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The relativistic bullet

  1. Jan 10, 2004 #1
    Lets say you take a 30 gram bullet and fire it out
    of a barrel that gives the bullet a linear velocity of 86.6% of c
    with no angular velocity.
    Now it's mass would be 60 grams. Next you
    accelerate the bullet angularly while it is in
    motion,with a force that in no way affects the
    bullet's linear motion.Assuming mechanical
    failure isn't allowed, the bullet is angularly
    accelerated until the bullet's mass becomes
    90 grams. The bullet's linear velocity of 86.6%
    of c won't be affected, since the angular
    acceleration force acted perpendicular to the bullet's
    linear motion.Relativistic momentum of the
    bullet is given by p=gamma*(rest mass bullet)*v.
    You now have a case where linear momentum of the
    bullet at 86.6% of c,before angular acceleration
    is(p=2*.03*.866*c),after angular acceleration of
    the bullet,the linear momentum is now
    (p=3*.03*.866*c).Why aren't these 2 momentums
    equal since linear momentum must be conserved?
  2. jcsd
  3. Jan 10, 2004 #2
    What do you mean by "accelerate the bullet angularly? To me this means seems to mean that you're forcing it to move in a circle. As such that linear momentum of the particle is not conserved.
    How can you not effect linear motion and yet accelerate it? What do you mean by linear motion?
    The mass of the bullet is a function of speed alone. If the speed is 86.6% of c (i.e. |v|/c = 0.866) then gamma ~ 2 no matter what the velocity is. As such the mass must be twice its rest mass and thus remains 60 grams.
  4. Jan 10, 2004 #3


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    I think carl means the bullet starts spinning.

    If the mass of an object in motion increases, and you keep its velocity constant, its linear momentum will increase.

    You can get an idea of why and how this applies to your relativistic bullet by considering energy. If your bullet initially was not spinning, and you spin it up, you have to give it energy somehow.

    If that energy comes from 'outside' the bullet, you need to consider how the spinning up was done; and however it was done you will need to consider the linear momentum of the whole system, not just the bullet.

    If you spin up the bullet 'from inside', again you need to consider how that was done. Most likely it would involve some loss of mass (e.g. firing small rockets tangentially on the rim of the bullet), and you need to consider the linear momentum of the whole system (e.g. include the linear momentum of the gas molecules in the rocket exhaust).
  5. Jan 10, 2004 #4
    relativistic bullet

    The angular velocity I am refering to, is the
    same angular velocity a bullet has when it
    it is fired out of a standard rifle barrel,it
    spins.It is possible to accelerate a bullet
    angularly when the bullet is travelling in a
    straight line and not effect it's trajectory.
    If you have a device that is travelling next
    to the bullet at 86.6% of c that applies an
    accelerating force evenly all around the
    outside surface of the bullet,perpendicular to
    it's line of travel, you will not effect it's
    trajectory or speed.When you accelerate
    angularly you are increasing the effective mass
    of the bullet,since energy=mass,but you are
    not changing it's linear speed.If you allow the
    bullet's angular velocity to remain constant
    when decelerating it from 86.6% of c to zero it
    feels heavier to the decelerating force than the
    accelerating force,when no angular velocity is
    present.It follows that linear momentum posessed by
    the bullet with angular velocity is
  6. Jan 10, 2004 #5
    Re: relativistic bullet

    Okay. I see now. You're not thinking of the bullet as a particle then. If the total momentum is zero but the bullet is spinning about its center of mass then the mass of the bullet has increased from m_o to m'_o. Now change to a frame of referance which is moving relative to the bullet. The momentum is then p = gamma*m'_o*v.

    Yes. Quite true.

    There is an easy way to look at this. Think of the moving spinning bullet as composed of particles. As the bullet moves down the barrel the particles are moving in a helix. The mass of the particles depends on the speed of the particles moving along the helix. The bullet speed then reflects only one component of the velocity of the particles. The mass of the bullet is then the sum of the masses of all the particles.
  7. Jan 10, 2004 #6
  8. Jan 10, 2004 #7
    Carl - I think I have the answer to your question.

    The transformation equation for the Lorentz funny factor is Eq. (2) at

    Think of the most simple example: Let S be an inertial frame of referance. Let a particle of proper mass m0 move in a circle in the yz-plane at constant speed u. Call S the rest frame.

    Then m = m0/sqrt(1-u^2/c^2) is the mass of the particle in frame S.

    Now transform to the frame S' which is moving in the +x direction with velocity v. Let gamma = 1/sqrt[1-v^2/c^2]. Let u' be the speed of the particle as measured in S'. Then the mass, m', of the particle in the frame S' is given by

    m' = m0/sqrt(1-u'^2/c^2)

    This can be expressed in terms of the mass of the particle as measured in frame S as

    m' = m/sqrt(1-v^2/c^2)

    It should be clear that we can now just add up all the masses of all the particles which make up the rotating bullet in S and obtain the equation

    M' = M/sqrt(1-v^2/c^2)

    where M is the total relativistic mass in S which then becomes the zero momentum frame. In that frame it is reasonable to call M the 'rest mass' of the bullet. Notice that the rest mass of the bullet is then the sum of the relativistic masses of the particles which make up the bullet

    The equation for the momentum is then trivial

    P = Mv

    The results above can formally be obtained by noting that mass, m is proportional to the time component of the 4-momentum vector P where

    P = (mc, p)

    where p is the momentum vector. Thus one can simply form this vector in S and transform to S' and you have m'
  9. Jan 11, 2004 #8
    relativistic bullet

