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Relativity and elementary physics

  1. Oct 10, 2011 #1
    I started reading about relativity before doing my first calculus problem, and knew relativity concepts before learning basic physics I concepts.

    It actually helped me simplify a few problems. I knew that two cars colliding at different speeds was just another description of one car colliding into a stationary one at the net speed.

    However, as I learn more about early physics, it also requires me to go back to my understanding of relativity.

    Learning about kinetic energy, for example. I can only assumr that the amount of kinetic energy an object possesses is relative. But, energy of a closed system is constant. How do I reconsile these two facts?

    Is the energy still there, but "where it is" is relative?

    If I observe a friend moving at 50 meters per second, he and I both have relative velocities, and either of us could possess that kinetic energy?

    But that leads me to my next problem.

    If my friend is 10kg heavier than me, we are both traveling at 50 m/s relative to one another, but relative to me he is traveling with a greater kinetic energy than I am relative to him. How does that follow a constant amount of energy?
     
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  3. Oct 10, 2011 #2

    DaveC426913

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    Yes. Within the closed system of the passenger compartment of a train, you and I have zero velocity wrt each other, and thus have zero kinetic energy wrt each other.
     
  4. Oct 10, 2011 #3
    But what about the case where we are moving relative to one another and have different masses? We move at the same speed relative to one another, but we have different kinetic energies relative to one another. Switching reference frames can't change the total amount of energy in the car, so what am I missing in this picture?
     
  5. Oct 10, 2011 #4

    pervect

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    The amount of energy a body has is frame-dependent even in classical Newtonian physics.

    If you chose a particular frame, the energy in that frame is constant, but the value of that constant varies depending on what frame you chose.
     
  6. Oct 10, 2011 #5
    So, how about this.

    Say I am outside of the system. If I measure the energy of the system, would anyone agree with me provided they are outside of the system?

    And conversely, those who are in the system can experience the amount of energy as relative?
     
  7. Oct 10, 2011 #6

    DrGreg

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    They would agree with you only if they are at rest relative to you. (I'm assuming you are inertial and we are using SR not GR, i.e. gravity effects are negligible.) Each agrees the total energy is constant but they disagree over the value of the constant.

    It doesn't matter if they are inside or outside, the answer is the same.
     
  8. Oct 10, 2011 #7
    Wait, it doesn't?

    If I'm driving by the train car at a speed, and you are at a different speed, wouldn't we still agree on the constant of energy is inside the car?

    Why would the energy inside the car differ just because the car itself is moving differently relative to us?
     
  9. Oct 10, 2011 #8

    DrGreg

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    The energy of the train includes its kinetic energy which depends on velocity relative to the observer. Different observers measure a different speed of the train relative to themselves.
     
  10. Oct 10, 2011 #9
    This one is tricky. First we must ask ourselves if different relative motion of one object will bring with it a different energy, as proven in a collision for example. Then we have to ask if this is true for uniform motion too.

    And if we think it is then we have something we call motion, no matter if it is uniform or a acceleration. And thats the trickiest part because in a black box scenario there is no way for you distinguish between different uniform motions as I know. They are all the same and perfectly possible to define as being non moving, 'at rest'.

    If we now assume a (black box) space train, uniformly moving with passengers, all 'at rest' relative their train, they represent a 'energy'. The total 'energy' for the system is everything making up that train, but, as long as it is moving uniformly there will be no way for you to decide what 'relative motion' your train have. So does this mean that a different uniform motion won't have a effect on a 'space collision'? Expressed in that collisions 'energy'?

    That can't be true. Uniform motion do express different energies relative different velocity's, and it's not 'relative', although its 'speed/velocity' always will be 'relative' what it meets, or what frame of reference you measure it against. It's easy to prove that one. You just pick a reference frame, send something of a defined mass into a collision, then repeat with the same mass, on the same object, but now accelerate it to a higher velocity relative your target (one kg hitting the moon at 10 km, then accelerate it to a 100, then a a 1000 km, and measure, assuming 'uniform motion' before the impact here.)

