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Spoonfed (www.spoonfedrel
science forum Guru Wannabe
Joined: 28 Apr 2006
Posts: 144
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 Posted: Thu Jun 22, 2006 4:02 am Post subject:
Challenging exercises in relativity theory
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Here are three fairly challenging but not impossible problems to try.
I think exercise 3 is the easiest, as there are no numerical
calculations to perform, though I expect some disagreement as to the
answer to 3a.
Exercise 1.
Two rocketships leave earth simultaneously at .866c, at 90 degree angle
from one another. (i.e. straight north and straight west.) Later, a
message is beamed from earth to the first rocket, to the second rocket,
and back to earth. Calculate the sum of the angles of this triangle by
considering the angle between the earth and the other rocket from the
frame of each rocket. Would the result suggest that the universe is
closed like a sphere (positive curvature), flat (0 curvature) or open
(negative curvature?)
Exercise 2.
Assume that you have n particles of mass m and the momentum of the nth
particle is equal to h*n, where h is constant.
a) Calculate the velocity of the nth particle.
b) Assuming these particles are all traveling in the same direction.
What is the difference in velocity between particle n and particle n+1?
c) Assume the particles all started at the same point in space. What
is the distance between particle n and particle n+1 at time t=1?
d) What is the limit of the distance between particles as n-->infinity?
What is the density in that region?
Exercise 3.
a) If you were to calculate the force of gravity at point P due to an
arbitrary distribution of masses, and you chose to set up a spherical
coordinate system, what point should you use to determine the center of
that coordinate system?
b) Read the following and determine what point Einstein used as the
center of this spherical coordinate system to find an infinite force at
point P.
Theory: "The stellar universe ought to be a finite island in the
infinite ocean of space." [1]
Proof: "According to the theory of Newton, the number of "lines of
force" which come from infinity and terminate in a mass m is
proportional to the mass m is proportional to the mass m. If on the
average, the mass-density rho_0 is constant throughout the universe,
then a sphere of volume V will enclose the mass density rho_0*V. Thus
the number of lines of force passing through the surface F of the
sphere into its interior is proportional to rho_0V. For unit area of
the surface of the sphere the number of lines of force which enters the
sphere is thus proportional to rho_0 * volume/(surface area) or to
rho_0*R. Hence the intensity of the field at the surface would
ultimately become infinite with increasing radius R of the sphere,
which is impossible.
[1] From Albert Einstein, Relativity: The special and General Theory,
translated by R.W.Lawson, 1920, Part III, Chapters XXX, XXXI, XXXII.
Reprinted in "Theories of the Universe" edited by Milton K. Munitz. |
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Sue...
science forum Guru
Joined: 08 May 2005
Posts: 2684
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| Posted: Thu Jun 22, 2006 5:00 am Post subject:
Re: Challenging exercises in relativity theory
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Spoonfed wrote:
| Quote: | Here are three fairly challenging but not impossible problems to try.
I think exercise 3 is the easiest, as there are no numerical
calculations to perform, though I expect some disagreement as to the
answer to 3a.
Exercise 1.
Two rocketships leave earth simultaneously at .866c, at 90 degree angle
from one another. (i.e. straight north and straight west.) Later, a
message is beamed from earth to the first rocket, to the second rocket,
and back to earth. Calculate the sum of the angles of this triangle by
considering the angle between the earth and the other rocket from the
frame of each rocket. Would the result suggest that the universe is
closed like a sphere (positive curvature), flat (0 curvature) or open
(negative curvature?)
|
Why would you be comparing displacement in Euclidian space
with a topology that maps energy density as a function of
volume ?
<< ... a general Lorentz transformation preserves the volume of
space-time. Since time is dilated by a factor gamma in a moving
frame, the volume of space-time can only be preserved if the volume
of ordinary 3-space is reduced by the same factor. As is well known,
this is achieved by length contraction along the direction of motion
by a factor gamma. >>
http://farside.ph.utexas.edu/teaching/jk1/lectures/node14.html
http://map.gsfc.nasa.gov/m_uni/uni_101bb2.html
http://www.astro.ucla.edu/~wright/cosmo_04.htm
http://arxiv.org/abs/physics/0204034
| Quote: |
Exercise 2.
Assume that you have n particles of mass m and the momentum of the nth
particle is equal to h*n, where h is constant.
a) Calculate the velocity of the nth particle.
b) Assuming these particles are all traveling in the same direction.
What is the difference in velocity between particle n and particle n+1?
c) Assume the particles all started at the same point in space. What
is the distance between particle n and particle n+1 at time t=1?
d) What is the limit of the distance between particles as n-->infinity?
What is the density in that region?
|
Something missing here? Like at rocket motor or a gun?
| Quote: |
Exercise 3.
a) If you were to calculate the force of gravity at point P due to an
arbitrary distribution of masses, and you chose to set up a spherical
coordinate system, what point should you use to determine the center of
that coordinate system?
|
Arbitrary? random? homogenous?
There are n number of spherical co-ordinate systems.
Baloons? Rasin cakes?
This one finds wide application.
http://hyperphysics.phy-astr.gsu.edu/hbase/forces/isq.html#isq
Walk on the wild side?
http://www.physics.nyu.edu/faculty/sokal/transgress_v2/node4.html
| Quote: | b) Read the following and determine what point Einstein used as the
center of this spherical coordinate system to find an infinite force at
point P.
Theory: "The stellar universe ought to be a finite island in the
infinite ocean of space." [1]
Proof: "According to the theory of Newton, the number of "lines of
force" which come from infinity and terminate in a mass m is
proportional to the mass m is proportional to the mass m.
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Hmmm... I am not sure that is deriveable from 1/r^2.
Did Newton really say that? It sounds more like a Mach musing.
| Quote: | If on the
average, the mass-density rho_0 is constant throughout the universe,
then a sphere of volume V will enclose the mass density rho_0*V. Thus
the number of lines of force passing through the surface F of the
sphere into its interior is proportional to rho_0V. For unit area of
the surface of the sphere the number of lines of force which enters the
sphere is thus proportional to rho_0 * volume/(surface area) or to
rho_0*R. Hence the intensity of the field at the surface would
ultimately become infinite with increasing radius R of the sphere,
which is impossible.
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So the lines have to taper by 1/r^2. Newton knew that.
We should check the good professors references where
he is quoting Newton. Maybe somthing was taken out
of context.
Sue...
| Quote: |
[1] From Albert Einstein, Relativity: The special and General Theory, |
translated by R.W.Lawson, 1920, Part III, Chapters XXX, XXXI, XXXII.
Reprinted in "Theories of the Universe" edited by Milton K. Munitz.
http://www.bartleby.com/173/ |
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Sue...
science forum Guru
Joined: 08 May 2005
Posts: 2684
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| Posted: Thu Jun 22, 2006 5:34 am Post subject:
Re: Challenging exercises in relativity theory
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Spoonfed wrote something nice but Sue... snipped it  ) :
What many theorists forget to do is look at the reciprocal
view of this:
http://hyperphysics.phy-astr.gsu.edu/hbase/forces/isq.html#isq
The night sky can fool your mental model because you see
varying luminance with no depth perception.
