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Why does light travel at the speed of light?


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it is constant 'locally'.

 

http://en.wikipedia....l_time_dilation

 

The speed of light in a locale is always equal to c according to the observer who is there. The stationary observer's perspective corresponds to the local proper time. Every infinitesimal region of space time may have its own proper time that corresponds to the gravitational time dilation there, where electromagnetic radiation and matter may be equally affected, since they are made of the same essence (as shown in many tests involving the famous equation E=mc2). Such regions are significant whether or not they are occupied by an observer. A time delay is measured for signals that bend near the Sun, headed towards Venus, and bounce back to Earth along a more or less similar path. There is no violation of the speed of light in this sense, as long as an observer is forced to observe only the photons which intercept the observing faculties and not the ones that go passing by in the depths of more (or even less) gravitational time dilation.

 

I already explained why a correct computation of the speed of light using the physical spacetime of general relativity gives that the speed of light is constant, in general relativity, along any path. If you use a fictitious background spacetime then you obtain a speed that varies.

 

A consistent computation shows that light does not bend because light is moving in a curved spacetime at constant speed.

 

http://math.ucr.edu/...d_of_light.html

 

Finally, we come to the conclusion that the speed of light is not only observed to be constant; in the light of well tested theories of physics, it does not even make any sense to say that it varies.

 

Check also: Why doesn't gravity change the speed of light?

Edited by juanrga
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No. I said that is constant globally and given by c.

Nope. granpa is correct. The speed of light is locally equal to c. Using the words global and general relativity in the same sentence doesn't make sense.

 

Look to special relativity first. Note that one of the postulates of special relativity is that the speed of light is the same to all inertial observers. Why the qualification? The answer is simple: The speed of light is not constant to non-inertial observers. This concept carries over to general relativity with one huge caveat. Inertial frames are global in special relativity, but they are local in general relativity.

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Nope. granpa is correct. The speed of light is locally equal to c. Using the words global and general relativity in the same sentence doesn't make sense.

The speed of light is locally c. But now consider Irwin Shapiro' work on radar ranging onf the inner plantes. The EM waves sent out from Earth and bounced off a planet or two (or whatever. I forgot the details over the years.) There was a time delay due to the path the EM wave took through the gravitational field thus making the light slow down. This is explained by saying that the speed of light varies as the light moves through a gravitational field. It is this notion in which it is said that the speed of light varies. To be exact, it's the coordinate speed of light that varies with gravity. That's the same sense in which Einstein proved that the speed of light varies in gravitational field in his 1911 paper.

 

See the derivation here - http://home.comcast.net/~peter.m.brown/gr/c_in_gfield.htm

 

Eq. (5) c' = (1 + Phi/c2)c

 

is the result that Einstein obtained in that paper. Similar derivations can be found in Ohanians GR text as mentioned above.

Edited by pmb
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The speed of light is locally c. But now consider Irwin Shapiro' work on radar ranging onf the inner plantes. The EM waves sent out from Earth and bounced off a planet or two (or whatever. I forgot the details over the years.)

Venus and Mercury.

 

I had a longer post that talked about how both the Sagnac and Shapiro effects exemplify that the local constancy of the speed of light is not global but decided to keep that post short and simple instead.

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Nope. granpa is correct. The speed of light is locally equal to c. Using the words global and general relativity in the same sentence doesn't make sense.

 

http://en.wikipedia....lobal_structure

 

http://en.wikipedia....etime_structure

 

Look to special relativity first. Note that one of the postulates of special relativity is that the speed of light is the same to all inertial observers. Why the qualification? The answer is simple: The speed of light is not constant to non-inertial observers. This concept carries over to general relativity with one huge caveat. Inertial frames are global in special relativity, but they are local in general relativity.

 

In SR, the speed of light is only constant among inertial observers. In 1911 Einstein considered an extension of SR where the speed of light was not constant in presence of gravitational fields but finally he took a geometrical route and in 1916 he developed general relativity where the speed of light is always constant.

 

In general relativity the speed of light is always constant, not only locally constant. c is always constant, because is a global property of the spacetimes considered in general relativity.

