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Correct formula for gravitational redshift

  1. Oct 11, 2006 #1
    Based on general relativity (see book Gravity; an introduction to Einstein's General relativity by James B. Hartle), the frequency of light (emitted at R) is modified by the factor SQRT (1 - 2 G M / c**2 R) (see page 191).

    On the other hand, based on the equivalence principle, one arrives at the factor (1- G M /c**2 R). It is clear that the former equation reduces to the latter for small values of G M /c**2 R.

    What I do not understand however is why the simple straightforward derivation, based on the equivalence principle, apparently gives only an approximation to the real red-shift. So, what could be wrong or missing in the original derivation (same book, page 116-119) ?
     
  2. jcsd
  3. Oct 11, 2006 #2

    pervect

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    It seems to me the main thing that is missing is the exact form of the gravitational field of a massive body according to GR.

    The equivalence principle only applies to nearby clocks. You can use the Newtonian gravitational field to get an approximate time dilation formula, but you need the exact GR formula for the gravitational field to get the exact formula.

    In fact, the problem is usually worked the other way. One starts with the Schwarzschild solution, which gives the time dilation formula explicitly, and then uses the Schwarzschild solution to find the exact GR formula for the "gravitational field".


    http://en.wikipedia.org/wiki/Frame_fields_in_general_relativity

    gives the exact answer, which is


    [tex] \frac{m/r^2}{\sqrt{1-2m/r}} \,[/tex]

    This is the force/unit mass, i.e gravitational acaceleration, as measured by a local observer using his local defintions of distance and time, i.e. local clocks and local rods - or a local accelerometer.

    You can see that the acceleration required goes to infinity at the event horizon, i.e. at r=2m.

    I should add that this expression has been written using geometric units.

    This is the outward acceleration required for the observer to remain stationary at a constant Schwarzschild r coordinate (in terms of local clocks and rods).

    Note that sometimes people use a reference frame "at infinty" rather than local clocks - this gives rise to the so-called "surface gravity" of a black hole, which is finite.

    http://en.wikipedia.org/wiki/Surface_gravity

    I personally tend use local clocks and rulers wherever possible.
     
    Last edited: Oct 11, 2006
  4. Oct 11, 2006 #3
    Thank you very much for this detailed answer. It has been puzzling me for some time. The correct expression for the gravitational acceleration is interesting (and was new to me). One final question : How should the "r" be interpreted or measured in this expression ?
     
  5. Oct 11, 2006 #4

    George Jones

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    Construct a spherical shell around the object, and measure the surface area [itex]A[/itex] of the shell. Then,

    [tex]r = \sqrt{\frac{A}{4 \pi}.[/tex]

    Now construct a spherical shell that has [itex]R[/itex], and with [itex]R[/itex] slightly larger than [itex]r[/itex]. Drop a tape measure straight down from the larger shell to the smaller shell. If spacetime were flat, then the distance measured by the tape would [itex]R - r[/itex]. However, a spacetime is curved, and the distance [itex]L[/itex] measured by the tape is approximately

    [tex]L = \frac{R - r}{\sqrt{1 - 2m/r}},[/tex]

    which is greater than [itex]R - r[/itex].
     
    Last edited: Oct 11, 2006
  6. Oct 12, 2006 #5

    pervect

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    As George said, the 'r' is the Schwarzschild radial coordinate. The set of points of constant r is a sphere around the central mass, and the coordinates are designed such that the area of this sphere using local rulers is 4*pi*r^2.

    For more details, see the Wikipedia article on the Schwarschild metric

    http://en.wikipedia.org/wiki/Schwarzschild_metric

    that will also explain (hopefully) George's computation of L above.

    To convert geometric units into non-geometric units, replace all occurences of m/r by Gm/r c^2
     
  7. Oct 12, 2006 #6
    So how do you reason you can use [tex] \inline \pi [/tex] in curved space?
     
    Last edited: Oct 12, 2006
  8. Oct 12, 2006 #7
    Thanks Pervect and George. Another question came to my mind which is related to this. Suppose I have two small bodies (say A and B) which are both located in a gravitational field (of a larger body C), such that A and B are at a different gravitational "potential". How would the distance between A and B be defined (or measured) to an observer which is far away from A,B and C ?
     
  9. Oct 13, 2006 #8

    George Jones

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    Why can't [itex]\pi[/itex] be used? It's just a number.
     
  10. Oct 13, 2006 #9

    George Jones

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    In a general spacetime, I don't think this is possible.
     
  11. Oct 13, 2006 #10
    It seems you are simply avoiding the question.

    You claim that:

    [tex]Distance_R_r > \sqrt{\frac{A_1}{4 \pi}} - \sqrt{\frac{A_2}{4 \pi}}[/tex]

    because spacetime is curved.
    But if spacetime is curved then how can you use [itex]\pi[/itex] to express the relationship between A and r?
     
    Last edited: Oct 13, 2006
  12. Oct 13, 2006 #11
    The experts are just too clever to understand your simple question. The reason to use Pi is, in my view, just to have correspondence with our classical definition of distance in flat spacetime. If it were another number (say 4), then, clearly, when going from curved to flat spacetime we would not find back the familiar formula for the volume of a sphere.
     
  13. Oct 13, 2006 #12

    pervect

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    In general, to measure the distance between two points, one measures the length of a curve connecting them. This can be done fairly easily if one has a metric, IF one has the exact curve.

    The problem that makes the notion of distance tricky is the problem of specifying the particular connecting curve.

    For instance, in cosmology, the comoving distance requires that all points on the curve have the same "cosmological time". One might in genaral say that the usual notion of distance require that the line connecting the points be "simultaneous" in some sense.

    Since the defintion of simultaneity is observer dependent, it's not really clear how to draw the curve between points A and B. In general, there isn't even any general defintion of "simultaneity" in GR for different events that I'm aware of, other than that imposed by adopting a specific coordinate system. Usually, in a specific coordinate system, events that have the same time coordinate are "simultaneous". One might also impose the requirement that the interval between simultaneous points must be spacelike, which will restrict the number of coordinate systems a bit.
     
  14. Oct 13, 2006 #13
    I do not see the problem.
    The rountrip time of a light signal from A to B divided by 2 could be an fine measure of distance from A to B.

    The "problem" comes when people desire a measure of distance in terms of some flat plane of simultaneity. Something that is utterly absurd in curved spacetime.
     
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