# Wavelength of Light Ray Affected by Gravity

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• GeniVasc
In summary, the conversation discusses Einstein's 1911 paper on the influence of gravitation on the propagation of light, where he derived equations for the frequencies and speed of light measured by two different observers on Earth's surface. The result was a gravitational redshift and a coordinate speed of light, which is different from the locally measured speed of light. The conversation ends with a question about the transformation of wavelength between the two observers, highlighting the difference between the coordinate and locally measured quantities.
GeniVasc
I was reading Einstein's 1911 paper named "On the Influence of Gravitation on the Propagation of Light" when stated the formula for frequencies measured by observers at different fixed positions (heights) on Earth surface. One observer is at the origin of some coordinate system and measures a frequency ##\nu_0## of a light beam. The second observer is at some position ##x## in the same chart and measures a frequency ##\nu## of the same light beam. Einstein obtais that

$$\nu = \nu_0 \left( 1 + \frac{\phi}{c^2}\right) \tag1$$

where ##\phi \le 0## such that ##\nu \le \nu_0##, the famous gravitational redshift.

However, at the end of section 3 of this paper, Einstein came to other result about speed of light measured by these observers:

$$c= c_0 \left( 1 + \frac{\phi}{c^2} \right) \tag2$$
and so ##c\le c_0##. Here ##c## is measured by the "##\nu##" observer an its called "coordinate speed of light" and ##c_0## is measured by the "##\nu_0##" observer, and it is the usual vacuum speed of light.

My question is: How the wavelength should transform from one observer to the other? I mean, if I take eq. (1), I could use

$$\frac{c}{\lambda} = \frac{c_0}{\lambda_0}\left( 1+ \frac{\phi}{c^2} \right)$$

Then by eq.(2) we obtain ##\lambda =\lambda_0##, which I think is non-sense because since the observers have two different frequencies of the same light beam, they certainly should disagree about its wavelength. Where is my mistake?

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Sorry for the various TeX mistypes. I think I've corrected all of them.

GeniVasc said:
since the observers have two different frequencies of the same light beam, they certainly should disagree about its wavelength.
Not necessarily if they also disagree about its speed.

You would normally use local measurements of ##c##, ##\nu## and ##\lambda##, in which case the speed of light is always ##c## and hence$$\frac{\nu}{\nu_0}=\frac{\lambda_0}{\lambda}$$as you appear to be expecting. But here you seem to be using the coordinate velocity of light, and hence you are defining a "coordinate wavelength", which isn't necessarily trivially related to what anyone would measure.

Ibix said:
Not necessarily if they also disagree about its speed.

You would normally use local measurements of ##c##, ##\nu## and ##\lambda##, in which case the speed of light is always ##c## and hence$$\frac{\nu}{\nu_0}=\frac{\lambda_0}{\lambda}$$as you appear to be expecting. But here you seem to be using the coordinate velocity of light, and hence you are defining a "coordinate wavelength", which isn't necessarily trivially related to what anyone would measure.

$$\frac{\lambda - \lambda_0}{\lambda_0} = \frac{\Delta \lambda}{\lambda_0} = \Delta \phi, \ c=1$$

where his ##\Delta \phi## is my ##\phi##. The above formula could be derived if I assume that the speed of light is the same for both observers. That's what bothering me.

GeniVasc said:
The above formula could be derived if I assume that the speed of light is the same for both observers. That's what bothering me.
I don't see the problem. You are using the coordinate speed of light. Carroll is using the locally measured speed of light. You're getting a "coordinate wavelength" and Carroll is getting the locally measured wavelength. They're different things.

I'm not sure I understand what you are asking.

Ibix said:
I don't see the problem. You are using the coordinate speed of light. Carroll is using the locally measured speed of light. You're getting a "coordinate wavelength" and Carroll is getting the locally measured wavelength. They're different things.

I'm not sure I understand what you are asking.
I think the "coordinate speed of Light" refers to the perspective of only one observer (i.e
coordinate system), when the Carrols formula, as you said, involve measurements at different locations(i.e from perspective of the two observers). I'm not sure if I get it, but I think that is a reasonable explanation.

