Is the Speed of Light Really Absolute?: An Amateur Astronomer's Musings

[meddyn]

I am sure this has been addressed many times but, is the speed of light really absolute? If a photon path is indeed "bent" in passing an object with an appropriate gravitational field then this field acted upon the photon. If the photon is exposed upon approach to the gravitational attraction of an object would this same attraction not serve to accelerate the photon on approach to the body and retard the velocity on passing the object as its course is also altered in a slightly different direction?

Now, let me dig an even deeper amateur hole (black hole). If the gravitational attraction of a black hole prevents photon emission then is the photon actually emitted, proceeds at C, slowed, reversed, and attracted back to the core?

Be easy on a semi-educated amateur astronomer please...

 

[Arcon]

Yes. The gravitational field does slow the light down as - as measured by a distant observer. In fact what you just mentioned was what Einstein used to do his first calculation of the bending of star light by the Sun. The speed of light is a function of the gravitational field and the direction the light is moving.

In a uniform gravitational field, aligned in the -z direction, it has the value c' which is related to the value c (c = 2.998x10² = speed of light as measured in an inertial frame in a vacuum) has the exact value

$$c' = (1 + 2\Phi(z)/c^{2})c$$

where

$$\Phi(z) = gz$$

g = gravitational acceleration as measured at z = 0. In this case there is no dependence on the direction, onlz on z

 

[DW]

Originally posted by meddyn
I am sure this has been addressed many times but, is the speed of light really absolute?

No, the local vacuum speed of light is invariant.

If a photon path is indeed "bent" in passing an object with an appropriate gravitational field then this field acted upon the photon. If the photon is exposed upon approach to the gravitational attraction of an object would this same attraction not serve to accelerate the photon on approach to the body and retard the velocity on passing the object as its course is also altered in a slightly different direction?

Not if you mean does it "speed up" the photon. As light approaches a body according the remote coordinate speed slows down. This is called the gravitational slowing of light and has been experimentally verified. You can not use the formula for the case of an accelerated frame in flat spacetime as pmb suggests. The field from a body will be nonuniform.

Now, let me dig an even deeper amateur hole (black hole). If the gravitational attraction of a black hole prevents photon emission then is the photon actually emitted, proceeds at C, slowed, reversed, and attracted back to the core?

The behavior of the light before it impacts the physical singularity of the Schwarzschild hole I believe you are considering depends on your choice of observer coordinate frame. From the perspective of Schwarzschild coordinates yes this is what happens for a photon that is emmited at a location inside the event horizon. From the perspective of Kruskal-Szekeres coordinates as another example the answer would be no. From the perspective of these coordinates the photon travels at constant speed c at a 45 degree angle to the right of the virticle until it impacts the physical singularity. According to these coordinates the physical singularity is not a point of zero length, but is an entire hyperbolic curve that may be intersected from various paths.

 

[HallsofIvy]

What happens when a beam of light "tries" to get out of a black hole is that it loses energy. Since the energy of a light wave is proportional to its frequency, its frequency decreases. The light does NOT slow down, its frequency goes to 0.

The speed of light is always c to any observer no matter how distant.

 

[Arcon]

Originally posted by HallsofIvy
What happens when a beam of light "tries" to get out of a black hole is that it loses energy.

That is incorrect. The total energy of a beam of light remains constant as it moves through a gravitational field.

Since the energy of a light wave is proportional to its frequency, its frequency decreases.

That is incorrect as well. The frequency of the light does not change as it moves through a gravitational field.

The light does NOT slow down, its frequency goes to 0. The speed of light is always c to any observer no matter how distant.

That is incorrect.

In all of the above comments you're referring to observations made by different observers who are comparing results. You're not considering obervations made by a single observer.


For a detailed explanation see
"On the Interpretation of the Redshift in a Static Gravitational Field," L.B. Okun, K.G. Selivanov , V.L. Telegdi, Am.J.Phys. 68 (2000) 115
This article is online at -- http://xxx.lanl.gov/abs/physics/9907017

 

[meddyn]

Thank you for the replies.

With regard to the nature of the bending of the light ray/photon passing close to a significant gravitational attraction, what is the nature of the forces involved in altering the path in laymen terms? Are there any recent observations/studies indicating an at rest "mass" property to the "photon" instead of the "zero mass" definition?

(Sorry for the "three dimensional fixation". Trying to work out of it.)

 

[Arcon]

Originally posted by meddyn
..what is the nature of the forces involved in altering the path in laymen terms?

The nature of the gravitational force is that it's an inertial force. An inertial force is one that is absent in an inertial frame of reference and exists only due to the choice of a non-inertial frame of reference.

