7
$\begingroup$

I only can think of kinetic energy as absolute. I know velocity is relative but I can't see kinetic energy as being relative because that would violate energy conservation. For example, if in some reference frame, the loss of kinetic energy is $60\mathrm{\ J}$, how can, in another reference frame, the loss be different? If kinetic energy is relative then its effects should also be relative. If $60\mathrm{\ J}$ produces an $X$ temperature and a $Y$ wave sound, how can this temperature and sound be different just because of changing the reference frame? It should be the same. Also, I thought that if the classical kinetic energy equation is true then the Galilean relativity is wrong because the more speed or velocity, the more loss of kinetic energy will there be in a collision, so what is true and what is wrong? Are the effects of kinetic energy equal no matter what reference frame is it? Is the kinetic energy equation wrong or is the Galilean relativity wrong?

$\endgroup$
4

6 Answers 6

15
$\begingroup$

Is the kinetic energy equation wrong or is the Galilean relativity wrong?

Neither is wrong, only the assumption that they are in any way incompatible is wrong. Kinetic energy is frame variant, but energy is conserved within each frame. Similarly with momentum. It is also different in different reference frames, but still conserved. Frame invariance and conservation are separate concepts.

Edit: to fully work out the example you proposed consider an inelastic collision with coefficient of restitution of 0, between a ball of mass $m_a=0.4 \mathrm{\ kg}$ and room of mass $m_b= 1000 \mathrm{\ kg}$ from a reference frame where the ball is initially moving at $u_a=20 \mathrm{\ m/s}$ and the room is initially at rest $u_b=0 \mathrm{\ m/s}$. In this frame after the collision $v_a=v_b=0.008 \mathrm{\ m/s}$ for a $\Delta KE_a= -80 \mathrm{\ J}$ and a $\Delta KE_b=0.032 \mathrm{\ J}$ for a total $\Delta KE = 79.968 \mathrm{\ J}$ of KE converted into thermal energy.

Now, consider the situation in the reference frame where the ball is initially moving at $u_a=40 \mathrm{\ m/s}$ and the room is initially moving at $u_b=20 \mathrm{\ m/s}$. In this frame after the collision $v_a=v_b=20.008 \mathrm{\ m/s}$ for a $\Delta KE_a=-239.936 \mathrm{\ J}$ and a $\Delta KE_b = 159.968 \mathrm{\ J}$ for a total $\Delta KE=79.968 \mathrm{\ J}$ of KE converted into thermal energy.

In both frames energy is conserved. The change in KE of each object depends on the reference frame, but the amount of energy converted into heat does not. I leave it as an exercise to show that the momentum is also frame dependent but conserved in both frames.

$\endgroup$
1
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – SuperCiocia
    Feb 2, 2022 at 5:53
5
$\begingroup$

For example, if in some reference frame, the loss of kinetic energy is 60j, How can in another reference frame the loss be different?

It can't. So in any inertial frame the difference between the initial and final states will be the same.

Similar to saying that if you drop a mass 1m, it doesn't matter if you consider it dropping from a height of 1m above the table to the table top, or if you consider it dropping from a height of 1.5m above the floor to the table top 0.5m above the floor. The difference is conserved even when the zero reference changes.

The same is true for KE. Each frame may have a separate reference for the amount of energy, but the difference between two events should be consistent.

Also i thought that if the classical kinetic energy equation is true then the Galileo relativity is wrong because the more speed or velocity, the more loss of kinetic energy will be in a collision

Collisions don't take you to zero in all reference frames. We have an incredible amount of KE if you measure our speed in the frame where the sun is at rest. But if you get in a car crash, the KE relative to the sun doesn't go to zero and isn't released to do mayhem.

$\endgroup$
3
  • $\begingroup$ I'm still confused because according to the K equation the loss in a collision on earth using the sun reference frame, will be considering earth velocity + objects velocity, this will give us a bigger loss than if we use earth reference frame, so which reference frame is correct? or is the K equation wrong? $\endgroup$
    – yass_the1
    Feb 2, 2022 at 1:25
  • 3
    $\begingroup$ "It can't. So in any inertial frame the difference between the initial and final states will be the same." -- this is not true. Let's say the difference b/w initial and final kinetic energy is $\sim(v_f^2-v_i^2)$ in a given frame. The difference b/w the initial and final kinetic energy in a frame moving at a velocity $v$ would be $\sim((v_f-v)^2-(v_i-v)^2)=v_f^2-v_i^2-2v(v_f-v_i)$. $\endgroup$
    – user87745
    Feb 2, 2022 at 1:27
  • 1
    $\begingroup$ Also, the analogy drawn to gravitational potential energy is wrong for obvious reasons. The gravitational potential energy near the surface of the earth is linear in height, the kinetic energy is quadratic in speed. $\endgroup$
    – user87745
    Feb 2, 2022 at 1:42
1
$\begingroup$

KE is frame dependent, as is momentum. If you stand on a street corner, your KE and momentum are each zero in your rest frame, but if I pass you in a car you have both energy and momentum in my frame.

