Energy Discrepancies in Changing Frames: Where Did I Make the Mistake?

[jacobrhcp]

edit: this problem is about energy discrepancies when changing frames and is no homework.

suppose I accelerate a car from 0 to 100. the energy difference is 1/2 m v^2 - 0 = 1/2 m 100^2 = 5000 m joules

now I accelerate another car from 0 to 50. the energy difference is 1/2 m v^2 - 0 = 1250 m joules

when this car reaches 50, I change the frame to the frame in which the car is in rest. In this new frame, I again accelarate to 50. The energy change is again 1250 m joules

Now I change back the frame to the starting frame. Both cars reached a 100, but the second one seems to have done it with half the energy. Where did I make the mistake?

I honestly don't know, if anyone could help it would be very much appreciated.

 

[ganstaman]

Ok, I'm not so good on the units (I'll assume you are right that m joules is right), but I can see that the math should work out.

When you say at first 1/2 m v^2 - 0 = 1/2 m 100^2, why do you put the "- 0" in there? I believe it's because what you are actually doing is:

(1/2)*m*(v)^2 - (1/2)*m*(v0)^2, where v0 is the initial velocity and v is the current velocity. If you work everything out like this, you'll see that when you accelerate from 50 to 100, it's not the same 1250m joules as going from 0 to 50.

 

[jacobrhcp]

the m is just a number representing the mass, the joules are the units.

the -0 is because I start out from rest in all cases, v0=0. The whole problem is that I changed the frame in which I was looking, so that v0 is always 0, and I get an energy discrepancy.

I don't get why I wouldn't be allowed to change frames if all I did was correct.

 

[russ_watters]

I change the frame to the frame in which the car is in rest.

Where did I make the mistake?

You arbitrarily changed reference frames in the middle of the problem!

 

[alvaros]

jacobrhcp:
When thinking about inertial frames or reference ( IFR ) you must think about something material not just three axis.

Your question:

suppose I accelerate a car from 0 to 100. the energy difference is 1/2 m v^2 - 0 = 1/2 m 100^2 = 5000 m joules

The Earth is your IFR, the force that accelerates the car is between the Earth and the car. The 5000 joules is the energy that you wasted in exerting that force.

now I accelerate another car from 0 to 50. the energy difference is 1/2 m v^2 - 0 = 1250 m joules

The same as above.

when this car reaches 50, I change the frame to the frame in which the car is in rest

. ( I put the car in a long train which runs at 50 )

In this new frame, I again accelarate to 50. The energy change is again 1250 m joules

The train is your IFR now and 1250 J are the energy the car wasted. The force now is between the car and the train.
At the end the car has wasted 2500 J. But:
Who pushed the train ? The car accelerates N ( say ) so the train accelerates S and if we want the train continues at 50 of velocity the motor of the train must have wasted those 2500 J that you missed.

 

[bel]

alvaros, I think russ is right, because the energy depended on the mass of the train then and that could be different than that of the mass and so the apparent energy loss is not lost to physically changing frames.

 

[ganstaman]

the m is just a number representing the mass, the joules are the units.

My problem wasn't with understanding what mass and joules are, it's that you gave no units for velocity, so I could only assume your conversion to joules was accurate.

 

[jacobrhcp]

gangstaman: I'm sorry I didn't get what wasn't clear to you, so I just said this to be sure.

alvaros: the train does solve the problem. But what if I don't want to look at the IFR as a car in a train? I want to look at it as saying: 'now we're going to look at this problem from a viewer outside the car that goes at 50 too and thinks he's standing still'. And for this guy, the startup velocity is zero and the ending velocity is 50. Now there is no train he can push back, but the energy discrepancy holds. I still don't really get it.

thanks for the input

 

[Dale]

Hi Jacob,

Don't get bogged down in the numbers and units etc. They are really irrelevant to the point of your question. Russ is correct, but maybe too brief. Here is the point: energy is conserved but energy is not invariant. Conservation and invariance are two completely different ideas.

What your problem demonstrates is that, in any single frame, the two cars will have the same energy, but any two frames will disagree about exactly how much energy that is.