 

[marcus]

Meddyn asked about the bending of a ray of light, passing alongside a massive object (e.g. the sun):

------------------
Originally posted by meddyn
..what is the nature of the forces involved in altering the path in laymen terms?
------------------

To this, Arcon replied as follows:

Originally posted by Arcon
The naturd of the gravitational force is that its an inertial force. An inertial force is one that is absent in an inertial frame and exists due to the choice of a non-inertial frame.

To me, this does not sound like an explanation "in layman terms". I seem to recall that in the past several PF people, mentors and the like, when responding to the same question, have given an answer at least superficially different from Arcon's.
If I remember correctly, their explanation was to the effect that light does not experience any gravitational force at all, but simply travels along a geodesic (the analog of a straight line in curved space).

 

[Arcon]

Originally posted by marcus

To me, this does not sound like an explanation "in layman terms".

Something is in layman terms when it does not require a background in math or physics to understand it. That description does not require a background in physics to understand it and it contains no math. It was the simplest explanation and the most precise that I know of in layman terms.

If I remember correctly, their explanation was to the effect that light does not experience any gravitational force at all, but simply travels along a geodesic (the analog of a straight line in curved space).

They are certainly entitled to their opinion. But I choose to describe things in what I believe to be a more precise manner.


marcus - You've referred to the article The Concept of Mass, Lev B. Okun, Physics Today, June 1989. Did you read that paper?

In that paper Okun used the expression for the gravitational force that I posted in the thread SR and the earth, sun, and galaxy.. If you have this paper its Eq. (16) on page 51. Okun states

The so-called gravitational mass of the photon falling vertically toward Earth is, incidentally, given by E/c². As you an see from equation 16, however, a horizontally moving photon ... is twice as heavy.

This of course means that he is speaking of the gravitational force on the photon.


Yes. Light does travel on a geodesic. There is no question about that. In fact any test particle (no charge, non-spinning etc.) which experiences only gravitational forces will move on a geodesic. In that sense it is incorrect to say that a particle in free-fall in a gravitational field experiences no gravitational force. To me that is like saying that a projectile fired from a battleship experiences no Coriolis force. Such a statement is highly inaccurate.

You're probably used to hearing inertial forces referred to as "fictituous forces." That is a very misleading name as has been pointed out in the physics literature. You can learn more about the Coriolis force and gravitational force as an inertial force in these lecture notes
http://www.whoi.edu/science/PO/people/jprice/class/aCt.pdf. . Note what the author says on this point

If we need to call attention to these special properties of the Coriolis force, then the usage Coriolis inertial force seems appropriate because it is free from the taint of unreality that goes with ’virtual force, fictitious correction force’, etc., and because it gives at least a hint at the origin of the Coriolis force.

Please take note that these notes are by no stretch of the imagination to be considered "old" since they were just written. This is the newest version which was updated two days ago.

It is only correct to say that the 4-force (i.e. non-inertial forces) on a particle in free-fall in a gravitational fied zero. However according to Einstein's Equivalence Principle the gravitational force is an inertial force. However inertial forces are not 4-vectors. They are related to the "affine connection" which is not a 4-vector. This is a well known fact in general relativity.

It's not as if I'm alone in this position. In fact this is how I learned it. Actually to be precise - this is how I learned it when I went beyond the layman's view that is always given that there is no such thing as a gravitational force. That is a view held by people who are stuck in the Newtonian viewpoint. The more I looked into it the more I realized how inaccurate that was since Einstein never held that view. Einstein held that since the gravitation force is similar in nature to the Coriolis force, which is an inertial force, then since the gravitational force is "real" then so too is the Coriolis force. People who say Einstein held/proved that "there is no gravitational force" or claim that the gravitational force is "fictitious" have it all backwards. These people claim that since the Coriolis force is not "real" then the gravitational force, being of the same nature, must also not be a "real" force. These people are being Newtonian and are refusing to view things as Einsteinian. They think that they are being Einsteinian but they are not. So be it.


However some of the best, most authoritative, GR texts out there agree with Einstein - as they should! In fact read what Kip Thorne and Roger Blandford say in their new text
http://www.pma.caltech.edu/Courses/ph136/yr2002/chap11/0211.1.pdf

This derivation of the wave equation is an elementary illustration of the Principle of Equivalence|the equivalence of gravitational and inertial forces, or gravitational and inertial accelerations|which we shall discuss in Part VI as an underpinning for general relativity
theory.

