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Does kinetic energy increase the inertial mass?

  1. Mar 14, 2015 #1
    In class we had a conceptual problem:

    There will be a race between two springs. Both will receive the same initial impulse, but one has been agitated and is therefore oscillating in some way, the other is just still. Who arrives first?

    The answer is that the still spring arrives first because it has less energy.

    However I don't really get this. I know of the mass energy equivalence, but my understanding is that you actually have to consider the relativistic total energy, not just ##E=mc^2##. The spring has more energy, right, but this because each particle in the system has a kinetic energy. The rest energy is left unaltered. Namely, the total energy of the oscillating system would be ##E_{tot}=\sum_i \frac{\gamma m_i c^2}{\sqrt{1-u^2/c^2}} = \frac{\gamma M c^2}{\sqrt{1-u^2/c^2}}## … ##M## is still the original inertial mass.

    An oscillating system hasn't more inertial mass, this is nonsense to me. The same analysis could be done in the case of a hot body. People say that a hot coffee is more massive than a cold one, but because of a similar analysis as above I would say that's wrong.

    Places where inertial mass could be changed, though, are nuclear processes and other things I have no knowledge of.

    Obviously I am the one who is wrong. But I want to know why.

    Thanks.
     
  2. jcsd
  3. Mar 14, 2015 #2

    Dale

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    The rest energy of each particle is unaltered, but the rest energy of the system is altered. The rest energy of the system includes the kinetic energy of the system in the frame where the center of mass is at rest.
     
  4. Mar 14, 2015 #3

    Orodruin

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    You need to consider the total 4-momentum of each spring. The excited spring will have a larger squared 4-momentum, and giving both the same initial impulse will therefore result in the non-excited spring having a larger velocity. You seem to be mixing inertia of a composite object with the sum of the rest masses of the constituents.
     
  5. Mar 14, 2015 #4
    Yes, at a first sight also I would think it's a nonsense. Yet the equation E=mc^2 just means the more you accelerated the particle giving it kinetic energy, the more mass it gets, being c constant.
    This is true in general but to observe this increase of mass actually the particle must be relativistic, that is its speed should be almost half c.
    So the oscillating spring has more energy and then more mass according to the equation.
     
  6. Mar 14, 2015 #5

    Orodruin

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    I suggest reading the following FAQs:
    What is the Mass–energy equivalence
    https://www.physicsforums.com/threads/what-is-relativistic-mass-and-why-it-is-not-used-much.796527/ [Broken]
     
    Last edited by a moderator: May 7, 2017
  7. Mar 14, 2015 #6
    I thought about this but I can't justify the importance of the center of mass here. The theorems I know relevant to center of mass depend on classical physics (ideas like momentum is mv and force is ma).

    I'm not sure I'm making sense on this question, but how do you justify that considering the system as a whole and considering only the center of mass are the same thing in relativity?

    Or if that question didn't make sense then: how do you prove that "The rest energy of the system includes the kinetic energy of the system in the frame where the center of mass is at rest" ?

    Thanks.


    Oroduin: I'm not familiar with 4-momentum, so I'll have to read about that. :S Thanks.
     
  8. Mar 14, 2015 #7

    Orodruin

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    You can consider each particle individually as well, you will obtain the same result for the motion of the center of mass.

    If you want to have a more "intuitive" answer, a system will have some particles going in the direction of the added momentum and some in the opposite direction. Accelerating these particles will be more difficult (i.e., they have more inertia) than if only considering particles at rest.
     
  9. Mar 14, 2015 #8
    How is ##\sum \gamma_i m_i \vec{u_i} = \gamma_{CM} M_{system} \vec{U_{CM}} ## ?

    I don't see how I can arrive there starting from

    ##\sum \frac{m_i \vec{r_i}}{M}=\vec{R_{cm}}##

    In any case I don't see how this will tell me that ## \Delta E_{system} = \Delta M c^2##

    If I consider each particle individually I would proceed this way:

    ##E_{system} = \sum( KE_i + m_i c^2) = ( \sum KE_i ) + \sum m_i c^2##

    If I have a spring and excite it I will get:

    ##\Delta E_{system} = \Delta (\sum KE_i) + \Delta (\sum m_i c^2) ##

    I would say the second term on the RHS is equal to zero.

    What DaleSpam seems to tell me (I think) is that if I go to a frame where the center of mass is at rest:

    ##\Delta E_{system} = \Delta (\sum m_i c^2) ## And therefore I would conclude the system has become more massive. But this is assuming that ##\Delta (\sum KE_i)=0## which doesn't seem true to me. Why is this so?

    Thanks.

    I like your intuitive argument but I want to make this idea mathematical, if its possible. Thanks.
     
  10. Mar 14, 2015 #9

    Orodruin

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    This is not how the CoM is defined in relativity, unless you consider the m_i to be the relativistic masses and M the total energy of the system, and I really would not like to.

    Nobody is saying that. What Dale is saying is that there is a difference in the rest mass of the system and the sum of the rest masses of the particles that make up the system. The kinetic energy of the individual particles do not manifest itself as motion of the system as (by definition) the CoM frame is the frame where the object has zero momentum.
     
  11. Mar 14, 2015 #10

    Dale

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    Consider a system composed of two particles, each of invariant mass 1 and in units where c=1. Consider their four-momenta in the frame where the center of mass is at rest (aka the center of momentum frame).

    The four momentum of one particle is given by ##\left( \frac{1}{\sqrt{1-v^2}},\frac{v}{\sqrt{1-v^2}} ,0,0\right)##
    The four-momentum of the other particle is given by ##\left( \frac{1}{\sqrt{1-v^2}},-\frac{v}{\sqrt{1-v^2}} ,0,0\right)##
    And the four momentum of the system is given by ##\left( \frac{2}{\sqrt{1-v^2}},0 ,0,0\right)##

    If you calculate the invariant mass you get that the first particle has an invariant mass of 1, the second has an invariant mass of 1, but the system as a whole has an invariant mass of ##\frac{2}{\sqrt{1-v^2}}## which is equal to 2 plus the KE in the COM frame. Thus, the rest energy of the system includes the kinetic energy of the system in the COM frame.
     
  12. Mar 15, 2015 #11
    I wish to know if a system formed by a dancer that during a rotation enlarges his arm to increase the inertia, slowing so his speed, is similar or if it has no relation with this inertia.
     
  13. Mar 15, 2015 #12

    Matterwave

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    That's moment of inertia, which is a very different concept than the inertia discussed in this thread. The reason the dancer slows in rotation is due to conservation of angular momentum.
     
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