In 1961 W. Rindler, a professor of mathematics at Cornell University, devised one of the cleverer problems in Relativity. He imagined a rod sliding over a flat surface toward a hole in that surface and examined that situation from the viewpoints of two observers, one stationary relative to the surface and the other stationary relative to the rod. The following year R. Shaw, a professor of mathematics at Hull University in England, modified the problem by lifting the rod off the surface and putting it onto a ballistic trajectory that passed through the hole. Shaw's version has the advantage of being simpler than Rindler's version in that there are no bending forces acting upon the rod, thus having one less complication in a problem that is already difficult.

Also known as the Meter-Stick Paradox or the Bar-and-Plate Paradox, its standard form is stated as follows: Observer Able sees a straight rod oriented parallel to the X-axis flying in the positive X-direction at 86.6% of lightspeed (actually 86.6025403% of lightspeed to the accuracy of my calculator) and a flat plate, also oriented parallel to the X-axis, moving in the positive Y-direction at 10% of lightspeed. The rod appears to Able to have a length of one meter (which means that its proper length, the length it has in the frame in which it is at rest, is two meters) and the plate has in it a hole one meter wide (which is the proper width of the hole because that width is oriented parallel to the X-axis and perpendicular to the motion of the plate). We must assume, of course, that the ends of the bar and the edges of the hole in the plate are so beveled that the two bodies will not strike each other if they pass with any clearance, however small. We so establish the trajectories of the two bodies that the center of the rod and the center of the hole in the plate coincide at some instant, so Able sees the rod pass through the hole with barely enough room to spare. Observer Baker sees the same situation from an inertial frame in which the rod moves entirely in the negative Y-direction and the plate moves entirely in the negative X-direction. In that frame the rod is presumed to have its proper length of two meters and the hole in the plate is Lorentz-Fitzgerald contracted in the X-direction. In Baker's view will the rod pass through the hole without striking the plate?

If the answer to that question is No (and that seems to be a reasonable first guess), then Relativity is false to Reality; that is, it does not give us a correct description of the relation between space and time. But the answer is Yes, as you should expect by now, though reaching that answer and the details that support it will take us on a difficult and intensely algebraic journey ("Oy, caramba!" exclaimed the Jewish vaquero, a man who has his own problems with different frames of reference.). To that end we will work out the full Lorentz Transformation in which two spatial dimensions are distorted and we will accomplish that task by using a little bit of simple analytical geometry to parlay the one-dimensional Lorentz Transformation into its two-dimensional counterpart.

Before we take that journey I want to view the situation from two inertial frames that will offer some insights into what we may expect of the Lorentz Transformation in two spatial dimensions. Those frames are the ones occupied and marked, respectively, by the bar and the plate. In those frames the relative motion distorts only one spatial dimension, the Y-direction in the plate's frame and the X-direction in the bar's frame; nonetheless, an examination of those two perspectives will give us useful insights into the more complex case of Baker's inertial frame.

To ease the journey a bit I want to restate the orientation of our coordinate grid on the imaginary map that we will use as an aid to visualization. I say that the positive x-direction points east, the negative x-direction points west, the positive y-direction points north and the negative y-direction points south. We read our coordinate grid as we would read a map, so the positive x-direction (east) points to our right.

In the bar's inertial frame of reference Able's frame and everything in it appears to move in the negative x-direction (west) with the speed vx. We want to calculate the northward speed of the plate as seen from the bar by using the Pythagorean Theorem. We can, indeed, do that, but we must remember one caveat before we do: the speed of the plate parallel to the Y-axis is slower in the bar's frame than it is in Able's frame in accordance with

(Eq'n 1)

Now we use the Pythagorean Theorem to calculate the relative speed between the bar and the plate and obtain

(Eq'n 2)

The Lorentz factor that corresponds to that speed is simply (and rather neatly)

(Eq'n 3)

which is to say that the Lorentz factor between the bar's frame and the plate's frame equals the product of the Lorentz factors between the bar's frame and Able's frame and between the plate's frame and Able's frame.

The situation as seen from the bar's frame is that the bar is a full two meters long while the hole in the plate has been Lorentz-Fitzgerald contracted to a width of one-half meter. If we remind ourselves of why that contraction happens, we will see that something truly strange happens to the plate. Because the plate is properly oriented parallel to the x-axis, clocks on opposite sides of the hole will be out of synchrony; that is, even though both clocks are perfectly synchronized in the plate's frame and in Able's frame, in the bar's frame the clock on the trailing edge of the hole will be "faster" than the clock on the hole's leading edge by an interval equal to the proper width of the hole (one meter) multiplied by vx and divided by the square of lightspeed. Because the plate is also moving in the y-direction, at any given instant measured on the bar's clocks, the trailing edge of the hole will have progressed farther along the y-axis than the leading edge of the hole will have done, so the plate will appear to an observer moving with the bar as though it had been tilted away from the x-axis in the counterclockwise sense.

