How Does Spacetime Curvature Affect Tangential Light Speed Measurements?

[Austin0]

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

 

[grav-universe]

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.

 

[pervect]

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 \tau, your coordinate velocity at some particular point would be the four-velocity: [dt/d\tau, dr/d\tau, d\theta/d\tau, d\phi/d\tau].

You might be thinking of [dr/dt, d\theta/dt, d\phi/dt] 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 dr,dt,d\theta, d\phi are numbers that don't represent distances because the metric isn't diagonal. dr_1, dt1, d\theta_1, and d\phi_1ARE 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 :).

 

[Austin0]

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.

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

You might be thinking of [dr/dt, d\theta/dt, d\phi/dt] 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 dr,dt,d\theta, d\phi are numbers that don't represent distances because the metric isn't diagonal. dr_1, dt1, d\theta_1, and d\phi_1ARE 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 .