In fact Weinberg has an entire section in his text Gravitation and Cosmology: Principles and Applications of the General Theory of Relativity entitled Gravitational Forces (page 70). The section which follows that section states, in the first paragraph

Our treatment of freely falling particles has shown that the field that determines the gravitational force is the "affine connection" $$\Gamma^{\lambda}_{\mu\nu}$$, whereas the proper time interval between two events with a given infinitesimal coordinate separation is determined by the "metric tensor" $$g_{\mu\nu}$$. We now show that $$g_{\mu\nu}$$ is also the gravitational potential; that is, its derivatives determine the field $$\Gamma^{\lambda}_{\mu\nu}$$.

There is an article on the gravitational force on light called
The gravitational interaction of light: from weak to strong fields, V. Faraoni, R.M. Dumse, Gen.Rel.Grav. 31 (1999) 91-105
http://xxx.lanl.gov/PS_cache/gr-qc/pdf/9811/9811052.pdf

This article uses the technique's of gravitomagnetism.

 

[deda]

c is const means:
$$c=\frac{L_1}{T_1}=\frac{L_2}{T_2}=...=\frac{L_n}{T_n}$$
where n is the frame.
therefore:
$$\frac{L_1}{T_1}=\frac{hL_2}{hT_2}$$
so the length and its elapsed time should shrink at the same time.
in gravity field (near black hole for instance) the length and the time shrink proportionally. that's why c remains const...

i'm expecting arguments on this as it's expected otherwise.
that time dilates while length contracts.

 

[Arcon]

Originally posted by deda
c is const means:
$$c=\frac{L_1}{T_1}=\frac{L_2}{T_2}=...=\frac{L_n}{T_n}$$
where n is the frame.
therefore:
$$\frac{L_1}{T_1}=\frac{hL_2}{hT_2}$$
so the length and its elapsed time should shrink at the same time.
in gravity field (near black hole for instance) the length and the time shrink proportionally. that's why c remains const...

i'm expecting arguments on this as it's expected otherwise.
that time dilates while length contracts.

That holds in special relativity. Not general relativity

 

[Arcon]

Originally posted by deda
c is const means:
$$c=\frac{L_1}{T_1}=\frac{L_2}{T_2}=...=\frac{L_n}{T_n}$$
where n is the frame.
therefore:
$$\frac{L_1}{T_1}=\frac{hL_2}{hT_2}$$
so the length and its elapsed time should shrink at the same time.
in gravity field (near black hole for instance) the length and the time shrink proportionally. that's why c remains const...

i'm expecting arguments on this as it's expected otherwise.
that time dilates while length contracts.

That holds in special relativity. Not general relativity when the frame of reference is not an inertial one, i.e. not when there is a gravitational field present.

 

[meddyn]

As expected, as understanding grows some additional questions arise:

In reading explanations of inertial reference frames a scenario comes to mind. Take a large tour bus full of Japanese tourist. Each equipped with the normal compliment of Minolta and Cannon camera gear. The tour bus has taken a wrong turn and has come under the influence of the black hole in Andromeda. As the bus plummets into the black hole it experiences awesome acceleration as expected with the corresponding increase in speed. An instant before reaching the singularity a tourist at the back of the bus stands up, looks to the front of the bus and snaps a picture of the event horizon he is about to merge with. The bus has attained a speed of .90 C under the influence of the gravitational attraction. Is the speed of the photons sourced from his camera toward the black hole C + .90 C?

And please allow another brief side note. If the photon in the initial question had an "odometer" would it indicate a greater distance traveled in traversing the gravitational field that it bent around than another photon transiting from an identical point A to point B distance without passing a large gravitational mass?

 

[Arcon]

Originally posted by meddyn
An instant before reaching the singularity a tourist at the back of the bus stands up, looks to the front of the bus and snaps a picture of the event horizon he is about to merge with. The bus has attained a speed of .90 C under the influence of the gravitational attraction. Is the speed of the photons sourced from his camera toward the black hole C + .90 C?

You seem to be confusing the event horizon with the singularity. If the black hole is large enough the one can safely pass through the event horizon and not even know it. In fact must pass through the event horizon before you can move onto the singularity. For a super-duper massive black hole one can actually live inside during your continual free-fall for years and never know you were inside unless you looked out the windows of the bus (or have super duper sensitive gradiometers).

The speed of light inside the bus, as measured by the tourists inside the bus, is c = 2.998x10²

 

[Arcon]

In my first post above I wrote

The speed of light is a function of the gravitational field ...

Note: Just to be clear on the definition here - Consider as an example a spacetime whose metric is time-orthgonal, i.e. g_0k = 0.

