GR Metric Meaning: Unpacking Confusion

[speeding electron]

I've been trying to work out the meaning of the metric in General relativity. I have a few ideas, but nothing's really come together.
These are what I think is right, from SR: the space-time distance is a quantity which is agreed upon by all observers, a fundamental property of the interval. The proper time is the time between two events measured by an observer who also measures the spatial interval to be zero. It is \frac{ds}{c^2}.
Now the book I'm reading says about the Schwarzschild metric that dt in the expression for the spacetime distance is the time measured by an observer who is in a region of spacetime so far away from the object that spacetime there can be considered flat. This is different from what I understood about the metric, namely that dt, dr, d\theta, d\phi were the coordinate differences as measured by anyone. Furthermore, from my understanding it seem that d\tau is not \frac{ds}{c^2} anymore but ds c^{-2} \left( 1 - \frac{r_0}{r} \right)^{-1}. This does seem wrong. Please could some kind person shed light on my confusion, in particular the meaning of the coordinates in the metric equation. Thanks in advance.

 

[donjennix]

You really should purchase "Exploring Black Holes" by J.A. Wheeler and E.F. Taylor (ISBN 0-201-38423-X). This is the most incredibly understandable book on GR I have ever seen -- in a few lessons you will be computing the effects of a black hole using Schwartzchild all by yourself.

BTW - Wheeler and Taylor agree with your author.

 

[pervect]

I've been trying to work out the meaning of the metric in General relativity. I have a few ideas, but nothing's really come together.
These are what I think is right, from SR: the space-time distance is a quantity which is agreed upon by all observers, a fundamental property of the interval. The proper time is the time between two events measured by an observer who also measures the spatial interval to be zero. It is \frac{ds}{c^2}.

Close - people usually write

ds^2 = c^2 dt^2 - (dx^2 + dy^2 + dz^2)

for the metric of flat space-time. Well, actually there are a number of different sign conventions, but this is a common sign convention which is the closest to your post.

It's easy to see that dt = ds /c when dx,dy,dz=0, not ds/c^2. I'm not sure how you got ds/c^2, I don't think it's quite right.

Now the book I'm reading says about the Schwarzschild metric that dt in the expression for the spacetime distance is the time measured by an observer who is in a region of spacetime so far away from the object that spacetime there can be considered flat. This is different from what I understood about the metric, namely that dt, dr, d\theta, d\phi were the coordinate differences as measured by anyone.

The book is talking about the coordinate time t, which is different from proper time tau. The coordinate time is indeed what the book describes.

A local observer can and does measure his own local times and distances which are different from the Schwarzschild coordinates.

The coordinate times are used just to have a "standard" coordinate system. There are actually many possible coordinate systems, the Schwarzschild coordinate system is just one of the most common. Given the path that a particular observer is taking through space-time, the more physical observed proper times can be derived from the coordinate times. But one must specify the path to be able to perform this calculation.

 

[speeding electron]

So what is signficant about the observer's position - is this considered a transformation. This is different from SR, where the metric was an invariant under transformations?
When you transform to a frame in free fall, what happens to the matric? Will it become minkowskian from the equivalence principle, even though the proper time is that measured in a frame not following a geodesic? Or it may stay the same, which I suspect, but will the position of the observer in the mass' potential well matter?

donjennix: I'll be on the look out for that book.

pervect: yes, sorry, it of course ds/c.

Thanks for the replies.

 

[donjennix]

So what is signficant about the observer's position - is this considered a transformation. This is different from SR, where the metric was an invariant under transformations?

Let me copy/paste a section of Wheeler and Taylor to see if it helps --"wristwatch time" is the observed time of the person in the GR space. I use "tau", "phi" etc. to keep from using weird fonts. phi is the angle around the plane of rotation about the black hole. I see that my superscript "2's" indicating squares have ceased being superscripts so please interpret. The "reduced circumference", r, is a radius of course -- it is just not the radius you would measure as you got closer to the black hole. It is the circumference (which is not affected by GR dilations) divided by 2pi.

