12
$\begingroup$

When doing some research for some of the resent questions proposed, I ran across an article that pointed out a danger of traveling at extreme speeds in normal space as a high percentage of c I was unaware of. (besides just running into a few grains of sand...)

The first problem after creating some shielding to protect against projectiles, appears to be to protect against ionizing radiation. One thing I never thought of was the blue shifting issue when traveling at near light speeds. So at what speed would say blue light shift to x-ray? What speed would near infra-red shift to x-ray? and what might be an effective shielding against this much larger amount of damaging radiation? Obviously you only need it on the 'front' of the ship.

On top of that it appears that the faster one goes getting closer to the speed of light, light itself starts to push back on the object, like air-resistance on a super-sonic jet. (I just found that fascinating and wanted to share it in this question!)

@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.

$\endgroup$
  • $\begingroup$ how fast does one have to go before the 'light pushing back' effect is noticeable to the degree air-resistance does? That's really interesting $\endgroup$ – Twelfth Nov 17 '14 at 23:07
  • $\begingroup$ the calculations they were using were 99.99995 percent c, added an excerpt $\endgroup$ – bowlturner Nov 18 '14 at 0:52
  • $\begingroup$ this is a cool question that I am unfortunately in no way qualified to answer...or fortunately since I was never much for physics. $\endgroup$ – James Nov 18 '14 at 15:23
  • $\begingroup$ @bowlturner - is this another one of the relations where the force exerted back on the ship approaches infinity as speed approaches c? Perhaps another reason why speed of light travel is impossible beyond mass approaching infinity. Or can we actually peg a value of resistance at the speed of light? $\endgroup$ – Twelfth Nov 18 '14 at 19:27
  • $\begingroup$ My understanding is that the force approaches infinity as you approach the speed of light. Maybe it's like the sound barrier? once you figure out how to break through (or go around it) the sky's the limit? $\endgroup$ – bowlturner Nov 18 '14 at 19:31
9
$\begingroup$

TL;DR - MATH INCOMING: Approximately 99.9996% the speed of light.

Let's take a trip in a spaceship, just to get away from all that silly air friction and other inconvenient matter. Assuming a frequency of light from our sun out of atmosphere preferably, only caring about the visible spectrum, assuming 500nm wavelength for simplicity:

By the Doppler effect, the frequency of light/sound observed is a ratio of speeds multiplied by the actual frequency. Note that the equations differ when considering the high speed of light. We use the Relativistic Doppler effect for this.

$f_o$ = Frequency observed
$\lambda_o$ = Wavelength observed
$f_s$ = Source frequency
$\lambda_s$ = Source wavelength
$c$ = Speed of light, ~$3.0 \cdot 10^8 m/s$ in vacuum or air.
$v_r$ = The velocity of the receiver (our imaginary spaceship) with respect to the source. It can also be the source with respect to the receiver.

Note that the frequency of an electromagnetic wave is $f = \frac{c}{\lambda}$. Thus we have $f_o = 6 \cdot 10^{14} Hz$

We have the following equation for the Doppler effect

$$f_s = f_o \cdot \sqrt{\frac{1+\beta}{1-\beta}}$$

Where $\beta = \cfrac{v_r}{c}$

Let's assume X-Rays (which range from 0.01 nm to 10 nm) are going to be 1 nm in wavelength for us, a frequency of $3 \cdot 10^{17} Hz$.

After some rearranging we find the following:

$$-v_r = \frac{({f_s}/{f_o})^2 - 1}{({f_s}/{f_o})^2 + 1} \cdot c ~ = \frac{({\lambda_o}/{\lambda_s})^2 - 1}{({\lambda_o}/{\lambda_s})^2 + 1} \cdot c$$

A negative speed (as we have here) indicates that the source and observer are approaching one another. In our hypothetical scenario, we have $\frac{f_s}{f_o} = 500$ The answer to this gives $0.999996 \cdot c$, quite fast for our imaginary spaceship. If we can move that fast, I would hope to have reasonable radiation shielding.

However...

A star can emit more than just ordinary light, and space is full of more potent parts of the EM spectrum. What if you were flying near a star emitting high-intensity X-Rays? Or a source gamma rays? Stars can also eject high energy particles, which could cause some problems if they hit you.

You also mentioned that light eventually "pushes back." I'm not 100% sure if this will be the correct interpretation of that, but I'll try my best. What happens when we approach the speed of light? Why can't we reach it? According to the theory of relativity, as we approach the speed of light, our mass increases more and more, until eventually we hit a point (at $v = c$) where our mass becomes infinite. The problem with this is that our speed is maintained (and increased) with energy, and energy requirements increase with both mass and speed ($E = \frac{1}{2}mv^2$). For more info see this answer on Physics.SE.

Alternatively (and probably an explanation for the effect in the question), as Kaine mentioned in the comments, light does indeed have momentum that will exert a force as you move against it! What's really interesting is that we don't necessarily need to move at high speeds to observe the effects. Solar sailing works this way, and there is the "Light Mill."

$\endgroup$
  • $\begingroup$ Are you sure you did your math right? The article implied they would be shifted into the x-rays while still going slower than the speed of light. $\endgroup$ – bowlturner Nov 17 '14 at 18:34
  • $\begingroup$ I have checked a few times, but I'm not immune to error. :) I'll check again. Do you have a link to that article, by the way? It sounds quite interesting. $\endgroup$ – Crabgor Nov 17 '14 at 18:34
  • $\begingroup$ I'll see if I can find it again $\endgroup$ – bowlturner Nov 17 '14 at 18:35
  • 1
    $\begingroup$ It think the formula might not take into relativistic effects. I think I made the same mistake. en.wikipedia.org/wiki/Relativistic_Doppler_effect $\endgroup$ – kaine Nov 17 '14 at 18:38
  • 1
    $\begingroup$ " while still going slower than the speed of light." 0.999996c is still slower than the speed of light... $\endgroup$ – PlasmaHH Nov 17 '14 at 22:32
7
$\begingroup$

The formula is:

$$f'=f\frac{\sqrt{1+\beta}}{\sqrt{1-\beta}}$$ where c is the speed of light, f is the frequency of the light normally, v is the velocity of the observer, $\beta$ is $\frac v c$ and f' is the freqyency of light observed by the observer.

For blue light (650THz) to xrays (1500000 THz), you would need $\beta=.99999962$ so you would need to be very very close to the speed of light.

$\endgroup$
  • $\begingroup$ Awesome! Thanks! That is a little closer to c than I thought it was going to be. $\endgroup$ – bowlturner Nov 17 '14 at 18:47
  • $\begingroup$ What's the typical thing done when we have two identical answers? Kaine's was first, so I'll delete mine if that's the convention. $\endgroup$ – Crabgor Nov 17 '14 at 18:52
  • 2
    $\begingroup$ @Cragor leave them both up. How the answerer explains it matters. $\endgroup$ – kaine Nov 17 '14 at 18:56
  • $\begingroup$ @bowlturner there is a big gap between xrays and visible light. If you asked about visible to ultraviolet or infrared to visible this would be simpler. Please note, the intensity of the light would increase several times as well so the superbright UV would be a massive threat. $\endgroup$ – kaine Nov 17 '14 at 18:59
  • 5
    $\begingroup$ "first answer gets upvotes" encourages short non-detailed answers....we prefer best answer gets upvotes even if they have the same math. $\endgroup$ – Tim B Nov 17 '14 at 20:05

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.