Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

High-speed bomb dilemma?

  1. Dec 16, 2012 #1
    Suppose there is a nuclear bomb orbiting around the earth at speed close to c. Some Bad Guy tries to set off the bomb by some remote control device? Wil he successfully destroy the earth in this way?

    From the perspective of the bomb, it seems it is possible. But from the earth, it seems the process would take very long and the bomb would not be set off.

    How can I resolve this dilemma?
    Thanks
     
  2. jcsd
  3. Dec 16, 2012 #2

    K^2

    User Avatar
    Science Advisor

    Normally, in Special Relativity this problem is resolved trivially, because object traveling at near speed of light will be in close proximity only briefly. Here, you've come up with a situation where object travels fast, and yet, isn't going anywhere. That should ring some alarm bells.

    Specifically, realize that an object that's traveling near speed of light around Earth is undergoing very high centripetal acceleration. That means, frame of reference relative to the orbiting body is going to be an accelerated one, and SR no longer applies. You need General Relativity to work in that frame. Earth, however, isn't accelerating, so you can use SR to describe situation from Earth's perspective. Therefore, any conclusions you arrive at (correctly) from Earth's perspective are the valid ones.
     
  4. Dec 16, 2012 #3

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Moving from Newton mechanics to Special Relativity doesn't make you suddenly forget how to make a change of variable. :grumpy:
     
  5. Dec 16, 2012 #4
    I did not realize one need to consider general relativity here. But this makes it even more confusing. I am not sure what earth people should see now. Could you elaborate?
     
  6. Dec 16, 2012 #5

    K^2

    User Avatar
    Science Advisor

    Then why don't you write down the correct time-dilation of Earth in the frame of reference attached to an object moving at some v~c in circular orbit around Earth.
    Earth is not an accelerated frame. (Well, it is due to gravity, but we can ignore that for the moment. It's a tiny effect compared to what's happening with this hypothetical satellite.) If you have an inertial frame of reference, you can just apply Special Relativity. Accelerating objects are not a problem in SR. You can use your normal time-dilation formulae to compute how quickly, say, a timer on the satellite ticks from perspective of people on Earth. It will run much slower, of course.

    It's describing things from perspective of satellite that is tough. There are ways to cheat, and compute all of the effects via inertial frame(s). But if you want to do things honestly from just the accelerated frame of the satellite, you would have to construct the coordinate system and metric tensor in which satellite is stationary, yet experiences acceleration, and in which Earth is in "free fall" while staying in circular orbit around the satellite. Alternatively, you can say that the satellite also rotates to keep same side towards Earth, and so Earth is also stationary in this frame of reference.

    P.S. I should probably note that if Earth is still taken as an origin of this new coordinate system, and stays at origin, while satellite has constant position relative to Earth, there is a very simple coordinate system to take care of it. It's the ordinary rotating coordinate system with a rotating polar metric. It's straightforward enough as far as GR goes, but still a bit messy, even in 2+1 dimensions.
     
    Last edited: Dec 16, 2012
  7. Dec 16, 2012 #6
    Thank! But is it possible to resolve this 'dilemma' without using any equations? Just qualitative explanation is enough. Thanks again!
     
  8. Dec 16, 2012 #7

    ghwellsjr

    User Avatar
    Science Advisor
    Gold Member

    Are you saying that because time is progressing so slowly that it would take "forever" for the bomb to finishing exploding? In other words, you're not questioning the fact that the Bad Guy's signal would quickly get to the bomb and initiate the explosion, but rather that the explosion would take place in slow motion and therefore would lose its explosive impact, correct?

    But have you considered the fact that in order for anything to orbit the earth at near light speed, there would have to be an expenditure of energy far greater than what is contained in a nuclear bomb? So there would hardly be any difference between the release of the vast amount of energy keeping the bomb in orbit and the bomb exploding rather quickly. Presumably, once the bomb explodes, the rockets that are required to keep it in orbit would be destroyed, ending the continual release of rocket energy and so the Bad Guy is really doing the earth a favor by setting off the bomb.
     
  9. Dec 16, 2012 #8

    K^2

    User Avatar
    Science Advisor

    Eh, I might as well just run through it. It's a good exercise. (P.S. If you can't follow this, don't worry about it. Just skip to the part after the quote.)

    The coordinate system of choice will be a rotating frame with angular velocity Ω. Naturally, we'll chose Ω such that the satellite isn't moving, but we can work it out later. I'll also completely ignore gravity. It's a minor effect. One can use all the same logic used here to add correction to Schwarzschild metric that would allow you to do gravity from a rotating frame, so long as center of rotation matches center singularity of Schwarzschild metric.

    Anyways, we need the distance element. Lets first work out what the distance is in spacial coordinates. We really need to only consider r and [itex]\small \varphi[/itex], so I'm going to drop the 3rd spacial coordinate all together. Because the frame is rotating, an object that did not move in [itex]\small \varphi[/itex] is actually Ωdt from where it started out. So the square of spacial distance between two points separated by dt is dr² + r²(Ωdt - d[itex]\small \varphi[/itex])². Separation in time is still just dt. So the line element works out to the following.

    [tex]ds^2 = c^2dt^2 - dr^2 - r^2(d\varphi^2 + \Omega^2 dt^2 - 2\Omega d\varphi dt)[/tex]

    And so we have the metric tensor.

    [tex]g_{\mu\nu}=\left( \begin{array}{ccc}
    1-\frac{\Omega^2 r^2}{c^2} & 0 & \frac{\Omega r^2}{c}\\
    0 & -1 & 0\\
    \frac{\Omega r^2}{c} & 0 & -r^2 \end{array} \right)[/tex]

    With 3-vector [itex]\small x^{\mu}[/itex] given by (ct, r, [itex]\small \varphi[/itex]). The inverse will come in handy.

