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Is curvature guaranteed if only one connection coefficient is 'large'

by avirab
Tags: entwurf, transformations
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avirab
#1
Feb16-10, 03:11 PM
P: 5
A (3-d or higher) metric which is flat except for one non-trivial metric function of a different coordinate - eg changing dx2 to f(y)dx2 in Euclidean or Minkowski metric [but not f(x)dx2] - is curved if f(y) has a non-zero second derivative; there is no way to make the f(y) 'disappear', ie to make the metric flat, via a coordinate transformation.
Einstein's 'entwurf' metric was of this type.
Would it be true to say that the metric of this type can't be made flat because there are no other metric functions to 'cancel' with this one during the coordinate transformations? And thus the presence of only one such non-trivial metric function guarantees curvature?

The classical limit (Newtonian regime) of the r-geodesic equation for the Schwarzschild (and entwurf) metric has one dominant connection coefficient, all the others are 'small' (or 'much smaller'), which is why it reduces to the Newtonian gravity acceleration equation. Would it be true to say that no coordinate transformation could make all the connection coefficients 'small', so that the presence of only one connection coefficient - or one very dominant one - in a particular coordinate system guarantees curvature?
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Altabeh
#2
Feb16-10, 05:04 PM
P: 665
Quote Quote by avirab View Post
A (3-d or higher) metric which is flat except for one non-trivial metric function of a different coordinate - eg changing dx2 to f(y)dx2 in Euclidean or Minkowski metric [but not f(x)dx2] - is curved if f(y) has a non-zero second derivative; there is no way to make the f(y) 'disappear', ie to make the metric flat, via a coordinate transformation. Einstein's 'entwurf' metric was of this type.
Would it be true to say that the metric of this type can't be made flat because there are no other metric functions to 'cancel' with this one during the coordinate transformations? And thus the presence of only one such non-trivial metric function guarantees curvature?
I think you mean that if there is no direct coordinates transformation to make a Minkowski spacetime out of a given metric [tex]g_{\mu\nu}(x^{\alpha})[/tex] with a non-vanishing second derivative, let n be 4 and [tex]\alpha=0,..,3[/tex], for the sake of convenience, then that spacetime is curved, this is true in the sense that [tex]g_{\mu\nu}(x^{\alpha})\rightarrow \eta_{\mu\nu}[/tex] cannot be obtained through an explicit coordinates transformation [tex]x^{\alpha}\rightarrow \bar{x}^{\alpha}[/tex]. But be careful about this. For example,

[tex]d\bar{s}^2 = d\bar{t}^2 +2\bar{x}^2d\bar{t}d\bar{x}-(1-\bar{x}^4)d\bar{x}^2,[/tex]

has a non-vanishing second derivatives wrt [tex]\bar{x}[/tex], but it can be made Minkowski through [tex]\bar{x}=x[/tex] and [tex]\bar{t}=t-x^3/3[/tex] at every point. This last bold-face is so important in our observation of flat spacetimes and distinguishing them with the curved ones.

But if [tex]g_{\mu\nu}[/tex] be a function of, say, an auxiliary parameter [tex]X[/tex] which does not count as a part of coordinate system by which the metric is described, then the first derivatives of metric vanish iff there is no relation between [tex]X[/tex] and coordinates! Such metric is already flat; for example,

[tex]d\bar{s}^2 = -X^2d\bar{t}^2 +(4X^4+6)d\bar{x}^2,[/tex]

is flat because there exists [tex]x^{\alpha}:= t= \bar{t}X,x=\bar{x}\sqrt{4X^4+6}.[/tex] Furthermore even if we assume that there is no such coordinate system, all Christoffel symbols vanish and so does Riemann tensor because X acts as a constant in the operation of differentiation.

Here is a very stringent point: You can look at the first metric and say: even if all first derivatives of a metric tensor do not vanish, yet the metric can be flat. This is so tricky because the statement

"if all first derivatives of a metric tensor vanish, then it is flat"

is only correct when there is no direct coordinates transformation to bring [tex]g_{\mu\nu}[/tex] to [tex]{\eta}_{\mu\nu}[/tex].

For example, the Schwarzschild metric is by no means flat because

a) there is no such direct coordinates transformation,
b) the first derivatives of its metric components do not vanish.

Only metric transformations can form a Minkowski version of Schwartzchild metric which in general are not considered coordinates transformation and that they just work locally not globally. The whole thing is known as "local inertia" or "local flatness" and as to how to get it, see here.

