# Objects move into gravity=shorter time?

## Main Question or Discussion Point

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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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:

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

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.
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.)

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Matterwave
Gold Member
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.

PeterDonis
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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).
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.

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?

(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?)

PeterDonis
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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?
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.

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?
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:

In the case of the apple in the instant before it leaves the tree, I'm picturing an initial velocity of zero.
Yes.

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?)
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.

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.

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).
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.

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Matterwave
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(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?)
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!)

"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.

Matterwave
Gold Member
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:

$$\frac{d^2 x^\mu}{d\lambda^2}+\frac{dx^\rho}{d\lambda} \Gamma^\mu_{\rho\tau} \frac{dx^\tau}{d\lambda} = 0$$

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.

PeterDonis
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But the next tangent vector to the falling-apple worldline is something like (1,0,0,-1).
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).

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.
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)

$$a = \sqrt{g_{rr}} a^{r} = \sqrt{g_{rr}} \Gamma^{r}_{tt} u^{t} u^{t} = \frac{M}{r^{2}} \frac{1}{\sqrt{1 - \frac{2M}{r}}}$$

PeterDonis
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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.
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.

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.
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.

If we toss a clock in this way we will find this clock shows a later time than the clock that stayed on Earth when it eventually meets the other clock back on Earth.
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).

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).
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

$$u = (\frac{1}{\sqrt{1 - \frac{2M}{r}}}, 0, 0, 0)$$

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?

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.

Matterwave
Gold Member
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.

PeterDonis
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But before I get lost in the Christoffel forest, can we try once more from a conceptual level?
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 $\tau$ assigned to it, which uniquely identifies that point among all the points on the curve. Obviously I can pick the "origin" of $\tau$ wherever I like, and I can also change the units of $\tau$ however I like, without affecting the actual physics; so given one parametrization $\tau$, I can construct another parametrization $\tau ' = A \tau + B$, where A and B are arbitrary constants, and $\tau '$ will work as well as $\tau$ for describing points on the worldline. The technical term for all this is that $\tau$ is an "affine parameter". We'll assume that we've fixed the origin and units of $\tau$ 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 $x^{a}$, 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 $\tau$. We express this mapping by writing each coordinate as a function of $\tau$, thus: $x^{a} = x^{a} \left( \tau \right)$.

Now that I have the coordinates as functions of $\tau$, 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 $u^{a}$ is just the derivative of the coordinate vector $x^{a}$ with respect to $\tau$, thus:

$$u^{a} = \frac{d x^{a}}{d \tau}$$

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:

$$a^{a} = \frac{D u^{a}}{d \tau} = \frac{d^{2} x^{a}}{d \tau^{2}} + \Gamma^{a}_{bc} \frac{d x^{b}}{d \tau} \frac{d x^{c}}{d \tau} = \frac{d^{2} x^{a}}{d \tau^{2}} + \Gamma^{a}_{bc} u^{b} u^{c}$$

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:

$$a^{a} = \frac{d^{2} x^{a}}{d \tau^{2}} + \Gamma^{a}_{bc} u^{b} u^{c} = 0$$

Matterwave's version had $\lambda$ instead of $\tau$, but $\lambda$ is just an affine parameter and we've already seen that any affine parameter we pick will work equally well. The reason $\lambda$ 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 $\tau$.

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 $g_{ab}$ (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 $\tau$ changes along a small differential element of a curve that I have specified as above. The general form of the line element is:

$$d \tau^{2} = g_{ab} dx^{a} dx^{b}$$

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 $g_{ab}$ 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.

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

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?
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:

$$a^{r} = \frac{d^{2} r}{d \tau^{2}} + \Gamma^{r}_{bc} u^{b} u^{c}$$

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:

$$| a | = \sqrt{g_{ab} a^{a} a^{b}}$$

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 $\Gamma^{r}_{rr}$ 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.)

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

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?
Hopefully the above clarifies that; $g_{rr}$ 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.

PeterDonis
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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.
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.

Peter: I really appreciate the time you're taking here.
Yes, I was confusing coordinate dr/dt with proper acceleration (or perhaps I should say I didn't know it was proper acceleration we should be talking about). (I should at least have realized that I was comparing a first derivative to a second!) Let's forget that bit.