It helps to apply some rigour to the concept of homogenity
by refering to illustration above. If you are standing at
point S, the deeper you look into space, the more objects
will be in the aperture area A. So as the radiaton decreases
by the square root, the number or radiators increases by
the square. You don't perceive that by spatially integrating
all points of light in the night sky but a massive object
does exactly that by integrating all the gravitational forces.
So... it becomes hard to see how you derive a cosmological
constant from something that can be replaced by a hollow
sphere of some significant wall thickness.
Hubble flows, redshift surveys excepted of course. They provide
a different type of information.
Sue... |
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Spoonfed (www.spoonfedrel
science forum Guru Wannabe
Joined: 28 Apr 2006
Posts: 144
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| Posted: Thu Jun 22, 2006 3:03 pm Post subject:
Re: Challenging exercises in relativity theory
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Sue... wrote:
| Quote: | Spoonfed wrote:
Here are three fairly challenging but not impossible problems to try.
I think exercise 3 is the easiest, as there are no numerical
calculations to perform, though I expect some disagreement as to the
answer to 3a.
Exercise 1.
Two rocketships leave earth simultaneously at .866c, at 90 degree angle
from one another. (i.e. straight north and straight west.) Later, a
message is beamed from earth to the first rocket, to the second rocket,
and back to earth. Calculate the sum of the angles of this triangle by
considering the angle between the earth and the other rocket from the
frame of each rocket. Would the result suggest that the universe is
closed like a sphere (positive curvature), flat (0 curvature) or open
(negative curvature?)
Why would you be comparing displacement in Euclidian space
with a topology that maps energy density as a function of
volume ?
|
The question is a valid one. What is the sum of the angles of the
triangle?
Hint: it is not 180 degrees.
I'm not looking for more sources. I'm asking a simple question with a
simple answer.
| Quote: |
Exercise 2.
Assume that you have n particles of mass m and the momentum of the nth
particle is equal to h*n, where h is constant.
a) Calculate the velocity of the nth particle.
b) Assuming these particles are all traveling in the same direction.
What is the difference in velocity between particle n and particle n+1?
c) Assume the particles all started at the same point in space. What
is the distance between particle n and particle n+1 at time t=1?
d) What is the limit of the distance between particles as n-->infinity?
What is the density in that region?
Something missing here? Like at rocket motor or a gun?
|
No.
| Quote: |
Exercise 3.
a) If you were to calculate the force of gravity at point P due to an
arbitrary distribution of masses, and you chose to set up a spherical
coordinate system, what point should you use to determine the center of
that coordinate system?
Arbitrary? random? hoogenous?
|
The common General Relativity assumption that this choice can be made
arbitrarily. This was the disagreement I expected for question 1a,
because outside GR, I can't imagine anyone giving that answer. Most
people would surely say that the force at point P should be calculated
relative to point P; not relative to some random position away from
point P.
Yet it is foundational to cosmological GR theory to presume that it is
OKAY to calculate the force at P from a reference point far away from
P. I would suggest that if you are to do something like this, you
should calculate the force from all reference points in the universe,
and then take the average. As a matter of fact, let's add that as
exercise 4, for anyone who gives your answer, Sue.
Exercise 4: For people who answered that you can place the center of
the spherical coordinate system at any point in the universe, calculate
the average force given by all such calculations.
| Quote: | There are n number of spherical co-ordinate systems.
Baloons? Rasin cakes?
This one finds wide application.
http://hyperphysics.phy-astr.gsu.edu/hbase/forces/isq.html#isq
Walk on the wild side?
http://www.physics.nyu.edu/faculty/sokal/transgress_v2/node4.html
b) Read the following and determine what point Einstein used as the
center of this spherical coordinate system to find an infinite force at
point P.
Theory: "The stellar universe ought to be a finite island in the
infinite ocean of space." [1]
Proof: "According to the theory of Newton, the number of "lines of
force" which come from infinity and terminate in a mass m is
proportional to the mass m is proportional to the mass m.
Hmmm... I am not sure that is deriveable from 1/r^2.
Did Newton really say that? It sounds more like a Mach musing.
|
This is apparently Einstein's interpretation of Newton. I agree that
it sounds like a Mach musing. From Albert Einstein, Relativity: The
Special and General Theory,
translated by R.W.Lawson, 1920, Part III, Chapters XXX, XXXI, XXXII.
Reprinted in "Theories of the Universe" edited by Milton K. Munitz.
His wording is odd, but it seems consistent with a spherical coordinate
system with a center at distance r from point P.
| Quote: | If on the
average, the mass-density rho_0 is constant throughout the universe,
then a sphere of volume V will enclose the mass density rho_0*V. Thus
the number of lines of force passing through the surface F of the
sphere into its interior is proportional to rho_0V. For unit area of
the surface of the sphere the number of lines of force which enters the
sphere is thus proportional to rho_0 * volume/(surface area) or to
rho_0*R. Hence the intensity of the field at the surface would
ultimately become infinite with increasing radius R of the sphere,
which is impossible.
So the lines have to taper by 1/r^2. Newton knew that.
We should check the good professors references where
he is quoting Newton. Maybe somthing was taken out
of context.
|
Actually, if you have a sphere of matter of radius R, it can be treated
as a point mass at distance R, so if the density of the sphere is the
same throughout, the mass is proportional to rho_0 * Volume =rho_0 *4/3
Pi R^3.
If you are at any point inside a uniform sphere of any radius, the
force is zero.
So what Einstein is doing is adding up the finite volume centered at
distance R from P, finding a force toward the center of that volume,
and then treating the rest of space as a series of concentric spheres
of uniform density which add nothing to the force.
| Quote: | Sue...
[1] From Albert Einstein, Relativity: The special and General Theory,
translated by R.W.Lawson, 1920, Part III, Chapters XXX, XXXI, XXXII.
Reprinted in "Theories of the Universe" edited by Milton K. Munitz.
http://www.bartleby.com/173/ |
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Sue...
science forum Guru
Joined: 08 May 2005
Posts: 2684
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| Posted: Thu Jun 22, 2006 4:06 pm Post subject:
Re: Challenging exercises in relativity theory
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Spoonfed wrote:
| Quote: | Sue... wrote:
Spoonfed wrote:
Here are three fairly challenging but not impossible problems to try.
I think exercise 3 is the easiest, as there are no numerical
calculations to perform, though I expect some disagreement as to the
answer to 3a.
Exercise 1.
Two rocketships leave earth simultaneously at .866c, at 90 degree angle
from one another. (i.e. straight north and straight west.) Later, a
message is beamed from earth to the first rocket, to the second rocket,
and back to earth. Calculate the sum of the angles of this triangle by
considering the angle between the earth and the other rocket from the
frame of each rocket. Would the result suggest that the universe is
closed like a sphere (positive curvature), flat (0 curvature) or open
(negative curvature?)
