Are Christoffel symbols measurable?


by waterfall
Tags: christoffel, measurable, symbols
Ben Niehoff
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Feb17-12, 12:48 PM
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Quote Quote by PAllen View Post
This does raise some interesting points. If one allows reflections in your diffeomorphism class, then chirality is clearly not invariant (and it could not be defined in terms diff invariants). Yet I would consider it an observation.
I'm not so sure we can "observe" chirality, anyway. It is a matter of whether some spinor lies in a particular subspace...I would say that the things we actually measure are projections, which are scalars (i.e., we can ask what are a spinor's projections onto subspaces R and L, which we have defined relative to some given frame).

It's a bit of double-talk, really. Measurements are always scalars, but we can use collections of scalars to reconstruct tensorial objects in a given frame.

Coordinates, however, are always irrelevant. The important thing is a frame. A system of coordinates can be used to define a frame, by taking the coordinate basis, but all we really care about is the frame.
PAllen
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Feb17-12, 01:04 PM
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Quote Quote by Ben Niehoff View Post
I'm not so sure we can "observe" chirality, anyway. It is a matter of whether some spinor lies in a particular subspace...I would say that the things we actually measure are projections, which are scalars (i.e., we can ask what are a spinor's projections onto subspaces R and L, which we have defined relative to some given frame).

It's a bit of double-talk, really. Measurements are always scalars, but we can use collections of scalars to reconstruct tensorial objects in a given frame.

Coordinates, however, are always irrelevant. The important thing is a frame. A system of coordinates can be used to define a frame, by taking the coordinate basis, but all we really care about is the frame.
Yeah, I take it back, it is not really an exception. I was actually thinking of chirality of a body (as we were mostly discussing classical field theory). But that really means (as an observation) shares a type of similarity to a reference object we call 'left handed'. If we do a reflection, it is still similar to the object (in the manifold) we have labeled left handed. So we still observe it to be left handed.
Naty1
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Feb17-12, 02:32 PM
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I'm really glad 'observables [measurements] in GR is being hashed out. I've been making notes for several months and still am uncertain.


[1] I am unable to keep track of all the claims, counter claims and retractions and some of the terminology as "observation" vs "observable" and varying "definitions" of scalars makes it more uncertain....
If anybody would summarize ANY points of consensus that would be great for us 'amateurs'.

[2] What do you think of this assessment:

Two fish says:

..Another observable that's not a scalar. color [post #65]
[and in post #68]...we have different definitions of what a "scalar" is. I'm defining it as a quantity that doesn't change when you change coordinate systems. I measure something in my coordinate system. You measure something in your coordinate system. We get the same number.
How about this explanation from the reference posted previously:

[bottom of page 6, http://brucel.spoonfedrelativity.com...ackground.pdf]

".... Suppose I measure the temperature (C) at a given point P at a given time. You also measure the temperature (C) at P at the same time but from a different location that is in motion relative to my location. Would it make any sense if we measured different values; for example, my thermometer measured 25 C and yours measured 125 C? ..... Only scalars that remain invariant between coordinate systems like this can be called “tensors of rank 0”.

Now let f be the frequency of light coming from a laser at P. Again, let two observers, K and K*, measure the frequency of the light at P at the same time using the same units (Hz). If I am stationary relative to the source, the light will have a certain frequency, for example f. If, on the other hand, you are moving toward or away from the source, the light will be red or blue shifted.... although frequency is a scalar, it is not a tensor of rank 0..."

So what I'm asking is if we agree some scalars ARE tensors of rank zero, and these are Lorentz invarient, while other scalars are NOT and these are NOT invariant.

If so, how do we tell them apart other than making measurements in different frames of reference and comparing results?
PAllen
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Feb17-12, 02:59 PM
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Quote Quote by Naty1 View Post
".... Suppose I measure the temperature (C) at a given point P at a given time. You also measure the temperature (C) at P at the same time but from a different location that is in motion relative to my location. Would it make any sense if we measured different values; for example, my thermometer measured 25 C and yours measured 125 C? ..... Only scalars that remain invariant between coordinate systems like this can be called “tensors of rank 0”.

Now let f be the frequency of light coming from a laser at P. Again, let two observers, K and K*, measure the frequency of the light at P at the same time using the same units (Hz). If I am stationary relative to the source, the light will have a certain frequency, for example f. If, on the other hand, you are moving toward or away from the source, the light will be red or blue shifted.... although frequency is a scalar, it is not a tensor of rank 0..."

