# Shapiro time delay?

1. Oct 9, 2007

### yuiop

Can someone clear up what is observed in Shapiro time delay experiments.
As I understand it a delay is seen in the round trip signal time of a radar signal sent to a distant planet on the opposite side of the sun due to gravitational effects. The signal path would be longer than the straight line path to the calculated position of the planet in its orbit due to deflection by the gravitational field of the sun. This longer path would introduce a geometric delay. Would it be right to assume the Shapiro time delay is a measured delay over and above the the simple geometric delay?

2. Oct 9, 2007

### pervect

Staff Emeritus
As I understand it, the Shapiro time delay will include both the longer geometric path, and the effects of gravitational time dilation.

3. Oct 9, 2007

### meopemuk

Yes, there is some additional delay due to the non-straight path of light. But it is much smaller than the main effect. In my opinion, the main origin of the Shapiro delay is the fact that light propagates slower in the vicinity of massive objects.

Eugene.

4. Oct 15, 2007

### yogi

From the vantage point of the earth observer, the speed of light passing near the sun's surface of radius r is c[(1-2GM/(c^2)r] but the local speed for an infinitely small physical distance and time is still c.

5. Oct 15, 2007

### meopemuk

This might be true, however I don't think the speed of light in different gravitational potentials was ever measured with sufficient accuracy. We know for sure that the rate of all physical processes (e.g., the frequency of atomic clocks) slows down in the gravitational potential according to f' = f[1-GM/(c^2)r]. This fact can be reconciled with the variation of the speed of light c' = c[1-2GM/(c^2)r] if we assume that all distances decrease according to d' = d[1-GM/(c^2)r].

Eugene.

6. Oct 15, 2007

### pervect

Staff Emeritus
While direct tests of the speed of light may or may not have been done to any particular degree of accuracy, it is a fundamental theoretical prediction of General Relativity that the local speed of light is constant.

If the local speed of light were not constant, General Relativity would be falsified.

General Relativity to date has a fairly impressive amount of experimental support, certainly it does a good job of modeling transit times and light paths in the solar system, including the shapiro effect.

Note that "the speed of light" near the sun as seen from Earth is not even measurable unless one devises some synchronization scheme. What Yogi computes above is the "coordinate speed" of light in the Schwarzschild coordinate system, i.e. dr/dt, where r is the Schwarzschild r coordinate, and t is the Schwarzschild t coordinate. This can be and usually is interpreted as the coordinate speed of light viewed from an "observer at infinity", (so Yogi is talking about the Earth as if it were infnitley far away from the sun here), but there are conceptual ambiguities about talking about speeds at distant events.

Measuring dr/dt is not really a very fundamental way to measure speed, because (among other reasons) r is not really a distance measure, it is just a coordinate. The situation is rather similar to someone who claims that the speed of naval ships increases near the poles of the earth, because d(longigutude)/dt is greater, i.e. the ship can cover one degree of longitude in a shorter time near the poles than it can at the equator. If you look at the actual local speed of the naval ship relative to the ocean, it is constant - the apparent increase in d(longitude)/dt near the poles is an artifact of the curvature of the Earth.

Philosphically, lattitude (usually denoted by $\phi$) and longitude (usually denoted by $\lambda$) on the Earth are coordinates, not distances. The metric of the curved Earth's surface converts small changes in these coordinates into actuall distances. Because the Earth's surface is curved, the metric coefficients are not constant, assuming the Earth's surface is constant one can write this metric as

$$ds^2 = r^2 d\lambda^2 + r^2 \cos^2 \, \phi d \phi^2$$

(I've taken the liberty of assuming one converts from degrees to radians rather than write out the conversion explicitly)

here ds is the distance on the Earth's surface, and $\phi$ and $\lambda$ are the lattitude and longitude as previously mentioned.

Compare this to the Schwarzschild metric: it's very similar as far as the angular terms go. (Note that equator of the earth is $\phi=0$ while the equator in Schwarzschild coordinates is $\theta=\pi/2$.)

Similarly, the local speed of light is a constant everywhere - what could be called the "change" of dr/dt with r is an artifact of the curvature of space-time, not an physical change in speed, just as the change in $d\lambda/dt$ for our naval ship as it got closer to the poles (i.e. as $\phi$ increased) was.

The confusion comes about in large part by not distinguishing between coordinates: on the Earth, lattitude and longitude, in space: Schwarzschild r, theta, phi, and t coordinates, and distances.

Distances are computed from small changes in the coordinates via the metric. Note that in relativity, one actually computes the value of the space-time interval.

