Re: Curvature and Redshift [Archive]

Tom Snyder

Apr13-04, 05:48 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>"Daryl McCullough" <daryl@atc-nycorp.com> wrote in message\nnews:c3nn6d0512@drn.newsguy.com...\n\n> Tom Roberts says...\n> >\n> >Consider a source sending a light ray to a detector. The proper way to\n> >compute the redshift is to take the 4-velocity of the source and\n> >parallel-transport it along the (null) path of the light ray to the\n> >detector, and dot the result into the 4-velocity of the detector.\n>\n> I\'m not sure if I understand what that recipe is giving you. Let\'s\n> take the simple case in which (1) spacetime is flat, (2) the source\n> is at rest (in an inertial coordinate system), (3) the detector is\n> travelling at constant speed v in the x-direction. Then\n>\n> 4-velocity of the source:\n> v^0 = 1\n> v^1 = v^2 = v^3 = 0\n>\n> 4-velocity of the detector:\n> v^0 = gamma\n> v^1 = gamma v\n> v^2 = v^3 = 0\n>\n> Parallel-transport is trivial in this case, so the dot-product gives\n>\n> v(source) . v(detector) = gamma\n>\n> But the ratio f(detector)/f(source) gives (by the relativistic\n> Doppler formula)\n>\n> f(detector)/f(source) = square-root(1-(v/c))/square-root(1+(v/c))\n>\n> So I\'m not sure that I understand how to compute this ratio from\n> the dot-product of the velocity 4-vectors.\n>\n\nYes, Tom\'s prescription doesn\'t seem to work. Perhaps he meant to say the\nfollowing:\n\nFirst, parallel transport the 4-velocity of the source along the null\ngeodesic of the light ray until you arrive at the detector. Then subtract\nthis parallel transported vector from the 4-velocity of the detector (at the\nlocation of the detector). Finally, take the scalar product of this\ndifference with the wave 4-vector (at the location of the detector). I\nthink this would give the frequency shift between the source and detector.\n\nBut I don\'t see any advantage of doing it this way. Why not just take the\nscalar product of the 4-velocity of the source with the wave 4-vector (at\nthe location of the source) and, likewise, take the corresponding scalar pro\nduct for the detector? The difference in these scalar products would be the\nfrequency shift of the light with respect to the source and detector. Why\nbother with the parallel transport business?\n\n----------------------\n\nTom Roberts also gave us some homework:\n\n>>Exercise: The metric at the event of detection is clearly used,\n>>as it is needed for that dot product. Explain how the metric at\n>>the event of emission is used in the computation. This has\n>>nothing to do with the relationship between metric and\n>>connection. Hint: the metric at the event of detection enters\n>>in two different ways.\n\nI think that Tom wants us to realize that the metric is needed to define the\n4-velocities of the source and detector. The 4-velocity of a particle is\nthe unit tangent vector to the worldline of the particle and you need the\nmetric to normalize to a unit vector.\n\nTom S.\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>"Daryl McCullough" <daryl@atc-nycorp.com> wrote in message
news:c3nn6d0512@drn.newsguy.com...

> Tom Roberts says...
> >
> >Consider a source sending a light ray to a detector. The proper way to
> >compute the redshift is to take the 4-velocity of the source and
> >parallel-transport it along the (null) path of the light ray to the
> >detector, and dot the result into the 4-velocity of the detector.
>
> I'm not sure if I understand what that recipe is giving you. Let's
> take the simple case in which (1) spacetime is flat, (2) the source
> is at rest (in an inertial coordinate system), (3) the detector is
> travelling at constant speed v in the x-direction. Then
>
> 4-velocity of the source:
> v^0 = 1
> v^1 = v^2 = v^3 =
>
> 4-velocity of the detector:
> v^0 = \gamma
> v^1 = \gamma v
> v^2 = v^3 =
>
> Parallel-transport is trivial in this case, so the dot-product gives
>
> v(source) . v(detector) = \gamma
>
> But the ratio f(detector)/f(source) gives (by the relativistic
> Doppler formula)
>
> f(detector)/f(source) = square-root(1-(v/c))/square-root(1+(v/c))
>
> So I'm not sure that I understand how to compute this ratio from
> the dot-product of the velocity 4-vectors.
>

Yes, Tom's prescription doesn't seem to work. Perhaps he meant to say the
following:

First, parallel transport the 4-velocity of the source along the null
geodesic of the light ray until you arrive at the detector. Then subtract
this parallel transported vector from the 4-velocity of the detector (at the
location of the detector). Finally, take the scalar product of this
difference with the wave 4-vector (at the location of the detector). I
think this would give the frequency shift between the source and detector.

