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Kinetic Energy and Frames of Reference

  1. Sep 29, 2011 #1
    From what I can glean, since kinetic energy = 1/2 mv^2, it follows that a doubling of velocity requires a quadrupling of energy.
    One joule is required to accelerate a 1 kg mass from zero to one meter per second per second.
    ie 1m/s2.

    Now, to further accelerate the mass to 2 meters per second requires an additional 3 joules... because at that velocity, the energy now possessed by the mass is 4 joules.

    Ok, let's say I use 1 joule of energy to accelerate the mass to 1m/sec.
    Now, say we have train passing by with velocity 1m/sec and going in the same direction as the mass. A person on that train grabs hold of the mass. Because it's and the trains speed are equal, the mass is effectively at rest from the train's frame of reference.
    Now let that person - on the moving train - apply 1 joule of energy to again accelerate the mass to 1 m/sec - but relative to HIM this time. So the actual velocity of the mass is now 2m/sec, relative to the first observer.
    So only TWO joules of energy have been required to get the mass to 2 m/sec instead of the FOUR implied by the KE equation.
    I have no difficulty in grasping that energy goes up as the square of velocity, but for the life of me, I cannot understand why both observers cannot use 1 joule of energy to effect an increase of 1 m/sec each
    :-p
    Where am I getting stuffed up here ?
     
  2. jcsd
  3. Sep 29, 2011 #2
    first your calculation of KE is little off. if m=1 kg , v = 1 m/s then K= 0.5 J. and the reason
    you seem to be confused is because of the concept of relative velocity. remember that KE depends on v and v is different according to different persons.
     
  4. Sep 29, 2011 #3

    A.T.

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    You are adding up frame dependent quantities measured in different frames. The result of such operations is meaningless because it doesn't apply in any reference frame.
     
    Last edited: Sep 29, 2011
  5. Sep 29, 2011 #4

    AlephZero

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    This is one of the "oddities" of classical mechanics. Once you have chosen a particular inertial reference frame, all the numbers work out fine within that reference frame.

    It doesn't matter which frame you choose, But the numbers representing quantites like KE are different in different frames. So are the numbers representing position, velocity, etc, so that should not really be a surprise.

    The "paradox" is mainly that the naive logic "energy is conserved, and physics is indepedent of the reference frame you use to describe it, therefore the number that measures energy should be the same in all reference frames" is false.

    Of course Einstein had some interesting things to say about this situation... :smile:
     
  6. Sep 29, 2011 #5
    Thankyou all for your patience... please forgive my thickness... I don't come across this sort of thing much in electronics :-)

    I figured that the change of reference frame was coming into it *somehow*....
    and yes, I did get the energy wrong, but whatever it is, and however the maths are getting skewed here, the case for the first observer is pretty well defined: he adds 0.5 joule of energy to the mass which is converted to it's velocity of 1m/sec.

    What I want to know is; why should the subsequent acceleration added by the second person be any different ? - it seems a fact that it requires that 0.5J for a 1m/sec velocity relative to that person. So if two of them put in 0.5J each, from their respective reference frames, adding to 1 Joule, then haven't we achieved the 2m/sec using only half of the energy ?

    I mean "a joule is joule", after all - from *anyones* frame.... isn't it ?
     
  7. Sep 29, 2011 #6

    A.T.

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    You are adding up frame dependent quantities measured in different frames. The result of such operations is meaningless because it doesn't apply in any reference frame.

    Kinetic energy and the change of kinetic energy are frame dependent quantities. The work done on the mass is force * distance, which is not the same in the two frames, because the distance traveled by the mass is different in each frame.
     
  8. Sep 29, 2011 #7

    BobG

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    Initially, the 1 kg object was at rest with 0 J of kinetic energy while the very small 1000 kg train was cruising along at 1 m/sec for a kinetic energy of 500 J

    You hurl the object forward increasing it's speed to 1 m/sec and it's kinetic energy 0.5 J and train is still cruising along a 1 m/sec with a kinetic energy of 500 J. The sum of your kinetic energy is now 500.5 J.

    The person on the train grabs the object and increases its speed by another 1 m/sec. Except, he has to push against something to do this and the something is the train itself. With conservation of momentum, the train's momentum has to decrease by 1 kg-m/sec, which essentially decreases its speed to .999 m/sec.

    At this point, the 1 kg object is moving at 2 m/sec with a kinetic energy of 2 J.

    The 1000 kg train is moving .999 m/sec with a kinetic energy of 499 J.

    I don't see a problem.
     