 

http://curious.astro....php?number=266

 

http://math.ucr.edu/...d_of_light.html

 

Yes I know that there is an old interpretation of general relativity based in an earlier misconception by Einstein that believes that the speed of light varies in the deflection of light or in the Shapiro effects but this is not true. From Baez link:

 

the modern interpretation is that the speed of light is constant in general relativity

 

Notice also that the word "local" is not even mentioned in the above link.

Edited by juanrga
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http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/speed_of_light.html

 

Einstein ...In the 1920 book "Relativity: the special and general theory" he wrote: . . . according to the general theory of relativity, the law of the constancy of the velocity of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity [. . .] cannot claim any unlimited validity. A curvature of rays of light can only take place when the velocity of propagation of light varies with position

 

In special relativity, the speed of light is constant when measured in any inertial frame. In general relativity, the appropriate generalisation is that the speed of light is constant in any freely falling reference frame (in a region small enough that tidal effects can be neglected). In this passage, Einstein is not talking about a freely falling frame, but rather about a frame at rest relative to a source of gravity

 

but a 'freely falling reference frame' cannot be global.

Edited by granpa
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In general relativity, the appropriate generalisation is that the speed of light is constant in any freely falling reference frame (in a region small enough that tidal effects can be neglected).

 

but a 'freely falling reference frame' cannot be global.

 

But neither Baez nor me are saying that there exists a global reference frame. We say something different.

Edited by juanrga
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juanrga: "Baez is right. The speed of light in general relativity is a constant."

pmb: "The speed of light is locally c."

granpa: "the speed of light is constant when measured locally."

etc..

 

So c = dr/dt

 

dr = cdt

 

0 = dr^2 - (cdt)^2

 

0 = dx^2 + dy^2 + dz^2 - (cdt)^2

 

So the constant speed of light implies that Lorentz Invariance exists (in the places that the posters claim c is constant).

 

This was my original point, that c is due to the geometrical form of the universe. Part of the answer to the question "Why does light travel at the Speed of Light?".

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juanrga: "Baez is right. The speed of light in general relativity is a constant."

pmb: "The speed of light is locally c."

granpa: "the speed of light is constant when measured locally."

etc..

 

So c = dr/dt

 

dr = cdt

 

0 = dr^2 - (cdt)^2

 

0 = dx^2 + dy^2 + dz^2 - (cdt)^2

 

So the constant speed of light implies that Lorentz Invariance exists (in the places that the posters claim c is constant).

 

This was my original point, that c is due to the geometrical form of the universe. Part of the answer to the question "Why does light travel at the Speed of Light?".

 

c = dr/dt and your 0 = dx^2 + dy^2 + dz^2 - (cdt)^2 are only valid in flat spacetimes.

 

Consider two frames which are not Lorentz invariant

 

ds2 = gab dxadxb = g'ab dx'adx'b

 

for both the primed and the non-primed frames the speed of light is constant and equals c. A light signal sent from a frame to the other travels with speed c.

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Consider two frames which are not Lorentz invariant

 

ds2 = gab dxadxb = g'ab dx'adx'b

 

for both the primed and the non-primed frames the speed of light is constant and equals c. A light signal sent from a frame to the other travels with speed c.

 

Locally the equation I gave is a reasonable approximation, I used it to show the sheer triviality of the fact that the constancy of "c" is a geometrical hypothesis. The standard form that you have given is correct generally. The point I was making is that "c" is a ratio of two, dimensional extents, ie: meters per second. Null geodesics given from 0 = gab dxadxb = g'ab dx'adx'b are geometrical entities that describe the path of light rays and express the geometrical nature of "c". Why is the speed of light constant? Because the universe is a four dimensional, pseudo Riemannian manifold where, in the general case, 0 = gab dxadxb = g'ab dx'adx'b gives the path taken by light photons.

 

The discussion about whether metric tensors are sewing together local Minkowskian spaces and coordinate systems or not etc. is non-sequitur to the essential point that the constancy of the speed of light is a geometrical phenomenon due to the nature of spacetime.

Edited by mindless
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I'll work in units where c=G=1 for simplicity.

