Is the Sun invisible at relativistic speeds? Part II

[Vanadium 50]

But the sun is a black body. And one of the characteristics of a black body is that the intensity at a fixed frequency (and therefore a fixed frequency band) always increases with increasing temperature.

 

[Russell E]

But the sun is a black body. And one of the characteristics of a black body is that the intensity at a fixed frequency (and therefore a fixed frequency band) always increases with increasing temperature.

Right, but that's the intensity per unit area of the emitter and per unit solid angle, which is decreased by the square of the Doppler factor due to aberration. Thus when you integrate to get the received power this reduces the result. Even in the far field limit, when the light can be treated as a plane wave, this still affects the intensity for a given stellar source.

There are two ways of approaching this problem - the simple way is to analyze it in terms of the frequency density in the rest frame of the Sun, which makes everything very clear and simple, and it's very obvious why the intensity drops. (See, for example, the link given earlier in this thread.) The complicated way is to analyze it in terms of the wavelength density and transform the Sun to a different frame, and then you have to be very careful to take all the relativistic geometrical effects into account, both when writing the transformed spectrum and when integrating it.

 

[pervect]

I still don't understand why the paper has the output turning over at large beta.

Suppose I am going towards a planet, and on that planet is a man with an Aldis Lamp pointing at me. Every second he flashes the light at my position. Eventually, the pulses reach me, and I start seeing them at one per second.

Now I start moving towards him, and therefore the pulses: I maintain that there is no velocity where I see them coming slower than once per second.

The total output always increases. The amount of energy collected by a receiver of constant cross sectional area increases as D^2, one factor of D comes from the pulses coming more rapidly, the other factor of D coming from the fact that the frequency, and hence the wavelength of the pulse is shifted upwards.

However, the question being asked is the amount of energy in the visual band. This requires the computations that are done in the paper. The short answer is that the amount of energy in the visual band will decrease in the case of moving towards a black body source. But it decreases very slowly.

 

[tionis]

The total output always increases. The amount of energy collected by a receiver of constant cross sectional area increases as D^2, one factor of D comes from the pulses coming more rapidly, the other factor of D coming from the fact that the frequency, and hence the wavelength of the pulse is shifted upwards.

^^ Is that the same as saying that both methods agree that in 2+1 dimensions the power would go to a constant, and in 1+1 it would grow?

 

[pervect]

^^ Is that the same as saying that both methods agree that in 2+1 dimensions the power would go to a constant, and in 1+1 it would grow?

This seems to me to be a giant leap from anything I've said.

 

[Russell E]

To achieve a drop of six magnitudes you need a Doppler factor of about 7160, which implies a speed of about 0.999999961.

Another comment: As mentioned before, the expression for the intensity given as equation 16 in the paper of McKinley & Doherty is correct, but they were unable to evaluate the integral directly, so they numerically plotted some points and made a least squares curve fit to those point (fitting points with DT only up to 10^5). As a result, their formulas beginning with 19 are not rigorous, especially for extrapolating far past the curve fit range.

However, we can evaluate their equation 16 directly to give reliable results. Just for fun, I did this, and the results agree nicely with my calculations. The only slight difference is due to a different model for the sensitivities to brightness of the human eye in the visible range.

For lack of any better model of the human eye's sensitivity profile, I've been basing my calculations on uniformly weighting the frequencies within the range perceptible (as light) to the human eye. The paper of McKinley & Doherty uses a normal distribution with a mean of 500 nm and standard deviation of 57 nm to weight the sensitivities of the eye at different wavelengths in the visible range. So their analysis gives more weight to the light around 500 nm, and less weight to the other frequencies in the visible range. If I apply that model of the eye's sensitivity to my calculations, the results agree exactly with McKinley & Doherty's equation 16 (although not with their later "curve fit" formulas, and their Figure 4, which don't agree with their equation 16).

Using this model of the sensitivity of the human eye, we need a Doppler factor of about 8680 to give a drop of 6 magnitudes in the visible brightness, which requires an approach speed of 0.999999973.

 

[tionis]

Russell, you are pretty invested in this thought experiment.. are you going to write a paper? That would be awesome!

