4
$\begingroup$

Suppose a rocket carrying a nuclear weapon is moving past our planet at very near the speed of light while a stationary observer watches from here on Earth. As the rocket passes directly overhead, an on-board mechanism detonates the weapon. What would the ensuing explosion look like to the stationary observer?

It's my guess that the chain reactions involved in nuclear fission would take longer to play out from the Earth-perspective so the weapon then releases its energy much slower than normal, and over a long distance in the direction the rocket was travelling? If an observer could "see" all energy being released I suppose it would like a long "streak" of an explosion but rather weak at each point?

EDIT 4/29/15: In userLTK's answer below he/she simplifies the situation to a single Uranium atom undergoing fission directly above the observer. Keeping with this spirit I think that it's far easier to forget the Rocket-Earth setup and just consider a single uranium atom moving along a particle accelerator with sensors that can detect the released fission products. In this, much nicer scenario, how would a fission event at relativistic speeds compare to one that is stationary with respect to the sensors in the particle accelerator? My guess is still basically the same - that the fission proceeds more slowly and that the atom will cover quite a bit of distance in the process. However I also do not know how much "time" nuclear decay processes takes or if it even makes sense to speak of these processes this way. Also, now considering mass-energy equivalence, I'm wondering if the fission event will be more energetic than normal in the sense that there is more relativistic mass available to convert to energy. However I'm no expert on special relativity (that's why I'm here) and my conceptualization of relativistic mass may be totally incorrect.

$\endgroup$
  • 2
    $\begingroup$ It would be brief $\endgroup$ – rob Apr 17 '15 at 22:21
3
$\begingroup$

Mass-energy equivalence (Rest energy = rest mass * c^2) can be applied to a moving body:

Total energy = relativistic mass * c^2

From the perspective of a stationary observer, the total energy in a moving body is greater than the rest energy in a stationary body. As velocity approaches c, it is much greater.

The ensuing explosion would be much bigger than if the payload were stationary with respect to the stationary observer. However, in space there would be no double pulse typical of nuclear explosions in the atmosphere. There would be a brief intense light, and then expanding and decaying radiation.

Time dilation would cause the explosion to proceed more slowly in its own inertial frame than in the earthbound observer's:

t = t0/(1-v^2/c^2)1/2

where: t = time observed in the other reference frame t0 = time in observers own frame of reference (rest time) v = the speed of the moving object c = the speed of light in a vacuum

The stationary observer would see an intense explosion initiate, ripen, and dissipate very quickly. There would be no streak.

See the Wikipedia articles "Mass-Energy Equivalence". Also see Wikipedia "Relativistic Kill Vehicle" for a science fiction application.

$\endgroup$
1
$\begingroup$

Well, a rocket traveling at close to the speed of light would be very hard to see at all cause it would go from a moon's distance in one direction to a moon's distance in the other direction in a little over 1 second, and seeing a rocket as far away as the moon would be difficult - but I'm thinking that's not what your asking, so lets pretend that we have a perfectly calibrated telescope that follows the rocket so you can watch and follow the explosion.

Assuming that you could watch the rocket approach, fly past and go boom in the process, then, yes, the blast would be slowed down by time dilation. Similar to watching a clock on a space ship flying past the earth, the clock would move slower than normal.

But the clock/rocket ship is a little bit more complicated than it appears. A ship traveling at, lets say, 99% the speed of light but moving towards us (a ship that flies overhead traveling in a straight line, is virtually traveling straight towards us until it gets very close).

Lets say that 2 clocks are synchronized 10 light seconds apart and the rocket is flying towards us at .9C, the rocket will cover those 10 light seconds in 11.11 seconds, but from the rockets perspective, space will be shortened and time dilation of about 2.3, the trip would appear to take 4.8 seconds. Now, consider what you see from earth watching the rocket. It would take 10 seconds to even see the rocket, cause it takes 10 seconds for light to travel 10 light seconds, but then, you'd see the rocket overhead 1.111 seconds later, so the clock would, from an earth perspective, move forward 4.8 seconds in 1.111 seconds, so it would appear to be moving faster, not slower (We're assuming you could see a clock from 10 light seconds away). Then, as it moves away, you would see it slow down.

As to the long streak of a slow explosion. Kind of, you'd see a slow explosion, releasing photons and exploding matter and neutrons and neutrinos happen much more slowly. The photons would be red or blue shifted depending on the relative velocity. so you would see that, and if the bomb took minutes or hours from your point of view to explode, the energy released would be spread out over that time. The matter and Neutrons would travel at less than the speed of light, so from a ship traveling close to the speed of light, you'd probably not see 1 single atom from the explosion, you'd only see photons bouncing off the atoms. The atoms would move at the combined vector velocity which would, from your point of view, be kind of light a flashlight pointing in the direction of the ship.

It's worth pointing out what what you could actually "see" on a space ship moving past you at near the speed of light is iffy anyway, but in theory, yes, if the ship was traveling fast enough you'd see a time-dilated slowed down explosion.

Edit - I read the answer by Ernie and I think I was wrong about the energy being released being the same. It helps, I think to simplify. Instead of a bomb, and a rocket, lets call it a single uranium atom flying past at 99.9999% of the speed of light and it undergoes fission right overhead.

at 99.9999% of the speed of light time dilation would be 707 times. So it would appear to split 707 times as slowly, but the mass of the Uranium would be 707 times greater than a non moving uranium atom, so the rest mass difference between the Uranium and the parts it splits into would be 707 times as great - so, Ernie is right. You wouldn't see a low energy long lasting explosion because the energy would increase in proportion with the time dilation. As to the rest - observing the particle flight path (mostly in the direction of the ship) and the red/blue shift of the photons, those parts I stand by. (sorry for the wordiness, it's how I write)

$\endgroup$
  • $\begingroup$ Your simplification to a single atom is probably how I should have asked this in the first place, but I'm confused when you say: "the rest mass difference between the Uranium and the parts it splits into would be 707 times as great". If I can tweak the scenario to ease things, say the Uranium is moving along a particle accelerator at 99.9999c, and say we just have sensors along the length of the accelerator that will detect any products released by fission. Is it true the sensors would detect fission products which are (in total) 707 times more energetic than in a stationary fission event? $\endgroup$ – Eric Emmons Apr 19 '15 at 2:12
  • $\begingroup$ Hmmm. I hate to say this, but I'll have to think about this. I'm not sure now. $\endgroup$ – userLTK Apr 19 '15 at 2:43
  • $\begingroup$ @userLTK Is this increased mass is the same mass that is needed for nuclear fission? $\endgroup$ – m8labs Jul 12 '18 at 18:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.