To Master Physics, First Master the Rotating Coordinate System

What's more dizzying than being spun really fast in a circle? Well, that would be trying to decipher the vertiginous mathematics behind the rotating coordinate frame. But given that rotating frames appear in many important engineering fields, such as fluid flows, weather patterns, or aviation systems, a solid conceptual understanding of them cannot be dispensed with. Moreover, the mathematics of rotation have a tendency to crop up in rather unexpected places. Including, notably, in the equations of Maxwell's electromagnetism. This is dialect. And indeed, if you want to master physics, then first you must master the rotating coordinate system. It's always best to start with a simple case before building to a more complex one. And here it's the simple case of the linearly translated frame that first and foremost requires our attention. Indeed, let's consider two separate coordinate systems, k and k-naught, that are in motion relative to one another. Within k-naught is a particle that, over a small amount of time dt, traverses a distance of dx-naught coordinate units. During that same time period, the entire k-naught frame itself traverses a distance of dx-f units in the k-frame. Now, for conceptual clarity, we're going to label the k-frame as the rest frame, and the k-naught frame as the motion frame. Next, it should be fairly apparent that the actual distance travelled by the particle is equal to the coordinate distance it travelled within the k-naught frame, plus the distance travelled by the k-naught frame itself. Now, by dividing through by dt, we also see that the velocity of the particle over this duration is simply equal to its coordinate velocity plus the frame velocity. This gives us our first equation of motion, the velocity equation of the particle. Next, let's find the acceleration equation of the particle. To do this, let's note that over the succeeding time period of duration dt, the particle travels a new coordinate distance dx-naught prime, while the frame travels a new distance dx-f prime. The velocity v-prime over this duration is, via our first equation of motion, expressed as the new coordinate velocity plus the new frame velocity. To find the actual acceleration of the particle over this duration, we must subtract the prior velocity v from this new velocity v-prime, and divide by dt. Doing this tells us that the actual acceleration of our particle is hence equal to its acceleration through the coordinate system, plus the acceleration of that coordinate system itself. Now, granted, this derivation may seem pretty obvious, as all we're doing is just adding accelerations and velocities here. But once we start working with vectors, we can find our equations of rotational motion, or any motion really, using this exact same procedure. That is, first we'll calculate the requisite spatial displacements over a small unit of time. Then we'll calculate the corresponding velocities over that time period and one immediately subsequent to it. And then lastly, we'll calculate the acceleration over this final period, before expressing it in terms of the coordinate and frame quantities. But before we go there, it's probably best to do a quick review of basic rotational motion. So let's now consider our k-naught frame to be rotating with a constant angular velocity omega about some origin. Now, this rotation means that, at any given moment, every point in the k-naught frame is subject to both a tangentially directed velocity, and an inwardly directed acceleration. The farther out along the coordinate frame we travel, the greater both the magnitudes of the velocity and the acceleration become. In precise terms, the tangential velocity has a magnitude of omega times the distance r to the center of the coordinate system, while the acceleration has a magnitude of omega times the tangential velocity, or omega squared r. This tangential velocity and inwards acceleration again describe the velocity v-f and the acceleration a-f of our coordinate frame at any point. Even more precisely, we can express these equations via the use of cross products, where the velocity equation becomes v-f equals omega cross r, and the acceleration equation becomes a-f equals omega cross v-f, or omega cross omega cross r. Here we interpret r as a vector reaching from the origin to our point of interest, and omega as a vector pointing along the axis of our rotation, whose magnitude corresponds to the angular velocity. Now, cross products have a really deep connection to rotation that's deserving of a video of its own. But for simple rotations like this which are confined to 2D coordinates, they can often be overkill. So let's simply note for now that these equations can alternately be expressed in non-cross product form as omega r times omega hat, and minus omega squared r times r hat. Where omega hat indicates the tangential unit direction, and r hat indicates the unit radial direction. All right, now we're ready to go ahead and pull up the first equation of motion for a rotating frame, which looks like this. What does this equation mean? Well, say we're given a particle that's moving through some k-naught frame with a coordinate velocity v-naught. Well again, that frame itself is rotating. Then this equation simply tells us that the true velocity of the particle equals its coordinate velocity through the frame, plus this second term, omega cross r. Which, as we've already seen, is nothing more than the velocity of the