So what I'm asking is if we agree some scalars ARE tensors of rank zero, and these are Lorentz invarient, while other scalars are NOT and these are NOT invariant.

If so, how do we tell them apart other than making measurements in different frames of reference and comparing results?
I don't think we will get philosophic consensus on the broader part of your question. For example, while I think the question of an observation or measurement can be well defined, the attempt to introduce 'observable' distinct from 'observation' leaves me cold. If one defines it broadly, e.g. a construct of theoretical formulation that aids calculation of observations, I would feel compelled to accept not only E and B, but also wave functions and 'possible histories' (in sum over histories qft). I really don't consider the latter observables, so, only slightly reluctantly, rule out E and B as well. Note, I have no problem with useful but unobservable constructs in a theory - with my definitions, you can't really have a theory without them.


Your second major question is easy to answer. Frequency is effectively the timelike component light's 4-momentum. You can't treat one component of a vector as a scaler, just because it is a number in a particular coordinate representation.

As to how you know, in GR and SR, there are primary objects defined as scalars, vectors, or tensors. Then, there are standard results in differential geometry to construct scalars out of these (contraction; integration of contractions; etc.). For example, while frequency is not a scalar, a measurement of frequency by a device is, as follows: you take the dot product of light's 4-momentum with the instrument's 4-velocity to compute the measurement of frequency by the instrument. This is contraction of two vectors with the metric tensor, and is definitely a scalar. (Thus, no matter who does the computation, and what coordinates they use, everyone will agree on the frequency measurement by the specified instrument).
twofish-quant
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Feb17-12, 10:37 PM
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Quote Quote by DaleSpam View Post
Once you have measured something the observation is a scalar. If you perform some experiment and the number 7.43 pops out on your measuring device then no change of coordinate systems can possibly change that number to anything other than 7.43. Therefore, the number measured is a scalar.
Doesn't work that way. I have a device that has a switch "metric" versus "English." If I flip the switch, the system uses different logic to come up with the number that gets displayed on the LCD. The fact that I can change the number displayed based on the system makes it a non-scalar.

It may be that you claim that 7.43 is a length and I disagree, but regardless of how we interpret the number in terms of physical quantities in our favorite coordinate system, we will agree that the number is the same. That makes it a scalar.
We don't. We can flip the switch in the measuring device and get different numbers.
twofish-quant
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Feb17-12, 10:47 PM
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Quote Quote by Ben Niehoff View Post
I'm not so sure we can "observe" chirality, anyway.
I have a pair of gloves. They look left-handed.

It's a bit of double-talk, really. Measurements are always scalars, but we can use collections of scalars to reconstruct tensorial objects in a given frame.
I think that you could come up with a definition of "measurement" that always gives you a scalar, but I think it would be a technical definition that wouldn't have any obvious relationship to the common definition or actual definition or process of "measuring" or "observing."

Part of the reason I think we are arguing is that we are at the boundary where the platonic world meets the real world. I'm pretty sure you could have a mathematically precise definition of "measurement" in which all "measurements" are scalars, but the trouble is that I have a physics background, and if those definitions don't fit the "real world" it's not going to make sense.

So I see a pair of gloves and establish that they are left-handed. If you argue that really isn't an "observation" or a "measurement" then I'm going to have severe problems with those definitions. Similarly, to specify "color" requires at least three numbers. If you argue that "measurements" can only result in one number, then color cannot be measured. You can choose whatever definition of measurement you want when you are proving theorems, but if you come up with a definition of "measurement" that excludes colors, then I'm going to have problems with it.

Funny thing. I looked up the word "scalar" on wikipedia and they have three different articles on the topic. Scalar (mathematics), scalar (physics), and scalar (computer science). Curiously, the definition that I've been using is closer to the math definition which is different from the physics one. Apparently the physics definition of scalar are invariants under subset of transformations whereas that restriction doesn't exist in the math definition. Probably the reason for that is that my main interactions with scalars and vectors is financial and computational.

But rather than arguing about which definitions is correct, it might be useful to just list the definitions of "scalar" and "observation". Once you have N definitions of scalars and M definitions of observations, then we can create a chart indicating whether all "observations" are "scalars".
PAllen
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Feb17-12, 11:59 PM
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Quote Quote by twofish-quant View Post
Doesn't work that way. I have a device that has a switch "metric" versus "English." If I flip the switch, the system uses different logic to come up with the number that gets displayed on the LCD. The fact that I can change the number displayed based on the system makes it a non-scalar.



