Back to ContentsWe have made the concept of the clock fundamental to the theory of Relativity, the touchstone against which we test our hypotheses about the phenomena that the theory describes. Thus, we have deduced the main features of the Lorentz Transformation through the imaginary study of moving clocks. Further, the principle of Relativity tells us that time dilation and the temporal offset occur in all clocks in exactly the same way, regardless of the mechanism by which those clocks count time: that fact tells us why we can use imaginary Feynman clocks to infer the rules of the Lorentz Transformation. What, then, can we make of any claim that some given clock mechanism does not conform to the rules of Relativity? We have come too far to dismiss Relativity easily, so we feel a strong temptation to dismiss the claim outright as false to Reality. But I know of a simple clock mechanism that should give us pause in yielding to that temptation.
I conceived a simple wigwag clock that consists of an elliptical frame, a laser, a photocell connected electrically to a counter, a set of rubber-band-like springs, and a wide, massive plate that has a narrow slot in it. When the clock has been assembled the frame lies in the Y-Z plane with its center at the origin of our coordinate grid. We attach the laser to the frame where it crosses the positive Z-axis at the point Z=+a and attach the photocell where the frame crosses the negative Z-axis at the point Z=-a. Thus, when we activate the laser its ray flies parallel to our Z-axis from Z=+a to Z=-a. We attach the plate to the springs, which we attach to the frame where it crosses the Y-axis, so the plate will oscillate in the Y-direction. Because the slot in the plate now lies oriented parallel to our X-axis, twice each cycle the slot lines up with the laser’s ray, allowing a short pulse of light to pass to the photocell, which then advances the counter by one unit.
We can use simple Newtonian mechanics to calculate how frequently the counter will advance. If we make the amplitude of the oscillations small relative to the length of the springs, then we can use Hooke’s law to describe the motion of the plate as though it were attached to the end of a single spring whose length we have oriented parallel to our Y-axis. Hooke’s law tells us that the force acting to bring the end of a stretched spring back to its equilibrium position is proportional to the distance that it has been pulled or pushed away from the equilibrium position (which in this case puts the slot at Y=0) and proportional to a constant (the "spring constant", usually represented by the lower-case letter kay) that represents the stiffness of the spring. If we imagine attaching a massive body to the end of the spring, then we can equate the rate at which that body’s linear momentum changes to the force described by Hooke’s law and thus devise a simple differential equation that we can solve for the distance that the body moves along the Y-axis as a function of the elapsed time. The solution simply equals the product of the amplitude of the oscillation and the sine of an angle that equals the product of the system’s inherent angular frequency and the elapsed time; that is, the body will go from one extreme distance from the equilibrium position to the opposite one and back again with a smooth sinusoidal motion. In fact, the motion of a mass on a spring mimics exactly the horizontal motion of the main drive rod of a steam locomotive whose drive wheels turn at a uniform rate.
Hooke’s law gives us for a spring displaced a distance Y from its equilibrium position
If we have a body of mass M attached to the end of that spring the equation of motion thus comes out as,
We have the conventional solution of that equation as the sinusoidal function,
in which Y0 represents the amplitude of the oscillation (the maximum displacement of the free end of the spring from the equilibrium position) andω represents the angular frequency (measured in radians per second) of the oscillation. If we substitute that expression for Y in Equation 2 and carry out the indicated differentiation, we obtain the equivalent algebraic equation
which has the solution
In our wigwag clock at rest in our inertial frame of reference M represents the mass of the plate and k represents the collective spring constant of the rubber-bands.
Now let’s imagine that a rocketship carrying a wigwag clock identical to ours flies past us at some speed V oriented parallel to our X-axis. The time dilation theorem leads us to expect that the moving wigwag will oscillate at a frequency slower than the frequency of our stationary wigwag in proportion to the Lorentz factor between the rocketship and us; that is, the frequency of the rocketship’s wigwag should be related to the frequency of our wigwag by the relation
the Lorentz factor. To make sure that Equation 6 stands true to Reality we can rederive the angular frequency from the equation of motion, albeit one modified to accommodate what we know about Relativity.