Last edited: Oct 15, 2007
7. Oct 15, 2007

### meopemuk

Hi pervect,

you are correct that my post about the varying speed of light is not consistent with the ideas of general relativity. I simply wanted to note that the Shapiro time delay can be explained without GR's assumption of the geometry change in gravitational fields. An alternative explanation - that the speed of light slows down according to

$$c' = c(1-\frac{2GM}{c^2 r})$$

also works.

Eugene.

8. Oct 15, 2007

### pervect

Staff Emeritus
http://xxx.lanl.gov/abs/astro-ph/0006423 takes a somewhat similar point of view, I think. We like to talk about peer-reviewed theories here. This is a peer reviewed theory that I think (hope) is simlar to what you are talking about.

It's also an example of a non-geometrical interpretation of something that is locally the same as GR. Unfortunately, this is not quite the same as being equivalent to GR.

It is a different thoery because it doesn't necessarily have the same global topology as GR does.

The above paper unfortunately doesn't go into or even mention the topology issues. There used to be some criticism of this theory, informally called "funky fields in a Minkowski space-time" that I would refer people to on the WWW, but it disappeared :-(. You can probably find some past PF discussions if you google. I think the idea has promise for getting rid of pesky time machines in GR, but that's a personal opinion, and there are still some unanswered questions as to how one deals with the topological issues, and it is also not quite clear how the theory deals with black holes. (Do said "funky fields" become infinite on the event horizon? Or what?).

9. Oct 30, 2007

### yuiop

Thanks for all the thoughtful replies and sorry for the delay replying (family issues).

We all seem to be agreed there would be a measurable time delay compared to a constant speed of light over the the coordinate distance. Yogi and Eugene interpret the cause as due to the speed of light slowing down in a gravitational field while Pervect has the more formal point of view that the speed of light is constant in a gravitational field when measured in 4 dimensional curved spacetime rather than the 3 dimensional space coordinates we are used to.

Shouldn't that be $$c = c'\sqrt(1-\frac{2GM}{c^2 r})$$ ?

Did you intend f' = f[1-GM/(c^2)r] (the excellent approximation of f' = f sqrt[1-2GM/(c^2)r] in a week field that is the first term of a binomial expansion) or f' = f[1-GM/(c^2)r] ?

This aproximation has the interesting property that the solutions for r < GM/(c^2) do not involve imaginary numbers. It also implies negative gravity and negative time for r < GM/(c^2). I wonder if any real experiments have been carried out that can distinguish f' = f[1-GM/(c^2)r] from f' = f sqrt[1-2GM/(c^2)r] ?

Assuming we all agreed that the speed of light as measured by a local observer is always c then a slowing down of light as observed by a distant observer would be more likely to be reconciled with an increase of horizontal distances? For example, say we had a fibre optic that took one second for a photon to traverse in a very weak gravitational field. If we lowered that optical fibre ruler into a strong gravity well where the gravitational time dilation factor was 2 then a photon would take 0.5 seconds to traverse that ruler as measured by a local observer with slow clocks (assuming no length change). If the ruler had "length dilated" to twice its proper length, then the local observer would still measure the time taken to traverse the ruler as one second and the speed of light would still appear to be c to the local observer.

I would like to suggest a thought experiment with 2 scenarios to explore the possibilities and further clarify what is thought to be happening in a strong gravitational field.

Thought experiment:

Suppose we had optical fibre ring that had a radius of $$\frac{1}{2\pi}$$ light seconds when the ring is far from any gravitational fields. A device is spliced into the ring that can inject a photon into the optical fibre ring and times the interval for the photon to complete one loop and this should presumably be one second. R is calculated by a distant observer from the angle subtended by the ring when viewed through a telescope and the distance of the observer from the ring. At this point both local and distant observers agree on the radius of the ring, the apparent speed of light and that the circumference is given by $${2\pi r}$$ .

Now suppose we had a convenient (non rotating) gravitational body which just happens to be a snug fit for our optical ring around the equator of the body. Assume the mass of the body is such that the gravitational time dilation factor is 1.25. Also assume there is atmosphere that would give rise to optical effects. Please ignore practical considerations such as this body is probably a neutron star and all local instruments and observers would be crushed ;) What would the local and distant observers measure?

Scenario A: (Assuming horizontal gravitational length dilation).

Here we assume the circumference of the ring has expanded by a factor of 1.25
The local observer measures the time for a photon to travel within the optical fibre ring as one second due to his slow clocks relative to the distant observer.
The distant observer measures the photon as taking 1.25 seconds to travel a distance of 1.25 light seconds. Both agree on the same number for the speed of light in metres per second so neither sees an apparent slowing down of the speed of light in a gravitational field.

