# Is the quantity $mv^{2}$ also conserved?

The following is my reasoning: Suppose $\vec{F}$ is a conservative force field. Then the total mechanical energy, $E$, of the isolated mechanical system is a conserved quantity which implies that $E=T+U=k_{0}$, where $k_{0}$ is a constant, $T$ is the kinetic energy and $U$ is the potential energy. Now, $\nabla{E} =\nabla{(T+U)}=\nabla{T}+\nabla{U}=\vec{0}$. Hence, $\vec{F}=-\nabla{U}=\nabla{T}$. Now, $U=\int_{C} \nabla{T}\cdot d\vec{r}= T+k_{1}$, where $k_{1}$ is a constant of integration. Now, $E=T+U=T+T+k_{1}=k_{0}$ which implies that $E=2T=k_{0}-k_{1}$ is constant and since $T=\frac{1}{2}mv^{2}$ and hence $E=mv^{2}=k_{2}$, where $k_{2}=k_{0}-k_{1}$ is a constant and therefore $mv^{2}$ is a conserved quantity.

The quantity $mv^2$ is not constant.

You can argue this by an example. Consider a ball in a uniform gravitational field. If you drop the ball, the quantity $mv^2$ changes as it falls, so it's not constant.

Or, more simply, $mv^2$ is twice the kinetic energy. Think of any system where the kinetic energy changes, and you've shown that the quantity can't be constant.

So what's wrong with your mathematical argument?

Well, you write $U=\int_C \nabla T\cdot d\vec r$, but that's not true. First, you have a sign error. (Try replacing $\nabla T$ with $-\nabla U$ and you'll see why.) If you correct the sign error, you'd get $U=-\int_C \nabla T \cdot d\vec r = -T + k_1$. Now if you add $T$ to both sides you get $U+T=k_1,$ which at least isn't wrong, but doesn't really gain you anything since $k_1 = E$.

The definition of change in potential energy for a conservative force $\vec F_c$ is $\Delta U = -\int \vec F_c \cdot d\vec r$. I suppose you might sometimes see this written without the negative sign, but in that case, the force in the integrand would be the "minimum" force exerted by an external agent in order to overcome the conservative force. (This is a sloppy explanation, and best left for a separate Physics.SE question.)

The bottom line is, if you're working with some conservative force field $\vec F_c$, you need the negative sign.

• There is no sign error. By definition $\vec{F}=-\nabla U$ and $-\nabla U=\nabla T$. Please check the definition of the potential energy and force relation. – user11937 Aug 13 '14 at 4:59
• $\Delta U=-\int \vec F\cdot d\vec r = -\int \nabla T\cdot d\vec r$. If the conservative force and displacement are in the same direction, then the change in potential energy is negative. (Consider a ball being dropped. It moves down, $\Delta U < 0$.) But do try replacing your $\nabla T$ with $-\nabla U$ in your integrand and you'll see a contradiction pop up immediately. – BMS Aug 13 '14 at 5:02
• NO! $U=\int_{C} \vec{F}\cdot d\vec{r}$. There is no minus sign as you have pointed out. How can the net force on a particle be opposite the direction of displacement. It does not make sense. Mathematically, we don't care about the displacement of the particle being opposite to the direction of displacement because I am using the fundamental theorem of line integrals. – user11937 Aug 13 '14 at 5:11
• You may find this helpful: hyperphysics.phy-astr.gsu.edu/hbase/pegrav.html – BMS Aug 13 '14 at 5:15
• "Consider a ball in a uniform gravitational field" the same thought experiment would 'prove' mv is not conserved. – Aron Aug 13 '14 at 5:55