# KE of system / different reference frames question

by jduffy77
Tags: frames, reference
P: 60
 Quote by DaleSpam Sure, I would be glad to clarify. Differences in velocity are frame invariant in Newtonian physics. So you are right to be concerned about this.
Differences in velocity are frame invariant? No, velocity is relative, so is frame dependent.

 Quote by DaleSpam In frame (a) $v_{f,e}-v_{f,c}=10^{-21} \ m/s - 0 \ m/s = 10^{-21} \ m/s$ In frame (b) note the number of 9's behind the decimal point $v_{f,e}-v_{f,c}= -9.999999999999999999999 \ m/s - (-10 \ m/s) = 10^{-21} \ m/s$ So the final velocity of the car relative to the earth is the same in both frames.
No, in the first case, the Earth is considered to be stationary, and the car changes velocity by 10m/s, and the Earth moves in the opposite direction - 1e-21m/s, so the difference, the relative velocity, is 10m/s + 1e-21m/s

In the second case the the car remains at 0m/s, and the Earth changes by -9.999999999999999999999m/s. which is not the same as the first case.

 Another concern, involves repetition of the event. The reclaimed energy could be used to accelerate the car back to almost 10m/s, ... But, in the second case, the Earth loses 100kJ each time the car is braked, and that is (approx) 10kW Over 200 cycles, a mere 2000s, the Earth has lost ( supplied) almost 1MW. it would seem that a cheap source of energy has gone unexploited.
 Quote by DaleSpam Just as the earth lost 100 kJ (50 kJ to the car's KE and 50 kJ to the battery), as the car went from 0 to 10 m/s the earth will gain 100 kJ (50 kJ from the car's KE and 50 kJ from the battery) as the car goes from 10 to 0 m/s. There is no free lunch here.
No, in the first case, the Earth gains 5e-18J.
That will hold if the acceleration is from 0m/s to 10m/s or 10m/s to 0m/s

Car loses p
Earth gains -p

Total energy = p^2/2mcar + p^2/2mearth
When the car accelerates the sign of the force changes, so the sign of the second term changes, but remains negligible.

Another worry is the mass of the Earth.
The math is circular, where a velocity of 10m/s is posited. The total p is calculated, and then 10000J subtracted, and then the KE is re-calculated using the mass, which was used to derive the momentum on the basis that the initial velocity of the Earth was 0m/s.
The result is always 10000J. Try it use m= 1e26kg.

ETA: That is in contrast with the case where actual values are used, and the same method.

Vearth ~ 29.8km/s = 29800m/s
mearth ~ 6e24kg
pi = 178800000000000000000000000000J
KEi = 2.66412e+33
subtract 10000kg.m/s
KEf 2.664119999999999999999999702e+33
dKE =298000000J
P: 60
 Quote by DaleSpam If the KERS works ideally then the remaining 50 kJ goes into the battery. Of course, nothing is 100% efficient, so some fraction will be lost to heat. It would only go all to heat if you were using traditional non-regenerative braking.
That's true, but I understood that it was idealised. The loss from the Earth of 50kJ, must be dissipated at the source, which is the Earth. Which does not happen in the first case.
PF Gold
P: 2,944
 Quote by Humber Car loses p Earth gains -p Total energy = p^2/2mcar + p^2/2mearth
This is incorrect, and will lead you astray. Changes in energy are not p2/2m, they are changes in p2/2m, so you must not treat p, and changes in p, as if they were the same thing. They aren't! Let the change in p be denoted dp, as in calculus. Then a change in p2/2m looks like (p+dp)2/2m - p2/2m, and if dp << p, this is very close to p*dp/m or v*dp. That's why when KERS kicks in and there is some dp across some relative v between the car and the Earth, the energy change of the Earth, in the car-stationary frame, is the same as the energy change of the car, in the Earth-stationary frame: it is because either way the v and the dp are the same numbers. Work it out, and spend no time on the irrelevant dp2 terms.
P: 60
 Quote by Ken G This is incorrect, and will lead you astray. Changes in energy are not p2/2m, they are changes in p2/2m, so you must not treat p, and changes in p, as if they were the same thing. They aren't! Let the change in p be denoted dp, as in calculus.
Momentum is frame independent, otherwise, there could be violations of conservation.
There would be no means of correcting after the fact, so violations simply don't occur.
That alone guarantees frame independence. If follows that changes are also frame independent.

