Doug Sweetser
May12-04, 02:40 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Hello:\n\nIn this post I will calculate the bending of light caused by a metric\nwith exponential coefficients. This metric first appeared in work by\nRosen and more recently by Watt and Misner. Frequent fliers at SPR\nknow I claim this metric may be a key to unified field theory, work\nthat still must address issues about the background structures and\ndetailed calculations in quantum field theory.\n\nThe calculation for light bending is guaranteed to succeed because the\nexponential metric has the same ten first order parameterized\npost-Newtonian coefficients as the Schwarzschild metric. The details\nof the calculation are different and worth noting. Oh, and it is fun\nto be able to do!\n\nI will be relying on Wald, Chapter 6, p. 138-146, if you wish to see\nthe professional presentation for general relativity. I will be\nrearranging everything because I like Paul Hewitt\'s approach that\nstarts where one needs to end up. For this problem, we want to know:\n\nphi = ?\n\nThis is how much light bends due to gravity. Let\'s draw an accurate\nASCII cartoon for the bending of light near the Sun:\n\n--------- - light path\no o the Sun\n\nThis ASCII representation is considerably more accurate than Wald\'s\nfigure 6.6, animations in "The Elegant Universe", or any textbook I\nhave seen on the topic. The point is gravity works reliably over the\ntime and distance scales it does because gravity does almost nothing.\nThe subtlety of gravity will justify an approximation step to be made\nlater.\n\nThe mass of the Sun stays constant. What varies is the distance R.\nThe light goes from minus infinity to positive infinity, getting as\nclose as the radius of the Sun. Draw a picture of the path in, and\ncompare it to the path out:\n\nin out\n\n----- -----\no o\n\nLooks similar! Figure out half the problem, then multiply by 2:\n\nphi = 2 R_Sun to Infinity ?\n\nProgress. We want to know the sum of all changes in phi with\nrespect to R as R goes from a minimum of R_Sun out to infinity. That\nsounds like and integral to me.\n\nphi = 2 Integral from R_Sun to infinity (d phi/dR) dR\n\nAdd up all the small changes in phi as R varies, and the result is\nthe total for phi.\n\nThe next task is to find an expression for d phi/dR that is related\nto gravity. First think about a path parameterized by tau:\n\nU^u = d x^u/dtau\n\nFor light, tau will be the affine parameter. Light does not have a\nproper time, but the affine parameter can keep track where it is on a\nnull geodesic. We will try and get to d phi/dR by calculating two\nother derivatives:\n\nd phi/dR = (d phi/dtau)/(d R/dtau)\n\nContract two paths of light:\n\ng_uv U^u U^v = 0 =\n\ng_00 (dt/dtau)^2 + g_11 (dx/dtau)^2\n+ g_22 (dy/dtau)^2 + g_22 (dz/dtau)^2\n+ g_01 (dt/dtau dx/dtau) + g_02 (dt/dtau dy/dtau)\n+ g_03 (dt/dtau dz/dtau) + g_12 (dx/dtau dy/dtau)\n+ g_13 (dx/dtau dz/dtau) + g_23 (dy/dtau dz/dtau)\n\nTen terms, too many, lets simplify! For the exponential metric, only\ng_00, g_11, g_22, and g_33 are non zero. Let\'s also work in spherical\ncoordinates, fixing theta to pi/2:\n\n0 = e^(-2 GM/c^2 R) (dt/dtau)^2 - e^(2 GM/c^2 R) (dR/dtau)^2\n- r^2 (dphi/dR)^2\n\nNotice two things about the exponentials: they do no depend on time,\nnor the angle phi. Come back at a later time, and the metric remains\nthe same. Look at it from a different angle, and the coefficients\nkeep their same value. This indicates that both energy and angular\nmomentum are conserved because there is no way to change them. Here\nare their definitions:\n\nE = e^(-2 GM/c^2 R) dt/dtau\n\nL = R^2 d phi/dtau\n\nGood, there is an expression for d phi/dtau right here:\n\nd phi/dtau = L/R^2\n\nNow to get d phi/dR. Plug E and L back into the parametrized path for\nlight, multiply through both sides by e^(-2 GM/c^2 R) in order to\nsolve for dR/dtau:\n\nd R/dtau = (E^2 - e^(-2 GM/c^2 R) L^2/R^2)^(1/2)\n\nThe effective potential is the last term in the above expression:\n\nV(R) = e^(-2 GM/c^2 R) L^2/R^2\n\nThe potential is an exponential decay, quite different from figure 6.5\nin Wald which starts near zero at