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Relativistic Star Brightness

  1. Jun 18, 2008 #1

    Say we had two identical stars one light year away from us, but one is stationary and the other is receding from us at 0.8c. What would be the apparent brightness of the receding star relative to the stationary star?

    I think at least two effects would have to taken into account.

    1) Time dilation would effectively reduce the power output of the receding star making it appear fainter.

    2) The light from the moving star would be beamed forward due to aberration so it would appear fainter from the rear.

    I think factor (1) would be a straight forward application of the time dilation gamma factor. Factor (2) might be a little more involved. Would anyone care to hazard a guess at the equation for the combined effect and are there any other effects that might be needed to be taken into account?

    Please assume insignificant gravitational effects, as I am only looking for SR effects at this point :)
  2. jcsd
  3. Jun 18, 2008 #2


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    Hi kev! :smile:

    Hint: first work out everything in the frame of the moving star, then Lorentz-transform it into the Earth frame.

    Imagine the light is being collected by a receiver of a fixed area, and that a stationary star emits a photon every k seconds into that area.

    Work out what is the larger area into which those same photons go, using x' = etc, y' = y. :smile:
  4. Jun 18, 2008 #3
    In my unwashed opinion, 1 and 2 are basically the same thing: doppler shift.

    Light from an aproacing star would be blue-shifted, light from a receding star would be red shifted - use E=hf as a measure of relative intensity in either direction.


  5. Jun 18, 2008 #4
    Hi Tim! Hi Bill!

    I calculated that if the observer is moving towards the source at v he would be closer to the source by a ratio of (c+v)/c and when that is squared the apparent brightness would be increased by (c+v)^2/c^2 because of the inverse square power law. In order that when the source is comoving with the observer the source would have to be diminished by the inverse of that factor to give 1/(1+v/c)^2 in order that the comoving observer sees the normal brightness in the rest frame of the light. When that is divided by the gamma factor to allow for time dilation I get:

    [tex]B' = \frac{B}{(1+v/c)} \frac{\sqrt{1-v/c}}{\sqrt{1+v/c}} [/tex]

    where B' is the transformed brightness or luminosity (Same thing?)

    Relativistic redshift (z) is given by:

    [tex]z= \frac{\sqrt{1+v/c}}{\sqrt{1-v/c}} -1[/tex]

    so the luminosity factor can be expressed as:

    [tex]B' = \frac{B}{(1+v/c)(z+1)}[/tex]

    Does that agree with what you get?
  6. Jun 18, 2008 #5


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    Hi kev!
    Sorry, but that makes no sense. :frown:
  7. Jun 18, 2008 #6
    Shouldn't there be a +/- and a -/+ in the relativistic doppler formula?


  8. Jun 18, 2008 #7
    Yes, but I am considering the specific case of a star going away from the observer to avoid confusion with signs ;)
  9. Jun 18, 2008 #8
    Hi Tim,

    yeah, it is kinda funny way to put it. I was considering what happens if a brief burst of light was emitted. The point where the light is emitted is stationary and the moving observer is moving towards the point where the light was emitted (according to the non moving observer).

    Alternatively you can consider how the brightness increases when an observer moves towards a stationary star. That is relativistically the same as a star moving towards a stationary observer. It would appear brighter whichever way you look at it.

    If you can figure out how much brighter a star appears when the reciever is moving towards the stationary star, then the star has to be that much dimmer when it is moving away to cancel out the excess "brightness" and make everything look normal when the emmiter and receiver are comoving. Anyway, the proof of the pudding is whether the equation is correct or not?

    A bit more detail of the calculation.

    Say the moving receiver is a distance L from the stationary emitter. In the time (t) that it take light to travel towards the receiver would have moved forward a distance of (vt). So we can say the reciever is distance (ct) from the emitter when the light hits it. ct=L-vt so t=L/(v+c). The ratio of the brightness when the receiver is distance L from the source compared to when the receiver is a distance ct from the source is (L/ct)^2. Substituting the value for t obtained above gives (v+c)^2/c^2.
    Last edited: Jun 18, 2008
  10. Jun 18, 2008 #9


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    Here's another approach: consider that the light coming from a distant monochromatic source is effectively an electromagnetic plane wave. The intensity (power per unit area) is proportional to the square of the amplitude (maximum value) of the electric field.