 

[Dale]

But the sun is a black body. And one of the characteristics of a black body is that the intensity at a fixed frequency (and therefore a fixed frequency band) always increases with increasing temperature.

This was my first thought also. However, note equation 14 and the comments immediately following "Apart from the factor D^-2, this is exactly the spectrum of a blackbody at the Doppler shifted temperature DT." (emphasis added). That factor out in front changes the outcome. For fixed D, as you increase T you increase the intensety at every frequency, but for fixed T increasing D does not always increase the intensity at every frequency.

 

[PAllen]

This was my first thought also. However, note equation 14 and the comments immediately following "Apart from the factor D^-2, this is exactly the spectrum of a blackbody at the Doppler shifted temperature DT." (emphasis added). That factor out in front changes the outcome. For fixed D, as you increase T you increase the intensety at every frequency, but for fixed T increasing D does not always increase the intensity at every frequency.

Another way to look at it is you have a race between competing effects. D increases effective T, increasing power at every frequency. However, D also shifts the visible range further and further away from peak power. To see how these balance needs the detailed analysis. Of course, total EM power (all frequencies) goes rapidly up, especially gamma rays (at the required speeds), so this will be the last thing you 'see'.

Note that for moving away from the star, these two effects reinforce rather than compete: T goes down, and visible range gets further and further from peak on the high frequency side. That's why fairly modest relativistic speeds away from the sun will cause it to disappear to the naked eye (even, say, as close as Jupiter's orbit). Also, moving away, you don't get fried, and total EM power goes down.

 

[Vanadium 50]

Dale, I think I understand the argument: the sun remains as a BB of increasing T but decreasing r, and at low frequencies these almost balance, with the intensity falling logarithmically.

Now that I understand it, I am not sure I agree with it: since the balance is so close, higher order effects are important. For example, the sun is not at fixed distance; the center is closer than the limbs. Getting this right is quite tricky.

 

[PAllen]

Dale, I think I understand the argument: the sun remains as a BB of increasing T but decreasing r, and at low frequencies these almost balance, with the intensity falling logarithmically.

Now that I understand it, I am not sure I agree with it: since the balance is so close, higher order effects are important. For example, the sun is not at fixed distance; the center is closer than the limbs. Getting this right is quite tricky.

The OP proposed 4 ly away from sun. Hard to see how this matters at that distance.

 

[Dale]

I wonder how bright the CMB is at speeds sufficient to make the sun "invisible".

 

[Vanadium 50]

The OP proposed 4 ly away from sun. Hard to see how this matters at that distance.

It's formally of the same order as the disk size.

 

[pervect]

An important part of the issue is what the detector can resolve.

Wiki puts the angular resolution of the human eye at 1 arcminute. The dark adapted pupil is about 1 cm, the theoretical resolution would be https://en.wikipedia.org/wiki/Dawes'_limit about 10 arceconds. Taking the more conservative estimate, we covert 10 arc seconds to radians and get 5*10^-5 radians.

THis means that optimistically, the human eye might be able to distinguish objects of $r \, \theta$ = where r = (4 light years) and $\theta$ = (5*10^-5). The actual answer is probaby closer to an arcminute that the theoretical maximum , though.

This optimisitc estimate gives ~ 2*10^12 meters of resolution. The radius of the sun is 7*10^8 meters. This is much lower than what the eye is capable of resolving

This justifies mathematically what I assumed earlier without a detailed proof (I rather thought it was obvious) - the human eye cannot resolve the sun at this distance, and that we are more than justified in using the plane wave approximation. The paper explicitly mentions they use the plane wave approximation.

To use the plane wave approximation simply need to normalize the radiated energy in terms of meter^2, rather than the "solid angle". We have to do this because our detector cannot resolve said angle - we can't use geometric optics, we need to take the diffraction limit into account.

Thus we multiply the power emitted / unit area of the black body, by the surface area of the sun (to get the total power), multiply it by (1 meter^2) [if we are using MKS units] area and divide by the area of a sphere of radius of 4 light years to get the power / meter^2

This gives us the intensity in power / meter^2.

