# How would we see a near-lightspeed object emitting light?

Consider an object travelling near the speed of light relative to us (let it be a spaceship or a star), which is emitting light (consider it monochromatic resulting from a two level electronic transition).

Regarding the different relative time in the object itself (since it's travelling at high speeds) and the time of an observer (say in Earth), what would happen to the number of photons that we would see emitted by that object, i.e. photons/second, as it increases speed ?

I believe that we would see that the object would be emitting less and less photons (per unit time) as its speed increases because the time for radiative recombination, i.e. the time required to produce a photon, would pass slower. Is this true?

If yes: It is possible that we may fail to detect matter when we look to the outer space and think that nothing is there? Relating the fact that no radiation detectable from that point and that the relativistic mass is especially high, could it, or why cannot, this be related to dark matter?

Note: I am NOT referring to change of energy caused by red-shift, I am referring to change of the number of photons per unit time!

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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 '13 at 14:07
But that's just some off-the-cuff thinking. Not really an answer –  Jim Aug 7 '13 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 '13 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 '13 at 14:34
You might be interested in this game called "A Slower Speed of Light": gamelab.mit.edu/games/a-slower-speed-of-light It shows you how would you see the world if you travelled near the speed of light. –  Zammbi Aug 30 '13 at 3:41

Yes, the rate at which we would receive photons from the emitter would slow down, and the photons be redshifted. No, this could not account for Dark Matter in the Universe.

### Rate of photons

If you think about relativistic Doppler shift of photons as a slowing down of frequency due to relativistic time dilation, rather than a change in wavelength, it is quite easy to see that the redshift of the photons and the slower rate at which they will be received by the observer is the exact same effect.

Remember that $$\Delta t = \gamma \Delta \tau, \\ \gamma = \frac{1}{\sqrt{1-v^2/c^2}},$$

with $\tau$ being the time as measured on a clock in the emitter's frame and $t$ the time on a clock in the observer's frame. The graph for $\gamma$ as a function of $v$ looks like this (from Wikipedia):

(The nice thing about $\gamma$ is that it is often just a multiplicative factor, so remembering its shape gives a good intuition for how many things in SR work.)

So, the time between two wave tops in a photon will be $\tau$ in the emitting frame, and $\gamma \tau$ in the observing frame, and looking at the figure we get a good grasp how that time - and thus the redshift of the photon - will develop as $v \rightarrow c$.

But the exact same effect will of course happen to the time span between the emission of two subsequent photons/light pulses, so yes, the frequency with which they arrive will approach zero as $v$ approaches $c$.

### Explanation for Dark Matter?

It is true that particles moving at relativistic speeds would have a larger mass than the same particles at rest relative to us. But we should remember that from symmetry arguments, there would be about as many emitting particles moving towards us as away from us at those speeds. Particles approaching us at relativistic speed will blueshift their photons emitted in our direction (remember that $v$ is negative at motion towards us and thus $\gamma$ gets smaller with higher speed of approaching), and so we would definitely observe if there was a lot of matter moving at relativistic speeds around us. There is of course the cosmologically redshifted photons emitted from distant parts of the Universe that are receding from us due to cosmic expansion, but the redshift is gradual as a function of distance (and we should also remember that SR does not apply on these scales).

Another problem with the Dark Matter scenario is that we can observe the effects of Dark Matter everywhere, even where we still observe lots of ordinary, luminous matter, e.g. in the halos of nearby galaxies. These haloes are making up the bulk ($\sim 90\%$) of the galactic mass, and can definitely not be receding from us at relativistic speed, without taking the luminous matter of the galaxy with it, which we can easily observe that it doesn't. Similarly, if we look at e.g. the Bullet Cluster, we can observe that Dark Matter also on a larger scale is holding together systems of luminous matter which are definitely not receding from us at anything resembling light speed.

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Nice. I was especially surprised with "symmetry arguments". Thanks! –  cinico Jan 25 at 15:13
Thanks for accepting my answer! –  Thriveth Jan 25 at 18:01

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 '13 at 15:22
Yep ...(12 more to go) –  JSQuareD Aug 7 '13 at 15:26
Would this affect other forces besides electromagnetism? Would an object traveling near light speed relative to me exert weakened nuclear, or even gravitational force on me as it passed? –  kbelder Jan 23 at 20:15

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.

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The intensity transforms too, and you're going to have to conserve momentum in the lab frame to see how. –  Jerry Schirmer Aug 29 '13 at 21:00
Let us imagine that we are in a one-dimensional world. There is an object moving away from the observer with velocity $v>0$. This object emits, in its reference frame, $n_1$ photons per second, of frequency $\nu_1$ during a time $t_1$.
In the observation frame, one observes photons $n_0$ photons per second, with frequency $\nu_0$ during a time $t_0$. The Lorentz transformation relates $t_0$ and $t_1$ as $$\frac{t_0}{t_1}=\gamma=\frac{1}{\sqrt{1-\beta^2}}\qquad\left(\text{with}\;\;\beta=\frac vc<1\right).$$ The number of photons is conserved, so $n_1t_1=n_0t_0$ and therefore $$\frac{n_0}{n_1}=\frac{t_1}{t_0}=\frac1\gamma=\sqrt{1-\beta^2}.$$ The number of photons per second decreases.
However for observational purposes, it is not the number of photons per second that matters but the power. The energy of a photon is $\hbar\nu$, so the power emitted by the object is $P_1=\hbar\nu_1n_1$. The power observed in the reference frame is $P_0=\hbar\nu_0n_0$. The red shift formula gives the relation $$\frac{\nu_1}{\nu_0}=1+z=\sqrt{\frac{1+\beta}{1-\beta}}$$ hence $$P_0=\hbar\nu_0n_0=\hbar\frac{\nu_1}{1+z}\frac{n_1}\gamma=\left(1-\beta\right)P_1.$$
So for velocities $v$ approaching $c$, the power will be quite small. This makes the detection of dark matter even more difficult. If there is such dark matter, anyway, its emission power is expected to be very small, the fact that we observe only a fraction $1-\beta$ of this power is not the biggest issue. Dark matter has, by definition, a small emission power which is a stronger obstacle for its observability.