Gravity inside mass
littlejoe
Ie. if there were a hole to the earth's center then
how much would the value of "g" be at point 3/4r, 1/2r, 1/4r ?
N:dlzc D:aol T:com (dlzc)
"littlejoe" <litt...@littlejoelittlejoe.inv> wrote in message
news:f6212e$h66$1...@aioe.org...
> How is the equation for calculating the gravity for inside the
> earth?
g = GM/r^2, where M is the mass contained in the sphere described
by r.
> Ie. if there were a hole to the earth's center then
> how much would the value of "g" be at point 3/4r, 1/2r, 1/4r ?
Unfortunately, the Earth is not homogeneous. But it is probably
close enough...
http://www.madsci.org/posts/archives/oct98/904616994.Ph.r.html
... and you can get the average density of the Earth here:
http://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html
David A. Smith
arvee
wrote:
> How is the equation for calculating the gravity for inside the earth?
> Ie. if there were a hole to the earth's center then
> how much would the value of "g" be at point 3/4r, 1/2r, 1/4r ?
If you imagine a spherical shell (centered at the earth's center)
inside the earth, the gravitational force at the surface of that
sphere is equal to the force that would be exerted by a point mass at
the center, with mass equal to that contained within the whole shell.
In particular, all the mass outside the shell has a total contribution
of ZERO. Assuming (somewhat unrealistically) that the mass density is
uniform throughout the interior of the earth, the mass contained
within a spherical shell of radius r is proportional to r^3, while the
force varies with distance by 1/r^2, so altogether the force varies as
r^3/r^2 = r. Therefore, at R/2 from the center, the force is 1/2 that
at the surface.
For more details, Google Gauss's Law.
R.G. Vickson
R.G. Vickson
littlejoe
Hmm, I must admit I had expected a somewhat non-linear relation.
Florian
BTW, did you know that the peak of gravity is in the mantle, not at the
surface? This is because the core is much denser.
--
Florian
"Tout est au mieux dans le meilleur des mondes possibles"
Voltaire vs Leibniz (1-0)
littlejoe
> littlejoe <litt...@littlejoelittlejoe.inv> wrote:
>
> > How is the equation for calculating the gravity for inside the earth?
> > Ie. if there were a hole to the earth's center then
> > how much would the value of "g" be at point 3/4r, 1/2r, 1/4r ?
>
> BTW, did you know that the peak of gravity is in the mantle, not at the
> surface? This is because the core is much denser.
Yes, that fact I already knew.
Martin Hogbin
"littlejoe" <litt...@littlejoelittlejoe.inv> wrote in message news:f62dnf$qjg$1...@aioe.org...
Yes, the result is somewhat surprising but that is indeed the
answer to your question.
--
Martin Hogbin
>
Pmb
"littlejoe" <litt...@littlejoelittlejoe.inv> wrote in message
What fact is it that you know?? The gravitational field is always larger at
the surface of a spherically centered body. If you have a example please
provide, e.g. let me know what the mass density is. Then I can calculate the
gravitational field strength (Usinbg Newtonian mechanics) as a function of
radius.
Thank you
Pete
PD
wrote:
> How is the equation for calculating the gravity for inside the earth?
> Ie. if there were a hole to the earth's center then
> how much would the value of "g" be at point 3/4r, 1/2r, 1/4r ?
Gauss' law tells you that (aside from any anisotropies), gravity will
increase linearly, from 0 at the center of the earth to full g at the
surface.
So at (1/2)R, the acceleration due to gravity is (1/2)g.
At (3/4)R, the acceleration due to gravity is (3/4)g.
Deriving why this should be the case takes a little algebra, but
that's the result.
PD
PD
> "littlejoe" <little...@littlejoelittlejoe.inv> wrote in message
>
> news:f632gi$5d5$1...@aioe.org...
>
> > "Florian" <firstn...@lastname.net> wrote in message
> >news:1i0grwo.mjwrsy1dewk7kN%firs...@lastname.net...
> >> littlejoe <little...@littlejoelittlejoe.inv> wrote:
>
> >> > How is the equation for calculating the gravity for inside the earth?
> >> > Ie. if there were a hole to the earth's center then
> >> > how much would the value of "g" be at point 3/4r, 1/2r, 1/4r ?
>
> >> BTW, did you know that the peak of gravity is in the mantle, not at the
> >> surface? This is because the core is much denser.
