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Mass and inertia at high speeds

  1. May 16, 2014 #1
    When a particle is traveling very close to c. Will its mass actually increase and therefore its ability to resist change in its state of motion increase, OR will JUST its ability to resist change in its state of motion increase and the mass will remain the same?

    Will the particle just behave as the mass has increased, or will the mass actually increase?

    (And if the mass actually increases, what would that really mean?)
     
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  3. May 16, 2014 #2

    HallsofIvy

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    The mass actually increases.
     
  4. May 16, 2014 #3
    Ok, now what does that mean;

    If the particle is moving in this state through a gravitational field, would the gravitational force on it be stronger? Do they actually account for this in particle accelerators?
     
  5. May 16, 2014 #4

    Doc Al

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    These days, the term "mass" refers to the invariant mass, which is a property of the particle that doesn't change with speed. (The old-fashioned concept of 'relativistic mass', which does increase with speed, has largely been dropped as having too many problems.)

    So to answer your question: Mass remains invariant. But resistance to change in motion depends on velocity (and direction!).
     
  6. May 16, 2014 #5
    Invariant means that its a constant?
     
  7. May 16, 2014 #6

    Doc Al

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    Yes, that it is frame independent. The same, regardless of speed.
     
  8. May 16, 2014 #7

    adjacent

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    Direction? How is that?
     
  9. May 16, 2014 #8

    Doc Al

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    The same force acting on the same particle will produce a different acceleration depending on the direction the force is applied. For example: Pushing something transverse to its motion requires less force that pushing something in the direction it's moving (to get the same acceleration).
     
  10. May 16, 2014 #9

    adjacent

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    I don't understand. Will not the same force always produce the same acceleration?
    Are you talking about the net acceleration?
     
  11. May 16, 2014 #10
    So they dont account for a gravitational increase in accelerators?
     
  12. May 16, 2014 #11

    Then why do they use the formula:

    F=((lorentz)mv^2)/r

    for circular accelerators? The force in these cases are always transverse to the motion...

    Read about it here:
    http://en.wikipedia.org/wiki/Centripetal_force
     
    Last edited: May 16, 2014
  13. May 16, 2014 #12

    Doc Al

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    No. As the speed of a particle increases, it takes more force to produce the same acceleration.
     
  14. May 16, 2014 #13

    Doc Al

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    What does that have to do with what I posted?

    In any case, for forces transverse to the motion: [itex]F_t = m\gamma a_t[/itex].
     
  15. May 16, 2014 #14

    adjacent

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    That contradicts ##F=ma##.
    Uhh. That's Newtonian Universe. So as the speed increases,the mass increases,so the same force produce a lesser acceleration.
    Oh. i understand.
     
  16. May 16, 2014 #15

    Doc Al

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    Yes, the Newtonian F = ma is approximately true for low speeds, but not relativistic speeds. But F = dp/dt works, even relativistically. (Of course you must use the relativistic momentum.)

    I would forget the part about mass increasing. Best to stick with invariant mass.
     
  17. May 16, 2014 #16

    xox

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    [tex]\vec{F}=\frac{d \vec{p}}{dt}[/tex]

    1. In Newtonian physics

    [tex]\vec{p}=m \vec{v}[/tex]

    so

    [tex]\vec{F}=m\frac{d \vec{v}}{dt}=m \vec{a}[/tex]

    2. In reality, SR teaches us that :

    [tex]\vec{p}=m \frac{\vec{v}}{\sqrt{1-(v/c)^2}}=m \gamma \vec{v}[/tex]

    Therefore:

    [tex]\vec{F}=m \gamma \frac{d \vec{v}}{dt}+m \vec{v} \frac{d \gamma}{dt}=\gamma m \vec{a}+\gamma^3 ma\frac{v\vec{v}}{c^2} [/tex]

    so, force has not only a component along the acceleration but also one aligned wit the velocity, unlike in Newtonian physics where force is aligned with the acceleration.

    The above complicates the equations of motion for particle accelerators because we have to solve a very complicated set of differential equations produced by the fact that

    [tex]\vec{F}=q(\vec{E}+\vec{v} \times \vec{B})[/tex]

    In other words we need to solve:

    [tex]q(\vec{E}+\vec{v} \times \vec{B})=m \gamma \frac{d \vec{v}}{dt}+m \vec{v} \frac{d \gamma}{dt}[/tex]

    If [itex]\vec{E} \ne 0[/itex] the resulting equation is very tough to solve (symbolic solutions still exist). On the other hand, if [itex]\vec{E} = 0[/itex] then the force is perpendicular on the speed and the equation is much easier to solve (for details, see Bill_K's post below).
     
    Last edited: May 16, 2014
  18. May 16, 2014 #17

    Bill_K

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    Force in SR can be defined in several different ways. Let's say we use F = dp/dt where p = γmv and γ = √(1 - v2/c2).

    If the force is transverse, |v| will be constant, hence γ will be constant, and we just get F = γma. But if the force is parallel to the motion, in performing the derivative we must include dγ/dt as well. After some algebra the result turns out to be F = γ3ma.

    Since γ > 1, the second result is greater than the first, and we see it takes more force to accelerate a mass in the parallel direction than transversely.
     
  19. May 16, 2014 #18

    DrGreg

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    It's worth pointing out that in relativity, acceleration (like many other things) is relative. Although all inertial observers agree whether something is accelerating or not, they may disagree over the numerical value of a non-zero acceleration. Proper acceleration is the acceleration measured in an inertial frame where the velocity is momentarily zero. It's also what accelerometers measure. Proper acceleration does not depend on the direction of the applied force, but coordinate acceleration does depend on the angle between the force and the (non-zero) velocity.

    So, to answer your question, the same force always produces the same proper acceleration (for a given mass -- i.e. rest mass), but not necessarily the same coordinate acceleration.
     
  20. May 16, 2014 #19

    Doc Al

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    Good clarification. My comments earlier were for coordinate acceleration, not proper acceleration.
     
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