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I Morris-Thorne Wormhole Metric Terms

  1. Jul 4, 2016 #1
    Here is the Morris-Thorne Wormhole line element:
    ds2 = - c2dt2 + dl2 + (b2 + l2)(dθ2 + sin2(θ)d∅2)

    Now my main question here (even though I've asked this before, but never quite understood) is:

    What exactly is l?

    I know that b is the radius of the throat of the wormhole. I know that the rest of the terms in this line element besides l itself are just the usual terms that appear in a line element that has a spherical basis. I just don't know what l is.

    From what I've gathered, l can range from - ∞ to ∞. Two different signs for l represent two different universes.. Apparently, large l absolute values represent flat Minkowski space-times. Finally, I've read that increasing l corresponds to either increasing or decreasing energy density,or something to that nature.

    Those are some properties about certain values of l, but I still don't know what l itself is. I don't know what l = 1 or l = 5 or any other random l value would mean. Please help me on this.

    On another note, based on the fact that the line element has a spherical basis that this wormhole is spherically symmetric. I can also tell that it is static since no functions of time are involved in the line element. Would this description of the wormhole be correct? I think it would really help me understand this wormhole if I could get a visual in my mind of what the wormhole looks like.
  2. jcsd
  3. Jul 4, 2016 #2


    Staff: Mentor

    Well, there's one obvious coordinate that usually appears in a spherically symmetric line element, which is missing; and ##l## certainly appears to be there in its place. Look at it this way: what happens as ##l## goes to infinity?
  4. Jul 5, 2016 #3
    As l goes to infinity, ds2 goes to infinity. I can see that l seems to replace the radial term r in the line element. However, even that is made somewhat ambiguous due to the fact that (b2 + l2) is where r2 would usually be, while dl2 simply replaces dr2. Also, even if you say that l really is just the radial term, there has to be some special physical meaning behind l (kind of like how the physical meaning of b is the radius of the throat of the wormhole). If there is no physical meaning for l that differs from the r in other metrics such as Schwarzchild's metric, then what is the point of replacing r with l? This is also confusing because if b is already the radius of the wormhole's throat, then what other radial aspect is there to account for? A wormhole in theory only consists of its openings and the inside (which is the throat). Is l perhaps the length of the throat (whereas b is the radius of it)? The only thing that I can think of as to what l is, is possibly a type of layer ranking that measures how deep or how far into the wormhole you've gone. In other words, perhaps l = 1 means you've just entered the wormhole, and l = 5 means you have gone 5 units of distance or have gone through 5 different sets of decreased energy density through the wormhole (or something like that)? Is that it? Please help me grasp what l is.
  5. Jul 5, 2016 #4


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    Staff: Mentor

    The choice of label is arbitrary, so you aren't going to find any solid reason for why Morris and Thorne used the letter ##l##. Maybe they thought that people would expect something named ##r## to range from zero to infinity, whereas the coordinate they named ##l## ranges from minus infinity to plus infinity? Maybe the ##r## key on their typewriter was broken?
    The metric is describing an entire spacetime that happens to contain a wormhole with throat diameter ##b## somewhere. One way or another, we're going to need three spatial coordinates and one time coordinate to describe that spacetime.
    1) What spacetime is this metric describing if we take ##b=0##? What is the physical significance of a surface of constant ##l## and ##t## in this case?
    2) Returning to the case in which ##b## is non-zero, is there a point in space that we could call the center of the wormhole? (Hint: if such a point exists, you will be able to label it with some coordinate values).
    3) Consider a three-dimensional space of constant ##t##. As ##l## ranges from plus infinity to zero and then towards minus infinity, how does the area of a surface of constant ##l## vary? Compare this result with the result for the case in which ##b## is zero.
  6. Jul 5, 2016 #5


    Staff: Mentor


    That's because ##b## is a constant while ##l## is a coordinate. So there isn't any ##db^2## because ##b## never changes.

    The real question is what range of values ##l## can take. See below.

    First you need to think carefully about what "the radius of the wormhole's throat" actually means. I suggest working on answering the questions Nugatory asked, particularly his question #3. Also, in addition, think carefully about the question I asked above: what range of values can ##l## take?
  7. Jul 6, 2016 #6


    Staff: Mentor

    To help with discussion, I have run the wormhole metric through Maxima and computed some key quantities.

