Copyright © 2002 jsd
Presumably you have heard of the first law of motion, which says that a free particle moves in a straight line at a uniform velocity. That’s true, but in order to make it useful, we need to be able to recognize straight paths and distinguish them from non-straight paths.
Tangential remark: If you think about things in spacetime, both parts of the first law of motion – the “straight line” part and the “uniform velocity” part – turn out to be exactly the same thing. A uniform velocity is a straight line in spacetime, nothing more, nothing less. A discussion of this, along with an interactive diagram, can be found in reference 1.
Also, you have probably heard something about general relativity, including the idea that gravitation is explained by the “curvature” of spacetime. The purpose of this document is to explain some of the important details such as the direction in which the space is curved, how much it is curved, and how this produces the effects we call gravitation.
It would be helpful to have some prior understanding of what we mean by spacetime, as discussed in reference 1 and elsewhere.
This is a bit of a digression, but it may be worth mentioning, because it is probably the simplest and easiest introduction to the idea of geodesics.
Start with an ordinary globe of the earth, preferably one with prominently marked lines of latitude and longitude. Stretch a string along the surface of the globe from point A to point B, where the points are not too close together.
We make use of the fact that in such a situation, the shortest path is also the straightest path. So a string under tension follows a geodesic. On a sphere, the geodesics are great-circle routes.
Note that the great-circle route from London to Los Angeles departs London on a northwesterly heading but approaches Los Angeles on a southwesterly heading. That is to say, it goes north for a while before going south. This looks perfectly normal when you see the string on the globe, but it would look mighty peculiar if plotted on a Mercator projection. The parallels of latitude are definitely not geodesics (except for the equator).
Masking tape has the wonderful property that it very non-stretchy, as you can verify by trying to stretch a piece. (If you have some stretchy tape, set it aside and get some non-stretchy tape.)
Next, note that it has definitely nonzero width. It is non-stretchy across its width, and also across innumerable diagonals, so it can hold itself straight, the way a triangular truss holds itself rigid, as in the bridge in figure 1. If the strut-lengths don’t change, the overall shape of the structure cannot change.
Figure 2 shows the kind of cross-bracing we are talking about. Lines drawn on the tape cannot stretch, and these define triangles that cannot change their shape. The length of line DB and line AC (shown in gray) cannot change, because the tape is non-stretchy. Similarly the length of line AD and line BC (shown in red) cannot change. In this way you can prove that all the triangles keep their shape. We know from high-school geometry that if the lengths stay the same, the angles must stay the same also. See reference 2 page 249 for a much more detailed and rigorous discussion.
You can draw triangles like this on your tape if you want, but the tape keeps its shape whether you draw the triangles or not.
This notion of straightness, defined by cross-bracing, has numerous good properties. For one thing, if you stick an initial piece of tape to a surface, it defines a unique way of laying down the next piece, and then the next. And it is reversible: You can retrace the such a tape-path in the reverse direction and get the same result.
It is intriguing that in mathematics, lines are defined to be straight and have zero width, but in physics, if you want to make sure it is straight, it needs to have nonzero width (so the cross-braces have some leverage). You can pass to the limit of infinitesimal width, but not zero width.
In the first paragraph of the introduction to the Principia, Isaac Newton said: “The description of right lines and circles, upon which geometry is founded, belongs to mechanics. Geometry does not teach us to draw these lines, but requires them to be drawn”.
A more modern way of framing the issue is this:
In your imagination, you can postulate straight lines and circles that exist in some formal, mathematical space. Such lines and circles are abstractions, devoid of physical meaning. | If you want to construct a “line” or anything else real life, making it straight requires physics, not just mathematics. |
Sometimes people try to “define” straightness by saying that a straight “line” is the shortest path between two points. That is, however, not the ideal definition. It would be better to say that any extremal path (either shortest or longest) is necessarily straight. You can show that the tape satisfies this definition, by the following argument: The two edges of the tape have equal length. The tape was made to have that property, and it retains that property because it is non-stretchy. If you choose a hypothetical path that is not a geodesic – not straight – then nearby paths to one side will be slightly longer, and nearby paths to the other side will be slightly shorter. You cannot make the tape follow such a path without buckling.
Once we have a good way to make straight paths, it is a simple matter to create curved paths, by forcing one edge of the tape to follow a longer path than the other.
Tangential remark: Similar notions of curvature play a role in the expansion of the universe, as discussed in reference 3.
Here is a convenient way to correctly demonstrate the motion of a free particle in curved space.
We use a two-dimensional surface as our model universe. To model the path of a particle, apply masking tape to this surface.
For reasons explained in section 3, the tape will follow a geodesic (straight line) in the two-dimensional universe. Meanwhile, masking tape is so thin that it can bend as necessary in the third dimension (the embedding dimension). Be sure you use masking tape, which is designed to be non-stretchy (as opposed to something stretchy like electrical tape).
1) Let’s start with an uncurved universe. Use a flat sheet of construction paper or (if you’re using an overhead projector) a blank acetate foil. Lay out some tape and see what happens. Put down a couple of inches of tape to get things started, and then lay down the rest bit by bit. Allow the most-recently attached bit to guide the placement of the next bit. Hold the supply of tape slightly slack; do not try to “force” the result to go in any particular direction. Observe that the result is automatically straight, to an excellent approximation.
As long as you don’t allow “crumpling” or “air pockets” under the tape, it should guide itself quite well.
2) Lay out another line of tape that is initially parallel to the first. Extend it, letting the tape itself do the guiding, and observe that the two lines remain parallel for a long ways.
