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Relative energy ?

  1. Dec 16, 2008 #1
    "Relative" energy ?

    Hi all,

    I have this quest :
    If a man jump from the earth surface to the h - height, we can calculate the work he's created, which equals to mgh (m : his weight, g : gravity acceleration).

    But if one considers a new frame which is mounted to the jumping man, he can calculate the energy the jumping man has created according to the new frame, then we can see he has pushed the earth to the distance h with a certain velocity. So, in the new frame, he has created certain energy which is very much bigger than mgh.

    Thanks for reading and any explanation.
     
  2. jcsd
  3. Dec 16, 2008 #2
    Re: "Relative" energy ?

    No, I think you get one wrong assupmtion. That's, if you take the frame on the man, the g will not be the g for gravity, like the 10ms-2. It will be very much smaller.

    Imagine the moon, so big and it only gets 1/6 gravity of the Earth, what about the humans?

    Anyway, it is a very interesting question.
     
  4. Dec 16, 2008 #3

    russ_watters

    User Avatar

    Staff: Mentor

    Re: "Relative" energy ?

    That's not it. The potential energy he has is mgh and the frame of reference is arbitrary. For the energy he created (transferred, really), you need to subtract the starting from the ending potential energy and you'll find it always to be the same regardless of where you locate your reference.
     
  5. Dec 16, 2008 #4
    Re: "Relative" energy ?

    Thanks for the answer.
    I think in the man's frame, we can see the earth move at first at some velocity, so the energy : (M*v^2)/2 can still be applied then it'd be very large, right?
    M: the earth's mass.
     
  6. Dec 16, 2008 #5

    Dale

    Staff: Mentor

    Re: "Relative" energy ?

    Energy is conserved. That is a completely different concept from energy being frame-invariant, which it is not. Additionally, in your scenario you have the added confusion that one of your chosen referenec frames is non-inertial. In that reference frame don't forget the work done by the fictitious force.
     
  7. Dec 16, 2008 #6
    Re: "Relative" energy ?

    It looks to me that you are using the same value of ‘h’ for both “reference frames” of the man and the earth. That is incorrect because you will get a much higher figure for the PE of the earth. IF you are going to keep the displacement the same in both frames, you will need to compute a new value of gravitational acceleration, g’ that the man exerts on the earth; you cannot use 9.8 m/sec in both cases. OR you can use the same value for g and different values for h. You need to do one or the other, but not both!

    I can best illustrate this with a simple example of a 100 kg man jumping 1 meter in the air. His potential energy is mgh = 980 J. But potential energy is mutual, he does not possess it any more than the earth does; it belongs to the earth-man system. So we can use this same value for the earth and now compute the gravitational acceleration that the man exerts on the earth by making 980 J = Mg’h. Where we keep the h the same at 1 meter and compute g’. So g’ = 980 J divided by the earth’s mass of 5.98 E24 kg = 1.64 E-22 m/s^2 a very small acceleration. OR you can use the alternate method of keeping g at 9.8 m/s and solve for the new value of h’.
    That is, 980 J = Mgh’ so h’ = 980 / (5.98 E24 x 9.8) = 1.67 E-23 meter a very small distance. So when you compute the PE of the earth, use either Mg’h or Mgh’ , NOT Mgh and it will always come out to be 980 J the same as the PE of the man. This is so because they form a system which has that much PE.
     
  8. Dec 16, 2008 #7
    Re: "Relative" energy ?

    Still, I am confused about the "relative" energy. Let's suppose another senario, which is about dynamic energy only.
    There were 2 objects flying in the space at velocity V (constant) toward each other. One is very large, say, a planet, and the other is just a spacecraft.
    If we use the formula for Dynamic Energy : E = (mv^2)/2 and calculate DE of one object from the other object's frame, the 2 "relative" DEs were different !!??
     