    Suppose the wave is traveling in the z-direction, with the electric field oscillating along the x-direction and the magnetic field oscillating along the y-direction. Then the amplitudes of the fields are [itex]{\vec E}_0 = E_0 \hat x[/itex] and [itex]{\vec B}_0 = B_0 \hat y[/itex].

    Transform the wave to another frame moving at constant velocity v along the z-axis, so that [itex]\vec v = v \hat z[/itex]. Use the Lorentz transformation for E and B fields given near the beginning of this page. The component of [itex]\vec E[/itex] parallel to [itex]\vec v[/itex] is zero, so all we need is the perpendicular component:

    [tex]{\vec E}_0^{\prime} = \gamma ({\vec E}_0 + \vec v \times {\vec B}_0)[/tex]

    [tex]{\vec E}_0^{\prime} = \gamma (E_0 \hat x + (v \hat z) \times (B_0 \hat y))[/tex]

    [tex]{\vec E}_0^{\prime} = \gamma (E_0 \hat x + v B_0 (\hat z \times \hat y))[/tex]

    [tex]{\vec E}_0^{\prime} = \gamma (E_0 \hat x - v B_0 \hat x)[/tex]

    For an electromagnetic wave, [itex]B_0 = E_0 / c[/itex] so this reduces to

    [tex]{\vec E}_0^{\prime} = \gamma (1 - v/c) E_0 \hat x[/tex]

    [tex]{\vec E}_0^{\prime} = \gamma (1 - v/c) {\vec E}_0[/tex]

    The intensity (power per unit area perpendicular to the wave direction) is proportional to the square of [itex]E_0[/itex], so

    [tex]I^{\prime} = \gamma^2 (1 - v/c)^2 I[/tex]

    [tex]I^{\prime} = \frac {(1 - v/c)^2} {(1 - v/c)(1+v/c)} I[/tex]

    [tex]I^{\prime} = \frac {1 - v/c} {1+v/c} I[/tex]

    Hmm, this looks like the square of the relativistic Doppler-shift formula!
    Last edited: Jun 18, 2008
  11. Jun 18, 2008 #10
    I note that if you keep track of the sign of v with respect to [itex]\hat{z}[/itex], you get both forms of the relativistic doppler shift (squared).


    Last edited: Jun 18, 2008
  12. Jun 18, 2008 #11


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    According tp PJE Peebles, and the surface brightness theorem which may be applicable here, the intensity ( at certain frequency) is

    [tex]i(\nu)_{obs} = i(\nu)_{emit}(1+z)^{-3}[/tex]

    and over all frequencies,

    [tex]i_{obs} = i_{emit}(1+z)^{-4}[/tex].

    See page 253 of 'Principles of Physical Cosmology' ( 1986).
  13. Jun 18, 2008 #12
    Out of curiosity, does that definition of intensity mention anything resembling "energy received over a finite/fixed time interval"? If not, I cannot see how those relations might follow from what jtbell worked out.


  14. Jun 18, 2008 #13
    Hi Mentz!

    Thanks for the input :)

    This is a lot like the result obtained by jtbell in post #9 which equates to:

    [tex]i(\nu)_{obs} = i(\nu)_{emit}(1+z)^{-2}[/tex]

    I have seen a lot of references to a similar equation stated as:

    [tex]i(\nu)_{obs} = i(\nu)_{emit}(1+z)^{-(2-S)}[/tex]

    where (s) is the spectral index which basically describes how the spectrum of wavelengths coming from the star are distributed. THe whole issue can be comlpicated by factors such as the an uneven spectrum of light in the star and how the human eye registers the relative brightness of different frequencies. Obviously with significant redshift a given frequency can shift right out the visible spectrum.

    I want to keep the whole thing simple and see if the relatavistic luminosity/intensity can be derived in a simple SR thought experiment considering only monochromatic light and considering only the total energy received per unit time per unit area by the observer and considering a source that only moves parallel to the line of sight.
    Last edited: Jun 18, 2008
  15. Jun 18, 2008 #14


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    I give Peeble's derivation in full.

    The phase-space relationship for the number of photons in a volume element is

    [tex]dN = f(x,p,t)d^3pd^3x[/tex] where f is a scalar.