Suppose now that the leading edge of the bar just brushes past the trailing edge of the hole in the plate at some instant. At some later instant corresponding to the temporal offset between the clocks on the plate the leading edge of the hole will reach the y-coordinate occupied by the bar. Will the x-coordinates of the trailing edge of the bar and of the leading edge of the hole then be the same? Since the bar is confined to one y-coordinate in this situation, we need only ask how far apart in the x-direction are the points where the edges of the hole cross that y-coordinate. There's already a distance of x/Lx between the edges of the hole, where we have

(Eq'n 4)

as the Lorentz factor that we must apply here. We must then add to that distance the distance that the plate moves in the interval represented by the temporal offset between the clocks on the edges of the hole. But that temporal offset appears on the plate's clocks in dilated time; the corresponding time in the bar's frame equals the calculated interval multiplied by Lx. The distance that the plate moves equals that interval multiplied by vx, so the total distance between the points where the edges of the hole cross a given y-coordinate is

(Eq'n 5)

But x = 1 meter and Lx = 2, so )x = 2 meters, which is just the length of the bar. Thus, if the east end of the bar just brushes past the east edge of the hole, then the west end of the bar will just brush past the west edge of the hole, which is exactly what we need to see. The only difference between what the observer moving with the bar sees and what Able sees is that the passages of the edges are not simultaneous. But that's entirely consistent with what we already know of Relativity, so we have no cause to dismiss the theory here.

Now how does the situation look to someone riding on the plate? As seen by Able, the plate moves in the positive Y-direction (north) at the speed vy, so an observer moving with the plate would see Able's frame and everything in it moving in the negative Y-direction (south) at the speed vy. In Able's frame the bar moves in the positive X-direction (east) at the speed vx, so the plate-borne observer calculates the speed of the bar parallel to the X-axis to be

(Eq'n 6)

That is the X-ward component of velocity that must be used in the Pythagorean equation to calculate the speed of the bar relative to the plate as seen from the plate's frame. Again we use the Pythagorean Theorem to calculate the relative speed between the bar and the plate and obtain, as we certainly should if the Principle of Relativity is to be true to Reality, Equation 2. So we have

(Eq'n 7)

I approached this aspect of the bar-and-plate problem with an expectation that had been raised in my mind by the previous aspect of the problem. I expected that in the plate's frame the bar, because it moves slower than vx, would be longer than it is in the Able frame and that it would be so rotated in the clockwise sense that it would pass through the hole in the plate as required. But when I tried to set up the algebraic description of the scenario I found that I couldn't figure out the right way to apply the Lorentz Transformation; nothing that I tried gave the right result.

Frustrated, I looked again at the imaginary experiment that I was trying to describe and modified it slightly to clarify what was happening. I imagined my Able-frame observer erecting a series of fences, each fence built parallel to the y-axis and each fence having drilled in it a hole just wide enough to pass the bar. I so positioned the holes that a straight line drawn through their centers lies parallel to the x-axis of our coordinate frame and marks the trajectory of the bar as the Able-frame observer sees it.

In the plate's frame those fences are shrunk in the y-direction, all by the same amount, so the holes are all displaced toward the x-axis by the same amount; thus, the line tracing the bar's trajectory remains parallel to the x-axis as I go from the Able frame to the plate's frame. In the Able frame every point on the bar's centerline lies on that trajectory, so the same must be true in the plate's frame: if it were not, then the bar would collide with the first fence to which it comes. That inference necessitates, then, that the bar be oriented parallel to the x-axis in the plate's frame as well as in the Able frame. But the plate is also oriented parallel to the x-axis in both frames, so I must infer that the length of the bar is the same in both frames if all observers are to agree that the bar goes through the hole in the plate without striking the plate.

I must make that inference seemingly in spite of my earlier statement (encoded in Equation 6) that the bar moves more slowly in the x-direction in the plate's frame than it does in the Able frame. But if the bar moves more slowly in the x-direction in the plate's frame, then surely the Lorentz factor by which its length is contracted must also be smaller and the bar's length must therefore be longer than it is in the Able frame. That statement implies that I have discovered a self-contradiction in the theory of Relativity, which means (as it has every other time that I seem to have devised a paradox or contradiction) that I have failed here to take proper account of some minor but nonetheless relevant feature of the situation. In this case that feature is the fact that in the plate's frame the bar is also moving in the negative y-direction. When I take that fact into account, I can prove and verify my above inference through a straightforward relativistic calculation.