We don't. We can flip the switch in the measuring device and get different numbers.
What definition of scalar are you using? I have never seen yours. Despite disagreements on other matters, Dalespam, Ben, Mentz, and myself seem to have a common definition different from yours. In the GR context, we simply use the one from differential geometry such that, for example, any contraction of a tensors or integral of differential contraction is a scalar. I have never, in any book on GR, seen any other definition. Yet you say proper time between two events is not a scalar because it could be expressed in seconds or years.

Later you say yours is the physicists definition. I disagree - I have GR books going back 100 years, including the originals by Einstein, Pauli, etc. None use a definition resembling yours.
twofish-quant
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Feb18-12, 12:12 AM
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Quote Quote by PAllen View Post
In the GR context, we simply use the one from differential geometry such that, for example, any contraction of a tensors or integral of differential contraction is a scalar.
It appears that in GR, something is a scalar if it is invariant to Lorenz rotations and translations, but if it changes as a result of scaling relationships, it doesn't matter. In differential geometry, you aren't confined to those specific transformations. You can define an arbitrary set of transformations for determining whether something is a scalar or not.

Later you say yours is the physicists definition. I disagree
I backed away from that statement. The way I was using scalar is certainly not the way it's used in relativity, and now that I've understood that, I don't have any problems with the statement that all observables are "scalar" using the definition in relativity (i.e. number changes due to scaling don't matter, whereas number changes due to rotation and translation do).

I think one reason I ended up with my definition is that my introduction to differential geometry was through mathematical finance where obviously Lorenz rotations don't matter. Illiski's "Physics of Finance: Gauge Modelling in Non-Equilibrium Pricing" has one of the most readable introductions to fiber bundles that I've ever seen, and I think that left me with definitions that are different from those that are used in GR.

What happens in mathematical finance is that in liquid markets, things are scale invariant, but what you are looking for are situations in which scale invariance breaks down. What I think is happening is that, in relativity, unit conversions are trivial (i.e. you can always multiply X feet by a constant factor to get X km) so in defining scalar and vector, it's natural to ignore those. However, in mathematical finance, you can't just multiply X dollars by a constant factor to get Y euros, so when differential geometry is used there, the definition of scalars and vectors are such that scaling relationships are not ignored since that's what you are interested in.
PAllen
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Feb18-12, 12:27 AM
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Quote Quote by twofish-quant View Post
I have a pair of gloves. They look left-handed.
To my mind, to make this a physical observation rather than a math definition, we need to incorporate a more complete definition of what this means. To me, it means identifying the class of objects in physical reality we label left handed rather than right handed. Then, if I do a diffeomorphism that includes a reflection, this class of objects remains matching in chirality, remaining the class I call 'left handed'. This is not so different than the way pullbacks preserve metrical quantities under diffeomorphism.
PAllen
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Feb18-12, 12:55 AM
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I'm not quite sure how to define observation with no circular aspect to its definition. I mean any type of instrument (including a camera, human eye+brain, a dial on a meter, etc) capturing a snapshot of information. The description of a device includes its state of motion or rotation.

Within a particular theory, when physically interpreting its mathematical constructs, I require that internal conventional changes do not affect what I match to physical measurement. What these features are, depends on the theory. In GR, it means diffeomorphisms. The subtlety, is that a mapping that changes, say, a value of (x1 -x0) from 1 to 3 will correspond with a change in metric that makes the length stay the same. And I don't consider units of measure part of this at all. Using the same example as chirality, I get to units as saying I have a reference object for meter, for foot, etc. Under diffeomorphism, any measurement referenced to these objects stays the same.

Finally, I note, that within GR, to have the right properties (as above) a computation deemed represent a measurement must be a collection of scalars. Really, this could be viewed as a collection of individual measurements. Thus, a color photo of something is really a measurement for each pixel on a CCD or grain on chemical film.
DaleSpam
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Feb18-12, 06:37 AM
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Quote Quote by twofish-quant View Post
Doesn't work that way. I have a device that has a switch "metric" versus "English." If I flip the switch, the system uses different logic to come up with the number that gets displayed on the LCD. The fact that I can change the number displayed based on the system makes it a non-scalar.