>
> > Yes, that fact I already knew.
>
> What fact is it that you know?? The gravitational field is always larger at
> the surface of a spherically centered body. If you have a example please
> provide, e.g. let me know what the mass density is. Then I can calculate the
> gravitational field strength (Usinbg Newtonian mechanics) as a function of
> radius.
>
> Thank you
>
> Pete
OK, consider a spherical body whose density falls off with radius as P/
r^3, up to an outer radius a. Find the acceleration due to gravity at
r=a and r<a and show that the acceleration due to gravity is always
greatest at r=a.
PD
OG
"littlejoe" <litt...@littlejoelittlejoe.inv> wrote in message
> How is the equation for calculating the gravity for inside the earth?
> Ie. if there were a hole to the earth's center then
> how much would the value of "g" be at point 3/4r, 1/2r, 1/4r ?
>
Because the density of the earth varies significantly from crust (typically
2.5 g/cm^3) to core (~ 13g/cm^3) there isn't a smooth variation with
increasing depth - rather the value of 'g' increases, then decreases, then
increases again to a maximum at about r/2 before reducing sharply to zero at
the centre of the earth
The following shows the relationship
http://splung.com/kinematics/images/gravitation/variation%20of%20g.png
which is an image from this webpage
http://splung.com/kinematics/gravitation.htm
Pmb
"PD" <TheDrap...@gmail.com> wrote in message
news:1183135227.9...@m36g2000hse.googlegroups.com...
Why? You're agreeing with me. I said that the gravitational field is largest
at the surface and when you say the outere radius is a that means the
surface is located at r = a where I said that the highest gravitational
field is.
Pete
Florian
> What fact is it that you know?? The gravitational field is always larger at
> the surface of a spherically centered body. If you have a example please
> provide, e.g. let me know what the mass density is. Then I can calculate the
> gravitational field strength (Usinbg Newtonian mechanics) as a function of
> radius.
>
> Thank you
help yourself:
http://nachon.free.fr/densityearth.gif
OG
"littlejoe" <litt...@littlejoelittlejoe.inv> wrote in message
In real life it is non linear - because the mass-density of the earth ISN'T
uniform .
PD
> "PD" <TheDraperFam...@gmail.com> wrote in message
You sure? Take a radius r=b<a. The increase in mass from b to a is
less than the decrease due to the 1/r^2 fall-off from b to a. I'm just
inviting you to be sure.
PD
Pmb
Yes.
> Take a radius r=b<a. The increase in mass from b to a is
> less than the decrease due to the 1/r^2 fall-off from b to a. I'm just
> inviting you to be sure.
The inner sphere of matter would act like it was comming from a point. Let
the field strength be g1. Now take the rest of the matter and make a sphere
out out of it such that the inner radius contains the mass which created the
g1 field. The matter due to the shell (Let's call that the matter added to
increase the radius of r) by itself has a gravitational field of g2 =
GM2/r^2. The total gravitational field at a point outside the sphere
increases from the value g1 to the value g1+g2 when you surround that with
the mass shell. Therefore the g-field must increase the further you get from
the center.
Pete
Burt...@gmail.com
wrote:
> Ie. if there were a hole to the earth's center then
> how much would the value of "g" be at point 3/4r, 1/2r, 1/4r ?
What is the space-time curvature there?
If space does not curve can we assume no time dilation?
Tom Roberts
>>>>> The gravitational field is always
>>>>> larger at the surface of a spherically centered body.
>>> largest
>>> at the surface
> the center.
Consider the earth plus its atmosphere as a "spherically centered body".
Your claim that the g-field is largest at the top of the atmosphere is
contradicted by measurements. It should be clear that if the density of
a spherically-symmetric object falls off rapidly enough with radius that
the maximum gravitational field [#] occurs inside the object, not at its
surface.
Exercise for the reader: determine the functional dependence
on radius required for this to occur.
[#] meaning the Newtonian gravitational force.
Tom Roberts
littlejoe
>
> "littlejoe" <litt...@littlejoelittlejoe.inv> wrote in message
> news:f632gi$5d5$1...@aioe.org...
> > "Florian" <firs...@lastname.net> wrote in message
> > news:1i0grwo.mjwrsy1dewk7kN%firs...@lastname.net...
> >> littlejoe <litt...@littlejoelittlejoe.inv> wrote:
> >>
> >> > How is the equation for calculating the gravity for inside the earth?