The classical limit (Newtonian regime) of the r-geodesic equation for the Schwarzschild (and entwurf) metric has one dominant connection coefficient, all the others are 'small' (or 'much smaller'), which is why it reduces to the Newtonian gravity acceleration equation. Would it be true to say that no coordinate transformation could make all the connection coefficients 'small', so that the presence of only one connection coefficient - or one very dominant one - in a particular coordinate system guarantees curvature?
As the existence of the non-vanishing Christoffel symbols always does not guarantee that spacetime is curved, so your claim can be locally true! Remember that if r is so small, then the first component of metric and consequently its first derivative in SM gets so large, and thus one cannot make a good realization of whether metric is curved or not. The inverse statement is also true; if r is so large, then SM tends to MM because the first derivatives of metric components are so small.

AB

Edit: Some errors were corrected.
avirab
#3
Feb17-10, 03:54 AM
P: 5
Thanks for the reply but it does not directly relate to my question : I wrote specifically about a flat metric with only ONE additional metric function, as in the example I gave, and in Einstein's entwurf metric. Calculating the Riemann tensor for such a metric (for the conditions stated) shows that it indeed is curved, so that part is not my question. I am asking about the legitimacy of the 'reasoning', and then asking something similar regarding connection coefficients, again where there is only ONE, or only one which is 'large'. (I also would like to know whether a statement to this effect can be made for a non-metric connection)
[BTW: The second metric you brought has something missing,dX?dx?]

Altabeh
#4
Feb17-10, 05:03 AM
P: 665
Is curvature guaranteed if only one connection coefficient is 'large'

Quote Quote by avirab View Post
Thanks for the reply but it does not directly relate to my question : I wrote specifically about a flat metric with only ONE additional metric function, as in the example I gave, and in Einstein's entwurf metric. Calculating the Riemann tensor for such a metric (for the conditions stated) shows that it indeed is curved, so that part is not my question. I am asking about the legitimacy of the 'reasoning', and then asking something similar regarding connection coefficients, again where there is only ONE, or only one which is 'large'. (I also would like to know whether a statement to this effect can be made for a non-metric connection)
I don't know what you mean by "Einstein's entwurf metric". In the years priror to his derivation of covariant field equations, Einstein eventually realized that his "Entwurf" field equations that were found during 1907-1912 are all wrong! I don't understand what you are trying to give away from something proven already wrong, but maybe it would be awesome to write exact metric you are referring to using Latex or put a typed version of it here or at least give us some article to know what it is all about!

[BTW: The second metric you brought has something missing,dX?dx?]
I fixed it!

AB
avirab
#5
Feb17-10, 05:45 PM
P: 5
da2 + db2 + dc2 is flat. f2(b)da2 + db2 + dc2 is not flat if f(b) has a non-zero second derivative. That's it. This form of metric is interesting for several reasons
1) because it is in some sense the simplest change one can make to a flat metric to make it curved.
2) This is basically the metric Einstein presented in 1913 (the 'entwurf' metric) as a model of gravity.
3) In 4-d spacetime, f''(x) = 0 can be for example something like a uniform field

If there are two metric functions, eg f(b)da2 + h(a)db2 + dc2, then in theory there can be coordinate transformations which make these 'cancel' each other, so that they are both 1. It of course depends on what functions f and h are. But if h = 1, since Rtrtr ~ f''/f , then if has a non-zero 2nd derivative, there's no way to make the metric flat.
I am looking at some intuitive 'reason' that one can give for this (that there's no way that this one metric function f(b) can be made to be 1 via a coordinate transformation), it can only be spread around to functions in the other parts of the metric.

And similarly for the connection, if there is only one non zero or only one 'much larger' than all the others, which are negligibly small, as is the case for the classical limit of Schwarzschild.
Altabeh
#6
Feb18-10, 07:56 AM
P: 665
I just realized that my early post had some error in it, so I fix it here:

Quote Quote by Altabeh View Post
I think you mean that if there is a direct coordinates transformation to make a Minkowski spacetime out of a given metric [tex]g_{\mu\nu}(x^{\alpha})[/tex] with a non-vanishing second derivative, let n be 4 and [tex]\alpha=0,..,3[/tex], for the sake of convenience, then that spacetime is not curved, this is true in the sense that [tex]g_{\mu\nu}(x^{\alpha})\rightarrow \eta_{\mu\nu}[/tex] can be obtained through an explicit coordinates transformation [tex]x^{\alpha}\rightarrow \bar{x}^{\alpha}[/tex]. For example,

[tex]d\bar{s}^2 = d\bar{t}^2 +2\bar{x}^2d\bar{t}d\bar{x}-(1-\bar{x}^4)d\bar{x}^2,[/tex]

has a non-vanishing second derivatives wrt [tex]\bar{x}[/tex], and it can be made Minkowski through [tex]\bar{x}=x[/tex] and [tex]\bar{t}=t-x^3/3[/tex] at every point. This last bold-face is so important in our observation of flat spacetimes and distinguishing them with the curved ones.
Now according to this, I think you are able to answer the questions huddling in your mind:

Quote Quote by avirab View Post
da2 + db2 + dc2 is flat. f(b)da2 + db2 + dc2 is not flat if f(b) has a non-zero second derivative. That's it. This metric is interesting for several reasons
Take the metric

[tex]ds^2=(1+ax)^2dt^2-dx^2-dy^2-dz^2,[/tex]

where [tex]a[/tex] is some constant. [tex](1+ax)^2[/tex] has a non-zero second derivative wrt x, but the metric is flat!!