"The Schwarzschild metric describes the separation between two neighboring events in the vicinity of a spherically symmetric, nonrotating center of gravitational attraction.
d(tau)2 = {1-2M/r}dt2 – dr2/{1-2M/r} – r2d(phi)2 [A]
• dtau is the wristwatch time between the two events as measured on a wristwatch that moves directly from one event to the other.
• dt is the time between the events measured on a clock far from the center (page 2-27)
• r is the reduced circumference: circumference divided by 2pi (page 2-7).
• M is the mass of the center of attraction measured in units of meters (page 2-13).
Equation [A] is called the timelike version of the Schwarzschild metric, useful when a clock can be carried between the two events at less than the speed of light. If this is not possible, then we use the spacelike version, equation [11] on page 2-19.

d(sigma)2 = -{1-2M/r}dt2 + dr2/{1-2M/r} – r2d(phi)2


Here d(sigma) is the proper distance between the two events: the distance between them recorded by a measuring rod moving in such a way that the two events occur at the same time in its rest frame."

 

[pervect]

So what is signficant about the observer's position - is this considered a transformation. This is different from SR, where the metric was an invariant under transformations?
When you transform to a frame in free fall, what happens to the matric? Will it become minkowskian from the equivalence principle, even though the proper time is that measured in a frame not following a geodesic? Or it may stay the same, which I suspect, but will the position of the observer in the mass' potential well matter?

In a non-rotating free fall frame, the metric will always be diagonal. With the correct choice of units, the diagonal metric will be Minkowskian. The correct unit choice represents measuring distances with respect to your local rulers, and time with respect to your local clocks.

So the answer is basically that a free-fall frame will look Minkowskian - if it doesn't rotate, and if you also scale the coordiantes correctly.

The metric transforms like any other tensor with respect to a change in coordinates. The explicit equations for this are

g_a'b' = g_ab L^a_a'L^b_b'

It may look intimidating, but like all tensor equations you just sum over the repeated indices as far as doing the calculation.

Here the primed coordinates are the transformed coordinates, and the tensor L describes the coordinate transformation. The same matrix L would describe how a "vector" x_a transformed, for instance, like this

x_a' = L^a_a' x_a

So if you know how to transform a vector, transforming the metric isn't much different in principle, except that you have to apply the transformation matrix twice.

With the convention that subscripts on the matrix L alwas run from north-west to south-easth, it's easy to remember this equation too, if you think of the superscripts as "cancelling out" a repeated subscript.

You might compare this tensor notation with the usual rule for how a matrix A transforms in standard linear algebera via the matrix L if you're ambitious, I should point out that g^a_b would transform just like a matrix would. Unfortunately, matrix notation is a bit of a dead end, it can handle mixed tensors of rank two (such as the previously mentioned g^a_b but not the more general tensors of rank two such as g_ab or g^ab, and higher order tensors are Right Out. Since many important tensors in GR are of rank greater than two, matrix notation can't handle deal with them.

 

[pmb_phy]

In a non-rotating free fall frame, the metric will always be diagonal. With the correct choice of units, the diagonal metric will be Minkowskian.

Note: That assertion assumes one is using Minkowski coordinates [i.e. (ct, x, y, z) and not somerthing like (ct, r, theta, phi)] and in the case of curved spacetime g_uv = diag(1, -1, -1, -1) only applies locally.

The explicit equations for this are

g_a'b' = g_ab L^a_a'L^b_b'

It may look intimidating, but like all tensor equations you just sum over the repeated indices as far as doing the calculation.

Here the primed coordinates are the transformed coordinates, and the tensor L describes the coordinate transformation.

L does not a represent a tensor.

The same matrix L would describe how a "vector" x_a transformed, for instance, like this

x_a' = L^a_a' x_a

Note: In SR x^{\alpha} is the prototype for the Lorentz 4-vector if L represents the Lorentz transformation. In such a case x^{\alpha} is a displacement vector and represents a spacetime displacement from an event which is chosen to be the origin of coordinates.

Pete

 

[pervect]

Note: That assertion assumes one is using Minkowski coordinates [i.e. (ct, x, y, z) and not somerthing like (ct, r, theta, phi)] and in the case of curved spacetime g_uv = diag(1, -1, -1, -1) only applies locally.

Good point

L does not a represent a tensor.

What would you call it then?

Note: In SR x^{\alpha} is the prototype for the Lorentz 4-vector

Note:

Here we are talking about the difference between x_a and x^a, in case it isn't obvious (which it probably isn't). I deliberately oversimplified my first remark slightly by calling x_a a "vector" (in scare quotes), precisely because I thought it would be far too confusing to go into a long drawn out explanation about the difference between x_a and x^a at this point.