Scenario B: (No length dilation)

The distant observer measures the circumnavigation time of the photon within the optical fibre ring on the equator of the gravitational body as 1.25 seconds. From the distant observers point of view the circumference is 1 light second so the speed of light seems to have slowed down to 0.8c

If the distant observer defines distance by light travel times then he could say that that the circumeferance of the body is not given by $${2\pi r}$$

The local observers on the surface of the body (with the slower clock) measures the photon travel time as 1 second and so the local speed of light appears to be c to them.

Note: Self gravitational lensing

I have not taken self gravitational lensing into account in either scenario. It should be apparent that the strong gravitational field of the body will magnify the apparent size of the body to a distant observer. The body and ring will appear to subtend a larger angle in the telescope the distant observer. I can find formulas for the amount light is bent by a gravitational field as it passes a massive body. However I can not find a formula for how much a body would be appear to be magnified by its own gravity. It may be possible that when the distance of the observer from the body, the mass of the body and the gravitational dilation factor on the surface of the body and at the location of the distant observer (assuming he is not at infinity) are all taken into account then the optical magnification would make the speed of light be c to both local and distant observers. For example, in scenario B the distant observer would see the radius and circumference of the body optically magnified by a factor of 1.25 so that the circumnavigation time of 1.25 seconds would be consistent with an unchanged speed of light even to the distant observer.

So, is scenario A or B (or neither) more consistent with the accepted wisdom?

Does anyone have a formula for self gravitational lensing of a body?

(My interpretation is that it would be something like $$s = s'\frac{\sqrt(1-\frac{2GM}{c^2 d}}{ \sqrt(1-\frac{2GM}{c^2 r}}$$ where s is the self gravitational optical magnification of the body and d is the distance of the observer from the body)

In measuring r and d I have assumed there is no dilation or contraction of length in the vertical direction in a gravitational field. Is that a reasonable assumption?

Thanks

10. Oct 30, 2007

### meopemuk

Hi kev,

I used f' = f[1-GM/(c^2)r] instead of f' = f sqrt[1-2GM/(c^2)r] simply because (as far as I know) all existing experimental tests have precision of not better than $c^{-2}$. So, keeping only the first order in $c^{-2}$ seems to be a prudent approach.

Your ideas about experimental tests for gravity effects on length and speed of light look interesting, however (you would probably agree) completely unrealistic. Is it possible to modify these tests, so that they become realizable with modern technology?

Eugene.

11. Nov 2, 2007

### yuiop

Hi Eugene,
In 2005 I posted a document on the web that conjectured "What if f' = f[1-GM/(c^2)r] if the true relationship rather than the conventional f' = f sqrt[1-2GM/(c^2)r] ?" As you point out the accuracy of existing tests can not rule out f' = f[1-GM/(c^2)r] (as far as I know). In strong gravity fields such as near a black hole the differences become significant and give rise to a different type of black hole. Using f' = f[1-GM/(c^2)r] the event horizon is half the conventional Schwarzschild radius. Below the event horizon gravity is negative giving rise to a solid shell with radius GM/(c^2) rather than the singularity of the conventional interpretation. Quantum fluctuations at the surface of the shell would allow radiation from the surface of black hole but that radiation would be at very long wavelengths and very faint due to extreme time dilation at the surface. (The alternative black hole would still be a very dark object) Gravitational lensing due to a black hole would probably be less than the conventional interpretation. One interesting thought is that if all the mass of the known universe was contained within a Planck radius there would extreme negative repulsive gravity giving rise to a simple model of the inflation/expansion of the big bang. The negative time dilation at such extremes in this model is not a problem as it is the cause of negative gravity. Objects fall away from each other instead of towards each other in negative time. Anyway, I have not posted a link to my document as I admit it is all idle speculation and conjecture. (Just an interesting thought)

I agree the thought experiments are unrealistic (as are most thought experiments) and being non local there is no possibility of testing them in a lab. However observations of X ray emissions from a black hole accretion disc and possibly measurements of red shifted emissions in the radio spectrum from nearer the event horizon of the black hole may give an indication of the actual "radius" of the black hole. Comparing these measurements to the black hole mass calculated from the orbit of a companion binary star may give some indications of physics at work in these extreme gravitational fields. Quite possibly, this has already been done.

In my last post I asked a lot of questions (perhaps too many).

To keep things simple the main question I am asking at his point is:

Do objects length contract or expand in a strong gravitational field as measured by a distant observer ...or not?

Any expert in GR should be able to answer that simple question :P

12. Nov 2, 2007

### Chris Hillman

A Few Pointers

What textbooks have you studied so far? I'll hopefully assume that you are familiar with the contents of a standard textbook such as O'Hanian and Ruffini, Gravitation and Spacetime, which has a nice exposition of the Shapiro time delay effect. For the reader's convenience I'll review the essential ideas here.