(as per http://en.wikipedia.org/wiki/Intensity_(physics))

AFTER we have done the above, we can , if we like, convert this to a classical EM wave of amplitude E in a vacuum (see the above wiki)

I = \frac{c \epsilon_o}{2} \left| E^2 \right|

The approach that Einstein used to transform this, and probably the least mathematically demanding one is to look at how E transforms. Because E&M radition is transverse, we need to look only at how the transverse component of E transforms.

The answer can be looked up in a classic E&M textbook or on Wiki (but probably not randomly guessed at):

${{E}_{\bot}}'= \gamma \left( \mathbf {E}_{\bot} + \mathbf{ v} \times \mathbf {B} \right)$

We can see right away that E DOES depend on our velocity. It's mistake to assume that E doesn't transform - E changes with our velocity in a manner described by the above equation.

The magnetic field B = (1/c)E (I believe, I took the trouble of working this out, but I don't have a reference) of the plane wave will contribute to our electric field in the new frame. if we work out the details one should get the standard result that E transforms as $\gamma \left( \beta + 1\right)$ which is the relativistic doppler factor.

This right away gives us the result that I (as expressed in power/meter^2) which is proportional to E, transforms as the square of the above factor.

If one wants to calculate when the sun is swamped out by black body radiation , one can probably estimate it by looking at the amount of energy radiated by CMB by the solid angle resolution of the human eye (this should be proportioal to the angular resolution squared, I'm not sure if there is any proportionality constant or not). One then compares this with the amount of energy emitted by the sun itself, which acts as a point source, that we calculated above.

 

[Russell E]

...and at low frequencies these almost balance, with the intensity falling logarithmically.

The intensity (ultimately) falls in direct proportion to the Doppler factor. The magnitude falls logarithmically, but that's only because the magnitude is the logarithm of the intensity.

Thus we multiply the power emitted / unit area of the black body, by the surface area of the sun (to get the total power), multiply it by (1 meter^2) [if we are using MKS units] area and divide by the area of a sphere of radius of 4 light years to get the power / meter^2. This gives us the intensity in power / meter^2.

Right. When you divide one square meter by the area of that big sphere you are calculating the solid angle. That's what the solid angle means. Unfortunately you seem to be claiming in the same breath that there is no need to account for the solid angle, so your statements are self-contradictory.

Remember that the "solid angle" in the usual expressions of the black body spectrum is the angle at the emitter subtended by the receiving entity. In other words, it isn't how big the Sun looks to your eye, it's how big your eye looks to the Sun, which is to say, what fraction of the big sphere centered on the Sun is subtended by your eye. That's exactly what you have talked about in the quote above, per unit area at the eye's location. What you've described is nothing but the solid angle. Needless to say, regardless of how far away you are, or whether you can resolve the disk of the Sun at that distance, the amount of energy your eye receives is determined by that solid angle (as you yourself have just said).

The only reason for this digression is that some people in this thread initially recalled that a Lorentz-boosted black body spectrum is simply another black body spectrum of the same form with a different temperature, and the intensity for a higher temperature is higher at every frequency. From this they mistakenly concluded that the light in the visible range must always be more intense for an approaching observer. They overlooked the fact that those text-book expressions for Lorentz-boosted spectra are expressed on a "per solid angle" basis, and the solid angle changes under a Lorentz boost, by the square of the Doppler factor, due to aberration.

An important part of the issue is what the detector can resolve.

No, that's irrelevant. For example, if you are at a distance R from the Sun, where R is so large that you can't resolve the Sun's disk (so it looks like a point), you will be receiving a certain intensity, and you may say it's virtually a plane wave. But if you locate yourself at 2R, what will be the intensity? If it was really a plane wave at R, the intensity at 2R would be the same, but it isn't. The intensity at 2R is just 1/4 the intensity at R - even though you can't resolve the disk of the Sun at R (let alone 2R). How do we know the intensity is lower by a factor of 1/4? Well, because the solid angle that your eye subtends at 2R is 1/4 of what it subtends at R. According to your reasoning, this is impossible, i.e., you claim that the solid angle no longer determines the intensity beyond the point when we can't visually resolve it, but that's obviously not true.

This justifies mathematically what I assumed earlier without a detailed proof (I rather thought it was obvious) - the human eye cannot resolve the sun at this distance, and that we are more than justified in using the plane wave approximation. The paper explicitly mentions they use the plane wave approximation.