Let someone standing on a platform at Schwarzschild coordinate r1 in Schwarzschild spacetime emit a light signal radially downwards at time t1. It reflects off a mirror at some height r2 and time t2, and returns to the initial height where it hits a detector. So, what does this person measure the speed of light to be?

 

The proper distance (distance he would measure with a ruler), s, between r1 andr2 is:

 

[math]s=\int_{r_1}^{r_2} \left (1-\frac{2M}{r} \right )^{-1/2}dr[/math]

 

Since we'll assume he's at a height where r>>2M, we can simplify this with a binomial expansion and discard higher order terms.

 

[math]s=\int_{r_1}^{r_2} \left (1+\frac{M}{r} \right )dr=\Delta r+Mln\left ( \frac{r_2}{r_1} \right )[/math]

 

 

The proper time the person measures between t1 and t2 is simply a time-dilated coordinate time:

 

[math]\tau =\sqrt{1-\frac{2M}{r_1}}\Delta t[/math]

 

But using the same simplifying approximation as before gives:

 

[math]\tau =\left (1-\frac{M}{r_1} \right )\Delta t[/math]

 

Δt can be found in terms of r by using the fact that light travels on null geodesics (i.e. ds2=0):

[math]ds^2=0~~\Rightarrow ~~dt=\left ( 1-\frac{2M}{r} \right )^{-1}dr\approx \left (1+\frac{2M}{r} \right )dr[/math]

[math]\Rightarrow ~~ \Delta t=\Delta r+2Mln\left ( \frac{r_2}{r_1} \right )[/math]

[math]\tau =\left (1-\frac{M}{r_1} \right )\left (\Delta r+2Mln\left ( \frac{r_2}{r_1}\right ) \right )[/math]

Multiplying the times and distances by two gives the round trip time and distance. So the guy on the platform measures the speed of light to be:

[math]v=s/\tau =\frac{\Delta r+Mln\left ( \frac{r_2}{r_1} \right )}{\left (1-\frac{M}{r_1} \right )\left ( \Delta r+2Mln\left ( \frac{r_2}{r_1}\right ) \right )}[/math]

As you can see, even to first order as M gets larger the speed of light slows. As M goes to zero, we get that v=1, which is what we would expect. Even changing the separation Δr will yield different speeds.

Conclusion: light doesn't globally travel at c. Only locally does it do so.

Edited by elfmotat
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The four-velocity (check any textbook) is

 

[math]v^\mu = \frac{dx^\mu}{d\tau}[/math]

 

The velocity v is

 

[math]v^i = \frac{dx^i}{d\tau}[/math]

 

The speed is

 

[math]|\mathbf{v}| = \sqrt{\mathbf{v}^2} = \sqrt{v_iv^i}= \sqrt{\gamma_{ij}v^iv^j}[/math]

 

with [math]\gamma_{ij}[/math] the spatial metric. For Schwarzschild spacetime, the spatial metric tensor coincides with spatial part of the spacetime metric tensor [math]\gamma_{ij} = g_{ij}[/math].

 

[math]\gamma_{ij}v^iv^j = g_{ii} \left( \frac{dx^i}{d\tau} \right)^2[/math]

 

Using the relation between [math]d\tau[/math] and [math]dt[/math] and the symmetries of the Schwarzschild metric

 

[math]\gamma_{ij}v^iv^j = g_{ii}^2 \left( \frac{dx^i}{dt} \right)^2[/math]

 

For a light signal [math]ds^2=0[/math] and using all we finally obtain

 

[math]|\mathbf{v}| = \sqrt{g_{ii}^2 \left( \frac{dx^i}{dt} \right)^2} = \sqrt{c^2}[/math]

 

which is the well-known result that the speed of light is always constant in general relativity and equals c.

 

Some people defines the speed of light to be anything that they like and then obtain any result that they like, including unphysical results.

 

Different people makes different mistakes and I am not going to revise all. Some people computes the (unphysical) speed associated to a fictitious flat-spacetime (light signals are moving in a curved spacetime) and then obtains an (unphysical) speed of light that varies with the gravitational field. Other people propose ad-hoc definitions of speed no compatible with the covariant equation of motion...

 

EDIT: I have used bold face to emphasize the difference between four-velocity, velocity, and speed.