negative infinity, climbs to a\npositive maximum at R=3M, then declines. There is no such maximum on\nan exponential curve! It strikes me as odd that this maximum plays a\nrole in Wald\'s calculation since it will apply to an extremely strong\nfield, which is not relevant around the Sun. In my opinion,\nan exponential function is the calling card for deep physics, so I\nlike this difference in the effective potential function :-)\n\nNow we can write out d phi/dR:\n\nd phi/dR = (d phi/dtau)/(d R/dtau)\n\n= L/R^2 1/(E^2 - e^(-2 GM/c^2 R) L^2/R^2)^(1/2)\n\nThe ratio of the constants of motion, L/E, is the apparent impact\nparameter b = L/E, which has units of distance as it should. Divide\nthe numerator and denominator of d phi/dR so to use the impact\nparameter b:\n\nd phi/dR = b/R^2 1/(1 - e^(-2 GM/c^2 R) b^2/R^2)^(1/2)\n\nYeah! A quick look at the dimensions should show this has the right\nform: it has units of 1/distance as d phi/dR should. When integrated\nover a distance, that will bring down a distance, leaving a\ndimensionless angle. Here is the integral:\n\nphi = 2 Integral from R_Sun to infinity\n\nb/R^2 1/(1 - e^(-2 GM/c^2 R) b^2/R^2)^(1/2) dR\n\nGood luck finding an exact solution. I asked Mathematica, a symbolic\nmath package with thousands of memorized opening moves, and it was\nunable to find a solution.\n\nRemember the earlier cartoon? There is hardly any bending going on.\nThis suggests a Taylor series should be used to simply the integrand\nenough that a solution can be found. One thing that is small is the\nexponent for the exponential. The mass of the Sun in geometric units\nis 1.5 km, compared to its radius of 5.96 x 10^5 km. The Taylor\nseries expansion for small M to first order in M is:\n\nd phi/dR =~ b/R^2 1/(1 - b^2/R^2)^(1/2)\n\n+ b^3 M/R^5 1/(1 - b^2/R^2)^(3/2) + O(M^2)\n\nThe exponential is gone. This can be integrated, but it still is not\nan easy integral to solve. I had to get Mathematica to assist:\n\nphi = 2 Integral from R_Sun to infinity\n\n( b/R^2 1/(1 - b^2/R^2)^(1/2)\n\n+ b^3 M/R^5 1/(1 - b^2/R^2)^(3/2)) dR\n\n= 4 GM/c^2 b + (4 other terms)/(b^3 - b R_Sun^2)\n\nIf one where to evaluate at b= R_Sun, the send term is singular. I\ndon\'t have the experience necessary to deal with a singularity like\nthis. It is nothing more than a hope that the singularity may be\nsafely ignored. If that is the case, then:\n\nphi = 4 GM/c^2 R_Sun\n\n= 4 1.5 km/5.96 10^5 km 180^o/Pi radians 60\'/^o 60"/\'\n\n= 1.78"\n\nThis is what has been measured.\n\nOne reason for me doing this calculation was I hoped I could determine\nthe M^2 contribution. All the terms are part of the singular-looking\nterm, so I am not confident with analyzing it.\n\n\ndoug\nquaternions.com\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Hello:
In this post I will calculate the bending of light caused by a metric
with exponential coefficients. This metric first appeared in work by
Rosen and more recently by Watt and Misner. Frequent fliers at SPR
know I claim this metric may be a key to unified field theory, work
that still must address issues about the background structures and
detailed calculations in quantum field theory.
The calculation for light bending is guaranteed to succeed because the
exponential metric has the same ten first order parameterized
post-Newtonian coefficients as the Schwarzschild metric. The details
of the calculation are different and worth noting. Oh, and it is fun
to be able to do!
I will be relying on Wald, Chapter 6, p. 138-146, if you wish to see
the professional presentation for general relativity. I will be
rearranging everything because I like Paul Hewitt's approach that
starts where one needs to end up. For this problem, we want to know:
\phi = ?
This is how much light bends due to gravity. Let's draw an accurate
ASCII cartoon for the bending of light near the Sun:
--------- - light path
o o the Sun
This ASCII representation is considerably more accurate than Wald's
figure 6.6, animations in "The Elegant Universe", or any textbook I
have seen on the topic. The point is gravity works reliably over the
time and distance scales it does because gravity does almost nothing.