  9. Dec 16, 2008 #8
    Re: "Relative" energy ?

    Yes, the kinetic energy of each body is calculated separately, depending on each body’s mass and velocity according to: KE = 1/2 mv^2 These kinetic energies are NOT related or relative to each other unless the two bodies are considered as part of a system, as in the previous example I gave. In that example, the potential energy, PE is mutual between the earth and the man and is the same; 980 J in that example. The kinetic energies are calculated separately and are different from each other, but they ARE related because their sum will be equal to 980 J. In that example, almost all of the 980 J will be due to the motion of the man and only a very tiny amount will be allotted to the earth as it will have a very small velocity.
    Since we know the PE already, and it will be equal to the man’s KE, it is easy to calculate the man’s initial jump velocity from : 980 = 1/2 *100*v^2 The velocity comes out to be 4.43 m/sec.
    Now use conservation of momentum to calculate the earth’s velocity: mv = MV where m is man’s mass v is man’s vel M is mass of earth and V is earth’s velocity. This will result in:
    100*4.43 = 5.98 E24*V So V = 7.4 E-23 m/sec a very very tiny velocity! But here, in classical physics, there is no limit on how slow something can move. Now, knowing this velocity, it is easy to compute the kinetic energy the earth has when the jump starts: KE = 1/2 MV^2 The result is: 5.98E24*(7.4 E-23)^2 /2 = 1.64 E-20 J a very small kinetic energy. Adding this to 980 J does not change it so the sum of the two kinetic energies is equal to 980 J. as it should be.
    So you can see that the two kinetic energies can be vastly different even though the PE is the same for both bodies in the system.
     
  10. Dec 16, 2008 #9

    Dale

    Staff: Mentor

    Re: "Relative" energy ?

    Yes, as I said above, energy is not frame-invariant. This is completely separate from the fact that energy is conserved.

    Do you understand the difference between a "frame invariant" quantity and a "conserved" quantity?
     
  11. Dec 16, 2008 #10
    Re: "Relative" energy ?

    You are right if we consider the original frame. But if we calculate the earth velocity according to the man's frame, it'd be V which is exactly the velocity of the man when he jumps.
    You said ".. the two kinetic energies can be vastly different even though the PE is the same for both bodies in the system", so we have something similar to "relative" energy?
     
  12. Dec 16, 2008 #11
    Re: "Relative" energy ?

    Thanks for your explanation.
    I am still confused about the fact : "conserved and not frame- invariant"

    I mysefl always believe that energy is conserve and frame invariant. And this query has been confusing me badly.

    I think someone will find the easy- to- understand resolution for the "paradox" soon.
     
  13. Dec 16, 2008 #12

    russ_watters

    User Avatar

    Staff: Mentor

    Re: "Relative" energy ?

    I think the scenario in your first post is the easiest one to understand. But to really get it, you need to explore it completely, not just gloss over it. So lets analyze that idea more fully, with an example:

    You live on the first floor of an apartment building and build yourself a little table-top, pump-powered waterfall, with a pool, a pipe, and a pump. It is 1m tall, from the surface of the pool to the top of the waterfall and pumps 1 kg of water in an arbitrary amount of time. PE=mgh=1*9.8*1=9.8 n-m, right?

    But I said it is on a table. Different frame of reference. The table is 1.5 meters tall. So PE = mgh = 1*9.8*2.5=24.5 n-m, right? Well, maybe... what energy are we talking about?

    24.5 n-m is the potential energy of the water at the top of the waterfall relative to the arbitrary reference frame of the floor. It is not the potential energy generated by the pump. The potential energy generated by the pump is 1*9.8*(2.5-1.5)=9.8 n-m. No matter what reference frame you select for your starting point, the potential energy generated by the pump does not change: Put your contraption on an elevator and take it to the 10th floor and that is still 9.8 n-m.

    The potential energy of any particular point is depend on a reference frame. The potential energy generated by the pump is not.
     
