# How would we see an near lightspeed object propulsed by laser?

Please forgive me if I'm not thinking right.

Imagine that we could put an object travelling near the speed of light and that it was powered by some sort of nuclear battery.

Regarding the different relative time in the laser itself (since it's travelling at high speeds) and the time of an observer (say in Earth), would we see that the laser would be emitting less and less photons as its speed increases because the time for radiative recombination of the laser would pass slower?

EDIT: If the answer is yes, is it possible that we are missing some celestial bodies because they are moving away at a near lightspeed comparatively to us?

EDIT2: The main problem here is the time that it takes to emmit a photon

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Excellent question. Just thinking intuitively (I know, not the best thing to do when discussing relativity), I know you should see a lower power output. There's two things to think about. 1) time is dilated, so less power meaning longer wavelength photons. 2) Length is contracted and the acceleration is blue-shifting the laser, so smaller wavelength photons. Since observer must see less power no matter what, there is a high chance this would mean less photons. Which makes sense because from the observer's perspective, the applied force should be decreasing since the acceleration is decreasing –  Jim Aug 7 at 14:07
But that's just some off-the-cuff thinking. Not really an answer –  Jim Aug 7 at 14:08
@Jim Thanks. I can follow your argument, but would that mean that the intensity (amount of photons), and of course red shifting, that we observe from a celestial body depends on the speed that the celestial body is moving? –  cinico Aug 7 at 14:32
OK, so we have 3 effects to consider with respect to the receding spacecraft. Doppler shift due to the movement of the spacecraft away from the observer, which should cause the light emitted by the laser to be red-shifted (regardless of relativitistic effects), time dilation aboard the spacecraft which should also contribute to the red shift, and contraction which should cause the light to be blue shifted. My common-sense thinking is that the contraction and time dilation effects on the laser will cancel each other out, so the Earth bound observer will only see red shift due to Doppler effect. –  Anthony X Aug 7 at 14:34
@AnthonyX but the doppler effect won't change the number of photons, just the wavelengths. So approaching or receding, the number is the same. But since the power output of the laser must be less from the observer's point of view, in the case of approaching, the only way to reduce it is to reduce the photon flux –  Jim Aug 7 at 14:45

Although photons are weird, not having a fixed wavelength, but somehow a fixed speed, their existence is not relative. Every frame of reference should agree about the existence of a photon. Therefore we can calculate the number of photons transmitted simply by using time dilation.

If the laser emits $N$ photons in $t$ seconds according to his own clock, the observer will also see $N$ photons being emitted, but in $t'=\gamma t$ seconds. Therefore the number of photons emitted per second is decreased by a factor $\gamma$.

In addition to this, the emitted light will be less energetic to the observer because of the relativistic Doppler effect ($f_{obs}=\sqrt{\frac{1-v/c}{1+v/c}}f_{src}$). Therefore, the observed energy output will drop quickly.

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So basically yes, we would see less photons coming out from the laser. And, as the object approaches speed of light, it will become darker and darker. Right? –  cinico Aug 7 at 15:22
Yep ...(12 more to go) –  JSQuareD Aug 7 at 15:26
If a photon source moves away from an observer at speed $v$ while sending out every second a photon of frequency $f$, at what rate would the observer see the photons coming in and what wavelength would they have?
First, forget about relativistic length contractions mentioned in some of the comments (the photon source being seen as shortened is not relevant) all that matters is the relativistic Doppler effect that answers both questions. The Doppler factor $\sqrt{\frac{c+v}{c-v}}$ gives the factor by which the photon wavelengths are observed to be increased, as well as the factor by which the time between subsequent observed photons is increased.
So if the photon source moves away from the observer at a speed $v=\frac{3}{5}c$, it follows that the observed wavelength is double the photon source wavelength, and the time between photons is also double the time interval as seen from the photon source. In other words, both the observed cycle time of the photons as well as the observed time interval between photons is double that as observed from the reference frame associated with the photon source.