Relativistic Angular Momentum
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Imagine floating at rest in an inertial frame of reference on which someone has imposed a four-dimensional Cartesian coordinate grid with an array of identical rulers and synchronized clocks. We call that entity the proper frame and represent any measurements that observers occupying that frame make with unprimed algebraic symbols. A second inertial frame moves past us in the positive direction parallel to both frames’ common x-axis at a speed v that puts a Lorentz factorã between the two frames. We represent any measurements made by observers occupying that second, moving frame with primed algebraic symbols. If, for relevant example, observers in the moving frame measure a force f’=(fx’,fy’,fz’) exerted upon some body, then we observers in the proper frame will measure that force (albeit indirectly) as
We want to parlay that knowledge into a description of how angular momentum appears to observers in different inertial frames. To that end we must examine the rotary analogue of force – torque.
Envision experimenters in the moving frame mounting a right-angle bracket on a pivot at the origin of their coordinate grid. For a simple pivot a narrow hole at the point where the bracket’s two arms meet each other slips over a nail that lies on the grid’s z’-axis. One of the bracket’s arms, of length y’, lies on the grid’s positive y’-axis and the other arm, of length x’=y’, lies on the grid’s positive x’-axis.
To exert a simple torque upon the bracket the experimenters exert a force fx’, acting only in the positive x’-direction, to the free end of the bracket’s y-arm. That force exerts on the bracket a clockwise torqueτ’(-)=fx’y’. The experimenters also apply a force fy’=fx’, acting only in the positive y’-direction, to the free end of the bracket’s x-arm, thereby producing a counterclockwise torque τ’(+)=fy’x’. Because of the equality in magnitude of the applied forces and of the lengths of the bracket’s arms, the moving observers have τ’(-)=τ’(+), which means that the torques balance each other and, thus, that the bracket does not rotate.
Observers in the proper frame also won’t see the bracket rotate. If that statement did not stand true to Reality, then the proper-frame observers would see the bracket’s arms undergoing continuous reorientation that does not conform at all to the Lorentz Transformation between the two frames. Thus, for observers in the proper frame the torques must transform in a way that makes rotation of the bracket not happen in their frame.
First, look at the clockwise torque. We haveτ’(-)=fx’y’=fxy=τ(-) when we make the appropriate substitutions from the first of Equations 1 and from the second equation of the standard Lorentz Transformation.
Second, look at the counterclockwise torque. Here we seem to encounter a contradiction. The second of Equations 1 tells us that the lateral force appears diminished in the proper frame and we know that the Lorentz-Fitzgerald contraction makes the x-arm appear shrunk in the same proportion, so the torque comes out less thanτ’(+). That diminished torque won’t balance τ(-), so we expect the bracket to rotate in the proper frame, though we know that it cannot do so.
To understand how our reasoning went wrong and to correct the error let’s look at the first equation of the Lorentz Transformation,
That equation tells us that if we have a situation in which t’=0, then we have a length dilation rather than a length contraction. The Lorentz-Fitzgerald contraction comes about because, in the usual analysis, t’ represents a temporal offset between opposite ends of a moving object, such as our bracket’s x-arm. Motion shoves the rear end of the moving object into the future relative to the front end, so the rear partially overtakes the front, thereby causing a uniform contraction of the object. In calculating the counterclockwise torque, then, our proper-frame observers must assert a proposition that the temporal offset does not come into play.
Multiplying the dilated length x=γx’ by the diminished lateral force fy=fy’/γ gives us the same torque τ(+)=fyx=τ’(+) that the moving observers measure. Thus, in the proper frame the two torques balance each other, so any observers occupying that frame would expect to see the bracket not rotating.
To complete our derivation of the transformation of torques we ask the experimenters in the moving frame to turn the bracket so that the nail about which it pivots lies on the two frames’ common x-axis. In that case both arms of the bracket have the same length in both frames (the x-arm of the bracket now lies on the z’-axis). The forces exerted on the ends of the bracket’s arms remain equal to each other in both frames, even though they emerge diminished in the proper frame, so the torques balance in both frames, even though the torques come out diminished in the proper frame,τ=τ’/γ.
