Recommendaton for Clarifying Special Relativity

[PAllen]

It seems part of this disagreement is terminological. To me, coordinates are system of labels covering a region; in contrast to measurements of distances, time intervals, and angles which I view as invariants. To me, stating that several instruments are each feeling no accelerations (measured with ideal accelerometers) and collectively maintaining fixed relative positions is part of the recipe for measurement; placing this in an inertial frame or coordinates is simply one way of describing the set up. Stating that a measurement comes out the same wherever, whenever, at any relative speed to something else, and in what orientation you do it is describing an invariant symmetry. [In this sense, I would correct my statement that you can't verify Lorentz invariance, meant as this collection of symmetries.] So, basically, all the things I define as geometric or physical invariants independent of coordinates or frames, Samshorn is bundling into an (to me) extended concept inertial frames.I don't see a constructive way to continue with such different definitions. I do see that the value in my approach is to separate concepts I consider invariant and coordinate independent from features of particular systems of labels, because I have seen much confusion in this area.

 

[PeterDonis]

Well, it's the property on which inertial coordinates are constructed, establishing a unique simultaneity, that can then be compared with the simultaneity given by isotropic light speed.

Yes, you can, *if you want to*, construct inertial coordinates using this property. That doesn't mean you *have* to do so, or that the coordinates are identical with the property. As I said before and as PAllen has pointed out, you appear to be conflating "inertial coordinates" with "physical properties that can be used to construct inertial coordinates".

The concept of speed has well-established meaning in terms of inertial measures of spatial distances and time intervals.

This is a fact about history, not physics. The fact that relative velocity was first defined in terms of these measures, and only later related to the Doppler shift, does not mean the Doppler shift is less fundamental.

The point is that if you define the (one-dimensional) speed of an object as whatever it must be to satisfy the relativistic Doppler equation, then you are simply defining the relativistic Doppler equation to be valid... it is no longer a falsifiable proposition... but we know it IS a falsifiable proposition. The reason we know this is because the concept of speed has meaning independent of Doppler shift.

But, as I said before, I can equally well turn this argument around: "the point is that if you define the speed of an object as whatever it must be to satisfy the inertial measures of distance and time, then you are simply defining those inertial measures to be valid; it is no longer a falsifiable proposition that relative velocity, so defined, is the "v" that appears in the Lorentz transformation equations. But we know it IS a falsifiable proposition. The reason we know this is because the concept of speed has meaning independent of those particular measures."

Once again, you appear to be privileging your definition of "speed" simply because it happened to be the one that was discovered first. But physically speaking, the fact that that "speed" happens to be equal to the "v" that appears in the Lorentz transformations is just as much a contingent, falsifiable proposition as the fact that the observed Doppler shift happens to be just right to make the relativistic Doppler equation valid. It could have turned out that the "speed" defined by inertial measures satisfied some other transformation equation, such as the Galilean transformation. The fact that it didn't is an empirical fact, not an a priori definition of speed.

 

[PeterDonis]

I am not aware of any tensor-based or coordinate-independent test theory of SR. Are you?

Sure, just write down all the equations the same way you can in GR, using only coordinate-free tensor expressions, and use the metric of Minkowski spacetime.

Or do you mean by "test theory" something like the Cartan geometric formulation of Newtonian gravity? In other words, casting Galilean spacetime in terms of coordinate-free tensor equations? I don't know that that has been done specifically, but I don't see why it couldn't be.

 

[Samshorn]

It seems part of this disagreement is terminological. To me, coordinates are system of labels covering a region; in contrast to measurements of distances, time intervals, and angles which I view as invariants.

Spatial distances and temporal intervals (and speeds, computed as the ratios of those two things), as well as angles are all frame dependent things (see length contraction, time dilation, and angular aberration), and when we refer to or make a measurement of these things, we are typically (implicitly or explicitly) invoking the measures corresponding to a particular system of inertial coordinates (just as MTW did in their allegedly "coordinate free" "calculation" - which wasn't coordinate free and wasn't a calculation).

