Speed of light same as light speed?

[Naty1]

My question is whether the speed of light, as defined below, is precisely the same as the speed of light in free space, without gravity.


DISCUSSION
I'm trying to figure out if the standard value for "c" is or is not corrected for the very,very slight influence of Earth's gravitational field (potential) for a stationary observer relative to Earth's surface.

The definition appears to occur at the surface of the Earth where there is a weak gravitational field; if the speed of light is defined from the frame of reference of a stationary observer on the Earth's surface, the gravitational field will make the speed appear slightly different (slower) than if the observer were in gravitational free fall...or the definition were in a gravity free location.

So it seems like that gravitational acceleration (crude estimate below) may be enough to change the last few integers in the definition of light speed...but I did not do any calculations. In any event, even if all nine digits of light speed remain the same, there is still a slight possible theoretical difference. (In other words, a fixed frame of reference at the Earth's surface is a coordinate type reference frame, right, where different speeds of light will generally be observed.)

DEFINITION
From http://math.ucr.edu/home/baez/physic..._of_light.html


Quote:

Is c, the speed of light in vacuum, constant?
At the 1983 Conference Generale des Poids et Mesures, the following SI (Systeme International) definition of the metre was adopted:

The metre is the length of the path traveled by light in vacuum during a time interval of 1/299 792 458 of a second.
This defines the speed of light in vacuum to be exactly 299,792,458 m/s. This provides a very short answer to the question "Is c constant": Yes, c is constant by definition!

from Peter Bergmann's THE RIDDLE OF GRAVITATION,:


Quote:

The magnitude of the gravitational acceleration on the Earth's surface is very roughly 10m/s..the acceleration a free falling refeence frame should have relative to earth...as the radius of the Earth is some 6,000 km distance from the center (of the gravitational source) the acceleration relative to the free falling frame changes very roughly at 1.6 x 10^-6/sec^2

...

 

[Jonathan Scott]

My question is whether the speed of light, as defined below, is precisely the same as the speed of light in free space, without gravity.

...

Wherever you are, the speed of light at your own location is the standard value. That is the key point in relativity.

The variation with gravitational potential means that if you look at the apparent speed of light at some other location, it may appear to be slightly faster or slower than the standard value. The exact result depends on the coordinate system used to compare the speed, but in isotropic coordinates (which are normally the most practical ones for astronomical purposes), the apparent speed of light varies by a fraction equal to twice the change in gravitational potential, -GM/rc^2, at the points being compared.

 

[MeJennifer]

Wherever you are, the speed of light at your own location is the standard value. That is the key point in relativity.

Exactly, that is the only factor to hold on in understanding spacetime.

 

[Naty1]

yes, you are both right...had I worded my question more carefully I would have been able to answer it myself...I should have asked:

My question is whether the speed of light, as defined below, is precisely the same as the speed of light in free space, without gravity, as viewed from a fixed Earth frame.

And Scott answered that clearly:

...The variation with gravitational potential means that if you look at the apparent speed of light at some other location, it may appear to be slightly faster or slower than the standard value...

ah,well, live and learn...or not...

 

[Naty1]

Rereading this I now have some uncertainty...

Do the above answers imply that even an accelerating observer, such as one standing "stationary" at a point on Earth's surface, always observes local light at "c"? It looks that way from the answers.

Then why is the speed of light postulated as c, locally, only in a freely falling (non accelerating) frame? It sounds like this frame is necessary to observe constant c locally irrespective of gravitational potential.

I'm now have difficulty reconciling these two ideas.

 

[RandallB]

Rereading this I now have some uncertainty...

Do the above answers imply that even an accelerating observer, such as one standing "stationary" at a point on Earth's surface, always observes local light at "c"? It looks that way from the answers. ...

Remember what Scott said about “the apparent speed of light at some other location” may be fast or slow.
The key word here is apparent.

All local locations will define a common physics where:

A meter is officially defined as the length of the path traveled by light in vacuum during a time interval of 1/299 792 458 of a second.

The second is officially defined as 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom.

Every local observer will use these same physics definitions and that results in “c” for local light.