k-naught frame at the point which the particle occupies. So exactly as with the linear case, the velocity of our particle at any moment is just its coordinate velocity plus the frame velocity. Next, let's take a look at the second equation of motion for a rotating frame. What does this equation tell us? Well, similar to the linear case, it's telling us that the true acceleration of our particle is equal to its coordinate acceleration plus the acceleration of the frame. Which, as we saw before, equals this omega cross omega cross r term. But now, there's this strange third term that's left over that isn't analogous to anything in the linear case. Indeed, this equation tells us that the true acceleration of the particle requires the addition of a final acceleration term whose magnitude equals twice omega times the coordinate velocity and whose orientation is perpendicular to that coordinate velocity. So where does this extra term come from? Why is it perpendicular to the coordinate velocity? And why is there a 2 there? To answer these questions, we'll need to walk through the derivation carefully, taking all the same steps we did for the linearly translated frame. So let's start by considering a single instance of time, dt, during which our coordinate frame rotates through an angle, omega dt. If our particle has no coordinate velocity within that frame, then both it and the coordinate frame traverse an approximate tangential distance, drf. The magnitude of which is found via the rule of small angle arc lengths. We can thus write drf as the vector quantity omega r dt omega hat, or equivalently, omega cross r dt. Meanwhile, if the particle does have a coordinate velocity through the k-naught frame, then as the frame rotates, it travels some extra distance, dr-naught, within that frame. The final displacement of the particle, dr, is thus simply the vector quantity dr-naught plus drf. Next, by dividing this expression throughout by dt, we obtain our first equation of motion, which once again tells us our particle velocity is equal to its coordinate velocity through the frame plus the velocity of the frame itself. Now, from here on out, we're going to normalize the size of our velocity vectors so that they are the same size as our displacement vectors. In reality, of course, the magnitude of these vectors should be much greater, but doing this will help ensure visual clarity, and allow us to retain a picture of what's actually happening physically. Alright, so next let's consider another interval of time, dt, during which the frame, as it again rotates through an angle, omega dt, traverses a new distance, drf-prime, while the particle traverses a new coordinate distance, dr-naught-prime. Setting the velocity vectors again equal to their respective displacement vectors, we now have two separate expressions representing an initial and final velocity of the particle. To find the particle's total acceleration, all we simply need to do is take the difference between these two expressions, and divide by dt. Visually, we can accomplish this quite easily, by essentially just subtracting these two vectors. But we want to be able to express our final acceleration in terms of the coordinate and frame quantities. So first, let's look exclusively at the change in our coordinate velocity vector, v-naught. Now, if the coordinate velocity remains the same from the first instant to the second, meaning the particle continues to follow the same trajectory within the k-naught grid, then during the second rotation, this vector gets rotated through a small angle, omega dt. By the property of small arc lengths, the magnitude produced by this rotation is equal to omega dt times v-naught. So in the case where the coordinate velocity remains constant, the difference between the v-naught and v-naught-prime vectors is a vector whose magnitude equals omega dt v-naught, and whose orientation is perpendicular to v-naught. In the language of cross products, this vector can be expressed as omega cross v-naught times dt. Next, in the case where the coordinate velocity does change from the first instant to the second, then the v-naught-prime vector will be altered from the rotated v-naught vector by some amount. Indeed, that amount of alteration will be precisely the change in coordinate velocity, dv-naught. We can thus write the total change in v-naught as dv-naught plus omega cross v-naught times dt. Now let's tackle the change in our frame velocity vector, v-f. If there's no coordinate velocity, then during the second rotation, v-f is rotated through an angle, omega dt, and the change due to that rotation is a vector with magnitude omega dt times v-f, and which is oriented perpendicularly to v-f. Since v-f equals omega r, we can further write this magnitude as omega squared r times dt. Or, in cross product notation, we can express the entire vector as omega cross omega cross r times dt. Now, if there is a coordinate velocity involved, then obviously v-f-prime must undergo an alteration from the rotated v-f by a certain amount. But what determines this amount? To start, let's note more precisely that we want to find the difference between the v-f-prime vector that would result were there no coordinate velocity, and the v-f-prime vector that does result when there is a coordinate velocity. To do this, let's rotate our initial vector setup through an angle, omega dt. And notice that the head of our v-naught vector traces out the v-f-prime vector, while its tail traces out our would-be v-f-prime