This spatial distance is defined as follows: The infinitesimal spatial distance between two closely spaced events is defined as

$$d\sigma^{2} = -g_{jk}dx^j dx^{k}$$

As such

$$v_{spatial} = \frac{d\sigma}{dt}$$

 

[meddyn]

Apologize for the confusion of terms. Been 30 + years since formal studies.

The bus experience would bring up a simple question utilizing an "Encapsulation Concept".

If we could make a spaceship 100,000 miles long and place another spaceship inside it that was 50 feet long and launched the "mother ship" toward a distant star and accelerated the mother ship to .6 the speed of light and then launched the "child ship toward the front of the mother ship at .6 the speed of light then to an outside observer the child ship would be traveling at 1.2 times the speed of light and, more importantly, the VMG (Velocity made good) toward the remote object would be 1.2 times the speed of light by the child ship thus exceeding the speed of light without attaining infinite mass. (Sorry, had caffeinated coffee this morning.)

 

[deda]

in reply to c=L_i/T_i=const

Originally posted by Arcon
That holds in special relativity. Not general relativity

How does it holds in special relativity?
SR claims different
L=L_0*Gamma length contracts
T=T_0/Gamma time dilates.
Gamma=sqrt(1-V^2/c^2)
It was a trickey shot.
but nevermind.

 

[Androcles]

As an amateur astronomer, perhaps you may be interested in this, which may help you with your question.

http://www.androc1es.pwp.blueyonder.co.uk/actual_data.htm

 

[Androcles]

Originally posted by deda
in reply to c=L_i/T_i=const

How does it holds in special relativity?
SR claims different
L=L_0*Gamma length contracts
T=T_0/Gamma time dilates.
Gamma=sqrt(1-V^2/c^2)
It was a trickey shot.
but nevermind.

There is a slight problem with your assertion
T=T_0/Gamma

In Einstein's paper, which you can find at
http://www.fourmilab.ch/etexts/einstein/specrel/www/

We read (section 3)
x' = x-vt.
and
tau = (t-vx/c^2)/sqrt(1-v^2/c^2)

Since x' = x - vt, it logically follows that x = x'+vt.
In order to differentiate, x' has been taken to be infinitessimally small by Einstein, x = vt

Substituting x for its value,
tau = (t-v^2t/c^2)/sqrt(1-v^2/c^2)
= t.sqrt(1-v^2/c^2)
and not
t/sqrt(1-v^2/c^2)
as you stated.

 

[David]

Originally posted by meddyn
I am sure this has been addressed many times but, is the speed of light really absolute? If a photon path is indeed "bent" in passing an object with an appropriate gravitational field then this field acted upon the photon. If the photon is exposed upon approach to the gravitational attraction of an object would this same attraction not serve to accelerate the photon on approach to the body and retard the velocity on passing the object as its course is also altered in a slightly different direction?

Now, let me dig an even deeper amateur hole (black hole). If the gravitational attraction of a black hole prevents photon emission then is the photon actually emitted, proceeds at C, slowed, reversed, and attracted back to the core?

Be easy on a semi-educated amateur astronomer please...

Einstein’s 1911 gravitational redshift theory, which is seldom read and often misunderstood, says that the local gravitational field controls the local speed of light. The light is not “attracted” by the sun as if the light has “mass”, it is deflected in a curve past the sun in a form of “gravitational refraction”, with the part of the light beam nearest the sun slowing down more than the part of the beam that is further from the sun.

The story that light “struggles” to climb out of a gravitational field is just not true, no more than sound “struggles” to travel through air, iron, or water. The gravitational field merely controls the speed of light through the field, at different speeds through different strengths of the field.

Einstein clearly said in his 1911 theory and restated it in his 1916 theory that light speed is NOT constant. It slows down while traveling through strong gravity and it speeds up while traveling in weak gravity. The reason this is not commonly known today is because of an unusual situation within atoms. Seems that an atom’s harmonic oscillation rate slows down in a strong gravity field and speeds up in a weak field, so atomic clock tick rates slow down in a strong gravity field and their tick rates speed up in a weak field. This causes local atomic clocks to always measure “c” as the local speed of light. I.E., the clocks slow down at the same rate as the light slows down inside a gravitational field.

But if we use a single atomic clock, that is constantly resting in one field, to measure the speed of light as it travels around different places in space, then we can notice and calculate the changing speeds of light. So the original 1905 “constancy” postulate was shown to be incorrect by the 1911 theory.