Thus we have the relativistic transformation of torques, the rotary analogue of Equations 1, as
in which the subscripts represent the axes about which the torques act.
The analogy between Equations 1 and Equations 3 runs deep. Torque bears the same relation to angular momentum that force bears to linear momentum. That fact tells us that we can calculate the angular momentum of a system by integrating the torque acting on the system with respect to the elapsed time;
The time t’ elapsed in the moving frame dilates in the proper frame to t=γt’, so we have the relativistic transformation of angular momentum as
We can confirm the first of those equations through a simple imaginary experiment. Imagine a free-turning flywheel so mounted on a vehicle occupying the moving frame that its axle aligns with the x-axis of our coordinate grid. An observer riding on the vehicle spins up the wheel, giving it a certain amount of angular momentum Lx’. If the wheel has a structure like that of a bicycle wheel, with almost all of the mass concentrated in its rim, the observer can calculate the amount of the wheel’s angular momentum by multiplying together the wheel’s mass, its radius, and the speed at which its rim moves. An observer occupying the proper frame would carry out the same calculation, but they would have to use the relativistically altered quantities; the increased mass m=γm’, the rim speed reduced by time dilation u=u’/γ, and the invariant radius r=r’. The calculation gives that observer
as required by the first of Equations 5.
Look at that experiment in another way. Imagine that we accelerate out of the proper frame until we find ourselves pacing the vehicle. During that acceleration we have done nothing to exert a torque upon the wheel, so the wheel must have the same angular momentum when we cease accelerating as it had before we began accelerating. Again we confirm the validity of the first of Equations 5.
But what can we say about the second and third of Equations 5? Could our acceleration into the vehicle’s frame actually exert a torque on a wheel spinning about an axis oriented in the y- or z-direction? To answer those questions imagine a wheel mounted on the vehicle so that its axle points in a direction parallel to the y-axis and spun up to a certain amount of angular momentum. Now imagine that we accelerate from the vehicle’s frame back into the proper frame.
Consider two small, identical segments of the wheel’s rim as they cross a line that runs parallel to the z-axis. In a Newtonian universe the masses of the segments neither differ from each other nor change as we accelerate. Likewise the velocities of both segments change by exactly the same amount, equaling the change that our acceleration puts between us and the vehicle. Thus the decrease in one segment’s angular momentum about the wheel’s axle precisely equals the increase in the other segment’s angular momentum, so the net angular momentum of the two segments taken together does not change. Symmetry considerations tell us that the entire wheel conforms to that latter statement: the wheel’s angular momentum doesn’t change as we accelerate in a universe in which the Lorentz factor does not exist.
But the Lorentz factor does exist in our real, relativistic universe. If the segments revolve about their axle at a speed u and if we move relative to the vehicle at the speed v, then at the instant that they cross the line parallel to the z-axis the segments move in our frame at the speeds
The Lorentz factors that correspond to those speeds take the form
If each segment has a rest mass of ½dm0, then the angular momentum of the segments at this instant come out as
Combining the two contributions gives us the total angular momentum of the two segments as
Although we have calculated that angular momentum for a given instant, the law of conservation of angular momentum necessitates that the segments’ angular momentum not change as the segments revolve about their common axis. That fact enables us to conceive the wheel as an array of such paired segments and, thus, to replace the mass dm0 with the rest mass m0 of the entire wheel. We thus have the wheel’s angular momentum in the vehicle’s frame as
in which we can see the relativistic mass due to the wheel’s rotation. Comparing that equation with Equation 10 tells us that we can write the description of the wheel’s angular momentum in our frame as
Thus we verify the validity of the second and third of Equations 5.
Now we have, in Equations 3 and 5, the relativistic transformations of torque and of angular momentum.
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