Why would you be comparing displacement in Euclidian space
with a topology that maps energy density as a function of
volume ?
The question is a valid one. What is the sum of the angles of the
triangle?
Hint: it is not 180 degrees.
|
Ahhh I just noticed compass points so I could have
assumed Euclidian coordinates.
1080.0 degrees.
1080.0 degrees.
| Quote: |
Exercise 2.
Assume that you have n particles of mass m and the momentum of the nth
particle is equal to h*n, where h is constant.
a) Calculate the velocity of the nth particle.
b) Assuming these particles are all traveling in the same direction.
What is the difference in velocity between particle n and particle n+1?
c) Assume the particles all started at the same point in space. What
is the distance between particle n and particle n+1 at time t=1?
d) What is the limit of the distance between particles as n-->infinity?
What is the density in that region?
Something missing here? Like at rocket motor or a gun?
No.
Sorry... I can't visualise your question. |
| Quote: |
Exercise 3.
a) If you were to calculate the force of gravity at point P due to an
arbitrary distribution of masses, and you chose to set up a spherical
coordinate system, what point should you use to determine the center of
that coordinate system?
Arbitrary? random? hoogenous?
The common General Relativity assumption that this choice can be made
arbitrarily. This was the disagreement I expected for question 1a,
because outside GR, I can't imagine anyone giving that answer. Most
people would surely say that the force at point P should be calculated
relative to point P; not relative to some random position away from
point P.
|
Not just GR but any system of mass you have to have some
notion when the system 'thinks' (locally) it is in equilibrium.
| Quote: |
Yet it is foundational to cosmological GR theory to presume that it is
OKAY to calculate the force at P from a reference point far away from
P. I would suggest that if you are to do something like this, you
should calculate the force from all reference points in the universe,
and then take the average. As a matter of fact, let's add that as
exercise 4, for anyone who gives your answer, Sue.
|
Notice the naughty word 'equilibrium'. There would be no
forces for a system in equilibrium. That is sort of a presumption
of GR anyway. Masses will move along conservative paths.
| Quote: |
Exercise 4: For people who answered that you can place the center of
the spherical coordinate system at any point in the universe, calculate
the average force given by all such calculations.
|
That sounds more like some BB conjecture has to invoked.
| Quote: |
There are n number of spherical co-ordinate systems.
Baloons? Rasin cakes?
This one finds wide application.
http://hyperphysics.phy-astr.gsu.edu/hbase/forces/isq.html#isq
Walk on the wild side?
http://www.physics.nyu.edu/faculty/sokal/transgress_v2/node4.html
b) Read the following and determine what point Einstein used as the
center of this spherical coordinate system to find an infinite force at
point P.
Theory: "The stellar universe ought to be a finite island in the
infinite ocean of space." [1]
Proof: "According to the theory of Newton, the number of "lines of
force" which come from infinity and terminate in a mass m is
proportional to the mass m is proportional to the mass m.
Hmmm... I am not sure that is deriveable from 1/r^2.
Did Newton really say that? It sounds more like a Mach musing.
This is apparently Einstein's interpretation of Newton. I agree that
it sounds like a Mach musing. From Albert Einstein, Relativity: The
Special and General Theory,
translated by R.W.Lawson, 1920, Part III, Chapters XXX, XXXI, XXXII.
Reprinted in "Theories of the Universe" edited by Milton K. Munitz.
His wording is odd, but it seems consistent with a spherical coordinate
system with a center at distance r from point P.
|
Ahhh... here it is:
http://www.bartleby.com/173/30.html
<< This conception is in itself not very satisfactory. It is
still less satisfactory because it leads to the result that the
light emitted by the stars and also individual stars of the stellar
system are perpetually passing out into infinite space, never to
return, and without ever again coming into interaction with other
objects of nature. Such a finite material universe would be
destined to become gradually but systematically impoverished. >>
Free space is 377 ohms. If light encountered 0 ohms or infinity ohims
ir would simply reflect back to the source. Since his premise is
false I don't quite know what you can make of the rest of the musings.
| Quote: |
If on the
average, the mass-density rho_0 is constant throughout the universe,
then a sphere of volume V will enclose the mass density rho_0*V. Thus
the number of lines of force passing through the surface F of the
sphere into its interior is proportional to rho_0V. For unit area of
the surface of the sphere the number of lines of force which enters the
sphere is thus proportional to rho_0 * volume/(surface area) or to
rho_0*R. Hence the intensity of the field at the surface would
ultimately become infinite with increasing radius R of the sphere,
which is impossible.
So the lines have to taper by 1/r^2. Newton knew that.
We should check the good professors references where
he is quoting Newton. Maybe somthing was taken out
of context.
Actually, if you have a sphere of matter of radius R, it can be treated
as a point mass at distance R, so if the density of the sphere is the
same throughout, the mass is proportional to rho_0 * Volume =rho_0 *4/3
Pi R^3.
If you are at any point inside a uniform sphere of any radius, the
force is zero.
|
I would say if matter in your local space is homogenous then
the force is zero and I'll offer a hollow at the center of the earth
as an example.
| Quote: |
So what Einstein is doing is adding up the finite volume centered at
distance R from P, finding a force toward the center of that volume,
and then treating the rest of space as a series of concentric spheres
of uniform density which add nothing to the force.
|
I would say he did it wrong. See following post about 1/r^2 and
and an object count increasing by the square or the distance.
Sue...
| Quote: |
Sue...
[1] From Albert Einstein, Relativity: The special and General Theory,
translated by R.W.Lawson, 1920, Part III, Chapters XXX, XXXI, XXXII.
Reprinted in "Theories of the Universe" edited by Milton K. Munitz.
http://www.bartleby.com/173/ (30 31 32) |
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Ben Rudiak-Gould
science forum Guru
Joined: 04 May 2005
Posts: 382
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| Posted: Thu Jun 22, 2006 6:06 pm Post subject:
Re: Challenging exercises in relativity theory
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Spoonfed wrote:
| Quote: | Exercise 1.
Two rocketships leave earth simultaneously at .866c, at 90 degree angle
from one another. (i.e. straight north and straight west.) Later, a
message is beamed from earth to the first rocket, to the second rocket,
and back to earth. Calculate the sum of the angles of this triangle by
considering the angle between the earth and the other rocket from the
frame of each rocket. Would the result suggest that the universe is
closed like a sphere (positive curvature), flat (0 curvature) or open
(negative curvature?)
|
I don't see why it would suggest anything at all. This is a very strange
calculation you're asking people to perform. It reminds me of Feynman's
anecdote about adding the temperatures of stars. Why would there be any
significance to angles obtained in this way? For one thing, it's not clear
why angles defined by projections of light beams (null lines) onto a 3-plane
would have any significance. For another, the light doesn't care about the
velocity of the rocketships which absorb and re-emit it, yet you're asking
us to calculate angles which depend on those velocities.
| Quote: | Assume that you have n particles of mass m and the momentum of the nth
particle is equal to h*n, where h is constant.
|
If you want a Lorentz-invariant notion of "evenly spaced velocities", this
isn't the way to do it. Exercise: calculate the momenta of particles k-1 and
k+1 with respect to the rest frame of particle k. (They won't be negatives
of each other.) Exercise 2: figure out the Lorentz-invariant distribution.
| Quote: | a) Calculate the velocity of the nth particle.
|
v = p/E and E = sqrt(p^2 + m^2), so v_n = 1 / sqrt(1 + (m/hn)^2).
| Quote: | b) Assuming these particles are all traveling in the same direction.