GeniVasc said:
I think the "coordinate speed of Light" refers to the perspective of only one observer (i.e
coordinate system), when the Carrols formula, as you said, involve measurements at different locations(i.e from perspective of the two observers). I'm not sure if I get it, but I think that is a reasonable explanation.
If we have a static spacetime then we can mark two points and measure the proper distance between them. We can then use a clock synchronisation convention to measure the time it takes for a light signal to travel from one point to the other. This gives you a "coordinate" speed of light for that path.

If, however, you measure the speed of light locally, you will get the invariant answer ##c##. Note that "locally" here is effectively the limit as the proper distance between the two points reduces to zero.

The coordinate speed for light that has traveled a large distance is irrelevant to the necessarily local (*) measurement of its wavelength: only the invariant local speed matters.

(*) Note that we are not talking here about measuring the distance between peaks of some extended physical wave. We are talking about a local interaction between the light and a detector.

In general, the measured wavelength of light depends on the relationship between the source and the detector. For example, in flat spacetime relative motion between source and detector leads to blueshift or redshift. More precisely, this has nothing to do with the light changing, per se, but means the source and detector measure different wavelengths for the light.

Similarly, for a source and detector separated by curved spacetime, each may measure a different wavelength for a given light signal.

PS although even in serious texts and articles redshift is often described as the light changing wavelength, a light signal has no inherent wavelength or energy. The wavelength relative to the source is, of course, important; but, once the light has left the source there is no internal variable that retains this value. The result of any subsequent measurement of the light depends on the relationship of the detector to the source.

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GeniVasc said:
I think the "coordinate speed of Light" refers to the perspective of only one observer (i.e
coordinate system)
A coordinate system is not one observer. Assuming it's not a null coordinate system (which is different), it's essentially an infinite number of observers filling spacetime who have agreed to label themselves with three numbers which they call their spatial coordinates. What the metric does is relate differences in coordinates to actual physical things that people would measure.

Coordinate speed is essentially the rate of change of spatial coordinate with respect to coordinate time. That does not necessarily have much to do with real measurements. A classic example would be a light ray passing through the origin of a polar coordinate system. The coordinate velocity isn't even defined as it passes through the origin, but reality doesn't care about our bad choice of labels - the light keeps on tickin' along at ##c##.

Coordinate velocity can certainly be useful mathematically, but it isn't usually what anyone actually measures. Carroll's formulation is more related to what any physics lab would measure given light emitted with a certain frequency.

It's important to realize that coordinate differences only relate simply to actual distances in Cartesian coordinates on Euclidean space. The metric is how you relate these things.

## 1. How does gravity affect the wavelength of light rays?

Gravity can cause the wavelength of light rays to change as they pass through a gravitational field. This is known as gravitational redshift, where the wavelength of light is stretched as it moves away from the source of the gravitational field.

## 2. Does the strength of gravity impact the wavelength of light rays?

Yes, the strength of gravity does impact the wavelength of light rays. The stronger the gravitational field, the greater the gravitational redshift will be. This means that light passing through a stronger gravitational field will have a longer wavelength compared to light passing through a weaker gravitational field.

## 3. Can light rays be affected by both gravity and other forces?

Yes, light rays can be affected by both gravity and other forces. For example, light passing through a medium such as water or glass can experience refraction due to the interaction between the light and the atoms in the medium. This can cause a change in the wavelength of the light.

## 4. Is the wavelength of light affected by gravity the same for all types of light?

No, the wavelength of light affected by gravity can vary depending on the type of light. For example, the wavelength of visible light may be affected differently than the wavelength of radio waves or gamma rays. This is due to the different properties and behaviors of these types of light.

## 5. Can the wavelength of light be used to measure the strength of gravity?

Yes, the wavelength of light can be used to indirectly measure the strength of gravity. By measuring the gravitational redshift of light from a known source, scientists can calculate the strength of the gravitational field it passed through. This has been used in experiments to confirm Einstein's theory of general relativity.

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