> >> > Ie. if there were a hole to the earth's center then
> >> > how much would the value of "g" be at point 3/4r, 1/2r, 1/4r ?
> >>
> >> BTW, did you know that the peak of gravity is in the mantle, not at the
> >> surface? This is because the core is much denser.
> >
> > Yes, that fact I already knew.
>
> What fact is it that you know??
That for a sphere body with uniformly distributed mass
g is highest at the crust.
> The gravitational field is always larger at
> the surface of a spherically centered body. If you have a example please
> provide, e.g. let me know what the mass density is. Then I can calculate the
> gravitational field strength (Usinbg Newtonian mechanics) as a function of
> radius.
A sphere with an exponential mass density function (ie. starting in the center).
How is g in this case calculated?
Pmb
"Tom Roberts" <tjrobe...@sbcglobal.net> wrote in message
news:QZjhi.36248$Um6....@newssvr12.news.prodigy.net...
> Pmb wrote:
>>>>>> The gravitational field is always larger at the surface of a
>>>>>> spherically centered body.
>>>> I said that the gravitational field is largest
>>>> at the surface
>> Therefore the g-field must increase the further you get from the center.
>
> Consider the earth plus its atmosphere as a "spherically centered body".
> Your claim that the g-field is largest at the top of the atmosphere is
> contradicted by measurements. It should be clear that if the density of a
> spherically-symmetric object falls off rapidly enough with radius that the
> maximum gravitational field [#] occurs inside the object, not at its
> surface.
Yep. You're right. I had a feeling that was wrong but I couldn't figure
what? Thanks.
Pete
Bilge
>
> "PD" <TheDrap...@gmail.com> wrote in message
> news:1183153339.6...@n60g2000hse.googlegroups.com...
>>> > OK, consider a spherical body whose density falls off with radius as P/
>>> > r^3, up to an outer radius a. Find the acceleration due to gravity at
>>> > r=a and r<a and show that the acceleration due to gravity is always
>>> > greatest at r=a.
>>>
>>> Why? You're agreeing with me. I said that the gravitational field is
>>> largest
>>> at the surface and when you say the outere radius is a that means the
>>> surface is located at r = a where I said that the highest gravitational
>>> field is.
>>>
>>
>> You sure?
>
> Yes.
Nyet. Surely you can imagine a mass with a radius R and a density
function \rho(r) that gives a maximum field for some r < R.
Pmb
"Bilge" <dub...@radioactivex.sz> wrote in message
news:slrnf8cea4....@iris.lebesque-al.net...
Given the example that Tom gave, and I now see the error of my ways, then
yes. I'm sure I could. My logic was flawed befoire and I know know what that
flaw is. Thanks Bilge
Best regards
Pete
Pete
Bilge
>
> "Bilge" <dub...@radioactivex.sz> wrote in message
> news:slrnf8cea4....@iris.lebesque-al.net...
>> On 2007-06-30, Pmb <som...@somewhere.net> wrote:
>>>
>>> "PD" <TheDrap...@gmail.com> wrote in message
>>> news:1183153339.6...@n60g2000hse.googlegroups.com...
>> [...]
>>>>> > OK, consider a spherical body whose density falls off with radius as
>>>>> > P/
>>>>> > r^3, up to an outer radius a. Find the acceleration due to gravity at
>>>>> > r=a and r<a and show that the acceleration due to gravity is always
>>>>> > greatest at r=a.
>>>>>
>>>>> Why? You're agreeing with me. I said that the gravitational field is
>>>>> largest
>>>>> at the surface and when you say the outere radius is a that means the
>>>>> surface is located at r = a where I said that the highest gravitational
>>>>> field is.
>>>>>
>>>>
>>>> You sure?
>>>
>>> Yes.
>>
>> Nyet. Surely you can imagine a mass with a radius R and a density
>> function \rho(r) that gives a maximum field for some r < R.
>
> Given the example that Tom gave, and I now see the error of my ways, then
> yes. I'm sure I could. My logic was flawed befoire and I know know what that
> flaw is. Thanks Bilge
>
You're welcome.
toxic
wrote:
> Dear litteljoe:
>
> "littlejoe" <little...@littlejoelittlejoe.inv> wrote in message
>
> news:f6212e$h66$1...@aioe.org...
>
> > How is the equation for calculating the gravity for inside the
> > earth?
>
> g = GM/r^2, where M is the mass contained in the sphere described
> by r.
