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Let's say you have rain hitting you evenly on all sides (not very realistic, I know). If you were to move forwards at a constant speed, there would be more droplets of rain hitting you per second on your front, since the relative speed of droplets in front of you has increased.

Now, if you were to have photons 'hitting' you evenly on all sides and you move forwards at a constant speed, surely the relative speed of the photons in front of you WON'T increase (since light travels at the same speed to all observers), and therefore photons will still be hitting you evenly on all sides (per second).

However the searchlight effect seems to disagree with my conclusion. What have I done wrong?

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You are right in that the speed of light doesn't change. It is a completely different effect to the rain drop analogy. If you had only light hitting you directly from the front and directly form the back, you would observe the same intensity in the moving frame (only blue/red shifted). But for light coming at you from an angle $\theta_s$ in the rest frame, the angle changes when you move with velocity $v$, the new angle in the moving frame is:

$ \theta_o=arccos(\frac{\cos \theta_s-\frac{v}{c}}{1-\frac{v}{c} \cos \theta_s})$

such that the angle shifts toward the front direction. And if you now had a uniform intensity in the rest frame, in the moving frame more light comes from the (general) forward direction and less from the (general) behind direction. Hope this helps.

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    $\begingroup$ Unless the raindrops are moving at relativistic speeds, in which case what-if.xkcd.com/1 applies $\endgroup$
    – Carl Witthoft
    Jul 13, 2015 at 14:40
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    $\begingroup$ "If you had only light hitting you directly from the front and directly form the back, you would observe the same intensity in the moving frame (only blue/red shifted)" - really? It seems to me that while you'd see the photons moving at the same speed, you'd see a lot more photons coming from the front. I never learned relativistic QM, so my analysis may be flawed. $\endgroup$
    – user2357112
    Jul 14, 2015 at 2:38
  • $\begingroup$ Thanks a lot John! I just found this website which includes what you said in images (as well as some more context) for anyone else interested. $\endgroup$
    – ToRQue
    Jul 14, 2015 at 11:00
  • $\begingroup$ John, I think you are making contradictory statements. In your first statement, you say, "you would observe the same intensity in the moving frame." In your second statement you say, "in the moving frame more light comes from the...forward direction..." Isn't the intensity (I) = number of photons / unit area)? $\endgroup$
    – Guill
    Jul 17, 2015 at 5:01
  • $\begingroup$ @Guill It is rather subtile. Notice that I say "general" forward direction. I mean all the photons that are coming at you with an angle less then $\frac{\pi}{2}$ w.r.t the direct forward direction. It is counter intuitive: A photon that in the rest frame has an angel of $\frac{\pi}{2}$ has an angle of $\theta = \text{arccos}(\frac{v}{c})$ in the moving frame. So for $v$ close to $c$ the light that previously came from your side no practically comes from the front. Does this answer you question or did I miss the point? $\endgroup$
    – john
    Jul 17, 2015 at 7:48
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What happens if you first drive on a road at speed 0.99 c, and then accelerate to speed 0.999 c?

You will pass milestones at much faster rate. That happens mostly because of length contraction, not because of larger speed, as the speed increase was only 0.009 c.

Roughly the same contraction as above will happen to a line of photons approaching you, if you change your speed by 0.009 c towards the photons.

The exact amount of contraction is the same as the contraction of the wave-length of the photons, which can be calculated using the relativistic Doppler-shift formula.

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  • $\begingroup$ @doetoe , why can't both answers be correct? $\endgroup$
    – ToRQue
    Jul 14, 2015 at 10:55
  • $\begingroup$ @ToRQue In this case I think they are contradictory (in the case that the movement is parallel to the direction of the source): I interpreted john's answer as stating that the number of photons doesn't change, only their frequency, while user7027's answer says it does. The part of the angles seems correct to me. Note by the way that while the actual value of the angle has to be computed using the relativistic formula, whether it hits you in the front or the back is the same classically or relativistically: it comes straight from above when $c\cos\theta = v$. $\endgroup$
    – doetoe
    Jul 14, 2015 at 12:07
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    $\begingroup$ @ToRQue Probably it would have been better to say that both provide parts of the answer. First there is the Doppler effect applied to the timings of emission of a photon by the "photon cloud", then there is the fact that photons, like rain drops, that would hit you in the back when you are at rest, hit you in front when you move. If the emission of photons is not periodic, you wouldn't speak of the Doppler effect anymore, but the perceived time between photon emissions would change in a similar way. $\endgroup$
    – doetoe
    Jul 14, 2015 at 12:14
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Let's say Bob is standing still while a one light second long photon formation flies past him. How long does the passing of the photon formation and Bob take according to us?

Answer: It takes one second. 1 light seconds / c = 1 seconds.

Let's say Jim is moving forwards at speed 0.1 while a one light second long photon formation moving to the opposite direction flies past him. How long does the passing of photon formation and Jim take according to us?

Answer: It takes 0.909 seconds. 1 light second / 1.1 c = 0.909 seconds. (Using relativistic velocity addition would be an error)

Because of time dilation according to Bob the passing takes a little bit less than 0.901 seconds.

This is one way to calculate at what rate photons pass a moving observer, or collide with him.

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