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Objects move into gravity=shorter time?

  1. Nov 21, 2011 #1
    Although I can follow many of the equations, given the Schwarzschild metric, I am stuck on what seems like a contradiction between two things.

    1) Objects follow geodesics, which exhibit the principle of extremal aging, meaning most (or least) time along the path between given start/endpoints.

    2) Decreasing distance from a center of gravitational mass corresponds to slower clocks (and increasing distances), as compared to far-away observers.

    My understanding of these two ideas would imply that things move away from gravity, not toward it. What do I have backwards?
    In trying to understand curvature at its most basic, I picture a particle released from a position at rest above the earth; then I imagine three nearby events in space time: 1) the particle is motionless during the time-interval, 2) the "next event" in the direction of gravitational mass (down), and 3) the "next event" in the opposite direction (up). Someone show me why event 2) is the natural one because the spacetime interval to it exhibits extremal aging.

    (Bonus: in what cases does "extremal aging" turn into minimal rather than maximal?)
     
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  3. Nov 21, 2011 #2

    PeterDonis

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    If you read your principle #1 carefully, you will see that it contains the answer to your dilemma. I'll quote it and emphasize the key phrase:

    In other words, the principle of extremal aging only tells you an object's trajectory if you already know the start and endpoints of the trajectory. It does not tell you what those start and endpoints are; you have to figure those out some other way.

    So, for example, if I am on the surface of the Earth and I throw a ball up into the air, and I stipulate that it must return to me exactly ten seconds after I throw it, then that fixes the start and endpoints of the trajectory, and the principle of extremal aging, combined with your principle #2, which tells how the "rate of aging" varies with height, tells me that the ball's trajectory will be a parabola (well, strictly speaking it will be the arc of an ellipse which can be approximated very closely by a parabola, but we'll ignore fine points of that sort here).

    However, if I am standing on the Earth and release an object and watch it fall, the principle of extremal aging by itself can't tell me its trajectory, because I only know its start point, not its end point. Only if I already know its end point by some other means can I apply the principle of extremal aging to calculate its trajectory. (Of course that invites the question, how do I figure out its end point? See below for further comments on that.)

    In summary, the principle of extremal aging does not say, categorically, that "objects will move to where they age faster". It only says that, given specified start and end points, the geodesic trajectory between them will be the one with extremal aging.

    Per the above, the principle of extremal aging does not say that "objects moving on geodesics will move to where they age faster". So that principle by itself cannot choose between these three alternatives. In order to see what *does* choose between them, we need a better way to think about geodesics.

    Try to visualize the trajectory of the particle in the above scenario as a curve in spacetime; the usual term for such a curve is "worldline". We know one point on this worldline, the point where the particle is momentarily at rest relative to the Earth. At that point, the particle not only has a "position" in spacetime; it also has a "velocity" in spacetime, called a 4-velocity, which is a 4-vector (a vector with four components, one time and three spatial components) that is tangent to the worldline at the given point. (I realize you can't visualize 4 dimensions, but we can make do with three or even two by suppressing one or two of the spatial dimensions. The key is to see the worldline as a curve with a tangent vector at each point.)

    The rule for a geodesic can now be stated as follows: a geodesic is a curve whose tangent vector does not change along itself. But we have to be careful about the meaning of "does not change", because we're in a curved spacetime. It turns out that, in the vicinity of a massive object, a curve whose tangent vector does not change along itself is a curve whose tangent vector bends inward towards the massive object. (Remember we're talking about curves in spacetime, so "bends inward" just means the curve's tangent vector is "pulled" in the spatial direction of the object as the curve goes forward in time.) That is what picks out alternative #2 of your three alternatives as the one that actually happens.

    Actually, a better way to state this would be: the worldline that appears to bend inward towards the object is actually "straight"; it's only our skewed perceptions, which think of curves that stay at the same distance from the object as "straight", that make us think the actual geodesic worldline is "curved". If we were able to perceive the full 4-dimensional reality directly, we would see that the actual geodesic path is the straightest possible one given the curvature of spacetime in the vicinity, just as a great circle, though it looks "curved" to us, is the straightest possible path on the curved surface of a 2-sphere.