And when any “other location” looks at what you are using for “Meter” and “Second” (Including IMO even if you are an “accelerating observer” relative to them) and see your defined “c” as apparently slow or fast wrt to what they know to be “c”. When they apply all appropriate transforms to redefine their time and length to your reference frame they will get the same results as you from making the above physical measurements to establish the standards for meter & second to define a local values and thus for “c” as well.
(Granted if you are an accelerating observer those transforms will be difficult, and not something I want to try)

 

[DrGreg]

Rereading this I now have some uncertainty...

Do the above answers imply that even an accelerating observer, such as one standing "stationary" at a point on Earth's surface, always observes local light at "c"? It looks that way from the answers.

Then why is the speed of light postulated as c, locally, only in a freely falling (non accelerating) frame? It sounds like this frame is necessary to observe constant c locally irrespective of gravitational potential.

I'm now have difficulty reconciling these two ideas.

All observers, accelerating or not, measure the local speed of light to be c.

Although the metre is officially defined as 1/299792458 light-seconds, that definition strictly works only in the infinitesimal limit. An observer measures the distance to an object by radar: time how long it takes for a there-and-back reflected signal and multiply the seconds by (299792458/2) to get metres. But, for an accelerating observer, or any observer in curved spacetime, that procedure is accurate only in the limit as the time interval $\Delta t \rightarrow 0$. In flat spacetime the true answer is defined to be the radar distance measured by a co-moving inertial observer (an inertial observer traveling at the same speed as the accelerating observer at the moment the measurement is made). In curved spacetime there is no unique answer because there are no globally-inertial frames, only locally-inertial frames. You could say distance is only a local concept, although it's nevertheless possible to integrate local distances to get longer distances, but you need to specify how the integration is performed.

What has that got to do with the question? Well, the local speed of light is c by definition. But if an accelerating observer, or any observer in curved spacetime, tries to measure the average speed of light over a distance, the answer might not be c, but the answer will converge to c as the distance converges to zero.

 

[Naty1]

"But, for an accelerating observer, or any observer in curved spacetime, that procedure is accurate only in the limit as the time interval . In flat spacetime the true answer is defined to be the radar distance measured by a co-moving inertial observer (an inertial observer traveling at the same speed as the accelerating observer at the moment the measurement is made). In curved spacetime there is no unique answer.."

I think I understand...BUT it seems to conflict with Einstein/s needd for inertial frames to view "c".

If even accelerating observers (standing till on Earth's surface) see light locally at "c" then why did Einstein specify local free falling frames as the inertial frames for measuring local light speed with gravity and inertial frames (generally) for special relativity. Is your description an "update" from Einstein's understanding?

My understanding is that when measuring light we observe "c" in flat spacetime, both locally and distant as space is uniform. That means when we freely fall in the presence of gravity locally: Standing on Earth's surface , we are not in free falling frame, but an accelerating frame (g)...a coordinate frame where "c" seems to depend on gravitational potential (curvature). The difference is minor because gravity is low, I know, but this is a question of principle.

Does anyone standing in any gravitational potential see light locally as "c"?? Apparently "yes" if it's instananeous, as you say. But Einstein said Only a free falling observer sees "c" locally.

Another way to phrase the question, I think, would be this: If we can observe a distant speed of light apparently slowing, say approaching a black hole event horizon, due to distant curved space and the fact we observe from a coordinate frame, why would we not also observe a different speed of light from "c" locally in curved space when in the same coordinate frame? I would think we'd measure two different values from "c", slower for the distant, closer to "c" for the local...

 

[Naty1]

Finally I got to one piece of my own puzzle...the test particle in GR:

(from http://math.ucr.edu/home/baez/einstein/node2.html )

... A `test particle' is an idealized point particle with energy and momentum so small that its effects on spacetime curvature are negligible. A particle is said to be in `free fall' when its motion is affected by no forces except gravity. In general relativity, a test particle in free fall will trace out a `geodesic'. This means that its velocity vector is parallel transported along the curve it traces out in spacetime.

 

[DrGreg]

"But, for an accelerating observer, or any observer in curved spacetime, that procedure is accurate only in the limit as the time interval . In flat spacetime the true answer is defined to be the radar distance measured by a co-moving inertial observer (an inertial observer traveling at the same speed as the accelerating observer at the moment the measurement is made). In curved spacetime there is no unique answer.."

I think I understand...BUT it seems to conflict with Einstein/s needd for inertial frames to view "c".