Correct. Consider the Earth, Venus, and the Sun. The experiment is best performed when all three are almost aligned, but Venus is on the other side of the Sun relative to the Earth. Then a radar pip is sent from Earth toward Venus, passing very near the Sun; it is reflected from Venus and returns to Earth, again passing very near the Sun.

Idealize Earth and Venus to have neglible mass, i.e. imagine that the ambient gravitational field is due to the Sun. This suggests treating the problem in the (exterior of the) Schwarzschild vacuum solution by considering two world lines representing the motion of the Earth and Venus as test particles. Since the round trip takes on the order of ten minutes we can even idealize them as static test particles. The problem is now reduces to studying null geodesics connecting the two world lines. By symmetry, we can consider the two legs of the journey to be almost identical.

The line element written in the standard Schwarzschild chart is
$$ds^2 = -(1-2m/r) \, dt^2 \, + \, \frac{dr^2}{1-2m/r} \, + \, r^2 \, \left( d\theta^2 + \sin(\theta)^2 \, d\phi^2 \right),$$
$$-\infty < t < \infty, \; 2 \, m < r < \infty, \; 0 < \theta < \pi, \; -\pi < \phi < \pi$$
The world lines of static observers in this chart have the form $r=r_0, \, \theta=\theta_0, \, \phi=\phi_0$. The spatial hyperslices orthogonal to such observers have the form $t=t_0$ and they are all identical (by the time translation symmetry of the Schwarzshild vacuum). The null geodesics are planar, so without loss of generality we can suppress the coordinate $\theta$ by setting $\theta=\pi/2$ (the locus of the equatorial plane).

If we project the null geodesic arc representing outward leg of the journey of the radar pip into $t=0$, we obtain a curve which I'll call "the path". It is the projection of a null geodesic in spacetime, but it is not a geodesic in the metric on t=0 which is induced by restricting the Schwarzschild metric tensor to this hyperslice (with $\theta$ suppressed):
$$d\sigma^2 = \frac{dr^2}{1-2m/r} \, + \, r^2 \, d\phi^2, \; 2 \, m < r < \infty, \; -\pi < \phi < \pi$$
This induced metric can be considered a perturbation of the usual polar coordinate chart for the euclidean plane, and also a good approximation to it, in the region we are considering (outside the surface of the Sun).

The path appears to be "bent" slightly as it passes near the origin (the location of the Sun). As everyone knows, extending the path to infinity in both directions we get a angle between two asymptotes, which Einstein computed ("light bending"). The two asymptotes in fact are rather close to the path itself, so to simplify the computation we can (subject to justification upon demand!) treat the path as two line segments.

We can write any straight line segment in a polar chart in the form $R = r\, \cos(\phi)$, where R is the distance of closest approach to the origin. Differentiating gives $r \, d\phi = \cot(\phi) \, dr$ and plugging this relation into the line element of the Schwarzschild vacuum (still working in the equatorial plane) and setting $ds^2 = 0$ (since we are working with a null geodesic arc) gives
$$dt = \frac{r}{r^2-R^2} \left( 1 + 2m/r - 2mR/r^3 \right) \, dr \; + \; O(m^2)$$
Now suppose that the Earth is at $r=R_1$ (angle irrelevant) and integrate from $R_1$ to $R$. Proceding similarly for $R$ to $R_2$, where Venus is at $r=R_2$ (angle irrelevant), and adding, we obtain an expression consisting of the flat spacetime "height" (travel time) of our broken null geodesic path, namely $\sqrt{R_1^2-R^2}+\sqrt{R_2^2-R^2}$, where of course $2 \, m \ll R \ll R_1, \, R_2$, plus some correction terms, of which the largest is a logarithmic term. (Closer examination would show that it overwhelms the small errors introduced by the broken line approximation.)

By symmetry, the return leg should give the same result, so twice the logarithmic correction term gives the desired time delay. This is the increased time it takes for the round trip journey, as measured by an ideal clock on Earth, in case the radar pip travels near the limb of the Sun, as compared to when it does not. It's a very concrete effect!

Here's a pointer for your original question: suppose we model bending a paper clip using some one parameter family of curves. The tip moves quite a bit as we vary the parameter, but the change of length due to the bending is comparatively modest.

To find out, and also to justify the broken line approximation, compute the length of
• a straight line path "from Earth to Venus" in $t=0$, using the euclidean metric or the induced metric,
• the broken line path, likewise,
• the actual path, likewise
(I'll let you attempt this before I say more.)

This is the same misconception which I just characterized as a FAQ in another thread! Time does not slow down and distances do not shorten! (Neither of these claims even makes sense.) Rather--- well, see [post=1181763]this post[/post]. Well, ditto Pervect generally

kev, I urge you to simply ignore what meopemuk said; IMO he's adding confusion, not clearing it up.