    At this point you may be wondering, where does the principle of extremal aging come in at all, since it seems like I can determine an object's geodesic trajectory without ever using it? The answer is that yes, you can determine the object's trajectory without ever using the principle of extremal aging. That principle is, at least in my view, more of a restatement of a particular characteristic of geodesic worldlines, than a rule that you can always apply to calculate what those worldlines are. But the principle still holds; for example, if we take the geodesic trajectory that we determine by the criterion I gave above (the curve's tangent vector does not change along itself), and pick a point on it as the "endpoint" (for example, the point at which the object hits the ground), then we will find that the geodesic curve is the curve of maximal aging between the given start and end point. (You can kind of see how it goes by observing that the geodesic curve accelerates towards the object, so the object spends more time at higher heights, where aging is faster, than at lower heights, where aging is slower, since it moves faster the lower it gets.)

    I've only given a very brief overview of how this question is answered; there's obviously a lot more that could be said, but it's hard to know what further things to focus in on. If you have further questions, please feel free to ask them. (Also, I'm holding off on addressing your bonus question until we've had some more discussion on the above.)
     
    Last edited: Nov 21, 2011
  4. Nov 21, 2011 #3

    Matterwave

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    I have 2 things to add to Peter's response.

    1) The Lagrangian method (i.e. the method to extremize the proper time in this case) is not so "bad" as it seems. You don't really need to "know" the endpoint to use this method, you only must posit that a certain endpoint exists and that the endpoint is not varied. So, one need not state "I know that at the end of the trajectory, the particle will be at (t,x,y,z)", one only needs to state "I know that there IS SOME end of the trajectory in the space-time, I don't know what it is, but it's already determined because I know the initial conditions". The Lagrangian method (specifically the Euler Lagrange equations) turn your global statement (minimization of action along the path) to a local statement (a second order differential equation).

    2) Peter's description of a Geodesic as one on which the tangent vector "does not change" is what is most often referred to as an "affine geodesic". More mathematically precisely, the affine geodesic is the curve which parallel transports its own tangent vector. Parallel transport is defined by some affine connection defined on the tangent bundles of your manifold (just a fancy way of saying a rule which helps you pick out what "parallel" means on a curved space-time).

    This description of the geodesic does not require a metric. The metric (time-like) geodesics are ones which extremize proper time. In order to define that, you obviously need a metric. In order that these 2 definitions do not contradict each other, you must limit your connection to be a metric connection.

    But you should note that conversely you don't need to define an affine connection in order to define metric geodesics. Thus, you don't need to posit anything about parallel transport in order to determine the geodesics of your space-time.
     
  5. Nov 21, 2011 #4

    PeterDonis

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    This raises two good points. First, you mention minimization of *action*, as distinct from *maximization* of *aging*. In the case of geodesic motion in a gravitational field, minimizing the action turns out to be the same as maximizing aging along the path, subject to the given initial conditions; but it's worth making the distinction because extremizing the action generalizes to cases that extremizing aging doesn't cover. The second good point is that, given a completely deterministic system, specifying the initial conditions (in this case, the initial position and velocity of the particle) is equivalent to specifying the start and end points of the trajectory (in this case, the starting and ending positions and times). I should have made that clear, since I switched from doing the latter to doing the former in the course of my post. :redface:
     
  6. Nov 21, 2011 #5
    PeterDonis: Thanks for your reply. I understand what you're saying about not knowing the endpoint, and I'm sure this is related to my confusion about how to understand the direction of curvature. But you've only alluded to the answer I'm looking for.

    "It turns out that, in the vicinity of a massive object, a curve whose tangent vector does not change along itself is a curve whose tangent vector bends inward towards the massive object. (Remember we're talking about curves in spacetime, so "bends inward" just means the curve's tangent vector is "pulled" in the spatial direction of the object as the curve goes forward in time.) That is what picks out alternative #2 of your three alternatives as the one that actually happens."