In your imagination look at the bar in its own frame. The bar has its proper length of x = 2 meters and that length is oriented parallel to the x-axis. The plate approaches the bar at a speed given by Equation 2 (and also Equation 7). If I imagine the x-axis as a horizontal straight line (positive toward the right) and the y-axis as a vertical straight line (positive toward the top of the page), then the motion of the plate lies on a line that slants across that diagram from lower right to upper left. That slanted line makes an angle qb with the x-axis and that angle can be determined from its sine, which is equal to the speed of the plate (as seen from the bar) in the y-direction divided by the overall speed of the plate (vplate = vbar = vo); that is, I have the equality

(Eq'n 8)

Likewise, I can calculate the cosine of the angle as being equal to the speed at which the plate moves in the negative x-direction (vx) divided by the overall speed, which gives me

(Eq'n 9)

Thus I have represented the overall velocity of the plate in the bar's frame as the hypotenuse of a right triangle whose sides are the components of the plate's velocity in the x- and y-directions.

The centerline of the bar and the plate's velocity vector also cross each other at the angle qb. That means that I can represent the length of the bar as the hypotenuse of a right triangle whose sides are parallel to the plate's velocity vector and perpendicular to the plate's velocity vector and that I can calculate the lengths of those sides by multiplying the bar's proper length by the cosine and the sine respectively of qb. The first of those lengths (x1) is given by

(Eq'n 10)

The other of those lengths (x2) is given by

(Eq'n 11)

Of those two component lengths, only the one oriented parallel to the velocity vector will be subject to the Lorentz-Fitzgerald contraction when I go from the bar's frame to the plate's frame; the other component length will be unchanged. The Lorentz factor that I must use in calculating the contraction is the one given by Equation 3, because it's based upon the relative velocity (vo) between the bar's frame and the plate's frame. The contracted length is thus

(Eq'n 12)

Now with that length and the unchanged component length I can use the Pythagorean theorem to calculate the length of the bar in the plate's frame and I obtain

(Eq'n 13)

which is just the length of the bar in the Able frame. Thus I have verified one part of my inference about the bar in the plate's frame.

The contraction of x1 while x2 remains unchanged makes the angle between the bar's centerline and the velocity vector different in the plate's frame from the corresponding angle in the bar's frame. I can determine that new angle, qp, from its sine, which I can calculate by dividing the unchanged length x2 by the bar's length in the plate's frame. I thus obtain

(Eq'n 14)

But that sine equals the sine of the angle that the velocity vector makes with the x-axis in the plate's frame. If you look again at the diagram that you have sketched in your imagination, you will see that those two angles must be equal to each other and that they lie on opposite sides of the line representing the velocity vector. Thus, with a nod of gratitude to Kyrios Euclid, simple geometry tells us that the bar is parallel to the x-axis in the plate's frame, as I inferred.

It is true to Reality, then, and true to Relativity as well that the orientation and length of the bar are the same in the plate's frame as they are in the Able frame. The only novelty that we have gained from this aspect of our imaginary experiment is the knowledge that the relative velocity between the bar and the plate makes different angles with the x-axis in the bar's frame and the plate's frame.

Now we want to look at the situation as Baker sees it; that is, we want to examine the situation in its maximum complexity. In so doing we will be deducing the form of the Lorentz Transformation in two spatial dimensions and looking at some of the complications that it entails.

Let's start with our two observers, Linda Able (who uses lower-case letters to represent her measurements of events) and Ursula Baker (who uses upper-case letters to represent her measurements of events), residents of our bizarre world in which the speed of light is equal to 100 miles per hour. Linda and Ursula are exploring their world's arctic region and find themselves far out on a perfectly flat, perfectly smooth glacier that seems to stretch to infinity in all directions. Mountain peaks poke up through the ice here and there and the ice has been scored with two sets of parallel lines that cross each other at right angles. Those lines have filled in with dark material blown off the mountains and then frozen over, thus preserving the smoothness of the ice while giving it the appearance of a vast bleached chessboard.

("Looks like an illustration from one of those 1950's science-fiction comics," Ursula comments. "Section, Town, and Range," Linda says. "Land Lot and District," Ursula replies in her Georgia drawl.)