Now, there is a sense in which these things ARE "invariant", once they are fully specified. There's a difference between (1) the spatial extent of an object, and (2) the spatial extent of an object in terms of a particular well-defined system of inertial coordinates. Item (1) is ambiguous and frame dependent, but item (2) is "invariant", because it includes the stipulation of both the object and the operational meaning of spatial extent that we intend (assuming you understand how inertial coordinate systems are operationally defined) - and that meaning is not defined circularly. This just highlights the fact that frame-dependent quantities are not physically meaningless. Ultimately every measurement is simply a comparison of something with something else.

For example, "the one-way speed of light" is ambiguous, but "the one-way speed of light in terms of a system of coordinates in which the equations of mechanics hold good" is a matter of empirical fact that can be tested and measured. It is not circular or tautological. It's vital to recognize and understand this, because it represents the empirical content of special relativity. Too often special relativity is presented as if it was just a bunch of unfalsifiable conventions, precisely because people don't grasp the independent physical significance of the inertial measures of space and time.

To me, stating that several instruments are each feeling no accelerations (measured with ideal accelerometers) and collectively maintaining fixed relative positions is part of the recipe for measurement; placing this in an inertial frame or coordinates is simply one way of describing the set up.

You're leaving out simultaneity. The crucial point is that the "coordinates in which the equations of mechanics hold good" (including equal action and reaction) entail a unique simultaneity, which is the basis for all measures of space, time, motion, angles, etc., within a given frame. Any time you refer to or measure those things, you are using the inertial coordinates implicit for that frame.

 

[WannabeNewton]

The whole point of special covariance is that if a family of observers make a complete set of measurements of the components of tensor fields on space-time (states of space-time) using the their measuring apparatuses (orthonormal frame field) and we act on the tensor fields by a diffeomorphism then the components of the images will correspond to a complete set of measurements made by a new family of observers (gotten by acting the diffeomorphism on the original orthonormal frame field) if and only if the diffeomorphism is an isometry. In the case of SR, these will be the representations of the proper Poincare group and tells us the above about the locally physically measurable quantities (components of tensor fields on Minkowski space-time).

 

[Samshorn]

But, as I said before, I can equally well turn this argument around: "the point is that if you define the speed of an object as whatever it must be to satisfy the inertial measures of distance and time, then you are simply defining those inertial measures to be valid; it is no longer a falsifiable proposition that relative velocity, so defined, is the "v" that appears in the Lorentz transformation equations. But we know it IS a falsifiable proposition. The reason we know this is because the concept of speed has meaning independent of those particular measures."

You got mixed up in your turn-around. The correct turn-around statement is this: If we define inertial coordinate systems as "systems of space and time coordinates in terms of which the mechanical inertia of every object is homogeneous and isotropic", then the homogeneity and isotropy of mechanical inertia is true by definition, i.e., not falsifiable. Well, this is perfectly correct (and well known - it was pointed out centuries ago that Newton's "laws" really constitute the definition of inertial coordinate systems), and indeed we could construct infinitely many more such tautologies. For example, if I define 'samshorn coordinates' as systems in terms of which all objects move in circles, then it is formally tautological that all objects move in circles in terms of samshorn coordinates. The difference is that, as an empirical matter, no samshorn coordinates exist, so the proposition has no applicability.

However, miraculously, there actually DO exist inertial coordinate systems (as defined). These kinds of definitions are hugely over-specified, and we have no right to expect that any single system of coordinates (let alone a whole family of them) could exist in which the mechanical inertia of every object is homogeneous and isotropic, any more than we should expect there to exist coordinates in terms of which every object moves in a circle. But based on enormous amounts of experience and observation, it appears (empirically) that inertial coordinates do exist - i.e., physical phenomena do exibit that amazing degree of coherence and uniformity. That coherence and uniformity, represented by the existence of inertial coordinate systems, is what makes the science of mechanics possible, and of course, living here on the moving Earth, we have always exploited this wonderful property to define our measures of space, time, motion, and simultaneity in terms of inertial coordinates (even before we realized we were doing it).

Now, when we examine the Doppler effect (for example), we find an equation that relates speed (in the inertial sense) to the shift in frequency, and we can test this equation to see how these frequency shifts of characteristic phenomena of emitting objects fits into our miraculous coherent and uniform physics of mechanics. Your position seems to be that the meaning of this equation has nothing to do with the ordinary inertial meaning of speed. You contend that we should simply interpret the Doppler equation as the definition of speed. My contention is that you are thereby discarding the entire physical meaning and significance of that equation. The content of that equation - and all the others representing Lorentz invariance - is precisely to show how the phenomena of emitting entities relates to the inertial measures of space and time.