I would naively think that my question should be answered in terms of proper acceleration, because I want to know how the object "feels" the curvature (which I crudely or maybe wrongly express as feeling "more time" in one direction). On the other hand, I know that the object doesn't "feel" anything in the sense that it is freely falling, that is, its proper acceleration is zero. (Maybe I should say that spacetime "guides" the object in the more-time direction?) So maybe my question pertains more to the coordinate sense.

The details of the acceleration equation for the fixed-radius case make some sense, although I don't immediately see how you got your equation (in #13). Is M/r^2 the magnitude (metric contraction) part? If we've simplified the problem enough, maybe you could show the Christoffel calculation for the r-acceleration with b=c=t that you alluded to.

I'm surprised that you still hesitate at my question about how "more time" implies a spatial direction, or how the bend in time, and corresponding bend in space, can be seen to guide objects the right way. This makes me think that there's something wrong with my question, whereas I assumed it had an easy answer.

"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."

I understand the physical point about the constant-height object being pushed outward, away from the geodesic. Yet I still feel like I'm missing something about how we were justified in choosing the sign convention that we know, physically, to be the right one.
Let me try this: suppose that in some bizarre (maybe impossible) kind of field, the coefficients of the dt and dr terms were reversed in the metric. (I think of this as: instead of "less time" with decreasing r, dt>dT, there would be "more time", dt<dT.)[I'm writing "T" for your "tau".] Would that by itself result in geodesics of increasing rather than decreasing radius?

"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)."

Can you specify this change a bit more? How do you describe the change in t-component, the divergence in time between the apple still on the tree and the apple just starting to fall?
Is the r/t tradeoff implied here related to stretch factors?

Matterwave: thanks for your input as well. Maybe what I'm missing is not being able to calculate the "shorter proper time" of the non-geodesic constant-height worldline. Peter's discussion of the individual Christoffel components sounds like it addresses this, but it might take awhile to sink in, or I might just need more calculus practice.

A.T.
But before I get lost in the Christoffel forest, can we try once more from a conceptual level?
Maybe you need some graphics that relate gravitational time dilation to free fall geodesics:

http://www.physics.ucla.edu/demoweb/demomanual/modern_physics/principal_of_equivalence_and_general_relativity/curved_time.gif

Here you can play around with an interactive version of that cone model:

This is the wider picture around a spherical mass:

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PeterDonis
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Peter: I really appreciate the time you're taking here.
No problem. This is fun for me or I wouldn't be doing it.

I would naively think that my question should be answered in terms of proper acceleration, because I want to know how the object "feels" the curvature (which I crudely or maybe wrongly express as feeling "more time" in one direction). On the other hand, I know that the object doesn't "feel" anything in the sense that it is freely falling, that is, its proper acceleration is zero.
Right, a freely falling object does not "feel" anything. And even for a non-freely-falling object, such as you or I standing on the surface of the Earth, although we feel something (our weight), what we feel is not "curvature" in the sense of curvature of spacetime. What we feel is curvature of our worldline--or, put another way, we feel the force that is pushing is out of the geodesic path we would otherwise follow. So curvature of spacetime itself is not really something that can be "felt" at all.

(Maybe I should say that spacetime "guides" the object in the more-time direction?)
No, this isn't quite right either, because as you've already observed, gravity pulls inward, which on the face of it appears to be the direction of "less time", not "more time". Hopefully we'll be able to come back to this at some point.

So maybe my question pertains more to the coordinate sense.
It seems like you are trying to understand what's going on in terms of a specific set of coordinates, yes (Schwarzschild coordinates). There's nothing wrong with that if that's where it's easiest to start from, and it often is because those coordinates are the "natural" ones for our intuition. But the physics does not depend on the coordinates, so it's important to try to get to a point where you are thinking about the invariants, the things that *don't* depend on the coordinates. Everything I wrote in my last post is independent of coordinates; even the formula for the actual proper acceleration experienced traveling on a worldline of constant r, though I used a specific coordinate chart to derive it, is independent of coordinates; you would get the same answer if you calculated it in any other coordinate chart.