Having a coordinate grid laid out for them, the women choose one line to be the x-axis and a line crossing it to be the y-axis. As before, the z-axis points in the vertical direction and can be ignored. Our explorers are aware of the bar and plate sliding frictionlessly across the ice, the bar sliding parallel to the x-axis at 86.6 miles per hour and the plate sliding parallel to the y-axis at 10 miles per hour. In this case the hole in the plate is a semicircular arch through which the bar can slide. In order to observe this phenomenon fully Ursula decides to take one of the expedition's propeller-driven sleds and go into a frame in which the motions of the bar and of the plate are interchanged as described above. The sled originally points along the x-axis, so Ursula must turn it to the left through an angle 2 before she accelerates to a speed U relative to Linda. One of the tracks that the sled gouges into the ice is defined by Linda to be the j-axis and Ursula defines the same line to be her J-axis. Linda then draws a straight line perpendicular to her j-axis and calls it her k-axis. Ursula so lays a straight beam across her sled that it is oriented perpendicular to her J-axis and uses it as the definition of her K-axis. Prior to her acceleration Ursula had also laid across her sled two beams parallel to Linda's x-axis and y-axis, labeling them respectively her X-axis and her Y-axis: because her acceleration proceeds in a straight line she assumes that those axes remain parallel to Linda's axes, a reasonable assumption since the angles between those axes and the J-axis don't change during the acceleration.

Linda starts the analysis of the situation by measuring the coordinates of an event that occurs at a point P = (Px, Py) at a time t as measured by a clock at that point (so that both Linda and Ursula will agree on the time of the event in Linda's frame). She wants to convert the spatial coordinates, which were measured parallel to the x- and y-axes respectively, into the equivalent coordinates measured parallel to the j- and k-axes. Analytic geometry gives her the rotation-of-coordinates equations that do the job:

(Eq'n 15)

(Eq'n 16)

Linda then communicates the coordinates (Pj, Pk, t) to Ursula, who applies the Lorentz Transformation to convert them into the equivalent coordinates (PJ, PK, T) in her J-K frame. Ursula thus obtains

(Eq'n 17)

(Eq'n 18)

(Eq'n 19)

in which the Lorentz factor

(Eq'n 20)

The negative signs in the first and the third of those transformation equations reflect the fact that in Ursula's frame Linda and her frame move in the negative J-direction, so Ursula must use transformation equations that are the inverse of the standard Lorentz Transformation.

She can now apply the inverse of the rotation equations that Linda used and convert the coordinates (PJ, PK) into the equivalent (PX, PY). She has thus

(Eq'n 21)

(Eq'n 22)

in which the angle theta is the same as the angle in Linda's rotation equations.

To expedite future such calculations Ursula uses Equations 17 and 18 to replace PJ and PK in Equations 21 and 22 with the corresponding Pj and Pk, then she uses Equations 15 and 16 to replace Pj and Pk with Px and Py in those equations and Equation 19. After gathering terms with like factors and writing them together, she obtains

(Eq'n 23)

(Eq'n 24)

(Eq'n 25)

Now if Linda gives Ursula the coordinates of any event measured in her frame, Ursula can use those equations to convert the coordinates directly into the equivalent coordinates that she would measure of the event in her frame. Equations 23, 24, and 25 thus comprise the Lorentz Transformation in two spatial dimentsions.

Of course Linda can use the same method to generate a trio of equations that will convert Ursula's measurements of events directly into the equivalent measurements that she would make in her own frame; those equations would be the inverse of the transformation embodied in Equations 23, 24, and 25. All she has to do is to switch the upper-case variables in Equations 15 through 22 and change the minus signs in Equations 17 and 19 into plus signs (thereby converting the Lorentz Transformation Ursula used into its inverse). When she works through the algebra and gathers together all terms with the same coordinates, she obtains

(Eq'n 26)

(Eq'n 27)

(Eq'n 28)

Linda can prove and verify the claim that those equations comprise the inverse of the transformation that Ursula devised in Equations 23 - 25. All that she must do is to substitute the formulae from Equations 23 - 25 for the equivalent coordinates in her equations and work through the algebra. At first she writes down what looks like an algebraic disaster area, but once she gathers together all terms having the same coordinate and recalls that sin2q + cos2q = 1 she readily collapses what she has written down into px = px, py = py, and t = t, which is exactly what she should get if her equations are the inverse of Ursula's.