3) Bend the sheet into a cylinder, in such a way that it doesn’t stretch the sheet. (Wrap it around an oatmeal carton if you need help maintaining a cylindrical shape.) This creates extrinsic curvature without creating any intrinsic curvature.
A cone is another shape that has extrinsic curvature but no intrinsic curvature (except at the single point at the tip of the cone).
Observe that rolling the sheet into a cylinder or cone has no consequences for the geodesics; things that start out parallel remain parallel, et cetera. This is important because it shows that extrinsic curvature has no effect on the path of particles in our two-dimensional model world. That contrasts with intrinsic curvature, which will be demonstrated next.
4) On top of a flat sheet, put a large bowl, upside down. I have a large salad bowl that works beautifully.
Lay out a geodesic that starts on the flat sheet and heads toward the bowl. When it reaches the bowl, it will refract.
5) Lay out a geodesic that starts out parallel to the previous one. It will hit the bowl with a different impact parameter, and refract differently.
*) Et cetera. You get the idea.
Note: If you are not using an overhead projector, choose the color of the paper and the color of the bowls to contrast with the color of the tape. If you can get multiple colors of non-stretchy tape, so much the better.
Another suggestion: You can pile smaller bowls onto the back of the larger bowl, to change the shape of the potential.
Remark: The embedding world’s gravity has no effect on the tape. This model would work perfectly in the weightless environment in a spaceship.
Related remark: The tape does not care whether the curvature is a bump or a pit. Consider what happens if you have a flat countertop with a bowl-shaped sink set into it. The tape runs straight along the countertop, but then drops down and follows the curvature of the inside of the sink. In all cases the geodesic will be bent toward the region of high intrinsic curvature.
A bump and a bowl both have positive intrinsic curvature, as discussed in section 8.5. By way of contrast, a saddle has negative intrinsic curvature.
Let us now consider a different model, namely a marble rolling in a bowl. This can be contrasted with the curvature-based model introduced in section 4.1.
Rolling in a bowl is a decent model of classical physics, i.e. Newtonian gravitation. | Rolling in a bowl is a false and deceptive model of the modern physics, i.e. general relativity. |
Rolling in a bowl depends on the fact that the bowl sits in the earth’s gravitational field. | The correct model, as introduced in section 4.1, works just fine in zero-gravity conditions. |
If the shape of the bowl is just right – a paraboloid – the height of the bowl faithfully represents the classical gravitational potential. At each point, the slope of the bowl represents the gravitational field. (The fact that the marble rolls – rather than sliding freely – introduces some nonidealities, but let’s ignore that.)
If you use a glass bowl, you can demonstrate this to the whole room using an overhead projector.
Although this models the classical physics to a fair approximation, it does not correctly model general relativity. In particular, the curvature of the bowl is not a good model of the spacetime curvature that general relativity uses to explain gravitation. Not at all.
There are several ways of seeing that it would be wrong to consider the bowl a model of general relativity.
To repeat: the alleged connection between “rolling in a bowl” and general relativity is essentially 100% wrong.
The model introduced in section 4.1 demonstrated the qualitative effects of curvature, but we have to refine it a bit if we want a really accurate model of, say, planetary orbits. It turns out that bowl-shaped potentials are not what we need. Not even close.
The world-line of a particle in orbit is best described as a helix. It goes around and around in two spacelike dimensions, while moving steadily forward in the timelike direction.
As pointed out by David Bowman, if you try to project that helix onto a two-dimensional model, there will be problems. Spacetime is curved (by gravity) in the timelike direction as well as the spacelike directions. So if you build a model that represents the two spacelike directions, suppressing the timelike direction, you can’t properly represent the sort of curvature that leads to planetary orbits.
We can do much better if we use our model to represent one timelike dimension and only one spacelike dimension. We will illustrate the gravitational field of a planet that exists for a long time (a long ways along the time axis). The gravitational field decreases as we move away from the planet in either direction along the X axis.
This can be modeled using darts, as shown in figure 3. The time axis runs vertically up the page, and the spacelike X axis runs horizontally across the page. The five ribs are made of two darts each, for a total of 10 darts. See section 9 for hints on how to fabricate darts.
I emphasize that the tape follows geodesics determined by the local curvature of spacetime at each point. Given an arrangement of darts, each trajectory is completely determined by its starting point and initial direction; there is no other choice involved.
Whenever the tape crosses a dart, the curvature of spacetime will deflect the trajectory toward the center. As it continues along, it will “orbit” the center, as you can see in the figure. In D=1+1, each trajectory looks like a sine-wave when viewed from above the page. One trajectory starts with an X value slightly left of center and completes a half-cycle before exiting at the top of the diagram. The next trajectory starts with an X value slightly right of center, and also completes a half-cycle. The third trajectory starts farther right of center, and completes only a small fraction of a cycle.
It is interesting to think about “what is the shortest path from point A_{1} to point A_{2}”. The path in the presence of the darts is very different from what it would be in the absence of the darts. Actually in the presence of the darts (spacetime curvature) there are a couple of different geodesics that connect A_{1} to A_{2}.
Hint: The stores around here sell masking tape that is 3/4” or 1” wide. Narrower tape is better for this demo. So you may want to divide the tape in half lengthwise. Do this while it was still on the roll, using a very sharp knife to cut through many layers.
If you attach the darts to a piece of acetate foil, you can set the whole thing on an overhead projector so everybody can see it. But letting people play with it hands-on is the best.
Figure 4 shows the same idea as figure 3. It shows the xt plane. The stripe along the contour of constant x=0 has no particular meaning; it just makes it easier to perceive the amplitude of the wiggles.
Figure 4 does not intersect the planet. It is a map of the xt plane at some nonzero y value, where y is big enough so that we don’t hit the planet.