 

[PeterDonis]

The correct turn-around statement is this: If we define inertial coordinate systems as "systems of space and time coordinates in terms of which the mechanical inertia of every object is homogeneous and isotropic", then the homogeneity and isotropy of mechanical inertia is true by definition, i.e., not falsifiable.

You are basically saying that "mechanical inertia" is coordinate-dependent. I'm not sure that's correct, but let's assume it is for the sake of argument. Then my response is that mechanical inertia, as you've defined it, is not a physical property! It can't be, because physical properties must be expressible in terms of invariants, and mechanical inertia, by your definition, is not; it depends on the coordinates you choose.

To put it another way: you're basically saying that inertial mass is not a scalar; its value will be different from its "inertial value" if I measure it relative to non-inertial coordinates. Again, I'm not sure that's correct, but let's assume it is for the sake of argument. Then inertial mass, as you've defined it, is not the real physical observable; the real physical observable would be a scalar describing the contraction of some geometric object describing "inertial mass" with one of the vectors in the 4-tuple describing the reference frame. Changing to non-inertial coordinates would change the 4-tuple, and therefore would change the contraction (assuming the "inertial mass" geometric object remained fixed).

 

[Samshorn]

You are basically saying that "mechanical inertia" is coordinate-dependent. I'm not sure that's correct...

One of the first and most important things one learns about special relativity is that energy (all forms of energy) has inertia - and this includes kinetic energy, which of course is frame dependent. Hence inertia is unavoidably frame dependent. This is the very cornerstone of special relativity.

...but let's assume it is for the sake of argument. Then my response is that mechanical inertia, as you've defined it, is not a physical property! It can't be, because physical properties must be expressible in terms of invariants, and mechanical inertia, by your definition, is not; it depends on the coordinates you choose.

So, you assert that kinetic energy is not a "physical property", because it is frame-dependent. I would say it differently: Kinetic energy is a frame-dependent physical property.

You're basically saying that inertial mass is not a scalar...

No, I'm saying all forms of energy - including kinetic energy - have inertia. (I wish I could take credit for this insight, but it's actually the well-known cornerstone of special relativity.) One consequence of this is that if we construct two inertial coordinate systems using mechanical inertia (i.e., such that mechanical inertia is homogeneous and isotropic), we get two different simultaneities. Of course, they are the very same simultaneities we get for light speed to be homogeneous and isotropic in the two frames. This works only because energy has inertia.

It might be helpful for you to think about how you would actually define an inertial coordinate system mechanically. This would make it clear how the inertia of (kinetic) energy comes into play. Hopefully when you've satisfied yourself that energy does indeed have inertia, you can make it past the first paragraph of my previous post. I think the next two paragraphs of that post explained things fairly well.

 

[PeterDonis]

One of the first and most important things one learns about special relativity is that energy (all forms of energy) has inertia - and this includes kinetic energy, which of course is frame dependent. Hence inertia is unavoidably frame dependent.

This is a good point, but note that it is *not* a cornerstone of Newtonian physics. In other words, "energy has inertia" is an empirical fact that helps to distinguish Newtonian/Galilean relativity from Lorentzian/Einsteinian relativity.

So, you assert that kinetic energy is not a "physical property", because it is frame-dependent. I would say it differently: Kinetic energy is a frame-dependent physical property.

Yes, I should have been more precise. I think I said in a previous post that total energy is the contraction of the object's 4-momentum with the observer's 4-velocity; kinetic energy is total energy minus the invariant length of the object's 4-momentum. Total energy and kinetic energy are therefore observer-dependent, yes; but we can write down an invariant expression for "energy of object X as measured by observer O", so in that sense "energy" is a physical property, yes.

However, I don't think any of this requires that the mechanical definition of relative speed be privileged over the Doppler definition.

 

[Samshorn]

I don't think any of this requires that the mechanical definition of relative speed be privileged over the Doppler definition.