What is the difference in velocity between particle n and particle n+1?
|
Subtract.
| Quote: | c) Assume the particles all started at the same point in space. What
is the distance between particle n and particle n+1 at time t=1?
|
Don't you want some units on that?
| Quote: | d) What is the limit of the distance between particles as n-->infinity?
What is the density in that region?
|
Zero and infinite respectively.
| Quote: | Exercise 3.
a) If you were to calculate the force of gravity at point P due to an
arbitrary distribution of masses, and you chose to set up a spherical
coordinate system, what point should you use to determine the center of
that coordinate system?
|
Er, it had better not matter, else your theory is inconsistent.
| Quote: | b) Read the following and determine what point Einstein used as the
center of this spherical coordinate system to find an infinite force at
point P.
[...]
|
He didn't find an infinite force at point P. He didn't even mention point P.
I'm still trying to figure out what /your/ point is. Presumably you think
that Einstein's argument is wrong. I tend to agree, but keep in mind that
he was writing a popularization and had to make a nonmathematical argument.
Here's a slightly more formal treatment.
Given a charge (i.e. mass) distribution rho(r), the Newtonian gravitational
field F satisfies
div F = -4 pi G rho.
When rho is constant (a universe filled uniformly with matter), this has the
family of solutions
F(r) = -4/3 pi G rho (r - r0)
where r0 is an arbitrary point in space. r0 is the center of a radial
gravitational field which increases without bound as you move farther from
the center.
So what's the significance of this? You could argue that the gravitational
field is not well-defined, since it depends on this arbitrary point, and you
could use this to make the case for a modification of Newton's theory
(perhaps an added exponential falloff). But there's another way out. The
difference between any two of these fields (with different values of r0) is
a constant field, i.e. you can move r0 around by adding a constant field.
What's the added effect of the constant field? Nothing! A constant
gravitational field filling the universe is the same as no field at all,
since it accelerates everything together, which preserves all spatial
relationships, which are all that you can actually measure. So the point r0
is not a physically measurable quantity, and the family of solutions turns
out to be a single solution after all, in some sense. What's the effect of
that field? Suppose the matter is initially all comoving; then clearly the
field will make it gravitate together. It'll gravitate toward the point r0,
but we just established that the point r0 isn't detectable, so the real
effect is just a general sort of collapse (increasing density) without any
particular center. It's easy to show that the collapse preserves the
constant-density property, and that infinite density is reached in a finite
time (another exercise). The time reversal of this is a Newtonian big bang.
The family of solutions I gave above is incomplete, since you can add any
divergenceless field (not just a constant field) and still satisfy the field
equation. This is sort of a Newtonian version of general covariance. The
Einstein field equations basically equate the matter and energy distribution
to a fancy sort of spacetime divergence, and just as in the Newtonian case
this doesn't completely constrain the field. There are still many solutions
with the same matter content, some of them meaningless reparameterizations
(like the change of r0) and others detectably different (gravitational waves).
Incidentally, I should clarify that Einstein's center point is independent
of r0. That is, you can pick any r0, and define the gravitational field
based on that, and then apply Gauss's law to spheres centered at any other
point, and get the right answer. Einstein's argument doesn't assume any sort
of symmetry of the field around his center point. The only part of the
argument that I disagree with is the bit at the end where he claims that the
mathematical result is physically absurd.
-- Ben |
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Spoonfed (www.spoonfedrel
science forum Guru Wannabe
Joined: 28 Apr 2006
Posts: 144
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| Posted: Thu Jun 22, 2006 8:59 pm Post subject:
Re: Challenging exercises in relativity theory
|
|
|
Sue... wrote:
| Quote: | Spoonfed wrote:
Sue... wrote:
Spoonfed wrote:
Here are three fairly challenging but not impossible problems to try.
I think exercise 3 is the easiest, as there are no numerical
calculations to perform, though I expect some disagreement as to the
answer to 3a.
Exercise 1.
Two rocketships leave earth simultaneously at .866c, at 90 degree angle
from one another. (i.e. straight north and straight west.) Later, a
message is beamed from earth to the first rocket, to the second rocket,
and back to earth. Calculate the sum of the angles of this triangle by
considering the angle between the earth and the other rocket from the
frame of each rocket. Would the result suggest that the universe is
closed like a sphere (positive curvature), flat (0 curvature) or open
(negative curvature?)
Why would you be comparing displacement in Euclidian space
with a topology that maps energy density as a function of
volume ?
The question is a valid one. What is the sum of the angles of the
triangle?
Hint: it is not 180 degrees.
Ahhh I just noticed compass points so I could have
assumed Euclidian coordinates.
1080.0 degrees.
... a general Lorentz transformation preserves the volume of
space-time. Since time is dilated by a factor gamma in a moving
frame, the volume of space-time can only be preserved if the volume
of ordinary 3-space is reduced by the same factor. As is well known,
this is achieved by length contraction along the direction of motion
by a factor gamma.
http://farside.ph.utexas.edu/teaching/jk1/lectures/node14.html
http://map.gsfc.nasa.gov/m_uni/uni_101bb2.html
http://www.astro.ucla.edu/~wright/cosmo_04.htm
http://arxiv.org/abs/physics/0204034
I'm not looking for more sources. I'm asking a simple question with a
simple answer.
1080.0 degrees.
|
That is incorrect.
| Quote: |
Exercise 2.
Assume that you have n particles of mass m and the momentum of the nth
particle is equal to h*n, where h is constant.
a) Calculate the velocity of the nth particle.
b) Assuming these particles are all traveling in the same direction.
What is the difference in velocity between particle n and particle n+1?
c) Assume the particles all started at the same point in space. What
is the distance between particle n and particle n+1 at time t=1?
d) What is the limit of the distance between particles as n-->infinity?
What is the density in that region?
Something missing here? Like at rocket motor or a gun?
No.
Sorry... I can't visualise your question.
|
It's really a plug and chug problem. momentum = mass *
velocity/sqrt(1-(v/c)^2). Except it may take a bit of work to solve
that for velocity.
| Quote: |
Exercise 3.
a) If you were to calculate the force of gravity at point P due to an
arbitrary distribution of masses, and you chose to set up a spherical
coordinate system, what point should you use to determine the center of
that coordinate system?
Arbitrary? random? hoogenous?
The common General Relativity assumption that this choice can be made
arbitrarily. This was the disagreement I expected for question 1a,
because outside GR, I can't imagine anyone giving that answer. Most
people would surely say that the force at point P should be calculated
relative to point P; not relative to some random position away from
point P.