If even accelerating observers (standing till on Earth's surface) see light locally at "c" then why did Einstein specify local free falling frames as the inertial frames for measuring local light speed with gravity and inertial frames (generally) for special relativity. Is your description an "update" from Einstein's understanding?

If you look at what I said, I first of all defined things for inertial (= free-falling) observers, and then extended the concept to a non-inertial observer defined in terms of a local co-moving inertial observer. The non-inertial point of view arises by definition.

I am working from memory and my own understanding and haven't looked up the exact words Einstein used. Just because we choose to use inertial observers to make certain measurements doesn't always mean we couldn't have chosen non-inertial observers in some circumstances. The theory is built around inertial observers but later we find that some (but not all) concepts may still work (locally) for non-inertial observers.

My understanding is that when measuring light we observe "c" in flat spacetime, both locally and distant as space is uniform. That means when we freely fall in the presence of gravity locally: Standing on Earth's surface , we are not in free falling frame, but an accelerating frame (g)...a coordinate frame where "c" seems to depend on gravitational potential (curvature). The difference is minor because gravity is low, I know, but this is a question of principle.

Does anyone standing in any gravitational potential see light locally as "c"?? Apparently "yes" if it's instananeous, as you say.

That's all correct. (But remember, someone at the top of a skyscraper will have their own frame that differs slightly from someone at street level, so each measures light at c at their own location, but slightly different at each other's location.)

But Einstein said Only a free falling observer sees "c" locally.

I don't know what Einstein actually said without looking it up, but I'd omit the word "only"!

Another way to phrase the question, I think, would be this: If we can observe a distant speed of light apparently slowing, say approaching a black hole event horizon, due to distant curved space and the fact we observe from a coordinate frame, why would we not also observe a different speed of light from "c" locally in curved space when in the same coordinate frame? I would think we'd measure two different values from "c", slower for the distant, closer to "c" for the local...

Sorry, I'm not sure I understand what you ask here.

 

[Kopachris]

EDIT: Please ignore this post. I hadn't had my coffee yet and said some stupid things.

c is defined as the speed (it's not usually a vector, but sometimes is) of light in a vacuum, 299,792,458 m/s. Special Relativity says that all velocities are relative to each other except the velocity of light. Light in a vacuum always goes at c to an outside observer. If you shoot a beam of light in front of you while you're going at 90% the velocity of light, you will see the light going 10% the speed of light while an outside observer will see it going at c. Light is always globally c, not locally c. Einstein concluded from that that time becomes dilated as a moving object approaches c.

 

[MeJennifer]

If you shoot a beam of light in front of you while you're going at 90% the velocity of light, you will see the light going 10% the speed of light.

You are mistaken, you will see light going at 100% of the speed of light.

 

[Kopachris]

You are mistaken, you will see light going at 100% of the speed of light.

Woops! I forgot about time dilation. Duh!

 

[Naty1]

Dr Greg posts:

If you look at what I said, I first of all defined things for inertial (= free-falling) observers, and then extended the concept to a non-inertial observer defined in terms of a local co-moving inertial observer. The non-inertial point of view arises by definition.

I appreciate your patience discussing this! Using your definition, I understand how you reach your conclusions. Everybody sees local light at "c" with your conventions.

Is this a generally accepted and understood convention? I can't reconcile other things I read with your perspective. Other descriptions are not as clearly defined as yours.


This thread is becoming painfully close to a prior one:
How does light slow in the presence of gravity?

I am basically still asking if all observers, inertial and non inertial as well , observe light locally at "c" in conventional physics discussions.

I'd like interpretations on the following, and then I'll stop asking this same question.

(http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/speed_of_light.html)
This includes consecutive paragraphs, no omissions, which to me appear inconsistent.

(1)
Einstein went on to discover a more general theory of relativity which explained gravity in terms of curved spacetime, and he talked about the speed of light changing in this new theory. 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. (It's results hold only so long as we are able to disregard the influences of the gravitational fields on the phenomena eg of light #) A curvature of rays of light can only take place when the velocity of propagation of light varies with position. Since Einstein talks of velocity (a vector quantity: speed with direction) rather than speed alone, it is not clear that he meant the speed will change, but the reference to special relativity suggests that he did mean so. This interpretation is perfectly valid and makes good physical sense, but a more modern interpretation is that the speed of light is constant in general relativity.