The details of the acceleration equation for the fixed-radius case make some sense, although I don't immediately see how you got your equation (in #13). Is M/r^2 the magnitude (metric contraction) part? If we've simplified the problem enough, maybe you could show the Christoffel calculation for the r-acceleration with b=c=t that you alluded to.
In the Schwarzschild coordinate chart, the relevant Christoffel symbol is:

$$\Gamma^{r}_{tt} = \left( 1 - \frac{2M}{r} \right) \frac{M}{r^{2}}$$

The derivation of this is somewhat laborious but straightforward, using the definition of the Christoffel symbols in terms of derivatives of the metric. Combining the above with the formula for the 4-velocity for a worldline that stays at constant r, which has only one non-zero component, as we saw before...

$$u^{t} = \frac{1}{\sqrt{1 - \frac{2M}{r}}}$$

...and the r-r metric coefficient, which can be read off the line element, as I said before...

$$\sqrt{g_{rr}} = \frac{1}{\sqrt{1 - \frac{2M}{r}}}$$

...we can plug these into the equation I wrote in an earlier post to obtain the final expression I gave.

I'm surprised that you still hesitate at my question about how "more time" implies a spatial direction, or how the bend in time, and corresponding bend in space, can be seen to guide objects the right way. This makes me think that there's something wrong with my question, whereas I assumed it had an easy answer.
It may be partly the fact that you're concentrating on time alone, as in experienced time along a worldline, when spacetime curvature is actually more complicated than that. The full description of spacetime curvature is the Riemann curvature tensor, which has twenty independent components in a general spacetime of 4 dimensions. So trying to reduce it to one parameter is always going to leave things out. (Schwarzschild spacetime has some symmetries that reduce the number of independent components, but there are still enough of them to make a one-parameter description insufficient to really capture what's going on.)

One way to think about the curvature of spacetime around a massive body, which may help a bit, is to consider two thin shells of "dust", where "dust" just means a swarm of particles that all move on geodesics--they don't interact with each other at all, so they don't affect each other's motion, but they do provide a way of "marking" various geodesics and seeing how they relate to each other. The two shells are as follows:

(1) A spherical shell of "dust" that surrounds the Earth on all sides, and starts out at rest relative to the Earth, centered on it, so that each dust particle is at the same radius r above the Earth at some instant of time t = 0.

(2) A spherical shell of "dust" that is centered at some radius r above the Earth and has no matter inside it; i.e., it surrounds only vacuum (so the actual center point of the sphere on whose surface the shell particles are is in vacuum).

The key difference between shells #1 and #2, of course, is that one has matter inside it (the Earth) and the other does not. So let's follow what happens to each shell:

Shell #1 will collapse inward; that is, its volume will decrease with time. More precisely, the first derivative of its volume at time t = 0 is zero, but the second derivative is negative. However, it will remain spherical (assuming a spherical Earth inside).

Shell #2, however, will elongate in the radial direction and will compress in the transverse directions, in such a way that its volume remains constant. More precisely, the first and second derivatives of its volume at time t = 0 are both zero, but the second derivatives of individual coordinates of dust particles (r in the radial direction, and theta, phi in the transverse directions) are nonzero at time t = 0 (the first derivatives of the coordinates are still zero).

In both cases, it is the *second* derivatives that manifest spacetime curvature; this is related to the fact that the components of the Riemann curvature tensor are built out of second derivatives of the metric (the Christoffel symbols are built out of first derivatives, and their derivatives are used to build the Riemann tensor).

We've been talking about the part that basically corresponds to shell #1 (gravity pulling inward), but it's important to recognize that there are other parts to the curvature as well, as manifested by shell #2.

I understand the physical point about the constant-height object being pushed outward, away from the geodesic. Yet I still feel like I'm missing something about how we were justified in choosing the sign convention that we know, physically, to be the right one.
The sign *convention*, as I am using the term, doesn't affect the physics at all, because if you choose one sign convention (say, that proper acceleration outward, in the positive r direction, is positive), you are implicitly choosing a bunch of other sign conventions as well (for example, the sign of the terms in the expression for the Christoffel symbols in terms of the metric). So focusing on the signs in isolation is probably not a fruitful way to look at it.

Let me try this: suppose that in some bizarre (maybe impossible) kind of field, the coefficients of the dt and dr terms were reversed in the metric. (I think of this as: instead of "less time" with decreasing r, dt>dT, there would be "more time", dt<dT.)[I'm writing "T" for your "tau".] Would that by itself result in geodesics of increasing rather than decreasing radius?
I can't really say because I don't know if this metric would be a solution of the Einstein Field Equation, so I'm not sure if it's consistent with the other mathematics that would need to be applied to calculate how the geodesics would go.

