# 1st Law of thermodynamics : moving reference frame

## Main Question or Discussion Point

I'd like to apply the 1st law of thermodynamics in a reference frame (RF) moving with constant velocity. We have:
##\Delta{}E = E_{in} - E_{out}##
I am limiting myself to rectilinear motion.
Suppose we are in a RF moving with a constant velocity ##V##.
Let the system consist of a mass ##m##. The absolute and relative velocities of ##m## are ##V_a## and ##V_r## with ##V_a = V + V_r##.
We burn a fuel releasing energy ##Q## which is completely utilized to do work on the mass ##m##. All this work goes into increasing the velocity of ##m##.
The absolute velocity of ##m## changes from ##V_{a1}## to ##V_{a2}##. The relative velocity of ##m## changes from ##V_{r1}## to ##V_{r2}##.
The change in kinetic energy (KE) as measured in the fixed and moving RFs are:
##\Delta{}KE_a = \frac{1}{2}m\left(V^2_{a2} - V^2_{a1}\right)## and ##\Delta{}KE_r = \frac{1}{2}m\left(V^2_{r2} - V^2_{r1}\right)##
Thus we have: ##\Delta{}KE = Q##.
I think that we measure identical ##Q## in both stationary and moving RFs. But ##\Delta{}KE_a \ne \Delta{}KE_r##.
So where am i going wrong?

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You are not doing anything wrong. Kinetic energy is frame dependent. The amount of damage a bullet will do if it hits you depends very much on whether you are standing still or traveling along with the bullet.

So you are converting one type of energy into kinetic energy and the kinetic energy is different in two different frames. Where did the energy go? Well, you aren’t considering the whole system. If you consider a complete system, the energy will balance. For example let’s say your example is talking about rockets. The rocket changes kinetic energy by throwing out a plume of gas in the opposite direction which also has kinetic energy. In a reference frame where the change in kinetic energy of the rocket appears smaller, the kinetic energy of the rocket plume will appear larger. In all reference frames the energy will balance.

• Dale and thinkingcap81
You are not doing anything wrong. Kinetic energy is frame dependent. The amount of damage a bullet will do if it hits you depends very much on whether you are standing still or traveling along with the bullet.

So you are converting one type of energy into kinetic energy and the kinetic energy is different in two different frames. Where did the energy go? Well, you aren’t considering the whole system. If you consider a complete system, the energy will balance. For example let’s say your example is talking about rockets. The rocket changes kinetic energy by throwing out a plume of gas in the opposite direction which also has kinetic energy. In a reference frame where the change in kinetic energy of the rocket appears smaller, the kinetic energy of the rocket plume will appear larger. In all reference frames the energy will balance.
Hi CK,

You are not doing anything wrong. Kinetic energy is frame dependent. The amount of damage a bullet will do if it hits you depends very much on whether you are standing still or traveling along with the bullet.
I am aware of this.

I do not dispute the ##\Delta{}KE## differences in stationary and moving RFs. However the energy, ##Q##, expended by burning a fuel is measured identically in both RFs (else are we measuring incorrect energy content of fuels on the Earth?). So how do i reconcile ##\Delta{}KE = Q## in both RFs? Is my equation incorrect?

In a reference frame where the change in kinetic energy of the rocket appears smaller, the kinetic energy of the rocket plume will appear larger. In all reference frames the energy will balance.
I don't know how to do a calculation to show the balance in KE in any reference frame. I'll try.

Chestermiller
Mentor
Hi CK,

I am aware of this.

I do not dispute the ##\Delta{}KE## differences in stationary and moving RFs. However the energy, ##Q##, expended by burning a fuel is measured identically in both RFs (else are we measuring incorrect energy content of fuels on the Earth?). So how do i reconcile ##\Delta{}KE = Q## in both RFs? Is my equation incorrect?

I don't know how to do a calculation to show the balance in KE in any reference frame. I'll try.
Your application of the first law of thermodynamics leaves something to be desired. You don't say anything about the change in internal energy of the fuel, when it is released and you don't say anything about how the work is done, knowing that, like kinetic energy, work is a frame-dependent entity. Maybe if you were less goofy-loosy in your problem definition, then analysis of the changes in the system might make more sense. And since when is the the symbol Q used to represent the energy released from burning the fuel? By convention, Q is the heat transfer between the system and surroundings.

