- #1
baw
- 9
- 0
Why, in lagrangian mechanics, do we calculate: ##\frac{d}{dt}\frac{\partial T}{\partial \dot{q}}## to get the (generalised) momentum change in time
instead of ##\frac{d T}{dq}##?
(T - kinetic energy; q - generalised coordinate; p - generalised momentum; for simplicity I assumed that no external forces are applied)
The kinetic energy is defined as the amount of work needed to accelerate some body to its velocity, so it is derived by ##\int \frac{dp}{dt} dx##.
Quite intuitive way of obtaining the time rate of change of the generalised momentum would be to take "the reverse": ##\frac{d T}{dq}##, but it doesn't seem to work. Why?
Such procedure would be based directly on the definition of T, while the actual, lagrangian approach looks (at least for me) like based on some property in cartesian coordinates. It's not obvious for me, that it would work in other coordinate systems etc. Could someone explain me why this property is so general? (I know it is just a definition but it's not completely arbitraty because it has to comply with Newtonian mechanics).
instead of ##\frac{d T}{dq}##?
(T - kinetic energy; q - generalised coordinate; p - generalised momentum; for simplicity I assumed that no external forces are applied)
The kinetic energy is defined as the amount of work needed to accelerate some body to its velocity, so it is derived by ##\int \frac{dp}{dt} dx##.
Quite intuitive way of obtaining the time rate of change of the generalised momentum would be to take "the reverse": ##\frac{d T}{dq}##, but it doesn't seem to work. Why?
Such procedure would be based directly on the definition of T, while the actual, lagrangian approach looks (at least for me) like based on some property in cartesian coordinates. It's not obvious for me, that it would work in other coordinate systems etc. Could someone explain me why this property is so general? (I know it is just a definition but it's not completely arbitraty because it has to comply with Newtonian mechanics).