If there are two metric functions, eg f(b)da2 + h(a)db2 + dc2, then in theory there can be coordinate transformations which make these 'cancel' each other, so that they are both 1.
This is not right unless one hits

It of course depends on what functions f and h are
.

But if h = 1, and f has a non-zero 2nd derivative, there's no way to make the metric flat.
I made it flat! Look at the example above!

I am looking at some intuitive 'reason' that one can give for this (that there's no way that this one metric function f(b) can be made to be 1 via a coordinate transformation), it can only be spread around to functions in the other parts of the metric.
You just need to have the Riemann tensor vanished to get a flat spacetime; this can't always be understood from the metric directly. For instance, the above metric has non-vanishing Christoffel symbols but is not curved since the Riemann tensor is non-zero. These exceptional cases do not let us distinguish flat spacetimes from curved ones. In GR, one uses the geodesic coordinates system to make sure that the metric gets flat locally, because once we understand that Christoffel symbols vanish, then the Riemann tensor is necessarily zero but the inverse is not true always, as you can see!

And similarly for the connection, if there is only one non zero or only one 'much larger' than all the others, which are negligibly small, as is the case for the classical limit of Schwarzschild.
Only locally ture! Why don't you draw a tiny attention to my notes? I said that Schwarzschild connections (Christoffel symbols) do not vanish so the spacetime is curved everywhere and tends to be Minkowski (flat) at large distances (r-->oo) from the source; leading connections to vanish locally because they are of the dimension 1/r outside the source of gravitational field!

AB
avirab
#7
Feb18-10, 08:40 AM
P: 5
Hi. I had written f in the metric instead of f2. See corrected version above. My points and question still stand.
And re the connections: I am not talking of BC's at large r, or vanishing locally, I am talking of the classical limit.
Altabeh
#8
Feb18-10, 12:07 PM
P: 665
Quote Quote by avirab View Post
Hi. I had written f in the metric instead of f2. See corrected version above.
Now the situation is way different! The answer is simple: Looking at the Ricci tensor of your metric, you can find

[tex]R_{00}=-f(b)\frac{d^2 f(b)}{db^2}, [/tex]
[tex]R_{11}=[f(b)]^{-1}\frac{d^2 f(b)}{db^2}[/tex] and [tex]R_{22}=0[/tex]. Other components are all zero.

So a space(time) of the species you've given, is flat if its Ricci tensor vanishes. Now the question is answered! But If [tex]g_{11}=-h(t)^2[/tex], then check that the space(time) is flat if

[tex]\frac{f \left( b \right) }{h\left( a \right)} = \frac{{\frac {d^{2}h \left( a \right)}{d{a}^{2}}}}{{\frac {d^{2}f \left( b \right)}{d{b}^{2}}}}. [/tex]

(Tip. For this case, as well, the vanishing of Ricci tensor imples the Riemann tensor = 0.)

My points and question still stand.
Yeah, in the case of changing the whole stuff again!!

And re the connections: I am not talking of BC's at large r, or vanishing locally, I am talking of the classical limit.
Back to the question,

Would it be true to say that no coordinate transformation could make all the connection coefficients 'small', so that the presence of only one connection coefficient - or one very dominant one - in a particular coordinate system guarantees curvature?
Not always! Because in the metric

[tex]ds^2=(1+ax)^2dt^2-dx^2-dy^2-dz^2,[/tex]

Assuming [tex]ax<<1[/tex], then both non-vanishing connections are equal to [tex]a[/tex], so for a large [tex]a[/tex], your assumption fails, because spacetime is flat! The only thing you have to remember is that the Riemann tensor always has to be non-zero to have a curved spacetime!