Your application of the first law of thermodynamics leaves something to be desired. You don't say anything about the change in internal energy of the fuel, when it is released and you don't say anything about how the work is done, knowing that, like kinetic energy, work is a frame-dependent entity. Maybe if you were less goofy-loosy in your problem definition, then analysis of the changes in the system might make more sense. And since when is the the symbol Q used to represent the energy released from burning the fuel? By convention, Q is the heat transfer between the system and surroundings.
Hi CM,

I meant that the fuel releases energy which gets transferred to the mass ##m## across a boundary. So can we take this energy transfer as ##Q##? The chemical composition of the fuel changes which i think is identically measured in both moving and stationary RFs.

You could ask for more clarity, which may be difficult for me to provide since we are not having a fact-to-face conversation. I hope you agree with me.

All i'm asking is this: we burn fuel on Earth to run various machines. The KE of the machines change due to this energy sypply. Energy released by the fuel is a scalar quantity and hence its value should not depend on the RF. But KE changes are frame dependent. So are we measuring incorrect calorific values of fuel on the Earth's surface.

Or could you suggest a proper analysis?

Dale
Mentor
So where am i going wrong?
You are neglecting the ##\Delta KE## of the exhaust.

The key point of this question is that it intuitively seems like conservation of energy is not working right. A rocket is powered by a chemical reaction that releases chemical energy at a constant rate. So how can a constant rate of energy release lead to a greater increase in KE when going fast?

To understand this it is useful to consider a “toy model” rocket that operates on the same principles, but is easier to analyze. Specifically, let’s consider a 10 kg ball (rocket) and a 1 kg ball (exhaust) which is attached to a massless spring (fuel).

Suppose this spring has enough energy stored that when the rocket is initially at rest it can propel it to 1 m/s, and by conservation of momentum the exhaust is propelled to -10 m/s. Conversely, if the rocket starts at 5 m/s then after “burning” the fuel the rocket is propelled to 6 m/s and the exhaust moves at -5 m/s.

So now let’s check energy. In the first case the KE of the rocket increased from 0 J to 5 J, while in the second case it increased from 125 J to 180 J. The spring stores the same amount of energy in both cases, so why does the KE increase by 5 J at the low speed and by 55 J at the high speed?

Notice that we forgot to calculate the energy that went into the exhaust. This is the pivotal mistake of most such analyses. In the first case the KE of the exhaust increased from 0 J to 50 J, while in the second case the KE was 12.5 J before and after. So in both cases the total change in KE (both the rocket and the exhaust) was 55 J.

At low speeds most of the fuel’s energy is “wasted” in the KE of the exhaust. At higher speeds more goes into the rocket and less into the exhaust. For a real rocket, the same thing happens on a continuous basis. Both energy and momentum are conserved, and in fact more power is delivered to the vehicle as the speed increases under constant thrust.

The absolute velocity
No such thing.

• russ_watters, thinkingcap81, vanhees71 and 1 other person
Nugatory
Mentor
We burn a fuel releasing energy ##Q## which is completely utilized to do work on the mass ##m##. All this work goes into increasing the velocity of ##m##.
As described, this process is not conserving momentum so is impossible; you can’t increase the speed of an object in one direction without pushing something else in the opposite direction. This “something else” and its kinetic energy must be included in the calculation as Dale does in the post above.

• thinkingcap81 and Dale
Hi Dale,

I'm still going to continue asking you and others though.

Let us have a simple adiabatic piston-cylinder arrangement with the cylinder moving at a constant velocity ##V##. Let a RF be attached to the cylinder.

There is a fuel in the cylinder which releases energy ##E## by a sudden 'explosion'. This ##E## changes the velocity of the piston from ##0## to ##V_r## in the moving RF. ##E## is measured equal is both RFs as it is a scalar.

Whether in a moving or stationary RF the ##\Delta{}KE_{fuel}## is the same (##0##) since it is contained in the cylinder (the velocity of the molecules of the fuel have the same mean velocity ##V##. At least this is what i think.)

However the ##\Delta{}KE_{piston}## is different in different RFs.

So is ##E = \Delta{}KE## in both cases?

Kindly point out the mistakes in my analysis.

Dale
Mentor
Let a RF be attached to the cylinder.
This RF is non-inertial with a time varying inertial force. Energy is not conserved in this type of RF.

Try analyzing from an inertial RF instead. Since there is no external force, the center of mass of the cylinder and piston is inertial.

Hint: you can use my analysis above as a guide. The piston is the “rocket” the cylinder is the “exhaust” and the fuel is the “fuel”.

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