# Is the energy conserved in a moving frame of reference?

Consider this situation:

When the box is at the bottom of the frictionless incline, it will have a velocity of $v_f$. The person is an inertial frame of reference that moves at a constant velocity of $v_f$.

From the person's frame of reference, the box has kinetic energy when it is at the top of the ramp. Even if it is moving in the negative direction, velocity would be squared in $KE = mv^2/2$, so now the box has both potiental energy and kinetic energy.

When the box is at the bottom of the incline it is going to have no kinetic energy from the person's frame of reference. How is it that the box had both potential energy ($mgh$) and kinetic energy ($mv^2/2$) at the top of the frictionless incline but had neither potential nor kinetic energy at the bottom?

$$mgh + \frac{mv^2}{2} = 0$$

The law of conservation of energy says energy is transferred, but not lost. Where has the energy gone?

Basically the law of conservation of energy should not be violated no matter the frame of reference, but the final formula does that. Where has the energy gone?

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The energy went into speeding up the incline and whatever it is attached to. The work on the incline is nonzero in the moving frame. –  Ron Maimon Apr 13 '12 at 21:19

Consider two reference frames $S$ and $S'$. Assume that $S'$ moves with a velocity $\mathbf{u}=u\mathbf{\hat{x}}$ relative to $S$ such that the origins of $S$ and $S'$ coincide at $t=0$ and their axes remain parallel. Further assume that the surface of the inclined plane is stationary in $S$ as shown in the figure (the dashed curve can be ignored till the end of the discussion).

There are two points to consider before proceeding further.

• The fundamental relation for energy considerations is the work-energy theorem which states that the change in the kinetic energy of a particle as it moves from an initial point $A$ to a final point $B$ is equal to the work done $$K_B - K_A = \int_{t_A}^{t_B} \mathbf{F}(t) \cdot \mathbf{v}(t) \, dt$$ where the force $\mathbf{F}(t)$ can arise from a constraint, be time-dependent or be non-conservative. The work-energy theorem is frame-independent. It is only in the special case of time-independent and conservative forces that one can identify a scalar potential energy $U$ such that $\mathbf{F} = - \nabla U$. Then the work energy theorem boils down to $K+U=\mathrm{constant}$.

• The block is confined to move on the surface of the inclined plane of constant inclination $\theta$. In the frame $S$, this static (scleronomic) constraint can be expressed as $$x \, \tan\theta + y - h=0.$$ In the frame $S'$, the surface of the inclined plane is non-stationary and this moving (rheonomic) constraint can be written as $$x \, \tan\theta + y - H(t)=0$$ where $H(t)=h - u \tan\theta \,\, t$. Forces arising from scleronomic constraints do no work since they are orthogonal to the velocity. However forces arising from rheonomic constraints can perform real work since the net velocity of the particle can have a component along the direction of the constraint forces. See section 2.1 of Jose and Saletan for a beautiful discussion of this concept.