Can you specify this change a bit more? How do you describe the change in t-component, the divergence in time between the apple still on the tree and the apple just starting to fall? Is the r/t tradeoff implied here related to stretch factors?
It's just relative motion; the two apples are moving relative to one another, so each sees the other as slightly time dilated. If we look at things from the falling apple's point of view, then the stationary apple's time appears to "run slow". The only reason the component change appears in the falling apple's 4-velocity and not the stationary apple's is that we chose coordinates in which the stationary apple is at rest. We could just as easily choose coordinates in which the falling apple is at rest and the "stationary" (or hanging) apple acquired an outward radial 4-velocity component (and had its time component adjusted accordingly).

It's true, however, that there is a difference between these two coordinate charts. One way to illustrate the difference is to observe that the spacetime as a whole has a "time translation symmetry", and the "stationary" apple happens to be moving on a worldline that is an "orbit" of that time translation symmetry; put another way, the stationary apple is moving in such a way that the metric in its vicinity does not change, because the metric coefficients depend on r, but not t; so as long as you stay at the same r, the metric doesn't change. (In fact, saying that the metric doesn't depend on t is another way of saying that the spacetime has a time translation symmetry.) The falling apple, however, changes its radius with time, so the metric in its vicinity changes; so any coordinate chart in which the falling apple was at rest, if it were going to cover more than a very small region of the spacetime, would have a time-dependent metric.

All of which just highlights again that it's important to try to express things in terms of invariants, things that don't depend on the coordinates.

I started this thread by saying that I knew I had something fundamentally backwards, which I expressed in the title as objects tending to move toward "shorter time". In subsequent discussion I talked of "more time" vs. "less time". I associated the latter with a gravitational center, because I knew that in the metric, far-away time is a larger number than proper time. I also associated "more time" with the length of a spacetime interval in a naive way.

Peter's discussion, especially in #19&20, has been very interesting and helpful to me.
But I think I have finally seen the light as a result of poring over A.T.'s visual aids.

Here's what they show me: time is stretched out near the earth. I would have thought on this basis that there would be "more time" there, and yet it is the ceiling clock that runs fast.
The fast ceiling clock corresponds to the fact that far-away time is greater (a larger number, more clock ticks) than proper time.
So I have to understand the apparently shorter distance around the upper part of the cone in the first diagram as corresponding to the larger numbers on higher-altitude clocks.

Well, if the diagram included coordinate lines, they would be bunched together more at the top, and spread out at the bottom. I might imagine it taking more energy to push through that higher density of coordinate lines than to take a path that crosses fewer lines per proper worldline distance.

So when time is stretched out, clocks show fewer ticks. (This seems obvious to you!)
What I now realize is that I've been confused for a long time about what "time dilation" means. I know that moving clocks run slow, but I thought that "dilation" meant larger numbers and so would apply to the stationary frame. Now I see that "stretching out" means smaller numbers, so I can finally understand "dilation" and "contraction" as both applying to the moving frame.

Instead of "more time", I should say "stretched-out time". You have "more time" (or maybe "more space") to get where you're going, because the clock is ticking more slowly. (Does that make sense?) The worldline can be longer, because there is "more space" in between parallel time-coordinate lines.

I'm not claiming to have understood everything that's been discussed. But I feel much better about the whole picture of curvature and free-fall motion now that I see how a longer worldline fits with the relative slowing of clocks.

Our discussion of what's going on in the "first instant" of free-fall motion might be modeled using the cylinder with time as the transverse dimension. It's a simple cylinder in flat spacetime, but curvature is represented by a fattening of the radius. Then the thing to see is how the geodesic bends in the direction of fattening. This is somewhat intuitive given that a geodesic on a sphere follows the biggest bulge, so to speak, rather than taking a less-than-great circle. (Maybe we can make this mathematically more obvious.)

One thing bothering me at this point is that A.T.'s diagram doesn't help me with the proper time defined according to t^2-s^2 metric. Is there a way to incorporate the indefinite signature metric into such a diagram? Or is it just something I should see arising out of Euler-Lagrange math? (Remember, one of my points of ignorance is why t^2-s^2 can play the role of (k.e. - p.e.).)

I appreciate Peter's point about the complexity of curvature not being reducible to the time (or perhaps I should say time-time) component. I think that the diagram showing "more space" between time-coordinate lines already suggests how tugging on one coordinate affects others. And reminding me of the difference between tidal acceleration in general and the special spherically symmetric case also helps.

A.T.