But I don't see any advantage of doing it this way. Why not just take the
scalar product of the 4-velocity of the source with the wave 4-vector (at
the location of the source) and, likewise, take the corresponding scalar pro
duct for the detector? The difference in these scalar products would be the
frequency shift of the light with respect to the source and detector. Why
bother with the parallel transport business?

----------------------

Tom Roberts also gave us some homework:

>>Exercise: The metric at the event of detection is clearly used,
>>as it is needed for that dot product. Explain how the metric at
>>the event of emission is used in the computation. This has
>>nothing to do with the relationship between metric and
>>connection. Hint: the metric at the event of detection enters
>>in two different ways.

I think that Tom wants us to realize that the metric is needed to define the
4-velocities of the source and detector. The 4-velocity of a particle is
the unit tangent vector to the worldline of the particle and you need the
metric to normalize to a unit vector.

Tom S.

ebunn@lfa221051.richmond.edu

Apr19-04, 01:52 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>In article <GqXdc.4048\\$xn4.17053@attbi_s51>, Tom Snyder <xxxxx@xxxx.com> wrote:\n\n>Yes, Tom\'s prescription doesn\'t seem to work. Perhaps he meant to say the\n>following:\n>\n>First, parallel transport the 4-velocity of the source along the null\n>geodesic of the light ray until you arrive at the detector. Then subtract\n>this parallel transported vector from the 4-velocity of the detector (at the\n>location of the detector). Finally, take the scalar product of this\n>difference with the wave 4-vector (at the location of the detector). I\n>think this would give the frequency shift between the source and detector.\n\nThat\'s right. To put it another way, Tom Roberts\'s recipe gives the\nratio of the frequencies (nu\'/nu) rather than the fractional change in\nfrequency (Delta nu / nu). To put it yet another way, his recipe\ngives the quantity astronomers call 1+z, rather than the quantity\ncalled z.\n\n>But I don\'t see any advantage of doing it this way. Why not just take the\n>scalar product of the 4-velocity of the source with the wave 4-vector (at\n>the location of the source) and, likewise, take the corresponding scalar pro\n>duct for the detector? The difference in these scalar products would be the\n>frequency shift of the light with respect to the source and detector. Why\n>bother with the parallel transport business?\n\nTo do this, you need to know the wave 4-vectors, which presumably\nmeans solving the wave equation or something. Tom Roberts\'s\nprescription is purely geometrical.\n\n-Ted\n\n\n--\n[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.]\n\n\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>In article <GqXdc.4048$xn4.17053@attbi_s51>, Tom Snyder <xxxxx@xxxx.com> wrote:

>Yes, Tom's prescription doesn't seem to work. Perhaps he meant to say the
>following:
>
>First, parallel transport the 4-velocity of the source along the null
>geodesic of the light ray until you arrive at the detector. Then subtract
>this parallel transported vector from the 4-velocity of the detector (at the
>location of the detector). Finally, take the scalar product of this
>difference with the wave 4-vector (at the location of the detector). I
>think this would give the frequency shift between the source and detector.

That's right. To put it another way, Tom Roberts's recipe gives the
ratio of the frequencies (\nu'/\nu) rather than the fractional change in
frequency (\Delta \nu / \nu). To put it yet another way, his recipe
gives the quantity astronomers call 1+z, rather than the quantity
called z.

>But I don't see any advantage of doing it this way. Why not just take the
>scalar product of the 4-velocity of the source with the wave 4-vector (at
>the location of the source) and, likewise, take the corresponding scalar pro
>duct for the detector? The difference in these scalar products would be the
>frequency shift of the light with respect to the source and detector. Why
>bother with the parallel transport business?

