# Do light and gravity take identical geodesics?

1. Jun 1, 2007

### duordi

Distant mass A, causes a gravitational field which travels along a geodesic to mass B.
Mass B experiences a force due to gravitational attraction along a specific vector.
Would light also travel along the same geodesic form mass A to B causing the vector of the visible image of distant mass A as viewed by mass B to coincide with the gravitational vector?

In other words, would the visible image of a distant mass object appear to be in the direction of its gravitational attraction?
Or would / could they have different vectors?

Duane

2. Jun 1, 2007

### cesiumfrog

In theory, sure, low energy photons should take the exact same geodesic trajectories as low energy gravitons would. So if there is a big supernovae nearby, we should hear gravitational waves from the same direction as we see the burst of light.

But I don't think the rest of your question made sense: if you already understand concepts underpinning GR, how can you define "the direction of gravitational attraction"? Remember, it's only the "changes in gravity" that propagate, just like real photons only represent *changes* in the electromagnetic field (and hence, a static/permanent magnet does not behave like a torch).

Last edited: Jun 1, 2007
3. Jun 1, 2007

### duordi

Are you saying high energy photons will take a slightly different geodesic?

Duane

4. Jun 1, 2007

5. Jun 1, 2007

### duordi

Thanks but I was not considering the path of a planet.

I was wondering if the gravity forces we experience from a particular distant star are always toward its visual immage?

Duane

6. Jun 1, 2007

### cesiumfrog

Why is it that you are wondering this? (Is this leading to or from something in particular?)
I suppose the answer is "approximately" (yes).

Obviously it isn't completely true (it seems highly unlikely that placing a mirror above a bed would make gravity pull up, even if the visual image of the earth is there).

To the extent that it makes sense to talk about this (a semi-Newtonian perturbative approximation), you would expect "mass charges" to behave analogously to electric charges. Consequently, despite that you would see the image of a star in the direction of "where it previously was" (back when the light was emitted), you would expect the gravitational field to point in the (different) direction of "where is should be now" (taking into account the velocity it had when the light was emitted, and ignorant of any intervening acceleration). And since the stars aren't wandering around very quickly, those two directions are going to almost match up.

In a static/stationary case where the image of the star is shifted due to gravitational lensing, assuming you could sense the accelerations accurately enough to determine the component due to the star as opposed to the lens, sure I'd intuitively expect the force to be in the direction of the lensed image. Unlike the case of gravitational waves, I don't yet imagine sufficient application of this to motivate the work of verifying the details.. hence, I wonder why the question?

7. Jun 2, 2007

### pervect

Staff Emeritus
Gravitational radiation travels along a null geodesic, just as light will. (More accurately, gravitational radiation is predicted to travel along the same null geodesics that light does - we havean't actually experimentally measured gravitational radiation yet, much less experimnetally determined its path).

However, the "force" of gravity (this requires some interpretation to be defined in GR) does not travel along such a null geodesic.

See for example:

http://math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html
which I will quote in part. To see the entire argument, I'd recommend reading the entire FAQ. It would be inappropriate for copyright reasons to quote the entire FAQ, but I find that short quotes are useful in motivating people to follow the links.

The amount of angular momentum being lost from the Earth's orbit due to gravitational radiation is negligible. In the Newtonian limit sort of language, this means that the "force" of gravity points almost exactly at the current position of the sun, and not towards the retarded position of the sun. (in a suitable Newtonian-like coordinate system, given the technical name of PPN coordinates).

Because of abberation due to the Earth's orbital velocity, light from the sun should point in a different direction than the gravitational pull of the sun.

People often ask at this point - "But what if the Sun were suddenly to disappear". The answer to this is basically "According to GR, the sun can't disappear" - because the local conservation of energy and momentum is built into GR.

However, while one can't make the sun disappear, one could blow up the sun, or better yet implode the sun. It turns out that blowing up the sun wouldn't produce much of a gravity wave signal unless the explosion were assymetrical and that even then imploding the sun into a black hole is actually a much better choice, producing a much larger gravity wave signal. In either case, the gravity wave signal produced by the explosion / implosion would according to GR never travel faster than light.