>
> > Ie. if there were a hole to the earth's center then
> > how much would the value of "g" be at point 3/4r, 1/2r, 1/4r ?
draw a line throo
x:6.37e+6 y:9.829
x:0 y:0
>
> Unfortunately, the Earth is not homogeneous. But it is probably
> close enough...http://www.madsci.org/posts/archives/oct98/904616994.Ph.r.html
>
> ... and you can get the average density of the Earth here:http://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html
>
> David A. Smith
now show him all that using geometry
not simple newtonians
wahaahaahaha
toxic
wrote:
> How is the equation for calculating the gravity for inside the earth?
> Ie. if there were a hole to the earth's center then
> how much would the value of "g" be at point 3/4r, 1/2r, 1/4r ?
amazingly no white man relativists around
here do tha right answar ta you
thay use simple newton in stead of more
right geometry, which has ta be used
in order ta derive gravity, which is
geometry thay say
ahaha AHAHA
toxic
> On 2007-06-30, Pmb <som...@somewhere.net> wrote:
>
>
>
>
>
> >news:1183153339.6...@n60g2000hse.googlegroups.com...
> [...]
> >>> > OK, consider a spherical body whose density falls off with radius as P/
> >>> > r^3, up to an outer radius a. Find the acceleration due to gravity at
> >>> > r=a and r<a and show that the acceleration due to gravity is always
> >>> > greatest at r=a.
>
> >>> Why? You're agreeing with me. I said that the gravitational field is
> >>> largest
> >>> at the surface and when you say the outere radius is a that means the
> >>> surface is located at r = a where I said that the highest gravitational
> >>> field is.
>
> >> You sure?
>
> > Yes.
>
> Nyet. Surely you can imagine a mass with a radius R and a density
> function \rho(r)
fine, what is that function
Pmb
>
> "Bilge" <dub...@radioactivex.sz> wrote in message
> news:slrnf8cea4....@iris.lebesque-al.net...
>> On 2007-06-30, Pmb <som...@somewhere.net> wrote:
>>>
>>> "PD" <TheDrap...@gmail.com> wrote in message
>>> news:1183153339.6...@n60g2000hse.googlegroups.com...
>> [...]
>>>>> > OK, consider a spherical body whose density falls off with radius as
>>>>> > P/
>>>>> > r^3, up to an outer radius a.
Actually I did try that last night and what I got was a divergent integral.
That's why I didn't do the problem outright. So I took another path which
was a false lead.
Can someone try using Gauss's law using a density proportional to 1/r^3
please?
Thanks
Pete
OG
Surely any such density/radius rule runs into problems as r tends towards
zero.
You can do it with calculations performed for discrete shells using the
tabulated min-max densities for crust, mantle and core.
Have a look at the tabulated densities provided by florian.
You'll end up with a graph similar to
http://splung.com/kinematics/images/gravitation/variation%20of%20g.png
The max value of 'g' is at about r/2 where 'g' is about 5% more than at the
surface.
adam.tre...@web.de
> What is the space-time curvature there?
Something like this:
http://www.adamtoons.de/physics/gravitation.swf
The Gaussian curvature is negative outside the mass, and positive
inside.
> If space does not curve can we assume no time dilation?
If by "space" you mean "spacetime", and by "time dilation" you mean
"gravitational time dilation of clocks at rest", then the answer is
"yes".
adam.tre...@web.de
> How is g in this case calculated?
If R is the radius of the sphere, and r (with r < R) the distance from
center at wich you want to calculate g(r), then you can divide the
sphere in two parts:
1) A core sphere with radius = r
2) The rest (an outer hull)
It turns out that 2) doesn't contribute to g(r) at all. So with your
mass density function you compute the mass of 1) and then the g on its
surface, as if 2) wasn't there.
Igor
Apparently, we've found one that can't imagine it.
wolfgang
imagine?
write it down fool, or do tha code, any code, also
pseudocode if you are ashamed of your self
but do it properly in tha right way using geometry
and relativity, not newton and shit
Igor
I don't do other people's homework for them. If you want to know what
a density function is you need to look it up.
