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tommac
2008-Apr-11, 09:45 PM
Would the gravity of a galaxy be the sum of all the mass in the galaxy?

Would it exert an external gravitational pull as a whole? Could it be treated similarly to a massive condensed object at its center of gravity?

01101001
2008-Apr-11, 09:58 PM
Would the gravity of a galaxy be the sum of all the mass in the galaxy?

Gravity is force.

Mass is mass.

StupendousMan
2008-Apr-11, 10:29 PM
Would the gravity of a galaxy be the sum of all the mass in the galaxy?

Would it exert an external gravitational pull as a whole? Could it be treated similarly to a massive condensed object at its center of gravity?

The gravitational pull of an entire galaxy on some test object is the same as the sum of the gravitational pulls from each little bit of mass in the galaxy on that test object .... if that's what you're asking.

If the test object is very far from the entire galaxy -- say, ten or twenty diameters away from the galaxy -- then the overall gravitational force of all the mass in the galaxy will be very similar to that of a single body with the entire mass, sitting at the center of the galaxy. As the test object approaches the galaxy, however, this simple approximation doesn't work very well.

You can read more about that simplification in any good physics book by looking in the index for words like "dipole" and "multipole expansion".

astromark
2008-Apr-12, 03:34 AM
Would the gravity of a galaxy be the sum of all the mass in the galaxy?

Would it exert an external gravitational pull as a whole? Could it be treated similarly to a massive condensed object at its center of gravity?

YES.

Cougar
2008-Apr-12, 03:50 AM
Would the gravity of a galaxy be the sum of all the mass in the galaxy?




Depends on what you're identifying as "all the mass in the galaxy." Of course you're counting the stars and planets and moons and asteroids and comets, etc. Then there's all the clouds of gases and dust, which make up a fair amount of a spiral's combined mass. That's all the baryonic stuff. Then there's the question of dark matter, which is inferred from the observation of too much gravitational effect coming from somewhere! And not just a little too much. So far, dark matter is undetectable except through its inferred gravity.

Then to get real nit-picky, are you including the effect of gravity that gravity has on itself? Yes, I believe that's in the equation there somewhere. It's not very big except in extreme circumstances, like in the neighborhood of an active galactic nucleus. (http://en.wikipedia.org/wiki/Active_galactic_nucleus)

Eroica
2008-Apr-12, 03:25 PM
Would it exert an external gravitational pull as a whole? Could it be treated similarly to a massive condensed object at its center of gravity?
Once consequence of Newtonian gravity is this:


When the mass of a system is distributed in a spherically symmetrical manner about some central point, then the net gravitational force on a point-like object at some radius is due only to the mass within that radius. Furthermore, the net gravitational force is the same as if the mass inside that radius was all located at the centre.In other words, the Sun's orbit about the centre of the Galaxy is (approximately) the Keplerian orbit you would get if all the mass in the Galaxy closer to the centre than the Sun were concentrated in a point mass at the centre, and all the mass in the Galaxy further from the centre than the Sun did not exist.

tommac
2008-Apr-14, 02:09 PM
I guess where I was going with this whole questioning is about the total gravitational effect on us ( the earth ).

Lets even take the moon for example. If I am standing on the moon. I am effected by the gravity of the moon ( most noticeable to me ) however I am greater effected by the gravity of the earth ( because it is fact is pulling me along with the moon ) , even greater is the gravity of the sun , and even greater is whatever is pulling the sun around, and I am not sure what the next step of that is ... whatever is pulling the galaxy around.

But I would think that in total compared to something in a true vacuum of space ( if this exists ), our space-time is hugely differerent. Much more compressed.

Does this make any sense?

formulaterp
2008-Apr-14, 07:08 PM
No, not really.

If you are standing on the Moon, the Moon has the greatest gravitational effect on you. Not the Earth or the Sun.