Newton's law in either frame is $$m \dot{\mathbf{v}} = -mg {\hat{\mathbf{y}}} + \mathbf{F}_c(t)$$ where $\mathbf{F}_c$ is the force of constraint. As done in proving the work-energy theorem, we multiply by $\mathbf{v}$ and integrate with respect to time to get $$K_B - K_A = mg \big[ y(t_A) - y(t_B) \big] + W_c$$ where the work done by the constraint forces is $$W_c = \int_{t_A}^{t_B} \mathbf{F}_c(t) \cdot \mathbf{v}(t) \, dt \qquad \textrm{in }S$$ and $$W_c' = \int_{t_A}^{t_B} \mathbf{F}_c'(t) \cdot \mathbf{v}'(t) \, dt \qquad \textrm{in }S'.$$ It is easily shown that $$\mathbf{F}_c(t) = \mathbf{F}_c'(t) = mg \cos\theta \, (\sin\theta \mathbf{\hat{x}} + \cos\theta \mathbf{\hat{y}} ).$$ Also it is easy to show that the velocity $\mathbf{v}$ and the position $\mathbf{r}$ of the block in $S$ are $$\mathbf{v} = g t \sin \theta \, (\cos\theta \mathbf{\hat{x}} - \sin\theta \mathbf{\hat{y}} )$$ $$\mathbf{r} = \frac{g t^2 \sin \theta \cos\theta}{2} \mathbf{\hat{x}} + \Big( h - \frac{g t^2 \sin^2 \theta}{2} \Big) \mathbf{\hat{y}}.$$ The corresponding quantities in $S'$ are obtained via the Galilean transformation $$\mathbf{v}' = \mathbf{v} - \mathbf{u}$$ $$\mathbf{r}' = \mathbf{r} - \mathbf{u} t.$$ We find $\mathbf{F}_c \cdot \mathbf{v} = 0$ and hence $W_c=0$ while $$W_c'= -\int_{t_A}^{t_B} \mathbf{F}_c(t) \cdot \mathbf{u} \, dt = -mug \sin\theta\cos\theta (t_B-t_A).$$

Therefore the work-energy theorem in $S$ is $$K_B - K_A = mg \big[ y(t_A) - y(t_B) \big]$$ while in $S'$ it takes the form $$K_B' - K_A' = mg \big[ y'(t_A) - y'(t_B) \big] - mug \sin\theta\cos\theta (t_B-t_A).$$

We can now finally come to the particular problem asked by the OP. Take the points $A$ and $B$ to be respectively those points where the inclined surface meets the $y$ and $x$ axes. The time $T$ taken by the block to slide down the incline from a height $h$ is obtained from solving $y(T) = 0$ as $$T = \frac{\sqrt{2h/g}}{\sin\theta}$$ and at this instant of time $\mathbf{v}(T) = \sqrt{2gh} \, (\cos\theta \mathbf{\hat{x}} - \sin\theta \mathbf{\hat{y}} ).$

We now verify the work-energy theorem

• in frame $S$ $$K_B-K_A=\frac{m}{2} \big[ v(T)^2 - v(0)^2] = mgh$$ while $$mg \big[ y(t_A)-y(t_B) \big] = mgh$$

• in frame $S'$ $$K_B'-K_A'=\frac{m}{2} \big[ v'(T)^2 - v'(0)^2] = \frac{mg^2T^2}{2} \sin^2\theta - mugT \sin\theta\cos\theta$$ while $$mg \big[ y'(t_A) - y'(t_B) \big] - mug \sin\theta\cos\theta (t_B-t_A) = mgh-mugT \sin\theta\cos\theta$$ which, using the relation between $h$ and $T$, is the same as $K_B'-K_A'$.

Thus the work-energy theorem is verified in both frames.

The OP asked for the particular case of $u=v_x(T)=\sqrt{2gh}\cos\theta=gT\sin\theta\cos\theta$ and considered energy conservation in $S'$. In this case - $$K_B'-K_A'= \frac{mu^2}{2} (\sec^2\theta-2)$$ while $$mg \big[ y'(t_A) - y'(t_B) \big] - mug \sin\theta\cos\theta (t_B-t_A) =\frac{mu^2}{2} (\sec^2\theta-2)$$ Notice that $K_B' \neq 0$ -- this is because the $v_y(T)\neq 0$ as the OP implicitly assumed. The reason is that the OP was looking at the case where no motion in the vertical direction would be allowed when $y=0$, i.e., for $t>T$. However this would require that the constraint force (normal reaction) change discontinuously. A better solution would be to consider the block sliding down a smooth curve (such as the dashed curve in the figure). In this case $v_y(T)=0$ and the reaction force would also vary smoothly. The work-energy theorem would, ofcourse, still be valid.