Careful now! I've seen even professional physicists fall into fallacy with this kind of thinking The basic problem is that it is rarely straightforward to compare two objects in two different spacetime manifolds and to declare these objects to be "equivalent". IMO you need to think much harder about your thought experiments. For example, your ring cannot remain rigid when you "lower" it, so you need to think about how it responds to the changing stresses.

Due to the difficulty of nonlinear mathematics, approximations are often neccessary to make even limited progress. Unfortunately, approximation is a tricky art form and this is where fallacies most often seem to creep into "mathematical physics discourse". Unfortunately for the present context, approximations which are somewhat tricky even in flat spacetime often become much trickier in curved spacetimes.

There are in fact a number of "fiber optic" thought experiments offered in arXiv eprints, some of which IMO are misleading and have led to incorrect conclusions. Some authors are much more careful than others; preprints posted to the gr-qc section seem to exhibit a particularly large range of quality

Yes, IMO meopemuk is confusing you and leading you away from your original question, which at least makes sense to me.

All that stuff meomepuk told you about length shortening and time slowing is just wrong. Gtr says no such thing. (See any good textbook.) The best answer to this question is that it doesn't make sense. It's like asking: "In the game of baseball, when the batter turns up an ace of spades, does that result in an automatic advancement of the player holding the office of second base to the office of first base?" It sounds like a reasonable question, but only if you know nothing about baseball!

Last edited: Nov 2, 2007
13. Nov 2, 2007

### meopemuk

Let me see if you would agree with the following logic:

The claim is that from the point of view of distant observer D objects in the gravity field (F) have the following properties:

1. All clocks go [1+GM/(c^2)r] slower, i.e., the duration of one F second is equal to [1+GM/(c^2)r] D seconds

2. All objects become shorted, i.e., what F observer perceives as 1-meter rod, the D observer sees as a [1-GM/(c^2)r] meter rod

3. Light is slower in the gravitational field, i.e., observer D measures the speed of light near observer F as c[1-2GM/(c^2)r], where c is the speed of light in the vicinity of D.

If these conditions are satisfied, then measurements of observer F would also yield the speed of light as c. If F takes a rod of 300.000.000m long (which, according to D is shorter than necessary) and sends a light pulse (which, according to D is slower than usual) along it, then signal would arrive to the other end exactly in one F-second (which, according to D is longer than normal). So, both D and F would measure the speed of light as 300.000.000 m/s in their local frames.

Eugene.

14. Nov 2, 2007

### Chris Hillman

Wrong, wrong (even by your own view since you forgot about radial versus nonradial orientations), wrong. GTR says nothing like this. Meopemuk, IMO you are not helping here.

kev, to state the obvious, it makes good sense to master the mainstream view before exploring/debunking fringe viewpoints (should you desire to do that).

Last edited: Nov 2, 2007
15. Nov 2, 2007

### meopemuk

I never claimed that what I said is a part of GTR. I simply looked at experimental observations and tried to understand them without prejudice in a simplest possible model.

I wouldn't say that "time slows down" in the gravitational potential (because I don't know what's the meaning of the expression "speed of time"). However, it is well established that identical clocks run slower in the gravitational potential than far away from it (e.g., the Pound-Rebka experiment, GPS,...). This difference in the clock rates is described well by formula

$$T' = T(1+\frac{GM}{c^2r})$$

where T' is the period of one clock tick in the potential.

It is also known that light takes longer time to propagate between two points (e.g., Earth and Venus) if there is a massive body (Sun) on its way. This Shapiro time delay can be explained within GTR, as you demonstrated. But one can offer a simpler explanation as well. One can simply assume that the speed of light depends on the gravitational potential as

$$c' = c(1-\frac{2GM}{c^2r})$$

(independent on the light direction) It can be shown that the numerical value of the Earth-Venus-Earth time delay comes out exactly as in GTR.

Another fact is that the speed of light appears the same to observers in different gravitational potentials. All these facts can be reconciled by the assumption that the length of any rod decreases in the potential as

$$d' = d(1-\frac{GM}{c^2r})$$

The slowing-down of clocks and the reduction of the speed of light can be explained within a simple Newton-like theory of gravity, where gravitational interactions are distance- and velocity-dependent. Of course, this theory has nothing to do with GTR, and doesn't belong to the mainstream. However, in contrast to GTR, it is perfectly compatible with quantum mechanics, so, in my opinion, it can't be dismissed lightly.

I have no idea how to explain the shortening of lengths in the field. I think it would be great to design an experiment to show whether this length shortening exists or not. This question makes sense no matter which theory of gravity is correct. And it would be a great independent check on the general theory of relativity. Here I agree with kev.