Also, vacuum does not imply lack of gravity. Just lack of stuff.

astromark
2008-Apr-14, 07:31 PM
The larger the mass the stronger the gravity, thats excepted.
With distance force is diminished.
Gravity might not be the strong force of the universe., but it is relentless.
Only that unknown force we name 'Dark Energy' would be stronger.
What drives it.? Unknown Force.

tommac
2008-Apr-14, 07:44 PM
No, not really.

If you are standing on the Moon, the Moon has the greatest gravitational effect on you. Not the Earth or the Sun.

Also, vacuum does not imply lack of gravity. Just lack of stuff.


Really ... anyone have the math behind this? Can someone quickly calculate :

gravitational pull from the moon from the surface of the moon
gravitational pull fom the earth from the surface of the moon
gravitational pull from the sun from the surface of the moon
gravitational pull from the galaxy from the surface of the moon

Jeff Root
2008-Apr-15, 12:55 AM
tommac,

This is something I thought about and tried to calculate 30 years ago.
The calculations should be simple but my math skills and my understanding
of how the math should be applied are poor, so I hope someone else will
tackle the question. I think your general idea is sound, but it is vaguely-
expressed, so far. To eliminate the vagueness, we need to know what it
is about the Earth's gravity that is acting more strongly on an astronaut
on the lunar surface than is the Moon's gravity: Is it the gravitational
field strength; the gravitational potential; the curvature of spacetime, or
what? And measured relative to what?

The progression you described doesn't necessarily continue, nor does each
step necessarily involve "stronger" gravity than the previous step. But that
does appear to be a general tendency.

-- Jeff, in Minneapolis

CodeSlinger
2008-Apr-15, 07:28 AM
Really ... anyone have the math behind this?

You can use Newton's equation for gravity (http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation) and get some approximations for the purpose of doing a comparison. First, decide on the mass for your test object (or you can omit this and compare instead the acceleration due to gravity exerted by the various objects). Then use the following numbers to see the magnitude of the force of gravity exerted by the various objects on your test object on the surface of the Moon.

G: 6.67 x 10^-11 (gnarly unit omitted for clarity)



gravitational pull from the moon from the surface of the moon


Mass, Moon: 7.35 x 10^22 kg
Distance, Moon surface to Moon center: 1.74 x 10^6 m



gravitational pull fom the earth from the surface of the moon


Mass, Earth: 5.97 x 10^24 kg
Distance, Moon to Earth: 3.84 x 10^8 m



gravitational pull from the sun from the surface of the moon


Mass, Sun: 1.99 x 10^30 kg
Distance, Moon to Sun: 1.49 x 10^11 m



gravitational pull from the galaxy from the surface of the moon


Mass, Milky Way: 1.15 x 10^42 kg
Distance, Moon to center of Milky Way: 2.5 x 10^20 m

In Newton's equation, m1 is the mass of your test object (which you decided on), and G will always have the same value (since it's a constant). To calculate the force of gravity the Moon exerts on your test object sitting on the surface of the Moon, use the mass of the Moon as m2, and distance from Moon's surface to Moon's center as r. To calculate the force of gravity the Earth exerts on your test object sitting on the surface of the Moon, use the mass of the Earth as m2, and distance from Moon to Earth as r (if you want an extra bit of accuracy, subtract the radius of the Moon from the distance from Moon to Earth, but it's ok if you don't since we're just approximating here). And so on and so forth.

Amber Robot
2008-Apr-15, 01:29 PM
Mass, Milky Way: 1.15 x 10^42 kg
Distance, Moon to center of Milky Way: 2.5 x 10^20 m

But, since we are inside the Milky Way, some of the forces will cancel out, so you can't just assume the Milky Way is a point mass at the center.

CodeSlinger
2008-Apr-15, 01:55 PM
This is true, and I would not have advocated treating the Milky Way as a point mass if this was a calculation where any semblance of accuracy and/or precision is required. However, for the purposes of this quick-n-dirty comparison, I believe it's acceptable to use this approximation.

tommac
2008-Apr-15, 01:56 PM
But, since we are inside the Milky Way, some of the forces will cancel out, so you can't just assume the Milky Way is a point mass at the center.

