa, in case it isn't obvious (which it probably isn't). I deliberately oversimplified my first remark slightly by calling xa a "vector" (in scare quotes), precisely because I thought it would be far too confusing to go into a long drawn out explanation about the difference between xa and xa at this point.

    The main points I wanted to make were:

    1) The proper choice of coordinates makes the metric "Minkowskian".

    2) Because space-time is curved, it's not generally possible to make the metric "Minkowskian" everywhere for everyone, even though it's always possible to make the metric "Minkowskian" at any given point.

    Actually, an example might help. Rather than GR, consider an application of the metric tensor to a sphere (the Earth's surface, approximated as being spherical).

    You have global coordinates on the Earth's surface, lattitude and longitude. We need symbols, let's call them A for lAtitude, and O for lOngitude.

    You note that the distance between any two very close points on the Earth's surface can be defined by the expression

    ds^2 = C^2 dA^2 + (C cos(A))^2 dO^2

    here C is a constant, which depends on the radius of the Earth. (If the coordinates are measured in degrees, C is 60 nautical miles - 1 minute of arc of the Earth's surface is 1 nautical mile, so 1 degree is 60 nautical miles).

    You can think of this expression as being the metric tensor defining the Euclidian geometry of the Earth's surface. A degree of latitude always points north or south, a degree of longitude always points east or west.

    The factors of C^2 and C^2 cos^2(A) convert the coordinate system (degrees of latitude, degrees of longitude) into distances.

    If you used different units (grads or radians) for the angular measure, the numerical value of "C" would change.

    This is an example of using the metric tensor with the familair Euclidian 2 dimensonal geometry, where ds^2 = dx^2 + dy^2 on a flat surface. On a curved surface, this formula needs to be modified. This is done by adding metric coefficients as above.

    Note that the metric coefficients vary with lattitude. At any point on the Earth's surface, one can set up an x-y coordinate system that's locally Euclidian. But because of the curvature of the Earth, it's not possible to set up coordiantes that make the metric coefficients constant for all observers.

    The situation with relativity is extremely similar, except that the geometry is "Minkowskian" rather than Euclidian.
  10. Dec 2, 2004 #9
    I'm beginning to get this. My main difficulty was in working out the analogy between the notion of different observers in space and that in space-time. In space it seems that where the observer is does not affect their metric. In space-time the metric is different for different observers and motion of observers - different motion means a transformation to a frame with a different metric equation, the way an observer in that coordinate system calculates the space-time distance.
    From the equivalence principle is a far way observer not the same as one in free fall? Does that mean that the Schwarzschild metric for such an observer can be made locally Euclidian? It seems that 1-2GM/rc^2 = 1 can only be true for M = 0 or when r goes to infinity, i.e. when the observed spatial distance is in flat spacetime also. Do both the observer and the observered have to be in freefall? That does not make sense.
    As you can see, my trouble lies in considering both the observer and the observed. In SR it was easy since we just had relative motion to deal with, but in GR both observers (the observer measuring the proper time and the remote observer) Have their positions etc. relative to the source of the spacetime curvature.
    I am relatively (no pun intended) familiar with tensor notation, co- and contravariant components and how they transform.
    Thank you for your time/patience.
  11. Dec 2, 2004 #10


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    You write:

    Let's go back to the round Earth for a bit.

    ds^2 = C^2 dA^2 + (C cos(A))^2 dO^2

    The anologous observation would be -- cos(A) is equal to 1 only on the equator. What's special about the equator?

    The answer is that there's nothing really special about the equator. It's a result of the particular coordinates chosen that cos(A) = 1 there.

    You don't seem to quite be getting the point I'm trying to make - that the value of the metric tensor at a given point depends on the coordinate system chosen. Just as you can consider the Earth's surface to be basically flat whever you go, you can consider space-time to be "Minkowskian" everywhere you go.