 Quote by Ken G Then a change in p2/2m looks like (p+dp)2/2m - p2/2m, and if dp << p, this is very close to p*dp/m or v*dp. That's why when KERS kicks in and there is some dp across some relative v between the car and the Earth, the energy change of the Earth, in the car-stationary frame, is the same as the energy change of the car, in the Earth-stationary frame: it is because either way the v and the dp are the same numbers. Work it out, and spend no time on the irrelevant dp2 terms.
That is not correct, because "change" in momentum dp/dt = Force. KE is not frame independent, and that is also accommodated.
KE = p^2/2m = m^2v^2/2m = 1/2mv^2.

Total energy = p^2/2mcar + p^2/2mearth is correct. If the energy change of the car were the same as the change in the ground, p^2/2mcar - p^2/2mearth = 0, then obviously the KERS could not store any energy.
Mentor
P: 16,300
 Quote by Humber No, in the first case, the Earth is considered to be stationary, and the car changes velocity by 10m/s, and the Earth moves in the opposite direction - 1e-21m/s, so the difference, the relative velocity, is 10m/s + 1e-21m/s In the second case the the car remains at 0m/s, and the Earth changes by -9.999999999999999999999m/s. which is not the same as the first case.
The quantity you are describing, $v_{i,c}+v_{f,e}$, is not the relative velocity. See: http://en.wikipedia.org/wiki/Relative_velocity

The fact that some random sum of velocities is not frame invariant is hardly surprising, nor is it troubling.

 Quote by Humber No, in the first case, the Earth gains 5e-18J. That will hold if the acceleration is from 0m/s to 10m/s or 10m/s to 0m/s Car loses p Earth gains -p Total energy = p^2/2mcar + p^2/2mearth When the car accelerates the sign of the force changes, so the sign of the second term changes, but remains negligible.
I recommend that you work through the math completely and post your work like I did. Unfortunately, I will be travelling the next couple of days so I won't be able to do it myself.

 Quote by Humber Another worry is the mass of the Earth. The math is circular, where a velocity of 10m/s is posited. The total p is calculated, and then 10000J subtracted, and then the KE is re-calculated using the mass, which was used to derive the momentum on the basis that the initial velocity of the Earth was 0m/s. The result is always 10000J. Try it use m= 1e26kg.
The fact that the result is independent of the mass of the earth is interesting, but not troubling. Many times you get a scenario where something cancels out. I will try to do it algebraically to see where and why it drops out, but not for a few days.

 Quote by Humber Vearth ~ 29.8km/s = 29800m/s mearth ~ 6e24kg pi = 178800000000000000000000000000J KEi = 2.66412e+33 subtract 10000kg.m/s KEf 2.664119999999999999999999702e+33 dKE =298000000J
Oh, that's a mess. The units of p should be kg m/s, not J. You cannot subtract a quantity in units of kg m/s from a quantity in units of J. I am not sure what you were trying to get at here.
Mentor
P: 16,300
 Quote by Humber The loss from the Earth of 50kJ, must be dissipated at the source, which is the Earth.
What makes you say that?
P: 60
 Quote by DaleSpam The quantity you are describing, $v_{i,c}+v_{f,e}$, is not the relative velocity. See: http://en.wikipedia.org/wiki/Relative_velocity The fact that some random sum of velocities is not frame invariant is hardly surprising, nor is it troubling.
Velocity is frame dependent, and sums linearly. Random velocities sum as the RMS of each, but none are random in this case.

 Quote by DaleSpam I recommend that you work through the math completely and post your work like I did. Unfortunately, I will be travelling the next couple of days so I won't be able to do it myself.
I did that. There are several problems, including independence of Earth's mass.
That is the result of simply assigning a velocity to the Earth that is the same as the car's.

 Quote by DaleSpam The fact that the result is independent of the mass of the earth is interesting, but not troubling. Many times you get a scenario where something cancels out. I will try to do it algebraically to see where and why it drops out, but not for a few days.
(Δpc=1e4kg.m/s)
pi,e = M.v
KEi,e = pi,e2/2M
KEf,e = (pi,e)2-Δpc/2M
KEi,e - KEf,e = 100kJ

 Quote by DaleSpam Oh, that's a mess. The units of p should be kg m/s, not J. You cannot subtract a quantity in units of kg m/s from a quantity in units of J. I am not sure what you were trying to get at here.
Vearth ~ 29.8km/s = 29800m/s (orbital velocity of the Earth)
mearth ~ 6e24kg (mass of the Earth)
pi,e = 178800000000000000000000000000 kg.m/s (initial momentum mv)
KEi,e = 2.66412e+33J ( Initial KE of the Earth)

subtract Δpc= 10000kg.m/s ( The applied impulse from the car)
KEf,e 2.664119999999999999999999702e+33J ( Final KE of Earth)
dKE =298000000J (change in KE of Earth)