    [tex]g_{\mu\nu}=\left( \begin{array}{ccc}
    1 & 0 & \frac{\Omega}{c} \\
    0 & -1 & 0 \\
    \frac{\Omega}{c} & 0 & \frac{\Omega^2 r^2}{c^2}-\frac{1}{r^2} \end{array} \right)[/tex]

    In this coordinate system, a stationary object will have velocity [itex]\small u^{\mu} = (u^t, 0, 0)[/itex].

    [tex]c^2 = u_{\mu}u^{\mu} = g_{\mu\nu}u^{\mu}u^{\nu} = (1-\frac{\Omega^2r^2}{c^2})(u^t)^2[/tex]
    [tex]u^t=c\left(1-\frac{\Omega^2r^2}{c^2}\right)^{-\frac{1}{2}}[/tex]

    This is actually sufficient for time dilation. If you note that in inertial frame of reference tied to Earth, v=Ωr for any circular orbit, and that c dt/d[itex]\small \tau[/itex] = [itex]\small u^t[/itex], you have the time dilation formula in hand. It results in no time dilation for Earth and time dilation consistent with SR for the satellite.

    Of course, having metric and 3-velocity, it might be interesting to find the acceleration. Because metric depends on radius only, it is not too difficult.

    [tex]a^\mu = \nabla_u u^{\mu} = \frac{\partial u^{\mu}}{\partial x^{\nu}}u^{\nu} + \Gamma^{\mu}_{\rho \sigma}u^{\rho}u^{\sigma}[/tex]

    That leaves just three relevant Christoffel Symbols.

    [tex]\Gamma^t_{tt} = 0[/tex]
    [tex]\Gamma^r_{tt} = -\frac{1}{2}g^{rr}\frac{\partial g_{tt}}{\partial r} = -\frac{\Omega^2 r}{c^2}[/tex]
    [tex]\Gamma^{\varphi}_{tt} = 0[/tex]

    And that means we only have one surviving term in acceleration, [itex]\small a^{\mu}[/itex] = (0, [itex]\small a^r[/itex], 0).

    [tex]a^r = \Gamma^r_{tt}u^tu^t = -\Omega^2r\left(1 - \frac{\Omega^2 r^2}{c^2}\right)^{-1}[/tex]

    Note that so long as Ωr << c, the term in parentheses is ~1. In which case, acceleration is simple centripetal acceleration from classical mechanics. As Ωr goes to c, acceleration becomes infinite.

    There is no dilemma. Special Relativity formulas are correct in Earth's frame of reference. The only reason you have "disagreement" is because you tried to apply SR formulas to satellite. They won't hold. The simple way way to figure out what happens to satellite is to simply consider it from Earth's perspective. You can use SR equations to describe accelerated objects, so long as your frame of reference isn't accelerating.

    If you compute satellite's time dilation and acceleration from Earth's perspective, you'll get exactly the same result as all that mess above gives you. That mess resolves the dilemma, but you already were able to compute the results using Special Relativity.
     
  10. Dec 16, 2012 #9
    All right. So let's assume it is not Earth but some extremely dense planet that a nuclear bomb can just orbit 'freely' around at speed close to c. Then what would happen?
     
  11. Dec 16, 2012 #10

    Nugatory

    User Avatar

    Staff: Mentor

    Can we start with a simpler case? Instead of having the bomb in orbit around some super-dense planet (in fact, to get an orbital speed near c, we'd be talking black hole here) so that we have to mess with gravitational effects and general relativity... Instead assume that the bomb is flying past the earth on a straight line at high speed.

    Now, as far the bomb and anyone moving with the bomb is concerned, they are rest while the earth is moving past them at a high speed. So the bomb explodes just as they'd expect; this planet rushing by outside the window doesn't change anything. All the energy of the bomb is released in a few milliseconds, and that's an explosion.

    However, we on Earth will see the energy released over a longer time, because of time dilation, and spread out over a longer distance, because the bomb is moving during the milliseconds that it's releasing that energy. That's still an explosion by any reasonable definition, just spread out a bit. You still don't want to be anywhere near it.

    You might also want to consider just how great the energies we're dealing with here. Bad guy doesn't have to detonate the bomb at all, he just has to aim it to hit the earth instead of flying by. Figure the bomb weighs about 100 kg... Well, 100 kg of feathers hitting the earth at near the speed of light will cause an explosion so enormous that any nuclear blast would look like a rounding error.
     
  12. Dec 16, 2012 #11

    Dale

    Staff: Mentor

    Hi K^2, SR can handle acceleration just fine, and even accelerating reference frames. You can even use tensors and all of the nice tensor features in SR. (Tensors are a part of math and are not "owned" by general relativity).

    The dividing line between SR and GR is the use of curved spacetime via the EFE. If there is no gravitation then it is SR, even if there is acceleration.

    See: http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html
     
  13. Dec 16, 2012 #12

    K^2

    User Avatar
    Science Advisor

    Then time-dilation due to gravity is no longer negligible, and the whole thing becomes a lot more complicated.
     
  14. Dec 16, 2012 #13

    Dale

    Staff: Mentor

    Basically, it depends how the detonator is constructed. The bomb will see some significant Doppler shifts in the signal. If the detonator is designed to expect a very specific frequency then it will not receive the proper detonation signal and therefore will not detonate at all. However, if the detonator is designed to compensate for the Doppler shift then it will receive the signal and detonate. It would, as you mention, take some time to detonate, but it would still detonate.

    Btw, it will take much more than a nuclear explosion in orbit to destroy the earth. Especially since relativistic beaming will focus most of the energy tangentially to the earth.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: High-speed bomb dilemma?
  1. Speed of light dilemma (Replies: 8)

Loading...