NOTE:

1. The point that moving constraints can do real work was mentioned in the answer by Pygmalion.

2. This problem of a block sliding on a moving inclined plane is considered in section 3.9 of Strauch.

3. Worrying about the mass of the wedge or of the earth is misleading. We can just consider a particle constrained to move on a mathematical surface subject to a uniform and constant body force in the $y$-direction. This surface could also be moving. For this particular case, think of the beads sliding down a tilted abacus made of thin massless wires. You look at this situation from a frame at rest w.r.t. the abacus and then in a moving frame.

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what did you use to draw the nice diagram? –  Physiks lover Apr 15 '12 at 15:25
Inkscape –  Vijay Murthy Apr 15 '12 at 15:32
Note point 3: I disagree on that. It is possible alternatively to draw the very same conclusions by calculating the difference in kinetic energy of wedge + Earth. Please see my calculation. I think downvoting is totally unapropriate here. And now guys tell me this: Ron Maimon is criticising my first answer (which is basically the same as yours), you and Physiks lover my second, while presenting only one of them seems to be fine. –  Pygmalion Apr 16 '12 at 8:25
@Pygmalion: Your second answer makes approximations about $M \gg m$. However, would it be possible to account for all energy changes without making this approximation? If yes, then that is a valid answer too. My approach was at looking whether we can in principle account for all the energy changes and work done -- not whether something is small compared to something else and so can be ignored. –  Vijay Murthy Apr 16 '12 at 10:27
+1 you've put a lot of work into this answer –  John McVirgo Apr 16 '12 at 13:00

In every inertial frame of reference, energy is conserved.

One could define inertial frame of reference as the frame of reference that moves with constant velocity, i.e. acceleration of such frame of reference is zero. In inertial frame of reference all Newton laws are valid, and since energy is defined as a work of the conservative force, energy should be conserved.

Basicly, there are several ways to explan why it seems that energy is not conserved, but in the end it all comes to the same: Not only that the incline acts on the box with normal force $\vec{N}$, but box also acts on the incline with the oposite force $-\vec{N}$ (third Newton law). As a result, kinetic energy of the incline (and in fact of the whole Earth) changes too.

A. First possible way to explain this paradox is seing incline and the Earth as some kind of external force ($\vec{N}$) that limits movement of the box. In the incline's frame of reference, this work is simply zero, because force is perpendicular to movement: $\vec{N} \cdot \textrm{d}\vec{s} = 0, \vec{N} \perp \textrm{d}\vec{s}$. However, in the person's frame of reference, work is no longer zero ($\vec{N} \cdot \textrm{d}\vec{s} \ne 0$), so you have additional work you have to calculate. Calculating this additional work is very complex and I'll rather skip to the next, easier explanation.

B. Second possible way of explaning this paradox is calculating the energy of the whole universe, that is energy of box as well as energy of the Earth+incline. In person's frame of reference the Earth is always moving left, so we obviously to calculate difference in the kinetic energy. It turns out that if you calculate kinetic energy of the Earth in person's frame of reference, you will find that it increases and compensates for smaller kinetic energy of box and smaller potential energy.

Why is calculating kinetic energy of Earth so important only in the person's frame of reference? Let's calculate the difference in its kinetic energy, if $M$ is mass of Earth, $m$ is mass of box, $V$ is velocity of Earth and $v$ velocity of box after the process in Earth+box frame of reference. Note that the velocity of person is also $v$. Obviously and $V \ll v$ and $M \gg m$ and from conservation of momenta $M V = m v$.

In Earth+box frame of reference:

$$\Delta KE'_\textrm{Earth} = \frac{1}{2} M V^2 - \frac{1}{2} M 0^2 = \frac{1}{2} M V^2 = \frac{1}{2} m v V \ll \frac{1}{2} m v^2$$

In person's frame of reference (using Galilean transformations):

$$\Delta KE_\textrm{Earth} = KE^\text{final}_\text{Earth} - KE^\text{init}_\text{Earth} = \frac{1}{2} M (V+v)^2 - \frac{1}{2} M v^2 \approx M V v = m v^2 = 2 (\frac{1}{2} m v^2)$$.