We use Newton’s second law of motion to determine how the relative motion of the rocketship affects Hooke’s law. We know that linear momentum enacted in a direction perpendicular to the direction of relative motion between us and the rocketship remains unchanged by the relative motion, so any change in the linear momentum of the plate would be the same for us as it is for the observers on the rocketship. But the rate at which that change happens depends upon whose clocks we use to measure how long it took to come manifest. Our clocks tick off more seconds between any two given events than the rocketship’s clocks do, so we calculate a proportionately weaker force in the rocketship’s rubber bands than do observers on the rocketship. Because the distance that the plate in the rocketship’s wigwag clock moves does not change in the Lorentz Transformation, we infer that the spring constant of the rubber bands in the rocketship’s wigwag is less than it is in our clock; that is, k’=k/L.
That result conforms to what I have written in "The Relativity of Force". In that essay we deduced the fact that forces exerted parallel to a relative motion don’t change over those in system at rest while forces exerted perpendicular to the direction of relative motion diminish by the Lorentz factor between the two frames,
We will need those equations in what follows.
Knowing that the mass of the plate in the rocketship’s clock has increased due to relative motion (M’=ML), we can then use Equation 4 in the form
to calculate the angular frequency of the rocketship’s clock as we would measure it. We get
as we require.
Now let’s rebuild the rocketship’s clock. We remove the rubber-band springs and attach the plate to the free end of a long, thin, flexible rod oriented parallel to the X-direction. In this device we make the maximum distance Y0 that the plate moves away from its equilibrium position so small relative to the length of the rod (X0) that we can make the approximation in both frames that Sinθ=θ, in which θ=Y0/X0 represents the maximum angle through which this pendulum swings away from equilibrium.
In this version of the clock a torque rather than a force acts to bring the plate back to its equilibrium position. To describe the operation of the clock in two different inertial frames we need to work out the Lorentz Transformation of torque. To that end we need to see how torque relates to force.
When the free end of a rod gets deflected away from its equilibrium position, the material on one side of the beam stretches and the material on the opposite side compresses. Those strains in the material of the beam generate forces that act to restore the beam to its equilibrium shape. That such restoring forces arise lies inherent in the understanding that the particles that comprise a solid body exist in a state of continuous isometric tension. The interplay between attraction and repulsion within the material has such a form that when the particles get pulled away from each other the attraction dominates and brings the particles back to their equilibrium positions and when the particles get pushed toward each other the repulsion dominates and also brings the particles back to their equilibrium positions. We see those attractions and repulsions manifested on the gross scale in a torque that acts to twist the beam back to its equilibrium state when someone displaces it from that state. Dividing that torque by the length of the beam tells us the force that acts on the end of the beam to resist displacement from its equilibrium position.
To assist in the analysis I want to employ several simplifying assumptions. First, I assume that I have used a rod so light that it effectively has no mass when compared to the plate attached to its free end. To simplify my mathematical description, then, I assume that, indeed, the rod has zero mass and, thus, that the plate carries all of the system’s mass M. And second, I assume that the strains in the rod due to the stresses caused by deflection of the rod away from equilibrium all occur near the base of the rod, the point where the rod attaches to a massive body that holds it steady. Thus I envision the rod tilting to and fro as a perfectly rigid body, with all bending occurring at the base.
Now we devise the angular analogue of Equation 2.
In this model the applied torque occurs at the base of the rod. If we measure y from the centerline of the rod, then a deflection of the rod by an angleθ causes a strain proportional to that distance, compression on one side of the centerline and stretch on the other. With a rectangular cross section of width z0 and height 2y0, the rod has a stiffness per unit area of
On any minuscule element of area the strain produces a force density of
in which we have the strainΔx=yθ. To calculate the torque we multiply that force density by the minuscule element of area dzdy, multiply that product by the distance y between that element and the rod’s centerline, then integrate over z from zero to z0 and over y from -y0 to +y0. We thus get
in which K represents a constant.