    The nonzero Christoffel symbols are (not including ones related to the following by symmetry in the lower indexes):

    \Gamma^{\theta}{}_{\theta l} = \Gamma^{\phi}{}_{\phi l} = \frac{l}{b^2 + l^2}

    \Gamma^l{}_{\theta \theta} = - l

    \Gamma^l{}_{\phi \phi} = - l \sin^2 \theta

    \Gamma^{\theta}{}_{\phi \phi} = - \cos \theta \sin \theta

    \Gamma^{\phi}{}_{\phi \theta} = \frac{\cos \theta}{\sin \theta}

    The latter two are just the standard Christoffel symbols in the usual angular coordinates on a 2-sphere.

    These are useful in computing the geodesic equation. Note in particular that an observer who stays at constant spatial coordinates is following a geodesic, since no nonzero Christoffel symbols have a ##t## index anywhere. Also, an observer who moves solely in the ##l## direction is also following a geodesic, since all of the nonzero Christoffel symbols have at least one angular coordinate in a lower index. This latter type of geodesic is worth careful consideration; for example, what happens when an observer following such a geodesic worldline reaches ##l = 0##?

    The Ricci scalar is:

    R = - \frac{2 b^2}{\left( b^2 + l^2 \right)^2}

    The Einstein tensor is ##G^a{}_b = \mathrm{diag} \left( \rho, -\rho, \rho, \rho \right)##, where

    \rho = \frac{b^2}{\left( b^2 + l^2 \right)^2}

    The Kretzschmann scalar is:

    K = R^{abcd} R_{abcd} = \frac{12 b^4}{\left( b^2 + l^2 \right)^4}

    I won't go into much detail about the last three above, but two points are worth noting:

    (1) The Einstein tensor is nonzero, so stress-energy is present (via the Einstein Field Equation), and this stress-energy has negative pressure (i.e., tension) in the ##l## direction equal to the energy density. This means it is a form of "exotic matter" and violates some of the energy conditions that ordinary matter and energy satisfy.

    (2) The scalar invariants are finite at all values of ##l##. (This is unlike the case of, for example, Schwarzschild spacetime, where the Kretzschmann scalar diverges as ##r \rightarrow 0##.) This fact is key in thinking about what range of values the ##l## coordinate can take.
  8. Jul 8, 2016 #7
    I know this reply is late, but thanks for the help. I believe I understand it now. I came up with and solved an example problem involving this metric, and I'd appreciate it if you could tell me if I am correct or not. Here is the problem:

    Trunks enters a Morris-Thorne wormhole of radius b = 20 Ls (light seconds). He enters the wormhole at the coordinates (l,θ,∅) = (20,0,0). His movement along the l coordinate with respect to coordinate time is denoted by the function l(t) = - 0.5t + 20, and dl/dt = - 0.5 Ls/s (so he is moving at half the speed of light through the l coordinate) . He moves along the θ-coordinate at a rate of dθ/dt = π/2 rad/s. He does not move along the ∅ coordinate. Now, Trunks' mother Bulma stays outside of the wormhole at a safe distance l = 30, and she counts a coordinate time t = 40 s. What is the proper time that Trunks experiences in the wormhole when Bulma counts t = 40 s? Assume c= 1 Ls/s.

    My proccess:
    Here's the line element: ds2 = c2dt2 - dl2 - (b2 + l2)(dθ2 + sin2(θ)d∅2)

    τ = ∫sqrt(ds2) which in this case = ∫sqrt(dt2 - dl2 - (b2 + l2)dθ2) = ∫sqrt(1 - (dl/dt)2 - (b2 + l2)(dθ/dt)2) dt = ∫40 0 sqrt[1 - 0.25 - ((800+ 0.25t2 - 20t) * (π2/4))] dt ≈ 1441.94i

    Now I want to clarify a few things:

    1. Just in case it is unclear due to formatting,the bounds on that last integral were 0 to 40.

    2. The imaginary number 1441.94i was what I got when I plugged this integral into an online definite integral calculator.

    3. The reason dl/dt is negative is because I deduced from your explanations that l = 0 would be the center of the wormhole. Therefore, if Trunks was entering the wormhole and approaching the center that connects the two universes, then his l-distance from the center would have to be decreasing rather than increasing.

    Now that I've showed my process, would this be correct?

    If this is correct, then would an imaginary result like this mean that at t = 40, Trunks and Bulma are actually space-like separated (rather than time-like separated), and that the proper distance in between them would be 1441.94 Ls?