The conservation laws apply to total energy and total momentum- how the overall energy and momentum is distributed between bodies is frame dependent.

If you play with a few examples of colliding objects or weights sliding down ramps and so on, you will find that Newton's laws apply in a consistent way regardless of which (inertial) frame of reference you adopt, although the KE and momentum of the individual objects involved will vary by frame.

There are many interesting examples that seem paradoxical at first until you figure them out. For example, consider the text-book case of a stationary block that then slides down a smooth ramp. You can calculate its speed at the bottom of the ramp by equating the PE lost with the KE gained. However, if you consider the experiment from the frame of someone walking past it with just the speed that the block attains in the frame of the ramp, the block will be moving at the top of the ramp and will end up stationary at the bottom, so the loss of PE seems to be accompanied by a loss of KE. Try to figure that one out (a hint- consider the recoil of the ramp, which is usually neglected).

$\endgroup$
1
  • $\begingroup$ $+1$: A nice example! :) Another tricky variation of the problem you mention is to consider a freely sliding ramp and analyze the system from the reference frame of the ramp. Both in the ground frame and the ramp frame, the change in the potential energy would be the same but the change in the kinetic energy would be different (of course, due to the ramp not being an inertial frame, one would have to take into account the work done by pseudo forces)! $\endgroup$
    – user87745
    Feb 2, 2022 at 15:07
1
$\begingroup$

There is a fundamental point in the interpretation of the concept of energy: even the kinetic energy (KE) exists in relation to other kinds of energy, whose exchange is realized through work. In this way, the existence of an absolute quantity, the KE does not make sense. On the other hand, differences that come from energy transformations (no matter what kind) are such that they conserve the overall energy (momentum) of the system.

Look for instance to the potential energy, in its most simple form: U = mgh. It is settled with the exception of a constant of height. This can be thought to be similar to the kinetic energy.

In both cases, for the transformations through work, the important concept is the difference from the initial/final KE (or potential) in a given process, not how each observer/experimenter would measure that energy.

$\endgroup$
7
  • 4
    $\begingroup$ -1: This answer is a word salad with no explanatory content. How does this actually address the specific question that OP asked? Also, many statements here are either misleading or wrong. For example, the arbitrariness of the datum of potential energy has nothing in similarity with kinetic energy, we don't fix the zero of kinetic energy arbitrarily -- it is zero when the speed is zero. $\endgroup$
    – user87745
    Feb 2, 2022 at 1:33
  • 1
    $\begingroup$ "It is zero when the speed is zero". This is the key to understand your conceptual mistake: there is no absolute velocity, as there is no absolute zero point for hight. $\endgroup$ Feb 2, 2022 at 2:04
  • 1
    $\begingroup$ I didn't say that there is an absolute velocity. Obviously, the velocity would be dependent upon the frame of reference. The point is that in a given frame of reference, the kinetic energy cannot be shifted by an arbitrary constant, whereas potential energy can. $\endgroup$
    – user87745
    Feb 2, 2022 at 2:07
  • $\begingroup$ In both cases, if you change the frame of reference, you can change the energy definition, and this is as simple as that, despite you most problem didn't understand it (or accepted). The ke attributed depends on the frame of reference... $\endgroup$ Feb 2, 2022 at 2:10
  • $\begingroup$ "The [KE] attributed depend on the frame of reference" -- Yes, obviously. This is not what I am contradicting. "In both cases, if you change the frame of reference, you can change the energy definition" -- Yes, but there is an inherent ambiguity in the definition of potential energy even within a given frame of reference. This is the distinction between kinetic energy and potential energy that I am referring to. $\endgroup$
    – user87745
    Feb 2, 2022 at 2:15
1
$\begingroup$

Kinetic energy is frame dependent by definition. In newtonian mechanics, the kinetic energy $T_{_{P,S}}$ of a point mass $P$ with respect to a frame of reference $S$ is $\mathit{defined}$ as:

$ T_{_{P,S}} := \frac{1}{2}m_{_{P}}|\vec{v}_{_{P,S}}|^2$

where $\vec{v}_{_{P,S}}$ is the velocity vector of $P$ relative to $S$.