The same method as you employed, but where the initial Earth velocity is not zero.
P: 60
 Quote by DaleSpam What makes you say that?
The change in the Earth's KE is 100kJ. The car gains 50kJ. The remainder must go somewhere, and heat at the source seems to be the most likely. The actual reason is that there is no 100kJ

Total energy = p^2/2mcar + p^2/2mearth
Applies to energy to and from the cart as the case may be.
 P: 60 When a car brakes, momentum is transferred from the car to the ground. From the conservation of momentum; Car gains momentum -p Earth gains momentum p The car loses KE = p2/2mcar The Earth gains KE = p2/2mearth The total energy is therefore; p2/2mcar+ p2/2mearth Because mearth is very large ,~6e24kg, the second term is negligible. All of that energy will have come from the fuel of the F1 car. It could be batteries, and the result is the same. In a conventionally braked vehicle, the KE of the car's mass is dissipated as heat in the brakes. KERS does not affect momentum transfer to the ground, but stores the KE that is otherwise lost to heat. That energy may be transferred to a flywheel; The angular momentum L, of a particle, and the moment of inertia, I L = r.p = r.mv L =I.ω KE = 1/2.I.ω2 There is no significant transfer of energy to or from the F1 and the ground, as a result of KERS.
Mentor
P: 16,300
 Quote by Humber Velocity is frame dependent, and sums linearly. Random velocities sum as the RMS of each, but none are random in this case.
They may not be random, but they are not relative velocity either.

 Quote by Humber I did that. There are several problems, including independence of Earth's mass.
I don't know why you think that is a problem.

 Quote by Humber (Δpc=1e4kg.m/s) pi,e = M.v KEi,e = pi,e2/2M KEf,e = (pi,e)2-Δpc/2M KEi,e - KEf,e = 100kJ Vearth ~ 29.8km/s = 29800m/s (orbital velocity of the Earth) mearth ~ 6e24kg (mass of the Earth) pi,e = 178800000000000000000000000000 kg.m/s (initial momentum mv) KEi,e = 2.66412e+33J ( Initial KE of the Earth) subtract Δpc= 10000kg.m/s ( The applied impulse from the car) KEf,e 2.664119999999999999999999702e+33J ( Final KE of Earth) dKE =298000000J (change in KE of Earth) The same method as you employed, but where the initial Earth velocity is not zero.
This is not the same method I employed. I never subtracted a momentum from an energy, it is an invalid operation. The units don't work out.
Mentor
P: 16,300
 Quote by Humber The change in the Earth's KE is 100kJ. The car gains 50kJ. The remainder must go somewhere, and heat at the source seems to be the most likely.
The remainder goes into the battery. Where else would the energy that goes into the battery come from?
 P: 96 I'd like to think that if the car's acceleration using the earth for traction makes the world spin faster, etc, when the car eventually stops, it will surely slow the earth back down by that same (infinitesimal) amount. ;)
P: 60
 Quote by DaleSpam They may not be random, but they are not relative velocity either.
 The quantity you are describing, vi,c+vf,e, is not the relative velocity. See: http://en.wikipedia.org/wiki/Relative_velocity
vi,c =initial velocity of car
vf,e = final velocity of earth
The difference is therefore the relative velocity of car and Earth.
They are not the same in each case.

 Quote by DaleSpam I don't know why you think that is a problem.
Then you don't have a case for your calculations.

 Quote by DaleSpam This is not the same method I employed. I never subtracted a momentum from an energy, it is an invalid operation. The units don't work out.
 (b) $p_{i,c}=m v_{i,c}= 0 \ kg \ m/s$ $\Delta p_c=f_c \Delta t= -10000 \ kg \ m/s$ $p_{f,c}=p_{i,c}+\Delta p_c = -10000 \ kg \ m/s$ $v_{f,c}=p_{f,c}/m= -10 \ m/s$ $KE_{i,c}=1/2 m v_{i,c}^2 = 0 \ kJ$ $KE_{f,c}=1/2 m v_{f,c}^2 = 50 \ kJ$ $p_{i,e}=M v_{i,e}= -10^{26} \ kg \ m/s$ $f_e=-f_c=1000 \ N$ This is the impulse momentum of the car $\Delta p_e=f_e \Delta t= 10000 \ kg \ m/s$ Here is where the momentum is decreased. $p_{f,e}=p_{i,e}+\Delta p_e = -9.999999999999999999999 \; 10^{25} \ kg \ m/s$ $v_{f,e}=p_{f,e}/M= -9.999999999999999999999 \ m/s$ $KE_{i,e}=1/2 M v_{i,e}^2 = 5\;10^{26} \ kJ$ $KE_{f,e}=1/2 M v_{f,e}^2 = 4.999999999999999999999\;10^{26} \ J$
The reduction of the Earth's momentum, is the basis of your KE claim.
It applies to the second case too. Which is why you will always get 100kJ, regardless of mass. It is an error.
The result is indenpendet of the Earth's mass because it is the equivalent of

Δpc*vf,c
= -10000kg.m/s*10m/s = 100000 kg.m2/s2
= -100kJ
P: 60
 Quote by DaleSpam The remainder goes into the battery. Where else would the energy that goes into the battery come from?
You have calculated 100kJ energy from the ground of which 50kJ is stored in the battery.
That leaves 50kJ unaccounted for.