You keep talking about one definition being privileged over another, but it isn't a question of privilege, it's a question of what the Doppler formula means. It means that if you plug in the ordinary speed v (based on the inertial definition), you get the relativistic Doppler shift. That's the non-trivial physical fact that the formula is expressing. If you refuse to recognize that the parameter v in that formula represents the inertial speed, then the formula doesn't mean anything, i.e., it doesn't connect something to something else. It's one hand clapping.

Your position seems to be that the meaning of v in the Doppler equation need not have anything to do with the ordinary inertial meaning of speed. You contend that we can just as well interpret the Doppler equation itself as the definition of the parameter v appearing in it. My contention is that you are thereby discarding the entire physical meaning and significance of that equation. The function of that equation is precisely to show how the frequencies of emitting entities relate to the inertial measures of space and time.

I hesitate to mention it, because it may just divert attention from the main point, but it is actually somewhat relevant to the main point: In more than one dimension there isn't even a one-to-one correspondence between speed and Doppler shift. For example, if you follow an equi-angular spiral path away from a source of light, with the right combination of angle and speed, you will have no Doppler shift at all, even though you obviously have speed (both radial and tangential) relative to the source. Of course, when we say this, we are talking about speeds defined in the inertial sense. We can't encode the full range of possible (3-dimensional) motions in terms of the Doppler shift from some emitter. (If you posit three or four stategically placed emitters, you are just constructing a coordinate system.)

Note that we called this motion "spiral", because we are changing the angular orientation of the line between the source and the object... but why can't we just define that line to be stationary, so that the object is simply moving away radially? Well, because the inertial sense of motion entails an absolute sense of rotation. So, again, the very statement of the problem unavoidably involves the use of inertial definition of motion. The same applies to the MTW exercise you cited, in which they smuggled in the "ordinary velocity v" without even blushing. This all just shows why we define motions in the sense of inertial coordinates, and we then determine formulas giving the Doppler shift for any specified motions. It wouldn't make sense to try to do the reverse, e.g., to try to infer the spiral motion from the (absence of) Doppler shift.

 

[PeterDonis]

it's a question of what the Doppler formula means. It means that if you plug in the ordinary speed v (based on the inertial definition), you get the relativistic Doppler shift. That's the non-trivial physical fact that the formula is expressing.

Yes, but that physical fact by itself doesn't tell you whether you should interpret it as telling you that, oh, look, the observed Doppler shift just happens to be exactly equal to what the formula tells you when you plug in speed v; or telling you that, oh, look, the speed v just happens to be exactly what you would expect when you plug the Doppler shift into the (inverted) formula.

Your position seems to be that the meaning of v in the Doppler equation need not have anything to do with the ordinary inertial meaning of speed.

No, my position is that the physical fact the formula is expressing can be interpreted in either direction, so to speak. It's telling you about a connection between two different sets of phenomena. It's not telling you which set of phenomena is "more fundamental". That's a matter of interpretation.

 

[Samshorn]

Yes, but that physical fact by itself doesn't tell you whether you should interpret it as telling you that, oh, look, the observed Doppler shift just happens to be exactly equal to what the formula tells you when you plug in speed v; or telling you that, oh, look, the speed v just happens to be exactly what you would expect when you plug the Doppler shift into the (inverted) formula.

In both of the "alternatives" you mentioned, the equation is empirically valid only if v equals the inertial speed, which is the point I've been making: The formula refers to inertial speed or else it's meaningless.

No, my position is that the physical fact the formula is expressing can be interpreted in either direction, so to speak. It's telling you about a connection between two different sets of phenomena. It's not telling you which set of phenomena is "more fundamental". That's a matter of interpretation.

I don't think it's a question of whether one thing is "more fundamental" than another (whatever that might mean). The point is just that both are physically meaningful, and the task of the theory is to describe how they are related. That's why we shouldn't say special relativity has nothing to do with inertial coordinate systems or inertial measures of space and time (or with the Lorentz transformations that relate them to each other). It has everything to do with these things.

