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A car in different frames

  1. Sep 9, 2007 #1
    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.
     
    Last edited: Sep 9, 2007
  2. jcsd
  3. Sep 9, 2007 #2
    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.
     
  4. Sep 9, 2007 #3
    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.
     
  5. Sep 9, 2007 #4

    russ_watters

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    You arbitrarily changed reference frames in the middle of the problem!
     
  6. Sep 9, 2007 #5
    jacobrhcp:
    When thinking about inertial frames or reference ( IFR ) you must think about something material not just three axis.

    Your question:
    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.

    The same as above.

    . ( I put the car in a long train which runs at 50 )
    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.
     
  7. Sep 9, 2007 #6

    bel

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    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.
     
  8. Sep 9, 2007 #7
    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.
     
  9. Sep 9, 2007 #8
    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
     
  10. Sep 9, 2007 #9

    Dale

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    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.
     
    Last edited: Sep 9, 2007
  11. Sep 10, 2007 #10

    Stingray

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    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 [itex] \Delta E_1 = 1/2 m v^2[/itex] and [itex] \Delta E_2 = 1/2 m [(v+u)^2 - u^2] \neq \Delta E_1[/itex].

    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.
     
  12. Sep 10, 2007 #11
    jacobrhcp:
    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.
     
  13. Sep 10, 2007 #12
    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
     
  14. Sep 10, 2007 #13

    Stingray

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

    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.
     
  15. Sep 10, 2007 #14

    rcgldr

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    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.

    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 space ship 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 space ship 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.
     
    Last edited: Sep 11, 2007
  16. Sep 11, 2007 #15

    Dale

    Staff: Mentor

    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.
     
  17. Sep 11, 2007 #16

    Stingray

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    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.

    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 [itex]M[/itex], and that the mass of the car is [itex]m= 1 ~\mathrm{kg}[/itex]. Also set [itex]v=100~\mathrm{m/s}[/itex]. In the first case considered by the OP, the change in kinetic energy of the system is

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

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

    [tex]
    \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
    [/tex].

    As I claimed, [itex]\Delta E = \Delta E_1 + \Delta E_2[/itex]. It didn't matter that frames were changed in the middle.
     
    Last edited: Sep 11, 2007
  18. Sep 11, 2007 #17

    Doc Al

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    Absolutely.

    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.)
     
  19. Sep 12, 2007 #18
    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
     
  20. Sep 12, 2007 #19
    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 ?
     
  21. Sep 12, 2007 #20

    Stingray

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    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).

    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.
     
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