Eugene.

16. Nov 2, 2007

### pervect

Staff Emeritus
It would be very helpful if you found peer-reviewed papers and/or respectable popularizations with a viewpoint similar to your own, and pointed them out.

http://xxx.lanl.gov/abs/astro-ph/0006423 might be a place to start. However, it's at a level that I wouldn't want to throw it at the average lay person. It also is arguably flawed, to boot, though I think it's interesting nonetheless for the advanced reader. It would unfortunately probably only confuse someone who didn't already know quite a bit about GR :-(.

It's problematic with respect to PF guidelines to promote ideas unless they've been published in a peer-reviewed journal. Because GR is very technical, I think that respectable popularizations are also acceptable.

It's also very handy to be able to point interested readers at respectable published sources. They can usually explain things in a lot more depth than a post can, and they are also something that interested readers can point to when *they* want to talk about the topic with someone else.

17. Nov 30, 2007

### mendocino

I agree that the non-straight path of light is much smaller than the main effect.
But the space can be curved (stretched or compressed) even if it is straight
for example, rubber band can be stretched without bending

You can say that space is flat but ruler shrinks in gravitational field
Anyway, if you measure the length of the path with rulers, you definitely need more rulers

In my opinion, if you ignore the curvature of space, you will get only half of the Shapiro delay

Last edited: Dec 1, 2007
18. Dec 1, 2007

### mendocino

It's perfectly OK to say the speed of light slows down and ruler shrinks in gravitational field. But the prediction must agree with measurement

In the book, Black Holes & Time Warps (page 400), Kip Thorne says:
"What is the real , genuine truth? Is spacetime really flat, or is it really curved?
To a physicist like me this is an uninteresting question because it has no physical consequences. Both viewpoints, curved spacetime and flat, give precisely the same predictions for any measurements"

19. Dec 2, 2007

### yuiop

Thought Experiment 2

As a layperson this is the sort of answer I am looking for. We can argue forever about the philosophy behind the physics, but supporters of a given theory should be able to agree on what would be measured in a given experiment. Some people (like me) might refer to "if you measure the length of the path with rulers, you definitely need more rulers" as length contraction (informally) while others like Chris Hillman would say "All that stuff meomepuk told you about length shortening and time slowing is just wrong. Gtr says no such thing. (See any good textbook.) "

(I do not know if Meomepuk is right or wrong but he is at least trying to discuss what would be measured which is what I am trying to find out in this post)

I am a little confused here. Are you now saying rulers do not shrink in a gravitational field? This seems to contradict your previous post. If you need more rulers to measure a given path in a gravitational field than you would require to measure the same path in a non gravitational context then surely one interpretation is that rulers have shrunk?

Anyway, in an early post I posed a though experiment that involved a fibre optic ring. Chris Hillman raised many objections to this thought experiment mainly due to the physical properties of a fibre optic in "different spacetime manifolds". I have now thought of a similar experiment that eliminates the requirement to use a fibre optic. It works like this:

Thought experiment 2:

Place a tape measure in space, far from any significant gravitational field. A clock is placed at one end and a mirror at the other end. The time for the two way trip of a light pulse from the clock to mirror and back to the clock is measured as 200 seconds. We can assume the tape measure is 100 light seconds long. We now wrap the tape measure around the equator of a convenient gravitational body and the two ends meet exactly in the same place. Conveniently we discover that the body has a mass exactly equal to R*(c^2)/3/G. In other words the radius of the body is the same as the photon sphere radius for that mass and density. A photon sent from the clock at the start of the tape measure will orbit the body without any need for a fibre optic to guide it and return to the clock.

a) What period will be measured by the clock on the surface of the body for one orbit of the light pulse around the body?

b) What period would be measured by the clock of a distant observer.

For the pedantic:

The gravitational body is non rotating and perfectly spherical. There is no atmosphere around the gravitational body. The original measurement of the tape measure was made in a inertial frame i.e. non-accelerating. The clocks are as near ideal as can be made. The "distant observer" is sufficiently far from the surface of the body that the gravitational field is assumed to be negligible.

Also, would the surface observer and distant observer agree on the measured geometrical and gravitational radius of the body, if we ignore strictly optical effects?

(In an early post I discussed how a distant observer would see a greater optical radius for the body due to self gravitational lensing)

Any decent theory should have testable predictions and therefore be able to predict what would be measured. Anyone claiming to understand GR should be able to give a precise answer in seconds to questions 2a and 2b.

Last edited: Dec 2, 2007
20. Dec 2, 2007

### pervect

Staff Emeritus
100 seconds. Note that ideal rigid bodies (ideal tape measures) don't actually exist, but since the tape measure isn't under any stress, it doesn't really matter in this case. If you wanted to be formal, I think that you could specify that the tape measure is "Born rigid.