You misunderstand what McKinley does (and doesn't do) with the plane wave approximation. He's just treating all the light as coming from the same direction and with the same Doppler, which is fine. But he certainly does not neglect the solid angle effect, as shown by the factor D^-2. He doesn't call this the solid angle factor, but that's only because he hasn't been asked to reconcile his result with the "well known facts" about Lorentz-boosted black body spectra. If you mentioned to him the mistaken reasoning that's been voiced in this thread about that, he would say "Well, obviously if you begin by taking the Lorentz boosted spectra on a per-solid-angle-basis, you then need to correct that for the aberration effect on the solid angle." He doesn't have to do this, because he never goes to the per-solid-angle basis in the first place. And for good reason... it is a pointless waste of time to jump to the per-solid-angle basis, only to immediately jump back to the per-unit area-at-this-location basis. You just apply the aberration factor, and then take it away. The only reason we're talking about it is because some people here wanted to understand how to reconcile the solution of this problem with the textbook expressions for Lorentz-boosted spectra.

To use the plane wave approximation we simply need to normalize the radiated energy in terms of meter^2, rather than the "solid angle".

But that "normalization" IS the solid angle factor, so your statement is a non-sequitur.

We have to do this because our detector cannot resolve said angle - we can't use geometric optics, we need to take the diffraction limit into account.

Again, that's a non sequitur. When you divide one unit area at the receiver by the area of the big sphere, you are using geometric optics to compute the solid angle to determine the intensity factor. It doesn't matter if you can visually resolve that angle.

 

[pervect]

The intensity (ultimately) falls in direct proportion to the Doppler factor. The magnitude falls logarithmically, but that's only because the magnitude is the logarithm of the intensity.

Right. When you divide one square meter by the area of that big sphere you are calculating the solid angle. That's what the solid angle means. Unfortunately you seem to be claiming in the same breath that there is no need to account for the solid angle, so your statements are self-contradictory.

Remember that the "solid angle" in the usual expressions of the black body spectrum is the angle at the emitter subtended by the receiving entity.

This wasn't clear to me - but if you specify "the solid angle in the emitter frame" explicitly, rather than implicitly I would agree.

I think many people have been calculating "the solid angle" in the receiver frame, which unfortunately, gives the wrong answer. (For instance, statements about the solid angle increasing because the sun is closer).

In other words, it isn't how big the Sun looks to your eye, it's how big your eye looks to the Sun, which is to say, what fraction of the big sphere centered on the Sun is subtended by your eye. That's exactly what you have talked about in the quote above, per unit area at the eye's location. What you've described is nothing but the solid angle. Needless to say, regardless of how far away you are, or whether you can resolve the disk of the Sun at that distance, the amount of energy your eye receives is determined by that solid angle (as you yourself have just said).

The only reason for this digression is that some people in this thread initially recalled that a Lorentz-boosted black body spectrum is simply another black body spectrum of the same form with a different temperature, and the intensity for a higher temperature is higher at every frequency. From this they mistakenly concluded that the light in the visible range must always be more intense for an approaching observer. They overlooked the fact that those text-book expressions for Lorentz-boosted spectra are expressed on a "per solid angle" basis, and the solid angle changes under a Lorentz boost, by the square of the Doppler factor, due to aberration.

Actually I would say the problem is even more complex:

We have at least three solid angles of interest. A) The solid angle subtended by the receiver from the source (which we seem to agree is the right way to work the problem), B) the solid angle subtended by the source from the receiver (computed geometrically from the source in the same manner as A) and C) the apparent angular diameter, which includes the effects of aberration, and isn't the same as B). Concept B is the geometrically computed size, ignoring aberration. Concept C is the visual angle subtended by an extended (non-point) source including aberration.

For instance, if you move towards the sun, B increases (by geometry), while C decreases (due to aberration, which is not included in concept B).

There might be some confusion about what to call all of these - in particular I'm not too happy about my name for the concept I call "C", but I can't think of a better one offhand and I don't have a reference to consult to give me a better suggestion as to what to call it.

In particular, I'm unhappy with my wording because I want C) to be measured in steradians, not radians.