Obviously, $\Delta KE'_\textrm{Earth}$ in negligible. On the other hand $\Delta KE_\textrm{Earth}$ is not negligible and equals twice final kinetic energy of box in Earth's frame of reference. This is exaclty starting potential energy of box plus final kinetic energy of box ($KE + PE$), that is the energy that went "missing" in your question:

$$KE^\text{init}_\text{Earth} + mgh + \frac{1}{2} m v^2 = KE^\text{final}_\text{Earth}$$

Ta-da!

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Basically I'm trying to prove that energy is conserved in a moving frame of reference view –  Fawaz Apr 13 '12 at 20:56
The person is the frame of reference and it is moving with the constant velocity of the box when it is at the bottom of the slide. –  Fawaz Apr 13 '12 at 21:02
Let's say this way: In both frames of references potential energy decreases, as bodies (box and Earth) come closer. In incline's frame of reference, in the beginning both bodies are at still in the beginning and both bodies have speed (Earth have very very small speed) in the end. In the person's frame of reference, speed of box goes to zero, but speed of Earth increases, very very little, but enough to compensate for smaller potential energy and smaller kinetic energy of the box. –  Pygmalion Apr 13 '12 at 21:08
In my previous comment, replace "incline's frame of reference" with "frame of reference of both bodies being at rest". –  Pygmalion Apr 13 '12 at 21:17
@Pygmalion: I wish you wouldn't say second "possible" explanation--- it is the correct explanation--- the box is doing nonzero work on the incline as it is sliding down in the moving frame, and this work shows up as extra energy in the Earth, or whatever the incline is attached to. –  Ron Maimon Apr 13 '12 at 21:19

Conservation of energy written as $\frac 1 2 mv^2 + \phi(\vec r) =const$ doesn't hold in the moving frame because the potential $\phi(\vec r)$ for the net force acting on the box now depends upon time, rather than just position.

If a force $F$ is conservative, then there is a potential $\phi$ such that $F = -\nabla\phi$ and the conservation of energy can be written as$$\frac 1 2mv^2 + \phi(\vec r) = const$$ For gravity, we can write $\phi = mgy$ so that $$F_g = -\left(\frac {\partial} {\partial x}mgy, \frac {\partial} {\partial y}mgy, \frac {\partial} {\partial z}mgy\right) = (0, -mg,0)$$ In the moving frame, the gravitational potential can also be written like this so that the conservation of energy for just the gravitational force acting on the box still holds $$\frac 1 2 mv^2 +mgy = const$$ There is also a $\phi_p$ for the reaction force $mg\cos\theta$ of the inclinded plane on the box since it's conservative in the frame of the plane $$\phi_p(\vec r) = -mg\cos\theta(y\cos\theta + x\sin\theta)$$so that the conservation of energy in the frame of the plane, including gravity, is finally $$\frac 1 2 mv^2 + mgy - mg\cos\theta(y\cos\theta + x\sin\theta) = const$$

However, in the moving frame, $\phi_p$ depends on time because of its dependence on $x$ in the inclined plane's frame, and so the conservation of energy cannot be written as above.

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Following is the way I figured out the answer. I hope this helps to simplify things a bit.

Let us first write the resultant force on the box in the frame in which the incline is at rest.

$$\mathbf{F_r} = mg \sin \theta \cos \theta \mathbf{\hat{x}} - mg \sin ^2 \theta \mathbf{\hat{y}}$$

This resultant force is the same in all inertial frames. Hence, acceleration is same in both the frames.$$\mathbf{a_r} = g \sin \theta \cos \theta \mathbf{\hat{x}} - g \sin ^2 \theta \mathbf{\hat{y}}$$