With Equations 23 - 25 in hand Ursula can solve a problem that has been bothering her as her sled flashes across the ice. She has gained the distinct impression that in her frame the grid that she and Linda are exploring is no longer rectangular. It occurs to her to think that the coordinate transformation she has devised may encode an effect that has distorted the grid and now she wants to decode that effect. To that end she makes T = 0 to represent the fact that, were it not for the delay due to the finite speed of light, she would see every point in her frame at the same instant (that is, all of the synchronized clocks in her frame would display the same time). She then solves Equation 17 for the times that she would thus observe on Linda's clocks at the points identified by the coordinates (Px, Py) and obtains

(Eq'n 29)

For her next step she makes Py = 0 and allows Px to take arbitrary values (that is, Px = x), which is simply a description of Linda's x-axis. Substituting those values and Equation 26 into Equations 23 and 24 then gives her

(Eq'n 30)

and

(Eq'n 31)

Because 1/L < 1, those equations show Linda's x-axis contracted and tipped "down" in Ursula's negative Y-direction: the axis appears to Ursula to be rotated clockwise. Those equations can be seen as the two-dimensional analogue of the Lorentz-Fitzgerald contraction of an object of length x oriented parallel to Linda's x-axis. On the left side of those equations the eks in square brackets is an index, not a multiplier; the equations are actually read "Pee sub upper-case eks, as a function of lower-case eks, is equal to...." and "Pee sub upper-case wye, as a function of lower-case eks, is equal to...." That ploy is necessary to distinguish those two equations from the ones that we will obtain in the next step. In visualizing the axis relative to Ursula's axes it helps to notice that the inverse of the Lorentz factor is less than one, so PY will be a negative number: that means that in Ursula's frame Linda's x-axis tilts "down" in the Y-direction; that is, it appears to have rotated clockwise from an orientation parallel to Ursula's X-axis.

Making Px = 0 and Py = y, which is the description of Linda's y-axis, and using them with Equation 18 as before, Ursula obtains the equations

(Eq'n 32)

and

(Eq'n 33)

In this case PX is a negative number, which means that in Ursula's frame Linda's y-axis appears to have rotated counterclockwise from an orientation parallel to Ursula's Y-axis.

Those four equations answer Ursula's question about the grid. Equation 27 tells Ursula that Linda's x-axis (and any lines parallel to it) is shrunk in her X-direction and Equation 28 tells her that Linda's positive x-direction is oriented partly in her negative Y-direction. Likewise, Equation 29 tells Ursula that Linda's positive y-direction is oriented partly in her negative X-direction and Equation 30 tells her that Linda's y-axis is shrunk in her Y-direction. Thus Ursula confirms that in her frame Linda's x- and y-axes are splayed farther away from their common j/J-axis than they are in Linda's frame; what appear to Linda to be rectangles scored into the ice appear to Ursula as diamonds. That fact is going to give us a significant complication of the bar and plate problem.

As a last step in this part of the process, Ursula makes q = 0 (which makes Sinq = 0 and Cosq = 1). That change makes Equations 23, 24, and 25 turn into the equations of the inverse Lorentz Transformation in one spatial dimension; that is, when the relative velocity between the two inertial frames is oriented entirely in the x-direction. This confirms that the equations deduced so far have the correct mathematical form.

For the next part of the process, Ursula wants to derive equations that will transform velocities measured in Linda's frame into the corresponding velocities in her frame. To accomplish that feat she simply does what we do in deriving the equation for compound velocities in one spatial dimension; in that derivation we divide the equation for spatial distortion by the equation of temporal distortion. In this case, though, Ursula must divide both Equations 23 and 24 by Equation 25 and then exploit the fact that Px/t = vx and Py/t = vy. The result gives her

(Eq'n 34)

and

(Eq'n 35)

Thus, if Linda measures a velocity with components vx in the x-direction and vy in the y-direction, then Ursula will calculate the corresponding velocity in her frame to have components VX in her X-direction and VY in her Y-direction in accordance with the above equations. I can now use those equations to determine the formulae from which I can calculate the values of U and Sinq for the bar-and-plate problem.

In Linda's frame I have been given the velocities of the bar (vx = vb, vy = 0) and of the plate (vx = 0, vy = vp). In Ursula's frame, I am told, the bar moves purely parallel to the Y-axis and the plate moves purely parallel to the X-axis, but I am not given the speeds of the bodies. It seems reasonable to think that the speed that Ursula measures of the bar is equal to the speed that Linda measures of the plate and that the speed that the speed that Ursula measures of the plate is equal to the speed that Linda measures of the bar. But, having been confounded before, I want to be careful to confirm what seems reasonable. I know, via the Principle of Relativity, that both Linda and Ursula will measure the same number for U (that is, the same number of kilometers per second, miles per hour, etc.). And both women will measure the same angle 2 between their respective x-axes and the direction of U. Do those two facts necessitate what seems reasonable?