I might add that both concepts B and C gives an intensity which is a delta function if you have a point source and/or a plane wave, because you have a finite amount of energy coming from _zero_ solid angle - solid angle as measured by the receiver. Hence my various remarks about why we DON"T want to normalize by the solid angle (but I should have specified, we don't want to normalize by the solid angle of the receiver. We don't want to do this because the solid angle is essentially zero, and/or not resolvable by our detecting insruments, thus it's a bad thing to normalize by!

I hope it's clearer now.

 

[pervect]

Why the visual angle of a rapidly moving body is different from the geometric angle.

Consider a frame of reference in which we are at rest at the origin, point O.

Consider a body that is "now" at AB, rapidly approaching us at relativistic speeds. At some time in the past, A was at A' and B was at B', and the light signals from A' and B' are just "now" reaching us at the origin, point O (using the notion of simultaneity of the frame in which we are at rest, the frame in which the diagram is drawn.)

From the diagram we can see that the visual angle, $\alpha'$ of the light reaching us, which is arriving "now" and was emitted "then", is smaller than the geometric angle "now" $\alpha$, due to the propagation delay of the light.

 

[pervect]

Another quick diagram - to illustrate why the emission of light, which is uniformly distributed by angle in the source frame, is NOT uniformly distributed in a frame in which the body is in motion.

We can see that the spherical sun, at rest in some frame, is not spherical in a moving frame, due to Lorentz contraction. The light emitted from the sun is also affected by the Lorentz transform, being squashed in a similar manner.

 

[pervect]

The Electromagnetic Plane Wave and its transformation:

Consider the E and B fields defined by:
E_y = f(x-c\,t) \quad B_z = \frac{1}{c} f(x-c\,t)

This is a transverse wave, because it moves in the x direction, and the E and B fields are in the y and z directions.

f(x-ct) is an arbitrary function. For a wave of a single pure frequency $\nu$ and wavelenght $\lambda$ we may write specifically

f = \cos \frac{2\pi}{\lambda} \left(x - c\,t \right)
http://en.wikipedia.org/w/index.php?title=Plane_wave&oldid=557456576

More generally, we can go from the time domian to the frequency domain by the Fourier transform.1_ Verify whether these fields as given above satisfy Maxwell equations:
\nabla \cdot E = 0 \quad \nabla \cdot B = 0 \quad \nabla \times E = -\frac {\partial B}{\partial t} \quad \nabla \times B = \frac{1}{c^2} \frac{\partial E}{\partial t}
https://en.wikipedia.org/w/index.php?title=Maxwell's_equations&oldid=560494962

The definition of curl $\nabla \times$ is given by:
http://en.wikipedia.org/w/index.php?title=Curl_(mathematics)&oldid=560793176

and divergence $\nabla \cdot$ by:
http://en.wikipedia.org/w/index.php?title=Divergence&oldid=560574862

2) Apply the transformation equations for the E and B fields from http://en.wikipedia.org/w/index.php...netism_and_special_relativity&oldid=542025190

{{E}_{\bot}}'= \gamma \left( \mathbf {E}_{\bot} + \mathbf{ v} \times \mathbf {B} \right) \quad {{B}_{\bot}}'= \gamma \left( \mathbf {B}_{\bot} -\frac{1}{c^2} \mathbf{ v} \times \mathbf {E} \right)

Because the EM wave is a transverse wave, the parallel components $E_{\parallel}$ and $B_{\parallel}$ are zero.
Show that the result of boosting the plane wave in the x direction is a plane wave whose amplitude is:

E' = \gamma \left(1 + \beta\right) E \quad B' = \frac{1}{c} E'

Depending on the sign convention used for the boost, one might get an equivalent result with the opposite sign of $\beta$.

[add]Shortcut

Because we've seen a plane wave is a function f of (x-ct), we can simply ask how (x-ct) transforms under a lorentz transform.

The answer is that (x' - ct') = k (x - ct). Light waves are always transformed into light waves by the Lorentz transform, thus light waves are eigenvectors of the transform. Because the transformation is linear, the most general transformation possible that maps a light wave into another light wave is multipliaton by a scalar.

The eigenvalue (k) corresponding to the eigenvector is just the relativistic doppler shift.

