Yiyou Chen Posted November 10, 2017 Posted November 10, 2017 When you are moving at a large velocity, v, the brightness of the stars changes. Does anyone know the ratio of the brightness of the stars in the primed frame to which in the unprimed frame?
Janus Posted November 10, 2017 Posted November 10, 2017 (edited) The light striking you that is coming from the direction of your relative motion will gain energy as measured by you, while the light coming from the opposite direction will lose energy. This will be expressed as an increased or decrease in frequency but not necessarily an increase in brightness, since light at both ends of the visible spectrum will be shifted into and out of the visible spectrum. A brown dwarf that emits heavily in the infrared might appear to brighten as the infrared is shifted into the visible spectrum for example. A star might appear to dim if the visible light is shifted out of the visible range and it emits less in those frequencies that shift into the visible range. The equation for determining the frequency shift is fo = sqrt(( 1+ Beta)/(1- \Beta)) fs Where fo is the observed frequency, fs is the frequency at the source. Beta = v/c and is positive if the source is approaching the observer Edited November 10, 2017 by Janus
Sensei Posted November 10, 2017 Posted November 10, 2017 You also need to take into consideration change of distance from point light source to observer using inverse-square law. https://en.wikipedia.org/wiki/Inverse-square_law
Yiyou Chen Posted November 10, 2017 Author Posted November 10, 2017 Sensei, could you please elaborate a little bit? Thanks!
Sensei Posted November 10, 2017 Posted November 10, 2017 (edited) Point light source, in this case star, is (typically) emitting photons in the all directions equally. Therefor power per area unit (in frame of reference of star) is given by equation: [math]P = \frac{P_0}{4\pi r^2}[/math] P0 is initial power of the star, P is power in watts per area unit at distance r. Spaceship, observer, is flying toward, or reverse, to star emitting these photons. In one second spaceship will travel some distance, but in the same time P in the above equation will change slightly because r will change (from r0 to r1). [math]P (r_0) = \frac{P_0}{4\pi r_0^2}[/math] [math]P (r_1) = \frac{P_0}{4\pi r_1^2}[/math] Periodical change of brightness can be used to measure distance to some distant star. Because of rotation of the Earth around the Sun, one day of year, Earth is ~ +150 mln km closer, half year later it's ~ 150 mln km further away from distant star. Giving you [math]\Delta r[/math] = ~ 300 mln km difference year-to-year. Edited November 10, 2017 by Sensei
Yiyou Chen Posted November 10, 2017 Author Posted November 10, 2017 (edited) I saw this solution a while ago: dN donates the number of photons in a certain space, f donates the occupation number, g donates the number of spin states, p donates the dimensional momentum, dt, dA, do, dE donate unit time, unit area, unit angle, and unit Energy. dN = f * g * (d3r * d3p) / (2*pi*h)3 Since d3r = c*dt*dA, d3p = p2*dp*do Intensity I = (E * dN) / (dt * dA * do * dE) = (c*f*g*p3)/(2*pi*h)3 Therefore, the change of the brightness of the stars when moving at velocity v is proportional to p3, which is the cubic of the doppler effect factor. Edited November 10, 2017 by Yiyou Chen
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