This model is in some ways faithful to reality, and in some ways not.
You can improve the model by imagining a larger number of smaller wiggles, as in Figure 5. In your imagination, continue this process (an ever larger number of ever smaller wiggles) until there is curvature at all times, or close enough.
Diagrams such as figure 6 are very common, but not very faithful to the physics of general relativity. Such a diagram might result from a “trampoline” model, i.e. a heavy bowling ball resting on a trampoline, causing it to sag and stretch, forming a pit like the one shown in the diagram.
The bowling ball produces curvature in the membrane. So far so good. The problem arises when they try to model the dynamic by putting a marble in “orbit” around the bowling ball.
Alas, in the “pit” or “trampoline” model, none of the circles of constant radius are geodesics. If you tried to drive around such a circle, your inside tires would travel a shorter distance than your outside tires. That’s a problem, because a defining property of a geodesic is that if you drive along one, your left tires travel just as far as your right tires.
For a static gravitational field, the curvature that matters is the curvature in the timelike direction, as shown in figure 3 and figure 4.
By way of background, suppose you travel from point A to point B along some chosen path. You can use an odometer to measure the length of the path. This length will not usually be the same as the length of some other path from A to B. In particular, an odometer is not like a rigid ruler, which measures the straight-line distance from A to B.
It turns out that ordinary clocks are more like odometers than like rulers. Suppose you travel from point A to point B along some chosen path, perhaps one of the paths shown in figure 3. The elapsed time along this path will not usually be the same as the elapsed time along some other path.
One way that the elapsed times can be different is if one of the paths has a lot of curvature in the time direction. You can see in figure 3 that a path that is deeper in the gravitational potential will have more curvature. Our simple model is quite faithful to the real gravitational physics in this regard: A clock deep in a gravitaitional potential will rack up more time than a clock not so deep in the gravitational potential.
You might be tempted to say that the deeper clock “runs fast” ... but you should resist this temptation. There is nothing wrong with either clock; both clocks run at the standard rate of 60 minutes per hour. (In our model, that is represented by saying the tape does not stretch.) The difference in elapsed time has nothing to do with the clocks. It has everything to do with the difference in path-length. Clocks are like odometers. They measure the length (in the time direction) of the path. This very much depends on which path you choose. The deeper clock is following a path that is more crumpled in the time direction.
Suppose you are helping a buddy with a plumbing project that involves cutting PVC pipes and gluing them into fittings. Your buddy sees no reason to buy a pipe cutter, since he has a perfectly good hacksaw. The problem is, if you want the joint to be strong, the pipe needs to be cut square, and this isn’t always easy to do. It’s not too hard under favorable conditions in the shop, but not so easy out in the field when the pipe is at a funny angle and there are other constraints.
The desired result can be obtained more easily and more reliably if the pipe is first properly marked ... then all you need to do is cut along the mark.
So now we come to the heart of the matter: What is the quick and easy way to create a “cut here” guide on the pipe?
Answer: Masking tape.
As discussed in previous sections, masking tape follows a geodesic. We assume the pipe is accurately cylindrical (otherwise the glue joint would fail anyway, so the whole exercise would be pointless). The geodesics of a cylinder are helices, including the zero-pitch helix which is a circle.
If you start the tape perpendicular to the axis of the pipe, the tape will come around and meet itself, making a zero-pitch helix, and you know you have succeeded. Cut parallel to the edge of the tape.
If you start the tape at an angle, it will come around and make a non-zero-pitch helix, and you know you need to re-do the taping.
This trick with the tape is super-accurate, super-quick, and super-easy.
The previous discussion has focused on what happens to a single point as it moves from place to place along a geodesic. In this section, we consider what happens to a vector as it moves from place to place along a geodesic. This is important, because it provides us with a particularly easy way to measure the curvature of the space.
Note the contrast:
When we were just talking about position-space, at each position there was a point, and that’s all there was. | Now are talking about a vector field. That means at each point in position-space there is a vector space. We have a space of spaces. |
Consider the scenario shown in figure 7. We start with the yellow arrow that is located just north of the eastern tip of Brazil, on the equator at 45 degrees west longitude. The vector points due north. We then construct another vector 18 degrees farther west, at a point near the mouth of the Amazon. We take care that the new vector is parallel to the old vector. It also points due north. We keep constructing new vectors, each parallel to the previous one, until we come to 90 degrees west longitude, in the ocean south of Guatemala.
The next vector is constructed at a position due north of the previous one. We are now moving northward along a line of constant longitude, namely 90 degrees west longitude. All the arrows point toward the celestial north pole, i.e. a point high above the earth’s geographic north pole. As we move north, each of the arrows is parallel to the previous arrow.
In this context moving north means moving closer to the geographic north pole, following the surface of the earth (as opposed to moving directly closer to the celestial north pole, which would take us away from the surface). Note the contrast: We are using geographic north for the positioning, but we are comparing the orientation against celestial north.
We continue around the triangular path until we get back to the starting point. The final arrow appears to be parallel to the original arrow. There is nothing surprising about this.
Actually, due to general relativity, the final arrow is not quite exactly parallel to the original arrow, but the difference is far too small to be perceptible in figure 7. The rest of this section is devoted to explaining how it is possible for these two arrows to wind up not exactly parallel.
Consider the scenario shown in figure 8. It starts out exactly the same as the previous scenario. However, in this case we suppose that the arrows exist in a two-dimensional space, namely a sphere, i.e. the surface of the earth.