    So what I'm trying to understand is why, in terms of the Schwarzschild metric, the tangent vector is "pulled" in the spatial direction of decreasing radius. I thought that the right way to express the "pull" was in terms of extremizing proper time, so that it would be longer in time to the natural "next event" in that direction. Is that wrong?
    I must admit I am a bit hazy as to how proper time plays the role of the Lagrangian, but I do know what a Lagrangian is. Also, feel free to talk Christoffel symbols if you have to. I should be able to follow a derivation involving tangent vectors. But in the end I am still hoping to boil it down to a "more time, less space" kind of an explanation that could be given without a detour through Euler-Lagrange.

    I still find this bit tantalizing:

    "…then we will find that the geodesic curve is the curve of maximal aging between the given start and end point. (You can kind of see how it goes by observing that the geodesic curve accelerates towards the object, so the object spends more time at higher heights, where aging is faster, than at lower heights, where aging is slower, since it moves faster the lower it gets.)"

    I can certainly see that the falling object spends more time at greater heights where aging is faster. But this doesn't help me see why it first moves into a slower-aging altitude. Again, I focus on that first instant of motion, when the object decides to move down rather than up or stay in place. Wouldn't it age faster if it didn't fall at all?
     
  7. Nov 21, 2011 #6
    (after reading #4) You guys both seem on top of it, so I have no doubt you'll get me to see the light. In the case of the apple in the instant before it leaves the tree, I'm picturing an initial velocity of zero. In flat spacetime, the tangent vector to the apple's worldline would point along the straight path to the same location at the next moment, but in the earth's curved spacetime the tangent vector no longer coincides with the line--it must be pointing down? (or maybe only in the next instant, after it already started to fall, does the tangent vector point down?)
     
  8. Nov 21, 2011 #7

    PeterDonis

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    In terms of trying to understand what determines the motion in a local sense, which is what you are trying to do, I don't think the principle of extremal aging is useful, because, as I said before, the principle does not say that objects moving on geodesics always move to where aging is faster. It only says that geodesics are worldlines of extremal aging, given the constraints (either start and end points, or initial conditions). But there may be multiple worldlines through a given event that satisfy the extremal aging property, just with different constraints.

    For example, to jump to the end of your post:

    It depends on the initial conditions. There is a set of initial conditions for which the object would age faster if it didn't fall at all: the initial conditions such that the object has exactly the right velocity to be in orbit about the Earth at its current height. But you specified a different set of initial conditions, that the object was at rest relative to the Earth at some instant of time. At any given event in spacetime, there are *many* geodesics passing through the event, corresponding to all the possible velocities that a freely moving particle could have there. So there isn't a unique "natural next event" for an object at that event; the "natural next event" depends on the object's velocity at the given event, i.e., on the initial conditions/constraints.

    I'm afraid I don't have a handy way of explaining why minimizing the action is equivalent to maximizing proper time for this scenario. I normally think about this type of situation in terms of spacetime curvature and which lines are "straight" lines in the curved spacetime. For example, in response to your next post:

    Yes.

    In curved spacetime, the "line" of constant height (or constant radius r) is no longer a straight line. It looks straight to our skewed perceptions, but it's really curved. The "straight" line (in spacetime, remember) is the worldline that the falling apple follows.

    A tangent vector always points "along" a curve at any given point; that's true for both geodesic and non-geodesic curves. The difference is that for a geodesic, the tangent vector doesn't have to "change" from point to point to keep pointing along it; for a non-geodesic curve, it does. In the case of the falling apple vs. another apple that stays on the tree, the apple that stays on the tree has to change its tangent vector from point to point to keep it pointing along its worldline. The falling apple does not.
     
  9. Nov 21, 2011 #8
    I understand that the worldline of the object at constant height is not straight, and that the worldline of the falling apple is straight. What I don't understand is how the space and time coordinates are stretched in such a way as to make these statements true. I understand how the stretch factor (1-2M/r) works, but I'm missing something about how it gets applied in the right direction.

    Maybe thinking about extremal aging was the wrong way to go. If I could just see how compressed time (and extended space?) leads to "straightness" being bent in the direction of the compression, I'd be happy.

