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chornedsnorkack
2011-Nov-03, 09:38 AM
Suppose you are approaching the Sun at hyperbolic deceleration of, say, 1g.

The distance (and square of distance) will be decreasing. But the approach speed and blueshift will be decreasing, too.

If Sun were viewed from a stationary observer at, say, the standard distance of 10 pc/32,6 ly, it would have its magnitude at standard absolute magnitude +4,85. But from an approaching observer?

Blueshift would shift the visible light of Sun into ultraviolet. But the observer should be seeing the infrared tail of Sun blueshifted into visible.

What will this to to the magnitude of the Sun? Will the Sun be brighter (because the blueshift amplifies the luminosity), or dimmer (because the infrared tail is so dim compared to visible light that even when amplified by blueshift, it still is dimmer than the visible light was)?

antoniseb
2011-Nov-03, 10:25 AM
The relative velocity to the Sun is what matters for what you're asking about, and so the brightness and color of the Sun will be changing during your deceleration.

Hornblower
2011-Nov-03, 02:47 PM
Suppose you are approaching the Sun at hyperbolic deceleration of, say, 1g.

The distance (and square of distance) will be decreasing. But the approach speed and blueshift will be decreasing, too.

If Sun were viewed from a stationary observer at, say, the standard distance of 10 pc/32,6 ly, it would have its magnitude at standard absolute magnitude +4,85. But from an approaching observer?

Blueshift would shift the visible light of Sun into ultraviolet. But the observer should be seeing the infrared tail of Sun blueshifted into visible.

What will this to to the magnitude of the Sun? Will the Sun be brighter (because the blueshift amplifies the luminosity), or dimmer (because the infrared tail is so dim compared to visible light that even when amplified by blueshift, it still is dimmer than the visible light was)?

For the Sun, the infrared tail has approximately as much energy as the visible band, with most of it in the range of 700 nanometers (nm) to 1400. If we blueshift the spectrum enough to make everything half the previous wavelength, that infrared band will be pushed into the visible range of about 350 to 700 nm. If my reasoning is correct, that will double the energy of each photon and also increase the number of photons per second, thus greatly increasing the apparent brightness and making it look like a type B star.

I am not ready to estimate what happens with further blueshifting. I lost my blackbody calculator in a hard drive failure recently and have not gotten around to restoring it. My guess from previous doodling is that it will continue to brighten somewhat. I welcome any correction from someone who can do the calculation at this time.

By the way I don't know what you mean by "hyperbolic deceleration" while approaching the Sun, but it is beside the point. As antoniseb pointed out, the blueshift or redshift at any given point in time depends on the velocity at that time, not on the rate of change of that velocity.

antoniseb
2011-Nov-03, 03:34 PM
The appearance can be determined by looking at the number of photons as a function of wavelength in the whole spectrum, don't worry about total energy output, except as is needed to get photon counts. That will give you an easy way to estimate apparent color. Maybe there is such a graph out there on the web... I haven't looked.

chornedsnorkack
2011-Nov-07, 06:37 PM
If you double the temperature of a black body, of a constant angular size, the total power across whole spectrum shall increase 16 times. The power of Rayleigh tail just doubles.

If you blueshift a black body doubling the temperature - how much does the total power increase?