# Speed of light an absolute?

1. Jan 16, 2004

### 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.....

2. Jan 17, 2004

### 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 dependance on the direction, onlz on z

3. Jan 17, 2004

### DW

No, the local vacuum speed of light is invariant.

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.

The behavior of the light before it impacts the physical singularity of the Schwarszchild hole I believe you are considering depends on your choice of observer coordinate frame. From the perspective of Schwarszchild 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.

4. Jan 17, 2004

### 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.

5. Jan 17, 2004

### Arcon

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

In all of the above comments you're refering 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

6. Jan 17, 2004

### 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.)

Last edited: Jan 17, 2004
7. Jan 17, 2004

### Arcon

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.

8. Jan 18, 2004

### 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:

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).

9. Jan 18, 2004

### Arcon

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 explaination and the most precise that I know of in layman terms.
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 refered 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
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 refered 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. [Broken]. Note what the author says on this point
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
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
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

Last edited by a moderator: May 1, 2017
10. Jan 18, 2004

### 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.

11. Jan 18, 2004

### Arcon

That holds in special relativity. Not general relativity

12. Jan 18, 2004

### Arcon

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.

13. Jan 18, 2004

### 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?

14. Jan 18, 2004

### Arcon

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

15. Jan 18, 2004

### Arcon

Re: Re: Speed of light an absolute?

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. g0k = 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

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

16. Jan 18, 2004

### 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.)

17. Jan 18, 2004

### deda

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.

18. Jan 21, 2004

### Androcles

Last edited by a moderator: Apr 20, 2017
19. Jan 21, 2004

### Androcles

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/

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.

20. Jan 21, 2004

### David

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.