 

[speso72]

Originally posted by meddyn
If the photon is exposed upon approach to the gravitational attraction of an object would this same attraction not serve to accelerate the photon on approach to the body and retard the velocity on passing the object as its course is also altered in a slightly different direction?

Hello, please forgive my ignorance, as I am only attempting to join this conversation in efforts to learn more and to share my thoughts and opinions (please feel free to correct me if I'm mistaken).

I think your questions are difficult to answer mainly because it does not seem to be agreed upon as to what a photon actually is. A quick look on the net shows me that there are several theories and debates discussing this.

I think, that if a photon has any amount of mass, then it has weight (no matter how small the amount of mass), which therefore can be influenced by the effects of a gravitational field. Therefore, I think its rate of acceleration or deceleration would be dependent upon the photons weight, velocity and angle of trajectory into/onto the gravitational field (in most cases the gravity field would act as friction, thus slowing it down, since the photon would have more kinetic energy than the gravity field strength itself - usually).

I do not believe that the speed or velocity of a photon determines the frequency at which its ionized energy resonates and oscillates (thinking that its ionized energy resonation and oscillation determines that photons radiated spectrum color), I think that the speed of the photon is more determined by the materialistic object which the photon interacted with (i.e. its original source, its reflected source, etc..), as well as the photon material itself. I think the speed of a photon is originally determined by the speed of the rf energy that ionized and radiated the photon, and rf energy travels at the speed of light (possibly acting as a carrier for the photon). But rather, the speed or velocity of a photon (determined by the amount of kinetic energy), determins the candle power of the photon.

With the case of a black hole, the gravitational field strength of the black hole is great enough to stop the motion of the materialistic photon itself (as with all other materialistic objects that enter into its event horizon). The gravitational field strength is usually greater than the energy level of a photon's kinetic energy, which would slow the photon down to the point that it can not escape the gravitational field of the black hole.

I think that some photons are accelerated as they travel directly into a black hole (we just can't see them to observe simply because they do not reflect) and the ones that we do see are the ones that neared its event horizon and bounced, bent or slung shot around, and we see those photons traveling at much slower speeds as they are trying to move away from the gravity pit (the closer to the event horizon the photon is the slower it moves – giving an illusion that time itself is slowing down, even though it is not, it is only the photons that we are observing time by in the form of light reflection that are being slowed down). Those photons that are close enough to the event horizon slows to the point that they stop and fall backward or sidewards into the black hole) . As for the energy of the photon, it is most likely treated as any other materialistic object that falls or is pulled into a black hole: consumed and added to the energy of the black hole as its mass is added to the mass of the black hole's.

Please let me know how off I am, Thank you.

This post was edited to add the following link (further discussion along similar topic) https://www.physicsforums.com/showthread.php?s=&postid=135614#post135614

 

[deda]

Originally posted by meddyn
I am sure this has been addressed many times but, is the speed of light really absolute? If a photon path is indeed "bent" in passing an object with an appropriate gravitational field then this field acted upon the photon. If the photon is exposed upon approach to the gravitational attraction of an object would this same attraction not serve to accelerate the photon on approach to the body and retard the velocity on passing the object as its course is also altered in a slightly different direction?

Now, let me dig an even deeper amateur hole (black hole). If the gravitational attraction of a black hole prevents photon emission then is the photon actually emitted, proceeds at C, slowed, reversed, and attracted back to the core?

Be easy on a semi-educated amateur astronomer please...

I think that now I have the right answer to that.
The photon travels at c.
c=const and Newton I => the force acting upon the photon is zero.
But does the SR support that?
Yes!
The force relative with the speed of motion is given with:
$$F=F_{rest}\sqrt{1-\frac{c^2}{c^2}}=0$$
if F_rest is not infinity.

I've outperformed my self.

 

[Androcles]

Having written a reply and then accidentally closing the browser, losing all, I have now written up a web page instead.

http://www.androc1es.pwp.blueyonder.co.uk/Radio Wave.htm

 

[DW]

Originally posted by Arcon
In my first post above I wrote

Note: Just to be clear on the definition here - Consider as an example a spacetime whose metric is time-orthgonal, i.e. g_0k = 0.

This spatial distance is defined as follows: The infinitesimal spatial distance between two closely spaced events is defined as

$$d\sigma^{2} = -g_{jk}dx^j dx^{k}$$

As such

$$v_{spatial} = \frac{d\sigma}{dt}$$

No that is specifically the rigid ruler distance and speed, not the coordinate distance and speed and the sign convention from wherever you got this is not the same as from wherever you got most of the rest of your matterial. It is actually the sign convention I use which leads to an easier transition into spinor calculus.