    So even if the motion always is relative some defined 'frame' and you not really can define a universal 'speed/velocity' with some universal scale telling you what 'null' is, you and I both know that there will be a different energy, relative a longer acceleration at the same G before that uniform motion. But, it is not measurable in a 'black box scenario'. So where and how do the universe keep count?
    ==

    To simplify it a little, think of it as accelerating 'rocket clones' into different uniform motions relative each other, and then let them hit the same object. They will express a different 'energy', but as you forgot windows in them, all passengers would swear to that they never moved, that as they slept over the acceleration :) and so blame the impact on the other objects 'motion'. And no experiments done inside would differ for them, before or after those accelerations, which makes all uniform motion, no matter their 'relative speed/velocity' into a sort of symmetry to me.
    ==

    Even simpler, what I'm stating here is that your rockets different uniform motions, in a otherwise empty space, must be coupled to a energy, even though not measurable locally. It's all connected to a 'acceleration' somehow, as I can see for now, but I'm also wondering how it express itself. A acceleration will always express itself as inertia/gravity, and you will find light changing energy, and clocks differ, inside your spaceship, but uniform motion is truly tricky.
     
    Last edited: Oct 10, 2011
  11. Oct 10, 2011 #10
    I'm not sure if this will help, but here is an example and some comparisons to maybe think about. If we consider two identical balls of mass m moving towards each other in 1D both at 3c/5 we can treat this in two different reference frames. The kinetic energy in SR is defined by
    [tex]
    T=E-m_0 c^2=(\gamma-1)m_0 c^2=\left(\frac{1}{\sqrt{1-\left(\frac{3}{5}\right)^2}}-1\right)m_0 c^2=\frac{1}{4}m_0 c^2
    [/tex]
    but then there are two balls so the total kinetic energy is [itex]T_{tot}=1/2 m_0 c^2 [/itex].

    Now lets change reference frames to be co-moving with one of the balls. Then Velocity additions says
    [tex]
    u'=\frac{\frac{3c}{5}+\frac{3c}{5}}{1+\left(\frac{3}{5}\right)^2}=\frac{30c}{34}
    [/tex]
    So in this picture only one of the balls is moving and it's bookin' it towards the other. Using the same kinetic energy we get
    [tex]
    T=(\gamma-1)m_0 c^2=\frac{9}{8}m_0 c^2
    [/tex]
    Thus the total kinetic energy is clearly different, just as it would be even without relativity, as was stated before. But you mentioned the total energy before also, here is a quantity that is the same regardless of reference frame, [itex]\sqrt{p_{\mu}p^{\mu}}=|\mathbf{p}|[/itex]. Lets compare. In the first reference frame
    [tex]
    p^{\mu}_{tot}=\left( \frac{5}{2}m_0 c,0,0,0 \right)
    [/tex]
    and
    [tex]
    |\mathbf{p}|=\sqrt{\eta_{\alpha\beta}p^{\alpha} p^{\beta}}=\frac{5}{2}m_0 c
    [/tex]
    Now lets look in the other reference frame. In this one
    [tex]
    p_{tot}^{\mu}=\left( \frac{25}{8}m_0 c,|p|,0,0 \right)
    [/tex]
    where [itex]|p|[/itex] is the momentum of the moving ball. We can get the momentum from
    [tex]
    E^2=(pc)^2+(m_0 c^2)^2 \implies |p|=\frac{15}{8}m_0 c
    [/tex]
    The
    [tex]
    |\mathbf{p}|=\sqrt{\left( \frac{25}{8}m_0 c \right)^2-\left( \frac{15}{8} m_0 c\right)^2}=\sqrt{\frac{25}{4}m_{0}^{2} c^2}=\frac{5}{2}m_0 c
    [/tex]
    as the situation before. I hope this helps and is correct...
     
  12. Oct 11, 2011 #11
    As the total energy expressed your example seems very clear Judah, and what I think that you say is that the total 'energy' must be the same, no matter which frame you choose to measure it from, that is if I get you right?

    Assume one rocket for this, in a 'deep space' of a 'even gravity', as found by the 'far observer' (unnecessary perhaps:). How do you define that rockets uniform motions 'energy', assuming that you have no knowledge of its former acceleration, being inside it? Is that energy only existent as a function of a interaction/observation relative another frame of reference? Or is it consistent in the uniform motion itself, although not measurable? Can I measure it inside?

    (assuming it all as a 'black box scenario', this is)

    What it boils down to for me is the idea of different 'energy levels' existing for different uniform motions, and why they aren't measurable?
     
    Last edited: Oct 11, 2011
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