Edited by juanrga
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http://en.wikipedia.org/wiki/Four_velocity

 

The magnitude of an object's four-velocity (in the sense of the metric used in special relativity) is always equal to c (it is a normalized vector). Only its direction can change. For an object at rest (with respect to the coordinate system) its four-velocity points in the direction of the time coordinate.

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Evidently, the speed |v| is not the "magnitude of an object's four-velocity".

 

[math]|\mathbf{v}| = \sqrt{v_iv^i} \neq \sqrt{v_\mu v^\mu}[/math]

 

From the Wikipedia link... Velocity v is a "a three-dimensional vector".

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The four-velocity (check any textbook) is

 

[math]v^\mu = \frac{dx^\mu}{d\tau}[/math]

 

The four-velocity of a photon is undefined! (Check any textbook.)

 

The velocity v is

 

[math]v^i = \frac{dx^i}{d\tau}[/math]

 

The speed is

 

[math]|\mathbf{v}| = \sqrt{\mathbf{v}^2} = \sqrt{v_iv^i}= \sqrt{\gamma_{ij}v^iv^j}[/math]

 

with [math]\gamma_{ij}[/math] the spatial metric. For Schwarzschild spacetime, the spatial metric tensor coincides with spatial part of the spacetime metric tensor [math]\gamma_{ij} = g_{ij}[/math].

 

[math]\gamma_{ij}v^iv^j = g_{ii} \left( \frac{dx^i}{d\tau} \right)^2[/math]

 

Okay, except whose proper time are we differentiating by? Certainly not the photon's!

 

Using the relation between [math]d\tau[/math] and [math]dt[/math] and the symmetries of the Schwarzschild metric

 

[math]\gamma_{ij}v^iv^j = g_{ii}^2 \left( \frac{dx^i}{dt} \right)^2[/math]

 

I can't follow this notation at all. Please explain exactly what you're doing here.

 

For a light signal [math]ds^2=0[/math] and using all we finally obtain

 

[math]|\mathbf{v}| = \sqrt{g_{ii}^2 \left( \frac{dx^i}{dt} \right)^2} = \sqrt{c^2}[/math]

 

Once again, I don't follow.

 

which is the well-known result that the speed of light is always constant in general relativity and equals c.

 

John Baez of UCR (among others - his link was just the first I found) disagrees:

 

"The problem here comes from the fact that speed is a coordinate-dependent quantity, and is therefore somewhat ambiguous. To determine speed (distance moved/time taken) you must first choose some standards of distance and time, and different choices can give different answers. This is already true in special relativity: if you measure the speed of light in an accelerating reference frame, the answer will, in general, differ from c.In special relativity, the speed of light is constant when measured in any inertial frame. In general relativity, the appropriate generalisation is that the speed of light is constant in any freely falling reference frame (in a region small enough that tidal effects can be neglected). In this passage, Einstein is not talking about a freely falling frame, but rather about a frame at rest relative to a source of gravity. In such a frame, the speed of light can differ from c, basically because of the effect of gravity (spacetime curvature) on clocks and rulers."

 

 

Einstein himself also disagrees:

 

". . . according to the general theory of relativity, the law of the constancy of the velocity of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity [. . .] cannot claim any unlimited validity. A curvature of rays of light can only take place when the velocity of propagation of light varies with position."

 

http://math.ucr.edu/...d_of_light.html

 

EDIT: I just looked over the thread and realized that both of these have been posted already, though you've so far failed to address them.

 

Some people defines the speed of light to be anything that they like and then obtain any result that they like, including unphysical results.

 

Different people makes different mistakes and I am not going to revise all. Some people computes the (unphysical) speed associated to a fictitious flat-spacetime (light signals are moving in a curved spacetime) and then obtains an (unphysical) speed of light that varies with the gravitational field. Other people propose ad-hoc definitions of speed no compatible with the covariant equation of motion...

 

 

That's the thing though - my example was completely physical! A man bouncing a light signal off a mirror in Schwarzschild spacetime would measure the speed of light to differ from c. If he measured the distance with a ruler and timed how long it took for the light to return and hit his detector, he would find a result that differs from c! Now, if he were freely falling and conducted the same experiment over a sufficiently small scale, then he would measure v=c. That's why we say the speed of light is locally c.