The subtlety of gravity will justify an approximation step to be made
later.
The mass of the Sun stays constant. What varies is the distance R.
The light goes from minus infinity to positive infinity, getting as
close as the radius of the Sun. Draw a picture of the path in, and
compare it to the path out:
in out
----- -----
o o
Looks similar! Figure out half the problem, then multiply by 2:
\phi = 2 R_{Sun}[/itex] to Infinity ?
Progress. We want to know the sum of all changes in \phi with
respect to R as R goes from a minimum of R_{Sun} out to infinity. That
sounds like and integral to me.
\phi = 2 Integral from R_{Sun} to infinity (d \phi/dR) dR
Add up all the small changes in \phi as R varies, and the result is
the total for \phi.
The next task is to find an expression for d \phi/dR that is related
to gravity. First think about a path parameterized by \tau:U^u = d x^u/dtau
For light, \tau will be the affine parameter. Light does not have a
proper time, but the affine parameter can keep track where it is on a
null geodesic. We will try and get to d \phi/dR by calculating two
other derivatives:
d \phi/dR = (d \phi/dtau)/(d R/dtau)
Contract two paths of light:
g_{uv} U^u U^v = =
g_{00} (dt/dtau)^2 + g_{11} (dx/dtau)^2+ g_{22} (dy/dtau)^2 + g_{22} (dz/dtau)^2+ g_{01} (dt/dtau dx/dtau) + g_{02} (dt/dtau dy/dtau)+ g_{03} (dt/dtau dz/dtau) + g_{12} (dx/dtau dy/dtau)+ g_{13} (dx/dtau dz/dtau) + g_{23} (dy/dtau dz/dtau)
Ten terms, too many, lets simplify! For the exponential metric, only
g_{00}, g_{11}, g_{22}, and g_{33} are non zero. Let's also work in spherical
coordinates, fixing \theta to \pi/2:= e^(-2 GM/c^2 R) (dt/dtau)^2 - e^(2 GM/c^2 R) (dR/dtau)^2- r^2 (dphi/dR)^2
Notice two things about the exponentials: they do no depend on time,
nor the angle \phi. Come back at a later time, and the metric remains
the same. Look at it from a different angle, and the coefficients
keep their same value. This indicates that both energy and angular
momentum are conserved because there is no way to change them. Here
are their definitions:
E = e^(-2 GM/c^2 R) dt/dtauL = R^2 d \phi/dtau
Good, there is an expression for d \phi/dtau right here:
d \phi/dtau = L/R^2
Now to get d \phi/dR. Plug E and L back into the parametrized path for
light, multiply through both sides by e^(-2 GM/c^2 R) in order to
solve for dR/dtau:d R/dtau = (E^2 - e^(-2 GM/c^2 R) L^2/R^2)^(1/2)
The effective potential is the last term in the above expression:
V(R) = e^(-2 GM/c^2 R) L^2/R^2
The potential is an exponential decay, quite different from figure 6.5
in Wald which starts near zero at negative infinity, climbs to a
positive maximum at R=3M, then declines. There is no such maximum on
an exponential curve! It strikes me as odd that this maximum plays a
role in Wald's calculation since it will apply to an extremely strong
field, which is not relevant around the Sun. In my opinion,
an exponential function is the calling card for deep physics, so I
like this difference in the effective potential function :-)
Now we can write out d \phi/dR:d \phi/dR = (d \phi/dtau)/(d R/dtau)= L/R^2 1/(E^2 - e^(-2 GM/c^2 R) L^2/R^2)^(1/2)
The ratio of the constants of motion, L/E, is the apparent impact
parameter b = L/E, which has units of distance as it should. Divide
the numerator and denominator of d \phi/dR so to use the impact
parameter b:
d \phi/dR = b/R^2 1/(1 - e^(-2 GM/c^2 R) b^2/R^2)^(1/2)
Yeah! A quick look at the dimensions should show this has the right
form: it has units of 1/distance as d \phi/dR should. When integrated
over a distance, that will bring down a distance, leaving a
dimensionless angle. Here is the integral:
\phi = 2 Integral from R_{Sun} to infinity
b/R^2 1/(1 - e^(-2 GM/c^2 R) b^2/R^2)^(1/2) dR
Good luck finding an exact solution. I asked Mathematica, a symbolic
math package with thousands of memorized opening moves, and it was
unable to find a solution.