vector. Next, we can parallel transport the original v-naught vector along the would-be v-f-prime vector, such that its head also now traces out the would-be v-f-prime vector. We can now clearly see that the difference between our two v-f-prime vectors is precisely the difference which arises when our v-naught vector is rotated through an angle, omega dt. That vector is of course, as we've already seen, described by omega cross v-naught times dt. The total change in v-f can then be written as omega cross v-naught times dt, plus omega cross omega cross r times dt. Putting this together with the total change in v-naught, we get this final expression for the total change in the particle velocity. Lastly, we can combine these two terms here, and then simply divide all the terms by dt to get the acceleration. And that's it! We're now looking at the complete equation of motion for a uniformly rotating frame. As a side note, if your frame isn't uniformly rotating, then there's an extra term which gets added to this equation that acts in the tangential direction, which just represents the additional acceleration imparted to the frame by the change in angular velocity. Now let's return to our original question about the meaning of this middle term. Where does it really come from? Well, we saw that one part of the term comes from rotating our coordinate velocity vector along with the frame. So this represents the additional acceleration required to ensure that our particle maintains its same coordinate velocity through the frame as that frame rotates. Then we saw the second part of the term comes from how the frame velocity of the new location of the particle deviates from the frame velocity of the location where the would have been without any coordinate velocity. Or, in other words, it accounts for the fact that different parts of the frame are moving at different velocities with respect to one another. We hence see that the breakdown in analogy between the linear and rotational case is twofold. The first is that the rotation or deflection of a single frame requires a particle moving within that frame to be pushed through a greater orthogonal distance than the frame itself over the same time span. Whereas in the linear case, no such extra push is required. The second breakdown is that a rotating frame is not truly a single frame, but rather many different frames all moving at many different velocities. And a particle moving between these frames must accelerate or decelerate in order to maintain the same coordinate velocity in its new frame as it had in its old. Thus, one extra acceleration term is required for the particle to maintain its coordinate velocity through the same frame as that frame rotates, while a second extra acceleration term is required for the particle to maintain its coordinate velocity between different frames. Alright, let's finish our exploration of rotational motion with some examples. First, let's say we have a particle that, in the rotating k-naught frame, is itself rotating counterclockwise with angular velocity omega. Then its coordinate acceleration in the k-naught frame is minus omega squared r, r-hat, while the acceleration of the k-naught frame itself is also minus omega squared r, r-hat. Its coordinate velocity, meanwhile, is omega r, omega-hat. And so the extra acceleration term equals minus two omega squared r, r-hat. The true acceleration of the particle is thus minus four omega squared r. This makes sense, because if the particle is rotating with angular velocity omega in a frame which itself is rotating with angular velocity omega, then the particle's true angular velocity is two omega. And plugging that value directly into the centripetal acceleration equation, we get an acceleration of minus four omega squared r, r-hat. This quadrupling is also evident if we take a look at the step-by-step infinitesimal picture of the particle's path. Next, let's consider a particle that is stationary in the rest frame, and as such is rotating clockwise with angular velocity omega in the rotating frame. Now, just as before, the coordinate acceleration of the particle and the acceleration of the frame are both minus omega squared r, times r-hat. But this time, the coordinate velocity is minus omega r, omega-hat. And so the extra acceleration term equals positive two omega squared r, r-hat. The true acceleration of the particle is thus naturally zero. And again, we can visualize this via the infinitesimal displacement picture. Lastly, we can consider a particle located at the origin and traveling with a constant coordinate velocity v-naught in the positive y-direction. Then there is zero coordinate acceleration and zero frame acceleration. And so the true acceleration of the particle is in the negative x-direction with magnitude two omega v-naught. Here, finally, we see this is also evident from the infinitesimal picture. Now, this conceptual understanding of rotating frames will provide us with a powerful tool for future use. Indeed, if we move these terms of our acceleration equation over to the other side, we'll get the expressions for the fictitious centrifugal and Coriolis accelerations, a proper understanding of which will prove crucial as we But our work here will also furnish us with some important insights into the interesting rotational math behind electromagnetism. As in future videos, we begin to ask some very important questions about the physical meaning of Maxwell's equations. So stay tuned. This has been Dialect. Thanks for watching.