    (By the way, one thing that got me going in this confused direction was thinking about the rubber sheet that all popular explainers use to illustrate curved spacetime. In that picture, there is more space around the ball on the sheet; but the picture of the sheet leaves out the fact that there is less, rather than more time, around the ball.)

    I read the first half of Taylor&Wheeler's book. I followed the derivation of initial acceleration for an object dropped from rest from a given radius. But at the end they said something like, "We choose the negative square root because we know the object is moving toward the gravitational mass." This seemed like cheating. Of course we know which direction the object will fall, but I want to see the direction emerge from the theory, or the metric, or something.
     
  10. Nov 21, 2011 #9
    I do not get that, if we ignore rotation and throw it straight up it would come straight down. In that case we have a line up and a line down.

    In solving this problem we have to take the 10 second proper time on the surface of the Earth and convert it to coordinate time. Then take half of this time to calculate the apogee. Then we can solve the velocity needed to throw the ball. No need for any (part of a) parabola or an ellipse.

    If we toss a clock in this way we will find this clock shows a later time than the clock that stayed on Earth hen it eventually meets the other clock back on Earth.
     
    Last edited: Nov 21, 2011
  11. Nov 21, 2011 #10

    Matterwave

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    I think you are still trying to visualize the vector in 3-D and that's what's throwing you off. The vectors in GR are 4 dimensional vectors. The 4 velocity vector of an object that is at fixed spatial coordinates is (1,0,0,0). This vector "points in the direction of time". It doesn't point ANYWHERE spatially (it's 3 spatial components are 0!)
     
  12. Nov 22, 2011 #11
    "The 4 velocity vector of an object that is at fixed spatial coordinates is (1,0,0,0). This vector "points in the direction of time". It doesn't point ANYWHERE spatially (it's 3 spatial components are 0!)"

    I know. That's the apple on the tree. But the next tangent vector to the falling-apple worldline is something like (1,0,0,-1). Curved spacetime is equivalent to acceleration or change in velocity. My question is how the curvature translates into motion toward higher gravity, which is motion into slower aging. I'm still trying to understand Peter's "the curve's tangent vector is "pulled" in the spatial direction of the object as the curve goes forward in time". I want to see the calculation of this "pull" and how it relates to the metric.
     
  13. Nov 22, 2011 #12

    Matterwave

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    As the object acquires speed, it's tangent vector is rotated from being purely in the time-like direction to having spatial components as well. The 4-velocity evolves according to the geodesic equation. The coordinates of the particle obey the differential equation:

    [tex]\frac{d^2 x^\mu}{d\lambda^2}+\frac{dx^\rho}{d\lambda} \Gamma^\mu_{\rho\tau} \frac{dx^\tau}{d\lambda} = 0[/tex]

    You solve that second order differential equation. As a second order ODE, the differential equation requires 2 initial values, namely the initial coordinates, and the initial 4-velocity.
     
  14. Nov 22, 2011 #13

    PeterDonis

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    It would actually be something like (1, 0, 0, -v) where v is some small velocity (in units where c = 1, v << 1 to start with, since the object initially falls slowly). A tangent vector of (1, 0, 0, -1) would describe a light ray moving radially inward (assuming that the last spatial component is the radial one).

    The "pull" is related to the Christoffel symbols; one way to see it is to calculate the 4-acceleration of a worldline that stays at a constant radius r above the gravitating body. For the Schwarzschild metric, this turns out to be (in units where G = c = 1)

    [tex]a = \sqrt{g_{rr}} a^{r} = \sqrt{g_{rr}} \Gamma^{r}_{tt} u^{t} u^{t} = \frac{M}{r^{2}} \frac{1}{\sqrt{1 - \frac{2M}{r}}}[/tex]
     
  15. Nov 22, 2011 #14

    PeterDonis

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    You're right, I was mixing up scenarios: the parabola (or ellipse if we're purists) applies to one person throwing a ball to another who is some distance away. For the case I actually described, the spatial trajectory is just a vertical line, ignoring rotation.

    True, I was only describing the spatial trajectory (and for a slightly different case, as above), not the full description of the motion. I really should have described a worldline in spacetime, which would, as you say, require solving for the coordinate time and thus the initial velocity, and using the initial position and velocity to determine the geodesic followed.

