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Kinetic energy was supposed to be simple

  1. Mar 29, 2009 #1
    I had a disturbing thought today and I'm hoping someone can silence it for me.

    A moving object has a certain amount of kinetic energy which is in proportion to the square of it's velocity.

    But velocity requires a frame of reference. There is no such thing as "absolute velocity", but does that mean that kinetic energy is relative?

  2. jcsd
  3. Mar 29, 2009 #2


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    Yes, kinetic energy is relative to a frame of reference.
  4. Mar 29, 2009 #3
    We can only measure the change in energy not the "absolute" energy.
  5. Mar 29, 2009 #4
    Yep K.E is relative.
  6. Mar 29, 2009 #5


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    Hi Archosaur! :smile:
    Yes, but it doesn't matter …

    The law of relativity applies to Newtonian mechanics as well as Einsteinian …

    if ∑ miui2 = ∑ mivi2,

    and if ∑ miui = ∑ mivi

    then in another frame:

    ∑ mi(ui + a)2 = ∑ mi(vi + a)2 :wink:
  7. Mar 29, 2009 #6


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    A simple example of the relative effect. If an object accelerates from -10 m/s to + 10 m/s there's no change in kinetic energy. But if the object accelerates from 0 m/s to 20 m/s the change in energy = 1/2 m (kg) (20 m/s)2. The same force over the same time could produce both effects.
  8. Mar 29, 2009 #7
    Thanks guys. I figured that, but the main thing that's bothering me is the conflict between that and the fact that we say energy is conserved. I know that kinetic energy isn't conserved, but still, what if I spin around at 5 radians/sec? Then, relative to me stars that are a million light-years away are revolving around me at 5million light-years/sec. Obviously that's a touch faster than light, so I surely wouldn't assign them ((huge # for M)((5million light-years)(conversion to meters))^2)/2 joules of KE.

    Unless, of course, I understand what tiny-tim is saying; that energy is conserved relative to any one frame of reference. is that right?

    That would make more sense, but it's still awkward to me.

    Like how that would mean that temperature is relative, too. If two object at the same temperature, relative to the earth, were flying in opposite directions, each would appear hotter to the other.

    What if those objects were bottles of water? What if, relative to the earth, they were just under their boiling point, such that, relative to one bottle, the other was OVER it's boiling point?

    Obviously, the boiling of water doesn't depend on a FoR, but if temperature is the average kinetic energy of a substance's particles, and kinetic energy is relative, then temperature is relative, so how can we say that any substance has a "boiling point"?

    This is just one of the many cogs that have been thrown into the wheels of my brain cause of this...
  9. Mar 29, 2009 #8

    Doc Al

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    The average internal kinetic energy--in the center of mass frame.
  10. Mar 29, 2009 #9


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    Yes, that is correct. A quantity is called "conserved" if its value is the same over time in a given reference frame. A quantity is called "invariant" if its value is the same in different reference frames. They are independent concepts.
  11. Mar 29, 2009 #10
    Fortunately, the objects don't care whether or not you are looking at them. Your reference frame choice is totally up to you, and it won't alter the actual behavior.

    At risk of throwing another cog... You could ask the same question about momentum, right?
  12. Mar 29, 2009 #11


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    Hi Archosaur! :smile:

    As DaleSpam says :smile:, there's a difference between conservation and invariance.

    Newtonian conservation of energy and of momentum still apply after change of velocity (ie, adding a fixed a to any velocity v in my example)

    but not in a rotating frame.

    In other words: they apply only in "inertial" frames (frames moving without rotation at uniform velocity).
  13. Mar 29, 2009 #12
    Kinetic Energy is a Lie!!!

    Yep. KE is not an intrinsic property that defines the behavior of a physical system. It's just a mathematical tool that helps in solving problems. As Doc Al said, the intrinsic, frame-invariant quantity you're looking for is the system's internal energy, which is equal to its KE in its own CoM frame. Other examples of frame-dependent mathematical tools are position, velocity, momentum, and potential energy :smile:
    Last edited: Mar 29, 2009
  14. Mar 29, 2009 #13
    Awesome guys, thanks. The "conserved" vs "invariant" clarification was very helpful. So was the re-defining of temperature.

    I just have one more kink to iron out:

    Since KE is proportional to the square of the velocity, more energy would be required to accelerate an already fast object than would be to accelerate a slower object at the same rate.