 

Your "definition" (you haven't really given one yet) of the speed of light in GR seems to be the unphysical one. My definition is simply that which we can measure!

Edited by elfmotat
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The four-velocity of a photon is undefined! (Check any textbook.)

 

Okay, except whose proper time are we differentiating by? Certainly not the photon's!

 

Where in my post did you read that [math]\tau[/math] is proper time? Some authors write [math]\tau[/math] other write [math]\lambda[/math], and others use another notation.

 

I can't follow this notation at all. Please explain exactly what you're doing here.

 

Once again, I don't follow.

 

I think that I explained what is each term.

 

John Baez of UCR (among others - his link was just the first I found) disagrees:

 

"The problem here comes from the fact that speed is a coordinate-dependent quantity, and is therefore somewhat ambiguous. To determine speed (distance moved/time taken) you must first choose some standards of distance and time, and different choices can give different answers. This is already true in special relativity: if you measure the speed of light in an accelerating reference frame, the answer will, in general, differ from c.In special relativity, the speed of light is constant when measured in any inertial frame. In general relativity, the appropriate generalisation is that the speed of light is constant in any freely falling reference frame (in a region small enough that tidal effects can be neglected). In this passage, Einstein is not talking about a freely falling frame, but rather about a frame at rest relative to a source of gravity. In such a frame, the speed of light can differ from c, basically because of the effect of gravity (spacetime curvature) on clocks and rulers."

 

Einstein himself also disagrees:

 

". . . according to the general theory of relativity, the law of the constancy of the velocity of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity [. . .] cannot claim any unlimited validity. A curvature of rays of light can only take place when the velocity of propagation of light varies with position."

 

http://math.ucr.edu/...d_of_light.html

 

This link was discussed before in this thread. Just after the Einstein's quote Baez adds, in the same paragraph, "but a more modern interpretation is that the speed of light is constant in general relativity."

 

About your other quote, Baez adds near the end of his page: "Finally, we come to the conclusion that the speed of light is not only observed to be constant; in the light of well tested theories of physics, it does not even make any sense to say that it varies."

 

Some people claims that the speed of light varies. I have just computed the speed and obtained that it is a constant.

Edited by juanrga
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Where in my post did you read that [math]\tau[/math] is proper time? Some authors write [math]\tau[/math] other write [math]\lambda[/math], and others use another notation.

 

Okay, so [math]\tau[/math] is just a random affine parameter in your case. Fair enough. Most people use [math]\tau[/math] to represent proper time.

 

You still haven't addressed the fact that the four-velocity of a photon is undefined.

 

I think that I explained what is each term.

 

You explained what every symbol is, but, like I said, I can't made heads or tails of your notation. I know what [math]\gamma_{ij}[/math] is, I know what [math]g_{ij}[/math] is, and I know what [math]\gamma_{ij}v^iv^j[/math] is, but I have absolutely no idea what [math]g_{ii}^2 \left( \frac{dx^i}{dt} \right)^2[/math] is.

 

You also didn't explain what you mean by "the relation between [math]d\tau[/math] and [math]dt[/math]" or "symmetries of the Schwarzschild metric," and I don't see how ds2=0 implies:

 

[math]|\mathbf{v}| = \sqrt{g_{ii}^2 \left( \frac{dx^i}{dt} \right)^2} = \sqrt{c^2}[/math]

 

 

This link was discussed before in this thread. Just after the Einstein's quote Baez adds, in the same paragraph, "but a more modern interpretation is that the speed of light is constant in general relativity."

 

Yes, LOCALLY!!! He explained that that was what he meant:

 

"In general relativity, the appropriate generalisation is that the speed of light is constant in any freely falling reference frame (in a region small enough that tidal effects can be neglected)."

 

About your other quote, Baez adds near the end of his page: "Finally, we come to the conclusion that the speed of light is not only observed to be constant; in the light of well tested theories of physics, it does not even make any sense to say that it varies."

 

See above.

 

Some people claims that the speed of light varies.

 

Yes, everyone in this thread besides you.