To do this, you need to know the wave 4-vectors, which presumably
means solving the wave equation or something. Tom Roberts's
prescription is purely geometrical.

-Ted

--
[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.]

Tom Snyder

Apr20-04, 02:35 AM

<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><ebunn@lfa221051.richmond.edu> wrote in message\nnews:c5mba7\\$1ch\\$1@lfa222122.richmond. edu...\n\n>To put it another way, Tom Roberts\'s recipe gives the\n> ratio of the frequencies (nu\'/nu) rather than the fractional change in\n> frequency (Delta nu / nu).\n>\n\nBut, it doesn\'t appear that Tom Roberts\' recipe yields the correct value for\nnu\'/nu. In flat spacetime with the source moving radially away from the\nreceiver with relative speed v, SR yields nu\'/nu = Sqrt[(1-v/c)/(1+v/c)].\nHowever, Daryl McCullough showed that Roberts\' recipe applied to this case\nyields the quantity 1/Sqrt[1-(v/c)^2]. How can these different expressions\nbe compatible?\n\nTom S.\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><ebunn@lfa221051.richmond.edu> wrote in message
news:c5mba7$1ch$1@lfa222122.richmond.edu...

>To put it another way, Tom Roberts's recipe gives the
> ratio of the frequencies (\nu'/\nu) rather than the fractional change in
> frequency (\Delta \nu / \nu).
>

But, it doesn't appear that Tom Roberts' recipe yields the correct value for
\nu'/\nu. In flat spacetime with the source moving radially away from the
receiver with relative speed v, SR yields \nu'/\nu = \Sqrt[(1-v/c)/(1+v/c)].
However, Daryl McCullough showed that Roberts' recipe applied to this case
yields the quantity 1/\Sqrt[1-(v/c)^2]. How can these different expressions
be compatible?

Tom S.

ebunn@lfa221051.richmond.edu

Apr21-04, 04:24 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>In article <kh0hc.30211\\$0b4.43177@attbi_s51>,\nTom Snyder <xxxxx@xxxx.com> wrote:\n><ebunn@lfa221051.richmond.edu> wrote in message\n>news:c5mba7\\$1ch\\$1@lfa222122.richmond .edu...\n>\n>>To put it another way, Tom Roberts\'s recipe gives the\n>> ratio of the frequencies (nu\'/nu) rather than the fractional change in\n>> frequency (Delta nu / nu).\n>>\n>\n>But, it doesn\'t appear that Tom Roberts\' recipe yields the correct value for\n>nu\'/nu.\n\nYou\'re right! Sorry about that. I misremembered how it all works.\n\nTom Roberts is right that you can understand all spectral shifts\n(gravitational and Doppler) in a unified way by parallel transporting\nthe source velocity vector to the observer (or vice versa) along the\npath of the light ray joining them. But you don\'t simply dot them\ninto each other. Rather, you parallel transport the vectors together\nand then use the special-relativistic Doppler shift formula on them.\n\nIn a bit more detail, to compute the spectral shift of light in any\nspacetime, you\n\n1. Parallel transport the source\'s 4-velocity to the observer\'s\nlocation along the null geodesic followed by the light ray.\n\n2. Calculate the spatial part of that 4-vector in the observer\'s rest\nframe.\n\n3. Use the resulting 3-vector in the ordinary special relativistic\nDoppler formula to determine the spectral shift. For instance, if\nv is radial, just use\n\n1+z = sqrt((1+v)/(1-v))\n\n(with c=1, of course).\n\nIn flat spacetime, this is obviously (vacuously) equivalent to the\nstandard procedure. The nice thing about it is that it works\nin curved spacetime as well. This recipe makes explicitly clear\nthe fact that there\'s no fundamental difference between gravitational\nshifts and Doppler shifts. In fact, whether you choose to describe\na spectral shift as gravitational or Doppler depends mostly on what\ncoordinates you\'re using.\n\n-Ted\n\n--\n[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.]\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>In article <kh0hc.30211$0b4.43177@attbi_s51>,
Tom Snyder <xxxxx@xxxx.com> wrote:
><ebunn@lfa221051.richmond.edu> wrote in message
>news:c5mba7$1ch$1@lfa222122.richmond.edu...
>
>>To put it another way, Tom Roberts's recipe gives the
>> ratio of the frequencies (\nu'/\nu) rather than the fractional change in
>> frequency (\Delta \nu / \nu).
>>
>
>But, it doesn't appear that Tom Roberts' recipe yields the correct value for
>\nu'/\nu.

