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we just had a couple of threads start about the effect of gravity on light----Jakrabb started "hubble and doppler" in this forum and in the Theoretical forum JSK333 started "light, black holes and gravity"
The two main effects to understand (two that are easy to estimate with very little arithmetic) are light bending and gravitational redshift. In common circumstances they are slight effects----e.g. the sun does not bend passing light rays very much, nor is the light coming to us from its surface redshifted very much by the effort of escaping from the sun's gravity----but it's not a bad idea to know HOW slight.
Any massive object has a Schwarzschild radius----proportional to its mass.
The sun's Schwarzschild radius is 3 km----that's the size of an equal mass Schw. black hole. A large galaxy like Andromeda or MilkyWay has a mass of about a trillion solar (E12) so its Schw. radius is 3 trillion kilometers. As far as concerns light-bending and grav redshift, an objects Schw. radius Rs tells it all.
If you know Rs then you can easily tell how much the star or galaxy will bend a passing lightray, and by how much light escaping from the star's surface or from some typical location inside the galaxy will appear depleted of energy or redshifted (it is usually not by much!)
A lightray passing within distance R from the sun's center gets bent by an angle (in radians) of 2Rs/R
Same for a ray of light passing by outside our galaxy at a distance R from the galaxy's center. It gets bent by an angle of 2Rs/R
All you need to do is double the Schw. radius of the thing and divide by the distance of closest approach.
The sun's radius is 700,000 km. So a grazing lightray gets bent by an angle of 6/700,000------very small number like a thousandth of a percent.
The redshift in light leaving the sun's surface is even less. An approximate but useful formula for it is Rs/2R
So you can see that for light from the sun's surface
zgrav = 3/1400,000-----on the order of a millionth.
It's incredibly easy to calculate--just take the sun's Schw. (3 km) radius and divide by twice its real (700,000 km) radius.
This formula is extremely accurate for the small gravitational redshifts associated with ordinary stars and galaxies.
The grav. redshift ordinarily amounts to only a broadening of a spectral line or a small correction to the main types of redshift (cosmological and doppler).
But still let's look at the exact formula and apply it to some more exotic object like a black hole or quasar. The formula for zgrav is still based on that versatile multipurpose parameter the Schwarzschild radius! There is just a square root in the formula so it is a tiny bit more complicated.
1+ zgrav = 1/sqrt(1 - Rs/R)
1 + z is the ratio by which wavelengths are lengthened in light escaping to infinity from an initial distance R from center.
So right off you can see that light trying to escape from the event horizon itself (R = Rs) is infinitely redshifted. It dies. All its energy is canceled. This is the basic meaning of the Schw. radius:
1/sqrt(1 - 1) equals infinity---the redshift is infinite for light trying to escape. Some people may pick nits with the figurative way of talking--light doesn't really "try", it's not a person, etc etc. I am more concerned with the simple formulas than with verbal niceness.
For a less extreme case, imagine stuff spiraling into a black hole as in a quasar: the accretion disk is heated essentially by compression and collisions as it is drawn into the hole, so it radiates. If temperatures as high as several million kelvin are reached then it can radiate x-rays.
This radiation will be gravitationally redshifted as viewed by a distant observer. If e.g. it is emitted at a distance of 100 Rs
then its Rs/R is equal to 0.01
and 1 - Rs/R = 0.99
and 1 + z = 1/sqrt(0.99) = 1/0.995 = 1.005
So the gravitational redshift for that particular quasar light is five thousandths----0.005 or half a percent.
Even light emitted from as close as 10 Rs to center
has Rs/R equal to 0.1
and 1 - Rs/R = 0.9
and 1 + z = 1/sqrt(0.9) = 1/0.95 = 1.054
so zgrav is only a little over 5 percent.
The overall redshift observed for quasars is much larger than that---it's commonly of order unity----numbers like 2, 3, 4 are common and one has been observed with z as high as 6.4. Evidently these observed redshifts are mostly cosmological in origin and are so treated by astronomers when they are discussed.
In such contexts the gravitational component of the redshift is best seen as a minor correction. But still one that is good to know about.
BTW the formula for Schw. radius has come up in several other threads and is 2GM/c2-----that's where the figure of 3 kilometers for the sun comes from.
Or 3 trillion km for a trillion-solar-mass galaxy.
I see I didnt yet do the grav redshift for a star like the sun which is about 3E17 km out from the center of a roughly trillion solar mass. The galaxy's Schw. radius is 3E12 km. So dividing 3E12 by twice 3E17 gives 0.5E-5. This is 5 ppm. Just to take a representative run-of-the-mill case,
the sun's light, seen by an observer well outside MilkyWay has a negligible grav redshift of around five parts per million.
