How Does Relativistic Motion Affect Star Brightness?

[yuiop]

Hi,

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 :)

 

[tiny-tim]

Hi kev!

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.

 

[Antenna Guy]

Hi,

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?

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.

Regards,

Bill

 

[yuiop]

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:

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

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

Relativistic redshift (z) is given by:

$$z= \frac{\sqrt{1+v/c}}{\sqrt{1-v/c}} -1$$

so the luminosity factor can be expressed as:

$$B' = \frac{B}{(1+v/c)(z+1)}$$

Does that agree with what you get?

 

[tiny-tim]

Hi kev!

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

Sorry, but that makes no sense.

 

[Antenna Guy]

Does that agree with what you get?

Shouldn't there be a +/- and a -/+ in the relativistic doppler formula?

Regards,

Bill

 

[yuiop]

Shouldn't there be a +/- and a -/+ in the relativistic doppler formula?

Regards,

Bill

Yes, but I am considering the specific case of a star going away from the observer to avoid confusion with signs ;)

 

[yuiop]

Hi kev!


Sorry, but that makes no sense.

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.

 

[jtbell]

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 ${\vec E}_0 = E_0 \hat x$ and ${\vec B}_0 = B_0 \hat y$.

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

$${\vec E}_0^{\prime} = \gamma ({\vec E}_0 + \vec v \times {\vec B}_0)$$

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

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

$${\vec E}_0^{\prime} = \gamma (E_0 \hat x - v B_0 \hat x)$$

For an electromagnetic wave, $B_0 = E_0 / c$ so this reduces to

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

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

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

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

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

$$I^{\prime} = \frac {1 - v/c} {1+v/c} I$$

Hmm, this looks like the square of the relativistic Doppler-shift formula!

 

[Antenna Guy]

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

$$I^{\prime} = \frac {1 - v/c} {1+v/c} I$$

I note that if you keep track of the sign of v with respect to $\hat{z}$, you get both forms of the relativistic doppler shift (squared).

Regards,

Bill

 

[Mentz114]

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

$$i(\nu)_{obs} = i(\nu)_{emit}(1+z)^{-3}$$

and over all frequencies,

$$i_{obs} = i_{emit}(1+z)^{-4}$$.

See page 253 of 'Principles of Physical Cosmology' ( 1986).

 

[Antenna Guy]

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

$$i(\nu)_{obs} = i(\nu)_{emit}(1+z)^{-3}$$

and over all frequencies,

$$i_{obs} = i_{emit}(1+z)^{-4}$$.

See page 253 of 'Principles of Physical Cosmology' ( 1986).

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.

Regards,

Bill

 

[yuiop]

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

$$i(\nu)_{obs} = i(\nu)_{emit}(1+z)^{-3}$$

and over all frequencies,

$$i_{obs} = i_{emit}(1+z)^{-4}$$.

See page 253 of 'Principles of Physical Cosmology' ( 1986).

Hi Mentz!

Thanks for the input :)

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

$$i(\nu)_{obs} = i(\nu)_{emit}(1+z)^{-2}$$

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

$$i(\nu)_{obs} = i(\nu)_{emit}(1+z)^{-(2-S)}$$

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.

 

[Mentz114]

Bill,

Out of curiosity, does that definition of intensity mention anything resembling "energy received over a finite/fixed time interval"?

I give Peeble's derivation in full.

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

$$dN = f(x,p,t)d^3pd^3x$$ where f is a scalar.

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

$$\delta u = i(\nu)\delta\nu\delta A\delta\Omega\delta t$$ 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 $$p^2f$$ and since the photon energy is p, we get

$$i(p) = p^3f$$

The observed energy $p_{obs}$ is related to the emitted energy by

$$p_{obs} = p_{emit}(1+z)^{-1}$$

and since f is unchanged,

$$i_{obs}(p) = p^3f = i_{emit}(p)(1 + z)^{-3}$$ for band p.

Over all frequencies this gives

$$i_{obs} = i_{emit}(1 + z)^{-4}$$

[edit] crossed posting with Kev.

 

[yuiop]

Hi,

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 missed one obvious factor:

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

(1) and (3) taken together amounts to the relativistic doppler factor which is what I think Bill was getting at.

 

[Antenna Guy]

I give Peeble's derivation in full.

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

volume?? Maxwell's equations say nothing about volume - I think we've identified the [fixed] time inverval I suspected.

Regards,

Bill

[BTW - Thanks for posting that Mentz. I'll see if I can digest it ]

 

[Mentz114]

Kev.

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.

I think JTBell's expression fits the case.