If I consider only the bar, I know that there is an infinite number of U-q combinations that will convert its given motion purely parallel to Linda's x-axis to a motion purely parallel to Ursula's Y-axis, each combination canceling the x-ward motion and imposing the Y-ward motion (each combination corresponding to a different VY). Likewise, if I consider only the plate, I know that there is an infinite number of U-q combinations that will convert its given motion purely parallel to Linda's y-axis into a motion purely parallel to Ursula's X-axis, each combination canceling the y-ward motion and imposing the X-ward motion (each combination corresponding to a different VX). But there is one and only one, perfectly unique U-q combination that will precisely cancel the bar's x-ward motion and at the same time precisely cancel the plate's y-ward motion, thereby converting the bar's given motion parallel to Linda's x-axis into a motion purely parallel to Ursula's Y-axis and also converting the plate's given motion parallel to Linda's y-axis into a motion purely parallel to Ursula's X-axis.

That unique U-q combination, representing the change of velocity needed to take an observer from Linda's frame into Ursula's frame, entails a transformation that converts the Given Motions of the bar and the plate in Linda's frame into New Motions in Ursula's frame. The reverse of that combination, a change of velocity in the opposite direction, entails an inverse transformation of the velocities of the bar and the plate that will convert the New Motions into the Given Motions. Because U and q are the same for the reverse shift of inertial frames as they are for the initial shift from Linda's to Ursula's frame, the inverse transformation must have the same mathematical form as does the initial transformation; that is, it must have the same algebraic form but with minus signs replacing some of the plus signs and vice versa.

Now I imagine temporarily rotating the scene by 180 degrees and interchanging the algebraic signs on the coordinate axes (e.g. the negative X-direction becomes the new positive X-direction, and so on). In this rotation the New Motions, which are oriented in the negative X- and Y-directions, are changed temporarily into their positive counterparts and the Given Motions are changed into their negative counterparts. Likewise, the transformation and its inverse must be interchanged because the U-q combination that gets us from Ursula's frame to Linda's frame is now indistinguishable from the U-q combination that got us from Linda's frame to Ursula's. That indistinguishability, in accordance with the Principle of Relativity, must go deep into the foundations of Reality; a simple rotation and reflection of coordinates cannot affect the laws of physics, because those laws know nothing of our coordinate systems, which are merely mathematical fictions that we employ merely to describe (and not to prescribe) the laws of physics. What that means in particular is that the two velocities that are precisely canceled by the U-q combination that takes an observer from Ursula's frame to Linda's frame must be equal to the negatives of the two velocities that are precisely canceled by the U-q combination that takes an observer from Linda's frame to Ursula's. Going back to the original orientation of the scene with that fact in mind, I now know that in Ursula's frame the velocity of the bar must be VX = 0, VY = -vp and the velocity of the plate must be VX = -vb, VY = 0.

To determine the value of U I substitute the velocities that both Linda and Ursula measure of the plate into Equations 34 and 35 to obtain

(Eq'n 36)

and

(Eq'n 37)

I multiply Equation 36 by Cosq, multiply Equation 37 by Sinq, and add the resulting two equations to obtain

(Eq'n 38)

I can solve that equation readily for U, which comes out as

(Eq'n 39)

For the corresponding Lorentz factor I have

(Eq'n 40)

Before I can determine a formula for the value of Sinq, I need to prepare two equations from the velocity transformation equations. First I multiply Equation 36 by Sinq, multiply Equation 37 by Cosq, and subtract the second so-modified equation from the first. After a little algebraic manipulation of the result, I obtain

(Eq'n 41)

Next I substitute Linda's and Ursula's measured velocities of the bar into the transformation equations to obtain

(Eq'n 42)

and

(Eq'n 43)

I multiply the first of those equations by Sinq, multiply the second equation by Cosq, and subtract the first so-modified equation from the second. A little algebraic manipulation of the result gives me

(Eq'n 44)

If I multiply Equation 41 by vb2Sin2q/vp2Cos2q, I can equate the result to Equation 44. Dividing that new equation by L and multiplying it by vp2Cos2q, I obtain

(Eq'n 45)

which I will solve for Sinq. I replace U by the equivalent formula from Equation 39, multiply the result by the denominator of that formula, and cancel the equal terms of opposite on each side of the equality sign to obtain

(Eq'n 46)

Finally, I replace Cos2q with the equivalent 1-Sin2q and solve the equation by the standard algebraic methodology (which includes, in this case, completing the square), obtaining

(Eq'n 47)

By way of the Pythagorean theorem I also obtain

(Eq'n 48)

I can now calculate numerical values for U, L, the sine, and the cosine that are involved in the bar-and-plate problem. My calculator displays nine decimal places, so I have displayed those places here. I have vp = 0.1 and vb = 0.866025403 and the equations actually let me use those fractions rather than their kilometers per second equivalents because the relevant coefficients involve ratios of velocities and not the velocities themselves. Thus, I have:

Sinq = 0.057928444,

Cosq = 0.998320736,

q = 3.320914888 degrees,

U = - 0.866026617c,

a negative number because Able frame is moving in the negative X- and Y-directions in Baker frame, and

L = 2.000008405.