Not just GR but any system of mass you have to have some
notion when the system 'thinks' (locally) it is in equilibrium.
Yet it is foundational to cosmological GR theory to presume that it is
OKAY to calculate the force at P from a reference point far away from
P. I would suggest that if you are to do something like this, you
should calculate the force from all reference points in the universe,
and then take the average. As a matter of fact, let's add that as
exercise 4, for anyone who gives your answer, Sue.
Notice the naughty word 'equilibrium'. There would be no
forces for a system in equilibrium. That is sort of a presumption
of GR anyway. Masses will move along conservative paths.
Exercise 4: For people who answered that you can place the center of
the spherical coordinate system at any point in the universe, calculate
the average force given by all such calculations.
That sounds more like some BB conjecture has to invoked.
|
Use the conjecture given in Einstein's proof. "If on the average, the
mass-density rho_0 is constant throughout the universe"
| Quote: |
There are n number of spherical co-ordinate systems.
Baloons? Rasin cakes?
This one finds wide application.
http://hyperphysics.phy-astr.gsu.edu/hbase/forces/isq.html#isq
Walk on the wild side?
http://www.physics.nyu.edu/faculty/sokal/transgress_v2/node4.html
b) Read the following and determine what point Einstein used as the
center of this spherical coordinate system to find an infinite force at
point P.
Theory: "The stellar universe ought to be a finite island in the
infinite ocean of space." [1]
Proof: "According to the theory of Newton, the number of "lines of
force" which come from infinity and terminate in a mass m is
proportional to the mass m is proportional to the mass m.
Hmmm... I am not sure that is deriveable from 1/r^2.
Did Newton really say that? It sounds more like a Mach musing.
This is apparently Einstein's interpretation of Newton. I agree that
it sounds like a Mach musing. From Albert Einstein, Relativity: The
Special and General Theory,
translated by R.W.Lawson, 1920, Part III, Chapters XXX, XXXI, XXXII.
Reprinted in "Theories of the Universe" edited by Milton K. Munitz.
His wording is odd, but it seems consistent with a spherical coordinate
system with a center at distance r from point P.
Ahhh... here it is:
http://www.bartleby.com/173/30.html
This conception is in itself not very satisfactory. It is
still less satisfactory because it leads to the result that the
light emitted by the stars and also individual stars of the stellar
system are perpetually passing out into infinite space, never to
return, and without ever again coming into interaction with other
objects of nature. Such a finite material universe would be
destined to become gradually but systematically impoverished.
|
Very nice. It appears you've located the article I was referencing.
| Quote: | Free space is 377 ohms. If light encountered 0 ohms or infinity ohims
ir would simply reflect back to the source. Since his premise is
false I don't quite know what you can make of the rest of the musings.
|
I'm not making anything of his musings. I'm just asking what point (or
set of points) Einstein used to calculate infinite force at point P in
his proof. It's a question with a definite answer.
| Quote: |
If on the
average, the mass-density rho_0 is constant throughout the universe,
then a sphere of volume V will enclose the mass density rho_0*V. Thus
the number of lines of force passing through the surface F of the
sphere into its interior is proportional to rho_0V. For unit area of
the surface of the sphere the number of lines of force which enters the
sphere is thus proportional to rho_0 * volume/(surface area) or to
rho_0*R. Hence the intensity of the field at the surface would
ultimately become infinite with increasing radius R of the sphere,
which is impossible.
So the lines have to taper by 1/r^2. Newton knew that.
We should check the good professors references where
he is quoting Newton. Maybe somthing was taken out
of context.
Actually, if you have a sphere of matter of radius R, it can be treated
as a point mass at distance R, so if the density of the sphere is the
same throughout, the mass is proportional to rho_0 * Volume =rho_0 *4/3
Pi R^3.
If you are at any point inside a uniform sphere of any radius, the
force is zero.
I would say if matter in your local space is homogenous then
the force is zero and I'll offer a hollow at the center of the earth
as an example.
|
That sounds reasonable to me. However, Einstein found a way to
mathematically show that the force was arbitrary, depending on the
choice of coordinate system.
| Quote: |
So what Einstein is doing is adding up the finite volume centered at
distance R from P, finding a force toward the center of that volume,
and then treating the rest of space as a series of concentric spheres
of uniform density which add nothing to the force.
I would say he did it wrong. See following post about 1/r^2 and
and an object count increasing by the square or the distance.
|
I would also say he did it wrong, but the problem is that he did what
you are saying right, following the 1/r^2 law, and got it wrong in an
entirely different way.
| Quote: | Sue...
Sue...
[1] From Albert Einstein, Relativity: The special and General Theory,
translated by R.W.Lawson, 1920, Part III, Chapters XXX, XXXI, XXXII.
Reprinted in "Theories of the Universe" edited by Milton K. Munitz.
http://www.bartleby.com/173/ (30 31 32) |
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Phineas T Puddleduck
science forum Guru
Joined: 01 Jun 2006
Posts: 759
|
| Posted: Thu Jun 22, 2006 9:41 pm Post subject:
Re: Challenging exercises in relativity theory
|
|
|
In article <h6Emg.8250$%q2.1441@reader1.news.jippii.net>, Henry
Haapalainen <kirppu@kolumbus.fi> wrote:
| Quote: | Is there any reason to challenge Einstein's relativity any more? Those who
still believe it in its entirety will keep on believing as long as they
live.
Henry Haapalainen
True science lives for challenge - the problem is no one person here |
claiming to challenge relativity even understands it. Some people
challenge it without knowing what proper time and distance is - for
example.
--
The greatest enemy of science is pseudoscience.
Jaffa cakes. Sweet delicious orangey jaffa goodness, and an abject lesson why
parroting information from the web will not teach you cosmology.
Official emperor of sci.physics. Please pay no attention to my butt poking
forward, it is expanding.
Relf's Law?
"Bullshit repeated to the limit of infinity asymptotically approaches
the odour of roses." |
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Henry Haapalainen
science forum Guru
Joined: 07 Jul 2005
Posts: 493
|
| Posted: Thu Jun 22, 2006 9:43 pm Post subject:
Re: Challenging exercises in relativity theory
|
|
|
Is there any reason to challenge Einstein's relativity any more? Those who
still believe it in its entirety will keep on believing as long as they
live.
Henry Haapalainen |
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| Back to top |
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|
Sue...
science forum Guru
Joined: 08 May 2005
Posts: 2684
|
| Posted: Thu Jun 22, 2006 9:51 pm Post subject:
Re: Challenging exercises in relativity theory
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|
|
Spoonfed wrote:
| Quote: | Sue... wrote:
Spoonfed wrote:
Sue... wrote:
Spoonfed wrote:
Here are three fairly challenging but not impossible problems to try.
I think exercise 3 is the easiest, as there are no numerical
calculations to perform, though I expect some disagreement as to the
answer to 3a.
Exercise 1.