Consider first the situation as seen in the frame in which the incline is at rest. Initial velocity $\mathbf{u_0} =0$ Considering the point of release as the origin, initial position $x_0 = 0;\ y_0 =0$. Solving Newton's laws we get

$$x = \frac{1}{2} g \sin \theta \cos \theta \ t^2 \space \ \ \ \ y= -\frac{1}{2}g \sin^2\theta \ t^2 \\ v_x = g\sin\theta\cos\theta\ t \ \ \ \ \ v_y =-g\sin^2\theta \ t$$

Work done is just

$$W = \int\ \mathbf{F_r \cdot dr} \ = -\Delta PE = \frac{1}{2}mg^2\sin^2\theta\cos^2\theta\ t^2\ + \frac{1}{2}mg^2\sin^4\theta\ t^2\\=\frac{1}{2}mv_x^2\ +\ \frac{1}{2}mv_y^2\ = \Delta KE$$

And we have energy conservation $\ \Delta KE\ +\ \Delta PE\ =\ 0$

Now to the situation which OP has asked. Let the time taken for the box to reach the bottom of the incline be $t_f$ (same in both frames). The relative velocity between the frames as posed by OP is $\mathbf{v_f} =g\sin\theta\cos\theta t_f \mathbf{\hat{x}}$. Let the origins coincide at $t=0$. We then have $x_0=0;\ y_0 =0$ and $\mathbf{u_0}= -\mathbf{v_f}$. Acceleration remains the same in both the frames. Solving Newton's laws we get

$$x = -g\sin\theta\cos\theta t_f t+\frac{1}{2} g \sin \theta \cos \theta \ t^2 \space \ \ \ \ y= -\frac{1}{2}g \sin^2\theta \ t^2 \\ v_x = -g\sin\theta\cos\theta t_f+g\sin\theta\cos\theta\ t \ \ \ \ \ v_y =-g\sin^2\theta \ t$$ Work done is just

$$W = \int\ \mathbf{F_r \cdot dr} \ = -\Delta PE = - mg^2\sin^2\theta\cos^2\theta t_f t+\frac{1}{2}mg^2\sin^2\theta\cos^2\theta\ t^2\ + \frac{1}{2}mg^2\sin^4\theta\ t^2\\$$ At $t=t_f$ $$W = -\Delta PE = - mg^2\sin^2\theta\cos^2\theta t_f^2+\frac{1}{2}mg^2\sin^2\theta\cos^2\theta\ t_f^2\ + \frac{1}{2}mg^2\sin^4\theta\ t_f^2\\ =-\frac{1}{2}mg^2\sin^2\theta\cos^2\theta\ t_f^2\ + \frac{1}{2}mg^2\sin^4\theta\ t_f^2\\ =-\frac{1}{2}mv_x^2(t=0)\ +\ \frac{1}{2}mv_y^2(t=t_f)\ = \Delta KE$$ Since $v_x(t=t_f)=0;\ v_y(t=0)=0$. And we have energy conservation $\ \Delta KE\ +\ \Delta PE\ =\ 0 \$!!!

Excuse me.. What was the problem again?

Well... the apparent catch was in posing the problem. While posing the problem OP has used $\Delta KE$ and $\Delta PE$ measured in different frames to verify Law of conservation of energy. The problem considers only change in $\Delta KE$ and not change in $\Delta PE$ when we change from the stationary frame to the moving frame. It is to be noted that both $\Delta KE$ and $\Delta PE$ change. Since the frames will have relative motion, the velocities of the particles change according to velocity addition formula and kinetic energy changes. Lets say in the initial frame for a given particle $$\Delta KE = \frac{1}{2} m (u_f^2 -u_i^2)$$If we now wish to see it in a frame with a relative velocity $-v$, the velocities get transformed into $u_f + v$ and $u_i +v$. Thus change in kinetic energy changes to $$\Delta KE = \frac{1}{2} m (u_f^2 -u_i^2) + mv(u_f-u_i)$$ To compensate for this gain in $\Delta KE$, Work done has to change to keep conservation of energy intact. We can show that it is exactly the same as the gain in $\Delta KE$. On changing the frame,$$\mathbf{dr}\longrightarrow\mathbf{dr}+\mathbf{v}dt$$ In the new frame,$$W=\int\mathbf{F}\cdot\mathbf{dr}+\int\mathbf{F}\cdot\mathbf{v}dt = \int\mathbf{F}\cdot\mathbf{dr}+\int m\dfrac{d\mathbf{u}}{dt}\cdot\mathbf{v}dt =\int\mathbf{F}\cdot\mathbf{dr}+mv(u_f-u_i)$$

Hence, both $\Delta KE$ and $\Delta PE$ have to be frame dependent.