The main points I wanted to make were:

1) The proper choice of coordinates makes the metric "Minkowskian".

2) Because space-time is curved, it's not generally possible to make the metric "Minkowskian" everywhere for everyone, even though it's always possible to make the metric "Minkowskian" at any given point.

Actually, an example might help. Rather than GR, consider an application of the metric tensor to a sphere (the Earth's surface, approximated as being spherical).

You have global coordinates on the Earth's surface, lattitude and longitude. We need symbols, let's call them A for lAtitude, and O for lOngitude.

You note that the distance between any two very close points on the Earth's surface can be defined by the expression

ds^2 = C^2 dA^2 + (C cos(A))^2 dO^2

here C is a constant, which depends on the radius of the Earth. (If the coordinates are measured in degrees, C is 60 nautical miles - 1 minute of arc of the Earth's surface is 1 nautical mile, so 1 degree is 60 nautical miles).

You can think of this expression as being the metric tensor defining the Euclidian geometry of the Earth's surface. A degree of latitude always points north or south, a degree of longitude always points east or west.

The factors of C^2 and C^2 cos^2(A) convert the coordinate system (degrees of latitude, degrees of longitude) into distances.

If you used different units (grads or radians) for the angular measure, the numerical value of "C" would change.

This is an example of using the metric tensor with the familair Euclidian 2 dimensonal geometry, where ds^2 = dx^2 + dy^2 on a flat surface. On a curved surface, this formula needs to be modified. This is done by adding metric coefficients as above.

Note that the metric coefficients vary with lattitude. At any point on the Earth's surface, one can set up an x-y coordinate system that's locally Euclidian. But because of the curvature of the Earth, it's not possible to set up coordiantes that make the metric coefficients constant for all observers.

The situation with relativity is extremely similar, except that the geometry is "Minkowskian" rather than Euclidian.

 

[speeding electron]

I'm beginning to get this. My main difficulty was in working out the analogy between the notion of different observers in space and that in space-time. In space it seems that where the observer is does not affect their metric. In space-time the metric is different for different observers and motion of observers - different motion means a transformation to a frame with a different metric equation, the way an observer in that coordinate system calculates the space-time distance.
From the equivalence principle is a far way observer not the same as one in free fall? Does that mean that the Schwarzschild metric for such an observer can be made locally Euclidian? It seems that 1-2GM/rc^2 = 1 can only be true for M = 0 or when r goes to infinity, i.e. when the observed spatial distance is in flat spacetime also. Do both the observer and the observered have to be in freefall? That does not make sense.
As you can see, my trouble lies in considering both the observer and the observed. In SR it was easy since we just had relative motion to deal with, but in GR both observers (the observer measuring the proper time and the remote observer) Have their positions etc. relative to the source of the spacetime curvature.
I am relatively (no pun intended) familiar with tensor notation, co- and contravariant components and how they transform.
Thank you for your time/patience.

 

[pervect]

You write:

It seems that 1-2GM/rc^2 = 1 can only be true for M = 0 or when r goes to infinity, i.e. when the observed spatial distance is in flat spacetime also. Do both the observer and the observered have to be in freefall? That does not make sense.

Let's go back to the round Earth for a bit.

ds^2 = C^2 dA^2 + (C cos(A))^2 dO^2

The anologous observation would be -- cos(A) is equal to 1 only on the equator. What's special about the equator?

The answer is that there's nothing really special about the equator. It's a result of the particular coordinates chosen that cos(A) = 1 there.

You don't seem to quite be getting the point I'm trying to make - that the value of the metric tensor at a given point depends on the coordinate system chosen. Just as you can consider the Earth's surface to be basically flat whever you go, you can consider space-time to be "Minkowskian" everywhere you go.

But just because the Earth is flat everywhere you go doesn't mean that in some specific coordiante system that the metric tensor doesn't vary. Only on the equator does g_oo, the metric tensor for the longitude component, have a value of unity. Contrast this to saying that only at spatial infinity does the mtric tensor for g_00, the time component of the Schwarzschild metric, have a value of unity. Both of these are statements about coordinate systems, not statements about geometry.

 

[pmb_phy]

What would you call it then?