-Regards
Dale

PS if you really want to do this kind of analysis you need to use the 4-momentum, not energy. Energy is a component of the 4-momentum. The 4-momentum as a whole is both frame invariant and conserved.

 

[Stingray]

This is a good question, and the answer is not that energy differences depend on which reference frame you choose. Since the original question was a little jumbled, I'll restate it more clearly (hopefully this is what the OP is trying to ask!):

Suppose that a car's engine burns a fixed amount of fuel for a given amount of expended mechanical energy. It should then be possible to figure out how much fuel will be consumed in accelerating between two given speeds. But the change in the car's kinetic energy depends on the frame you use: say \Delta E_1 = 1/2 m v^2 and \Delta E_2 = 1/2 m [(v+u)^2 - u^2] \neq \Delta E_1.

The solution to this "paradox" is in noticing that a car can only accelerate by "pushing" on the earth. In so doing, you add energy to both the Earth and the car. Somewhat surprisingly, the change in the Earth's kinetic energy can be neglected only if you work in a frame where it originally vanishes. It is a nice little exercise to prove this to yourself.

 

[alvaros]

jacobrhcp:

alvaros: the train does solve the problem. But what if I don't want to look at the IFR as a car in a train? I want to look at it as saying: 'now we're going to look at this problem from a viewer outside the car that goes at 50 too and thinks he's standing still'. And for this guy, the startup velocity is zero and the ending velocity is 50. Now there is no train he can push back, but the energy discrepancy holds. I still don't really get it.

The car is at 50 and you, the viewer ( the IFR ) are also at 50.
You want to accelerate the car. You need a force.
This force must be exerted between the car and another thing ( this is the key point ).
Which thing do you choose to exert that force ?
You can choose:
- a thing in the IFR at 50 ( the train ) ( A )
- a thing in the IFR at 0 ( the Earth ) ( B )
- other more complicated
The case ( A ) has been resolved.
In case ( B ) you must apply Work ( done by the force ) = F * distance ( path the force runs ).
The result must be the same.

 

[jacobrhcp]

DaleSpam: your words sound very intelligent, but to my shame I must admit I do not know what you mean by 4-momentum

StingRay: "The solution to this "paradox" is in noticing that a car can only accelerate by "pushing" on the earth. In so doing, you add energy to both the Earth and the car"

this kinda helps for the problem. How would that work with a spaceship though *thinks* (seriously, as I'm typing this, how doés that work for a spaceship!?)

alvaros: you're right

thanks for input all

 

[Stingray]

DaleSpam: your words sound very intelligent, but to my shame I must admit I do not know what you mean by 4-momentum

It's a relativistic construct which is actually not relevant to this problem.

StingRay: "The solution to this "paradox" is in noticing that a car can only accelerate by "pushing" on the earth. In so doing, you add energy to both the Earth and the car"

this kinda helps for the problem. How would that work with a spaceship though *thinks* (seriously, as I'm typing this, how doés that work for a spaceship!?)

You have to consider the change in kinetic energy of both the rocket and its propellant. The details are more difficult to verify in this case, but the same concept applies.

 

[rcgldr]

The issue here is that the frame of reference should correspond to the point of application of force, in this case the road that the car is traveling on.

First case, the car goes from 0 to 100 m/s. From your numbers, the apparently model car weighs 1 kg. Change in energy is 1/2 kg (100 m/s)^2 = 5000 Joules.

Using the second frame of reference which is +50m/s relative to the road, the car accelerates from -50m/s to +50m/s. Change in velocity is +100m/s and so the work done is the same 1/2 kg (100 m/s)^2 = 5000 Joules.

From the second frame of reference, the initial Kinetic Energy is 1/2 kg (-50m/s)^2 = 1250 Joules, and the final Kinetic Energy is 1/2 kg (+50 m/s)^2 = 1250 Joules. This is a bad choice for a frame of reference because it implies zero work done. Again the issue is the point of application of force.