The reaction torque occurs at the opposite end of the rod. We calculate that torque by multiplying the reaction force (the Newtonian eM-ay) of the plate by the plate’s distance (x0) from the base of the rod, getting
In that equation I’ve made use of the fact that y=x0θ.
Equating the torques of Equations 13 and 14 and then rearranging the equality gives us
which has the algebraic form of Equation 2. That fact tells us that Equation 15 has a solution algebraically identical to the solution of Equation 2; specifically, we have
in which we have the angular frequency as
Thus we obtain the mathematical description of the operation of this particular wigwag clock when it occupies the proper frame, the inertial frame in which the clock remains at rest.
Next we assign an observer to occupy an inertial frame moving parallel to the proper frame’s x-axis at a speed that puts a Lorentz factor L between the frames. That observer analyzes the clock’s timing mechanism through the angular analogue of Equation 9, the moving-frame version of Equation 15,
That equation has the solution
How does that angular frequency relate to the angular frequency calculated from Equation 17, the proper angular frequency? To answer that question we simply transform the variables in Equation 20.
We know immediately that x0'=x0/L due to the Lorentz-Fitzgerald contraction and that M’=LM due to the increased mass of moving bodies. As for the constant K, a look at Equation 13 shows us that it consists of the product of a numerical constant, y- and z-extents (which remain invariant under the Lorentz Transformation), and the kappa factor from Equation 12. Because force exerted in the x-direction (the direction of relative motion) remains invariant under the Lorentz Transformation, we have the force densities asφ’=φ, so we get κ’=Lκ, which leads to K’=LK for the angular spring constant. When we make the appropriate substitutions from those equivalents into Equation 20 and compare the result with Equation 17, we get
which is clearly wrong. The time dilation theorem tells us that in the moving frame the wigwag’s frequency will decrease, not increase, over the wigwag’s proper frequency.
I need to find my errors and correct them. I assume that I did, in fact, make errors, because the logic of Special Relativity has become too compelling at this stage of development for us to question the validity of the theory.
We can find one error when we re-examine Equation 12. In devising the transformation of the kappa factor I tacitly assumed that the strainΔx would shrink in accordance with the Lorentz-Fitzgerald contraction. But the equation Δx=yθ, properly understood, tells us something different. The lateral distance y remains invariant, certainly, but the angular deflection increases in proportion to the contraction of the rod; thus, the strain must also increase. That fact necessitates that we rewrite the transformation of the kappa factor as κ’=κ/L, which, in turn, makes the spring constant transform as K’=K/L.
The other error comes clear when we re-examine Equation 14 in light of a simple conceptual experiment. Consider the right-angle bracket problem.
Imagine a bracket that consists of two arms that meet each other at a right angle. We so establish the bracket in our proper frame that it remains free to pivot about the origin of the frame’s coordinate grid. One arm, of length y0, extends along the positive y-direction and at the free end we apply a force of magnitude fx oriented in the positive x-direction; thus, we exert a torqueτcw=fxy0 in the clockwise sense. The other arm, of length x0, extends along the positive x-direction and at the free end we apply a force of magnitude fy oriented in the positive y-direction; thus we exert a torque τccw=fyx0 in the counterclockwise sense. If we specify that x0=y0 and fx=fy, then the two torques have equal magnitudes and, thus, cancel each other out: now we know that the bracket doesn’t rotate.
In the moving frame we have fx’=fx and y0'=y0, so the clockwise torque remains invariant under the Lorentz Transformation. For the x-ward arm the moving observer writes fy’=fy/L and x0'=x0/L, which makesτ’ccw=τccw/L2. That torque won’t balance the clockwise torque, so the moving observer may feel a temptation to declare that the bracket will rotate in the clockwise sense. But if the bracket doesn’t rotate in the proper frame, then it certainly can’t rotate in the moving frame. Where, then, did the moving observer go wrong?