8. Jun 2, 2007

### MeJennifer

Strictly speaking there is no difference between a gravitational field and gravitational radiation. The whole universe is one gravitational field. This gravitational field determines the shape of geodesics, there is no "traveling" just constantly changing equivalent forms of this gravitational field.

Last edited: Jun 2, 2007
9. Jun 2, 2007

### pervect

Staff Emeritus
Strictly speaking, the term "gravitational field" is highly ambiguous. The sort of "gravitational field" being discussed here, which treats a "gravitational field" as if it were a vector (necessary in order for it, the "gravitational field", to have a direction) is by no means equivalent to gravitational radiation.

This is an example of the equivocation fallacy, the misleading use of a word with more than one meaning.

10. Jun 2, 2007

### Chris Hillman

Reformulating Duane's Question so That it Makes Sense

Hi, Duane,

The gravitational field "travels along a geodesic"? Not really. I guess that you're thinking of gravitational radiation, in which case you should have said that "gravitational radiation propagates at the speed of light as a wave, and the wave vector of a pp-wave is a null geodesic vector field".

By no means all gravitational fields are radiative, which are neccessarily dynamic and pure Weyl. Indeed it is likely that the gravitational fields you are probably thinking of are not radiative.

What, you ask, is a "radiative field"? It's defined pretty much the same way as in electromagnetism, where the field of an isolated charged object consists, roughly speaking, of a quasistatic Coulomb piece which decays like $1/r^2$, plus a radiative piece which is time varying and decays like $1/r$. In the same way, in gtr the tidal tensor of an isolated massive object consists, roughly speaking, of a quasistatic Coulomb piece which decays like $1/r^3$, a time varying "radiative piece" which represents transverse gravitational radiation and decays like $1/r$, and some other pieces which need not concern us here. The informal term "Coulomb piece" is meant to suggest the $\operatorname{diag} \, (-2,1,1)$ pattern shared with electrostatics and Newtonian gravitostatics.

(Pedantic caveat: I am deliberately oversimplifying quite a bit above, as advanced physics students will recognize! For example, I am referring to the algebraic classification of fields in EM/gtr and the multipole expansion; the term "quasistatic" is meant to suggest that the time derivative of the mass doesn't appear in an expression, so that any time variation is due to the time variation of the mass parameter itself.)

And what, you ask is a "Weyl field"? Well, the curvature splits into Weyl and Ricci pieces; the Ricci piece vanishes identically in a vacuum region because it is proportional to the mass-energy tensor, as stipulated by the Einstein field equation. The Weyl piece is the part of the curvature which can be nonzero in a vacuum, for example outside a static spherically symmetric massive object. As we saw above, the Weyl curvature decomposes further into a Coulomb piece (the technical term is a type D piece), a radiative piece (the technical term is a type N piece), and some others.

In particular, the Schwarzschild vacuum solution is the static Coulomb field corresponding to the spherically symmetric static solution; it is pure Weyl but static. Indeed, it is pure Coulomb (pure type D). This solution models the static exterior gravitational field of an isolated spherically symmetric object.

And the FRW models have pure Ricci curvature. These solutions (dust for "matter dominated epoch" and radiation fluid for "radiation dominated epoch"; there are also combined models and nonzero Lambda models) are the simplest decent cosmological model in gtr. The FRW models are highly idealized (homogeneous and isotropic); needless to say, one can construct more elaborate cosmological models which allow for inhomogeneities. Any good student of gtr will immediately recognize that the isotropic assumption rules out the possibility of gravitational radiation in the FRW models, by the way, but it's not hard to add linearized gravitational radiation to an FRW model, a proceeding which is valid if this radiation is everywhere weak.

Note that to generate gravitational radiation, you need to have mass-energy moving around someplace in the right way, and that doesn't happen in the Schwarzschild vacuum or the FRW dust.