Two rocketships leave earth simultaneously at .866c, at 90 degree angle
from one another. (i.e. straight north and straight west.) Later, a
message is beamed from earth to the first rocket, to the second rocket,
and back to earth. Calculate the sum of the angles of this triangle by
considering the angle between the earth and the other rocket from the
frame of each rocket. Would the result suggest that the universe is
closed like a sphere (positive curvature), flat (0 curvature) or open
(negative curvature?)
Why would you be comparing displacement in Euclidian space
with a topology that maps energy density as a function of
volume ?
The question is a valid one. What is the sum of the angles of the
triangle?
Hint: it is not 180 degrees.
Ahhh I just noticed compass points so I could have
assumed Euclidian coordinates.
1080.0 degrees.
... a general Lorentz transformation preserves the volume of
space-time. Since time is dilated by a factor gamma in a moving
frame, the volume of space-time can only be preserved if the volume
of ordinary 3-space is reduced by the same factor. As is well known,
this is achieved by length contraction along the direction of motion
by a factor gamma.
http://farside.ph.utexas.edu/teaching/jk1/lectures/node14.html
http://map.gsfc.nasa.gov/m_uni/uni_101bb2.html
http://www.astro.ucla.edu/~wright/cosmo_04.htm
http://arxiv.org/abs/physics/0204034
I'm not looking for more sources. I'm asking a simple question with a
simple answer.
1080.0 degrees.
That is incorrect.
|
2D infered from use of compass points.
360 deg emitter 1 exploring all paths
360 deg emitter 2 exploring all paths
360 deg emitter 3 exploring all paths
--------
1800 total of all angles
http://www.physics.yorku.ca/undergrad_programme/highsch/Feynm4.html
Hmmm... maybe I made an error in addition. I'm not
very good with bank books and such. )
| Quote: |
Exercise 2.
Assume that you have n particles of mass m and the momentum of the nth
particle is equal to h*n, where h is constant.
a) Calculate the velocity of the nth particle.
b) Assuming these particles are all traveling in the same direction.
What is the difference in velocity between particle n and particle n+1?
c) Assume the particles all started at the same point in space. What
is the distance between particle n and particle n+1 at time t=1?
d) What is the limit of the distance between particles as n-->infinity?
What is the density in that region?
Something missing here? Like at rocket motor or a gun?
No.
Sorry... I can't visualise your question.
It's really a plug and chug problem. momentum = mass *
velocity/sqrt(1-(v/c)^2). Except it may take a bit of work to solve
that for velocity.
|
I don't see a physics problem. That is 3D Pythagoras.
| Quote: |
Exercise 3.
a) If you were to calculate the force of gravity at point P due to an
arbitrary distribution of masses, and you chose to set up a spherical
coordinate system, what point should you use to determine the center of
that coordinate system?
Arbitrary? random? hoogenous?
The common General Relativity assumption that this choice can be made
arbitrarily. This was the disagreement I expected for question 1a,
because outside GR, I can't imagine anyone giving that answer. Most
people would surely say that the force at point P should be calculated
relative to point P; not relative to some random position away from
point P.
Not just GR but any system of mass you have to have some
notion when the system 'thinks' (locally) it is in equilibrium.
Yet it is foundational to cosmological GR theory to presume that it is
OKAY to calculate the force at P from a reference point far away from
P. I would suggest that if you are to do something like this, you
should calculate the force from all reference points in the universe,
and then take the average. As a matter of fact, let's add that as
exercise 4, for anyone who gives your answer, Sue.
Notice the naughty word 'equilibrium'. There would be no
forces for a system in equilibrium. That is sort of a presumption
of GR anyway. Masses will move along conservative paths.
Exercise 4: For people who answered that you can place the center of
the spherical coordinate system at any point in the universe, calculate
the average force given by all such calculations.
That sounds more like some BB conjecture has to invoked.
Use the conjecture given in Einstein's proof. "If on the average, the
mass-density rho_0 is constant throughout the universe"
|
OK that is homogenous distribution so any mass
will serve as a point of reference. Connect your]
fish scale and measure up.
"The "chair" used on Skylab to measure astronaut mass. "
http://www-spof.gsfc.nasa.gov/stargaze/Smass.htm
| Quote: |
There are n number of spherical co-ordinate systems.
Baloons? Rasin cakes?
This one finds wide application.
http://hyperphysics.phy-astr.gsu.edu/hbase/forces/isq.html#isq
Walk on the wild side?
http://www.physics.nyu.edu/faculty/sokal/transgress_v2/node4.html
b) Read the following and determine what point Einstein used as the
center of this spherical coordinate system to find an infinite force at
point P.
Theory: "The stellar universe ought to be a finite island in the
infinite ocean of space." [1]
Proof: "According to the theory of Newton, the number of "lines of
force" which come from infinity and terminate in a mass m is
proportional to the mass m is proportional to the mass m.
Hmmm... I am not sure that is deriveable from 1/r^2.
Did Newton really say that? It sounds more like a Mach musing.
This is apparently Einstein's interpretation of Newton. I agree that
it sounds like a Mach musing. From Albert Einstein, Relativity: The
Special and General Theory,
translated by R.W.Lawson, 1920, Part III, Chapters XXX, XXXI, XXXII.
Reprinted in "Theories of the Universe" edited by Milton K. Munitz.
His wording is odd, but it seems consistent with a spherical coordinate
system with a center at distance r from point P.
Ahhh... here it is:
http://www.bartleby.com/173/30.html
This conception is in itself not very satisfactory. It is
still less satisfactory because it leads to the result that the
light emitted by the stars and also individual stars of the stellar
system are perpetually passing out into infinite space, never to
return, and without ever again coming into interaction with other
objects of nature. Such a finite material universe would be
destined to become gradually but systematically impoverished.
Very nice. It appears you've located the article I was referencing.
Free space is 377 ohms. If light encountered 0 ohms or infinity ohims
ir would simply reflect back to the source. Since his premise is
false I don't quite know what you can make of the rest of the musings.
I'm not making anything of his musings. I'm just asking what point (or
set of points) Einstein used to calculate infinite force at point P in
his proof. It's a question with a definite answer.
|
There doesn't seem to be a lot of support for the musings or
the methodolgy. Who cares about an answer that is defiintly
wrong. Further development on Schordinger's cat might
do more for world peace and it has great posibilities for
reducing the population of vermin. At least a 50% chance
anyway. )
| Quote: |
If on the
average, the mass-density rho_0 is constant throughout the universe,
then a sphere of volume V will enclose the mass density rho_0*V. Thus
the number of lines of force passing through the surface F of the
sphere into its interior is proportional to rho_0V. For unit area of
the surface of the sphere the number of lines of force which enters the
sphere is thus proportional to rho_0 * volume/(surface area) or to
rho_0*R. Hence the intensity of the field at the surface would
ultimately become infinite with increasing radius R of the sphere,
which is impossible.
So the lines have to taper by 1/r^2. Newton knew that.
We should check the good professors references where
he is quoting Newton. Maybe somthing was taken out
of context.
Actually, if you have a sphere of matter of radius R, it can be treated
as a point mass at distance R, so if the density of the sphere is the
same throughout, the mass is proportional to rho_0 * Volume =rho_0 *4/3
Pi R^3.