Remember the earlier cartoon? There is hardly any bending going on.
This suggests a Taylor series should be used to simply the integrand
enough that a solution can be found. One thing that is small is the
exponent for the exponential. The mass of the Sun in geometric units
is 1.5 km, compared to its radius of 5.96 x 10^5 km. The Taylor
series expansion for small M to first order in M is:
d \phi/dR =~ b/R^2 1/(1 - b^2/R^2)^(1/2)+ b^3 M/R^5 1/(1 - b^2/R^2)^(3/2) + O(M^2)
The exponential is gone. This can be integrated, but it still is not
an easy integral to solve. I had to get Mathematica to assist:
\phi = 2 Integral from R_{Sun} to infinity
( b/R^2 1/(1 - b^2/R^2)^(1/2)+ b^3 M/R^5 1/(1 - b^2/R^2)^(3/2)) dR= 4 GM/c^2 b + (4 other [itex]terms)/(b^3 - b R_{Sun}^2)
If one where to evaluate at b= R_{Sun}, the send term is singular. I
don't have the experience necessary to deal with a singularity like
this. It is nothing more than a hope that the singularity may be
safely ignored. If that is the case, then:
\phi = 4 GM/c^2 R_{Sun}= 4 1.5 km/5.96 10^5 km 180^o/\Pi radians 60'/^o 60"/'
= 1.78"
This is what has been measured.
One reason for me doing this calculation was I hoped I could determine
the M^2 contribution. All the terms are part of the singular-looking
term, so I am not confident with analyzing it.
doug
quaternions.com
In this post I will calculate the bending of light caused by a metric
with exponential coefficients. This metric first appeared in work by
Rosen and more recently by Watt and Misner. Frequent fliers at SPR
know I claim this metric may be a key to unified field theory, work
that still must address issues about the background structures and
detailed calculations in quantum field theory.
The calculation for light bending is guaranteed to succeed because the
exponential metric has the same ten first order parameterized
post-Newtonian coefficients as the Schwarzschild metric. The details
of the calculation are different and worth noting. Oh, and it is fun
to be able to do!
I will be relying on Wald, Chapter 6, p. 138-146, if you wish to see
the professional presentation for general relativity. I will be
rearranging everything because I like Paul Hewitt's approach that
starts where one needs to end up. For this problem, we want to know:
\phi = ?
This is how much light bends due to gravity. Let's draw an accurate
ASCII cartoon for the bending of light near the Sun:
--------- - light path
o o the Sun
This ASCII representation is considerably more accurate than Wald's
figure 6.6, animations in "The Elegant Universe", or any textbook I
have seen on the topic. The point is gravity works reliably over the
time and distance scales it does because gravity does almost nothing.
The subtlety of gravity will justify an approximation step to be made
later.
The mass of the Sun stays constant. What varies is the distance R.
The light goes from minus infinity to positive infinity, getting as
close as the radius of the Sun. Draw a picture of the path in, and
compare it to the path out:
in out
----- -----
o o
Looks similar! Figure out half the problem, then multiply by 2:
\phi = 2 R_{Sun}[/itex] to Infinity ?
Progress. We want to know the sum of all changes in \phi with
respect to R as R goes from a minimum of R_{Sun} out to infinity. That
sounds like and integral to me.
\phi = 2 Integral from R_{Sun} to infinity (d \phi/dR) dR
Add up all the small changes in \phi as R varies, and the result is
the total for \phi.