 

[Russell E]

Why the visual angle of a rapidly moving body is different from the geometric angle.

Both of the angles are geometric. I think you meant that the geometrical angles of a pulse of light are different in terms of relatively moving systems if coordinates, which is called aberration. The "explanation" you've shown in your illustration is isomorphic (although slightly less sophisticated) to Bradley' "falling rain" explanation proposed in 1727 based on a ballistic model of light. However, to get the correct answer (relativistically) you must account for the relativity of simultaneity between the two frames (so the meaning of what you called "Now" is different depending on the frame). This is why, relativistically, there is aberration of every angle, not just of the angles of incidence of moving objects (contrary to what your pre-relativistic explanation would predict).

Another problem with your explanation is that it applies only to cases when the source and emitter have been in unaccelerated motion for the entire transit time of the light, whereas this is hardly ever true for stellar light, so it leads to the standard crackpot misunderstanding about the aberration of light from the two components of a binary star system, with different directions of motion (relative to the earth) but the same aberration.

If you work out the actual relativistic aberration, you'll find tan(a'/2) = D*tan(a/2), where "a" is the angle relative to the direction of v, and for very small angles this is a' = Da, so the solid angle for things "straight ahead" goes as D^2. This is why there is a factor of D^2 difference in the expression for spectral densities quoted on a per solid angle basis.

 

[tionis]

We can see that the spherical sun, at rest in some frame, is not spherical in a moving frame, due to Lorentz contraction. The light emitted from the sun is also affected by the Lorentz transform, being squashed in a similar manner.

https://www.youtube.com/watch?v=JQnHTKZBTI4


Pervect, on this^^ video, from 5:08 - 5:18, they say that:

another property of aberration is that it preserves circles, that is - a sphere will always present a spherical outline to any observer regardless of their relative motion.

Is that wrong?

 

[Russell E]

On this video they say that: "another property of aberration is that it preserves circles, that is - a sphere will always present a spherical outline to any observer regardless of their relative motion." Is that wrong?

The video is correct, and what pervect said is also correct. They are describing two different things. The video is talking about the visual image of a moving object, which consists of light reaching the observer's eye that has traveled different amounts of time to reach the eye (because different points on the sphere are at slightly different distances), so the visual image is not an "instantaneous" representation of the object (where "instantaneous" is defined in terms of the observer's rest frame inertial coordinates). In contrast, Pervect was describing the true shape of the sphere at a single instant (where, again, "instant" is defined in terms of the observer's rest frame coordinates). It is an interesting fact (first discussed by Terrell and Penrose back in the 1950's) that although a spherical object is fore-shortened into an ellipsoid in terms of relatively moving coordinates, the optical image at any point always remains circular.

 

[tionis]

OK, thanks lol. So, are you guys both in agreement that the Sun does eventually disappear from view given enough speed?

 

[tionis]

Hey, I have a question: what does a Doppler shift greater or less than unity means?

 

[Dale]

By the way, tionis, we shouldn't be talking about "invisible" or "disappear". It simply goes black, but it remains opaque and not invisible.

 

[DrGreg]

Hey, I have a question: what does a Doppler shift greater or less than unity means?

That should, strictly speaking, be called "Doppler factor" rather than "Doppler shift": the ratio of observed frequency to source frequency.

So a Doppler factor greater than unity means blue shift.

A Doppler factor less than unity means red shift.

 

[tionis]

By the way, tionis, we shouldn't be talking about "invisible" or "disappear". It simply goes black, but it remains opaque and not invisible.

Yes, but for the purpose of our animation, we need to know if the Sun becomes indistinguishable (invisible to the human eye) from the background space once we accelerate the ship close to c. Bear in mind that we start from a distance of 4ly, but by the time we are done with our trip, we'll be a lot closer to the Sun. Do you maintain that as soon as we reach the desired speed (incredibly close to c) the Sun will remain ''opaque'' all throughout our trip even though we are getting closer and closer?

That should, strictly speaking, be called "Doppler factor" rather than "Doppler shift": the ratio of observed frequency to source frequency.

So a Doppler factor greater than unity means blue shift.

A Doppler factor less than unity means red shift.