Imagine instead that the model car is traveling at -50m/s towards a spring, which is totally elastic, and changes the velocity of the model car from -50m/s to + 50m/s. While decelerating from -50m/s to 0m/s the distance moved is negative, and while accelerating from 0m/s to +50m/s, the distance moved is positive. The average force in both cases is the same, but the distances have opposite signs, so the work (force x distance) done in this case is actually zero, and the kinetic energy ends up the same at 1250 joules, but in the opposite direction.

How would that work with a spaceship

For a spaceship an appropriate frame of reference is the spaceship itself, since that it the point of application of force. In the case of a space ship, part of the spaceship's mass, burnt fuel, is accelerated to high velocity, with the result of burnt fuel moving at high velocity in one direction, and the spacehip and remaining unburnt fuel moving with moderate velocity in the other direction. Note that the unburnt fuel within the spaceship experiences the same relative increase in kinetic energy that the rest of the spaceship does, which effectively increases it's potential energy relative to the initial frame of reference for the spaceship. Also the total mass of the spaceship and fuel decrease due to the fact that the fuel component of the spaceship is being burnt and ejected over time. This means that the spaceship can end up with more velocity than than the thrust velocity. If the initial frame of reference is used instead of the space ship, then the increase of kinetic energy of the unburnt fuel needs to be taken into account, effectively increasing the overall potential energy of the unburnt fuel.

 

[Dale]

It's a relativistic construct which is actually not relevant to this problem.

I am sorry, but this is completely wrong. The 4-momentum is the only way I know to solve the problem as posed. If you want to change reference frames in the middle of a problem you must use frame-invariant constructs. Otherwise the answers you get are meaningless. You cannot add 1250 J in frame A to 1250 J in frame B and come out with anything other than nonsense.

In fact, all of your discussion is ignoring the fact the the OP is not just doing the analysis in 2 different frames, but is doing the analysis twice and changing the frame in the middle of his second analysis. You are describing how to solve the following problem:

(Frame A) Vi = 0, Vf = 100, KEcar = ?, KEearth = ?
(Frame B) Vi = -50, Vf = 50, KEcar = ?, KEearth = ?

But the OP's question was

(Frame A) Vi = 0, Vf = 100, KE = ?
(Frame A) Vi = 0, Vf = 50, KEa = ? (change to Frame B) Vi = 0, Vf = 50, KEb = ?, KEa + KEb = ?

What you are proposing does not allow the OP to do the change of frames in the middle of the analysis. What I am proposing does. I will post a follow-up showing how it works.

 

[Stingray]

The issue here is that the frame of reference should correspond to the point of application of force, in this case the road that the car is traveling on.

As long as you're consistent, Newtonian mechanics will give you the same differences in kinetic energy in any inertial reference frame. Some will make a problem easier, of course, but none are preferred in any natural sense.

In fact, all of your discussion is ignoring the fact the the OP is not just doing the analysis in 2 different frames, but is doing the analysis twice and changing the frame in the middle of his second analysis. You are describing how to solve the following problem: [...]

Yes, the OP's question was a little more complicated. The answer I gave still solves it, though. As I said before, the difference in kinetic energy is the same in any inertial reference frame. It doesn't matter that frames were changed in the middle as long as each difference was computed in a single frame.

The 4-momentum is a very important concept in relativity, but I don't see how it has any real value other than conceptual elegance in Newtonian spacetime. It certainly doesn't add any information.

Just to prove that this works in the specific case discussed here, say that the mass of the Earth is M, and that the mass of the car is m= 1 ~\mathrm{kg}. Also set v=100~\mathrm{m/s}. In the first case considered by the OP, the change in kinetic energy of the system is

<br /> \Delta E = 1/2 [ m v^2 + M (mv/M)^2 ] \simeq 1/2 m v^2 = 5000~J<br />

In the second case, there are two frames. For the first, it's obvious that \Delta E_1 \simeq 1/2 m (v/2)^2 = 1250~J = \Delta E/4. The second calculation is more interesting:

<br /> \Delta E_2 = 1/2 [m (v/2)^2 + M (-mv/2M-v/2)^2 - M (-v/2)^2 ] \simeq 1/2 m [(v/2)^2 + 2 (v/2)^2] = 3750~J = 3 \Delta E/4<br />.