We note that the Lorentz-Fitzgerald contraction is not a fundamental part of the Lorentz Transformation. An examination of the first of the four equations of the Lorentz Transformation shows us that the raw transformation of lengths oriented in the direction of relative motion consists of a dilation, not a contraction. The contraction comes from applying the temporal offset combined with the relative motion of an object to the object’s fundamentally dilated length: relative motion shoves the rear of the object into the future relative to the front of the object, so the rear partly overtakes the front, thereby producing the contraction. If the temporal offset does not come into play in calculating torque or the associated moment of inertia, then the moving observer must write x0'=Lx0 instead of x0'=x0/L.
With that change the moving observer recalculates the counterclockwise torque and getsτ’ccw=fy’x0'=(fy/L)(Lx0)=fyx0=τccw. Thus the two torques balance in the moving frame and the moving observer asserts that the bracket does not rotate, in agreement with the stationary observer. In addition, the moving observer has demonstrated, as a general rule, that any torque exerted about an axis oriented perpendicular to the direction of relative motion remains invariant under a Lorentz Transformation.
That result obliges me to revisit my first error. Given now that the rod dilates in the x-direction, we must infer that the angular displacement of the rod contracts in the same proportion (because the lateral displacement of the rod’s free end remains invariant); thus the strain at the rod’s base contracts (though not for the reason we originally asserted) and Equation 12 leads the moving observer to recalculate the rod’s kappa factor asκ’=Lκ, which leads to the spring constant becoming K’=LK.
Making the appropriate substitutions into Equation 20, the moving observer recalculates the angular frequency of the rod’s vibrations in his frame as
which, as required, conforms to the time dilation theorem.
Thus we see how analyzing clock mechanisms in light of the time dilation theorem leads us to improve and refine our knowledge and understanding of Special Relativity. We have here a minor example of how we can parlay prior knowledge into new knowledge, in accordance with the Rationalist plan of the Map of Physics. I have full confidence that pursuing that plan will eventually reveal to us the full nature of Reality.
Appendix: Solving the Oscillator Equation
In Equation 2 we have an example of a second-order differential equation, mathematical entities that we encounter frequently in physics. Generally we can’t solve differential equations of second or higher order directly, as we would with an algebraic equation, but must resort to various tricks. However, we can use operator algebra to convert the problem of solving the oscillator equation into a problem of solving a first-order differential equation, which we know how to do directly.
If we have a spring of force constant k attached to an immovable wall or other support and oriented in the y-direction and if we attach a body of mass m to the spring’s free end, then if we displace that body away from the spring’s equilibrium point, we have the force equation applicable to that body’s motion as
If we define a new constantω2=k/m, then we can rewrite that equation as
In that equation I have rewritten the differentiation operator as∂t=d/dt and written it as a square to indicate the fact that the second derivative of a function comes from applying the first derivative operator on the function twice.
Next I factor the operator, so Equation A-2 becomes
Because we apply the two new operators, the expressions in parentheses, in sequence and because they differ from each other, we infer that one zeroes out the equation and the other doesn’t. Assume that the operator on the right zeroes out the equation. We thus have
To solve that equation add iωy to both sides, multiply by dt, divide by y, and integrate. That procedure yields
which gives us
In that latter equation I have tacitly assumed that C, the constant of integration, equals the natural logarithm of the amplitude of the oscillation. If we reverse the operators in Equation A-3 and apply them to that solution, we get
as we require. That calculation validates our assumption that one of the operators in Equation A-3 zeroes out the equation and the other doesn’t.
If we had chosen (∂t+iω) as the operator that zeroes out Equation A-2, then our procedure would have given us
as the solution. As with any quadratic equation, a second-order differential equation has two fundamental solutions and the actual solution for a specific system equals some linear combination of those two solutions. Thus we go from Equation 2 to Equation 3 in the above essay.
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