The TimeLord
> On Jul 2, 3:46 pm, wolfgang <e6k8s...@registerednurses.com> wrote:
>> On Jul 1, 8:19 pm, Igor <thoov...@excite.com> wrote:
>>> On Jun 30, 4:18 pm, toxic <e6k8s...@registerednurses.com> wrote:
>>>> On Jun 30, 1:06 pm, Bilge <dubi...@radioactivex.sz> wrote:
>>>>> On 2007-06-30, Pmb <som...@somewhere.net> wrote:
>>>>>> "PD" <TheDraperFam...@gmail.com> wrote in message
>>>>>> news:1183153339.6...@n60g2000hse.googlegroups.com...
>>>>> [...]
>>>>>>>>> OK, consider a spherical body whose density falls off with radius as P/
>>>>>>>>> r^3, up to an outer radius a. Find the acceleration due to gravity at
>>>>>>>>> r=a and r<a and show that the acceleration due to gravity is always
>>>>>>>>> greatest at r=a.
>>>>>>>> Why? You're agreeing with me. I said that the gravitational field is
>>>>>>>> largest
>>>>>>>> at the surface and when you say the outere radius is a that means the
>>>>>>>> surface is located at r = a where I said that the highest gravitational
>>>>>>>> field is.
>>>>>>> You sure?
>>>>>> Yes.
>>>>> Nyet. Surely you can imagine a mass with a radius R and a density
>>>>> function \rho(r)
>>>> fine, what is that function
>>> Apparently, we've found one that can't imagine it.
>> imagine?
>>
>> write it down fool, or do tha code, any code, also
>> pseudocode if you are ashamed of your self
> I don't do other people's homework for them. If you want to know what
> a density function is you need to look it up.
So much written, yet so little said. Ok, here we go:
Case 1 - Homogeneous Sphere
The density is
rho = M / (4/3 Pi R^3)
So the mass at distance r from the center is
m(r) = rho * (4/3 Pi r^3) = M (r/R)^3
So the classical gravitational acceleration is
a(r) = G*m(r)/r^2 = G*M/R^3 * r
since by Gauss' Theorem any mass outside r has no effect.
Note that as r->0, a->0. So that at the center of a homogeneous sphere
the gravity is zero.
Case 2 - Inhomogeneous Sphere
We can follow everything above up to
a(r) = G*m(r)/r^2
At this point we have to stop, since the density formula might not
apply if the mass is not homogeneous. However, if we know dm/dr for
the sphere, then we can use L'Hospital's rule to get a rate of
convergence for a(r):
a(r) = -2*G*(dm/dr)/r^3
Generally dm/dr will converge to zero faster than r^3 as r->0 even in
inhomogeneous spheres. So we end up with the same result that gravity
in an inhomogeneous sphere is also zero at the center.
Hope this helps, everyone.
--
// The TimeLord says:
// Pogo 2.0 = We have met the aliens, and they are us!
The TimeLord
> "Pmb" <som...@somewhere.net> wrote in message news:0qOdneHUEbO2rxjb...@comcast.com...
>> "littlejoe" <litt...@littlejoelittlejoe.inv> wrote in message
>> news:f632gi$5d5$1...@aioe.org...
>>> "Florian" <firs...@lastname.net> wrote in message
>>> news:1i0grwo.mjwrsy1dewk7kN%firs...@lastname.net...
>>>> littlejoe <litt...@littlejoelittlejoe.inv> wrote:
>>>>
> A sphere with an exponential mass density function (ie. starting in the center).
> How is g in this case calculated?
>
This seems like an unrealistic situation, but I assume you mean
something like
m(r) = (M/R)*e^(-r/R)
where M is total mass and R is some sort of characteristic radius for
the spherical form.
Then
a(r) = G*m(r)/r^2
since it is spherical in shape and the simplification from Gauss'
Theorem applies. So then
a(r) = g(r) = G*(M/R)*e^(-r/R) / r^2
The reason this is so simple, is that we are dealing with spherical
shapes. If it weren't spherical, then you'd have to do it the long
(and hard) way.
The TimeLord
> On Jun 28, 9:16 pm, "littlejoe" <little...@littlejoelittlejoe.inv>
> wrote:
>> How is the equation for calculating the gravity for inside the earth?
>> Ie. if there were a hole to the earth's center then
>> how much would the value of "g" be at point 3/4r, 1/2r, 1/4r ?
>
> What is the space-time curvature there?
I had understood that the whole thread assumed weak-field. If the
gravitation is strong enough to do significant space-time curvature,
then you have to resort to much, much more complicated analyses for
anything inside the event horizon, since Newtonian stuff won't apply
any more.
