This might be a very simple question. I read one previous post Can the kinetic energy be a function of the position vector?

I know that in Cartesian coordinates, the kinetic energy $T=\frac{1}{2}mv^2$. And $T$ is not an explicit function of position. So $\frac{\partial T}{\partial x}=0$, where we suppose $x$ is a coordinate.

But I got confused by one example, we have a ball move vertically from the origin O, with a velocity of $\vec V_0$.

enter image description here

then when the ball reach $y_1$ in the positive $y$ axis, we have $$mgy=\frac{1}{2}mv_0^2-\frac{1}{2}mv_1^2$$ so the velocity at y is $$v_1=\sqrt {\frac{1}{2}mv_0^2-mgy}$$ Does it means that velocity and kinetic energy both are explicit function of position y in this case? I know this is a special case, but the statement that $\frac{\partial T}{\partial x}=0$ in Cartesian coordinates seems to be quite general. So where have I missed so far? Thanks guys!

  • 3
    $\begingroup$ Related: physics.stackexchange.com/q/885/2451 and links therein. $\endgroup$
    – Qmechanic
    Feb 1 '16 at 13:23
  • $\begingroup$ Thanks Qmechanics! I have read that post too. But there seems to be many different versions of the answer. So I made an example here. I wonder if someone can help me with it. $\endgroup$
    – Wu Jeremy
    Feb 1 '16 at 13:27
  • $\begingroup$ Besides the statement $\frac{\partial T}{\partial x}=0$ in Cartesian lies before introducing the lagrangian. So I wonder whether there is some explanation without refer to the lagrangian. $\endgroup$
    – Wu Jeremy
    Feb 1 '16 at 13:33
  • $\begingroup$ Kinetic energy is a parameter/feature of an system, not an object's position. Do the math yourself for two colliding (Newtonian) objects in any reference frame: particle A stationary; particle B stationary; both moving relative to an external coordinate system. See what you get. $\endgroup$ Feb 1 '16 at 13:52
  • $\begingroup$ Velocity is a function of time (or position, which is itself a function of time), the thing is KE in the lagrangian is written as $\frac{1}{2}mv^2$ so when you take the partial derivative as there is no explicit dependence on position it's just 0. $\endgroup$
    – Suriya
    Feb 1 '16 at 13:56

In classical mechanics calculate the evolution of a particle means to know its position and its velocity for any time.

In general if I say that a particle is in position $x$ and has velocity $v$ and ask you about the kinetic energy. That is $\frac{1}{2}mv^2$ and this is independent of the system you have. The answer is always correct.

Solving the equations of the system means finding, for each moment of time, what is the position and the velocity. Once you solve the equations you will have a position and velocity for each time $t$. Therefore it means you can for any system, not only the one you described, write the velocity as a funtion of the position.

The kinetic energy, however, will never depend explicity on the position but will depend implicity. Informally this means that it does not depend on the position directly but if you can associate to each position a speed then it will depend indirectly. In fact taking the partial derivative as you mentioned $\frac{∂T}{∂x}$ means "how much the kinetic energy change if I vary the position while I keep the velocity constant" and this is always zero. If you do however $\frac{dT}{dx}$ you are saying "how much the kinetic energy change if I vary the position while I varying the velocity accordingly" and that is non-zero.

I think the mixing of the two concepts is the source of your question.


For your particular example, the kinetic energy of the ball does indeed depend on the height of the ball. The total energy of the ball is constant and the potential energy of the ball is different at every different height y. As a consequence, the kinetic energy of the ball must adjust at every different y to keep the total energy constant. You can see this from your equation $v_1=\sqrt {\frac{1}{2}mv_0^2-mgy} $.


Not the answer you're looking for? Browse other questions tagged or ask your own question.