    But just because the Earth is flat everywhere you go doesn't mean that in some specific coordiante system that the metric tensor doesn't vary. Only on the equator does g_oo, the metric tensor for the longitude component, have a value of unity. Contrast this to saying that only at spatial infinity does the mtric tensor for g_00, the time component of the Schwarzschild metric, have a value of unity. Both of these are statements about coordinate systems, not statements about geometry.
  12. Dec 3, 2004 #11
    A transformation matrix.
    I'm not sure what you mean by [itex]x^{\alpha}[/itex]. Is this supposed to be the component of a 4-vector or is it supposed to be a coordinate. If its a coordinate then it is not a component of a 4-vector in general relativity. As such [itex]x_{\alpha}[/itex] is not the component of a 1-form. In GR I'm not even sure if [itex]x_{\alpha}[/itex] is defined. Although I suppose that if it is defined then it'd be [itex]x_{\alpha} = gr_{\alpha\beta}x^{\beta}[/itex].

    Way above you wrote
    speeding electron made an error in that c^2 should have been just c. By definition

    [itex]ds^2 = g_{\alpha\beta}dx^{\alpha}dx^{\beta}[/tex]

    This is the spacetime interval between two events [Note to speeding_electron - I prefer to call this an "interval" since "distance" brings to mind things you can measure with a rod. Just my personal opinion though. But I think its not an uncommon opinion.]

    If the interval is timelike then there exists a frame of reference in which the events occur at the same place. The time interval measurd by such an observer is then labeled [itex]\tau[/itex]. The relationship between the spacetime interval and the proper time is then

    [tex]ds^2 = c^2d\tau^2[/tex]

    As such [itex]ds = cd\tau[/itex] and therefore [itex]d\tau = ds/c[/itex]

    The geometric meaning is as discussed above. The physical meaning is that the [itex]g_{\alpha\beta}[/itex] is a set of ten potentials called the gravitational potentials. It replaces Newton's scalar potential [itex]\Phi[/itex]. This is why [itex]g_{\alpha\beta}[/itex] is often refered to as the tensor potential. In at least one place Misner, Thorne and Wheeler call [itex]g_{\alpha\beta}[/itex] the Einstein potentials.

    Its interesting to note that Einstein didn't think of GR as geometerizing the gravitational field anymore than he thought that SR geometerized the EM field. Einstein once wrote in a letter to Lincoln Barnett
    Last edited: Dec 3, 2004
  13. Dec 3, 2004 #12


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    L isn't usually called a tensor because when you shove a vector into it you're changing the basis of the vector, not the vector itself (though arguably it's a tensor of type (1,1) as it maps a vector onto itself, but that's not really what it's there to do).
  14. Dec 4, 2004 #13
    A tensor doesn't map a vector onto itself. And to be a tensor a quantity must have all the properties of a tensor, not just one of them.

  15. Dec 4, 2004 #14


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    There is a tensor that maps a vector onto itself, i.e. the (1,1) identity tensor, but as I siad that the transformation matrix performs a slightly different function from this, so it is not regarded as a tensor.
  16. Dec 5, 2004 #15


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    Just to close the loop, MTW (one of my textbooks) also refers to L as a transformation matrix. If there were an actual vector that were being linearly transformed (for example, rotated), then it would be fair to call L a tensor, as per jcsd's point - but there isn't.
  17. Dec 7, 2004 #16
  18. Dec 8, 2004 #17


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  19. Dec 8, 2004 #18
    There is a debate in GR as to whether curvature is a local phenomena. Some say that curvature can be detected locally,i.e. in an arbitrarily small region of a manifold (e.g. Ohanian/Rufini, Taylor/Wheeler, MTW). Some say that "local" refers to regions that are so small that our instruments are not precise enough to detect curvature (Rindler).

    Each makes sense.

  20. Dec 8, 2004 #19
    When you decompose curvature R by trace, you get 3 terms.

    R = G + E + C ...(1)

    where G depends only on the curvature scaler and has 2 non zero traces, E depends only on the trace free piece of the Ricci tensor and has only one nonvanishing trace, and C is Weyl's conformal tensor- the totally traceless remains of the curvature. Applying Einstein's equations to these pieces puts the curvature scaler proportional to the trace of the energy-momentum tensor and the trace free part of the Ricci tensor proportional to the trace free piece of the energy-momentum tensor. So the traces of curvature are locally determined by the energy-momentum tensor and Einstein's equations. These equations make no demand on C. To find C one must solve the equations with appropriate boundry conditions to find the metric and the build the curvature from the second partials of the metric and finally subtract the traces out of the curvature. C is merely what it has to be. People claim that C is the non-local part of the curvature, but this is only true in the empty space outside the matter. The second Bianchi identity can be written as the vanishing divergence of the double dual of the curvature:

    Div *R* = 0 ...(2)

    Calculating the double duals of G, E, and C yields

    *G* = - G, *E* = E, *C* = - C ...(3)

    Combining these three equations gives

    Div E = Div G + Div C ...(4)

    Outside the matter G and E vanish so there (4) becomes

    Div C = 0 ...(5)

    This is the equation for the non local piece of curvature.