A transformation matrix.

Here we are talking about the difference between x_a and x^a, in case it isn't obvious (which it probably isn't). I deliberately oversimplified my first remark slightly by calling x_a a "vector" (in scare quotes), precisely because I thought it would be far too confusing to go into a long drawn out explanation about the difference between x_a and x^a at this point.

I'm not sure what you mean by x^{\alpha}. Is this supposed to be the component of a 4-vector or is it supposed to be a coordinate. If its a coordinate then it is not a component of a 4-vector in general relativity. As such x_{\alpha} is not the component of a 1-form. In GR I'm not even sure if x_{\alpha} is defined. Although I suppose that if it is defined then it'd be x_{\alpha} = gr_{\alpha\beta}x^{\beta}.

Way above you wrote

Close - people usually write

ds^2 = c^2 dt^2 - (dx^2 + dy^2 + dz^2)

speeding electron made an error in that c^2 should have been just c. By definition

ds^2 = g_{\alpha\beta}dx^{\alpha}dx^{\beta}[/tex]<br /> <br /> This is the spacetime interval between two events [Note to speeding_electron - I prefer to call this an &quot;interval&quot; since &quot;distance&quot; brings to mind things you can measure with a rod. Just my personal opinion though. But I think its not an uncommon opinion.]<br /> <br /> If the interval is timelike then there exists a frame of reference in which the events occur at the same place. The time interval measurd by such an observer is then labeled \tau. The relationship between the spacetime interval and the proper time is then<br /> <br /> ds^2 = c^2d\tau^2<br /> <br /> As such ds = cd\tau and therefore d\tau = ds/c<br /> <br /> <blockquote data-attributes="" data-quote="speeding_electron" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> speeding_electron said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I&#039;ve been trying to work out the meaning of the metric in General relativity. </div> </div> </blockquote>The geometric meaning is as discussed above. The <i>physical</i> meaning is that the g_{\alpha\beta} is a set of ten potentials called the <i>gravitational potentials</i>. It replaces Newton&#039;s scalar potential \Phi. This is why g_{\alpha\beta} is often referred to as the <i>tensor potential</i>. In at least one place Misner, Thorne and Wheeler call g_{\alpha\beta} the <i>Einstein potentials</i>.<br /> <br /> Its interesting to note that Einstein didn&#039;t think of GR as <i>geometerizing</i> the gravitational field anymore than he thought that SR <i>geometerized</i> the EM field. Einstein once wrote in a letter to Lincoln Barnett<br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I do not agree with the idea that the general theory of relativity is geometerizing physics or the gravitational field. The concepts of Physics have always been geometrical concepts and I cannot see why the g_ik field should be called geometrical more than f.i. the electromagnetic field or the distance of bodies in Newtonian Mechanics. The notion comes probably from the fact that the mathematical origin of the g_ik field is the Gauss-Riemann theory of the metrical contiuum which we are won&#039;t to look at as a part of geometry. I am convinced, however, that the distinction between geometrical and other kinds of fields is not logically founded[/tex]<br /> <br /> For the history buffs - That is from the same letter where Einstein told Barnett that its not a good idea to introduce a velocity dependent mass.<br /> <br /> <br /> Re - <i>Exploring Black Holes</i> - Part of that text is online at<br /> <br /> <a href="http://www.eftaylor.com/download.html" target="_blank" class="link link--external" rel="nofollow ugc noopener">http://www.eftaylor.com/download.html</a><br /> <br /> as are a bunch of other good stuff.<br /> <br /> Pete </div> </div> </blockquote>

 

[jcsd]

L isn't usually called a tensor because when you shove a vector into it you're changing the basis of the vector, not the vector itself (though arguably it's a tensor of type (1,1) as it maps a vector onto itself, but that's not really what it's there to do).

 

[pmb_phy]

L isn't usually called a tensor because when you shove a vector into it you're changing the basis of the vector, not the vector itself (though arguably it's a tensor of type (1,1) as it maps a vector onto itself, but that's not really what it's there to do).

A tensor doesn't map a vector onto itself. And to be a tensor a quantity must have all the properties of a tensor, not just one of them.

Pete

 

[jcsd]

A tensor doesn't map a vector onto itself. And to be a tensor a quantity must have all the properties of a tensor, not just one of them.