Uh oh! You are thinking of the Newtonian force law, but gtr does not treat gravitation in terms of "gravitational force"; indeed according to gtr this is a "fictitious force" analogous to "centrifugal force". Rather, gtr treats gravitation via the curvature of spacetime itself, the geometric arena for all physical interactions, as per Wheeler's slogan "[the curvature of] spacetime tells matter [or rather test particles] how to move, and matter tells spacetime [via the Einstein field equation] how to curve".

There is no "gravitational force vector" in gtr. However, there is a way of reformulating your question so that it makes sense.

In Newtonian gravitation, the tidal tensor is a symmetric second rank three-dimensional tensor which represents the relative accelerations of nearby test particles. In gtr, such relative accelerations are not fictitious, so there must be a corresponding concept in gtr, and there is. But in gtr, the tidal tensor is defined wrt a specific observer, or better, a family of them whose world lines form a timelike congruence (not neccessarily geodesic or irrotational), and it is a three-dimensional second rank symmetric tensor, just like the Newtonian tidal tensor. Therefore at each event it has an orthogonal basis of eigenvectors. In the case of static observers (or even many free-falling observers) outside an isolated object, one of the these will correspond to negative eigenvalue (tidal tension) and the other two will correspond to a positive eigenvalue (tidal compression). Similarly for the Newtonian tidal tensor, where in the static spherically symmetric field, the direction of maximal tidal tension is parallel with the radius vector from the center of mass of the massive object. Thus, in both theories, near an isolated massive object, a typical observer will be stretched in one direction and compressed orthogonally to that direction, and this direction will (approximately, in gtr) point at the massive object.

So you can ask about the direction of this tidal tensile eigenvector, with respect to various observers. Once you've figured out how this works, you can compare with the direction from which gravitational radiation is received, if the source of the field is something like a massive barbell which is oscillating in such a way as to produce radiation.

In past posts to various forums, I have studied test particles in stable circular orbits in the Schwarzschild and Kerr vacuums, using various alternative frame fields which are convenient for different purposes, including a gyrostabilized frame field and one locked with distant stars. Here we want the second:
$$\vec{e}_1 = \frac{1}{\sqrt{1-3m/r}} \; \left( \partial_t + \frac{\sqrt{m/r}}{r \, \sin(\theta)}} \, \partial_\phi \right),$$
$$\vec{e}_2 = \sqrt{1-3m/r} \, \partial_r,$$
$$\vec{e}_3 = \frac{1}{r} \, \partial_\theta,$$
$$\vec{e}_4 = \frac{1}{\sqrt{1-3m/r}} \; \left( \frac{-\sqrt{m/r}}{\sqrt{1-2m/r}} \, \partial_t + \frac{\sqrt{1-2m/r}}{r \, \sin(\theta)} \, \partial_\phi \right)$$
The dual coframe field
$$\sigma^1 = -\frac{1}{\sqrt{1-3m/r}} \; \left( (1-2m/r) \, dt + \sqrt{m/r} \, r \, \sin(\theta) \, d\phi \right),$$
$$\sigma^2 = \frac{1}{\sqrt{1-2m/r}} \, dr,$$
$$\sigma^3 = r \, d\theta,$$
$$\sigma^4 = \frac{\sqrt{1-2m/r}}{\sqrt{1-3m/r}} \; \left( -\sqrt{m/r} \, dt + r \, \sin(\theta) \, d\phi \right)$$
defines a metric tensor
$$g = -\sigma^1 \otimes \sigma^1 \; + \; \sigma^2 \otimes \sigma^2 \; + \; \sigma^3 \otimes \sigma^3 \; + \; \sigma^4 \otimes \sigma^4$$
yielding the familiar line element
$$ds^2 = -\left( 1-2m/r \right) \, dt^2 + \frac{1}{1-2m/r} \, dr^2 + r^2 \; \left( d\theta^2 + \sin(\theta)^2 \, d\phi^2 \right)$$
In GRTensorII speak this coframe is (I should do this as a matter of course, huh?)
Code (Text):