  14. Dec 16, 2008 #13
    Re: "Relative" energy ?

    The law of conservation of energy is book keeping. It is easy to violate if you only view part of a situation, but eventually if you consider the entire situation and keep adding up the energies the numbers will eventually come out right. Remember, it is the *sum* of kinetic and potential energies that is ultimately conserved. Take your situation for example:

    Scenario 1 - Earth static, man jumps

    A 1kg man is attached to the earth set at height = 0m

    He leaps to a height 1 m so that mgh, and busts out a calculator and records that his increase in energy is = (1kg)(9.8m/s)(1m) = 9.8 J

    He then is accelerated back towards the earth and measures his final velocity just before he hits the ground to be:

    vf^2 = vi^2 + 2(9.8m/s)(1m)

    vf = 4.427 m/s

    He then calculates his kinetic energy just before he lands as 1/2(1kg)(4.427m/s)^2 = 9.8 J

    He notices that the energy he put into raising his potential energy is equal in magnitude to the energy returned as kinetic energy when the two objects came back together. Energy is conserved, so far so good.

    Scenario 2 - Man static, Earth (err...) jumps

    A 65,000 kg earth is attached to the man set at height = 0m

    The earth leaps to a height 1 m so that mgh, and the man busts out a calculator and records that the increase in energy of the earth is = (65,000kg)(9.8m/s)(1m) = 637,000 J

    The earth then is accelerated back towards the man and the man measures the earth's final velocity just before the two come back together as:

    vf^2 = vi^2 + 2(9.8m/s)(1m)

    vf = 4.427 m/s

    He then calculates the earth's kinetic energy just before it lands as 1/2(65,000kg)(4.427m/s)^2 = 637,000 J

    Thus, this man ultimately reaches the same conclusion in either scenario. Notice that the velocity of the man and the earth are indeed the same in both examples as you mentioned and example #2 does indeed give a higher kinetic energy than the first, BUT it is the *sum of both potential and kinetic energies* that is conserved. Even though the kinetic energy in example #2 was measured to be much bigger so was the potential energy that created it.
     
  15. Dec 16, 2008 #14

    Dale

    Staff: Mentor

    Re: "Relative" energy ?

    Hi pixel01,

    There is no paradox here, just a question of definitions.

    A quantity is "frame invariant" if its value is the same in all different inertial reference frames. Examples of frame invariant quantities are mass and proper acceleration. Examples of frame variant quantities are distance, energy, and momentum.

    A quantity is "conserved" if its value is the same across time or across some specific interaction like a collision, this always refers to calculations made over time in a single reference frame. For example, the total momentum is the same before and after a collision, or the potential energy plus the kinetic energy of a satellite is constant.

    In relativity we use a convenient notation called four-vectors to keep track of physical quantities and make sure that you always transform them correctly between reference frames. Four-vectors and their Minkowski norms are frame invariant, but the components of a four-vector are frame variant, or "relative".

    If you compare the energy-momentum four-vector with the space-time four-vector you will see that energy is "relative" in the same way that time is, and momentum is "relative" in the same way that distance is, and that energy and momentum have the same relationship to each other as time and space do. None are individually frame-invariant, but taken together the conservation of the four-momentum captures in one tidy package the concepts of the conservation of mass, energy, and momentum.
     