The two main effects to understand (two that are easy to estimate with very little arithmetic) are light bending and gravitational redshift. In common circumstances they are slight effects----e.g. the sun does not bend passing light rays very much, nor is the light coming to us from its surface redshifted very much by the effort of escaping from the sun's gravity----but it's not a bad idea to know HOW slight.
Any massive object has a Schwarzschild radius----proportional to its mass.
The sun's Schwarzschild radius is 3 km----that's the size of an equal mass Schw. black hole. A large galaxy like Andromeda or MilkyWay has a mass of about a trillion solar (E12) so its Schw. radius is 3 trillion kilometers. As far as concerns light-bending and grav redshift, an objects Schw. radius Rs tells it all.
If you know Rs then you can easily tell how much the star or galaxy will bend a passing lightray, and by how much light escaping from the star's surface or from some typical location inside the galaxy will appear depleted of energy or redshifted (it is usually not by much!)
A lightray passing within distance R from the sun's center gets bent by an angle (in radians) of 2Rs/R
Same for a ray of light passing by outside our galaxy at a distance R from the galaxy's center. It gets bent by an angle of 2Rs/R
All you need to do is double the Schw. radius of the thing and divide by the distance of closest approach.
The sun's radius is 700,000 km. So a grazing lightray gets bent by an angle of 6/700,000------very small number like a thousandth of a percent.
The redshift in light leaving the sun's surface is even less. An approximate but useful formula for it is Rs/2R
So you can see that for light from the sun's surface
zgrav = 3/1400,000-----on the order of a millionth.
It's incredibly easy to calculate--just take the sun's Schw. (3 km) radius and divide by twice its real (700,000 km) radius.
This formula is extremely accurate for the small gravitational redshifts associated with ordinary stars and galaxies.
The grav. redshift ordinarily amounts to only a broadening of a spectral line or a small correction to the main types of redshift (cosmological and doppler).
But still let's look at the exact formula and apply it to some more exotic object like a black hole or quasar. The formula for zgrav is still based on that versatile multipurpose parameter the Schwarzschild radius! There is just a square root in the formula so it is a tiny bit more complicated.
1+ zgrav = 1/sqrt(1 - Rs/R)
1 + z is the ratio by which wavelengths are lengthened in light escaping to infinity from an initial distance R from center.
So right off you can see that light trying to escape from the event horizon itself (R = Rs) is infinitely redshifted. It dies. All its energy is canceled. This is the basic meaning of the Schw. radius:
1/sqrt(1 - 1) equals infinity---the redshift is infinite for light trying to escape. Some people may pick nits with the figurative way of talking--light doesn't really "try", it's not a person, etc etc. I am more concerned with the simple formulas than with verbal niceness.
For a less extreme case, imagine stuff spiraling into a black hole as in a quasar: the accretion disk is heated essentially by compression and collisions as it is drawn into the hole, so it radiates. If temperatures as high as several million kelvin are reached then it can radiate x-rays.
This radiation will be gravitationally redshifted as viewed by a distant observer. If e.g. it is emitted at a distance of 100 Rs
then its Rs/R is equal to 0.01
and 1 - Rs/R = 0.99
and 1 + z = 1/sqrt(0.99) = 1/0.995 = 1.005
So the gravitational redshift for that particular quasar light is five thousandths----0.005 or half a percent.
Even light emitted from as close as 10 Rs to center
has Rs/R equal to 0.1
and 1 - Rs/R = 0.9
and 1 + z = 1/sqrt(0.9) = 1/0.95 = 1.054
so zgrav is only a little over 5 percent.
The overall redshift observed for quasars is much larger than that---it's commonly of order unity----numbers like 2, 3, 4 are common and one has been observed with z as high as 6.4. Evidently these observed redshifts are mostly cosmological in origin and are so treated by astronomers when they are discussed.
In such contexts the gravitational component of the redshift is best seen as a minor correction. But still one that is good to know about.
BTW the formula for Schw. radius has come up in several other threads and is 2GM/c2-----that's where the figure of 3 kilometers for the sun comes from.
Or 3 trillion km for a trillion-solar-mass galaxy.
I see I didnt yet do the grav redshift for a star like the sun which is about 3E17 km out from the center of a roughly trillion solar mass. The galaxy's Schw. radius is 3E12 km. So dividing 3E12 by twice 3E17 gives 0.5E-5. This is 5 ppm. Just to take a representative run-of-the-mill case,
the sun's light, seen by an observer well outside MilkyWay has a negligible grav redshift of around five parts per million.
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