As to why we ordinarily conceive of phenomena as existing in space and time, rather than in some abstract "Doppler realm", I think that's a complicated subject. There are always naive positivists urging us to shed our hide-bound notions of a common space and time, and just focus on the raw uninterpreted sense impressions impinging on our own individual world line. The natural tendency of this line of thought is toward solipsism, but even if you succeed in avoiding that trap, most people find that there are good reasons for retaining the conceptual framework of space and time. (I mentioned some in a previous message.) On the other hand, quantum phenomena for n particles seem to reside in 3n-dimensional phase space, suggesting that our concept of 3+1 dimensional spacetime is not fundamental. Nevertheless, we somehow still make the conversion back to space-time representations for most purposes.

 

[bobc2]

...There are always naive positivists urging us to shed our hide-bound notions of a common space and time, and just focus on the raw uninterpreted sense impressions impinging on our own individual world line. The natural tendency of this line of thought is toward solipsism, but even if you succeed in avoiding that trap, most people find that there are good reasons for retaining the conceptual framework of space and time. (I mentioned some in a previous message.)...

Well said.

 

[PeterDonis]

In both of the "alternatives" you mentioned, the equation is empirically valid only if v equals the inertial speed

Yes, but "inertial speed" means "the actual measured speed according to a given procedure". You can describe that procedure and its results without defining inertial coordinates.

That's why we shouldn't say special relativity has nothing to do with inertial coordinate systems or inertial measures of space and time (or with the Lorentz transformations that relate them to each other). It has everything to do with these things.

Once again, you are conflating inertial coordinates with the physical properties and measurements that make inertial coordinates useful. SR does have everything to do with those physical properties and measurements, yes. But you can describe them without defining inertial coordinates.

As to why we ordinarily conceive of phenomena as existing in space and time, rather than in some abstract "Doppler realm", I think that's a complicated subject.

Yes, it is, because it's not just about physics; it's about our cognitive systems, which are much more complicated than the simple physical systems we're talking about here.

There are always naive positivists urging us to shed our hide-bound notions of a common space and time, and just focus on the raw uninterpreted sense impressions impinging on our own individual world line.

I have not said anything like this. Once again, you're conflating coordinates with physical properties that are conveniently described using coordinates. Spacetime, as a geometric object, is certainly the natural outcome of reconciling our common notions of space and time with the other physical facts we have discussed. But spacetime can be described without coordinates. That's all I'm saying. That's not the same as saying spacetime doesn't exist, only our sense impressions do.

 

[Samshorn]

Yes, but "inertial speed" means "the actual measured speed according to a given procedure". You can describe that procedure and its results without defining inertial coordinates.

If someone defines a speed (for example) using a procedure operationally equivalent to defining an inertial coordinate system and dividing the space difference by the time difference, then I'd count that as using an inertial coordinate system, which is to say, using inertial measures of space, time, speed, angles, etc. For example, the MTW exercise you cited as a coordinate-free "calculation" used an inertial coordinate system by referring to the "ordinary speed v".

Once again, you are conflating inertial coordinates with the physical properties and measurements that make inertial coordinates useful.

No, I'm conflating the use of inertial coordinates with the use of measures of distance, time, speed, angles, etc., based on the concept of inertial coordinates. If, for example, someone says an object is moving at speed v, and they mean this as the ordinary speed in terms of an inertial coordinate system, then I would say they have invoked an inertial coordinate system. (The v in the Doppler formula is just such a speed, and hence it refers to inertial coordinates.)

SR does have everything to do with those physical properties and measurements, yes. But you can describe them without defining inertial coordinates.

If someone asks you what special relativity predicts for the Doppler shift of a light source moving away from us at the speed v, where v is an "ordinary speed" defined in terms of inertial coordinates, how would your application of coordinate-free reasoning erase the fact that what you're doing is explicitly answering a question whose parameters are defined in terms of inertial coordinates?

Spacetime can be described without coordinates. That's all I'm saying.

Ohanian and Ruffini had it right: "We must not forget that the physicist who wishes to measure a tensor has no choice but to set up a coordinate system, and then measure the numerical values of the components. Thus, to carry out the comparison of theory and experiment, the physicist cannot ultimately avoid the language of components; only a pure mathematician can adhere exclusively to the abstract coordinate-free language of differential forms..." You see, this is the point: The epistemological foundations of a physical theory rest entirely on "the comparison of theory and experiment", and this decisive link is provided by coordinates. Without this link, any formal mathematical expressions are devoid of physical content.

