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# Equation of the Day #6: The Lagrangian

**Dr. Akano**, in Math/Physics Apr 22 2013 · 334 views

Classical Mechanics Physics Lagrangian

**You may have learned once that classical mechanics all stems from Newton's laws of motion, and while that is true, it is not necessarily the best way to solve a given physical problem. Often when we look at a physical system, we take note of certain physical parameters: energy, momentum, and position. However, these can be more generalized to fit the physical situation in question better. This is where Lagrange comes in; he thought of a new way to formulate mechanics. Instead of looking at the total energy of a system, which is the potential energy plus the kinetic energy, he instead investigated the**

*difference*in those two quantities,**where**

*T*is the kinetic energy and*V*is the potential energy. Since the kinetic and potential energy, in general, depend on the coordinate position and velocity of the particle in question, as well as time, so too does the Lagrangian. You're probably thinking, "okay, what makes that so great?" Well, if we were to plot the Lagrangian and calculate the area under the curve with respect to time, we get a quantity known as the action of the particle.**where**

*t*_{1}and*t*_{2}are the starting and ending times of interest. Usually if the motion is periodic, the difference between these times is one period. Now, it turns out that for classical motion, the action is minimized with respect to a change in the path along which the particle moves for the physical path along which the particle actually moves. This sounds bizarre, but what it means is that there is only one path along which the particle can move while keeping the action minimized. Physicists call this the Principle of Least Action; I like to call it "the universe is inherently lazy" rule. When you do the math out, you can calculate an equation related to the Lagrangian for which the action is minimized. We call these the Euler-Lagrange Equations.**These are the equations of motion a particle with Lagrangian**

*L*in generalized coordinates*q*with velocity components denoted by_{i}*q*with a dot above the q (the dot denotes taking a time derivative, and the time derivative of a coordinate is the velocity in that coordinate's direction). This is one of the advantages of the Lagrangian formulation of mechanics; you can pick any coordinate system that is best-suited for the physical situation. If you have a spherically symmetric problem, you can use spherical coordinates (altitude, longitude, colatitude). If your problem works best on a rectangular grid, use Cartesian coordinates. You don't have to worry about sticking only with Cartesian (rectilinear) coordinates and then converting to something that makes more sense; you can just start out in the right coordinate system from the get go! Now, there are a couple of special attributes to point out here. First, the quantity within the time derivative is a familiar physical quantity, known as the conjugate momenta._{i}**Note that these do not have to have units of linear momentum of [Force × time]. For instance, in spherical coordinates, the conjugate momentum of longitude is the angular momentum in the vertical direction, which has units of action, [Energy × time]. The Euler-Lagrange equations tell us to take the total time derivative of these momenta, i.e. figure out how they change in time. This gives us a sort of conjugate force, since Newton's second law reads that the change in momentum over time is force. The other quantity gives special significance when it equals zero,**

**This is just fancy math language for saying that if one of our generalized coordinates,**

*q*, doesn't appear at all in our Lagrangian, then that quantity's conjugate momentum is conserved, and the coordinate is called "cyclic." In calculating the Kepler problem – the physical situation of two particles orbiting each other (like the Earth around the Sun) – the Lagrangian is_{i}**Note that the only coordinate that doesn't appear in the Lagrangian is**

*ϕ*, the longitude in spherical coordinates. Thus, the conjugate momentum of*ϕ*, which is the angular momentum pointing from the North pole vertically upwards, is a conserved quantity. This reveals a symmetry in the problem that would not be seen if we used the Lagrangian for the same problem in Cartesian coordinates:**That just looks ugly. Note that all three coordinates are present, so there are no cyclic coordinates in this system. In spherical coordinates, however, we see that there is a symmetry to the problem; the symmetry is that the situation is rotationally invariant under rotations about an axis perpendicular to the plane of orbit. No matter what angle you rotate the physical situation by about that axis, the physical situation remains unchanged.**

Is it good or bad thing that I knew this already? =P