As I claimed, \Delta E = \Delta E_1 + \Delta E_2. It didn't matter that frames were changed in the middle.

 

[Doc Al]

As long as you're consistent, Newtonian mechanics will give you the same differences in kinetic energy in any inertial reference frame. Some will make a problem easier, of course, but none are preferred in any natural sense.

Absolutely.

The 4-momentum is a very important concept in relativity, but I don't see how it has any real value other than conceptual elegance in Newtonian spacetime. It certainly doesn't add any information.

I agree--I don't see how dragging in special relativity helps.

Just to back you up, here's a post I made a while back analyzing the same example of an accelerating car from several reference frames: https://www.physicsforums.com/showpost.php?p=1006609&postcount=12

(In that post I'm concerned with the "work"-energy theorem, but it's the same idea.)

 

[jacobrhcp]

I afraid my question was hard to answer simple, seeing all these long posts. I will read them all later, for I am sick and can't bring myself up to concentration right now.

Thanks all

 

[alvaros]

Stingray:
In the first case considered by the OP, the change in kinetic energy of the system is:
1/2 mv^2 + 1/2 M ( mv/M )^2

Where did you get 1/2 M ( mv/M )^2 ?

Where is the Inertial frame of reference after the car is accelerated ?
What does v mean ? velocity of the car respect to the IFR or respect to the Earth ?

 

[Stingray]

Where did you get 1/2 M ( mv/M )^2 ?

That came from momentum conservation. The car gains velocity v. The Earth has to go mv/M in the opposite direction (I'm idealizing it as a giant brick in saying that, but you get basically the same thing if you do it properly).

Where is the Inertial frame of reference after the car is accelerated ?
What does v mean ? velocity of the car respect to the IFR or respect to the Earth ?

The inertial reference frame is "in" the same place it started at. In the first case, it is such that the car looks like it's moving at speed v in one direction. The Earth is moving at speed -mv/M in the opposite direction.

You could define v as the final speed of the car with respect to this frame, but it's probably better to just say that it's the amount by which its speed has changed. That's invariant.

 

[alvaros]

Stingray:
The inertial reference frame is "in" the same place it started at.

The inertial reference frame remains at the center of masses ?
( or at any fixed location respect to the center of masses )

 

[Stingray]

The inertial reference frame remains at the center of masses ?
( or at any fixed location respect to the center of masses )

Yes. How could it change velocity? The center-of-mass is not accelerating if we ignore the outside universe. I would also not say that the reference frame is "at" any particular location, by the way.

 

[Dale]

Hi Jacob

Sorry about the delay in following up here, but at least it gave me something to do on the plane! Before I begin, the best way to work this problem is to simply pick any frame and stick with it and use Newtonian physics, there is no good reason to change frames in this problem.

That said, the way Stingray worked the problem is very good and only requires Newtonian physics, but it does require the introduction of another object in order to explain the discrepancy in energy. His approach is probably the best one if your question is actually about conservation of energy as calculated in different frames.

Below is the 4-momentum approach. It does not require the introduction of another object, but does require a little relativity. Relativity reduces to Newtonian mechanics at low velocities, so it can still be used here; I tend to use it any time I need to change frames regardless of the relative velocities. This approach may be better if your question is, as I understand it, more a kinematic one about how energy is defined in different frames and transforms between frames.

The 4-momentum can be written γm(c,v) = (E/c,p). So, in the first frame (frame A) since we know γ, m, c, and v we can simply write the 4-momentum at the beginning and middle as follows*:

p1 = (3.00E8, 0.) kg m/s
p2 = (3.00E8, 50.) kg m/s

The difference is:

dp1 = p2 - p1 = (4.17E-6, 50.) kg m/s

Multiplying the first component of dp1 by c gives:

dE1 = c 4.17E-6 kg m/s = 1.25E3 J

Now, in frame B (the “primed” frame indicated by ’ marks below) we can write the 4-momentum at the middle and end similarly:

p2’ = (3.00E8, 0.) kg m/s
p3’ = (3.00E8, 50.) kg m/s

The difference is also similar:

dp2’ = p3’ - p2’ = (4.17E-6, 50.) kg m/s

Multiplying the first component of dp2’ by c gives:

dE2’ = c 4.17E-6 kg m/s = 1.25E3 J

The Lorentz transform from frame A to frame B is

|1. , -1.66782E-7|
|-1.66782E-7, 1. | = L

So, to check our work up to this point we can verify that

p2’ == L.p2

Note that dE1 and dE2’ are measured in different frames, so they cannot be added together to get anything meaningful. Instead, we would like to calculate dE1’ and dE2. So:

p1’ = L.p1 = (3.00E8, -50.) kg m/s
dp1’ = p2’ - p1’ = (-4.17E-6, 50.) kg m/s
dE1’ = c -4.17E-6 kg m/s = -1.25E3 J

p3 = L^-1.p3’ = (3.00E8, 100.) kg m/s
dp2 = p3 – p2 = (1.25E-5, 50.) kg m/s
dE2 = c 1.25E-5 kg m/s = 3.75E3 J

Note that
dE1’ ≠ dE1
dE2’ ≠ dE2
dE1 + dE2 = 5.00E3 ≠ 0. J = dE1’ + dE2’

So, if the energy is different in the two frames, what is the same? It turns out that the magnitude, also called the norm or the spacetime interval**, of a 4-vector is the same in every frame even though all of the components may be different. In this case

|dp1| = |dp1’| = 50. kg m/s
|dp2| = |dp2’| = 50. kg m/s

So both frames agree about the magnitude of the change, but they disagree about how it is partitioned up into the energy and momentum components.

This may be a strange concept to grasp at first, but it is not really too difficult. For an analogy, say two individuals were describing where the statue of liberty was in relation to the Empire State building and one was using a compass based on magnetic north while the other was using one based on true north. Both would agree on how far the distance was, but they would disagree how much of that distance was partitioned into the North-South direction and how far it was in the East-West direction. Both views would be equally valid, and both would get you to from one to the other, but you couldn’t interchange or mix them without ending up in the middle of the Hudson. Similarly in this problem. Each frame agrees on the same “magnitude” of change in the 4-momentum, but they disagree about how much is energy and how much is momentum, each view is valid but you cannot mix them.

-Regards
Dale

*I have written the 4-momentum with only two components, the other two spatial components are each 0 for this problem. This is a pretty common short-hand.

**A traditional vector magnitude is sqrt(x²+y²+z²), but a spacetime interval is sqrt(-t²+x²+y²+z²). So the time component is different from the three spatial components.

 

[jacobrhcp]

hi, wow... thanks a lot DaleSpam and others for your elaborate efforts

I hope you did it because it was fun, not for me >_> because apart from small explanations I don't see myself giving these really long summaries to random guys from the internet like me. Anyways, I think the answer was already covered but I find this quite interesting so I'll read it anyways.

regards, Jacob

*starts reading*

 

[jacobrhcp]

this concept is a bit strange indeed... I did special relativity and passed well, but we didn't really cover this part >_>. It's very interesting and I'm sure I'll see it when I do general relativity next year. I'm still a bit unfamiliar with notations like γm(c,v) = (E/c,p) (I think they're about to introduce it in my QM course). Thanks for the explanation though, it was nice=)

anyways, thanks all, again.

edit: sorry about the double post thingy

edit2: In this expression: |-1.66782E-7, 1. | = L... where did that number -1.66etc come from?

 

[alvaros]

To all who write big formulas:
Could you give an real example of the post of the OP ? ( the way I did with the train )

 

[rcgldr]

Just to prove that this works in the specific case discussed here, say that the mass of the Earth is M, and that the mass of the car is m= 1 ~\mathrm{kg}. Also set v=100~\mathrm{m/s}. In the first case considered by the OP, the change in kinetic energy of the system is

<br /> \Delta E = 1/2 [ m v^2 + M (mv/M)^2 ] \simeq 1/2 m v^2 = 5000~J<br />

In the second case, there are two frames. For the first, it's obvious that \Delta E_1 \simeq 1/2 m (v/2)^2 = 1250~J = \Delta E/4. The second calculation is more interesting:

<br /> \Delta E_2 = 1/2 [m (v/2)^2 + M (-mv/2M-v/2)^2 - M (-v/2)^2 ] \simeq 1/2 m [(v/2)^2 + 2 (v/2)^2] = 3750~J = 3 \Delta E/4<br />.