So ''unity'' means shift?

 

[Doc Al]

So ''unity'' means shift?

Unity means 1.

 

[tionis]

Unity means 1.

Doc Al, so when someone says a ''Doppler shift greater or less than unity,'' he means a shift < or> than 1 what? The speed of light?

 

[Doc Al]

Doc Al, so when someone says a ''Doppler shift greater or less than unity,'' he means a shift < or> than 1 what? The speed of light?

The Doppler factor, as used by DrGreg, is a ratio of frequencies. It has no units. So when he says that that factor is greater or less than unity, he means greater or less than 1. Period.

 

[Russell E]

By the way, tionis, we shouldn't be talking about "invisible" or "disappear". It simply goes black, but it remains opaque and not invisible.

There's a valid distinction between "invisible" in the sense of "transparent", versus in the sense of "completely dark". However, in this context it's a moot point, because in the limit as the speed approaches c the apparent size of the Sun goes to zero, so the distinction between transparent or opaque doesn't matter. (Also note that on a per-area basis it actually gets brighter.)

 

[pervect]

Yes, but for the purpose of our animation, we need to know if the Sun becomes indistinguishable (invisible to the human eye) from the background space once we accelerate the ship close to c. Bear in mind that we start from a distance of 4ly, but by the time we are done with our trip, we'll be a lot closer to the Sun. Do you maintain that as soon as we reach the desired speed (incredibly close to c) the Sun will remain ''opaque'' all throughout our trip even though we are getting closer and closer?

I'm not sure what you mean by "opaque". And the answer is - probably not for the entire trip, no. This is the first time in the thread that a "trip" has even been mentioned. It's a good thing we dragged this minor little detail out of you, finally.

The details of what you'd see on a trip depends on how fast you are moving - which hasn't been specified either. There is also the issue of whether or not the sun acts as a black body. Consulting wiki, http://en.wikipedia.org/wiki/File:Solar_Spectrum.png, the sun's spectrum appears to be a good fit to a black body out to 2500nm. Since the visual range is around 500 nm, we can say there shouldn't be any issues up to a doppler factor of 5:1.

There's no reason I can think of that there would be issues at higher doppler factors, but the matter should be investigated. The first step of the investigation would be to specify exactly what the speed you're moving at (so as to get the corresponding doppler factor).

 

[tionis]

The first step of the investigation would be to specify exactly what the speed you're moving at (so as to get the corresponding doppler factor).

Rusell E mentioned a v = 0.999999973. Is that the magic number to make the Sun invisible?

 

[Dale]

we need to know if the Sun becomes indistinguishable (invisible to the human eye) from the background space once we accelerate the ship close to c.

Nobody has addressed the question of distinguishing the sun from the background. The cosmic microwave background radiation is also a blackbody. As you reach ultra relativistic speeds it will become bright also.

 

[pervect]

Nobody has addressed the question of distinguishing the sun from the background. The cosmic microwave background radiation is also a blackbody. As you reach ultra relativistic speeds it will become bright also.

It's been alluded to, nobody has worked out the details. We can say that at a doppler factor as low as about 2200, the temperature of the CMB in the directon of motion will be the same as that of the sun. Which is a lot lower doppler factor than some of the rather extreme numbers that have been thrown around already.

The CMB transforms differently than the light from a star. I suspect we could have another thread as long and confused as this one working out (or failing to work out, or working out via means that are unfortunately not familiar to most readers of the thread and thus not convicing them) the details.

 

[tionis]

Nobody has addressed the question of distinguishing the sun from the background. The cosmic microwave background radiation is also a blackbody. As you reach ultra relativistic speeds it will become bright also.

That is an interesting question. I hope someone delves into it. Maybe there is a breakdown in the continuum description at sufficiently high speeds. Since you can only see integer number of CMB photons, then at some point you either see a photon or not.

 

[Vanadium 50]

Nobody has addressed the question of distinguishing the sun from the background. The cosmic microwave background radiation is also a blackbody. As you reach ultra relativistic speeds it will become bright also.

Yes, but not as bright. More specifically, I believe there is no frame where the temperature of the CMB just past the limb of the sun and the limb of the sun are radiating as if they were at the same T.