Isn't the conservation of energy just a trivial result which arrives due to the way kinetic energy and potential energy are defined?

The kinetic energy of a body of mass $m$ and velocity $v$ is defined to be $\frac{1}{2}mv^2$ and its potential energy due to some force $F$ is defined as a function of its position $r$ with respect to some reference position $r_0$ as $PE=-\int_{r_0}^rFdr$

So, the absolute total energy of a body can be calculated as: $$E=\frac{1}{2}mv^2-\int_{r_0}^rFdr$$

Now, this $E$ remains unchanged even if the position and velocity of the body changes if all the forces are conservative. But let's see how the change in $E$ is defined:

The change in $E$ is the change in $PE$ + the change in $KE$.

Now, the change in $KE$ is just equal to the work done.

$$\delta KE= \int_{r_1}^{r_2}Fdr......(1)$$ By the changing variable into $v$ $$=\int_{v_1}^{v_2}mvdv$$ $$=\frac{1}{2}m(v_2^2-v_1^2)$$

And, change in PE as the body moves from $r_1$ to $r_2$ is: $$\delta PE=-\int_{r_1}^{r_2}Fdr....(2)$$

Note that the integrals $(1)$ and $(2)$ just differ by sign. So, it's no surprise that the net change remains $0$. To evaluate the expression of change in $KE$, we change the variables of the integral to $v$. And, to get the expression for change in $PE$, we don't change the variable and we evaluate the negative of the same integral in terms of position. If change in $PE$ is just defined to be negative of change in $KE$, then isn't it trivial that the net energy remains unchanged?

Even I can attribute an unchanged quantity to a body. Suppose a body is acted upon by a force $F$ and I attribute this quantity to the body:

$$A=m\ln(v)-\int_{t_0}^t\frac{F}{v}dt$$ $t_0$ is any arbitrarily chosen time. This quantity is made up of two quantities: $B=m\ln(v)$ and $C=-\int_{t_0}^t\frac{F}{v}dt$.

The change in $C$ as the time changes from $t_1$ to $t_2$ is:

$$\delta C=-\int_{t_1}^{t_2}\frac{F}{v}dt$$

Suppose the velocity of the body at time $t_1$ is $v_1$ and at time $t_2$ is $v_2$. Now, the change in $B$ as the velocity of the body changes from $v_1$ to $v_2$ is:

$$\delta B=m(\ln(v_2)-\ln(v_1))$$ $$=\int_{v_1}^{v_2}\frac{mdv}{v}$$ $$=\int_{v_1}^{v_2}\frac{mdv}{vdt}dt$$ $$=\int_{t_1}^{t_2}\frac{F}{v}dt$$ $$=-\delta C$$

So, $\delta B+\delta C=0$. Hence, $\delta A=0$. So, $A$ is also an invariant quantity. Then, what's so special about conservation of energy if numerous invariant quantities like $A$ can be attributed to a body?

In fact, defining $PE$ requires the force to be conservative because there can be infinitely many paths joining point $r_0$ to $r$ and we require $PE$ to be independent of path followed. But for defining the quantity $C$ there is no such restriction on the nature of force because time can only flow along a unique path from $t_0$ to $t$. $PE$ relates every point in space to a number whereas $C$ relates every time instant to a number.

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    $\begingroup$ That the change in KE is equal to the work done is already the statement of energy conservation (in this mechanical context where there seems to be no other forms of energy). Of course it is trivial to derive energy conservation from energy conservation... $\endgroup$
    – ACuriousMind
    Commented Apr 9, 2017 at 11:17
  • $\begingroup$ @ACuriousMind Oh, thanks. I didn't think about that. Is the work-energy theorem the same as energy conservation? $\endgroup$
    – Dove
    Commented Apr 9, 2017 at 11:19
  • $\begingroup$ @ACuriousMind If we take the whole universe as our system, then, in classical mechanics there are absolutely no other forms of energy . People talk about chemical and heat energy, but they too are just kinetic and potential energies. $\endgroup$
    – Dove
    Commented Apr 9, 2017 at 11:29
  • $\begingroup$ if $t_1=0$ then C diverges for a particle under a constant force that is at rest a $t_1=0$. And the main problem to me is that A does not depend on the particle's state (position and speed) at a given time, but also on its previous story $\endgroup$
    – user126422
    Commented Apr 9, 2017 at 21:45
  • $\begingroup$ @ArmandoEstebanQuito I think $A$ is almost analogous to energy. Here, $B$ is a function of speed just like $KE$. The only difference is that $C$ is a function of time whereas $PE$ is a function of position. $\endgroup$
    – Dove
    Commented Apr 10, 2017 at 1:35

3 Answers 3


It definitely is ! It may be interesting to look at some cases where it is not so.

One example that springs into mind would be friction but as we all know energy is still conserved, it is just being stored as heat which is typically not taken into account.

A more interesting example can be found in quantum field theories in curved spaces, where it is not always trivial to even define an energy let alone show that it is conserved !

Let me give an example: (I am sorry if this is not the level of answer you were hoping for...)

Typically we define the energy momentum tensor as:

$$T^{\mu\nu} = \partial^\mu \phi \pi^\nu - g^{\mu\nu} \mathcal{L}$$

Typically we say that the energy is simple the 00 component of this tensor namely $T^{00}$ and it is assumed that this is conserved. But let us think about this statement !