Where I was trying to ultimately go with this is to figure out an estimate of a total gravity well for say the moon. A few of the thoughts that I was having is that universal red shift could be caused by us either entering a deeper well ( maybe by moving closer to the center of the galaxy ??? or something ) or us leaving a well. My idea was that for us to conclude that all rulers are experiencing the same redshift phenomenon as us seems a bit of a stretch. I understand that we are not really the center of the universe but relativistically we are ... All of the redshifting galaxies means that all galaxies are moving away from us, relativistically this means that we are the center and what all things are measured against. ( we are 0,0,0,0 ). Now I also understand why people jump to the next conclusion but I am not sure if I fully understand that enough evidence exists to support that postulate.

CodeSlinger
2008-Apr-15, 02:11 PM
I understand that we are not really the center of the universe but relativistically we are ... All of the redshifting galaxies means that all galaxies are moving away from us, relativistically this means that we are the center and what all things are measured against. ( we are 0,0,0,0 ).

We are not the center of the universe, relativistically or not. In fact, the very notion that we are at the center of the universe goes directly against what relativity actually says; one of relativity's postulates is that there are no preferred/special frames of reference. Yes, all the redshifting galaxies are moving away from us, but there are also some blueshifting galaxies moving towards us. But aliens in any other galaxy would also observe almost everything receding away from them, with the appearance that they are the center of the universe. This is because of the way the expansion is happening, which speedfreak explained to you HERE (http://www.bautforum.com/against-mainstream/72240-distant-light.html#post1215843).

tommac
2008-Apr-15, 02:15 PM
OK ... quickly did some of the math and basically ... I get it now. The moon is the strongest gravity, the earth next, the sun next, the galaxy next. After that I felt stupid because without calculation this is clear because if you throw a rock on the moon it clearly drops back to the surface. so space-time is fairly curved right at the surface. Then the arch of the moon around the earth compared to a straight line is much tigher than the arch moving the moon around the sun or the sun around the galaxy. If the earth was stronger the rock would be pulled off the surface and pulled toward the earth. If the sun was stronger then the moon would circle the sun and not the earth. If the galaxy was stronger the moon would circle the galaxie and not the sun or the earth.

For the net effect, there is a signifigant drop off mostly from distance. As distances get large the r^2 component of the formula becomes larger quicker than the growing masses.

Now home much of a time dilation do we have when we are in space rather than on the surface of the earth? Lets say we go out to the distance of the moon but not near the moons gravitational field ( yes I realize that the field is infinite but becomes irrelevant fairly quickly )



You can use Newton's equation for gravity (http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation) and get some approximations for the purpose of doing a comparison. First, decide on the mass for your test object (or you can omit this and compare instead the acceleration due to gravity exerted by the various objects). Then use the following numbers to see the magnitude of the force of gravity exerted by the various objects on your test object on the surface of the Moon.

G: 6.67 x 10^-11 (gnarly unit omitted for clarity)



Mass, Moon: 7.35 x 10^22 kg
Distance, Moon surface to Moon center: 1.74 x 10^6 m



Mass, Earth: 5.97 x 10^24 kg
Distance, Moon to Earth: 3.84 x 10^8 m



Mass, Sun: 1.99 x 10^30 kg
Distance, Moon to Sun: 1.49 x 10^11 m



Mass, Milky Way: 1.15 x 10^42 kg
Distance, Moon to center of Milky Way: 2.5 x 10^20 m

In Newton's equation, m1 is the mass of your test object (which you decided on), and G will always have the same value (since it's a constant). To calculate the force of gravity the Moon exerts on your test object sitting on the surface of the Moon, use the mass of the Moon as m2, and distance from Moon's surface to Moon's center as r. To calculate the force of gravity the Earth exerts on your test object sitting on the surface of the Moon, use the mass of the Earth as m2, and distance from Moon to Earth as r (if you want an extra bit of accuracy, subtract the radius of the Moon from the distance from Moon to Earth, but it's ok if you don't since we're just approximating here). And so on and so forth.