The yellow arrows along the equator are exactly the same here as in the previous scenario. Even though the arrows are the same, we have to describe them differently. We say they all point north along the earth’s surface. They point toward the earth’s geographic pole, not toward the celestial north pole, because the latter does not exist in the two-dimensional space we are using.
As we move northward along the leg of the triangle that goes through North America, the arrows in figure 7 continue to point north toward the geographic north pole. Relative to the arrows in figure 8, these arrows must pitch down so that they remain within the two-dimensional space. They are confined to be everywhere tangent to the surface of the earth. As we move north, each of the arrows is parallel to the previous arrow, as parallel as it possibly could be.
Let’s be clear: Each new arrow is constructed to be parallel to the previous one, as parallel as it possibly could be. What we mean by “parallel” is discussed in more detail in section 8.3.
After we get to the north pole, we start moving south along a the prime meridian. We move south through Greenwich and keep going until we reach the equator at a point in the Gulf of Guinea. As always, each newly constructed vector is parallel to the previous arrow. All the arrows on this leg point due east.
Finally, we move west along the equator until we reach the starting point. Again each arrow is parallel to the previous one. All the red arrows on this leg point due east.
At this point we see something remarkable: The final arrow is not parallel to the arrow we started with.
From this we learn that in a curved space, there cannot be any global notion of A parallel to B. We must instead settle for a notion of parallel transport along a specified path. That is: the notion of parallelism is path-dependent. It also depends on whether you go around the path clockwise or counter-clockwise.
If you start with a northward-pointing vector in Brazil and parallel-transport it to the Gulf of Guinea, you get a northward-pointing vector. | If you start with the same vector and transport it clockwise around two legs of the triangle as shown in figure 8, you get an eastward-pointing vector. |
Creatures who live in the curved space can perceive this in a number of ways. Careful surveying is one way. Gyroscopes provide another way. That is, a gyroscope that is carried all the way around a loop will precess relative to a gyroscope that remains at the starting point.
It must be emphasized that this precession has got nothing to do with the spin of the earth or the peculiarities of the latitude/longitude coordinate system. In figure 9 the sphere is completly abstract, with no spin, no latitude, no longitude, and no poles. As we shall see in section 8.4, the area of the loop is what matters, not the shape or orientation.
You can also see in figure 9 that the initial arrow does not need to be parallel or perpendicular to the path. Any initial orientation is allowed. The orientation of each successive arrow is dictated by the requirement that it be parallel to the previous arrow.
In all these diagrams, each new arrow is constructed to be parallel to the previous one, as parallel as it possibly could be. If you don’t see each successive arrow in this part of the diagram as being exactly parallel to the previous one, that’s because you live in three dimensions. Creatures who actually live in the curved space – the two-dimensional curved surface in this model system – cannot detect the third dimension.
This is related to the idea that the tape used in figure 3 is non-stretchy in two dimensions but is free to bend in the third dimension. We are modeling the physics as seen by creatures who live in the two-dimensional space. The third dimension – the embedding dimension – is not part of their world.
In particular, in the discussion associated with figure 2 we said that all the triangles keep their shape, whatever that shape might be. Now, if we impose the further requirement that the midpoint of line DB coincides with the midpoint of line AC, then we have similar triangles (indeed congruent triangles) and this guarantees that line BC is parallel to line AD. We use this to define what we mean by parallel transport. We use this construction – called Schild’s Ladder – to transport vector AD along the path AB and thereby prove, by construction, that it AD is parallel to BC.
Here is another argument that leads to the same conclusion: The situation in figure 10 is very similar to the situation in figure 8. The main difference is that rather than having a single vector at each point, we have two vectors. This is one way of representing a bivector. (If the term “bivector” doesn’t mean anything to you, don’t worry about it.)
In particular, let’s look at the six bivectors on the leg of the triangle that goes through North America. Each of the gray vectors here points due west, along a line of constant latitude. Such lines are called parallels of latitude – as in the 49th parallel – and they are called that for a reason. The gray arrows are undoubtedly parallel. If you treat them as existing in the embedding space they are parallel, as surely as the arrows in figure 7 are parallel. Also if you treat them as existing in the D=2 surface they are parallel.
Once you are convinced that these six gray arrows are parallel, the rest of the argument is easy. The yellowish arrows are perpendicular to the gray arrows. In two dimensions, this implies that the yellowish arrows must be parallel to one another. In two dimensions, there is no other possibility.
The amount of precession is proportional to the area enclosed by the loop, times the amount of curvature. In two dimensions we can write:
| (1) |
where dA is an element of area, and K denotes something we call the Gaussian curvature.
We can always define the average curvature (averaged over some area) as follows:
| (2) |
Then as an immediate corollary to equation 1 we obtain:
| (3) |
This gives us a convenient way of measuring the average curvature. Let’s see how it works for the spherical triangle shown in figure 8. The loop comprises one octant, i.e. one eighth of the area of the sphere. A sphere of radius R has curvature K = 1/R^{2} everywhere.
| (4) |
which is nicely consistent with equation 1 and equation 3.
Figure 11 is similar to Figure 8 except that the path encompasses a smaller area. You can see that there is correspondingly less precession.
Occasionally, we choose to imagine that our curved space is embedded in some higher-dimensional space. For example, the spherical surface shown in figure 8 is intrinsically a two-dimensional space. We only need two numers (e.g. latitude and longitude) to span the space. In this subsection, we imagine that it sits in a three-dimensional embedding space.
This allows us to write the Gaussian curvature in the form
| (5) |
where k_{1} is the principal curvature in one direction and k_{2} is the principal curvature in the other direction. For example, a sphere of radius R has k_{1} = k_{2} = 1/R everywhere, which is consistent with our previous assertion that K = 1/R^{2}.