Ndim_ := 4:
x1_ := t:
x2_ := r:
x3_ := theta:
x4_ := phi:
eta11_ := -1:
eta22_ :=  1:
eta33_ :=  1:
eta44_ :=  1:
bd11_ := -(1-2*m/r)/sqrt(1-3*m/r):
bd14_ :=  (sqrt(m/r)/sqrt(1-3*m/r))*r*sin(theta):
bd22_ :=  1/sqrt(1-2*m/r):
bd33_ :=  r:
bd41_ := -(sqrt(m/r)/sqrt(1-3*m/r))*sqrt(1-2*m/r):
bd44_ := (sqrt(1-2*m/r)/sqrt(1-3*m/r))*r*sin(theta):
Info_ := Mode-locked Hagihara coframe for Schwarzschild vacuum:
# Coframe covers
#  -infty < t < infty, 3*m < r < infty, 0 < theta < pi, -pi < phi < pi
# Static chart covering right exterior region only
# Comoving with Hagihara observers orbiting ccwise
# in eq. plane (theta = Pi/2, 6*m < r < infty)

Here, notice that the frame field is only defined in the region $r>3m$. Furthermore, it turns out that these yield stable circular orbits only in the region $r>6m$.

The timelike congruence given by $\vec{e}_1$ is nongeodesic in general but gives the world lines of observers moving in circular orbits (they are latitude circles on coordinate spheres concentric with the central massive object). But in the equatorial plane $\theta=\pi/2$, the orbits are timelike geodesics, and with a bit of work we can confirm that these are the stable circular orbits (which only exist in the region noted above).

In the equatorial plane, the tidal tensor evaluated wrt our frame is
$$E \left[ \vec{e}_1 \right]_{ab} = \frac{m}{r^3} \; \left[ \begin{array}{ccc} -\frac{2 - 3m/r}{1-2m/r} & 0 & 0 \\ 0 & \frac{1}{1-3m/r} & 0 \\ 0 & 0 & 1 \end{array} \right]$$
Therefore, compared with either the radially freely infalling Lemaitre observers or the static observers, these Hagihara observers experience a slightly diminished maximal tension and a slightly increased maximal compression (the direction of maximal compression is orthogonal to the direction of motion along their circular orbits, which is given by the spacelike unit vector field $\vec{e}_4$).

Compare static observers who use their rocket engines to hover over the hole. They have the frame field
$$\vec{f}_1 = \frac{1}{\sqrt{1-2m/r}} \, \partial_t,$$
$$\vec{f}_2 = {\sqrt{1-2m/r}} \, \partial_r,$$
$$\vec{f}_3 = \frac{1}{r} \, \partial_\theta,$$
$$\vec{f}_4 = \frac{1}{r \, \sin(\theta)} \, \partial_\phi$$
and their tidal tensor is simply
$$E \left[ \vec{f}_1 \right]_{ab} = \frac{m}{r^3} \; \left[ \begin{array}{ccc} -2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]$$
The orbital frequency of these Hagihara observers is $d\phi/dt = \sqrt{m/r^3}$ (notice that Kepler would approve!) and you can compare the appropriate boost of the tidal tensor $E \left[ \vec{f}_1 \right]$ in the $\vec{f}_4$ direction with the tidal tensor $E \left[ \vec{e}_1 \right]$.

I am repeating myself at great length, but having said all this, I suppose I should also add that the tidal tensor, aka electrogravitic tensor, taken wrt a timelike unit vector field $\vec{X}$ is simply
$${E \left[ \vec{X} \right]}_{ab} = R_{ambn} \, X^m \, X^n$$
This piece of the Riemann tensor includes the so-called "electric" sectional curvatures. The magnetogravitic tensor contains the so-called "magnetic" sectional curvatures and is constructed analogously from the right Hodge dual of the Riemann tensor. The topogravitic tensor contains the "spatial" sectional curvatures and is constructed from the double Hodge dual of the Riemann tensor. These are three-dimensional tensors and they are respectively symmetric, traceless, and symmetric (6, 8, 6 algebraically independent components, making up the 20 algebraically independent components of the Riemann tensor). In a vacuum, the topogravitic tensor is not independent from the electrogravitic tensor, and the electrogravitic and magnetogravitic tensors are both traceless symmetric (5,5, components, making up the 10 algebraically independent components of the Weyl tensor).