Eroica
2008-Apr-15, 02:18 PM
Mass, Milky Way: 1.15 x 10^42 kg
Distance, Moon to center of Milky Way: 2.5 x 10^20 m
For the mass of the Milky Way you should use only the mass that is closer to the centre of the Galaxy than the Sun: about 2 x 1041 kg.

tommac
2008-Apr-15, 02:18 PM
We are not the center of the universe, relativistically or not. In fact, the very notion that we are at the center of the universe goes directly against what relativity actually says; one of relativity's postulates is that there are no preferred/special frames of reference. Yes, all the redshifting galaxies are moving away from us, but there are also some blueshifting galaxies moving towards us. But aliens in any other galaxy would also observe almost everything receding away from them, with the appearance that they are the center of the universe. This is because of the way the expansion is happening, which speedfreak explained to you HERE (http://www.bautforum.com/against-mainstream/72240-distant-light.html#post1215843).

Yes yes yes. I get that ... but what I am saying is that when I measure stuff from my perspective I am at 0,0,0,0 which realatively means that I am the center of everything. If I go to the moon I am still at 0,0,0,0. All I am saying is that the center of the universe is at the point of the observer, I really cant see it any other way. I guess what you are saying is that all points are the center of the universe. but relative to me I am the center of the universe.

CodeSlinger
2008-Apr-15, 03:04 PM
Eroica: Thank you for the number. As I explained above, I thought it was acceptable to approximate the galaxy as a point mass for the purpose of a rough comparison, but that number will be good to have for future reference.


Now home much of a time dilation do we have when we are in space rather than on the surface of the earth? Lets say we go out to the distance of the moon but not near the moons gravitational field ( yes I realize that the field is infinite but becomes irrelevant fairly quickly )

Time dilation relative to an observer located where? Once you decide on that, I think you can answer this question for yourself using the equations for time dilation, which are very Google-able.

tommac
2008-Apr-15, 03:12 PM
surface of the earth vs in space at roughly the distance of the moon.

Eroica: Thank you for the number. As I explained above, I thought it was acceptable to approximate the galaxy as a point mass for the purpose of a rough comparison, but that number will be good to have for future reference.



Time dilation relative to an observer located where? Once you decide on that, I think you can answer this question for yourself using the equations for time dilation, which are very Google-able.

grant hutchison
2008-Apr-15, 04:08 PM
surface of the earth vs in space at roughly the distance of the moon.The gravitational time dilation at the surface of the Earth is about one part per billion compared to "flat space", remote from any gravitational influence (calculation given here (http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/gratim.html#c5)).

Grant Hutchison

John Mendenhall
2008-Apr-15, 04:38 PM
If the earth was stronger the rock would be pulled off the surface and pulled toward the earth. If the sun was stronger then the moon would circle the sun and not the earth. If the galaxy was stronger the moon would circle the galaxie and not the sun or the earth.



You've got a hiccup here in your logic. Loose rocks on the surface of the Moon are in orbit around the Earth, just as the Moon is. The rocks will not be pulled off the surface of the Moon by the Earth's gravity, they are much more strongly attracted to the Moon, since their orbital velocity with respect to the Moon is about 28 days per revolution, the same as the Moon's rotation. This also applies to the oceans on the Earth; the tides do not pull the oceans off the Earth, because the water is in orbit around the Sun just as are the solid parts of the Earth, with a period of 365 days per orbital revolution. Another example is the orbiting astronaut who releases a pencil in front of him; it feels the Earth's gravity just fine, and it floats there, because it is in orbit around the Earth with the astronaut.

As an aside, you might try looking up the Roche limit. Inside the limit, things operate a little differently.

Jeff Root
2008-Apr-15, 06:49 PM
tommac,

Your original idea about the gravity wells we are in causing the cosmic
redshift doesn't work, but it does have merit.