The next task is to find an expression for d \phi/dR that is related
to gravity. First think about a path parameterized by \tau:U^u = d x^u/dtau
For light, \tau will be the affine parameter. Light does not have a
proper time, but the affine parameter can keep track where it is on a
null geodesic. We will try and get to d \phi/dR by calculating two
other derivatives:
d \phi/dR = (d \phi/dtau)/(d R/dtau)
Contract two paths of light:
g_{uv} U^u U^v = =
g_{00} (dt/dtau)^2 + g_{11} (dx/dtau)^2+ g_{22} (dy/dtau)^2 + g_{22} (dz/dtau)^2+ g_{01} (dt/dtau dx/dtau) + g_{02} (dt/dtau dy/dtau)+ g_{03} (dt/dtau dz/dtau) + g_{12} (dx/dtau dy/dtau)+ g_{13} (dx/dtau dz/dtau) + g_{23} (dy/dtau dz/dtau)
Ten terms, too many, lets simplify! For the exponential metric, only
g_{00}, g_{11}, g_{22}, and g_{33} are non zero. Let's also work in spherical
coordinates, fixing \theta to \pi/2:= e^(-2 GM/c^2 R) (dt/dtau)^2 - e^(2 GM/c^2 R) (dR/dtau)^2- r^2 (dphi/dR)^2
Notice two things about the exponentials: they do no depend on time,
nor the angle \phi. Come back at a later time, and the metric remains
the same. Look at it from a different angle, and the coefficients
keep their same value. This indicates that both energy and angular
momentum are conserved because there is no way to change them. Here
are their definitions:
E = e^(-2 GM/c^2 R) dt/dtauL = R^2 d \phi/dtau
Good, there is an expression for d \phi/dtau right here:
d \phi/dtau = L/R^2
Now to get d \phi/dR. Plug E and L back into the parametrized path for
light, multiply through both sides by e^(-2 GM/c^2 R) in order to
solve for dR/dtau:d R/dtau = (E^2 - e^(-2 GM/c^2 R) L^2/R^2)^(1/2)
The effective potential is the last term in the above expression:
V(R) = e^(-2 GM/c^2 R) L^2/R^2
The potential is an exponential decay, quite different from figure 6.5
in Wald which starts near zero at negative infinity, climbs to a
positive maximum at R=3M, then declines. There is no such maximum on
an exponential curve! It strikes me as odd that this maximum plays a
role in Wald's calculation since it will apply to an extremely strong
field, which is not relevant around the Sun. In my opinion,
an exponential function is the calling card for deep physics, so I
like this difference in the effective potential function :-)
Now we can write out d \phi/dR:d \phi/dR = (d \phi/dtau)/(d R/dtau)= L/R^2 1/(E^2 - e^(-2 GM/c^2 R) L^2/R^2)^(1/2)
The ratio of the constants of motion, L/E, is the apparent impact
parameter b = L/E, which has units of distance as it should. Divide
the numerator and denominator of d \phi/dR so to use the impact
parameter b:
d \phi/dR = b/R^2 1/(1 - e^(-2 GM/c^2 R) b^2/R^2)^(1/2)
Yeah! A quick look at the dimensions should show this has the right
form: it has units of 1/distance as d \phi/dR should. When integrated
over a distance, that will bring down a distance, leaving a
dimensionless angle. Here is the integral:
\phi = 2 Integral from R_{Sun} to infinity
b/R^2 1/(1 - e^(-2 GM/c^2 R) b^2/R^2)^(1/2) dR
Good luck finding an exact solution. I asked Mathematica, a symbolic
math package with thousands of memorized opening moves, and it was
unable to find a solution.
Remember the earlier cartoon? There is hardly any bending going on.
This suggests a Taylor series should be used to simply the integrand
enough that a solution can be found. One thing that is small is the
exponent for the exponential. The mass of the Sun in geometric units
is 1.5 km, compared to its radius of 5.96 x 10^5 km. The Taylor
series expansion for small M to first order in M is:
d \phi/dR =~ b/R^2 1/(1 - b^2/R^2)^(1/2)+ b^3 M/R^5 1/(1 - b^2/R^2)^(3/2) + O(M^2)
The exponential is gone. This can be integrated, but it still is not
an easy integral to solve. I had to get Mathematica to assist:
\phi = 2 Integral from R_{Sun} to infinity
( b/R^2 1/(1 - b^2/R^2)^(1/2)+ b^3 M/R^5 1/(1 - b^2/R^2)^(3/2)) dR= 4 GM/c^2 b + (4 other [itex]terms)/(b^3 - b R_{Sun}^2)
If one where to evaluate at b= R_{Sun}, the send term is singular. I
don't have the experience necessary to deal with a singularity like
this. It is nothing more than a hope that the singularity may be
safely ignored. If that is the case, then:
\phi = 4 GM/c^2 R_{Sun}= 4 1.5 km/5.96 10^5 km 180^o/\Pi radians 60'/^o 60"/'
= 1.78"
This is what has been measured.
One reason for me doing this calculation was I hoped I could determine
the M^2 contribution. All the terms are part of the singular-looking
term, so I am not confident with analyzing it.
doug
quaternions.com