You're right! Sorry about that. I misremembered how it all works.

Tom Roberts is right that you can understand all spectral shifts
(gravitational and Doppler) in a unified way by parallel transporting
the source velocity vector to the observer (or vice versa) along the
path of the light ray joining them. But you don't simply dot them
into each other. Rather, you parallel transport the vectors together
and then use the special-relativistic Doppler shift formula on them.

In a bit more detail, to compute the spectral shift of light in any
spacetime, you

1. Parallel transport the source's 4-velocity to the observer's
location along the null geodesic followed by the light ray.

2. Calculate the spatial part of that 4-vector in the observer's rest
frame.

3. Use the resulting 3-vector in the ordinary special relativistic
Doppler formula to determine the spectral shift. For instance, if
v is radial, just use

1+z = \sqrt((1+v)/(1-v))

(with c=1, of course).

In flat spacetime, this is obviously (vacuously) equivalent to the
standard procedure. The nice thing about it is that it works
in curved spacetime as well. This recipe makes explicitly clear
the fact that there's no fundamental difference between gravitational
shifts and Doppler shifts. In fact, whether you choose to describe
a spectral shift as gravitational or Doppler depends mostly on what
coordinates you're using.

-Ted

--
[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.]

Ken S. Tucker

Apr24-04, 12:18 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>ebunn@lfa221051.richmond.edu wrote in message news:<c63ka7\\$e3f\\$1@lfa222122.richmond.edu>...\ n>In article <kh0hc.30211\\$0b4.43177@attbi_s51>,\n>Tom Snyder <xxxxx@xxxx.com> wrote:\n>><ebunn@lfa221051.richmond.edu> wrote in message\n>>news:c5mba7\\$1ch\\$1@lfa222122.richmon d.edu...\n>>\n>>>To put it another way, Tom Roberts\'s recipe gives the\n>>> ratio of the frequencies (nu\'/nu) rather than the fractional change in\n>>> frequency (Delta nu / nu).\n>>But, it doesn\'t appear that Tom Roberts\' recipe yields the correct value for\n>>nu\'/nu.\n>\n>You\'re right! Sorry about that. I misremembered how it all works.\n>Tom Roberts is right that you can understand all spectral shifts\n>(gravitational and Doppler) in a unified way by parallel transporting\n>the source velocity vector to the observer (or vice versa) along the\n>path of the light ray joining them. But you don\'t simply dot them\n>into each other. Rather, you parallel transport the vectors together\n>and then use the special-relativistic Doppler shift formula on them.\n>In a bit more detail, to compute the spectral shift of light in any\n>spacetime, you\n>\n>1. Parallel transport the source\'s 4-velocity to the observer\'s\n> location along the null geodesic followed by the light ray.\n>2. Calculate the spatial part of that 4-vector in the observer\'s rest\n> frame.\n>3. Use the resulting 3-vector in the ordinary special relativistic\n> Doppler formula to determine the spectral shift. For instance, if\n> v is radial, just use\n> 1+z = sqrt((1+v)/(1-v)) (with c=1, of course).\n>In flat spacetime, this is obviously (vacuously) equivalent to the\n>standard procedure. The nice thing about it is that it works\n>in curved spacetime as well. This recipe makes explicitly clear\n>the fact that there\'s no fundamental difference between gravitational\n>shifts and Doppler shifts. In fact, whether you choose to describe\n>a spectral shift as gravitational or Doppler depends mostly on what\n>coordinates you\'re using.\n>-Ted\n\nThat\'s true, the tough part is making that work in GR,\nwhen g-fields are present, ie. TR\'s "dot product" must\ninvolve the metric "g_uv", which is an evolved treatment\nof the "dot product" concept.