Bill,

volume?? Maxwell's equations say nothing about volume

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

$$\delta u = i(\nu)\delta\nu\delta A\delta\Omega\delta t$$.

 

[Antenna Guy]

$$\delta u = i(\nu)\delta\nu\delta A\delta\Omega\delta t$$.

One thing that strikes me as odd about that relation:

Is it not true that:

$$\frac{\delta A}{4\pi r^2}=\frac{\delta\Omega}{4\pi}$$

Regards,

Bill

 

[yuiop]

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

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

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 $(\theta/ \theta ')^2$ and the intensity is further reduced by the relatavistic doppler factor to give a total reduction in brightness of a receding source as

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

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.

 

[Mentz114]

Kev,
I'm convinced.

Bill,
you raised a point about about the differentials $$\delta A \delta\Omega$$.

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

M

 

[tiny-tim]

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).

Hi kev!

Sorry, still wrong.

If the distance is L, then the distance is L, and that's that!

(Unless you use another frame, in which case it's the Lorentz-Fitzgerald contraction, which isn't c/(v+c).)

Decide which frame you're using, and then make all calculations in that frame … no wistfully thinking that the next frame looks greener, and if only I was there …

hmm … I can't believe that the last ten posts have been so complicated for such a simple problem … electromagnetic waves, surface brightness theorem, phase-space, volume elements …​

You actually spotted the right method in your original post:

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

and finally got round to it in Derivation 2 in your last post:

Derivation 2

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

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

Very good!

Now, all you need do is consider the limit of this as θ approaches zero (because the pupil of the eye is extremely small compared with the radius of the star ) … put cosθ = 1, sin θ = θ, and your B'/B becomes … ?

EDIT: btw, did you copy that θ' formula from somewhere, or did you work it out from the Lorentz equations? if the former, then try the latter.

 

[yuiop]

Hi kev!

Sorry, still wrong.

If the distance is L, then the distance is L, and that's that!

(Unless you use another frame, in which case it's the Lorentz-Fitzgerald contraction, which isn't c/(v+c).)

Decide which frame you're using, and then make all calculations in that frame … no wistfully thinking that the next frame looks greener, and if only I was there …

Hey! Don't knock it! It works for me .. its much simpler and less obscure than the relativistic aberration method, but I am sure someone could come up with a much cleaner derivation...

Now, all you need do is consider the limit of this as θ approaches zero (because the pupil of the eye is extremely small compared with the radius of the star ) … put cosθ = 1, sin θ = θ, and your B'/B becomes … ?

Yep..it looks like it goes to (1+z)^(-3) when you look at it like that.

EDIT: btw, did you copy that θ' formula from somewhere, or did you work it out from the Lorentz equations? if the former, then try the latter.

Ummm... I have a pretty good idea how to derive that, but I was being lazy and got it from somewhere else I only wanted it as a check on the first derivation...

By the way, I put a small typo in the equation to see if you were paying attention at the back there. You failed the test! hehe... Actually the aberration part of the equation was for a source that is approaching the observer. Here is the corrected and tidied up version.

Derivation 2

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

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

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 ratio $(\theta/ \theta ')^2$ and the intensity is further reduced by the relativistic doppler factor to give a total reduction in brightness of a receding source as

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

Although it looks nothing like derivation (1) it plots the same curve as $1/(Z+1)^3$ to within about 1% for any value of theta.

 

[Mentz114]

Tiny-tim,

hmm … I can't believe that the last ten posts have been so complicated for such a simple problem … electromagnetic waves, surface brightness theorem, phase-space, volume elements …

Hah ! to you, mate. What's complicated about Peebles' derivation ?

Anyhow, we were having fun ...

M

 

[shadowpuppet]

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

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

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 $(\theta/ \theta ')^2$ and the intensity is further reduced by the relatavistic doppler factor to give a total reduction in brightness of a receding source as

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

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.

You are right to think that length contraction would affect brightness, but you can only measure light that is shining directly at you (or that is reflected off of a mirror).

Because objects whose absolute displacement is not directly perpendicular or parallel to the velocity $\dot{x}$ have a radial component of velocity $\dot{r} = \dot{x} - \dot{\theta}$.

