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Relativity - Sphere flattening due to relativistic speed

  1. Jan 20, 2014 #1
    Relativity -- Sphere flattening due to relativistic speed

    1. The problem statement, all variables and given/known data

    Gum balls are spherical, and about 1.5 cm in diameter. Smarties are circular in one cross
    section, with the same diameter, but perpendicular to this circular cross section, they
    are flattened, with the smallest diameter being half the largest diameter. The mass of a
    smartie is 15 g.

    a) How fast does a gum ball have to be moving with respect to an observer for this
    observer to mistake it for a smartie? Give your answer as a fraction of the speed of
    light.

    (b) Purple and red smarties reflect incident sunlight predominantly at wavelengths of
    380 nm and 750 nm respectively. If one colour of smartie is moving towards you,
    it can look like the other colour when it is stationary. Which smartie has to be
    moving, and at what fraction of the speed of light does it have to travel?

    (c) If it were possible to convert [itex] 10^{-6} [/itex] of the mass of the smartie into energy, for how long coul the released energy power a 100MW power station?

    (d)If the energy released as described in part 2(c) was converted instead into kinetic
    energy of the smartie, by how much would the smartie final velocity differ from
    that of light? Give your answer in m/s

    2. Relevant equations



    3. The attempt at a solution
    (a) - Erm.... :cry:
    (b) - Not much better. I know that it is the red smartie that must be moving towards us.
    I know for non-relativistic doppler shift [itex] \frac{\Delta\lambda}{\lambda_{emitted}}=\frac{v}{c} [/itex]
    and for relativistic doppler shift I know [itex]\frac{\lambda_{observed}}{\lambda_{emited}}=\sqrt{\frac{1+\beta}{1-\beta}} where \beta=\frac{v}{c} [/itex]
    Do I simply rearrange this formula? Doing this I got [itex] v=\frac{59}{100}c[/itex]
    (c)
    [itex] m=0.015*10^{-6}[/itex]
    [itex] E=mc^{2}; c=3*10^{8} [/itex]
    [itex] so E=1350MJ [/itex]
    [itex] Power = \frac{Energy}{time} ; so time=\frac{Energy}{Power}[/itex]
    [itex] time = 13.5 seconds [/itex]
    I'm not too confident with this, it seems a bit too simple :/ , however it is only 2 marks

    (d)Again, I think it would be far too simple to use [itex] E=\frac{1}{2}mv^{2} [/itex] with E=1350 MJ

    Thanks
     
  2. jcsd
  3. Jan 20, 2014 #2

    vela

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    The difference between the two is the shape, right? So why does the gum ball change shape when it moves relativistically?

    Yup. I didn't check your answer, but your method is correct.
    This is correct too.

    Are you justified in using the non-relativistic formula for kinetic energy? If not, you need to use the relativistic formula for kinetic energy. But otherwise, yeah, it is as simple as you're thinking.
     
  4. Jan 20, 2014 #3
    I think I've got it. My biggest problem was being able to picture it. I was thinking about it the wrong way. I imagined it sort of as the gum ball 'stretching' into a smartie, and was confused as I thought it should appear to shrink as it moves faster. As I was a bit frustrated I overlooked the simpler explanation where the 'horizontal diameter' shrinks, but the 'vertical diameter' remains the same. I also remember finding relativity so difficult when I was first introduced, that I always expect my method to be far too simple :/

    I can use [itex] L=\frac{L_0}{\gamma}[/itex] with [itex] L=0.0075m; L_{0}=0.015m; \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} [/itex]

    Solving for v, I get [itex] v=0.866c [/itex]
     
    Last edited: Jan 20, 2014
  5. Jan 20, 2014 #4

    vanhees71

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    It's
    $$\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$$
     
  6. Jan 20, 2014 #5
    Sorry, that's what I meant :S
     
  7. Jan 20, 2014 #6

    mfb

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    Just one comment on (a), not related to the intended answer here:
    You can calculate the length, and the value will get smaller due to length contraction. You cannot see it like that, however, due to the different times light needs to reach you from the front and back. A sphere can appear longer or shorter, depending on its direction (towards you / away from you). At its point of closest approach, it appears just rotated.
     
  8. Jan 21, 2014 #7
    for (d), using non- relativistic formula I got a speed greater than c, so obviously that's wrong.
    Using [itex] E_{k}=(\gamma-1)mc^{2} [/itex] and
    [itex] \epsilon=\frac{c}{2\gamma}[/itex] where [itex]\epsilon[/itex]=difference in velocities
    I get the difference in velocities to be [itex]7.5*10^{-7} ms^{-1}[/itex], the answer sheet, however, simply states [itex](d) 3.0*10^{8} m/s [/itex]
    I have also emailed my tutor about this, but he always seems to take a very long time to reply :S
     
  9. Jan 21, 2014 #8

    vela

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    You must have made a mistake because you should have gotten the correct result using the non-relativistic formula for the kinetic energy. Show us your work.
     
  10. Jan 22, 2014 #9
    Ok I know what I did wrong. Tried to rush it and used the [itex] 10^{-6} [/itex] value for the mass.
    So: [itex] E_{k}=\frac{1}{2}mv^{2} [/itex]
    [itex] E_{k}=(0.0015*10^{-6})*c^{2}=(1.5*10^{-9})c^{2}=1.35*10^{9}J[/itex]
    Solving for v; [itex] v=424264ms^{-1}[/itex]
    So the difference in velocities would be 299575735.9; which is [itex]3.0*10^{8}m/s[/itex] to 2 s.f.

    How do I know when I need to use relativistic formula for the kinetic energy?
     
  11. Jan 22, 2014 #10

    mfb

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    The low velocity looks realistic.
    That depends on the precision you want.
    If the calculation gives more than 25% of the speed of light, relativistic corrections are significant, and below 5% they are usually negligible. In between, they could be relevant, but don't have to be.
    Those are not fixed numbers - the GPS satellites need relativistic corrections for velocities of ~0.001% the speed of light as they need an extreme precision.
     
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