    Arcon I'm not high up the ladder in theoretical
    physics it's a hobby of mine so I didn't
    really fully follow your explination, but that
    doesn't matter.Here's an example I might better
    make my point with. You start with a x-y axis
    and place a particle at the origin and accelerate
    it in the y direction to 61.2% of c,then
    accelerate it in the x direction to 61.2% of c so
    the net velocity of the particle is 86.6% of c
    at 45 degrees.Since the gamma factor at 61.2% of
    c is 1.265 the change in gamma factor of the
    particle in the
    y axis acceleration is 1.265-1 or .265 the change
    in the gamma factor in the x acceleration is
    2-1.265 or .735 . Roughly 3 times the work
    was required to accelerate in the x direction
    than in the y direction since the gamma factor
    is 3 times greater in the x direction. The integral
    of force as a function of time acting on the x or
    y axis during acceleration is not equal when a particle is
    accelerated in this way.If particle
    is now moving at 86.6% of c is decelerated to zero with
    equal forces on the x and y axis it will
    decelerate in a straight line.You now have
    a situation where the forces of acceleration
    and deceleration are not balanced for the x and y
    axis.In a nonrelativistic situation these forces
    will be balanced when a particle is accelerated
    and decelerated in this way.
  10. Jan 11, 2004 #9
    Re: relativistic bullet

    gamma1 = 1.26
    gamma2 = 2.00

    What you did was to do an amout of work in the y-direction. That amount of work equals the change in kinetic energy of the particle. The change in kinetic energy was

    dK = (gamma1-1)m0c2 = 0.26 m0c2

    Then you did more work on the particle by exerting a force on it in the x-direction. Note that this was perpendicular to the initial velocity. The final kinetic energy was

    K = (gamma2-1)m0c2 = m0c2

    The change in kinetic energy was

    K - dK = 0.74 m0c2

    The reason for this is that the larger the velocity the more force you have to exert to obtain the same changes in velocity. In the first part the initial velocity was zero. You were then exerting a force which was at all times parallel to the velocity. The force is related to the acceleration by

    Fy = gamma3m0a = mLay


    mL = gamma3m0

    is called the Longitudinal mass. Notice that when you first applied the force the velocity was zero and therefore

    Fy = m0ay

    When you then exerted a force in the x-direction the particle had a non-zero velocity. So the moment you applied the force it was already moving and as such had a greater inertia. However the force/acceleration relation for force perpendicular to velocity is different than it was for the parallel case.

    Fx = gamma m0a = mTax


    mT = gamma m0

    is called the Transverse mass.

    Yes. That is correct. A very astute observation on your part in fact. Nice work!

    Hope that helps!
  11. Jan 11, 2004 #10


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    No, it is still 30g.

    You mean you want to give it spin?

    P = gamma*m*v

    By putting spin on it m, not gamma, changes. The momentum would be greater by the factor that m changed by. If you add spin in such a way that the speed did not change then there is a back reaction on the sorce of the torque on the bullet conserving momentum.

    No, mass is invariant.

    That is your own site and is wrong as mass is invariant to velocity transformation.

    Instead of introducing more and more mass definitions, just do it the correct and easy way. Use modern relativity in which there is only one mass m which is invariant.
  12. Jan 13, 2004 #11
    gamma 2 or 3

    Is gamma 2 or 3?
    This is my view on the situation:
    To keep things simple let say you put the 30 grams
    of the bullet on the outside circumference of the
    bullet at radius R.Next you would start the bullet revolving
    along it's central axis at a speed where it's
    total energy would now be the equivalent 45 grams.
    If you didn't know that bullet was spinning
    it would now have a rest mass of 45 grams. If
    you then accelerated the bullet to 86.6% of c
    then Gamma would be 2 and the bullet would have
    a total energy of 90 grams.
    If you knew the bullet was revoling then it must
    have a velocity with respect to a stationary
    observer so therefore it must already have
    a Gamma factor, if you knew it's angular velocity
    and that it's total energy is 45 grams you could
    calculate that it's stationary mass would be
    30 grams and the Gamma factor would be 1.5.Now
    when you accelerate the bullet to 86.6% of c
    the molecules of the bullet would have a helical
    trajectory with a velocity that would produce
    a Gamma factor of 3 and a total energy of the
    bullet of 90 grams.So you could say technically
    that the gamma factor could be 2 or 3 in my view.
  13. Jan 13, 2004 #12
    Re: gamma 2 or 3

    That is 100% correct.
    That is 100% correct.
    That is 100% correct. This is a result of the way mass transforms from one frame to another. Assume the x-component of velocity in the frame S is zero. There is a gamma factor for the bullet in this frame. Go to the inertial frame S' moving in the x-direction. There is a gamma factor due to this motion. The overall gamma of the bullet in S' is the product of these two gammas. Thus M = Gamma(s') gamma(bullet)m_o = (2)(1.5)(30 grams) = 90 grams.

    For proof of this relationship between gamma's please see

    The relevant equation is Eq. (15)
  14. Jan 13, 2004 #13


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    Re: gamma 2 or 3



    Absolutely not. Mass is invariant to translational velocity boosts.

    This is your own site pmb, not a valid referrence. Instead see
  15. Jan 19, 2004 #14


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    Arcon, DW - Enough with Waite & pmb. Discuss the issues at hand. Lose the baggage. Your discussions are welcome. The harassing is not. My patience is coming to an end.
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