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Commits (3)
 ... ... @@ -76,16 +76,21 @@ The inverse relation is given as $$r=\sqrt{x^2 + y^2};$$ $$\varphi=\begin{cases} \arctan(y/x) & \text{x>0,}\\ \pi + \arctan(y/x) & \text{x<0 and y>0,}\\ -\pi + \arctan(y/x) & \text{x<0 and y<0.} \end{cases}$$ \arctan(y/x) & \text{$x>0$,}\\ \pi + \arctan(y/x) & \text{$x<0$ and $y>0$,}\\ -\pi + \arctan(y/x) & \text{$x<0$ and $y<0$.} \end{cases}$$The last formula for \varphi warrants a closer explanation: It is easy to see that \tan(\varphi)=y/x - but this is not a unique relation, due to the fact that the \tan has different branches. Convince yourself that the expression above is correct for all the four sectors! ## Distances and areas Now suppose we want to calculate the distance between two points, one with polar coordinates (r_1, \varphi_1), and the other with (r_2, \varphi_2). This looks like a difficult exercise. A convenient (r_2, \varphi_2). This looks like a difficult exercise. One possible way to perform this is by translating the polar coordinates into Cartesian coordinates and using the expression given above for this distance:$$\Delta s^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2,$$so ... ... @@ -103,32 +108,34 @@ arc of radius r_1 and then we move a small distance radially outward from r_1 to r_2. Provided the difference between the angles \varphi_1 and \varphi_2 is (very) small, these paths are approximately perpendicular and we can use Pythagoras’ theorem to find the distance \Delta s. Note that the arc is approximately straight – it has a length r_1 \Delta \varphi, where \Delta \varphi = \varphi_2-\varphi_1. So we have:$$\Delta s^2 = (\Delta r)^2 + (arc~length)^2 = (\Delta r)^2 + r_1^2 (\Delta \varphi)^2 .$$For r\equiv r_1 \approx r_2, replacing the \Delta’s by d’s to emphasise the infinitesimal character of the differences, we finally have:$$ds^2 = dr^2+ r^2 d\varphi^2 .$$You should recall this formula! If we traverse a curved path in the plane, we must sum up all the ds, which actually boils down to calculating an integral:$$Path~length = \int_{start}^{end} ds = \int_{start}^{end} \sqrt{dr^2 + r^2 d\varphi^2}.$$Here, the boundaries start and end stand for the starting and end point of the path. Note that the path can often be written in the form r(\varphi), i.e. we write r as a function of \varphi. We can then write the integral as:$$Path~length = \int_{start}^{end} \sqrt{dr^2 + r^2 d\varphi^2} = \int_{\varphi_{start}}^{\varphi_{end}} \sqrt{\left(\frac{dr}{d\varphi}\right)^2 + r^2} \; d\varphi.$$(Note: this parametrisation is strictly speaking only possible when r is a uni-valued function of \varphi; a meandering path generally does not have this property.) **Exercise:** write the same expression as an integral over dr instead of d\varphi. Please sketch two paths, one which is more suitable for the expression above (the integral over \varphi) and the other more convenient for the expression you obtained in this problem. the distance d s. Note that the arc is approximately straight – it has a length r_1 d \varphi, where d \varphi = \varphi_2-\varphi_1. So we have:$$d s^2 = (d r)^2 + (arc~length)^2 = (d r)^2 + r_1^2 (d \varphi)^2 .$$We can use the same arguments also for the area: since the different segments are approximately perpendicular, we find the area by simply multiplying them:$$dA = r dr d\varphi.$$This is an important formula to remember for integrating in polar coordinates! The extra r that appears here can be intuitively understood: the area swept by an angle difference d\varphi *increases* as we move further away from the origin. !!! check "Example: Integrating over a circular area" To check the area element we just derived, let us compute a simple integral. We compute the integral over a circle with radius r_0 with a very simple function that equals to 1. In this case, we expect to get as a result the are of the region we integrate over. We find:$$ \int_0^{2\pi} d\varphi \int_0^r_0 r dr = \\ 2\pi \int_0^r_0 r dr = 2 \pi \frac{1}{2} r_0^2 = \pi r_0^2$$which is indeed the area of a circle with radius 0. ## Cylindrical coordinates ... ... @@ -142,10 +149,10 @@ choose the coordinate system ourselves - this