Pete

There is a tensor that maps a vector onto itself, i.e. the (1,1) identity tensor, but as I siad that the transformation matrix performs a slightly different function from this, so it is not regarded as a tensor.

 

[pervect]

Just to close the loop, MTW (one of my textbooks) also refers to L as a transformation matrix. If there were an actual vector that were being linearly transformed (for example, rotated), then it would be fair to call L a tensor, as per jcsd's point - but there isn't.

 

[Rob Woodside]

You can think of this expression as being the metric tensor defining the Euclidian geometry of the Earth's surface. A degree of latitude always points north or south, a degree of longitude always points east or west.

Reply:
Sorry to nit pick, but the geometry of the Earth's surface is not Euclidean; but, as with all Riemannian manifolds, the tangent space at any point is. A "straight line" confined to a spherical surface is the shortest distance between points and these are the great circle routes. The lines of longitude are great circles and they are parallel at the equator. If the surface was Euclidean these longitudes would never meet, but due to the real curvature of the surface they meet at the poles.

Picture a flat plane tangent to a sphere at some point P. On the flat plane one can set up rectangular coordinates (x,y) with origin at P. Starting at P on the sphere you can carry a basis vector parallel to itself and trace out a curve on the sphere. Keeping the units on the sphere the the same as in the tangent plane allows you to map out a coordinate patch on the sphere.These are the Riemann normal coordinates.The farther you are from P the worse the fit, just like rectangular prairie road systems. Away from P the metric on this coordinate patch becomes a trigonometric nightmare, but at P it is just the flat space metric.

In general relativity the Riemann normal coordinates are the local inertial frames and along any timelike geodesic the units can be chosen so that the metric is flat. Off that geodesic the Christoffel symbols fail to vanish, but the coupling constant in Einstein's equations is so small that physicists approximate the region close to the geodsic as flat. While the metric is exactly flat along the geodesic, the curvature tensor does not vanish along the geodesic, which is why you will feel tidal forces in free fall.

 

[pervect]

You can think of this expression as being the metric tensor defining the Euclidian geometry of the Earth's surface. A degree of latitude always points north or south, a degree of longitude always points east or west.

Reply:
Sorry to nit pick, but the geometry of the Earth's surface is not Euclidean; but, as with all Riemannian manifolds, the tangent space at any point is.

Right - I should have said "locally Euclidean". The geometry of the Earth is locally Euclidean (because the tangent space is), the geometry of space-time in relativity is locally Lorentzian (sometimes called Minkowskian) for the same reason.

 

[pmb_phy]

Right - I should have said "locally Euclidean". The geometry of the Earth is locally Euclidean (because the tangent space is), the geometry of space-time in relativity is locally Lorentzian (sometimes called Minkowskian) for the same reason.

There is a debate in GR as to whether curvature is a local phenomena. Some say that curvature can be detected locally,i.e. in an arbitrarily small region of a manifold (e.g. Ohanian/Rufini, Taylor/Wheeler, MTW). Some say that "local" refers to regions that are so small that our instruments are not precise enough to detect curvature (Rindler).

Each makes sense.

Pete

 

[Rob Woodside]

There is a debate in GR as to whether curvature is a local phenomena. Some say that curvature can be detected locally,i.e. in an arbitrarily small region of a manifold (e.g. Ohanian/Rufini, Taylor/Wheeler, MTW). Some say that "local" refers to regions that are so small that our instruments are not precise enough to detect curvature (Rindler).

Each makes sense.

Pete

Reply:
When you decompose curvature R by trace, you get 3 terms.

R = G + E + C ...(1)

where G depends only on the curvature scaler and has 2 non zero traces, E depends only on the trace free piece of the Ricci tensor and has only one nonvanishing trace, and C is Weyl's conformal tensor- the totally traceless remains of the curvature. Applying Einstein's equations to these pieces puts the curvature scaler proportional to the trace of the energy-momentum tensor and the trace free part of the Ricci tensor proportional to the trace free piece of the energy-momentum tensor. So the traces of curvature are locally determined by the energy-momentum tensor and Einstein's equations. These equations make no demand on C. To find C one must solve the equations with appropriate boundary conditions to find the metric and the build the curvature from the second partials of the metric and finally subtract the traces out of the curvature. C is merely what it has to be. People claim that C is the non-local part of the curvature, but this is only true in the empty space outside the matter. The second Bianchi identity can be written as the vanishing divergence of the double dual of the curvature:

Div *R* = 0 ...(2)

Calculating the double duals of G, E, and C yields

*G* = - G, *E* = E, *C* = - C ...(3)

Combining these three equations gives

Div E = Div G + Div C ...(4)

Outside the matter G and E vanish so there (4) becomes

Div C = 0 ...(5)

This is the equation for the non local piece of curvature.