Note:-

1) The above discussion assumes that the forces in the problem are time independent and conservative. It is only then that we can define $\Delta PE$ as $W = -\Delta PE$. Which is not true if it were time dependent or non-conservative. In which case one cannot define potential energy. We will then be proving work kinetic energy theorem instead of $\Delta KE+\Delta PE=0$.

2) One might argue that the work done here was by the constraint force. Frankly, as I have proved, it should be true for any force. In this case it happens to be generic to the problem that the component of the resultant force in the direction of relative motion (x direction) happens to be contributed only by the constraint force.

Consider for instance the following situation. A $2 kg$ weight is made to fall freely under the influence of gravity from rest for a distance of $5 m$. Using $g=-10 m/s^2$ we find that the velocity after $5m$ is $-10 m/s$.$\Delta KE = 100 kgm^2/s^2$. Now let us observe this from a frame moving downward with a velocity $-5 m/s$. In this frame, initial velocity is $5m/s$ and initial kinetic energy is $25 kgm^2/s^2$. After falling for $5m$, the velocity of the particle will be $-5m/s$ and final kinetic energy will be $25 kgm^2/s^2$.$\Delta KE = 0 kgm^2/s^2$. Oh... there was no change in potential energy even as the particle fell for $5 m$ !!!

The resolution for this goes in the similar lines as in the situation above. It is just for gravity in this case.

3) It is to be noted that the change in work done on changing frames depends only on the initial and final velocities of the particle irrespective of what it has gone through in between!!

4) Throughout the calculation, I have considered real conditions of experiment i.e. a typical box and typical inclined plane on earth whose masses are small compared to earth. Acceleration due to gravity is constant. If people wish to consider this as an approximation, people are welcome to solve the problem in all detail to their satisfaction. But the result I have proved above is a very profound truth independent of the case at hand.

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Resultant Force on the box is mg sin theta along the inclined plane. The motion of the particle is due to this force. I have just written it in a co-ordinate system with horizontal direction x hat and vertical direction y hat –  Forsudee May 4 '12 at 20:52
Oh I am sorry there was a mistake with the sign in the y direction. I have edited it. The rest was anyways correct. –  Forsudee May 4 '12 at 21:03
Resultant force is the vector sum of the forces acting on the box. Which are... gravity = -mg \hat{y} and normal reaction of the inclined plane = mg sin theta cos theta \hat{x} + mg cos ^2 theta \hat{y} –  Forsudee May 4 '12 at 21:05

The issue is that you're not taking into account the motion of the wedge, and thus you're not completely addressing the full problem. When you change to a moving frame, the wedge gains kinetic energy:$\frac{1}{2} M v_w^2$

As the block is accelerated to the right, the wedge becomes accelerated to the left. So, in the final state of the moving frame, the block has no velocity (by construction of the moving frame) but the wedge's velocity toward the left has increased. This increase in velocity causes the wedge to gain kinetic energy, which makes up for loss in energy of the block.

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"This increase in velocity perfectly cancels out the loss in energy of the block" huh? you're comparing velocity and energy! Also, momentum conservation only works in the x direction, because you have an external force gravity acting along y. –  Physiks lover Apr 14 '12 at 20:54
when doing momentum conservation you must take initial and final situation only. Initial situation - no speed. Final situation, speeds only along $x$ axis. Intermediate situations are not important. –  Pygmalion Apr 16 '12 at 11:37