As for your question regarding $“\mathit{energy\; conservation}”$ I’m going to assume that you mean the law of conservation of $\mathit{mechanical}$ energy (as has been assumed by the other answers since the OP appears to be wanting an answer within the framework of Newtonian mechanics and Galilean relativity). The law of conservation of mechanical energy according to Wikipedia is:

“The principle of conservation of mechanical energy states that if an isolated system is subject only to conservative forces, then the mechanical energy is constant”

Here is a proof for the law of conservation of mechanical energy when all kinematic quantities are measured from an $\mathit{arbitrary}$ intertial frame $S$:

I am going to assume you are familiar with the work-energy theorem for an $n$-particle system $P_{1},P_{2},P_{3},...,P_{n}$:

In any inertial frame of reference, The sum of the works done by every force(internal as well as external to the system) on the system is equal to the change in the system’s kinetic energy:

$$\sum_{k=1}^n W_{\vec{F_{k}},P_{k},(S)} = \Delta{\sum_{k=1}^n T_{_{P_{k},S}}}$$$\;$

where $\vec{F_{k}}$ is the resultant force on the $k^{th}$ particle so $W_{\vec{F_{k}},P_{k},(S)}$ is the work done by $\vec{F_{k}}$ on $P_{k}$ relative to $S$ (yes, work is relative too as $W_{\vec{F},P,(S)}:= \int_{\vec{r}_{_{P,S,(i)}}}^{\vec{r}_{_{P,S,(f)}}} \vec{F}\cdot d \,\vec{r}_{_{P,S}}$) or equivalently, the sum of the works done by every force acting on $P_{k}$ All of these forces can be split into $3$ categories: external forces, conservative internal forces and non-conservative internal forces. Using this:

$\sum W_{\vec{F}_{ext},(S)} + \sum W_{\vec{F}_{int. cons},(S)} + \sum W_{\vec{F}_{int.noncons},(S)} = \Delta{\sum_{k=1}^n T_{_{P_{k},S}}}$

There are no external forces on an isolated system by definition and if there are no non-conservative internal forces either, the first and third sums vanish.

$\implies{\sum W_{\vec{F}_{int. cons},(S)} = \Delta{\sum_{k=1}^n T_{_{P_{k},S}}}}$

The change in potential energy of an $n$-particle system due due to an internal conservative force vector $\vec{F}$ acting on $P_{k}$ relative to any frame S is defined as:

$ V_{f,(S)} - V_{i,(S)} = -\,W_{\vec{F},P_{k},(S)}$

$\implies{\sum W_{\vec{F}_{int. cons},(S)} = V_{tot,i,(S)} - V_{tot,f,(S)}}$

where $V_{tot,(S)}$ denotes the total potential energy of the system due to all the conservative internal forces. Plugging this back into our equation gives us the desired result:

$V_{tot,i,(S)} + \sum_{k=1}^n T_{_{P_{k},i,S}} = V_{tot,i,(S)} + \sum_{k=1}^n T_{_{P_{k},f,S}}$

$\implies{V_{tot,(S)} + T_{tot,(S)}} = c$

for some constant $c$ (since $i$ and $f$ can be arbitrarily close configurations) for any inertial frame $S$.$\;\;$$Q.E.D$

Note: If you would like a proof based purely on first principles, I would be more than happy to add the proof of the Work-energy theorem.

$\endgroup$
0
$\begingroup$

The confusion may be a result of a very common misconception about the conservation of energy:

People often think that the energy conservation states that potential energy equals kinetic energy ($E_{\mathrm{kin}} = E_{\mathrm{pot}}$).

But this is only a special case, where the reference frame and the reference point for the potential energy are chosen such that the problem becomes very easy. Which is the case for most textbook problems.

At first, energy is a quantity that only decribes the state of a system. If a process occures and we want to describe the change of state, we have to use the physical quantity work.

The full law of energy conservation states that the sum of all energy forms is a constant. The value of this constant may depend on the chosen reference frame.

(Beyond galilean relativity, in the special theory of relativity this only works if energy and momentum as a whole are considered.)

$\endgroup$
4
  • $\begingroup$ "People often think that the energy conservation states that potential energy equals kinetic energy ($E_{\mathrm{kin}} = E_{\mathrm{pot}}$). But this is only a special case, where the reference frame and the reference point for the potential energy are chosen such that the problem becomes very easy. Which is the case for most textbook problems." -- Can you clarify this part of your answer? I don't see how energy conservation relates to potential energy and kinetic energy being equal. $\endgroup$
    – user87745
    Feb 2, 2022 at 15:01
  • $\begingroup$ Maybe what you mean is that during a process the change in kinetic energy is the negative of the change in potential energy (which would ensure, for an isolated system with no dissipative forces, that the net change in the energy of a system is zero, i.e., energy is conserved)? The rest of your answer looks great! Welcome to Physics StackExchange :) $\endgroup$
    – user87745
    Feb 2, 2022 at 15:02
  • $\begingroup$ A simple example: dropping a ball out of rest. The question is how fast is the ball when it reaches the ground. The easiest solution is obtained with the conservation of energy. Students now argue that the ball starts with only potential energy and when it reaches the ground it only has kinetic energy. This IS what the computation says but it is not what is going on. When you now ask them about the energies in an intermediate state in the motion, they are completely lost. $\endgroup$
    – stephanp
    Feb 2, 2022 at 16:00
  • $\begingroup$ Yes, constructing a simplified model system with no dissipative forces like friction, as you mentioned, unfortunately lead to this misconception. $\endgroup$
    – stephanp
    Feb 2, 2022 at 16:05

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