By way of constrast, a cylinder of radius R has zero Gaussian curvature, because k_{1} = 1/R but k_{2} = 0. The cylinder is curved in one direction but not the other. We can apply similar reasoning to a cone. We conclude that a cone has zero intrinsic curvature (except at the tip).
Each principal curvature is related to the corresponding radius of curvature:
| (6) |
It must be emphasized that r_{1}, r_{2}, k_{1}, and k_{2} are extrinsic, and cannot be measured by the creatures who live in the two-dimensional space. In contrast, they can measure the intrinsic curvature, i.e. the Gaussian curvature K, by carrying out parallel-transport experiments and applying equation 3.
To demonstrate parallel transport using the same tabletop model shown in figure 3, proceed as follows:
From a strip of masking tape at least one inch wide, cut out a parallelogram. The exact shape doesn’t matter.
Draw the long diagonal on the parallelogram, as shown in figure 12. Use a pen with a moderate width, not too fine, for reasons that will become obvious in a moment. Label the corners {A, B, C, D}. Add another label, C′}, so that there are labels on both sides of the diagonal at corner C. For strain relief (aka crimp relief), make a keyhole-shaped cut, as shown in figure 12. That is, make a round hole in the middle, and then make a cut from the middle to corner C, running right down the middle of the line you marked. Try to cut the line in half. I found it easy to make the crimp-relief cuts by putting the tape on a firm board and using a scalpel (aka “hobbyist razor knife”). Using a scissors is possible but probably less convenient.
Lay the parallelogram on the model, so that the tip of a dart falls into the crimp-relief hole. I did it starting from point B and proceeding counterclockwise to point C′, then re-starting at point B and proceeding clockwise to point C. There are probably other satisfactory tactics. The result is shown in figure 13.
The parallel transport story goes like this: Start at point C. The initial vector that we wish to transport is the line that is drawn on the tape, the line that runs from the center to point C. We transport that via B, A, and D all the way to C′. The transported vector is now parallel to the line from the center to C′. Since the space between point C′ and point C is flat, you can easily use your imagination to transport the vector the rest of the way, all the way back to the starting point C. As you can see in figure 13, the vector has precessed by about 7 degrees.
If you look closely, you find that the direction of precession is the opposite of what we saw in figure 8. There if you go around the loop clockwise the precession is also clockwise, but here if you go around the loop clockwise the precession is counterclockwise. This is the hallmark of negative Gaussian curvature. A sheres has positive intrinsic curvature, while a saddle has negative intrinsic curvature.
Also: If you think about it for a while, and/or do some experiments with the model, you discover that for transport around any path that encompasses the tip of a dart, the amount of precession is the same, namely −7 degrees. The only way to describe this is to say that there is a Dirac delta-function of curvature, located at the tip of the dart.
If you don’t know what a delta-function is, don’t worry about it too much. In general, it’s just a fancy way of saying something is hugely concentrated, with a high density in one place and zero density elsewhere. Imagine a very tall, very narrow spike. In the present contect, when we talk about curvature, we are not talking about the curvature of the spike; the spike is not what’s curved. The space is curved, and the spike is telling us where the curvature is.
We can formalize this concentration of curvature as follows:
| (7) |
where θ is −7 degrees in this model, and the tip of the dart is located at the point (x_{0}, y_{0}). When we integrate this curvature in accordance with equation 1, we get the correct total precession. Note that a δ-function has units of inverse length, so the dimensions in equation 7 are correct.
Meanwhile for any path that encompasses the base of a dart, the precession is +7 degrees. In the model, there is no way to encompass the base of one dart without encompassing the base of its partner, so we see a total precession of +14. For any path that does not encompass the tip or base of any dart, there is no precession.
Tangential remark: If you carefully measure the angle in figure 13, you find that the angle is very close to −8 degrees, rather than the −7 that we were expecting based on the specifications of the darts. I don’t know whether this has to do with imperfections in the darts, or imperfections in the way I laid down the tape.
To repeat: To us, living in the embedding space, the darts may “look” like they have curvature all along their length, but really the darts are like cones: zero intrinsic curvature except at the tips and bases. To say the same thing another way: the dart has one extrinsic curvature (k_{1}) that is nonzero all along its length, but the other extrinsic curvature (k_{2}) is zero, so the intrinsic Gaussian curvature is zero (except at the tip and the base).
As a consequence, the gravitational field that we are modeling in figure 8 is a piecewise-constant field. On the left side of the midline there is a field of constant strength pointing to the right, and on the right side of the midline there is a field of constant strength pointing to the left. This is similar to the field you would find in a three-plate parallel-plate capacitor, with charges of −Q, +2Q, and −Q (respectively) on the plates.
In particular, in the main region of the diagram, between the tips and the bases of the darts, there is no geodesic deviation. As seen by creatures who live in the two-dimensional space, there is no spacetime curvature in this region. We who live in the embedding space see geodesics that seem to curve, but in this region they all curve together. A pair of geodesics that starts out parallel will remain parallel. That is to say, they do not deviate from each other, so there is no geodesic deviation.
The parallel geodesics remain parallel unless and until the pair straddles the tip or the base of a dart, whereupon there will be geodesic deviation. This is the fundamental mechanism of general relativity: Curvature causes geodesic deviation.
The analogy to real-world gravitation goes like this: Any region where the gravitational acceleration g is essentially uniform does not have any appreciable spacetime curvature. In particular, over the typical laboratory lengthscale, spacetime curvature is negligible. This is related to Einstein’s principle of equivalence, which says that a uniform gravitational field is indistinguishable from an acceleration of the reference frame. To say the same thing the other way, you can make a uniform gravitational field disappear by choosing a different reference frame. Therefore it is obvious that a uniform gravitational field has got nothing to do with spacetime curvature, because you can’t make curvature go away by choosing a different reference frame.