You forgot one rather important point. How do you define the "period according to a distant observer"?

The purist POV is that this is a coordinate dependent quantity, it depends on what coordinates the distant observe adopts. The period according to a distant obserer would just be the difference in the time coordinates he adopts for the two events. This could be any number.

Now, there are some logical choices of coordinate system, or operational methods, that one can use in this particular case to define what one might mean by "the period of a distant observer". One could, for example, have an omnidiriectional light flash emitted which reaches the distant observer along a radial path.

Then, when this light pulse returns to its starting point, a second omindirectional light flash is emitted which also reaches the distant observer.

The interval along the distant observers worldline, according to his clock, is now a proper time measurement rather than a coordinate time measurement. This makes the answer to this problem independent of the coordinate system. It's not just the difference between two coordinates, it's an actual proper time measurement of a particular clock.

The reason that this approach works is that the "trip time" of the photon from the surface to the distant observer is expected to be constant, because the geometry is static. If the geometry wasn't static, this approach wouldn't work.

I've sort of skipped over the fact that there could be multiple paths (multiple flashes) observed from the single original flash that take different paths to the remote observer. The argument about the stationary aspect of the geometry still applies, however, as long as we can make sure by some means that the comparison process involves comparing two flashes that take the same path through the stationary geometry.

In this particular problem many of these side issues are not fundamentally important, but if/whenyou try to extend the result to non-static geometries, they become crucial.

BTW, the answer to your second question making these assumptions is 173 seconds, unless I've made a numerical error.

Last edited: Dec 2, 2007
21. Dec 3, 2007

### yuiop

Yes
Yes, yes
Agreed
Hi Pervect,
Thanks for keeping the conversation at a level us mere mortals can understand (really, no sarcasm intended ) I agree with your thoughts on how a distant observer would measure the orbital period. Perhaps the method could be refined by having a laser timing signal aimed directly at the distant observer to avoid the multiple path/ signal problem. I also agree that the period measured by the distant observer would be greater by a factor of 1.73 as calculated from the relativistic gravitational gamma factor of $$\gamma= \frac{1 }{ \sqrt{1-\frac{2 G M}{R c^2}}}$$

However, without knowing what radius the distant observer measures, it is still not possible to deduce if the speed of light as measured by the distant observer is slower by a factor of $$\gamma$$ or (as suggested by Meopemuk). $$\gamma^2$$

The simplest solution is that there is no length contraction (at least horizontally) and that light appears to move slower by a factor of just gamma. However I suspect it not as simple as that and it will be necessary to factor in some gravitational calculations to see which interpretation is most consistent.

If rulers shrink in a gravitational field then stating the expression for the photon orbit as

$$R = \frac{3 G M}{c^2}$$

begs the question “R as measured by whom?”

22. Dec 3, 2007

### mendocino

I think what Kip Thorne says is this:
In a gravitational field, either a) rulers do not shrink but space is curved/warped or b) space is flat but rulers shrink.
Both viewpoints make the same prediction such that you will need more rulers when you measure the length between two points in a gravitational field

Last edited: Dec 3, 2007
23. Dec 3, 2007

### yuiop

I agree that both both viewpoints give the same prediction, but I am not sure that either viewpoint gives the correct prediction. In other words I am not yet certain that more rulers would be required. An alternative explanation of the Shapiro time delay could be simply that light slows down in a gravitational field from the point of view of a distant observer. A local observer would not notice the slow down of light as the clocks of the local observer would tick slower than the clocks of the distant observer. In other words all the observed effects in a strong gravitational field could be accounted for by warping of just the time dimesion in the 4 dimensional spacetime.

I am also not sure if the "length contraction" (if any) is the same in the vertical and horizontal directions. (Vertical and horizontal have a clear definition in a gravitational field)

Pervect gave the answer 173 seconds for the orbital period of a photon as observed by a distant observer in the thought experiment. Since the photon takes 173 seconds to travel a distance of 100 light seconds as measured by the tape measure we can conclude that light appears to move slower in a deep gravity well from the POV of a distant observer. The local observer measures 100 seconds for the orbital period because his clock is advancing slower by a factor of 1.73 relative to the clock of the distant observer. No shrinking of rulers or warping of the spatial dimensions is required, at least as far as that thought experient is concerned.

24. Dec 3, 2007

### pervect

Staff Emeritus
I think that's a good way of looking at it. Note that the idea of rulers shrinking or growing can be traced back at least as far as Einstein, who proposed considering rulers on a non-uniformly heated disk as an example of a non-euclidean geometry. Based on the local temperature, the rulers would expand or contract, making the measured geometry of the disk non-Euclidean.