    Inside the matter where G or E do not vanish equation (4) holds. Trivially all three divergences in (4) could vanish, but generally this will not happen. Since G is merely the curvature scaler times an asymmetric kronecker delta, its non zero divergence can only have a trace and no traceless piece. However, C is traceless and so its non zero divergence can only be traceless. Since E has a single trace, its non zero divergence will generally have a trace and a trace free piece ensuring (4). Thus the piece of C that has a non zero divergence must also be a local piece of the curvature. Invariantly splitting C into a divergence free non local piece CNL and a local piece CL then allows the decomposition of the full curvature into

    R = G + E + CL + CNL ...(6)

    where G, E and CL are the locally determined pieces of curvature satisfying

    Div CL = Div E - Div G and Div CL non zero ,,,(7)

    and CNL is the non local piece of curvature arising from distant matter and obeying

    Div CNL = 0 ...(8)

    Although G, E, and CL are very different algebraically and geometrically, they are made of from the same functions describing the matter which ensures their common support. Just as one can add in a source free electromagnetic field to an electromagnetic field with sources (while adjusting the boundry conditions), one can add in a divergence free conformal tensor (also adjusting the boundry conditions) to mimic a passing gravitational wave.
    Last edited: Dec 8, 2004
  21. Dec 8, 2004 #20
    Can someone tell me the key differences between a frame in free-fall and one at rest in a gravitational field. For example: if monochromatic light is sent radially outward in the field described by the Schwarzschild metric, how will the observed frequency differ between an observer higher up at rest and an observer at that same higher position but in free-fall?
  22. Dec 8, 2004 #21


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    If the two observers (one free-falling, one fixed to the coordinate system) happen to be at the same point in space with the same velocity when they receive the light signal, they will not detect any difference in frequency.

    If the two observers have different velocities at the same point, the difference in frequencies will be given by the usual SR doppler shift equation.

    The relative acceleration won't matter as far as the measured value of doppler shift goes if the observers are at EXACTLY the same point in space.
  23. Dec 8, 2004 #22


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    Have you seen my side comment in the "Gravitomagnetism and GR" thread
    https://www.physicsforums.com/showthread.php?p=388636#post388636 ,
    which references an equation in Hawking/Ellis... effectively Div C=J, where J is formed from the gradient of a linear combination of Ricci and its trace. (It essentially arises from the decomposition of Riemann and the Bianchi Identity.)

    Do you have any insight into this equation?
  24. Dec 8, 2004 #23
    Please be more specific. What do you mean by "decompose curvature R by trace"? What is this "curvature scalar" that you speak of?

    If you know of somwhere which proves these assertions you've made please post tex reference or URL. Thanks.

    I see no relationship between what I posted and what you've posted here. You have spoken entirely mathematically and have not addressed the physical aspect I just spoke to. What I referred to had nothing to do with energy-momentum at a given point in spacetime.

    I see no point in what followed relating to what I said. Sorry.

  25. Dec 8, 2004 #24
    Yes I saw it and liked it and almost replied, but my reply didn't seem to fit the thread. I think I might have some insights- see my paper gr-qc/ 0410043.
  26. Dec 8, 2004 #25

    For a reference on this decomposition see Stephani et al , Exact Solutions of Eintein's Field Equations, 2nd ed. pg 37, C.U.P. (2003). I was unaware of this ref and so had to work it out myself. The curvature scaler is just the trace of the Ricci tensor.

    Relevant equations can be found in my paper gr-qc/ 0410043.

    Ultimately the curvature tensor is determined by the matter distribution, boundary values and Einstein's equation and I thought you were asking how much of the curvature at a point was determined by the matter at that point (the local curvature) and how much of the curvature at any point was determined by distant matter (non local curvature). I'm sorry I misunderstood you. Please tell the physical point I missed.
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