    The energy flux differential through an area A, solid angle Omega, time t and frequency \nu is

    [tex]\delta u = i(\nu)\delta\nu\delta A\delta\Omega\delta t[/tex] and i(\nu) is the brightness at \nu.

    The first equation in Minkowski space and applied to photons says that the photon number flux density per unit interval of p and \omega is [tex]p^2f[/tex] and since the photon energy is p, we get

    [tex]i(p) = p^3f[/tex]

    The observed energy [itex]p_{obs}[/itex] is related to the emitted energy by

    [tex]p_{obs} = p_{emit}(1+z)^{-1}[/tex]

    and since f is unchanged,

    [tex]i_{obs}(p) = p^3f = i_{emit}(p)(1 + z)^{-3}[/tex] for band p.

    Over all frequencies this gives

    [tex]i_{obs} = i_{emit}(1 + z)^{-4}[/tex]

    [edit] crossed posting with Kev.
  16. Jun 18, 2008 #15
    I missed one obvious factor:

    3) The frequency would be classically doppler shifted reducing the effective energy recieved per unit time.

    (1) and (3) taken together amounts to the relativistic doppler factor which is what I think Bill was getting at.
  17. Jun 18, 2008 #16
    volume?? Maxwell's equations say nothing about volume - I think we've identified the [fixed] time inverval I suspected.



    [BTW - Thanks for posting that Mentz. I'll see if I can digest it :smile:]
    Last edited: Jun 18, 2008
  18. Jun 18, 2008 #17


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    I think JTBell's expression fits the case.

    the volume element is in phase ( position-momentum) space.

    I don't follow this very well. I think the same answer is got by boosting the differentials dt and d\nu in

    [tex]\delta u = i(\nu)\delta\nu\delta A\delta\Omega\delta t[/tex].
  19. Jun 18, 2008 #18
    One thing that strikes me as odd about that relation:

    Is it not true that:

    [tex]\frac{\delta A}{4\pi r^2}=\frac{\delta\Omega}{4\pi}[/tex]


  20. Jun 19, 2008 #19
    OK, I think I have it now :)

    Derivation 1

    Consider the increased brightness perceived by the receiver moving towards the point where the light is emitted then assume the power output of the moving emitter is diminished by the inverse of that factor so that the normal brightness is perceived in the comoving frame.

    When the pulse of light is emitted the moving receiver is already closer due to length contraction by a factor of gamma (y) and due to the motion of the receiver towards the emitter during the photon light travel time (1+v/c).

    Since brightness is proportional to the inverse of the distance squared, the increase in brightness at the receiver due to reduced distance is (1+v/c)^2/(1-v^2/c^2)

    The brightness is increased by relativistic doppler shift ((1+v)/(1-v))^(0.5)

    Together the total increase in light at the receiver is (1+v)^(1.5)/(1-v)^(1.5) = (z+1)^3

    The reduced brighntness at the emitter must therefore be (z+1)^(-3)

    Derivation 2

    The relativistic aberration equation which describes how the light of a moving source is focussed is given as

    [tex] \theta ' = \arctan \left({sin(\theta)\sqrt{1-v^2/c^2} \over cos(\theta)+v/c }\right)[/tex]

    where theta is the solid angle of light emitted from the stationary source and theta' is the the wider angle that the light "fans out" to when the source is moving away from the observer. The reduced intensity due to the wider angle of emmision by aberration when the source is moving away is proportional to the [itex] (\theta/ \theta ')^2[/itex] and the intensity is further reduced by the relatavistic doppler factor to give a total reduction in brightness of a receding source as

    [tex] \frac{B '}{B} = \theta^2 \arctan \left({sin(\theta)\sqrt{1-v^2/c^2} \over (cos(\theta)+v/c) }\right)^{(-2)} \left(\frac{1-v}{1+v} \right)^{(0.5)}[/tex]

    Although it looks nothing like derivation (1) it plots the same curve as (z+1)^(-0.3) and is very insensitive to the angle theta.
  21. Jun 19, 2008 #20


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    I'm convinced.

    you raised a point about about the differentials [tex]\delta A \delta\Omega [/tex].

    We have the energy flux travelling normal to the surface element. If the surface element is curved, the flux fans out by [itex]\delta\Omega[/itex]. I think it is the aberration that Kev speaks of.

    Last edited: Jun 19, 2008
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