It seems ultimately more productive to me to deal with the geometry itself - this is the "modern" approach to general relativity. However, in very early works on relativity and also in popularizations aimed at a lay audience, one still does see the idea of rulers shrinking and expanding used, as a way to avoid considering non-euclidean geometries and instead think of expanding/contracting rulers in a familiar Euclidean geometry. This sort of explanation of GR via "expanding rulers" becomes somewhat problematical for extreme cases such as black holes (in my opinion, this sort of formulation might have helped mislead some of the earlier scientists into their false notions of the nature of the event horizon as being singular, when in fact it is not). However, if one stays away from such extreme examples, the idea can be made to work.

25. Dec 3, 2007

### yuiop

Black hole orbital period

Hi again Pervect,

While looking around the forums on this subject I came across this old post of yours here (Black hole orbital period ) https://www.physicsforums.com/showthread.php?t=117893

In that thread you derived the following expression for the orbital period around a black hole as measured by a clock co-moving with the orbiting object (proper time).

$$\tau = \frac{2 \pi r}{c} \sqrt{\frac{2r}{r_s} -3}$$

Here $r_s$ = 2GM/c^2 is the Schwarzschild radius of the black hole.

---------------------------------------------------------------------------------

I thought you might be interested in this alternative derivation obtained by adapting Newtonian physics:

We start with the classic Keplerian equation for the orbital period

(Eq 1) $$T = 2 \pi \sqrt{\frac{R^3}{G M} }$$

The period $$T_L$$ measured by a local stationary clock at the same altitude as the circular orbit will be shorter by a factor of $$\sqrt{1-\frac{2 G M}{R c^2}}$$ giving

(Eq 2) $$T_L = 2 \pi \sqrt{\frac{R^3}{G M} (1-\frac{2 G M}{R c^2})}$$

which simplifies to

(Eq 3) $$T_L = \frac{2 \pi R }{c}\sqrt{(\frac{R c^2}{G M}-2)}$$

The time $$T_p$$ measured by a co-moving orbiting clock is subject to a further time dilation factor of $$\sqrt{1-\frac{v^2}{c^2}}$$

(Eq 4) $$T_p = \frac{2 \pi R }{c} \sqrt{(\frac{R c^2}{G M}-2)(1-\frac{v^2}{c^2})}$$

To simplify this further we need to express v/c in terms of G, M and R.

In Newtonian terms centripetal acceleration is defined as $${v^2}/{R}$$ and gravitational acceleration is defined as $${G M}/{R^2}$$. In a stable orbit we can equate the two as:

(Eq 5) $$\frac{v^2}{R} = \frac{G M}{R^2}$$.

Assuming that everything in a gravitational field moves slower by the gravitational time dilation factor of
$$\sqrt{1-\frac{2 G M}{R c^2}}$$ we can modify Eq 5 to allow for gravitational time dilation and obtain

(Eq 6) $$\frac{v^2}{R}(1-\frac{2 G M}{R c^2}) = \frac{G M}{R^2}$$

Rearrange

(Eq 7) $$v^2=\frac{G M c^2}{R c^2-2 G M}$$

Divide both sides by c^2 to obtain a new expression for $${v^2}/{c^2}$$

(Eq 8) $$\frac{v^2}{c^2}=\frac{G M}{R c^2-2 G M}$$

Substitute this new expression for $${v^2}/{c^2}$$ into Eq 4 and multiply out the brackets to obtain

(Eq 9) $$T_p = \frac{2 \pi R }{c} \sqrt{\frac{R c^2}{G M}+\frac{2 G M-R c^2}{R c^2- 2 G M}-2}$$

Since $$\frac{2 G M-R c^2}{R c^2- 2 G M} = -1$$ Eq 9 simplifies to

(Eq 10) $$T_p = \frac{2 \pi R }{c}\sqrt{(\frac{R c^2}{G M}-3)}$$

which is the result we are looking for.

I hope that is correct as I worked that lot out by hand on paper.

Not that this obtained using only the gravitational and inertial time dilation factors and that distance (radius and circumference) and mass are measured the same from the point of view of local and distant observers.

We can also obtain the radius of the photon orbit by substituting $$c^2$$ for $$v^2$$ in Eq 6 to get

(Eq 11) $$\frac{c^2}{R}(1-\frac{2 G M}{R c^2}) = \frac{G M}{R^2}$$

Rearrange

(Eq 12) $$R (1-\frac{2 G M}{R c^2}) = \frac{G M}{c^2}$$

Simplify

(Eq 13) $$R = \frac{3 G M}{c^2}$$

which is the accepted equation for the photon orbit.