 

[Dale]

Sure, just write down all the equations the same way you can in GR, using only coordinate-free tensor expressions, and use the metric of Minkowski spacetime.

No, if you use the metric of Minkowski spacetime then you are already assuming SR and not making a test theory for SR.

Or do you mean by "test theory" something like the Cartan geometric formulation of Newtonian gravity? In other words, casting Galilean spacetime in terms of coordinate-free tensor equations?

In order to test a theory you cannot assume it, so the best approach is to assume a test theory. A test theory is a general theory which has a set of one or more unknown parameters. Various competing theories (such as SR or Newtonian physics) are then reproduced through specific choices of the unknown parameters. You then propose an experiment to put constraints on the unknown parameters and see how closely they match the parameters corresponding to the various theories.

So, in this case, a coordinate-free test theory would be one which reproduces either Minkowski spacetime or Galilean spacetime with some set of tensors and scalars.

I don't know that that has been done specifically, but I don't see why it couldn't be.

I also don't see why it couldn't be, but as far as I know it has not been done, so I would be reluctant to make claims about it.

 

[Dale]

Ohanian and Ruffini had it right: "We must not forget that the physicist who wishes to measure a tensor has no choice but to set up a coordinate system, and then measure the numerical values of the components. Thus, to carry out the comparison of theory and experiment, the physicist cannot ultimately avoid the language of components; only a pure mathematician can adhere exclusively to the abstract coordinate-free language of differential forms..."

Ohanian and Ruffini are wrong on this point ("no choice but to set up a coordinate system"). You can set up a set of basis vectors without setting up a coordinate system. Components are then contractions with one of the basis vectors, which is still coordinate-free. So just because you are dealing with components doesn't mean that you are dealing with coordinates. Their conclusion ("no choice ...") doesn't follow from their argument ("cannot ultimately avoid ... components").

Coordinates always imply a unique vector field called the coordinate basis, but a basis does not imply a unique coordinate system.

 

[Phy_Man]

Ohanian and Ruffini are wrong on this point ("no choice but to set up a coordinate system"). You can set up a set of basis vectors without setting up a coordinate system. Components are then contractions with one of the basis vectors, which is still coordinate-free. So just because you are dealing with components doesn't mean that you are dealing with coordinates. Their conclusion ("no choice ...") doesn't follow from their argument ("cannot ultimately avoid ... components").

Coordinates always imply a unique vector field called the coordinate basis, but a basis does not imply a unique coordinate system.

Ummmm, nope! I'm afraid that you missed a very fundamental fact here. Ohanian and Ruffini are absolutely correct. They're experts in their field and know precisely what they're talking about. Setting up a system of basis vectors is identically the same thing as setting up a coordinate system. Contracting components on a basis is just another name for measuring components. Dealing with components is identical to dealing with a coordinate system. From your response it appear as if you might have a flawed notion of what a coordinate system is.

 

[Dale]

Setting up a system of basis vectors is identically the same thing as setting up a coordinate system.

See the attached image for a system of basis vectors. What are the coordinates of the blue point?

 

[Phy_Man]

See the attached image for a system of basis vectors. What are the coordinates of the blue point?

I'm sorry my good man but I don't see a blue point. I assume that you understand that when I said that setting up a coordinate system is identical to settiing up a basis that it can't be taken to mean that a set of basis vectors is a coordinate system, right?

Regarding your question, I assume you meant to place a point in that diagram somewhere to represent the position vector which is a displacement vector from a point chosen as the origin? While you're at it please make it clear as to whether you're asking me for the components of the displacement vector (which is what a position vector is) from the origin that the point represents or whether you have something else in mind. Thank you.

When someone sets up a basis then wants to express the position vector in terms of that basis they have to locate the two points which define the displacement vector; one point represents the point represented by the origin and the point of interest. When you tell me that there is a point in there and I should tell you its coordinates then what you're telling me is that you have chosen a point to serve as the origin. If not then you haven't mentioned a displacement vector and I have no response as a result.

If you missed that point then don't feel bad. It's no big deal. A lot of people seem to forget what it means to represent a point or speak of its coordinates. I.e. if you wish to speak of coordinates then what you've implied by your question is that you have chosen a point to serve as the origin and then the point is the vector displacement from the reference point (i.e. origin) to the point in question.