Inside the matter where G or E do not vanish equation (4) holds. Trivially all three divergences in (4) could vanish, but generally this will not happen. Since G is merely the curvature scaler times an asymmetric kronecker delta, its non zero divergence can only have a trace and no traceless piece. However, C is traceless and so its non zero divergence can only be traceless. Since E has a single trace, its non zero divergence will generally have a trace and a trace free piece ensuring (4). Thus the piece of C that has a non zero divergence must also be a local piece of the curvature. Invariantly splitting C into a divergence free non local piece CNL and a local piece CL then allows the decomposition of the full curvature into

R = G + E + CL + CNL ...(6)

where G, E and CL are the locally determined pieces of curvature satisfying

Div CL = Div E - Div G and Div CL non zero ,,,(7)

and CNL is the non local piece of curvature arising from distant matter and obeying

Div CNL = 0 ...(8)

Although G, E, and CL are very different algebraically and geometrically, they are made of from the same functions describing the matter which ensures their common support. Just as one can add in a source free electromagnetic field to an electromagnetic field with sources (while adjusting the boundary conditions), one can add in a divergence free conformal tensor (also adjusting the boundary conditions) to mimic a passing gravitational wave.

 

[speeding electron]

Can someone tell me the key differences between a frame in free-fall and one at rest in a gravitational field. For example: if monochromatic light is sent radially outward in the field described by the Schwarzschild metric, how will the observed frequency differ between an observer higher up at rest and an observer at that same higher position but in free-fall?

 

[pervect]

If the two observers (one free-falling, one fixed to the coordinate system) happen to be at the same point in space with the same velocity when they receive the light signal, they will not detect any difference in frequency.

If the two observers have different velocities at the same point, the difference in frequencies will be given by the usual SR doppler shift equation.

The relative acceleration won't matter as far as the measured value of doppler shift goes if the observers are at EXACTLY the same point in space.

 

[robphy]

Reply:
When you decompose curvature R by trace, you get 3 terms.

R = G + E + C ...(1)

[snip]
Outside the matter G and E vanish so there (4) becomes

Div C = 0 ...(5)

This is the equation for the non local piece of curvature.
[snip]

Have you seen my side comment in the "Gravitomagnetism and GR" thread
https://www.physicsforums.com/showthread.php?p=388636#post388636 ,
which references an equation in Hawking/Ellis... effectively Div C=J, where J is formed from the gradient of a linear combination of Ricci and its trace. (It essentially arises from the decomposition of Riemann and the Bianchi Identity.)


Do you have any insight into this equation?

 

[pmb_phy]

Reply:
When you decompose curvature R by trace, you get 3 terms.

R = G + E + C ...(1)

where G depends only on the curvature scaler and has 2 non zero traces, ..

Please be more specific. What do you mean by "decompose curvature R by trace"? What is this "curvature scalar" that you speak of?

If you know of somwhere which proves these assertions you've made please post tex reference or URL. Thanks.

I see no relationship between what I posted and what you've posted here. You have spoken entirely mathematically and have not addressed the physical aspect I just spoke to. What I referred to had nothing to do with energy-momentum at a given point in spacetime.

I see no point in what followed relating to what I said. Sorry.

Pete

 

[Rob Woodside]

Have you seen my side comment in the "Gravitomagnetism and GR" thread
https://www.physicsforums.com/showthread.php?p=388636#post388636 ,
which references an equation in Hawking/Ellis... effectively Div C=J, where J is formed from the gradient of a linear combination of Ricci and its trace. (It essentially arises from the decomposition of Riemann and the Bianchi Identity.)


Do you have any insight into this equation?

Reply:
Yes I saw it and liked it and almost replied, but my reply didn't seem to fit the thread. I think I might have some insights- see my paper gr-qc/ 0410043.