Light falling into a gravity well appears blueshifted, not redshifted, to
observers in the well. As Grant showed, the gravitational time dilation
which is partially responsible for this blueshift is extremely tiny.

However, light coming to us from distant galaxies comes from a time
when the Universe was more dense than it is now. As a result, the
light is redshifted by, in effect, rising out of the gravity well of the past
toward the less-deeply situated observers now. But this effect has
been calculated to be only about 1% of the total cosmic redshift.
I don't have a reference for this calculation, but it must be somewhere
on Ned Wright's website. It has been many years since I read this, and
I have not confirmed that it was correct.

One simple and obvious disproof of your original notion that gravity wells
we are in causes the cosmic redshift is the fact that the redshift depends
on distance. Light from the nearby Andromeda Galaxy, for example, is not
redshifted at all. It is part of the local cluster, so it and the Milky Way
are gravitationally bound together. As it happens, Andromeda and the
Milky Way are currently moving closer together, so the light we see from
Andromeda is blueshifted. Light from galaxies much farther away from us
than Andromeda is redshifted. As you look at galaxies farther and farther
away, the redshift becomes greater and greater. The greatest redshift,
by far, is that of the cosmic microwave background radiation.

If your original idea were valid, most of the cosmic redshift we see would
take place within the Solar System and within the Milky Way. Instead,
none of it does. It all takes place over distances of millions of light-years,
and by far most of it occurs over distances of billions of light-years.

However, it is true that an astronaut on the Moon's surface is deeper in
Earth's gravity well than he is in the Moon's gravity well. He is deeper in
the Sun's gravity well than he is in Earth's. And he is deeper in the Milky
Way's gravity well than he is in the Sun's. So that part of your idea is
correct. And important. I'm not quite sure what it implies, though.

Maybe you can figure it out. You can look up "Mach's principle".

-- Jeff, in Minneapolis

tommac
2008-Apr-15, 07:55 PM
Maybe I meant that we are moving out of a deep gravity well. So that everything is appearing red shifted. Why would local things appear shifted in my theory? If they are also escaping a deep gravity well?

Also are you sure that things would appear blue shifted as a fell into a deep gravity well? That would be counterintuitive to me. As as I went deeper into a gravity well I would think that things would appear to get further and further away as my time space shrunk ... Just before I passed into the event horizon of a black hole I wouldnt be able to see much of the galaxy as my space-time would be receeding to quickly. Hmmm ... I guess though that light is also getting pushed in ... OK ... I think I understand the blue shift a little. Because as we contract light moves closer to me, while it source moves further, right?




tommac,

Your original idea about the gravity wells we are in causing the cosmic
redshift doesn't work, but it does have merit.

Light falling into a gravity well appears blueshifted, not redshifted, to
observers in the well. As Grant showed, the gravitational time dilation
which is partially responsible for this blueshift is extremely tiny.

However, light coming to us from distant galaxies comes from a time
when the Universe was more dense than it is now. As a result, the
light is redshifted by, in effect, rising out of the gravity well of the past
toward the less-deeply situated observers now. But this effect has
been calculated to be only about 1% of the total cosmic redshift.
I don't have a reference for this calculation, but it must be somewhere
on Ned Wright's website. It has been many years since I read this, and
I have not confirmed that it was correct.

One simple and obvious disproof of your original notion that gravity wells
we are in causes the cosmic redshift is the fact that the redshift depends
on distance. Light from the nearby Andromeda Galaxy, for example, is not
redshifted at all. It is part of the local cluster, so it and the Milky Way
are gravitationally bound together. As it happens, Andromeda and the
Milky Way are currently moving closer together, so the light we see from
Andromeda is blueshifted. Light from galaxies much farther away from us
than Andromeda is redshifted. As you look at galaxies farther and farther
away, the redshift becomes greater and greater. The greatest redshift,
by far, is that of the cosmic microwave background radiation.

If your original idea were valid, most of the cosmic redshift we see would
take place within the Solar System and within the Milky Way. Instead,
none of it does. It all takes place over distances of millions of light-years,
and by far most of it occurs over distances of billions of light-years.