    Yes. In fact, the clock that follows the free-fall, geodesic trajectory will show a longer time elapsed than any other clock that travels between the same two events on some other worldline (which will, of course, be a non-geodesic worldline, like the worldline of the clock that stays on the surface of the Earth).
     
  16. Nov 22, 2011 #15

    PeterDonis

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    I should also note that these 4-velocities assume an orthonormal frame; 4-velocities referred to standard Schwarzschild coordinates will *not* look like this. For example, in standard Schwarzschild coordinates, the 4-velocity of a worldline that stays at a constant radius r is

    [tex]u = (\frac{1}{\sqrt{1 - \frac{2M}{r}}}, 0, 0, 0)[/tex]
     
  17. Nov 22, 2011 #16
    Well, I'm starting to think that either my question is asking too much, or I'm too dense. I appreciate that we've gotten as far as bringing in the Christoffel symbols; maybe we're almost there… But before I get lost in the Christoffel forest, can we try once more from a conceptual level?

    Peter's result, (M/r^2)(1-2M/r)^-1/2, looks similar to but different from a result for acceleration from rest given by Taylor&Wheeler (p.3-13): -(1-2M/r)(2M/r)^1/2
    The biggest difference I see is that one is positive and one is negative! Maybe you can say that the minus sign is "understood", but this is my main point of interest: how does curvature tell you the correct sign to take? And can we relate this conceptually to the time stretch factor, that is, to there being "more time" in one direction and "less time" in the other?

    I should admit that I'm unclear about the role of the "g<sub>rr" in your equation. Is this g already understood to be Schwarzschild, or is it what you're solving for?
     
  18. Nov 22, 2011 #17
    I wrote my #16 before reading #15. I do understand that my (1,0,0,-1) would refer to a light ray, I was just being sloppy about the units. But I do not understand the point about orthonormal vs. Schwarzschild coordinates. Perhaps it would help, though, if we compared the vector given in #15 with the vector for the apple in its first instant of falling.
     
  19. Nov 22, 2011 #18

    Matterwave

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    The orthonormal (tetrad) basis versus the coordinate basis issue is somewhat technical. I don't think it's going to help you all that much conceptually.

    Perhaps this will help you think about this problem. Suppose that I have my geodesic equation. Suppose it's given. I start off at a point a on my manifold with some 4-velocity, and I use the geodesic equation to get that after some proper time has ticked by on the clock I am carrying with me, I arrive at a point b on my manifold.

    The statement of "extremal proper time" is simply that if I took ANY OTHER path from a to b than the path I took by going along my geodesic, I would register a shorter proper time on my clock. (Remember that points "a" and "b" are EVENTS in space-time, and not points in space).

    Does that help at all.
     
  20. Nov 22, 2011 #19

    PeterDonis

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    Sure; at least, I think I can be more conceptual to some extent.

    First, for any curve in spacetime, I can construct a mapping from points on the curve to real numbers; this is called "parametrizing" the curve. For timelike curves, which are possible worldlines for objects with nonzero rest mass, the obvious parameter to choose is proper time; thus, each point on the worldline has a unique proper time [itex]\tau[/itex] assigned to it, which uniquely identifies that point among all the points on the curve. Obviously I can pick the "origin" of [itex]\tau[/itex] wherever I like, and I can also change the units of [itex]\tau[/itex] however I like, without affecting the actual physics; so given one parametrization [itex]\tau[/itex], I can construct another parametrization [itex]\tau ' = A \tau + B[/itex], where A and B are arbitrary constants, and [itex]\tau '[/itex] will work as well as [itex]\tau[/itex] for describing points on the worldline. The technical term for all this is that [itex]\tau[/itex] is an "affine parameter". We'll assume that we've fixed the origin and units of [itex]\tau[/itex] in what follows.