 

[Rob Woodside]

Please be more specific. What do you mean by "decompose curvature R by trace"? What is this "curvature scalar" that you speak of?

If you know of somwhere which proves these assertions you've made please post tex reference or URL. Thanks.

I see no relationship between what I posted and what you've posted here. You have spoken entirely mathematically and have not addressed the physical aspect I just spoke to. What I referred to had nothing to do with energy-momentum at a given point in spacetime.

I see no point in what followed relating to what I said. Sorry.

Pete

Reply:

For a reference on this decomposition see Stephani et al , Exact Solutions of Eintein's Field Equations, 2nd ed. pg 37, C.U.P. (2003). I was unaware of this ref and so had to work it out myself. The curvature scaler is just the trace of the Ricci tensor.

Relevant equations can be found in my paper gr-qc/ 0410043.

Ultimately the curvature tensor is determined by the matter distribution, boundary values and Einstein's equation and I thought you were asking how much of the curvature at a point was determined by the matter at that point (the local curvature) and how much of the curvature at any point was determined by distant matter (non local curvature). I'm sorry I misunderstood you. Please tell the physical point I missed.

 

[pmb_phy]

Ultimately the curvature tensor is determined by the matter distribution, boundary values and Einstein's equation and I thought you were asking how much of the curvature at a point was determined by the matter at that point ..

No. Not at all. In fact what I was speaking about applies to regions of spacetime where there is no matter as well as regions where there is matter.

..(the local curvature) and how much of the curvature at any point was determined by distant matter (non local curvature). I'm sorry I misunderstood you. Please tell the physical point I missed.

Some people say that if you're in a curved spacetime then no matter how small the region of spacetime is one can detect the spacetime curvature and hence can tell if there is a g-field which can't be completely transformed away (i.e. "I can tell there is a relativive acceleration of particles in free-fall no matter how confined they are in spacetime"). Hence according to this view "curvature" can be detected locally. Then there are others who say that "local" means that you've chosen a region in which your instruments are not sensitive enough to detect spacetime curvature.

There are two articles I have in mind if you're interested. One is in both Ohanian's text and an article he published in Am. J. Phys. There is also Rindler
s new text in which Rinlder states his view, which is opposite to Ohanian.

Pete

Ps - If I hadn't said it before - Welcome to the forum. Nice to have another knowledgeble person here.

 

[pervect]

There is a debate in GR as to whether curvature is a local phenomena. Some say that curvature can be detected locally,i.e. in an arbitrarily small region of a manifold (e.g. Ohanian/Rufini, Taylor/Wheeler, MTW). Some say that "local" refers to regions that are so small that our instruments are not precise enough to detect curvature (Rindler).

Each makes sense.

Pete

It sounds like a never-ending debate about semantics to me. Personally I'll take a pass, I've never liked arguing semantics that much (a certain amount of clarification is sometimes necessary though).

The basic point I'm making about the geometry of space-time being locally Lorentzian (and the geometry of the Earth being locally Euclidean) is very simple - if you want to calculate distances to some accuracy (say, one part in a million), and if the metric coefficients don't vary more than that amount over the region that you are measuring the distances in, you can calculate the distances in the tangent space - as opposed to trying to find the equations of the geodesic connecting the two points, and integrating (which is the most general procedure).

When the tangent space is Euclidian (resulting from a matrix that is a unit matrix), that means you can just use the Pythagorean theorem to calculate the distances. When the tangent space is Lorentzian, you can use the special relativity formula for the Lorentz interval.

 

[pervect]

Reply:
When you decompose curvature R by trace, you get 3 terms.

R = G + E + C ...(1)

where G depends only on the curvature scaler and has 2 non zero traces, E depends only on the trace free piece of the Ricci tensor and has only one nonvanishing trace, and C is Weyl's conformal tensor- the totally traceless remains of the curvature.

I usually just see the decomposition of the Riemann into two parts, the Ricci and the Weyl. I'm not positive what E is here, I assume that G+E = the Ricci and that G is the Einstein, which makes E = R g_ab, IIRC.

Anyway, if I'm following you correctly, the point you are making is that Riemann tensor (the curvature of space-time), at a distant point away from the sun, is entirely due to the Weyl curvature terms, because R is equal to zero where there is no matter density. And you are calling this the "non-local" component of curvature. This is interesting and useful to know, but it's considerably more advanced then the very simple point I was trying to make about how space-time (and the Earth's surface) can be considered to be locally flat. See my previous post for a fuller explanation.