 So in the second reference frame the energy for charging the battery comes from the KE of the earth which goes down by 100 kJ. The car gains 50 kJ of kinetic energy.
Which does not allow 50kJ for the car and 50kJ for the battery, because the system is a Kinetic Energy Recovery System. And you would still have a difference of 50kJ, when compared with the first case.
P: 60
 Quote by Tea Jay I'd like to think that if the car's acceleration using the earth for traction makes the world spin faster, etc, when the car eventually stops, it will surely slow the earth back down by that same (infinitesimal) amount. ;)
Infinitesimal it is. With the given numbers, 5e-18J and a velocity change of 1e-21m/s. That is well wide of the 100kJ claimed, and of the 100W of the o.p.
In reality, the Earth is not an infinitly rigid sphere, and will dissipate energy, just as it does for Earthquakes and nuclear weapons tests.
Mentor
P: 16,300
 Quote by Humber vi,c =initial velocity of car vf,e = final velocity of earth The difference is therefore the relative velocity of car and Earth.
No, the initial relative verlocity of the earth wrt the car would be $v_{i,e}-v_{i,c}$ the final relative velocity would be $v_{f,e}-v_{f,c}$. Your quantity $v_{f,e}+v_{i,c}$ is not a relative velocity of anything. See the link I posted.

 Quote by Humber Then you don't have a case for your calculations.
Sure I do, Newton's laws are the basis for all of my calculations. That and the standard definitions for the various quantities.

 Quote by Humber The reduction of the Earth's momentum, is the basis of your KE claim. It applies to the second case too. Which is why you will always get 100kJ, regardless of mass. It is an error.
What makes you think it is an error? A small rock and a big rock fall at the same acceleration despite the difference in the mass. Not everything is a function of every mass.

 Quote by Humber Δpc*vf,c = -10000kg.m/s*10m/s = 100000 kg.m2/s2 = -100kJ
If you are trying to calculate the change in the KE of the car then you are missing a factor of 1/2. Otherwise I don't know what you are trying to calculate.
Mentor
P: 16,300
 Quote by Humber You have calculated 100kJ energy from the ground of which 50kJ is stored in the battery. That leaves 50kJ unaccounted for.
I don't know if you are being deliberately irritating or you really can't grasp something this basic. I have said this multiple times already:

Earth loses 100 kJ KE
Car gains 50 kJ KE
Battery gains 50 kJ electrochemical energy
All energy is accounted for.

There is no missing energy 50 kJ + 50 kJ = 100 kJ, if you don't believe me then get a calculator and confirm.

 Quote by Humber Which does not allow 50kJ for the car and 50kJ for the battery, because the system is a Kinetic Energy Recovery System. And you would still have a difference of 50kJ, when compared with the first case.
The KERS recovers 50 kJ of electrochemical energy from the KE of the earth and the car gains 50 kJ of KE from the KE of the earth. The earth loses 100 kJ of KE.

Why don't you ponder that a bit until you get it.
P: 60
 Quote by DaleSpam No, the initial relative verlocity of the earth wrt the car would be $v_{i,e}-v_{i,c}$ the final relative velocity would be $v_{f,e}-v_{f,c}$. Your quantity $v_{f,e}+v_{i,c}$ is not a relative velocity of anything. See the link I posted.
They are different on each case.

 Quote by DaleSpam Sure I do, Newton's laws are the basis for all of my calculations. That and the standard definitions for the various quantities. What makes you think it is an error? A small rock and a big rock fall at the same acceleration despite the difference in the mass. Not everything is a function of every mass.
And the change in momentum that you said you did not make, though it is quite clear that you did.

 Quote by DaleSpam If you are trying to calculate the change in the KE of the car then you are missing a factor of 1/2. Otherwise I don't know what you are trying to calculate.
The primary error remains. The result of the elaborated calculation that eliminates the Earth's mass in the process is, and always is;

iΔpc*vf,c
= -10000kg.m/s*10m/s = 100000 kg.m2/s2
= -100kJ

 Related Discussions Introductory Physics Homework 1 Classical Physics 3 Introductory Physics Homework 0 General Physics 2 Special & General Relativity 3