If you are at any point inside a uniform sphere of any radius, the
force is zero.
I would say if matter in your local space is homogenous then
the force is zero and I'll offer a hollow at the center of the earth
as an example.
That sounds reasonable to me. However, Einstein found a way to
mathematically show that the force was arbitrary, depending on the
choice of coordinate system.
|
Ahhh! Invariant transformations. Try this:
http://en.wikipedia.org/wiki/Noether's_theorem
Actually Emmy is the original inventor of 'time dilation'
Click this URL and see if your watch doesn't stop. )
http://www.sdsc.edu/ScienceWomen/images/noether.JPG
| Quote: |
So what Einstein is doing is adding up the finite volume centered at
distance R from P, finding a force toward the center of that volume,
and then treating the rest of space as a series of concentric spheres
of uniform density which add nothing to the force.
I would say he did it wrong. See following post about 1/r^2 and
and an object count increasing by the square or the distance.
I would also say he did it wrong, but the problem is that he did what
you are saying right, following the 1/r^2 law, and got it wrong in an
entirely different way.
|
Yeah... he counted the particles but didn't diminish the fields.
Most folks screw up by diminishing the field and not counting
the particles. What do you expect from someont who tho't the
station comes to the train?
Ya have to do both. 1/r^2 x r^2 = 1
So any where you go in a homogenous universe, you
correct the inertial field by multiplying by 1. )
Sue...
| Quote: |
Sue...
Sue...
[1] From Albert Einstein, Relativity: The special and General Theory,
translated by R.W.Lawson, 1920, Part III, Chapters XXX, XXXI, XXXII.
Reprinted in "Theories of the Universe" edited by Milton K. Munitz.
http://www.bartleby.com/173/ (30 31 32) |
|
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Spoonfed (www.spoonfedrel
science forum Guru Wannabe
Joined: 28 Apr 2006
Posts: 144
|
| Posted: Thu Jun 22, 2006 10:29 pm Post subject:
Re: Challenging exercises in relativity theory
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|
|
Ben Rudiak-Gould wrote:
| Quote: | Spoonfed wrote:
Exercise 1.
Two rocketships leave earth simultaneously at .866c, at 90 degree angle
from one another. (i.e. straight north and straight west.) Later, a
message is beamed from earth to the first rocket, to the second rocket,
and back to earth. Calculate the sum of the angles of this triangle by
considering the angle between the earth and the other rocket from the
frame of each rocket. Would the result suggest that the universe is
closed like a sphere (positive curvature), flat (0 curvature) or open
(negative curvature?)
I don't see why it would suggest anything at all. This is a very strange
calculation you're asking people to perform.
|
It's challenging, but it has a definite answer.
| Quote: | It reminds me of Feynman's
anecdote about adding the temperatures of stars. Why would there be any
significance to angles obtained in this way? For one thing, it's not clear
why angles defined by projections of light beams (null lines) onto a 3-plane
would have any significance.
|
Indeed it is not clear at all. However, I have read five or six books
where the journey from the equator to the north pole, back to the
equator and back to the original position is described as a triangle
with three 90 degree angles in it. This is used to explain the
ephemeral concept of positively curved space. However, I have never
seen a single book containing the calculation I am suggesting.
If you prefer, you can use any object traveling fast enough to catch up
with the three rockets. Or you could just calculate the angle between
the two rockets as viewed from earth + the angle between the earth and
rocket from each of the two rockets.
After you do this calculation, we can consider the question of whether
this has more or less validity than adding the temperature of stars.
| Quote: | For another, the light doesn't care about the
velocity of the rocketships which absorb and re-emit it, yet you're asking
us to calculate angles which depend on those velocities.
|
No, but what is the appropriate place to measure the angle? Who would
be in the position to measure that angle? The people in the
rocketships, and the people on earth. Just like it is the ants on a
ball that must measure the angles on the surface of a sphere, it is
people with protractors that measure angles in space.
| Quote: | Assume that you have n particles of mass m and the momentum of the nth
particle is equal to h*n, where h is constant.
If you want a Lorentz-invariant notion of "evenly spaced velocities", this
isn't the way to do it. Exercise: calculate the momenta of particles k-1 and
k+1 with respect to the rest frame of particle k. (They won't be negatives
of each other.) Exercise 2: figure out the Lorentz-invariant distribution.
|
You're way ahead of me here. I had done this a different way,
calculating the density from the assumptions and arrived at a perfect
hyperbolic sine curve, which is Lorentz invariant as far as I know.
But I shall have to try your way, and see if I can resolve the
difference in our results.
Just to make sure, you did apply the Lorentz transformation at t=x=0,
right? Otherwise, of course, the Sinh curve will not be preserved.
| Quote: | a) Calculate the velocity of the nth particle.
v = p/E and E = sqrt(p^2 + m^2), so v_n = 1 / sqrt(1 + (m/hn)^2).
Yes, that appears alright. I have: |
v=(c*h*n)/Sqrt[c^2*m^2 + h^2*n^2]
| Quote: | b) Assuming these particles are all traveling in the same direction.
What is the difference in velocity between particle n and particle n+1?
Subtract.
|
Good enough.
| Quote: |
c) Assume the particles all started at the same point in space. What
is the distance between particle n and particle n+1 at time t=1?
Don't you want some units on that?
|
Good enough.
| Quote: | d) What is the limit of the distance between particles as n-->infinity?
What is the density in that region?
Zero and infinite respectively.
|
That all looks correct
| Quote: | Exercise 3.
a) If you were to calculate the force of gravity at point P due to an
arbitrary distribution of masses, and you chose to set up a spherical
coordinate system, what point should you use to determine the center of
that coordinate system?
Er, it had better not matter, else your theory is inconsistent.
|
Shouldn't you use the point P to calculate the force at point P?
| Quote: | b) Read the following and determine what point Einstein used as the
center of this spherical coordinate system to find an infinite force at
point P.
[...]
He didn't find an infinite force at point P. He didn't even mention point P.
I'm still trying to figure out what /your/ point is. Presumably you think
that Einstein's argument is wrong. I tend to agree, but keep in mind that
he was writing a popularization and had to make a nonmathematical argument.
Here's a slightly more formal treatment.
Given a charge (i.e. mass) distribution rho(r), the Newtonian gravitational
field F satisfies
div F = -4 pi G rho.
When rho is constant (a universe filled uniformly with matter), this has the
family of solutions
F(r) = -4/3 pi G rho (r - r0)
where r0 is an arbitrary point in space. r0 is the center of a radial
gravitational field which increases without bound as you move farther from
the center.
|
Correct. This is essentially what Einstein calculated. To get
infinite force at point P, then, he used r=infinity.
| Quote: | So what's the significance of this? You could argue that the gravitational
field is not well-defined, since it depends on this arbitrary point, and you
could use this to make the case for a modification of Newton's theory
(perhaps an added exponential falloff).
|
.... which is essentially what Einstein did?
| Quote: | But there's another way out. The
difference between any two of these fields (with different values of r0) is
a constant field, i.e. you can move r0 around by adding a constant field.