As I claimed, \Delta E = \Delta E_1 + \Delta E_2. It didn't matter that frames were changed in the middle.

I don't understand this. In the second case, starting at the second frame where 1250 joules of work have already been done, the car only ouputs 3750 joules more of work, not 5000. So starting at the second frame, \Delta E_{car} + \Delta E_{earth} = 3750\ J.

Using the second frame of reference for the entire sequence, then

\Delta E_{car} = 1/2 kg (+50 m/s)^2 - 1/2 kg (-50 m/s)^2 = 0\ J

Keeping the second frame of reference at it's original velocity, instead of relative to the earth, the engine does slightly more than 5000 Joules of work to compensate for the slight negative velocity of the Earth when the car reaches +50m/s relative to the second frame of reference.

Mean radius of Earth = Re = 6371010 m
Mass of Earth = Me = 5.9736×10^24 kg
Velocity of surface of Earth = Ve
Angular velocity of Earth = We = Ve/Re
Angular KE of Earth = 1/2 (2/5) Me (We)^2 = 1/5 Me (Ve)^2
Slight change in Earth velocity = Vs
Slight amount of work done in excess of 5000 J = Es

\Delta E_{earth} = 1/5 Me ((-50-V_s) m/s)^2 - 1/5 Me ((-50)m/s)^2 = (5000+E_s)\ J

 

[Stingray]

I don't understand this.

Which part? You seem to have essentially repeated back my post in different words.

 

[rcgldr]

I don't understand this.

Which part? You seem to have essentially repeated back my post in different words.

You're point was valid, but the numbers didn't match up with either reference frame mentioned by the OP:

The OP mentioned two reference frames, and in neither of these is:
\Delta E_{car} = 1250\ J \ {and}\ \Delta E_{earth} = 3750\ J

In the OP's first frame of reference:
\Delta E_{car} = 5000\ J\ {and}\ \Delta E_{earth} = 0\ J

And in starting with the OP's second frame of reference where the car peforms 3750 joules of work:
\Delta E_{car} = 1250\ J\ {and}\ \Delta E_{earth} = 2500\ J



Or using the second frame of reference and starting with the car at -50m/s.
\Delta E_{car} = 0\ J\ {and}\ \Delta E_{earth} = 5000\ J

To get your numbers, the frame of reference is moving at +37.5m/s (as opposed to +50m/s) relative to the initial starting point.
\Delta E_{car} = 1/2\ kg\ (+62.5\ m/s)^2\ -\ 1/2\ kg\ (-37.5\ m/s)^2 = 1250\ J

 

[Stingray]

The OP mentioned two reference frames, and in neither of these is:
\Delta E_{car} = 1250\ J \ {and}\ \Delta E_{earth} = 3750\ J

I never claimed that. \Delta E_1 just meant the total change in kinetic energy in the first frame (for the first half of the acceleration). Similarly, \Delta E_2 was the total change in kinetic energy over the second half of the acceleration calculated using the second reference frame.

 

[rcgldr]

I never claimed that. \Delta E_1 just meant the total change in kinetic energy in the first frame (for the first half of the acceleration). Similarly, \Delta E_2 was the total change in kinetic energy over the second half of the acceleration calculated using the second reference frame.

Ok, I understand now, I mistook E1 and E2 to be Ecar and Eearth.

 

[jacobrhcp]

I've analysed my original problem and have come to the following answer:

The thing with this paradox is that it assumes you can change frames in the middle of a problem. You can look at things in one frame, or another. And this does not mean that because they both are valid that it's still valid if you look at it in two different ways.

You can summarize all kinds of ways to change the frames legally, and make it two separate problems. But the real thing is just that physics is not in such a way you can think of it from two frames at the same time, and still see something sensible. You will need conversions. And that's okay, because no one ever said you could do without.