Many of the following fabrication ideas are due to Paul Fuoss.
Maple is an excellent material for making darts. (Any hard, fine-grained wood will do.) The grain should run down the long axis of the dart; this makes fabrication harder but makes the final result nicer. We found that Paul’s compound miter saw was the smart way to make them. We tried making them with a table saw but the compound miter saw was a much better choice.
Attach the hardwood piece (from which you are cutting the darts) to a much larger carrier piece, using wood screws, so you can hold it very securely without endangering your fingers. (Holding the piece securely is vastly more important for miter cuts than for regular cuts, where slight vertical motions are harmless.)
Each dart is a thin pyramid. The base of the pyramid is an equilateral triangle, 1/2 inch on a side. The pyramid is about 4 inch long in the other direction.
Set the saw so that the blade is inclined 30 degrees to the vertical. Then set it so that the arm is angled 3.6 degrees away from perpendicular to the fence. That gives a slope of 1 part in 16, i.e. 1/4 inch (half the base of the pyramid) per 4 inch.
Make the first cut. Cut just enough off one end of the stock so that the shape of the edge is determined by the cut. Then flip the stock over. If the stock was to the left of the blade to begin with, it will be to the right of the blade now. This may require unbolting the stock from the carrier and rebolting it (although you might avoid this step by using an extra-fancy carrier). Mark the starting point for the second cut. The base of the pyramid should be against the fence. Make the second cut. The first dart should fall free.
Jessica MacNaughton pointed out that you can make darts that are just as functional (but perhaps not as beautiful) out of clay. If you don’t have access to a compound miter saw, this may be your best option.
For an explanation of the concepts and notation used here, see reference 2.
We consider the motion of a free particle on the surface of the sphere. If all we cared about was spheres, there are lots of easy ways of describing the motion. We know the geodesics of a sphere are the great circles. However, let’s pretend we didn’t know that. The techniques used in this section work for spheres and non-spheres alike, even when the curvature varies from place to place in complicated ways. So this section is a warm-up exercise, intended to demonstrate the use of powerful tools by applying them to a relatively simple situation.
Figure 14 shows two trajectories on the sphere, one blue particle and one red particle. The trajectories look a bit odd, because this is a Mercator projection. Normally one should use something more sophisticaed than a Mercator projection, but in this section part of the goal is to show how to keep track of what’s going on, even in slightly non-ideal situations.
In the figure, each magenta tie-line connects the red particle at a given time with the blue particle at the same time. Both particles are moving at the same constant velocity. The spacing between tie-lines appears to change, but that’s just an artifact of the Mercator projection. Things near the pole get stretched sideways. A given east-west distance corresponds to less longitude near the equator and more longitude near the poles.
Let’s figure out how to calculate such things. Suppose a free particle starts out at a given position x with a given velocity v. At each time t, it is easy compute the new position a short time later:
| (8) |
Note that position component x^{1} is synonymous with θ, and component x^{2} is synonymous with φ.
In a curved space, it is nontrivial to calculate the new velocity. From the particle’s point of view, it feels like the velocity is unchanged, which stands to reason, since we are talking about a free particle. However, when we try to express this in components, the components of the velocity have changed.
To be specific, let’s assume ordinary spherical coordinates, namely longitude θ and latitude φ. For a free particle, θ· and φ· are constantly changing.† In contrast, maintaining constant θ· and φ· would correspond to following a rhumb line, not a great circle,† not a free-particle trajectory.†
Note † indicates statements that are true except on a set of measure zero. The exceptions correspond to following the equator or following a great circle through the poles, i.e. a line of longitude.
One way to make progress is to use the methods of classical mechanics, starting from the Lagrangian. As usual, the Lagrangian knows all and tells all. (A fancier solution to the same problem is discussed in section 10.3.)
For a free particle, the Lagrangian is just the kinetic energy. The mass doesn’t matter (assuming it is nonzero). It will drop out of the geodesic equation, so for simplicity we set it equal to 1. In polar coordinates the kinetic energy is:
| (9) |
assuming the particle speed is small compared to the speed of light. A slightly more general way of writing this is:
| (10) |
where g is the metric tensor, which will be discussed later; see equation 23.
In the usual way, the Lagrangian allows us to identify the momentum variable that is dynamically conjugate to each of our chosen position variables. For the free particle on a sphere, the momenta are:
| (11) |
Raising an index gives us an unsurprising result:
| (12) |
We can check these equations in various ways. For example we can check that the kinetic energy is:
| (13) |
Recall that we set m=1. Also in accordance with basic ideas of classical mechanics, the Hamiltonian is
| (14) |
So the Hamiltonian is equal to the Lagrangian, which comes as no surprise for a free particle.
For any Lagrangian, we can invoke the principle of least action to obtain the Euler-Lagrange equation, which is a nice way to express the equations of motion. The general form is:
| (15) |
It is instructive to check this by applying it to a well-understood system, such as a simple harmonic oscillator.
For the free particle on a sphere, we get:
| (16) |
and
| (17) |
That is all we need to calculate the curves in figure 14. See section 10.4 for some hints about numerical methods.
Here’s another solution to the same problem. This is a simplified version of the approach you see in books on general relativity. The fancy machinery used here is useful for a wide range of curvature-related and gravitation-related applications, not just geodesics, and not just spheres.
One simplification here is that the number of components in equation 19 is only 2×2×2 instead of 4×4×4 for general relativity. Furthermore, as we shall see, only three of the components are nonzero.