(Warning! Having said that, I had better add that the gravitational and electromagnetic fields are very different in character. The terms "electric components" and "magnetic components" are based on formal analogies meant to help one remember the definition of the above Bel decomposition, and to suggest the close relationship with a better known formulation of "gravitoelectromagnetism" (GEM), which is only valid for weak fields. The Bel decomposition is valid whenever we have a timelike congruence.)

The point in the present context is that the spatial unit vector field $\vec{e}_2$ points in the direction of maximal tidal tension; we want to compare this with the directions in which the central object appears to be located, as seen by our Hagihara observer. (Directions plural because the object must have radius larger than a certain limit; in the case of a black hole, we could ask about the location of the dark disk $r=3m$ as seen by our Hagihara observers.)

(Pedantic aside: I should note that the frame $\vec{e}_1, \dots \vec{e}_4$ is not gyrostabilized. That is, the spatial vectors of observers in the equatorial plane are actually rotating (parallel to the equatorial plane) in the sense measured by evaluating Fermi derivatives. This is a very small effect in the solar system context. It is called the Lense-Thirring precession, and this predicted effect has been tested by the Gravity Probe B experiment, whose results are still being analyzed.)

I think the reformulated question is actually pretty interesting and I don't want to spoil the fun of anyone who hasn't seen the answer. So an exercise: carry out the comparision for the Schwarzschild vacuum. Repeat for the Kerr vacuum. (The second half is much more challenging!) Qualitative results are acceptable if supported by careful reasoning.

[EDIT: Oops, I didn't notice that pervect already gave away the answer!]

It turned out that I had to say quite a bit just to explain the restated question! Explaining the answer would no doubt call for a post of even greater length...

Last edited: Jun 2, 2007
11. Jun 2, 2007

### MeJennifer

Perhaps I misunderstand this statement but if that were the case then it seems to me there would not be much of a point searching for gravitational waves in our universe.
Clearly our universe is neither pure Weyl nor pure Ricci.
Now how the two couple in a problem free way is perhaps another issue.

12. Jun 3, 2007

### duordi

Thank you for the answers, they obviously took a lot of typing and thought.
I will not address all of it as some of it was beyond my complete understanding, however I will save your answers and review them as my comprehension will allow.

If I am understanding your answers correctly the direction of a resultant gravitational force toward the sun is very close to the actual location of the sun and not towards the direction of the visible image of the sun which will lag behind the actual image of the sun by about 8.3 minutes of travel.

This is not what I was expecting however I must say GR is constantly giving me answers I do not expect.

If you have time...

Do you agree with this site, it seems to agree with you but it is hard to know when someone has an agenda.

http://www.ldolphin.org/vanFlandern/gravityspeed.html

Duane

13. Jun 3, 2007

### pervect

Staff Emeritus
That's the right idea, it is convenient to report this as the actual angular difference as measured from Earth, which by my calculations is 20 arc-seconds, In radians, this would be a angular difference of (v/c) radians, where v is the orbital velocity of the Earth.

Note that v_orbital / c is equal to sqrt(m/r) in geometric units, i.e. in Chris Hillman's analysis.

See http://scienceworld.wolfram.com/physics/StellarAberration.html or other websites on stellar aberration if you want more background on aberration.

In a PPN coordinate system, one can say that the direction of gravity points towards the instantaneous position of the sun, while the direction of light is aberrated 20 arc seconds as per the above.