A statement of conservation can only be made if there is some 4-vector current in this case, the energy current with zeroth component the energy density and its other components the energy flow let us call this object $E^\mu$.

If this exists and it can be extracted from $T^{\mu\nu}$ than we must have $E^{\mu} = T^{\mu\nu}k_\nu$ let us check what conditions need to be satisfied for $E^\mu$ to be conserved.

Conservation of a current means that $\nabla_\mu E^\mu = 0$ where $\nabla$ is the spacetime covariant derivative

$$\nabla_\mu T^{\mu\nu}k_\nu = 0$$

Assuming that $\nabla_\mu T^{\mu\nu} = 0$ this becomes:

$$T^{\mu\nu} \nabla_\mu k_\nu = 0 \rightarrow \nabla_\mu k_\nu=0$$

This is a very strict demand that is typically not satisfied ! If $T^{\mu\nu}$ were to be symmetric we could still rewrite it as:

$$T^{\mu\nu}\frac{1}{2} (\nabla_\mu k_\nu + \nabla_\nu k_\mu)=0 \rightarrow (\nabla_\mu k_\nu + \nabla_\nu k_\mu)=0 \rightarrow k = \text{killing vector}$$

Which is satisfied in more cases !

Conclusion: $T^{00}$ is corresponds to the energy density of some vector current if and only if $T^{\mu\nu}$ is symmetric, covariant constant and there is some killing vector in our spacetime !

Obviously this is not always the case. For example the Krasner Universe does not have such a killing vector !

Even more relevant, the Friedman universe in which we live does not have such a killing vector either ! And indeed in cosmology it is very hard to come up with a satisfactory defintion of energy.

To round up, in classical simple cases, no it is not amazing. If we go to quantum systems it is very amazing an defenitly not always true !

Hopefully this helped you :) ! Feel free to ask some questions as this is obviously quite a bit to understand at once...


What you did is to derive and integrate a constant (mass, in your example), the conserved quantity that you find that way is trivial and is not a conserved quantity, it is an invariant scalar by definition.

You started with:


Then differentiate:

$0=\frac{mv}{v^2}dv-\frac{mv^2}{v^3}dv=\frac{m }{v }dv-\frac{m }{v }dv$

Then integrate, to get


  • $\begingroup$ +1, but still, how is it not a conserved quantity? You just showed that it remains constant. If we consider an object moving through space whose velocity changes as time passes, then my quantity is also a strange number which you can calculate at different instants and find it to be a constant. If $a$ and $v$ both are functions of $t$, then call $\frac{ma}{v}=g(t)$. Then my quantity is: $$A=m\ln(v_2)-m\ln(v_1)-\int_{t_0}^tg(t)dt=constant$$. Whereas the conservation of energy is: $$E=\frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2-\int_{r_0}^rF(r)dr=constant$$. $\endgroup$
    – Dove
    Commented Apr 11, 2017 at 1:54
  • 1
    $\begingroup$ Your conserved quantity is guaranteed to be true regardless of the laws of physics. It is only a trivial identity. You can always write, for any f(x,v,t), f-f=0, then F-int{f}=c, where F is the primitive of f. It is not about physics, it is a mathematical identity $\endgroup$
    – user126422
    Commented Apr 11, 2017 at 2:04

I think the reason why it seems unimpressive is that you're looking at a special case that was developed long after the more impressive general case arose.

In the case you mention, you start from the fact that changes in potential energy are the opposite of changes in kinetic energy. But you start from the assumption that potential energy exists in the first place.

The part that was truly amazing about the conservation of energy was discovering that we could write terms, such as kinetic energy potential energy, which caused energy to be conserved. Consider gravitational potential energy as an example. We discovered that there was a number, proportional to the height of an object and the mass of an object which could wholly predict the velocity of a an object if it fell from that height. Think about it. Some strange connection between a height and a velocity, that was incredibly repeatable? That's an impressive statement!

Then we started to discover other forms of energy. We discovered electromagnetic (light) energy. We discovered sound energy. We discovered thermal energy. We discovered chemical energy. With every one of those types of energies, we found that if you converted one energy to another and back, the result was always a net of no change at all in this strange and powerful scalar value.

The part that's amazing is that I can show you a motorcycle that "runs on water, with almost no gas at all," by using the turning of the motor to run a generator (mechanical->electric) which is then used to electrolyze water (electric->chemical), then use a magnetic field generated by rare earth magnets to align the molecules produced so that they re-combine more efficiently, and use that to drive the piston (chemical->mechanical) to create energy with no waste -- just a cycle using water! I can even show you my prototype which is "almost there:" it still needs a little gas, but it's "mostly running on water now." And after all of that, you can just smile and say "that's nice, but it doesn't really work." Because you can see that I just did a round trip from the mechanical energy in a piston, through all these fancy steps, right back into mechanical energy in a piston and be very confident that I did not, in fact, end up with more energy than I started.

In fact, you can be so certain, that the US patent office no longer accepts patents for any machine which claims to create energy in conflict with the conservation of energy laws.

All those steps. Think of all the equations you could write, and all the empirical curve fits you could make regarding how explosions in pistons work and what my rare earth magnets could be doing. And you can skip all of them because you can be confident that energy is conserved.

Now that's impressive.


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