> If space does not curve can we assume no time dilation?
>
Well, yeah, as far as gravitation is concerned. That is not to say
that there may be time dilation from fast motion past the source of
gravity, as predicted by SR.
Tom Roberts
> I had understood that the whole thread assumed weak-field. If the
> gravitation is strong enough to do significant space-time curvature,
> then you have to resort to much, much more complicated analyses for
> anything inside the event horizon, since Newtonian stuff won't apply
> any more.
Yes. Indeed it's much more complicated everywhere.
> Burt...@gmail.com wrote in sci.physics.relativity:
>> If space does not curve can we assume no time dilation?
>
> Well, yeah, as far as gravitation is concerned.
Not true. In the Newtonian approximation to GR, the metric components in
the usual Newtonian coordinates are just the Minkowski values, EXCEPT
for g_tt = (1 + 2 \phi), where \phi is the Newtonian gravitational
potential, and I'm using units with G=c=1.
So there is "gravitational time dilation" present. This should be
obvious, as numerous experiments have measured it.
Tom Roberts
Ahmed Ouahi, Architect
Obviously!
However, that a gravity makes anything to move, along a fastidious
attraction, as for instance, a molecule which is a mass, which contains
atoms reacting together, along whatever phase.
Therefore, specifically as an atoms of hydrogen added to one atom of oxygen,
does makes a molecule of a water, but along that matter, would remains the
following question.
However, that would be, is it the gravity the source of the speed or the
source of the stability as in an either case, it has to be paid an attention
to which is really needed, along an appropriate regulations along the nature
all along, whether the one does need the other, and this is what is all
about, a definitely as a matter a fact.
--
Ahmed Ouahi, Architect
Best Regards!
"Tom Roberts" <tjrobe...@sbcglobal.net> wrote in message
news:bDPii.20524$RX....@newssvr11.news.prodigy.net...
The TimeLord
>> Burt...@gmail.com wrote in sci.physics.relativity:
>>> If space does not curve can we assume no time dilation?
>>
>> Well, yeah, as far as gravitation is concerned.
>
> Not true. In the Newtonian approximation to GR, the metric components in
> the usual Newtonian coordinates are just the Minkowski values, EXCEPT
> for g_tt = (1 + 2 \phi), where \phi is the Newtonian gravitational
However, then the space-time would be curved. The question was, if
space-time is _not_ curved. If it is not curved, then in any sensible
metric g_tt would be a constant.
> potential, and I'm using units with G=c=1.
>
> So there is "gravitational time dilation" present. This should be
> obvious, as numerous experiments have measured it.
I agree for curved space-time, but the question was if there was _no_
space-time curvature. In that case, we are back to flat metric as in
SR. Then the only time dilation could only be due to motion.
Tom Roberts
> Tom Roberts wrote:
>>> Burt...@gmail.com wrote in sci.physics.relativity:
>>>> If space does not curve can we assume no time dilation?
>>> Well, yeah, as far as gravitation is concerned.
>> Not true. In the Newtonian approximation to GR, the metric components in
>> the usual Newtonian coordinates are just the Minkowski values, EXCEPT
>> for g_tt = (1 + 2 \phi), where \phi is the Newtonian gravitational
>
> However, then the space-time would be curved. The question was, if
> space-time is _not_ curved. If it is not curved, then in any sensible
> metric g_tt would be a constant.
OK. If spacetime is not curved, there is no time dilation, and no
gravity. I misread.
Tom Roberts
Pmb
"The TimeLord" <mathnphy...@bellsouth.net> wrote in message
news:NvWdncXevL5jphHb...@comcast.com...
> Tom Roberts wrote:
>>> Burt...@gmail.com wrote in sci.physics.relativity:
>>>> If space does not curve can we assume no time dilation?
>>>
>>> Well, yeah, as far as gravitation is concerned.
>>
>> Not true. In the Newtonian approximation to GR, the metric components in
>> the usual Newtonian coordinates are just the Minkowski values, EXCEPT
>> for g_tt = (1 + 2 \phi), where \phi is the Newtonian gravitational
>
> However, then the space-time would be curved. The question was, if
> space-time is _not_ curved. If it is not curved, then in any sensible
> metric g_tt would be a constant.
Not necessarily. Take for instance a uniformly accelerating frame of
reference in flat spacetime. g_tt is then a function of position, namely the
value of z if the acceleration is in the z-direction.
Pete