Correct formula for gravitational redshift

In summary: A, B...B), each with a mass of m, that a certain distance away from a third (large) body (C...C) with a mass of M. Suppose the bodies are small enough and the distance between them is big enough so that the two small bodies are essentially in the same gravitational field. I'm not sure what you mean by "essentially in the same gravitational field".If you mean that the local gravitational forces at A and B are the same, then this is the definition of the statement that the two small bodies are in the same gravitational field. If you mean that the small bodies are sufficiently close together that you can consider them as a single object, that's different - you can do
  • #1
notknowing
185
0
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) ?
 
Physics news on Phys.org
  • #2
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:
  • #3
pervect said:
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.

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 ?
 
  • #4
notknowing said:
How should the "r" be interpreted or measured in this expression ?

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:
  • #5
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 Schwarzschild 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
 
  • #6
George Jones said:
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].
So how do you reason you can use [tex] \inline \pi [/tex] in curved space?
 
Last edited:
  • #7
pervect said:
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 Schwarzschild 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

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 ?
 
  • #8
MeJennifer said:
So how do you reason you can use [tex] \inline \pi [/tex] in curved space?

Why can't [itex]\pi[/itex] be used? It's just a number.
 
  • #9
notknowing said:
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 ?

In a general spacetime, I don't think this is possible.
 
  • #10
George Jones said:
Why can't [itex]\pi[/itex] be used? It's just a number.
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:
  • #11
MeJennifer said:
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?
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.
 
  • #12
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.
 
  • #13
pervect said:
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.
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.
 

1. What is gravitational redshift?

Gravitational redshift is the phenomenon where light from an object appears to have a longer wavelength and lower frequency when observed from a region with a strong gravitational field, such as near a massive object like a planet or star.

2. How does gravitational redshift occur?

According to Einstein's theory of general relativity, massive objects cause a curvature in space-time. When light travels through this curved space-time, its path is bent, causing a change in its wavelength and frequency.

3. What is the correct formula for calculating gravitational redshift?

The formula for gravitational redshift is Δλ/λ = GM/Rc^2, where Δλ is the change in wavelength, λ is the original wavelength, G is the gravitational constant, M is the mass of the object, R is the distance from the object's center of mass, and c is the speed of light.

4. How can gravitational redshift be observed?

Gravitational redshift can be observed through spectroscopy, where the change in wavelength of light from an object with a strong gravitational field can be measured and compared to the expected value from the correct formula.

5. Are there any practical applications of gravitational redshift?

Yes, gravitational redshift has practical applications in the field of precision timekeeping. The effect must be taken into account when using GPS satellites, as the clocks on the satellites experience a slight time dilation due to their position in a strong gravitational field.

Similar threads

  • Special and General Relativity
Replies
27
Views
2K
  • Special and General Relativity
Replies
5
Views
447
  • Special and General Relativity
Replies
4
Views
1K
  • Special and General Relativity
2
Replies
43
Views
2K
  • Special and General Relativity
Replies
28
Views
2K
  • Special and General Relativity
Replies
2
Views
831
  • Special and General Relativity
3
Replies
82
Views
4K
  • Special and General Relativity
Replies
9
Views
1K
  • Special and General Relativity
Replies
19
Views
1K
  • Special and General Relativity
3
Replies
75
Views
5K
Back
Top