    Next, suppose I have some coordinate chart on a given spacetime. A "chart" is just a mapping of events in the spacetime to 4-tuples of real numbers, which are called "coordinates". Abstractly, the coordinates are given "indexes" from 0 to 3 (or sometimes 1 to 4, depending on whether you like your "time" index to be 0 or 4; I prefer 0), so a particular coordinate is written as [itex]x^{a}[/itex], where "a" is the index. The 4-tuple of coordinates at each event can be thought of as a vector; more precisely, it is a vector in a vector space that is "attached" to the particular event on the worldline at which we are evaluating the coordinates. This vector space is called the "tangent space", and there is a separate tangent space at each event.

    Given my coordinate chart, I can construct a mapping between coordinate 4-tuples and points on my worldline, which is equivalent to a mapping between coordinate 4-tuples and values of [itex]\tau[/itex]. We express this mapping by writing each coordinate as a function of [itex]\tau[/itex], thus: [itex]x^{a} = x^{a} \left( \tau \right)[/itex].

    Now that I have the coordinates as functions of [itex]\tau[/itex], I can talk about derivatives of those functions, without having to talk about the specific form of the functions in terms of a particular coordinate chart or metric expression. The 4-velocity [itex]u^{a}[/itex] is just the derivative of the coordinate vector [itex]x^{a}[/itex] with respect to [itex]\tau[/itex], thus:

    [tex]u^{a} = \frac{d x^{a}}{d \tau}[/tex]

    The 4-acceleration, or "proper acceleration" as it is often called, is then the covariant derivative of the 4-velocity. This is where the Christoffel symbols come in:

    [tex]a^{a} = \frac{D u^{a}}{d \tau} = \frac{d^{2} x^{a}}{d \tau^{2}} + \Gamma^{a}_{bc} \frac{d x^{b}}{d \tau} \frac{d x^{c}}{d \tau} = \frac{d^{2} x^{a}}{d \tau^{2}} + \Gamma^{a}_{bc} u^{b} u^{c}[/tex]

    You will note, by the way, that the above expression looks very similar to the geodesic equation that matterwave wrote down; that's because the geodesic equation is just a special case of the above, where the 4-acceleration is zero:

    [tex]a^{a} = \frac{d^{2} x^{a}}{d \tau^{2}} + \Gamma^{a}_{bc} u^{b} u^{c} = 0[/tex]

    Matterwave's version had [itex]\lambda[/itex] instead of [itex]\tau[/itex], but [itex]\lambda[/itex] is just an affine parameter and we've already seen that any affine parameter we pick will work equally well. The reason [itex]\lambda[/itex] is often used is that the above expressions are actually not limited to timelike curves; they apply to any curve in the spacetime. We're only considering timelike curves here so I'll stick with [itex]\tau[/itex].

    You'll note that all of the above is perfectly general; I haven't made any assumptions at all about the coordinate chart (except for assumptions about continuity, differentiability, etc., that are necessary in order to meaningfully talk about this stuff at all). But suppose I also have an expression for the metric in my chosen coordinate chart; the metric is a tensor [itex]g_{ab}[/itex] (the tensor is symmetric, so it has ten independent components in 4-D spacetime), which acts on the coordinates to produce a "line element" that describes how [itex]\tau[/itex] changes along a small differential element of a curve that I have specified as above. The general form of the line element is:

    [tex]d \tau^{2} = g_{ab} dx^{a} dx^{b}[/tex]

    where I have used the Einstein summation convention (repeated indexes in an expression are summed over). The Schwarzschild metric is a particular case of this general expression; looking at the line element, you should be able to read off the components [itex]g_{ab}[/itex] by looking at the coefficients of each combination of coordinate differentials. (The Schwarzschild line element is particularly simple because only the diagonal metric coefficients, where a = b, are nonzero.)

    Hopefully the above will help to make more sense of some of the previous posts; I'll comment on one particular aspect further below.

    Are you sure the latter expression is for proper acceleration? It looks like a coordinate velocity dr/dt for an infalling observer.

    The full answer will require digging into those Christoffel symbols and such in more detail. But one general comment to make is that before interpreting any signs, you have to make sure that the sign conventions being used are consistent. There are a number of different ones in the literature, and many times what appears to be a difference in physics is actually just a difference in sign convention.

