I Fuel paradox arising from Galilean transformation?

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The discussion revolves around a perceived paradox in the context of Galilean transformations, where two observers, O and O', measure different power outputs from the same engine due to their relative motion. Observer O calculates that the engine operates at a higher power than observer O', leading to the conclusion that fuel depletion rates differ between the two frames. However, the resolution lies in recognizing that momentum is not conserved in this scenario, as energy is lost to the exhaust stream and the reactive acceleration of the Earth. The participants clarify that the change in mechanical energy remains invariant under Galilean transformations, ensuring consistency across frames. Ultimately, the paradox is resolved by accounting for unconsidered energy losses in the system.
SeniorGara
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Does the fuel paradox arise when the same engine operates with different power in different frames of reference?
I have encountered a problem related to the Galilean Transformation. Let's consider two observers who will be referred to as ##O## and ##O^{'}##, with their corresponding coordinates ##(t,x,y,z)## and ##(t^{′},x^{′},y^{′},z^{'})## respectively. They are initially at the same location, at time zero. Furthermore, observer ##O^{'}## moves away from observer ##O## as shown in the picture.
geogebra-export.png


Any point ##P## that does not move in relation to ##O## will be described by ##O^{'}## with the following equations: $$x_{P}'=x_{P}-vt,\quad{y'_{P}=y_{P},}\quad{z'_{P}=z_{P},}\quad{t'=t}.$$ If an object (for example a car) moves in relation to ##O## according to the equation $$x(t)=vt+\frac{1}{2}at^2,$$ ##O'## will describe this movement in the following way: $$x'(t)=x(t)-vt=\frac{1}{2}at^2.$$ Assuming that no resistance force is present, the resultant force is equal to the force of engine thrust. Of course, ##F=F'=ma## because ##\ddot{x}(t)=\ddot{x'}(t)=a##. Observer ##O## claims that the work done by the force of engine thrust is equal to $$W(t)=F\cdot{x(t)}=ma\left(vt+\frac{1}{2}at^2\right)=mavt+\frac{1}{2}ma^2t^2\mathrm{,}$$ whereas ##O'## observes that the engine has performed work equal to $$W'(t)=F'\cdot{x'(t)}=ma\left(\frac{1}{2}at^2\right)=\frac{1}{2}ma^2t^2.$$ Thus, ##O## concludes that the engine is operating at power $$P(t)=\frac{dW}{dt}(t)=mav+ma^2t,$$ while ##O'## considers that the engine power is equal to $$P'(t)=\frac{dW'}{dt}(t)=ma^2t.$$ Here is my question: If the same engine works with different power in two frames of reference, wouldn't it lead to the "fuel paradox"? In other words, according to ##O##, the fuel will be depleted faster than according to ##O'##. Of course, it can't be true. So, where is the mistake?
 
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SeniorGara said:
In other words, according to ##O##, the fuel will be depleted faster than according to ##O'##. Of course, it can't be true. So, where is the mistake?
What about the work being done on the exhaust stream? [Always the answer in this flavor of paradox]
 
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Notice that momentum is not conserved in your example, so you don't have a closed system and energy is slipping out unaccounted for. Account for the reactive acceleration of the Earth and you will find your missing energy.

Edit: scooped by mere seconds!
 
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You don't need anything as elaborate as your calculations. If a ##2 \ kg## object increases its velocity from ##0## to ##1 \ m/s## in one frame, then it gains ##1J## of energy. But, in a frame where it changes from ##1## to ##2 \ m/s## it gains ##3J## of energy. This gives the same potential paradox when considerihg the energy supply.
 
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PeroK said:
You don't need anything as elaborate as your calculations. If a ##2 \ kg## object increases its velocity from ##0## to ##1 \ m/s## in one frame, then it gains ##1J## of energy. But, in a frame where it changes from ##1## to ##2 \ m/s## it gains ##3J## of energy. This gives the same potential paradox when considerihg the energy supply.
To carry this scenario through, let us consider that this ##2 \text{ kg}## object gets its velocity increment by pushing off at ##1 \text{ m/s}## from an equally massive object. Our object moves off to the right at ##+1 \text{ m/s}## and the other object moves off to the left at ##-1 \text{ m/s}##.

In the original rest frame of the two objects, that is ##2J## of total energy increment, ##1J## for each.

In a frame where the two objects start at ##1 \text{m/s}##, that is ##-1J## for the left hand object and ##+3J## for the right hand object. The total is ##2J##. Same as before.

The change in mechanical energy is invariant under a Galilean transformation to a new inertial frame. Also under a Lorentz transform as it turns out, though the formula for mechanical energy needs to be corrected for that to work.
 
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Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
This has been discussed many times on PF, and will likely come up again, so the video might come handy. Previous threads: https://www.physicsforums.com/threads/is-a-treadmill-incline-just-a-marketing-gimmick.937725/ https://www.physicsforums.com/threads/work-done-running-on-an-inclined-treadmill.927825/ https://www.physicsforums.com/threads/how-do-we-calculate-the-energy-we-used-to-do-something.1052162/

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