Can a spaceship traveling close to light speed be knocked off course by a gamma ray burst?

This was spawned from this question about Blueshifting (BS) when traveling near the speed of light.

@99.99995 percent c

And interestingly, the students also realized that, when traveling at such an intense speed, a ship would be subject to incredible pressure exerted by X-rays — an effect that would push back against the ship, causing it to slow down. The researchers likened the effect to the high pressure exerted against deep-ocean submersibles exploring extreme depths. To deal with this, a spaceship would have to store extra amounts of energy to compensate for this added pressure.

So then I began to wonder, I am assuming the higher the EM energy to begin with the more 'pressure' is exerted back, starting with X-Rays, they'd be BSed to Gamma and they would cause more pressure than visible light pushed back to soft X-Ray.

If that assumption is correct would a star going nova with a gamma ray burst cause enough pressure to throw a ship way off course? And would it be (reasonably) possible to cause a 'burst' large enough to push a ship off course (assuming we already have the tech to go near the speed of light).

On top of that, if you have a gamma ray machine and point it at a ship going at .9999995 light speed would Newton's 3rd law kick in with an equal and opposite reaction? Because if so, then we have an other issue, Mass increases with speed and near the speed of light it would be downright dangerous for anything in its path.

• I assume you assume "If it is possible to use enough radiation energy to do this, the ship will remain intact". No calculations, but I can very well imagine that at the energies required the ship would disintegrate under the radiation bombardement. – Jan Doggen Nov 21 '14 at 15:09
• Possible, but I was guessing that the ship was already built to deal with the pressure, I was more assume that it would be an unexpected 'pressure' from an angle knocking it off course, like a ricochet. – bowlturner Nov 21 '14 at 15:11
• Don't forget your radiation-proofing! I recommend a lead umbrella. – Crabgor Nov 21 '14 at 16:55
• A ship being knocked directly backwards along its line of flight will not be "off course". – Oldcat Nov 21 '14 at 18:47
• I don't know enough about relativistic speeds to post an answer, but what you describe sounds very much like wave drag which plagues sailing vessels and trans-sonic aircraft. It might be worth looking into that – Cort Ammon Nov 22 '14 at 4:23

Let's first do some math, the first part taken from this pdf regarding (solar) radiation pressure (the formulas should be applicable from any source of electromagnetic radiation).

The intensity $I$ depends on the power $P$ and the distance from the source $r$. We can write the expression $$I=\frac{P}{4 \pi r^2}$$

The force $F$ depends on the intensity and the area $A$ of the object being buffeted. It is $$F=\frac{2I}{c}A$$ and, substituting, we get $$F=\frac{2P}{4 \pi r^2c}A$$ which is $$F=\frac{P}{2 \pi r^2c}A$$ Given the mass of the ship $M$, we find that the acceleration $a$ is $$a=\frac{P}{2 M\pi r^2c}A$$

How much power does a gamma-ray burst emit? From Wikipedia:

Because their energy is strongly focused, the gamma rays emitted by most bursts are expected to miss the Earth and never be detected. When a gamma-ray burst is pointed towards Earth, the focusing of its energy along a relatively narrow beam causes the burst to appear much brighter than it would have been were its energy emitted spherically. When this effect is taken into account, typical gamma-ray bursts are observed to have a true energy release of about 10^44 J, or about 1/2000 of a Solar mass energy equivalent—which is still many times the mass energy equivalent of the Earth (about 5.5 × 1041 J).

We have to divide this number by two to account for the fact that only one of the two beams is hitting the ship - and this is still a little inaccurate, as it assumes the ship is hit by the whole beam. Anyway, gamma-ray bursts, on average, can last from anywhere between less than a second to 30 seconds. Let's say ours lasts for 10 seconds. Because the definition of power is $\frac{E}{t}$, where $E$ is work and $t$ is time, we can say that the power here is $$P=\frac{E}{t}$$ $$P=\frac{5 \times 10^{43}}{10}$$ $$P= 5 \times 10^{42} \text{ Watts}$$ Plugging into our earlier equation for acceleration, we get $$a=\frac{\frac{E}{t}}{2 M\pi r^2c}A$$ $$a=\frac{ 5 \times 10^{42}}{2 M\pi r^2c}A$$ Assuming a mass of the ship similar to the Space Shuttle Orbiter (109,000 kilograms) which is admittedly an unlikely comparison, we make this $$a=\frac{5 \times 10^{42}}{2 \times 1.09 \times 10^5 \pi r^2c}A$$ $$a=\frac{5 \times 10^{37}}{2.18 \pi r^2c}A$$ If you want, you can plug in the area of the underside of the Space Shuttle Orbiter (a stat I can't find, at the moment) and discover that the space shuttle would take quite a hit if it was near a gamma-ray burst.

