How Does Spacetime Curvature Affect Tangential Light Speed Measurements?

In summary, the speed of light at infinity can be calculated as (1-2M/r)c, and locally measured as c for radial measurements and sqrt(1-2M/r)c for tangential measurements. The coordinate velocities can be converted into velocities with physical significance by introducing local coordinates near a specific point. The view is generally held that coordinate results have no physical significance.
  • #1
Austin0
1,160
1
The coordinate speed of light relative to its speed at infinity is calculated as (1-2M/r)c (at infinity) yes??

SO it makes sense that local measurements made with local rulers and clocks which are contracted and dilated respectively would still be c for radial measurements.
But how does that work with tangential measurements where the clocks are dilated but the distance ruler is not?
Thanks
 
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  • #2
For light traveling in the tangent direction, it is measured as c locally and calculated as sqrt(1 - 2m/r) c by a distant observer at infinity.
 
  • #3
Austin0 said:
The coordinate speed of light relative to its speed at infinity is calculated as (1-2M/r)c (at infinity) yes??

Thanks

If you use the usual formulation for 4-velocity, where you use the proper time [itex]\tau[/itex], your coordinate velocity at some particular point would be the four-velocity: [itex] [dt/d\tau, dr/d\tau, d\theta/d\tau, d\phi/d\tau] [/itex].

You might be thinking of [itex] [dr/dt, d\theta/dt, d\phi/dt] [/itex] I suppose, for your coordinate velocitiy? I can't tell from your post what you're thinking, the four-velocity approach is the most commonly used in textbooks and papers.

I don't know what are you thinking of as "the coordinate speed relative to its speed at infinity". If you want to know how to convert coordinate velocities into velocities that (unlike coordinate velocities) have some physical significance, I can try to tell you how to do that, if that's the real question.

I"ll do a short verison now, but if it's not your question I suppose it's wasted.

Theshort answer is that you introduce some local coordinates (t1, r1, theta1, phi1) near some point P where t1 is a linear function of t, r1 is a linear function of r, theta1 is a linear function of theta, and phi1 is a linear function of phi.

It's simplest if t1=r1=theta1=phi1=0 at your point P - so your linear function includes a scale factor and an offset, basically.

Then, when you make the metric at point P a unit diagonal in geometric units, or [c,0,0,0] in non-geometric units, you can say that dr1 corresponds to a change of distance. If your metric isn't diagonal, dr1 can't be physically interpreted as a change of distance.

So [itex]dr,dt,d\theta, d\phi[/itex] are numbers that don't represent distances because the metric isn't diagonal. [itex]dr_1, dt1, d\theta_1, and d\phi_1 [/itex]ARE numbers which DO represent distance because the metric IS diagional, at least at point P.

That's it in a nutshell. If this addresses your question and you want more let me know. If you don't think it addresses your question, you can either hope someone else guesses more accurately what it is that you want, or expand on your question a bit so that I know what you're asking :).
 
  • #4
grav-universe said:
For light traveling in the tangent direction, it is measured as c locally and calculated as sqrt(1 - 2m/r) c by a distant observer at infinity.

Thanks this makes complete sense and answers my question.

pervect said:
If you use the usual formulation for 4-velocity, where you use the proper time [itex]\tau[/itex], your coordinate velocity at some particular point would be the four-velocity: [itex] [dt/d\tau, dr/d\tau, d\theta/d\tau, d\phi/d\tau] [/itex].

You might be thinking of [itex] [dr/dt, d\theta/dt, d\phi/dt] [/itex] I suppose, for your coordinate velocitiy? I can't tell from your post what you're thinking, the four-velocity approach is the most commonly used in textbooks and papers.

I don't know what are you thinking of as "the coordinate speed relative to its speed at infinity". If you want to know how to convert coordinate velocities into velocities that (unlike coordinate velocities) have some physical significance, I can try to tell you how to do that, if that's the real question.

I"ll do a short verison now, but if it's not your question I suppose it's wasted.

Theshort answer is that you introduce some local coordinates (t1, r1, theta1, phi1) near some point P where t1 is a linear function of t, r1 is a linear function of r, theta1 is a linear function of theta, and phi1 is a linear function of phi.

It's simplest if t1=r1=theta1=phi1=0 at your point P - so your linear function includes a scale factor and an offset, basically.

Then, when you make the metric at point P a unit diagonal in geometric units, or [c,0,0,0] in non-geometric units, you can say that dr1 corresponds to a change of distance. If your metric isn't diagonal, dr1 can't be physically interpreted as a change of distance.

So [itex]dr,dt,d\theta, d\phi[/itex] are numbers that don't represent distances because the metric isn't diagonal. [itex]dr_1, dt1, d\theta_1, and d\phi_1 [/itex]ARE numbers which DO represent distance because the metric IS diagional, at least at point P.

That's it in a nutshell. If this addresses your question and you want more let me know. If you don't think it addresses your question, you can either hope someone else guesses more accurately what it is that you want, or expand on your question a bit so that I know what you're asking :).

grav-universe cleared up the question.

I am still trying to understand the reasoning behind the generally held view that coordinate results have no physical significance but that is a question for another thread.
Thanks for your response .
 
  • #5
for sharing this information! The concept of coordinate speed of light can be a bit confusing, but essentially it refers to the speed of light as measured by an observer at a specific coordinate in a gravitational field. In the equation you provided, M represents the mass of the object causing the gravitational field and r represents the distance from that object.

To answer your question about tangential measurements, it is important to consider the concept of spacetime curvature. In a gravitational field, spacetime is curved, meaning that the distance between two points is not constant. This can result in the dilation of clocks, as you mentioned, but also in the contraction or expansion of distance rulers.

For tangential measurements, the distance ruler may appear to be the same length, but due to the curvature of spacetime, the actual distance between two points may be different. This can result in a difference in the measured speed of light compared to radial measurements.

Overall, the coordinate speed of light takes into account the effects of spacetime curvature on measurements of light, and is an important concept in understanding the behavior of light in a gravitational field.
 

1. What is "Local c" in Sc coordinates?

"Local c" in Sc coordinates refers to the local curvature of a surface at a specific point, measured in units of inverse length (e.g. meters^-1). It is a fundamental concept in differential geometry and is used to describe the shape of surfaces in three-dimensional space.

2. How is "Local c" calculated in Sc coordinates?

The calculation of "Local c" in Sc coordinates involves determining the curvature of a surface at a specific point using mathematical equations such as the Gaussian and Mean curvatures. These equations take into account the changes in the surface's tangent planes and are dependent on the surface's first and second derivatives.

3. What is the significance of "Local c" in Sc coordinates?

"Local c" in Sc coordinates is an important measure of the shape of a surface at a specific point. It helps to quantify the degree of bending or curving of the surface and is used in various fields such as physics, engineering, and computer graphics to study and model surfaces.

4. How does "Local c" differ from global curvature?

"Local c" refers to the curvature of a surface at a specific point, while global curvature refers to the overall curvature of a surface. While "Local c" is dependent on the surface's local features, global curvature takes into account the entire surface's shape and can be used to compare the curvature of different surfaces.

5. Can "Local c" be negative in Sc coordinates?

Yes, "Local c" can be negative in Sc coordinates. This indicates that the surface is bending in the opposite direction at that point. A positive "Local c" value indicates a convex curvature, while a negative value indicates a concave curvature.

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