  16. Dec 17, 2008 #15
    Re: "Relative" energy ?


    No. Scenario number two is completely wrong! This is a conserved system of two masses in a gravitational field. The total energy is equal to the total PE since it is all conserved. (We are ignoring any losses due to heat from jumping) In your scenario one, you calculated a PE of 9.8 J for a 1 kg man jumping. That will be equal to the total energy available in the system. You cannot possibly have 637,000 J in the system! When you calculate the KE of the earth, you cannot use the g of the earth on the man of 9.8, but must use the g’ of the man on the earth. That is calculated by Gm/s^2 which is 6.67 E-11*1/ (6.38E6)^2 = 1.64 E-24 m/s^2 a very small acceleration. Now you can use that g’ to calculate the PE of the earth Mg’h as being:
    5.98 E24*1.64 E-22*1m = 9.8 J This is exactly the same as the PE of the man as it must be in the conserved system The kinetic energies will be vastly different, but they cannot add up to more than 9.8 J showing that the figure of 637,000 J is ridiculous. The problem comes from assuming that both the earth and the man have the same velocity. That is not true as velocity is a vector quantity. The man and the earth each move oppositely away from their center of mass, which remains stationary throughout. The man moves one meter at the velocity of 4.43 m/sec one way, while the earth moves an incredibly small distance of 1.67 E-23 meter the other way at the velocity of 7.4 E-23 m/sec. You cannot apply constant velocities and distances because they are in a system with a fixed center of mass. That is the point that pixel01 is missing also. There is no paradox here, no “relative” energies but the energies are related according to the principle of conservation of mechanical energy in a conservative system.
     
  17. Dec 20, 2008 #16
    Re: "Relative" energy ?

    In the scenario you described above, the *center of mass* was stationary while both the man and the earth could move. That is a different sort of analysis than the one the original post wanted as far as I can tell. What I think the original post was asking is what if two people viewed the same event and one person "considered" the earth to be completely motionless the entire time (v=0m/s) and the jumping man to move, while the other person "considered" the earth moving and the jumping man completely motionless the entire time (v=0m/s). If both of them made "honest" measurements from either perspective they would measure the same distances traveled between the two bodies and the same time of jump, and therefore the same velocity and acceleration in either case.

    You can try this experiment easily enough. Go outside (err...or stay inside because it is cold) and jump in the air. Measure the distance between the ground and the bottom of your feet at the maximum height of the jump and record the time it takes to jump off the floor and land again. Have someone else independently measure the distance from the earth to the bottom of your feet and the time it takes you to jump up and down. Since the velocity difference between you and that someone else is small, the measurements from both people should agree. However, suppose you then calculated energy from *your perspective* assuming that you were completely motionless and everything else was moving. You would be making the assumption that you just kicked the earth up into the air and that you were watching it fall back to your feet (this is the ludicrous conclusion that creates the paradox as it is assuming that the body responsible for g *is the man*...but if you think about it, *how would the man know* it was a ludicrous conclusion from his perspective alone?).

    Further suppose that someone else calculated your energy from their perspective assuming you were moving and the earth was static (more or less the "correct" approach we are taught in school). The calculated numbers at the end of their experiments wouldn't agree, but both could rightfully insist that they made no errors in their measurements or in the calculations made *from their perspective* using the physical laws. Would the higher numbers calculated from the jumping mans perspective violate the conservation of energy? The answer is no, you still end up with energy being conserved no matter which frame of reference you consider to be "true" (even "ridiculous" ones that go against our common sense).
     
    Last edited: Dec 20, 2008
  18. Dec 21, 2008 #17
    Re: "Relative" energy ?

    If the man has studied any physics, he would know from the two great conservation laws, that the total energy and total momentum are both conserved and limited. Of course, if you really want to calculate ridiculous numbers for energy, you can do so as long as you realize they are ridiculous. :biggrin:
     
  19. Dec 21, 2008 #18
    Re: "Relative" energy ?

    I thought it was a pretty interesting question even if it was more philosophical than scientific.
     
  20. Dec 21, 2008 #19

    Dale

    Staff: Mentor

    Re: "Relative" energy ?

    I think it is a good question because it is an important distinction to learn and one that is not often emphasized in classrooms.

    Do you feel that you now understand the difference between the concepts "frame-invariant" and "conserved"? And how energy is a conserved but frame-varying (aka relative) quantity? Or was my explanation still unclear?
     
  21. Dec 21, 2008 #20
    Re: "Relative" energy ?

    Potential energy transfer in this case is between the Earth and the man...so it doesn't matter how you shape your reference frame within the Earth-man system.
     
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