Note that this is only valid for a ship traveling at a slow speed. At near-light speeds, the relativistic mass would increase (although I don't know if this is valid perpendicular to the direction of its initial motion). This would impact your calculations; I'll try to figure out the corrections later.

Belated Conclusion

The answer is a definitive yes. A ship moving at "normal" speed (i.e. something we could make today - think a successor to the space shuttle) would meet some severe buffeting if it was anywhere near a gamma-ray burst and was hit by one of the beams emitted from the progenitor. If it was hit full-on, it would be severely pushed back; if it was hit partially on, it could be sent spinning. Either way, things wouldn't turn out well.

Your ship, though, is a bit more advanced and is traveling at a speed pretty close to $c$. This means that it is highly unlikely that it would get hit with the beam for an extended period of time if it was anywhere close to the source. If it was further out, the cross-section of the beam would be a lot larger, though (eventually, on the order of hundreds of thousands of miles), and the ship could continue to travel through it for the duration of the burst. The downside is that the energy would be greatly dissipated over the beam's cross-section.

But the answer is yes, the sip would be buffeted if it was reasonably close (i.e. about an AU away, though that's an estimate) to the source, and would most likely be impacted in some way if it was further out.

SJuan76 and Oldcat pointed out that Lorentz contraction would impact the area of the side of the ship receiving radiation pressure if the ship was moving tangentially to the beam. At speeds nearing $c$, this phenomenon would have huge implications for the pressure on the ship. This answer is already math-heavy, so I figure adding a few more equation can't hurt. Haters of algebra beware.

The length of an object due to Lorentz contraction can be found by $$L=L_0 \sqrt{1-v^2/c^2}$$ This means that the area (previously the height times the length, $A=H \times L$) is now written as $$A=H \times L_0 \sqrt{1-v^2/c^2}$$ and so the original equation becomes $$a=\frac{5 \times 10^{37}}{2.18 \pi r^2c} \times H \times L_0 \sqrt{1-v^2/c^2}$$

• Also, to note, at relativistic speeds Lorentz contraction would make the (lateral) area smaller – SJuan76 Nov 21 '14 at 21:55
• Re length contraction: objects travelling across your vision at relativistic speeds don't look contracted; they actually appear rotated because the object moves out of the way of light emitted from the back side. Would this work in reverse? Incoming light would not just be diluted due to foreshortening, but would hit the front surface as well as it hit the light crossing in front on the ship's path. – JDługosz Jul 5 '15 at 9:47
• But what about the ship's reference frame? There the gamma ray burst's area is contracted. I think you need to consider en.wikipedia.org/wiki/Ladder_paradox. The expression you are using for intensity is only for isotropic (spherically symmetric) sources. The two in the numerator of your force equation means that the ship is reflecting 100% of the gamma rays (which I guess is reasonable as, if it wasn't, it wouldn't be a ship anymore). Flying within 1 AU of a GRB source at .99c doesn't make very much sense. If you want to continue flying at .99c for very long. It's rather soupy. – alessandro Jul 6 '15 at 5:24

I suppose I should take my comments and make them a response.

First - if a gamma ray burst has no effect on changing the course of a ship at rest, it will not have any greater effect on a ship at near-light speed. So lets assume this burst would not shove a ship at rest at all. What about all the blue-shifting?

The blue shifted extra momentum will all be in a direction that opposes the flight of the near-C ship. When you change frame to the rest frame of the ship, the energy and direction added will be directly opposing the flight path. So if this blue shift does anything, it merely slows down the ship and does not 'knock it off course'.

Second Case - The burst would add a significant kick to an at-rest ship.

When the ship is traveling at near C, the impact of this kick is reduced in a number of ways: first the mass is greatly increased, reducing the sideways velocity resulting. The Lorentz contraction reduces the apparent area of the ship, which might reduce the ability of the storm to push the ship off course.

It would take an amazing impact to still push the ship off course, but it could possibly happen. But the near light speed does not make it worse.

The one thing that would be massively increased is the in-line "braking" momentum of anything that the fast ship ran into in flight. But I wouldn't consider a slowing, or a requirement to fire the engines more to overcome it "knocking the ship off course".

• Relativistic mass isn't mass, it's a (rather cludgy I think) way to think about how momentum's work in special relativity. See: en.wikipedia.org/wiki/…. Note the bit about transverse and longitudinal mass. In reality, as you quite correctly put in the last paragraph of your first case, only the components of the GRB which hit the ship parallel to the axis of travel need to have any relativistic corrections applied to them. The perpendicular kick in your second case would be the same as if the ship was at rest. – alessandro Jul 6 '15 at 5:30