We expect that in general, the geodesic equation will tell us that the acceleration is bilinear in the velocity. The most general form for such an expression is:
| (18) |
where Γ embodies the information we have about the local curvature.
The components of Γ are called the Christoffel symbols. More specifically, the components Γ_{ijk} are called the Christoffel symbols of the first kind, while the components Γ^{i}_{jk} (with a raised index) are called the Christoffel symbols of the second kind. We can display the components as follows:
| (19) |
In other words, the first component specifies which plane, the second component specifies which row, and the third component specifies which column. Imagine the two planes are stacked vertically to make a 2×2×2 cube, but we offset them for better visibility.
As we shall see, the components of Γ for a sphere are:
| (20) |
Plugging this into equation 18 we find:
| (21) |
which is consistent with equation 16 and equation 17. However at this point you should be wondering where we got equation 20. It turns out there is a general expression for the Christoffel symbols in terms of the metric, namely:
| (22) |
where the comma means to take the derivative. This expression can be derived using the notions of parallel transport as discussed in section 8. The basic idea is that the velocity vector is transported along the world-line in such a way that it stays parallel to itself and stays tangent to the world-line.
The metric for the 2-sphere (in spherical coordinates) is
| (23) |
The metric defines what we mean by dot product, which in turn defines what we mean by angle and distance. For example, equation 9 or equivalently equation 10 is just the dot product of the velocity with itself. An even simpler example uses plain old distance:
| (24) |
where ds is the element of arc length. If we combine the two previous equations it gives us another way to describe the metric of the sphere, namely:
| (25) |
You can verify that this metric is correct by thinking about the geometry of the sphere: how long is each line of constant longitude, and how long is each line of constant latitude.
If we differentiate this metric, the only nonzero component is:
| (26) |
It is then a matter of bookkeeping to figure out which three components of Γ_{ijk} are nonzero, in accordance with equation 22. Then raise an index to obtain Γ^{i}_{jk}.
As a check, note that our Γ^{i}_{jk} in equation 20 is symmetric with respect to interchange of j and k. This is as it should be, as a consequence of equation 22 and the fact that g_{ij} itself is symmetric.
Let’s talk about what’s going on in terms of physics.
In figure 14, each particle is moving at constant speed along a great-circle route. Each particle, according to its own reckoning, is moving in a straight line. Each particle feels zero acceleration. Each particle is oblivious to the curvature. It is impossible in principle to measure the curvature using only one pointlike particle. Even the fact that each trajectory closes on itself does not demonstrate curvature. You could perfectly well have a flat space with periodic boundary conditions.
Things get much more interesting when we compare the two particles. They start out moving parallel to each other, but they don’t remain parallel. Each particle considers that the other is subject to a mysterious acceleration. In some sense it looks like gravitation. In particular, it looks like a tidal stress δg (not a simple acceleration g). The physical fact is that they accelerate relative to each other. This phenomenon is known as geodesic deviation.
For the spreadsheets used to do the calculations and prepare figure 14, see reference 4.
Hint #1: When doing the numerical integrations, first-order Euler is not good enough. Second-order Runge-Kutta (RK2) is good enough, if the stepsize is reasonably small (one degree or so) and the latitude is not too high (less than 70 degrees). One could imagine using fourth-order Runge-Kutta (RK4), but for the purposes of figure 14 it’s not worth the effort. Keep in mind that the coordinate system is singular at the poles, so there are limits to what you can hope to do.
Tangential remark: If you switch to a better representation, e.g. Clifford algebra aka bivectors aka quaternions, you can get decent results near the poles (and everywhere else). However, for present purposes we are sticking with polar coordinates.
Hint #2: As usual when applying Runge-Kutta to a system of differential equations, a useful trick is to reduce it to a single differential equation with more components. In this case the crucial variable is a vector with four components:
| (27) |
where we define
| (28) |
Loosely speaking, X is a point in phase space. Non-loosely speaking, remember that the coordinates in phase space are (position, momentum) not (position, velocity), and one of the momenta has a factor of cos^{2} in accordance with equation 11.
When we differentiate equation 27 and plug in equation 28 and equation 21 we find:
| (29) |
The first two rows in equation 29 may appear trivial, but they are an important part of the game. Overall equation 29 gives us a nice simple expression for X· in terms of X, which we can integrate using RK.
Beware that integrating equation 29 is probably not optimal. It would probably be smarter to build a symplectic integrator, but that’s more work than I feel like doing at the moment. It’s work, because the Lagrangian is not easily separable. This stands in contrast to the Lagrangian for simple mechanics problems, where the force depends only on position, not on velocity. Even though it’s not easily separable, the geodesic equation is still undoubtedly conservative and reversible, so we “should” be able to build a symplectic integrator somehow. See reference 5.
Hint #3: As always, the rule is: Show the work / check the work / show the checks. In that spirit, the spreadsheet in reference 4 calculates the speed of the particle at each step. If this is appreciably non-constant, it indicates a defect in the calculation. You can generally tell by looking at the plot that there’s something wrong, but the speed provides a way of quantifying the problem.
Here’s another instructive exercise, even simpler than the previous one: Calculate the geodesics in a plane, i.e. a flat two-dimensional plane. To make it interesting, use polar coordinates.
In this situation there is no tidal stress and no geodesic deviation. However, the geodesic equation is nontrivial, because the coordinate basis {dr, dθ} means different things in different places, and the equation of motion must take this into account.
One advantage here is that it is super-easy to graph the results, and there’s no doubt about how to interpret the graph.
If you want to understand gravitational waves, the first step is to have a clear understanding of what gravitation is. See reference 6.