However, it is true that an astronaut on the Moon's surface is deeper in
Earth's gravity well than he is in the Moon's gravity well. He is deeper in
the Sun's gravity well than he is in Earth's. And he is deeper in the Milky
Way's gravity well than he is in the Sun's. So that part of your idea is
correct. And important. I'm not quite sure what it implies, though.

Maybe you can figure it out. You can look up "Mach's principle".

-- Jeff, in Minneapolis

grav
2008-Apr-19, 04:09 AM
Just a couple of quick nitpicks. First, as StupendousMan mentioned earlier, the gravity of a disk galaxy is not the same as that of a point with the same mass which is located at the center. That is only true for a spherical body of mass. A disk would only come close to that of a point mass from a very large distance outside the galaxy. Also, within the disk of a galaxy, we wouldn't just take that amount of mass that lies from that point toward the center, since that also applies only to a sphere, except perhaps discounting the central bulge if roughly spherical.

Second, the gravity of the sun on the moon is actually greater than that of the Earth on the moon, which would almost make it seem as though the moon should orbit the sun independently of the Earth, then. But since the Earth and moon are at the same distance from the sun, the same gravity acts upon both, so that they orbit the sun together at the same rate, almost inertially in this respect to one another. This negates any greater pull of the sun of one over the other, and the Earth and moon then become an independent system of their own as well, due to the gravity acting directly between them.

Eroica
2008-Apr-19, 02:57 PM
Just a couple of quick nitpicks. First, as StupendousMan mentioned earlier, the gravity of a disk galaxy is not the same as that of a point with the same mass which is located at the center. That is only true for a spherical body of mass. A disk would only come close to that of a point mass from a very large distance outside the galaxy. Also, within the disk of a galaxy, we wouldn't just take that amount of mass that lies from that point toward the center, since that also applies only to a sphere, except perhaps discounting the central bulge if roughly spherical.
But don't forget that most of the mass of the Galaxy is dark matter, which is distributed in an oblate spheroid - not quite spherical, but certainly not a disc.

mugaliens
2008-Apr-19, 05:26 PM
Would the gravity of a galaxy be the sum of all the mass in the galaxy?

Would it exert an external gravitational pull as a whole? Could it be treated similarly to a massive condensed object at its center of gravity?

The sum of a galaxy's mass would contribute to the galaxy's gravitational pull, but only if you're outside the galaxy's edge. The rest of the galaxy's pull would be a result of the energy (radiated) within the galaxy.

And yes, you could treat it as a single point, but again, only if you're beyond the galactic rim.

mugaliens
2008-Apr-19, 05:34 PM
Really ... anyone have the math behind this? Can someone quickly calculate :

gravitational pull from the moon from the surface of the moon

If you were standing on the moon at the point intersecting the line between the centers of the Earth and Moon, you would weigh less than if you were standing at a point where the line between you and the center of the Moon was perpendicular to the Earth-Moon line.

Why? Tidal force. You're closer to the Earth than is the center of the Moon, therefore, the Earth does exert a slight force on you.


gravitational pull fom the earth from the surface of the moon
gravitational pull from the sun from the surface of the moon
gravitational pull from the galaxy from the surface of the moon

The same for these three, although the further you get away from the source, the shallower the graviational gradient and thus the effect becomes smaller.

We can measure the Earth and Sun tidal effects quite accurately, but I doubt we could measure the tidal force from the galactic center.

grant hutchison
2008-Apr-19, 06:58 PM
The galactic tide probably has a significant effect on the solar system as a whole, however. The small forces involved are nevertheless large enough to modify the orbits of Oort Cloud objects, both during the formation of the Oort and in its subsequent evolution.

Grant Hutchison

Jeff Root
2008-Apr-19, 11:41 PM
Grant,

Can you comment on what tommac asked about in post #7, and I talked
about in post #11 and the end of post #24?