  9. Sep 29, 2011 #8

    cepheid

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    Electrohead,

    As you've no doubt figured out by now, kinetic energy is NOT an example of what physicists refer to as an "invariant" quantity. An invariant quantity is one that is measured to have the same value in all inertial reference frames. Kinetic energy depends on the speed of the object, which will be different for two observers in relative motion. As a result, the object has different kinetic energies to those two different observers. So, your reasoning that "a joule is a joule" is what is in error. Clearly, in the example you gave above, the object had to have gained more energy in one reference frame than it did in the other. The question is, can we explain why more energy is imparted to it in one frame than in the other? The answer, yes, we can explain that by considering each reference frame in turn:

    Train Reference Frame:

    In the train reference frame, the ball is initially stationary, because both it and the train are moving at 1 m/s relative to the track. For convenience, let's assume there's no gravity outside the train and the 1 kg ball is just hovering in mid-air outside the window. A passenger in the train sticks his hand out the window and makes contact with the ball just behind it. He then walks forward in such a way that he applies a constant 1 N force to the ball for the duration that his hand is in contact with it. F = ma says the ball then accelerates at 1 (m/s)/s. You wanted the ball to change speed until it reached a final velocity of 1 m/s relative to the train (2 m/s relative to the track). Clearly, at this acceleration it will take 1 s for it to gain this speed:

    v = v0 + at
    = 0 + 1 (m/s)/s *1 s

    = 1 m/s

    So, the force is applied for a duration of 1 s. The question is, how far does the ball move forward relative to the train during this 1 s period of constant acceleration? In other words, how far forward does the guy who's pushing on the ball have to walk? That's easy:

    d = v0t + (1/2)at2

    = 0 + (0.5)*(1 (m/s)/s)*(1 s)2

    = 0.5 m

    So the ball moves over a distance of 0.5 m while the force is being applied. Therefore, the work done on the ball is:

    W = Fd = (1 N)(0.5 m) = 0.5 J

    This is entirely consistent with the work-energy theorem, since the ball's change in kinetic energy is (0.5)(1 kg)(1 m/s)2 - (0.5)(1 kg)(0 m/s)2 = 0.5 J.

    Track Reference Frame:

    We can do the exact same analysis as above for the track reference frame. Since the ball's speed must increase from 1 m/s to 2 m/s, and the force and hence the acceleration are the same, we conclude once again that the force must be applied for a time interval of 1 s. How far forward does the ball move relative to the track while the force is being applied? I'll use the prime symbol (') to represent quantities in the track reference frame:

    d' = v0't' + (1/2)a'(t')2

    = (1 m/s)(1 s) + (0.5)(1 (m/s)/s)(1 s)2

    = 1 m + 0.5 m = 1.5 m

    So, relative to the track, the ball moves forward by a distance of 1.5 m while it is being accelerated. As a result, the work done on the ball according to the track observer is:

    W' = F'd' = (1 N)(1.5 m) = 1.5 J

    This is entirely consistent with the work-energy theorem, since the ball's change in kinetic energy is (0.5)(1 kg)(2 m/s)2 - (0.5)(1 kg)(1 m/s)2 = 2 J - 0.5 J = 1.5 J.

    So, although the gain in kinetic energy is different according to the two different observers, the work-energy theorem is being obeyed in both reference frames. The track observer sees the ball gain more KE because, from his perspective, more work is done on the ball. This is because, relative to him, it moves over a larger distance while the force is being applied than it does relative to the train observer.
     
  10. Sep 29, 2011 #9
    electrohead: a few simple illustrations why anything with velocity depends on the relative speed of the observers:

    Suppose you and are are standing together, motionless...we observe a passing object with uniform velocity as having the same speed (and energy).

    but now suppose I start walking toward the object and you remain in our original position...Now I measure you (in my own reference frame) to have some relative velocity, you measure me as having some, and we no longer "agree" on the observed velocity of the object!!!
     
  11. Sep 29, 2011 #10
    Thankyou Cepheid, for your efforts and great explanation :-)

    ok, I still have trouble with one little bit of this:

    You say that the second observer needs to input more energy to effect the same velocity increase and yes, I can see how that applies with reference to the TRACK. What I am having trouble with is that, as far as THAT observer is concerned, his "world" is only the train and the actual velocity of it is "transparent" to him... let's say, for example, it was a spaceship with no track... his "experience" will be that he can apply the same force over the same distance (ie same energy) to get the same velocity increase - of course, relative to HIM.
    Sure, KE is frame-variant, but I didn't think energy was... if the train observer has to input more energy than I did, then how can that reconcile from his frame ? In other words, why would the energy be different just because the frame is in motion - it is ONLY "in motion" relative to ME. So take ME out of the picture (and along with it, the stationary track) and we end up with the conditions being different for him as they were for me - and THAT is where I have the trouble...

    First observer (me): 1 joule (1kg @1N over MY 1m) yields 0.5 m/sec (relative to me)
    Second observer: 1 joule (1kg @1N over HIS 1m) yields 1.0 m/sec (again, relative to me)

    You said "The track observer sees the ball gain more KE because, from his perspective, more work is done on the ball. This is because, relative to him, it moves over a larger distance while the force is being applied than it does relative to the train observer.

    Yes, it moves over a larger distance - but ONLY with respect to a track observer - who cannot "influence" any outcomes simply by BEING there... as far as the train observer is concerned, he is only applying the force over a distance of 1 meter, relative to HIM...

    I was of the understanding that there really is no such thing as "absolute velocity" anyway, as pointed out by Naty1 (thankyou!) so the velocity as measured by someone else is immaterial - so the same laws of physics must apply to both.... (?)