In particular, the distinction between the framative acceleration g and the barogenic tidal stress δg is often underappreciated. Note that g is completely frame-dependent, whereas δg is completely frame-independent. If you’re looking for gravitational waves, you need to look for tidal stress.
By way of analogy, it is not possible to explain electromagnetic radiation in any direct way starting from the electrostatic interaction. There are lots of introductory textbooks that claim to do this, but it’s nonsense. In the Liénard-Wiechert potentials, the term that is needed to explain radiation does not appear in Coulomb’s law. See reference 7.
By the same token, it is not possible to explain a gravitational wave in terms of the Newton’s law of gravitation. If you try, you’ll get the wrong polarization, the wrong dependence on frequency, the wrong dependence on distance, et cetera. You might be able to make something happen in the near-field region, but it’s not really a wave. It doesn’t propagate properly. In the far field, it does not exist.
Here’s the concept map:
Electrostatics | Electrodynamics | Electromagnetic Waves | ||
(Coulomb) | (Maxwell) | (Larmor) | ||
Static Gravitation | Dynamic Gravitation | Gravitational Waves | ||
(Newton) | (Einstein) | (section 11.3) |
The point is, the interesting analogy moves vertically from electromagnetic waves to gravitational waves (not starting from static Newtonian gravity). The Larmor formula is discussed in section 11.2 and the corresponding idea is applied to gravitational waves in section 11.3.
The canonical Gedankenexperiment for transmitting gravitational waves is shown in figure 15. It is a mechanical oscillator, with two masses connected by a spring. The masses move 180 degrees out of phase, so that the center of mass remains stationary.
The distribution of mass in the transmitter has a time-dependent quadrupole moment. (This is analogous to an electromagnetic transmitter with a time-dependent dipole moment.)
The canonical receiver is shown in figure 16. It has two beads that are free to slide along a stick. The beads are what’s important; the stick is just there to give us a robust operational definition of distance, so we don’t go crazy.
By way of contrast, if we just had a loose collection of dust particles, it would be very unwise to lay out an X axis by marking some of the particles. When the wave came along, each and every particle would move as a free particle, oblivious to the wave. None of the particles would move relative to the bogus local X axis. The stick in figure 16 is not an abstract axis but rather a sturdy mechanical stick. When the wave comes along, it sets up some tidal stress in the stick, but the atoms in the stick do not respond as free particles. To a very good approximation they do not move at all relative to one another. The atoms resist the stress and maintain their size, as set by the laws of atomic physics.
When the wave comes along, the beads accelerate relative to one another. If we add a tiny amount of friction, the stick heats up, leaving indisputable evidence of the effect of the wave.
Tidal stress on the stick produces an exceedingly small strain; tidal stress on beads, dust, and other free particles produces a much larger strain.
It is an amusing exercise to derive the Larmor formula for the total radiated power using little more than scaling arguments. You can then obtain the gravitational wave formula in the same way.
Suppose we have an oscillating electric dipole. The radiated power cannot depend on the dipole moment itself; otherwise a static dipole would radiate, which would not make sense physically. Similarly the power cannot depend on the time derivative of the dipole moment, because two charges moving past each other in uniform straight-line motion cannot possibly radiate. Neither one radiates by itself, so (by superposition) they cannot radiate together ... yet this configuration has a time-varying dipole moment. So the lowest term that makes sense is the second time derivative of the dipole moment.
Step 2: The radiated power must depend on the square of that, for any number of reasons including symmetry with respect to flipping the definition of positive versus negative charge. Also we know that the fields are proportional to the dipole moment, and the energy goes like the square of the field.
Step 3: We need a factor of 1/є_{0} out front, to get something with dimensions of energy (as opposed to field strength). Then we need a factor of 1/c^{3} to make the dimensions come out right.
Step 4: If you want the received power per unit area at some distance r from the source, divide by 1/r^{2}. Also at this point you need to worry about polarization and other directional effects, but let’s not worry about that.
Conceptually, that’s all there is to it. There are some factors of π running around, but let’s not worry about that. We have a perfectly reasonable feel for what’s going on.
We can replay the ideas from section 11.2 and apply them do gravitational waves.
You can’t have a mass dipole, so the lowest-order thing we need to worry about is the quadrupole moment. (If you want to get fancy, it’s the reduced quadrupole moment, but let’s not worry about that.)
It can’t be the quadrupole moment, or the first time derivative, or even the second time derivative thereof, because otherwise two masses moving past each other in uniform straight-line motion would radiate, and we know that’s not going to happen. So the object of interest is the third time derivative of the reduced quadrupole moment.
Step 2: The radiated gravitational-wave power must depend on the square of that, for any number of reasons including time-reversal symmetry. Also we should be very unsurprised to find that the energy goes like the square of the field strength.
Step 3: We need a factor of capital G out front, to get something with units of energy (as opposed to field strength). Then we need a factor of 1/c^{5} to make the dimensions come out right.
Step 4: If you want the received power per unit area at some distance r from the source, divide by 1/r^{2}. Also at this point you need to worry about polarization and other directional effects, but let’s not worry about that.
Conceptually, that’s all there is to it. There are some factors of 2 running around, but let’s not worry about that. We have a perfectly reasonable feel for what’s going on.
Note that G is small, 1/c^{5} is small, and ω^{6} is small for typical events involving massive objects. So we are not surprised to hear that the gravitational waves are weak.
"The description of right lines and circles, upon which geometry is founded, belongs to mechanics. Geometry does not teach us to draw these lines, but requires them to be drawn."
– Sir Isaac Newton
(preface to the Principia)
Copyright © 2002 jsd