So for a sphere of radius $R$ situated at the origin $x^2 + y^2 + z^2 = R$, because of the similarity of right triangles

$\frac{\dot{r_{i \rightarrow j}}}{\dot{x_{i \rightarrow i+{\delta}i}}} = \frac{x_{j}}{R}$

points $j(R,{\theta}_j)$ on the surface will seem to a point-like observer $i(x_i,0,0)$ passing the origin along the $x$ axis as

$r_{i \rightarrow j}(x_j) = \frac{R}{{\gamma}_{\dot{r}_j}} = \sqrt{R^2-{\frac{{\dot{x_i}^2}{x_j}R}{c^2}}}$

because length contraction is dependent on velocity and not displacement, $R$ can be replaced by any point on the line that coincides with both the origin and $j$.

 

[shadowpuppet]

This maps the sphere to a flatter object (pancake) that becomes pinched in the middle as the speed of electromagnetic radiation in vacuo is approached (donut).

 

[tiny-tim]

… here we go … !

Anyhow, we were having fun ...

Hi Mentz!

Yes … physics is fun!

… and it is better to travel hopefully than to arrive … ! ​

 

[yuiop]

I think it is good to be able to work it out by as many methods as possible, and they should all come to the same answer. Any differences are educational. For example I am sure that jtbells method is basically correct but it seems to be missing a factor of 1/(1+z) and it would be interesting to know what that missing factor is.


Oh ... and we have not seen Tiny Tim's derivation yet ...hint, hint


...and thanks to ShadowPuppet for the additional contributions

 

[shadowpuppet]

...and thanks to ShadowPuppet for the additional contributions

You're welcome.

 

[tiny-tim]

Oh ... and we have not seen Tiny Tim's derivation yet ...hint, hint

Hi kev!

erm … mine (see post #2) was the same as your "Derivation 2" (with the correct sign)!

 

[Antenna Guy]

Bill,
you raised a point about about the differentials $$\delta A \delta\Omega$$.

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

Treating A and $\Omega$ as separable variables strikes me as something less than correct in this application.

I read "the flux fanning out" as analogous to the power density falling off with 1/r^2 (which does not seem applicable here). I have seen (Balanis) radiation intensity defined as watts per unit solid angle (I interpret solid angle as square radians) - I think this is roughly what we have as I in jtbell's derivation of the instantaneous (relativistic?) poynting vector (since there is no 1/r^2 factor to attenuate power per unit area). However, since the instantaneous poynting vector still has units of Watts/m^2, it seems plausible that a factor of $\frac{\lambda^2}{4\pi^2}$ could be applied to convert to watts/radian^2 since $\lambda$ has the units of distance per wavelength (wavelength being defined as $2\pi$ radians worth of waveform).

But what to use for frequency when you have (arbitrary) doppler effects to consider...

IMO, if we were to construct a spherical surface of radius ct within the ineterial frame of the moving star, the power density integrated over that surface should equal the power radiated by the star at t=0. This integrated power should hold for the same surface transformed into any other inertial reference frame.

Regards,

Bill

 

[yuiop]

...

IMO, if we were to construct a spherical surface of radius ct within the ineterial frame of the moving star, the power density integrated over that surface should equal the power radiated by the star at t=0. This integrated power should hold for the same surface transformed into any other inertial reference frame.

Regards,

Bill

The attenuation of power radiated per unit area of the star only occurs at the rear of the star as seen by an observer who sees the star as receding from him. This is compensated by the amplification of the power radiated per unit area at the front of the star, as seen by an observer that sees the star as aproaching him. However, the total power radiated per unit time by the star is attenuated by time dilation and the stars burns for longer according to obserers that see the star moving relative to them. The focusing and concentrating of power at the front is seen in blazars that eject double jets of luminous material at relativistic velocities with one jet coming towards us. The time dilation effect is seen in 1a type supernovae. They typically shine brightly for one week, but at high recession velocities they shine brightly for two weeks or more.

After working out the power radiated per unit time by the star by allowing for time dilation, the radiated energy is simply redistributed and more concentrated at the front and more diluted at the rear by relativistic aberration but the total radiated power per unit time remains the same.

 

[Mentz114]

Bill,
regarding your last post, I can only say that Peebles is working in the most general background, where gravity can cause lensing, in which case $\Omega$ and A are independent. In the purely SR background they can probably be elided into one differential.

Unfortunately I haven't had time to study this thread very closely - just dropping in and out.

M

 

[Antenna Guy]

Bill,
regarding your last post, I can only say that Peebles is working in the most general background, where gravity can cause lensing, in which case $\Omega$ and A are independent. In the purely SR background they can probably be elided into one differential.

Unfortunately I haven't had time to study this thread very closely - just dropping in and out.

M

If you find the time to think about it: If $\Omega$ and A are separable, what two independent surfaces might they relate to?

Regards,

Bill

P.S. I'll be off-line for ~ a week, so there's no rush...