Chris Hillman has done the analysis by using pure GR (no PPN approximation) and only local coordinate systems (rather than imposing a global PPN coordinate system on the solar system), but of course the two methods should come up with the same numerical answer for the case of our solar system where the PPN approximation is justified.

By my calculations, if we consider the rotating Earth, which rotates at 15 degrees per hour or 15 arc seconds per second, that should imply a little over a second difference between a reading of maximal tension and the actual direction of light. Thus if we had a very sensitive and fast acting gravimeter, the tidal effect of gravity from the sun would show up as a small difference in the timing of the minimum gravimeter reading and noon. The peak tidal gravity deflection should occur slightly over one second apart from noon, when the sun is visually straight overhead. I believe it is correct to say that the peak reading on the gravimeter occurs first, and high noon occurs one second afterwards, but I could be making a sign error.

Van Flanderen's ideas are not mainstream at all. His self-published website doesn't really meet the PF guidelines for peer-reviewed articles. If you check out the literature, you'll see that some (not all) of Van Flanderen's idea have been published in peer reviewed journals, for instance

T. Van Flandern, Phys. Lett. A250 (1998) 1.

but if you check further you also find that many critical responses pointing out errors in Van Flandern's analysis have also been published, for instance

Phys.Lett. A267 (2000) 81-87 available online at http://arxiv.org/abs/gr-qc/9909087

and I believe that there have been several other criticisms of Van Flandern's pulbished work as well.

14. Jun 4, 2007

### pervect

Staff Emeritus
BTW, here are what should be the final steps in getting the above result using Chris Hillman's method:

Light following a geodesic travels along the null radial vector. In the equatorial plane, this will be just:

$$v^a = \frac{1}{\sqrt{1-2 m / r}} \, \frac{\partial}{\partial t} \pm \sqrt{1 - 2 m / r} \, \frac{\partial}{\partial r}$$

One can confirm this is the right vector by noting that it is a) radial and b) is null, i.e. it's square-norm is equal to zero.

Taking the spatial projection of this null vector in Chris Hillman's basis gives the apparent direction of propagation of light in terms of his Hagihara basis vectors:

$$\pm 1 \, \hat{r} -\frac{\sqrt{m \, / \, r}}{\sqrt{1-3\,m\,/\,r}} \, \hat{\phi}$$

where we've ignored the time part of the projection and

$$\hat{r} = \vec{e_1}$$ is a unit vector pointing in the 'r' direction in the Hagihara frame specified by Crhis Hillman and

$$\hat{\phi} = \vec{e_4}$$ is a unit vector pointing in the $\phi$ direction.

I hope I haven't confused anyone by my different notation: specifically $\hat{r}$ is to be interpreted as the radial vector as seen by a Hagihara observer, not in terms of $\frac{\partial}{\partial r}$.

(Note also that I've cut and pasted the GRTensor code to get these basis vectors, and would recommend that anyone else do likewise: I was having some annoying difficulties in getting them correct from the latex version. You'll know they are right when the metric comes out correct, and wrong if the metric isn't correct.)

We see that in the Newtonian limit where r >> m, the sine of the aberration angle, which is essentially equal to the aberration angle as $sin \theta \approx \theta$ for small $\theta$ is

$$sin \theta \approx \theta = -\sqrt{\frac{m}{r}}$$

which in non-geometric units would be

$$\theta \approx \sqrt{\frac{G\,m}{r}} \, / c$$

which is equivalent to the orbital velocity divided by c as I mentioned. So this another alternate way of demonstrating that the "direction" of gravity, experimentally interpreted in terms of something that one could measure as the direction of maximum tension does not aberrate due to orbital velocity (see CH's post for this demonstration), while the geodesic path of light (a null radial vector) does aberrate (as the above calculations illustrate).

Last edited: Jun 4, 2007
15. Jun 4, 2007

### Voltage

Interesting. I've printed and will study. Thanks all.

16. Jun 4, 2007

### duordi

Does this mean that if the earth and the sun were on a collision course we would see the sun as farther away then the gravitational force toward the sun would indicate?