What's the added effect of the constant field? Nothing!
|
That is similar to my idea of taking the average field which I
mentioned to Sue.
| Quote: | A constant
gravitational field filling the universe is the same as no field at all,
since it accelerates everything together, which preserves all spatial
relationships, which are all that you can actually measure. So the point r0
is not a physically measurable quantity, and the family of solutions turns
out to be a single solution after all, in some sense. What's the effect of
that field? Suppose the matter is initially all comoving; then clearly the
field will make it gravitate together. It'll gravitate toward the point r0,
but we just established that the point r0 isn't detectable, so the real
effect is just a general sort of collapse (increasing density) without any
particular center. It's easy to show that the collapse preserves the
constant-density property, and that infinite density is reached in a finite
time (another exercise). The time reversal of this is a Newtonian big bang.
The family of solutions I gave above is incomplete, since you can add any
divergenceless field (not just a constant field) and still satisfy the field
equation. This is sort of a Newtonian version of general covariance. The
Einstein field equations basically equate the matter and energy distribution
to a fancy sort of spacetime divergence, and just as in the Newtonian case
this doesn't completely constrain the field. There are still many solutions
with the same matter content, some of them meaningless reparameterizations
(like the change of r0) and others detectably different (gravitational waves).
Incidentally, I should clarify that Einstein's center point is independent
of r0. That is, you can pick any r0, and define the gravitational field
based on that, and then apply Gauss's law to spheres centered at any other
point, and get the right answer. Einstein's argument doesn't assume any sort
of symmetry of the field around his center point. The only part of the
argument that I disagree with is the bit at the end where he claims that the
mathematical result is physically absurd.
-- Ben
|
Ah, but in his proof he is actually picking r0 to be infinite. Of
course, you can pick r0 to be any arbitrarily high value, but picking
infinity is absurd, and of course that leads to absurd results. |
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PD
science forum Guru
Joined: 03 May 2005
Posts: 4363
|
| Posted: Fri Jun 23, 2006 4:19 am Post subject:
Re: Challenging exercises in relativity theory
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Henry Haapalainen wrote:
| Quote: | Is there any reason to challenge Einstein's relativity any more?
|
Sure there is, and there are plenty of talented physicists that do.
| Quote: | Those who
still believe it in its entirety will keep on believing as long as they
live.
|
Nah, it's just that "Seems silly to me" is not really a compelling
challenge. More serious challenges are levied every day.
Consider the parable of the fly and the gladiator.
The gladiator was in the middle of training for a serious contest
against a Visigoth challenger and leopards. A fly landed on the
gladiator's shoulder. The gladiator did not notice the fly and
continued with his practice with the sword and club. The fly buzzed
relentlessly but the gladiator paid no notice. Finally the fly fleed,
muttering, "I guess that big muscley guy is really a wimp and won't
take on a challenger even when it bites him on the shoulder."
PD |
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dda1
science forum Guru
Joined: 06 Feb 2006
Posts: 762
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| Posted: Fri Jun 23, 2006 4:33 am Post subject:
Re: Challenging exercises in relativity theory
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Henry Haapalainen wrote:
| Quote: | Is there any reason to challenge Einstein's relativity any more? Those who
still believe it in its entirety will keep on believing as long as they
live.
Henry Haapalainen
|
Luckily your type will die soon and no more cretins of your calliber
will be born. |
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Sorcerer
science forum Guru
Joined: 09 Jun 2006
Posts: 410
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| Posted: Fri Jun 23, 2006 7:00 pm Post subject:
Re: Challenging exercises in relativity theory
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"Henry Haapalainen" <kirppu@kolumbus.fi> wrote in message
news:h6Emg.8250$%q2.1441@reader1.news.jippii.net...
| Is there any reason to challenge Einstein's relativity any more? Those who
| still believe it in its entirety will keep on believing as long as they
| live.
| Henry Haapalainen
One can only try to prevent the infection and spread of Insanitititis.
Androcles. |
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Spoonfed (www.spoonfedrel
science forum Guru Wannabe
Joined: 28 Apr 2006
Posts: 144
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| Posted: Sat Jun 24, 2006 3:30 am Post subject:
Re: Challenging exercises in relativity theory
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Spoonfed wrote:
| Quote: | Ben Rudiak-Gould wrote:
Spoonfed wrote:
Exercise 1.
Two rocketships leave earth simultaneously at .866c, at 90 degree angle
from one another. (i.e. straight north and straight west.) Later, a
message is beamed from earth to the first rocket, to the second rocket,
and back to earth. Calculate the sum of the angles of this triangle by
considering the angle between the earth and the other rocket from the
frame of each rocket. Would the result suggest that the universe is
closed like a sphere (positive curvature), flat (0 curvature) or open
(negative curvature?)
I don't see why it would suggest anything at all. This is a very strange
calculation you're asking people to perform.
It's challenging, but it has a definite answer.
It reminds me of Feynman's
anecdote about adding the temperatures of stars. Why would there be any
significance to angles obtained in this way? For one thing, it's not clear
why angles defined by projections of light beams (null lines) onto a 3-plane
would have any significance.
Indeed it is not clear at all. However, I have read five or six books
where the journey from the equator to the north pole, back to the
equator and back to the original position is described as a triangle
with three 90 degree angles in it. This is used to explain the
ephemeral concept of positively curved space. However, I have never
seen a single book containing the calculation I am suggesting.
If you prefer, you can use any object traveling fast enough to catch up
with the three rockets. Or you could just calculate the angle between
the two rockets as viewed from earth + the angle between the earth and
rocket from each of the two rockets.
After you do this calculation, we can consider the question of whether
this has more or less validity than adding the temperature of stars.
For another, the light doesn't care about the
velocity of the rocketships which absorb and re-emit it, yet you're asking
us to calculate angles which depend on those velocities.
No, but what is the appropriate place to measure the angle? Who would
be in the position to measure that angle? The people in the
rocketships, and the people on earth. Just like it is the ants on a
ball that must measure the angles on the surface of a sphere, it is
people with protractors that measure angles in space.
Assume that you have n particles of mass m and the momentum of the nth
particle is equal to h*n, where h is constant.
If you want a Lorentz-invariant notion of "evenly spaced velocities", this
isn't the way to do it. Exercise: calculate the momenta of particles k-1 and
k+1 with respect to the rest frame of particle k. (They won't be negatives
of each other.) Exercise 2: figure out the Lorentz-invariant distribution.
You're way ahead of me here. I had done this a different way,
calculating the density from the assumptions and arrived at a perfect
hyperbolic sine curve, which is Lorentz invariant as far as I know.
But I shall have to try your way, and see if I can resolve the
difference in our results.
|
Actually the last time I worked on this, the value I got for the linear
density was
(c^3*m)/(h*t*(c^2 - x^2/t^2)^(3/2))
and I was too busy worrying about the fact that the units came out to
be meter^-2 instead of meter^-1. This I have more-or-less resolved by
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