AB
avirab
#9
Feb18-10, 05:29 PM
P: 5
Firstly, thanks, your point about the metric helped me spot the typo of f instead of f2.
Also: re what you wrote: for ax <<1, a << 1/x, so x <<<<.
But all this is irrelevant to my question: I am asking a subtle question, please try to understanding the issue even if I am not being sufficiently precise, I value your input: I am talking of a curved spacetime not a flat one, where there is effectively only one connection coefficient in the geodesic equation, as is the case for the r-geodesic equation in the classical limit of the Schwarzschild metric. For this case it is perhaps impossible to make a coordinate transformation from this situation of only one dominant/surviving connection coefficient, all the others terms are tiny in comparison due to the factors of c2, to another metric with only all very tiny connection coefficient contributions in the geodesic equation, because then the Newtonian limit acceleration would be too small, after all we know there is gravitational acceleration near the Earth surface; so the fact of the existence/survival of only one (dominant or large connection coefficient) in the geodesic equation is significant for curvature, and the functional form of this one surviving connection coefficent perhaps encodes some 'reliable' information about curvature, even without directly computing Riemann. Indeed, the GM/r2 of the connection coefficient is a real phenomenon - although the acceleration disappears in free fall, GM/r2 is certainly the correct relative acceleration between the source rest frame and the falling particle, it is not some arbitrary funciton arising from strange choice of coordinates. (Of course a relative acceleration of this type is a form of Riemann curvature measure, so it is not impossible that this connection coefficient gives information about Riemann curvature.)
Maybe if there are more than one large connection coefficients they can in theory be made to cancel via a coordinate transformation, but if there is only one (or one very dominant one) it cannot cancel, and so it demonstrates something real, ie the presence of Riemann curvature. That's the point or question. And similarly for the metric: the fact that it is flat with a very simple modification, ie changing dx to f(y)dx in the metric, so there is only this one fuinction, perhaps that is why there's no coordinate transformation which can make the metric flat, as opposed to if there are two functions for example, their contributions to RIemann can cancel. Thanks.
Altabeh
#10
Feb19-10, 04:03 PM
P: 665
Quote Quote by avirab View Post
Firstly, thanks, your point about the metric helped me spot the typo of f instead of f2.
Also: re what you wrote: for ax <<1, a << 1/x, so x <<<<.
But all this is irrelevant to my question: I am asking a subtle question, please try to understanding the issue even if I am not being sufficiently precise, I value your input: I am talking of a curved spacetime not a flat one, where there is effectively only one connection coefficient in the geodesic equation, as is the case for the r-geodesic equation in the classical limit of the Schwarzschild metric. For this case it is perhaps impossible to make a coordinate transformation from this situation of only one dominant/surviving connection coefficient, all the others terms are tiny in comparison due to the factors of c2, to another metric with only all very tiny connection coefficient contributions in the geodesic equation, because then the Newtonian limit acceleration would be too small, after all we know there is gravitational acceleration near the Earth surface; so the fact of the existence/survival of only one (dominant or large connection coefficient) in the geodesic equation is significant for curvature, and the functional form of this one surviving connection coefficent perhaps encodes some 'reliable' information about curvature, even without directly computing Riemann. Indeed, the GM/r2 of the connection coefficient is a real phenomenon - although the acceleration disappears in free fall, GM/r2 is certainly the correct relative acceleration between the source rest frame and the falling particle, it is not some arbitrary funciton arising from strange choice of coordinates. (Of course a relative acceleration of this type is a form of Riemann curvature measure, so it is not impossible that this connection coefficient gives information about Riemann curvature.)
Maybe if there are more than one large connection coefficients they can in theory be made to cancel via a coordinate transformation, but if there is only one (or one very dominant one) it cannot cancel, and so it demonstrates something real, ie the presence of Riemann curvature. That's the point or question. And similarly for the metric: the fact that it is flat with a very simple modification, ie changing dx to f(y)dx in the metric, so there is only this one fuinction, perhaps that is why there's no coordinate transformation which can make the metric flat, as opposed to if there are two functions for example, their contributions to RIemann can cancel. Thanks.
First of all, the largeness or smallness of a connection does not have anything to do in determining a curved or flat spacetime. So talking continuously about that the only connection is perhaps dominant or large, so the spacetime is not flat without wanting to check the Riemann tensor directly is a huge errancy! You have to know that if there is only one non-zero and non-constant connection, then the Riemann tensor is always non-zero!

But if there is a couple of non-zero and non-constant connections, then the situation is a little bit different: the Riemann tensor could vanish (as in my example [tex]ds^2=(1+ax)^2dt^2-dx^2-dy^2-dz^2[/tex] because the function [tex]f=1+ax[/tex] has a vanishing second derivative)!! Note that for spacetimes with only f^2 function, there are always two non-zero connections so the vanishing of second derivative of f would imply the metric is flat otherwise it is curved. But if we have the h^2 function, too, then there are always four non-zero connections and this calls for more pedantism: Deep down, if the criterion [tex]\frac{f \left( b \right) }{h\left( a \right)} = \frac{{\frac {d^{2}h \left( a \right)}{d{a}^{2}}}}{{\frac {d^{2}f \left( b \right)}{d{b}^{2}}}}[/tex] holds, then the spacetime will be flat. Otherwise it is curved! In either cases, the spacetime being curved has nothing to do with the magniude of connections!

And finally that seems so bizarre to me that you are talking about "...gravitational acceleration near the Earth surface..." while you push me away from local discussion of the problem!!

AB


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