I am assuming that if there is an angular difference between the light and gravity in rotation then there is a distance difference between light and gravity in linear motion.

Duane

17. Jun 4, 2007

### duordi

If mass A is falling or orbiting mass B.

Is the following a good way to look at it?

The gravitational field ‘a’ is moving independently but identically to the mass ‘A’ which caused gravitational field 'a'.
Gravitational field ‘a’ is orbiting or falling with respect to mass ‘B’ and gravitational field ‘b’
So gravitational field ‘a’ arrives fully formed at the same time that mass ‘A’ arrives instead of gravitational field ‘a’ having to form after A arrives.

I had to read that last paragraph three for four times before I was sure it meant what I though I was saying.

Duane

Last edited: Jun 4, 2007
18. Jun 4, 2007

### pervect

Staff Emeritus
The best way of describing this situation is to say that the components of tidal force measured by an observer stationary with respect to the sun measured in his local orthonormal basis frame are the same as the components of tidal force measured by a radially moving observer at the same location in space-time.

I.e. assume we have two observers, momentarily at the same point in space-time, one stationary with respect to the sun, one moving towards the sun. Each of them measures the same numerical value for the components of the tidal force with respect to their individual basis vectors (note that each of them has different basis vectors used to define their frames!), i.e. each measures the same maximum value X of meters/sec^2 of tidal acceleration per meter, using their local meter sticks and clocks, and each finds that this maximum value of tension, X, occurs when the meter stick is pointed radially, directly towards the sun.

Of course they won't agree on how long their meter sticks are, each will say that the other's meter stick is short. Nor will they agree on how fast their clocks tick.

Also, they will disagree on the distance to the sun, though it's hard to define distances correctly in GR because the notion of "distance in the large" is ambiguous in GR. If this were SR, we could say that the observer moving towards the sun would see it as being closer - but it's not SR.

There's an exercise in MTW that works out the tidal forces for radially infalling observers that I could post a reference to if it is needed. This exercise is motivated by considering the tidal forces experienced by someone falling radially into a black hole at nearly the speed of light (and it turns out they are just the same as the tidal forces experienced by someone who is stationary).

Note that as per Chris Hillman's post, orbital motion DOES change the tidal force components, though radial motion does not.

Last edited: Jun 4, 2007
19. Jun 4, 2007

### NateTG

20. Jun 4, 2007

### Chris Hillman

Please avoid needless mention of irrelevant controversies!

Whoa!

First, you misread the New Scientist article, which describes controversial claims by Fomalent and Kopeikin, who claim to have confirmed that the "speed of gravity" equals the speed of light, a feature which is more or less built in to gtr and most other relativistic gravitation theories (in fact, it is not a simple task to create a theory in which this doesn't hold.)

Second, New Scientist has (sad to say) become more of pseudoscience organ than a science magazine, and many prominent physicists have denounced its uncritical promotion of numerous pseudoscientific claims over the past ten years.

Third, you have confused the claims of Fomalent-Kopeikin with some well known crank claims which would be inappropriate to discuss in this forum. (Anyone curious should Google, and anyone wanting to discuss those claims further should trot on over to the Scepticism and Debunking forum).

Fourth, numerous cranky claims which have promoted by New Scientist in recent years have been problematical at Wikipedia, where NPOV has been difficult to maintain in part because the protagonists of several of the controversies edited articles on the controversies in partisan ways. Again, discussion of this would be off topic here, but I feel I must warn students to be very careful about accepting uncritically what they read in the Wikipedia. At various times the relevant WP articles on all these topics have presented viewpoints quite divergent from mainstream science.

For some accurate accounts of the Fomalent-Kopeikin controversy, see
http://news-info.wustl.edu/tips/page/normal/119.html
www.sciencemag.org/cgi/content/full/299/5605/323a
http://www.nature.com/news/2003/030...l;jsessionid=E0E96C52D9C388DE839FD1F06D7EC175
http://www.cyberpunks.org/display/987/article/