 

[pmb_phy]

It sounds like a never-ending debate about semantics to me. Personally I'll take a pass, I've never liked arguing semantics that much (a certain amount of clarification is sometimes necessary though).

The important thing about an argument which seems entirely about semantics is to understand each side completely. In the particular debate it was a question on measurement and the interpretation of measurements etc. But the point was significant enough to make an entire article in Am. J. Phys. and it was well worth the read. I urge you not to entirely avoid such topics when you think its entirely about semtanics. Rather look at each side and thouroughly understand it. Then ignore it at will.

The basic point I'm making about the geometry of space-time being locally Lorentzian (and the geometry of the Earth being locally Euclidean) is very simple - if you want to calculate distances to some accuracy (say, one part in a million), and if the metric coefficients don't vary more than that amount over the region that you are measuring the distances in, you can calculate the distances in the tangent space - as opposed to trying to find the equations of the geodesic connecting the two points, and integrating (which is the most general procedure).

When the tangent space is Euclidian (resulting from a matrix that is a unit matrix), that means you can just use the Pythagorean theorem to calculate the distances. When the tangent space is Lorentzian, you can use the special relativity formula for the Lorentz interval.

Think about this. It may appear that if you're restricted to an arbitrarily small region of an infinitely smooth manifold then you can't distinguish it from a flat manifold. However when you think about it then you'll see that the smaller region of the manifold that you stick to then the more specific and more exactly you can specify the curvature of the manifold. E.g. curvature of a sphere is defined in a limit as the open ball neighborhood goes to zero radius. The smaller the open ball the more precisely you can specify the curvature. Yet the smaller the open ball the more the metric becomes the metric for a flat manifold.

Pete

 

[Rob Woodside]

No. Not at all. In fact what I was speaking about applies to regions of spacetime where there is no matter as well as regions where there is matter.

Some people say that if you're in a curved spacetime then no matter how small the region of spacetime is one can detect the spacetime curvature and hence can tell if there is a g-field which can't be completely transformed away (i.e. "I can tell there is a relativive acceleration of particles in free-fall no matter how confined they are in spacetime"). Hence according to this view "curvature" can be detected locally. Then there are others who say that "local" means that you've chosen a region in which your instruments are not sensitive enough to detect spacetime curvature.

Sure, Riemann's idea was that in a tiny region around a point would look nearly flat and the smaller the region the better the approximation. This is what Pervect has been talking about. It turns out that along any timelike geodesic you can find coordinates (Riemann normal = local inertial) such that the components of the metric tensor are those of flat spacetime and the first partials of the metric components along the geodesic vanish. However the second partials do not. At points close to the geodesic the metric components depart slightly from the flat spacetime values, the first partials don't quite vanish and the second partials certainly do not vanish. The gravitational force (Christoffel symbols) depend on the first partial partials of the metric and so the gravitational force along the timelike geodesic vanishes in these coordinates. So in this local inertial frame one is in free fall along the geodesic. The curvature components depend on the second partials of the metric components. So even though things look pretty flat along the geodesic the curvature survives there. If this is happening inside matter, the curvature has both Ricci and Weyl pieces. It is not clear to me how you could slip a curvature detector inside matter without disrupting it. Outside matter only the Weyl piece of curvature survives and it is divergence free by the full second Bianchi identity. Usually the curvature is small as the coupling constant in Einstein's equations is so small. Mathematically that is what happens. However physically what happens depends on how accurately you can measure gravitational forces and curvature. With perfect accuracy you have the mathematical situation. If the gravitational force and curvature are lost in the error bars a physicist will conclude that the tiny region around the geodesic is flat, just as you report.

There are two articles I have in mind if you're interested. One is in both Ohanian's text and an article he published in Am. J. Phys. There is also Rindler
s new text in which Rinlder states his view, which is opposite to Ohanian.

Yes I would be interested and I'm trying to get Rindler's new book.

Ps - If I hadn't said it before - Welcome to the forum. Nice to have another knowledgeble person here.

Thanks for the kind words, Pete. Things seem a little less Raunchy here than over at SPR. What do all the letter icons and padlocks icons on the message board mean?