    (In this particular case, as I said in my previous comment above, I think the expression you've taken from Taylor & Wheeler is for the coordinate velocity dr/dt of an infalling observer, which will be negative because r is decreasing with increasing t.)

    The sign convention I was using for my expression for the 4-acceleration (or proper acceleration, as it is often called) of a worldline that stays at a constant radius r for all time, is that positive acceleration means the 4-velocity vector is being pushed "outward", i.e., in the positive r-direction, relative to what it would be if the worldline were a geodesic starting from the same event. However, we can see how this works in a bit more detail by looking specifically at the "r" component of the equation above for the 4-acceleration:

    [tex]a^{r} = \frac{d^{2} r}{d \tau^{2}} + \Gamma^{r}_{bc} u^{b} u^{c}[/tex]

    Now for a worldline where r stays constant for all time, the first term on the RHS above is zero; so all we have is the second term. And, since for such a worldline the only nonzero component of the 4-velocity is the "t" component, all that survives of the second term on the RHS above is the b = c = t component. That gives most of the equation I wrote down in my earlier post; the only other piece we need is that the actual physical magnitude of a 4-vector is given by contracting it with the metric:

    [tex]| a | = \sqrt{g_{ab} a^{a} a^{b}}[/tex]

    Since the only non-zero component of the 4-acceleration is the r-component (I haven't shown this, but it can be seen by working out the other Christoffel symbols and seeing that there are no other nonzero ones which have both "downstairs" indexes of t, which are the only ones that would survive after being contracted with the 4-velocity), the actual physical magnitude of the 4-acceleration, which is what an observer following the given worldline would actually measure with an accelerometer, is the expression I wrote down.

    I should have clarified, though, that that acceleration is *not* quite what you were asking for. What you were asking for was the *coordinate acceleration* of a freely falling object (i.e., one traveling on a geodesic), relative to an observer who is at a constant height. But it should be apparent that that is just the inverse of the acceleration that I wrote down; i.e., if you put a minus sign in front of the expression I wrote, you have the expression for the acceleration that you were looking for. (Strictly speaking, it's only "apparent" for the particular event where the free-falling object is momentarily at rest at some radius r; once the free-falling object is moving inward, the [itex]\Gamma^{r}_{rr}[/itex] Christoffel symbol also comes into play since the r-component of the 4-velocity is now nonzero. But it only takes a little more work to see that the final answer still comes out the same; the t-component of the 4-velocity changes in just the right way to offset the additional term coming in from the r-component.)

    I'm not sure. At any rate, I don't think we're ready to bring that factor in yet.

    Hopefully the above clarifies that; [itex]g_{rr}[/itex] is one of the metric coefficients, and is given once you've determined what spacetime you're in and what coordinate chart you're using.
     
  21. Nov 22, 2011 #20

    PeterDonis

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    Answering this will expand somewhat on the last part of my previous post.

    At the instant when the apple is momentarily at rest at radius r (i.e., it has *just* been released), its 4-velocity is the same as the vector I wrote down in #15. What is different is its rate of change:

    (1) For the stationary object (say an apple still hanging on the tree, right next to the one that's just started to fall), the *coordinate* acceleration at this instant is zero; but the *proper* acceleration is positive (it's the expression I wrote down).

    (2) For the freely falling object (the falling apple), the *proper* acceleration is zero, but the *coordinate* acceleration is the negative of the expression I wrote down.

    (Putting these two statements together with the expression for proper acceleration I wrote down in my last post should make it clearer why the second part of #2 is true.)

    So "an instant later", so to speak, the freely falling apple will have acquired a small r-component to its 4-velocity, in the *coordinate* sense, and the t-component will have changed so as to keep the overall length of the 4-velocity equal to 1 (since the 4-velocity is always a unit vector). But the apple still hanging on the tree will not have changed its 4-velocity at all, in a *coordinate* sense. I say "in a coordinate sense", because physically, the falling apple is the one whose 4-velocity has not changed, and the hanging apple is the one whose 4-velocity has changed. But the coordinates are skewed so that in coordinate terms it "looks" the other way around.
     
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