We don't. We can flip the switch in the measuring device and get different numbers.
That is a different measurement. Yes, different measurements can have different results.

The question about whether or not a quantity is a scalar has nothing to do with flipping switches, just about changing coordinates. Regardless of the coordinate system used to analyze the experiment the number does not change, it is therefore a scalar.
twofish-quant
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Feb18-12, 07:10 AM
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As a result of this discussion, I've become convinced that within relativity, all observed quantities must be Lorenz invariant and hence are scalar quantities as the term scalar is used in relativity.

Whew.....

I'm still not altogether convinced that only scalar quantities can be observed in contexts outside of relativity. For example, you look at a wind tunnel with a wing and then you have little flags pointing in different directions. It seems to me that you are in fact observing a vector field. Also, if you look at a weather vane pointing the in the direction of the wind, that seems like observing a vector. Part of this is that my other exposure to differential geometry has to do with data visualization and the whole point there is to observe vector fields.

Again there is this problem with definitions, but if you tell an aerospace engineer or graphics visualization guru that they aren't really observing a vector but a collection of scalars, they will look at you funny, and if you tell them that those are really scalars because they are invariant under a Lorenz transform, they will really look at you funny.

So it seems to be that when you apply differential geometry to fluid dynamics or graphic visualization, you can indeed observe vectors.

Now if you have a CFD flow that is around a black hole, it seems to me that in that situation you'd have quantities that are "scalar" in the relativistic sense (i.e. the measurement does not change when you change the reference frame) but vector in the CFD sense (you need multiple components to describe the situation).

So if suppose you have a fluid flow around a black hole, and you have a field that describes the velocity of the fluid in the local reference frame of each point, I'd guess that a GR person would describe the fluid flow as a "collection of scalars" because the components of that flow do not change when you do a Lorenz transform, but the CFD person would describe the fluid flow as vector since you need multiple components to describe the velocity field. At this point I suppose we bring in fiber bundles.

Thoughts?

Also if it is the situation that different areas of physics are using different terminology then analogies aren't going to work.
DaleSpam
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Feb18-12, 07:19 AM
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Quote Quote by twofish-quant View Post
For example, you look at a wind tunnel with a wing and then you have little flags pointing in different directions. It seems to me that you are in fact observing a vector field.
Hmm, it does seem hard to cast that as a collection of scalars. The measuring device doesn't produce a number nor a set of numbers, so my previous statements either don't apply at all or their applicability is not clear.
PAllen
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Feb18-12, 09:28 AM
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Quote Quote by DaleSpam View Post
Hmm, it does seem hard to cast that as a collection of scalars. The measuring device doesn't produce a number nor a set of numbers, so my previous statements either don't apply at all or their applicability is not clear.
I agree it is a little more strained, but not fundamentally. Each flag can be treated as reading out an angular direction relative to gyroscope providing reference (conceptually). The key point for me about type of geometric object is that the wind measurement of the 'wind field' requires specification of the position and states of motion (including rotation) of the collection of flags. Change these, and you have a different measurement. You cannot talk about observing the wind field without specifying information about each and every flag. Having done so, each flag's read out is, indeed, described as (say) a pair of angles (assuming it has full range of motion) in time, each of which would be computed in GR as a Lorentz scalar function.
JDoolin
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Feb18-12, 09:39 AM
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Quote Quote by twofish-quant View Post
As a result of this discussion, I've become convinced that within relativity, all observed quantities must be Lorenz invariant and hence are scalar quantities as the term scalar is used in relativity.

Whew.....

I'm still not altogether convinced that only scalar quantities can be observed in contexts outside of relativity. For example, you look at a wind tunnel with a wing and then you have little flags pointing in different directions. It seems to me that you are in fact observing a vector field. Also, if you look at a weather vane pointing the in the direction of the wind, that seems like observing a vector. Part of this is that my other exposure to differential geometry has to do with data visualization and the whole point there is to observe vector fields.

Again there is this problem with definitions, but if you tell an aerospace engineer or graphics visualization guru that they aren't really observing a vector but a collection of scalars, they will look at you funny, and if you tell them that those are really scalars because they are invariant under a Lorenz transform, they will really look at you funny.

So it seems to be that when you apply differential geometry to fluid dynamics or graphic visualization, you can indeed observe vectors.