Or perhaps you thought you had me? Nope. Sorry. :D

 

[WannabeNewton]

A tensor is a purely algebraic object. Let $V$ be a real finite dimensional vector space and let $T:V^{*}\times...\times V^{*}\times V\times...\times V\rightarrow \mathbb{R}$ be a tensor over $V$ (there are $k$ products of $V$ and $l$ products of $V^{*}$). If $\{e_{i}\}$ is a basis for $V$ and $\{e^{i*}\}$ is the dual basis then any tensor over $V$ can be written in terms of the simple tensors formed out of this basis i.e. $T = \sum T^{i_1...i_k}{}{}_{v_1...v_l}e_{i_1}\otimes...\otimes e^{v_l*}$. The $T^{i_1...i_k}{}{}_{v_1...v_l}$ are the components of $T$ with respect to the above basis.

Now let $M$ be a smooth manifold and let $T$ be a smooth tensor field (a section of a tensor bundle) on $M$. We can define a smooth basis field for $M$ (a section of the tangent bundle that assigns a basis to each $T_p M$) with an associated smooth dual basis field. Then the components of the tensor field with respect to this basis field are defined point-wise as above for each $T_p M$. If this is an orthonormal basis field then it is often called a frame field (in 4 dimensions a vierbein).

If we now have a coordinate chart $(U,\varphi)$ on $M$ then we can choose to use the coordinate basis fields $\{\frac{\partial }{\partial x^{1}},...,\frac{\partial }{\partial x^{n}}\}$ so that for each $T_p M$ with $p \in U$, the components of $T(p)$ are with respect to $(\frac{\partial }{\partial x^{i}}|_p)$ and the associated covector field $(dx^{i}|_p)$. This is a special case of a basis field (and in general not an orthonormal one) that is associated with a coordinate chart.

One can however always associate a given orthonormal basis for $T_p M$ with some coordinate basis field evaluated at $p$ (and in general this association will only be valid at p). This is how Riemann normal coordinates are constructed.

 

[PeterDonis]

So, in this case, a coordinate-free test theory would be one which reproduces either Minkowski spacetime or Galilean spacetime with some set of tensors and scalars.

Ah, ok, so it would be something like the PPN formalism, and assuming Minkowski spacetime would be like assuming all the PPN parameters have their GR values, instead of figuring out how to test the parameters experimentally. Fair point.

As far as that goes, there is one variable parameter, so to speak, that has been tested: the speed of light itself, or rather the fact that it is independent of the speed of the source. A PPN-like theory of spacetime would have the invariant speed as a variable parameter; its Galilean value would be infinity, and its Minkowski value would be something finite. However, standard SR gives no way of predicting exactly *which* finite speed should be invariant, so this still isn't quite the same as the PPN tests of GR.

 

[Phy_Man]

A tensor is a purely algebraic object.

That is incorrect. Tensors are geometric objects. MTW explain this on page 49. Schutz explains it on page 36. They are referred to as geometrical objects because they can be defined without referring to a specific coordinate system. A point in spacetime (i.e. an event) is also a geometric object. Vectors and 1-forms (which are examples of tensors) are also geometric objects.

 

[WannabeNewton]

A tensor is algebraic as it only requires a vector space to define. It lies within algebraic categories. The tensors obtained by evaluating a section of a tensor bundle over a smooth manifold fiber by fiber of the tangent bundle are simply special cases as they are with respect to the tangent space, which is just a specific type of vector space, defined in the category of smooth manifolds.

MTW and others' use of the word "geometric object", while warranted, is not a mathematical classification of tensors in the language of categories.

As a side note, MTW is about as accurate with mathematics as Fox News is with actual news ;) Schutz is awesome though, got to love that guy.

 

[Jorriss]

That is incorrect.

He's not incorrect. Tensors are elements of products of vector spaces, that's completely algebraic. They are multilinear maps.

 

[Bill_K]

If you missed that point then don't feel bad. It's no big deal. A lot of people seem to forget what it means to represent a point or speak of its coordinates. I.e. if you wish to speak of coordinates then what you've implied by your question is that you have chosen a point to serve as the origin and then the point is the vector displacement from the reference point (i.e. origin) to the point in question.