    If so, doesn't this violate the first postulate of SR?
  15. Mar 29, 2009 #14
    What about the mass of the object? Are we talking about relative to an inertial frame?
  16. Mar 29, 2009 #15
    Not a problem since this transfer of energy is governed by conservation of momentum, and hence the frame-dependent amount of energy transferred to the object is balanced by a frame-dependent loss of kinetic energy from whatever pushed it. (It may help you to go through the numbers yourself for a specific example.)
  17. Mar 31, 2009 #16
    Well...yes this can infringe with the law of conservation of energy if a special sort of component is made (I made that actually, and later on realize this fact; it was defying the law of conservation of energy).

    A classic scenario here is that, consider a sort of substance which undergoes plastic deformation after collision.

    Now imagine substance mounted on a moving cart which has track on which another mass moves.

    Consider this mass has motion relative to the cart as well as earth.

    Now this mass (call it M) when collides with the plastic substance will make it undergo deformation.

    In the same scenario, when this mass (with the same velocity) is made to collide this same substance which is on earth, the plastic deformation will be different; what will this mean?...work done on the plastic body is relative to the frame; and finally K.E is relative to the frame.

    Suppose the impact is higher when M hits the plastic body on the cart...so where did this extra energy come from?...it came form the motion of the cart, the cart will too slow down after the impact and so its energy was used too.

    It depends more on the collision.

    Considering classical mechanics only, it's conserved relative to all frames.

    Temperature is not relative..............I guess :tongue2:

    Point is that this will change your point of view...you went wrong concluding the fact that energy is only conserved relative to frames.


    I suppose I'm wrong then. :confused:

    I mean how can the energy will be conserved relative to one frame only?? I mean, by this law of conservation of energy can be defied easily, energy can be stored from one frame and taken to the another, and the work done on the energy storage component will be different in both the frame. :tongue:
  18. Mar 31, 2009 #17
    I dont know but I think this is strictely related to some quantities and some laws not all laws.

    I mean can you pls state an example where the net energy possessed by a body, which can be used is relative to frames?
  19. Mar 31, 2009 #18


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    An example: rest-mass is invariant.

    Total energy never is.
  20. Mar 31, 2009 #19


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    Certainly. Let's consider the perfectly http://en.wikipedia.org/wiki/Elastic_collision#One-dimensional_Newtonian" of a 1 kg (A) mass with a 2 kg (B) mass which collide with a speed of 1 m/s (A to the left of B, moving directly towards each other, i.e. one dimensional collision). Let's analyze the collision from the rest frame of each mass, and I will use u to denote the velocity before the collision and v to denote the velocity after the collision.

    Rest frame of A
    uA = 0 m/s
    uB = -1 m/s
    vA = -4/3 m/s
    vB = -1/3 m/s
    KE(pre) = 1 J
    KE(post) = 1 J

    Rest frame of B
    uA = 1 m/s
    uB = 0 m/s
    vA = -1/3 m/s
    vB = 2/3 m/s
    KE(pre) = 1/2 J
    KE(post) = 1/2 J

    So you see, in the two cases, KE is conserved (i.e. it is the same before and after the collision), but frame variant (i.e. it is 1 J in frame A and 1/2 J in frame B). Both frames agree that energy is conserved, but they disagree about how much energy there is. Note that you can get the vA and vB for the rest frame of B either by plugging in the initial conditions for B or by "boosting" the answers from A by 1 m/s as you would expect.
    Last edited by a moderator: Apr 24, 2017
  21. Mar 31, 2009 #20


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    Here the choice of frame of reference isn't so free. The frame of reference should correspond to the point of application for things to make sense. For example, a car acceleraing west at the equator verus a car accelerating south from the north pole. The point of application is between tire contact patch and the surface of the road. In this case the road or the car are the appropriate frames of references, especially when calculating power, which is force times speed. The source of that power would typically be chemical potential energy, which is frame independent, ignoring relativistic effects.

    A rocket can appear to be confusing if you ignore the mass of the spent fuel. With no outside forces, then the momentum of rocket and fuel is conserved. The center of mass of rocket and fuel doesn't accelerate in any direction.
  22. Apr 1, 2009 #21
    We're talking about energies right?

    I meant there'll be no case when the the work done will be different for 2 frames.

    Ok yeah...the energy might be different for 2 frames, but if that very energy possessed by the body is used, the work done will be same for all frames.
  23. Apr 1, 2009 #22
    I thought the frame you were considering in this case was of A, so vA should be 0 right?

    I mean after the collision, the velocity of B relative to A should be 1 (subtracting both the velocity's modulus)
  24. Apr 1, 2009 #23


    Staff: Mentor

    No, uA (the initial velocity of A) is 0. The velocity of A after the collision (vA) is obviously not going to be 0). I linked to a page with the formula for elastic collisions.

    (-4/3) - (-1/3) = -3/3 = -1
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