Now if you have a CFD flow that is around a black hole, it seems to me that in that situation you'd have quantities that are "scalar" in the relativistic sense (i.e. the measurement does not change when you change the reference frame) but vector in the CFD sense (you need multiple components to describe the situation).

So if suppose you have a fluid flow around a black hole, and you have a field that describes the velocity of the fluid in the local reference frame of each point, I'd guess that a GR person would describe the fluid flow as a "collection of scalars" because the components of that flow do not change when you do a Lorenz transform, but the CFD person would describe the fluid flow as vector since you need multiple components to describe the velocity field. At this point I suppose we bring in fiber bundles.

Thoughts?

Also if it is the situation that different areas of physics are using different terminology then analogies aren't going to work.
The goal seems to be (and correct me if I'm wrong), describing "the universe as it really is." The idea is that if you collect all the local observations of all of the observers in the environment, you can patch together a picture of the whole thing.

While that may well be so, but do you think a patching together of an image from different perspectives is an accurate representation of the thing as it really is?

That if you take the observations of many observers and patch them together in some well-defined fashion, then you have a picture of the universe "as it really is."

And why not? As long as your picture contains every event that ever happened, and every event that ever will happen, what do you think? Is that an accurate picture of the universe, or is it flawed?

Is it better to represent the universe from the perspective of a single observer; a single observer looking at non-local phenomena? With a single observer, your picture can only contain the events which the single observer observed.

By contrast, with local observations of all the observers in the environment, your picture contains every event in the environment.
Naty1
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Feb18-12, 09:52 AM
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[two fish posts immediately above clear up some ambiguities for me....]


[1] PAllen posts:
For example, while frequency is not a scalar, a measurement of frequency by a device is, as follows: you take the dot product .....
That was at least hinted at elsewhere, and I did not 'get it'...good insight, thanks.


[2]The referenced paper says:

...although frequency is a scalar, it is not a tensor of rank 0..."
PAllen says:

Frequency is effectively the timelike component light's 4-momentum. You can't treat one component of a vector as a scaler, just because it is a number in a particular coordinate representation.

Although I believe I do understand that components of a vector are themselves vectors...[I had never thought of frequency as a vector component]....I have to think more about this answer......meantime: so what is the referenced paper claiming....Are they wrong, do they have a different definition of scalar, or are they really taking about the 'measurement' ?


[3] I also did some searching and found this comparison of classical and relativistic scalars which I did not realize [it seems obvious after reading it though] :

In the theory of relativity, one considers changes of coordinate systems that trade space for time. As a consequence, several physical quantities that are scalars in "classical" (non-relativistic) physics need to be combined with other quantities and treated as four-dimensional vectors or tensors. For example, the charge density at a point in a medium, which is a scalar in classical physics, must be combined with the local current density (a 3-vector) to comprise a relativistic 4-vector. Similarly, energy density must be combined with momentum density and pressure into the stress-energy tensor.

Examples of scalar quantities in relativity:
electric charge
spacetime interval (e.g., proper time and proper length)
invariant mass
http://en.wikipedia.org/wiki/Scalar_...ativity_theory

No problem with these ideas, right??
TrickyDicky
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Feb18-12, 10:10 AM
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I think the source of confusion here came from the different meaning each poster attributes to "observations", in fact this concept is broader and more ill-defined than the more strict concept of measurement of a physical quantity although some physicists use them indistinctly to refer to the latter meaning. When used strictly in the sense of measurement it is clear all of them are scalars in the physical sense as has been explained in this thread.
So can the affine connection of GR be measured? It is obvious that in the stricter, invariant sense referred to above, it can't.
Does this mean it is not "physical"? No. We are certainly feeling their consequences and therefore "observing" it as a force. But what we measure is not so much the connection but the EM resistance of the ground against our natural tendence to follow our geodesic.
JDoolin
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Feb18-12, 10:26 AM
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Quote Quote by Naty1 View Post


http://en.wikipedia.org/wiki/Scalar_...ativity_theory

No problem with these ideas, right??
The article lists three examples of quantities that are scalars in Relativity: electric charge, spacetime interval, invariant mass, but where is the definition of scalar in Relativity?

The article has three examples of scalars, but no clear examples of what are NOT scalars.

It lists several quantities:
  • charge density
  • current density
  • momentum density
  • pressure
  • stress-energy tensor

but does not specify whether these things are considered to be scalars or not.


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