    I guess the crux of the matter is: WHY does the second person (ON the train) observe a different outcome from doing exactly the same as I did ? If HE was to measure the energy needed for that same velocity increase, what would it be ?
     
  12. Sep 29, 2011 #11
    btw, A.T., Thankyou and yes, I can understand THAT part entirely - from the stationary observer's viewpoint - sure, a bigger distance means more work... but from the TRAIN observers viewpoint, there IS no "bigger distance" as it only applies when viewed from a different reference frame... as far as HE is concerned, we are still talking about the same distance, same force, same energy...
     
  13. Sep 30, 2011 #12
    You can all blame Tom Bearden for this with his damn "regauging" concept... since he told me (directly, through email) that overunity WAS possible - but only between systems NOT in thermdoynamic equlibrium, I have been looking for a way to "cheat" - and extract of that wonderful ZPE he always raves about... but then, a few months ago, I saw him on a video where he stated that overunity was "complete nonsense"... yet he still harps on about ZPE and extracting this energy...
    Just from my limited experience in electronics, it has always seemed that one has to put more energy INTO extracting any "FREE" energy from some system than what one gets back... so even IF a cubic centimeter of space contains enough energy to boil an ocean, what use is THAT, if you need to invest even MORE to access it ???
    I guess I should just be happy with the law of energy conservation and be done with it... but I just cannot be satisfied with all this - like, if energy can never be created, then WHERE did it all come from in the first place ?
    No need to answer this... I just felt like opening up to you guys :-)
     
  14. Sep 30, 2011 #13
  15. Sep 30, 2011 #14
    Indeed, Isaac.N,

    It is amazing how ANYONE can become a "Nuclear Physicist" in this great world of ours... all u need is Google :-)
    Serves me right for even entertaining entertaining this twaddle... it's funny - I can watch a thousand "overunity" videos and simply SEE that one might as well try to make a motor out of springs that runs itself, rather than magnets... but it still doesn't stop me from hoping somebody cracks it!
     
  16. Sep 30, 2011 #15

    A.T.

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    Kinetic energy is not an "outcome". It is just an abstract number that each observer can assign to objects.
     
  17. Sep 30, 2011 #16
    I was referring to the OUTCOME of the experiment to see what the KE result was... and I can only hope that will "stop looking at my finger and look where I am pointing"...

    * Observer A verifies does an experiment and verifies that 2 joules, 1kg yields 1 m/sec
    * Observer B also verifies the same outcome from his frame.
    * This *appears* to be inconsistent with the KE formula... I just want to know why...

    I admit I am not up with this stuff and that's why I came here for help, but one thing I am sure of is that citing the shortcomings in my grammatical or terms usage is not going to help me understand...
     
  18. Sep 30, 2011 #17

    A.T.

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    What experiment would that be?
     
  19. Sep 30, 2011 #18
    Apparently you have difficulty with the (misleading) suggestion that point of view can affect how much work must be done. However, this has more to do with definitions and references than with "observation". The point is that kinetic energy is defined wrt a reference system. If you use an engine that is fixed to the system in which you are rest, then for a certain result that engine will need to do a different amount of work than an identical engine that is in rest in the other system.

    The opposite is also true, and perhaps much quicker to understand: a car in rest will not damage anything, while a car in motion can crush a wall. So, as measured in a system in which the car is in rest, its kinetic energy is zero; while as measured in a system in which the car is moving, it does have kinetic energy.

    PS. concerning the non-linearity: I remember that once I was also bugged by the non-linearity, as it is counter-intuitive that the predicted required energy will the same for different observers. The remedy for me was to work out one simple example as calculated from two reference frames. The results were the same - as it should.
     
    Last edited: Sep 30, 2011
  20. Sep 30, 2011 #19
    Thankyou Harry!

    I just finished writing about 2 pages to try to explain myself and then lost the whole lot because I got logged off... but in the process of doing so, I now understand what is going on;

    This whole thing is analagous to using a planet's gravitational field to "fling" a spacecraft around it and gain some energy. The missing energy is actually contributed by the planet.

    Thankyou all for your patience and attention!
     
  21. Sep 30, 2011 #20
    Errm, 1kg at 1 m/sec is 0.5 joules, not 2.
    From the point of view of observer A the force exerted by the observer B is 1N, acceleratrion is 1m/s^2, duration 1s, initial speed 1 m/s, final speed 2 m/s, distance travelled during this time is 1.5m and the total energy transferred to the mass is 1.5 joules.

    At the same time observer B pushes back on the train with his bum with the same force 1N. In B's frame the train is stationary and this force does not do any work. But in A's frame, the train travels 1m against this force, transferring 1 J of energy to the observer B. As a result, in A's frame initial KE is 0.5 J, the train adds 1 J, observer B adds another 0.5 J, brinning the total KE to 2 J.

    Your problem seems to be applying energy conservation law to a system which is not closed. Somethimes it can be done but the assumptions must be understood. For example, if you have your device bolted to the floor, you can assume that the forces at the attachment points do not do any work and so the energy is conserved. But this trick only works in the reference frame of the floor where the attachment points are stationary.
     
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