# My addition, from Einsteins book.
My Comment: It seems that historically, constant "c" holds only without gravitational curvature;The boldface seems to change that and appears inconsistent with (2) and (3) but consistent with (5))

(2)
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.

My comment: Is this referencing local measures? It seems to contradict (1) boldface "modern interpretation".

(3)
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.

My Comment: The boldface interpretation is different from say Brian Greene Fabric of Cosmos.

(4)
If general relativity is correct, then the constancy of the speed of light in inertial frames is a tautology from the geometry of spacetime. The causal structure of the universe is determined by the geometry of "null vectors". Travelling at the speed c means following world-lines tangent to these null vectors. The use of c as a conversion between units of metres and seconds, as in the SI definition of the metre, is fully justified on theoretical grounds as well as practical terms, because c is not merely the speed of light, it is a fundamental feature of the geometry of spacetime.

My Comment: good grief, now we are back to inertial frames... free falling ones!

Like special relativity, some of the predictions of general relativity have been confirmed in many different observations. The book listed below by Clifford Will is an excellent reference for further details.

(5)
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.

My comment:This sounds like all observerssee light at "c"...inertial and noninertial.


Thanks to all for input!

 

[DrGreg]

Is this a generally accepted and understood convention? I can't reconcile other things I read with your perspective. Other descriptions are not as clearly defined as yours.

I am basically still asking if all observers, inertial and non inertial as well , observe light locally at "c" in conventional physics discussions.

The answer should be "yes" but there is a catch. It depends what coordinate system you use. If you use your own proper time as the 0th coordinate (locally) and three orthogonal "physical distances" as your 1st 2nd & 3rd coordinates (locally), the answer is "yes" (locally). But in GR you are free to use any coord system you like, so you could choose the "wrong" coords and the answer could be "no".

I think in discussions people sometimes "play safe" and discuss inertial observers only to avoid complications with non-inertial frames.

I'm in a hurry and haven't had time to properly read the quote you gave, but I think the differences you found in your quote are probably referring to non-local measurements when they say the speed of light might not be c.

 

[Naty1]

For those trying to understand frames of refernce, an excellent thread and rather detailed, can be seen at

https://www.physicsforums.com/showthread.php?t=255769

 

[JesseM]

My comment:This sounds like all observerssee light at "c"...inertial and noninertial.

Why do you say that? The sentence immediately after the one you put in bold mentions that in the frame of the observer "at rest relative to a source of gravity" (i.e. not a freely-falling frame and therefore not locally inertial), "the speed of light can differ from c". Only in an inertial frame is the speed guaranteed to be c.

 

[Naty1]

Hi JesseM: MY post:

My comment:This sounds like all observers see light at "c"...inertial and noninertial.

was intended to addresses my interpretation of item #5, post 14:

(5)
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.

I am unsure what to make of (5)...seems like there are no restrictions.

 

[Naty1]

Here is what clears this situation up for me:

Under another thread, Light Velocity Measurements, I posted my conclusion:

George's post, # 2:

An accelerating observer measures the local speed of light to be c in all directions.

(and this was confirmed by MeJennifer)

is by convention measured according to DrGreg's definition, post #7

This is true by definition. Q: How does a non-inertial observer measure local distance and local time? A: by asking a co-moving (i.e. relatively stationary) inertial observer to make the measurement for him/her. This is how local distance and time are defined for a non-inertial observer

.


In the above thread, DrGreg explained:

So, to summarise, "spacetime curvature*" refers to the curvature of the graph paper, regardless of observer, whereas visible curvature in space is related to the distorted, non-square grid lines drawn on the graph paper, and depends on the choice of observer

* meaning physical curvature

The thing that had me stymied all this time was my strong suspicion that an accelerating observer would encounter apparent curvature when making observations (called visible by DrGreg,above) due to the equivalence principle; in the popular physics books I have been reading the past few years they only say inertial observers measure "c", not accelerating observers, but the context of their assertion was never made clear. I now suspect that others who have read books by Greene,Smolin, Kaku, etc, are, like me, the source of so many repeated questions on this forum about the speed of light. If we all use DrGregs "summary" above on this forum I'm all set...
and really aprpeciate the feedback to help me.

I'd like to find the FAQ for constant light speed (an item in discussion under general physics at the moment) and see if this is addressed there...if somebody will confirm the above explananation I'll try to include it in the FAQ...