-- Jeff, in Minneapolis

grant hutchison
2008-Apr-19, 11:53 PM
Grant,

Can you comment on what tommac asked about in post #7, and I talked
about in post #11 and the end of post #24?I don't understand what tommac means, I'm afraid.

Grant Hutchison

Jeff Root
2008-Apr-20, 12:57 AM
For me, a big part of this question is to determine what the question is.

An astronaut standing on the surface of the Moon is in the Moon's gravity
well. He is at some particular depth in that well. If I'm not mistaken, at
the same time, he is even deeper in Earth's gravity well than he is in the
Moon's gravity well. He is certainly deeper in the Sun's gravity well than
he is in either the Moon's or the Earth's. You and I on the surface of the
Earth are deeper in the Sun's gravity well than we are in Earth's. And we
are all deeper in the Galaxy's gravity well than we are in the Sun's.

To me, that suggests a basis for applying Mach's principle.

-- Jeff, in Minneapolis

tommac
2008-Apr-20, 02:06 AM
At the point that I asked this question I was trying to determine if the red shift that we see from most galaxies could be from us falling into a deep gravitational well. The thought that I had was that if we circle the galaxy in a eliptical orbit there would be periods where we would fall into a deeper and deeper gravitational well. However I think the question was that the gravity that we are effected by from the galaxy itself is somewhat weak ??? is that true?

Amber Robot
2008-Apr-20, 02:40 AM
I think you could calculate what the mass would be to produce a gravitational well so deep that the cosmological redshifts were caused by gravitational redshift. Then you could check to see if that amount of mass were consistent with any other observations.

Furthermore, you'd be stuck trying to explain why redshift is dependent on distance for external galaxies.

tommac
2008-Apr-20, 02:45 AM
I think you could calculate what the mass would be to produce a gravitational well so deep that the cosmological redshifts were caused by gravitational redshift. Then you could check to see if that amount of mass were consistent with any other observations.

Furthermore, you'd be stuck trying to explain why redshift is dependent on distance for external galaxies.


Because they would be moving away from us the quickest ...
my thought was that as space-time compressed for us things would move farther away because our ruler says so. Since they were moving further away ( farthest ones would be effected more ) they would be more redshifted. At least that was where I was going with this

grant hutchison
2008-Apr-20, 05:57 PM
For me, a big part of this question is to determine what the question is.

An astronaut standing on the surface of the Moon is in the Moon's gravity
well. He is at some particular depth in that well. If I'm not mistaken, at
the same time, he is even deeper in Earth's gravity well than he is in the
Moon's gravity well. He is certainly deeper in the Sun's gravity well than
he is in either the Moon's or the Earth's.Doing the sums, it seems your potential energy is more negative if you stand on the moon in an otherwise massless Universe, than if you float in free space at the moon's distance from Earth, in an otherwise massless Universe. So the progression doesn't seem to hold as you describe.


To me, that suggests a basis for applying Mach's principle.How?

Grant Hutchison

Jeff Root
2008-Apr-20, 09:36 PM
Doing the sums, it seems your potential energy is more negative if you
stand on the moon in an otherwise massless Universe, than if you float
in free space at the moon's distance from Earth, in an otherwise massless
Universe. So the progression doesn't seem to hold as you describe.
Okay, I indicated that I wasn't sure about that step, and said as much
to tommac in my first post:


The progression you described doesn't necessarily continue, nor does
each step necessarily involve "stronger" gravity than the previous step.
But that does appear to be a general tendency.




To me, that suggests a basis for applying Mach's principle.
How?
If the inertia of a mass is the result of some interaction between the
mass and its environment, it is clearly not the nearby environment
which is most significant. But it might be the result of an interaction
between the mass and the "total" gravitational potential of the entire
Universe, as measured at that location in space. Local contributions
to the total could be relatively minor, accounting for the apparent
invariant nature of inertia.

I haven't given this speculation much thought in the last 30 years.
In particular, I haven't tried to connect it to the Pioneer anomaly.

-- Jeff, in Minneapolis