    If this isn't correct then where did I go wrong?
  9. Jul 8, 2016 #8


    Staff: Mentor

    The first thing to do is to check whether this results in his worldline being timelike, i.e., with him always moving slower than light. He is obviously moving slower than light in the ##l## direction, but he's not just moving in the ##l## direction. You have to check that he is moving slower than light when all of his velocity components are taken into account.

    Here's how you would check this: first write his spacetime position ##X^{\mu}## as a 4-vector, parameterized by ##t##, since we know all of the other coordinates as functions of ##t##. That 4-vector will look like this (leaving out the fourth, ##\phi## component since it is always zero):

    X^{\mu} = \left(t, -0.5t + 20, \frac{\pi}{2} t \right)

    Now we take the derivative of this with respect to ##\tau## to get the 4-velocity ##U^{\mu}##. However, we have things as functions of ##t##, not ##\tau##. So we'll take derivatives with respect to ##t## and stick a factor of ##dt / d\tau## in front:

    U^{\mu} = \frac{dt}{d\tau} \left(1, -0.5, \frac{\pi}{2} \right)

    Now we want to compute the squared norm of this 4-vector and confirm that it is timelike. That means a squared norm of ##-1## with the metric signature convention we are using. The squared norm is ##g_{\mu \nu} U^{\mu} U^{\nu}##, which is simple since the metric is diagonal: we have, using the formula ##b^2 + l^2 = 400 + 40 \left( -0.5t + 20 \right) + \left( -0.5t + 20 \right)^2 = 400 - 20t + 800 + 0.25t^2 - 20t + 400 = 1600 - 40t + 0.25 t^2##:

    g_{\mu \nu} U^{\mu} U^{\nu} = \left( \frac{dt}{d\tau} \right)^2 \left[-1 + (-0.5)^2 + \left( 1600 - 40t + 0.25t^2 \right) \frac{\pi^2}{4} \right] = \left( \frac{dt}{d\tau} \right)^2\left[ -0.75 + \frac{\pi^2}{4} \left( 1600 - 40 t + 0.25 t^2 \right) \right]

    You can see already that this is not going to always be negative; for example, at ##t = 0## it is obviously positive. If you want you can derive a quadratic equation in ##t## from the above that will tell you at what value of ##t## the term inside the brackets is exactly zero, so you can see where the boundary is where the worldline you defined actually can be timelike. You will see that that boundary is way outside the range of ##t## coordinates you are using.

    See further comments below.

    Yes, this part is fine.

    The integral itself is fine; but you are starting from a premise that isn't physically correct, because the worldline you are having your observer follow isn't timelike in the range you are using.

    No. The imaginary result means you aren't using a timelike worldline for your observer; in other words, you made an unphysical assumption and so you got an unphysical answer.

    As far as whether Trunks and Bulma are spacelike separated, at any events on their worldlines with the same ##t## coordinates, once they are not co-located, they will be spacelike separated, because any surface of constant ##t## in this metric is a spacelike surface, so any two events with the same ##t## but different spatial coordinates must be spacelike separated.

    However, you still need to go back and rework your assumption about Trunk's worldline so that it is always timelike.
  10. Jul 9, 2016 #9
    I think one of us made some arithmetic errors. You got that b2 + l2 = 1600 - 40t + 0.25t2

    I got b2 + l2 = 800 - 20t + 0.25t2

    This is because l(t) = - 0.5t + 20 and b = 20 Ls

    Do you care to explain why you added 40(- 0.5t + 20) in the middle of your b2 + l2 calculation?

    Other than that, I believe I get the concept now, so thank you.
    Last edited by a moderator: Jul 9, 2016
  11. Jul 9, 2016 #10


    Staff: Mentor

    Ah, you're right; I was squaring ##b+l## instead of ##b## and ##l## separately. So my formula for the squared norm of ##U## should be:

    g_{\mu \nu} U^{\mu} U^{\nu} = \left( \frac{dt}{d\tau} \right)^2 \left[-1 + (-0.5)^2 + \left( 800 - 20t + 0.25t^2 \right) \frac{\pi^2}{4} \right] = \left( \frac{dt}{d\tau} \right)^2\left[ -0.75 + \frac{\pi^2}{4} \left( 800 - 20 t + 0.25 t^2 \right) \right]

    That doesn't change the key point I made, which is that this is not negative for the range of ##t## values you are using, so the worldline you specified for Trunks is not timelike.
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