is not imposed by the problem. Cylindrical coordinates are defined straightforwardly: we use polar coordinates \rho and \varphi in the xy plane, and the distance z coordinates r and \varphi in the xy plane, and the distance z along the symmetry-axis as the third coordinate. If the axis system is chosen in physical space, we have two coordinates which have the dimension of a distance: \rho and z. The other coordinate, dimension of a distance: r and z. The other coordinate, \varphi, is of course dimensionless. What is the distance travelled along a path when we express this in ... ... @@ -153,15 +160,15 @@ cylindrical coordinates? Let’s consider an example (Figure). ![image](figures/Coordinates_13_0.svg) We want to find the length of the (small) red segment \Delta s. By We want to find the length of the (small) red segment d s. By inspecting the figure, we see that the horizontal (i.e. parallel to the xy-plane) segment \Delta l is perpendicular to the vertical segment dz. Using for \Delta l the length we obtained before for a line xy-plane) segment d l is perpendicular to the vertical segment dz. Using for d l the length we obtained before for a line segment in the xy plane, expressed in polar coordinates, we immediately find:$$\Delta s^2 = \Delta l^2 + \Delta z^2 = \Delta \rho^2 + \rho^2 \Delta \varphi^2 + \Delta z^2.$$Using d instead of \Delta for infinitesimal changes, this reads:$$ds^2 = dl^2 + dz^2 = d\rho^2 + \rho^2 d\varphi^2 + dz^2.d s^2 = d l^2 + d z^2 = d r^2 + r^2 d \varphi^2 + d z^2.$$The volume element is consequently given as$$dV = r dr d\varphi dz.$$## Spherical coordinates ... ... @@ -178,12 +185,11 @@ The position of a point on the sphere is specified using the two angles \theta and \phi indicated in the figure. !!! warning Note that in mathematics, usually the angles are labelled the other way Note that in mathematics, often the angles are labelled the other way round: there, \phi is used for the angle between a line running from the origin o the point of interest and the z-axis, and \theta for the angle of the projection of that line with the x-axis. The convention used here is custom in physics. IT IS IMPORTANT TO BE AWARE OF THIS!! convention used here is custom in physics. The relation between Cartesian and coordinates is defined by$$x = r \cos \varphi \sin \vartheta$$... ... @@ -215,6 +221,10 @@ displacement. ![image](figures/Coordinates_19_0.svg) From these arguments we can again also find the volume element, it is here given as$$dV = r^2 \sin\theta dr d\theta d\varphi.$$### summary We have discussed four different coordinate systems: ... ... @@ -232,12 +242,14 @@ We have discussed four different coordinate systems: given in terms of these coordinates. Infinitesimal distance:$$ds^2 = dr^2 + r^2 d\phi^2.$$Infinitesimal area:$$dA = r dr d\varphi.$$3. *Cylindrical coordinates*:$${\bf r} = (\rho, \phi, z).$$Can be 3. *Cylindrical coordinates*:$${\bf r} = (r, \phi, z).$$Can be used in three dimensions. Suitable for systems with axial symmetry or functions given in terms of these coordinates. Infinitesimal distance:$$ds^2 = d\rho^2 + \rho^2 d\phi^2 + dz^2.$$Infinitesimal distance:$$ds^2 = dr^2 + r^2 d\phi^2 + dz^2.$$Infinitesimal volume::$$dV = r dr d\varphi dz.$$4. *Spherical coordinates*:$${\bf r} = (r, \theta, \phi).$$Can be used in three dimensions. Suitable for systems with spherical ... ... @@ -245,6 +257,8 @@ We have discussed four different coordinate systems: Infinitesimal distance:$$ds^2 =r^2 (\sin^2 \theta d\phi^2 + d\theta^2) + dr^2 .$$Infinitesimal volume:$$dV = r^2 \sin(\theta) dr d\theta d\varphi.$$Problems ======== ... ... @@ -279,11 +293,3 @@ Problems \frac{\partial}{\partial \vartheta}\left( \sin\vartheta \frac{\partial\psi(r,\vartheta, \varphi)}{\partial \vartheta}\right).$$ This is however even more tedious (you do not have to show this). 4. [:smirk:] A particle moves over a cone which is characterized by the condition $rho = z$. Calculate the kinetic energy expressed in terms of the cylindrical coordinates $z$ and $\varphi$ and their first order time derivatives. 5. [:smirk:] The same question for a particle moving on a spherical shell of radius $R$. (Use coordinates $\varphi$ and $\vartheta$.)