\n\nLet E_u and E^ v be photon energy components,\nthen GR requires the association,\n\nE_u = g_uv E^v {uv =0,1,2,3}\n\nto work in all circumstances.\n\nI use, E_0 = hf = E^i {i,j=1,2,3}\n\nand E_i =0\n\nwhere h= Planck\'s constant and f= frequency.\n\nExpanding over t,x\n\nE_0 = g_00 E^0 + g_0i E^i\n\nI use g_0i = - g_ij dx^j/dx^0 generally.\n\nSimplifying to SR by setting g_00=1\nand g_0i = -V in direction "i" gives,\n((I use g_11 =+1)),\n\nE_0 = E^0 - (V*E^i = V*E_0)\n\nThen the the Doppler Effect is found by\nalgebraic rearrangement to produce,\n\nE^0 = E_0 (1+V)\n\nEvery calculation above relies upon,\nand is consistent with,\n\ndx_i = 0.\n\nI\'m prepared to stake my reputation\non the theory of relativity being defined\nby dx_i =0.\n\nRegards\nKen S. Tucker\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>ebunn@lfa221051.richmond.edu wrote in message news:<c63ka7$e3f$1@lfa222122.richmond.edu>...
>In article <kh0hc.30211$0b4.43177@attbi_s51>,
>Tom Snyder <xxxxx@xxxx.com> wrote:
>><ebunn@lfa221051.richmond.edu> wrote in message
>>news:c5mba7$1ch$1@lfa222122.richmond.edu...
>>
>>>To put it another way, Tom Roberts's recipe gives the
>>> ratio of the frequencies (\nu'/\nu) rather than the fractional change in
>>> frequency (\Delta \nu / \nu).
>>But, it doesn't appear that Tom Roberts' recipe yields the correct value for
>>\nu'/\nu.
>
>You're right! Sorry about that. I misremembered how it all works.
>Tom Roberts is right that you can understand all spectral shifts
>(gravitational and Doppler) in a unified way by parallel transporting
>the source velocity vector to the observer (or vice versa) along the
>path of the light ray joining them. But you don't simply dot them
>into each other. Rather, you parallel transport the vectors together
>and then use the special-relativistic Doppler shift formula on them.
>In a bit more detail, to compute the spectral shift of light in any
>spacetime, you
>
>1. Parallel transport the source's 4-velocity to the observer's
> location along the null geodesic followed by the light ray.
>2. Calculate the spatial part of that 4-vector in the observer's rest
> frame.
>3. Use the resulting 3-vector in the ordinary special relativistic
> Doppler formula to determine the spectral shift. For instance, if
> v is radial, just use
> 1+z = \sqrt((1+v)/(1-v)) (with c=1, of course).
>In flat spacetime, this is obviously (vacuously) equivalent to the
>standard procedure. The nice thing about it is that it works
>in curved spacetime as well. This recipe makes explicitly clear
>the fact that there's no fundamental difference between gravitational
>shifts and Doppler shifts. In fact, whether you choose to describe
>a spectral shift as gravitational or Doppler depends mostly on what
>coordinates you're using.
>-Ted

That's true, the tough part is making that work in GR,
when g-fields are present, ie. TR's "dot product" must
involve the metric "g_{uv}", which is an evolved treatment
of the "dot product" concept.

Let E_u and E^ v be photon energy components,
then GR requires the association,

E_u = g_{uv} E^v[/itex] {uv [itex]=0,1,2,3}

to work in all circumstances.

I use, E_0 = hf = E^i {i,j=1,2,3}

and E_i =0

where h= Planck's constant and f= frequency.

Expanding over t,x

E_0 = g_{00} E^0 + g_{0i} E^i

I use g_{0i} = - g_{ij} dx^j/dx^0 generally.

Simplifying to SR by setting g_{00}=1
and g_{0i} = -V in direction "i" gives,
((I use g_{11} =+1)),E_0 = E^0 - (V*E^i = V*E_0)

Then the the Doppler Effect is found by
algebraic rearrangement to produce,

E^0 = E_0 (1+V)

Every calculation above relies upon,
and is consistent with,

dx_i = .

I'm prepared to stake my reputation
on the theory of relativity being defined
by dx_i =0.

Regards
Ken S. Tucker