Because of rounding errors in the calculator, the angle that my calculator inferred from the sine and from the cosine differed slightly from each other (3.320914459 degrees and 3.320915316 degrees respectively), so I used the median value. That ploy will not affect my solution of the bar-and-plate problem, because I won't actually use the numbers in devising the solution.

Now I can solve the bar-and-plate problem; that is, I can see whether a proper relativistic description of the situation in Ursula's frame has the bar passing through the hole in the plate without striking the plate.

I begin by recalculating the velocities of the bar and of the plate in Ursula's frame. Because we chose the relative motion between Ursula's and Linda's frames specifically to produce an interchange of speeds between the bar and the plate, I can express those objects' velocities as functions of that relative motion.

For the bar I have

(Eq'n 49)

The first term on the right of the second equality sign represents the motion that the bar shares directly with Linda's frame in Ursula's negative Y-direction and the second term represents the additional velocity due to the combination of the slope of the line that the bar follows in Linda's frame and the motion of that line in Ursula's negative X-direction. That slope is calculated directly as the ratio of Equation 31 to Equation 30; that is,

(Eq'n 50)

Thus, I have for the bar

(Eq'n 51)

For the plate I have similarly

(Eq'n 52)

In this case the second term on the right side of the second equality sign is a combination of the motion of Linda's frame in Ursula's negative X-direction and the slope of the line that the plate follows in Linda's frame, a line parallel to Linda's y-axis. That slope is calculated directly as the ratio of Equation 32 to Equation 33; that is,

(Eq'n 53)

Thus, I have for the plate

(Eq'n 54)

The ratio of Equation 54 to Equation 51,

(Eq'n 55)

is the criterion by which I will know whether I have correctly solved the bar-and-plate problem. After I work out a description of the bar and of the plate in Ursula's frame, I will assume that at some instant the right end of the bar just brushes past the right edge of the hole in the plate. I will calculate the X-component (X') and the Y-component (Y') of the distance separating the left end of the bar from the left edge of the hole in the plate at that instant and then calculate their ratio. If the ratio of X' to Y' is equal to R(X:Y), then I will know that my solution is correct, that my description of the bar and of the plate is true to Reality in Ursula's frame.

In my earlier discussion of the problem I described Linda (the Able-frame's observer) erecting a series of fences parallel to her y-axis and leaving in each fence a hole through which the bar can pass. A line drawn through the centers of the holes was parallel to Linda's x-axis. Because the fences are motionless in Linda's frame, they will appear to Ursula to be distorted in the same way that Linda's axes appear distorted; that is, the fences themselves will still be parallel to Linda's y-axis and the centerline through the holes will still be parallel to Linda's x-axis. Because the bar may be passing through more than one fence at any given instant, I know that it must be oriented along the centerline, so it will remain parallel to Linda's x-axis in all frames. I can now quantify the orientation of the bar by calculating the slope of Linda's x-axis in Ursula's frame. That slope is simply the ratio of Equation 31 to Equation 30; that is,

(Eq'n 56)

The Lorentz factor is always equal to or greater than one, so the numerator of that ratio is always a negative number and the ratio itself, consequently, is a negative number. A negative slope indicates a line slanting from upper left to lower right, so Ursula would see the bar rotated clockwise from the orientation naively inferred from the orientation that Linda describes ("perfectly horizontal").

I assume now that sliding gates cover the holes in the fences and that each opens when the right end of the bar reaches it and closes when the left end of the bar passes it. What I'm setting up her is the analogue, in two spatial dimensions, of the rails-through-the-car-barn problem in the essay on paradoxes. Of course, the gates all have clocks that record the times of their openings and closings. In Linda's frame, at some instant, one gate is opening and another is closing. The clocks on those two gates display the same time (call it tg) and the distance x between the gates is simply the length of the bar in Linda's frame.