 

I have just computed the speed and obtained that it is a constant.

 

Your "computation" looks quite nonsensical to me.

 

 

You also completely ignored the most important part of my last post. I'll state it again so that you'll be sure to see it:

 

"That's the thing though - my example was completely physical! A man bouncing a light signal off a mirror in Schwarzschild spacetime would measure the speed of light to differ from c. If he measured the distance with a ruler and timed how long it took for the light to return and hit his detector, he would find a result that differs from c! Now, if he were freely falling and conducted the same experiment over a sufficiently small scale, then he would measure v=c. That's why we say the speed of light islocally c. Your "definition" (you haven't really given one yet) of the speed of light in GR seems to be the unphysical one. My definition is simply that which we can measure!"

Edited by elfmotat
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Okay, so [math]\tau[/math] is just a random affine parameter in your case. Fair enough. Most people use [math]\tau[/math] to represent proper time.

 

You still haven't addressed the fact that the four-velocity of a photon is undefined.

 

You explained what every symbol is, but, like I said, I can't made heads or tails of your notation. I know what [math]\gamma_{ij}[/math] is, I know what [math]g_{ij}[/math] is, and I know what [math]\gamma_{ij}v^iv^j[/math] is, but I have absolutely no idea what [math]g_{ii}^2 \left( \frac{dx^i}{dt} \right)^2[/math] is.

 

You also didn't explain what you mean by "the relation between [math]d\tau[/math] and [math]dt[/math]" or "symmetries of the Schwarzschild metric," and I don't see how ds2=0 implies:

 

[math]|\mathbf{v}| = \sqrt{g_{ii}^2 \left( \frac{dx^i}{dt} \right)^2} = \sqrt{c^2}[/math]

 

Yes, LOCALLY!!! He explained that that was what he meant:

 

"In general relativity, the appropriate generalisation is that the speed of light is constant in any freely falling reference frame (in a region small enough that tidal effects can be neglected)."

 

See above.

 

Some people claims that the speed of light varies.

 

Yes, everyone in this thread besides you.

 

And my replies to some of them contained phrases as "No. I have just said the contrary." Other of those posters wrote (about the same link that you are using now to defend your point): "Baez doesn't make a lot of sense to me"... I explained to another poster that that speed is the norm of the velocity, and that velocity is a three-vector not a four-vector.

 

And so on. Therefore, I am not going to accept your argument by evident reasons.

 

Your "computation" looks quite nonsensical to me.

 

You also completely ignored the most important part of my last post. I'll state it again so that you'll be sure to see it:

 

"That's the thing though - my example was completely physical! A man bouncing a light signal off a mirror in Schwarzschild spacetime would measure the speed of light to differ from c. If he measured the distance with a ruler and timed how long it took for the light to return and hit his detector, he would find a result that differs from c! Now, if he were freely falling and conducted the same experiment over a sufficiently small scale, then he would measure v=c. That's why we say the speed of light islocally c. Your "definition" (you haven't really given one yet) of the speed of light in GR seems to be the unphysical one. My definition is simply that which we can measure!"

 

In my above demonstration of that the speed of light is globally c, I only forgot to say that I am using the standard four-velocity. Therefore v is the standard velocity and |v| the standard speed. I am so accustomed to use standards that I do not always emphasize it in my writings.

 

I ignored "the most important part" of your last post by reasons explained in the last part of my #67.

 

A discussion of equations of motion, clocks, etc. in general relativity is given in the section "The physical interpretation of the equations of point mechanics. Standard equations of motion. Standard simultaneity" found in Möller's classic textbook. Also relevant is the section "Propagation of light signals. Fermat's principle".

 

In both sections Möller computes the standard velocity and speed of a light particle. He uses another notation but his equation [math]\hat{w} = |\hat{\mathbf{w}}| = c[/math] is the equivalent of my last equation, |v| = c, in the demonstration #67.

 

Of course, Möller says the same in words:

 

we find that the standard speed of light [math]\hat{w}[/math] is c in all directions and everywhere.

 

This result was waited in pure physical terms because v, the standard velocity, is a gauge-invariant three-vector.

Edited by juanrga
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