In Ursula's frame those clocks appear to be offset, each from the other, by an interval equal to xUCos2/c2, the clock on the right advanced over the clock on the left. If Ursula sees the right end of the bar as it reaches the gate on the right, then she will see that the left end of the bar has not yet reached the gate on the left. If she photographs the bar at that instant, how far will she measure the left end of the bar to be from the right end?

Two simultaneous events that occur a distance x apart in Linda's frame are displaced each from the other in Ursula's frame by

(Eq'n 57)

(Eq'n 58)

and

(Eq'n 59)

But that's not the description of the bar that Ursula would derive from her measurements. Because the bar is not moving in the X-direction in Ursula's frame, the X-coordinates of its ends can be described quite simply: they are the X-coordinates of the points that Linda's gates occupy when their clocks display tg. In Ursula's photograph of the bar the clock at the right end of the bar displays tg and the left clock is a distance X to its left in the X-direction and a distance Y to the north of it (above it) in the Y-direction. Linda's frame moves to the left (in the negative X-direction) at a speed UCosq, so in the time T (until it displays tg) the left clock will move a distance UCosqT. The X-ward distance between the ends of the bar is thus

(Eq'n 60)

(In that derivation I exploited the fact that 1/L2 = 1-U2/c2).

Because I know how the bar tilts, I have a straightforward way to calculate the relative Y-coordinate of its ends. I multiply Xb, from Equation 60, by the slope in Equation 56 and I thereby obtain

(Eq'n 61)

As I did with the bar, I can calculate the relative coordinates of the extreme left edge and the extreme right edge of the hole in the plate by coordinating those points with objects that are stationary in Linda's frame. The objects that I choose to use are a pair of rails oriented parallel to Linda's y-axis and set a distance x apart. The edges of the hole in the plate slide along the rails. At the points where the rails cross Linda's x-axis a pair of catches mounted on the rails would impede the motion of the plate but for the fact that they swing open at the instant the plate reaches them. Clocks attached to the catches record the time that they open and, because the plate is parallel to Linda's x-axis, they both record the same time.

In Ursula's frame the relative coordinates of the events defined by the opening of the catches are described by Equations 57, 58, and 59. The rails remain parallel to Linda's y-axis, so in Ursula's frame they slope at a rate that I calculate by dividing Equation 33 by Equation 32 to obtain

(Eq'n 62)

The rails slide through the hole in the plate, moving in the negative Y-direction at the speed USinq. The right catch reaches the plate and opens, then a time xLUCosq/c2 later the left catch reaches the plate and opens. I calculate the relative Y-coordinate of the edges of the hole by adding the distance the rails travel in the Y-direction in that interval to the negative distance taken from Equation 58 and obtain

(Eq'n 63)

In order to calculate the relative X-coordinate I must subtract from the )X given by Equation 57 a number obtained by multiplying the inverse of the slope given in Equation 62 by the distance the rails travel in the interval xLUCosq/c2. In the calculation I am envisioning the length of the right rail between the right catch and the point adjacent to the right edge of the plate when the left catch opens as the hypotenuse of a right triangle whose sides are parallel to Ursula's X- and Y-axes. The length of the X-side of that triangle can be calculated by multiplying the length of the Y-side by the inverse slope of the hypotenuse. Thus I obtain

(Eq'n 64)

in which I have exploited the fact that U2/c2 = 1-1/L2 and I have changed to in order to make the length of the X-side of my imaginary triangle a positive number (though I could just as easily have done the calculation as addition with a negative number).

Yp will be a positive number and Xp will be positive or negative, depending upon the magnitude of the Lorentz factor. That means that, as Ursula sees it, the plate appears to be rotated counterclockwise from the orientation that would be inferred from a naive interpretation of Linda's description of its orientation ("perfectly horizontal").

At some instant, I assume, the right end of the bar just brushes past the right edge of the hole in the plate. At that instant the left end of the bar lies some distance from the left edge of the hole in the plate. I want to calculate the X- and Y-components of that distance. Because of the way in which the bar and the plate are tilted I calculate the Y-component by adding the Y-values in Equations 61 and 63 and I calculate the X-component by subtracting the X-value in Equation 64 from the X-value in Equation 60. I thus obtain

(Eq'n 65)

and

(Eq'n 66)

The ratio of X' to Y' is thus

(Eq'n 67)

That ratio is, as required, equal to the ratio shown in Equation 55. That means that if I divide Equation 65 by Equation 51 or divide Equation 66 by Equation 54, I will obtain an interval

(Eq'n 68)

In that interval the plate, moving at the speed vb, will cross the distance X' and the bar, moving at the speed vp, will cross the distance Y' and, consequently, the left end of the bar will just brush past the left edge of the hole in the plate;