Certainly one of us has forgotten. A coordinate system does not have to have an origin, let alone a point origin. What point is the origin of Schwarzschild coordinates?

 

[Phy_Man]

A tensor is algebraic as it only requires a vector space to define.

Recall that you said that a tensor is only a algebraic object. That is not true. Tensors are clearly geometric objects by definition. I referred to those texts because I assumed that you have them. A more precise definition is found in Differential Geometry by Erwin Kreszig, Dover Pub., page 92. Mind you, this is a definition given in this text and as such it's not wrong. There's really no room for debate on this. I know this as a fact. I suspect you'll have to search to find a text with the definition. I admire that by the way. Nothing wrong with challenging what you have a feeling may be wrong. But that's why I posted the reference to the source of the definition.

Please go to those websites I showed you and download the book and read the definition. I'd post it but I have this kink in my neck that is very painful right now.

Do you really believe that MTW would say those things were geometric objects if they really weren't? It seems as though a lot of people here have zero faith in the textbooks that people are learning this subject from. Please think long and hard before you assume that the experts can't get a simple definition wrong. I've yet to see that happen in my lifetime.

Actually I have to question the wisdom of arguing over definitions. I mean really! Where does that get us? Are these things being questioned because someone doesn't know how they're defined? Is someone saying that how things are defined today shouldn't be defined that way? I'm curious, that's all. Although I don't believe it's worth the effort posting on these subects in cases like this. If you folks are happy with what you believe than please let me know. I'm not trying to be a wisenheimer or anything like that. It's just that I live with chronic pain and all thus typing is getting to be very painful. I can't see the usefulness of posting defintions if the only response to them is denial of the definition. So my question is; what do you think is useful in cases like this? I mean after all, you sure don't need someone to look things up for you. Thank you in advance. I appreciate it. My neck needs a rest like you wouldn't believe.

 

[WannabeNewton]

What is your mathematical definition of geometric in the language of category theory? If you are using physics/math books which don't mathematically define the terms they are using as references then that is of no real use. Again, a tensor is purely algebraic as it only requires a vector space structure to define. Nowhere in its definition is there a requirement of a topological structure, a smooth structure, a Riemannian structure, a symplectic structure, or a metric structure. By your argument a group is geometric just because there are specific examples of groups which pertain to Platonic solids or to smooth manifolds but a group is an over-arching algebraic object.

 

[Phy_Man]

He's not incorrect. Tensors are elements of products of vector spaces, that's completely algebraic. They are multilinear maps.

Oy! I'd like to ask you folks to please read my posts more carefully in the future. It's too painful to have to repeat myself.

Jorriss - I did not say that they weren't algebraic. I explicity stated that they are more than that. They are, by defiition, geometric objects. Multilinear maps are geometric objects. I'd appreciate it if you'd look up the definition of geometric object before you claim that the defintion is wrong or that what I said is the definition is wrong. I'm in too much pain to have to repeat myself anymore in this thread. If there are any questions for me on this subject then please send them to me in PM. Thank you.

I don't mean to come off like I'm irritated. I'm just in a ton of pain from all this typing.

 

[Phy_Man]

What is your mathematical definition of geometric in the language of category theory? If you are using physics/math books which don't mathematically define the terms they are using as references then that is of no real use. Again, a tensor is purely algebraic as it only requires a vector space structure to define. Nowhere in its definition is there a requirement of a topological structure, a smooth structure, a Riemannian structure, a symplectic structure, or a metric structure. By your argument a group is geometric just because there are specific examples of groups which pertain to Platonic solids or to smooth manifolds but a group is an over-arching algebraic object.

Sorry but you have all the information you need to answer these questions yourself. I can no longer stand the pain and I'm not into repeating myself for no apparent reason. If you reject the definition then that's not my problem.

 

[Jorriss]

Jorriss - I did not say that they weren't algebraic. I explicity stated that they are more than that. They are, by defiition, geometric objects. Multilinear maps are geometric objects. I'd appreciate it if you'd look up the definition of geometric object before you claim that the defintion is wrong or that what I said is the definition is wrong. I'm in too much pain to have to repeat myself anymore in this thread. If there are any questions for me on this subject then please send them to me in PM. Thank you.

Fair enough, but the modern definition of a tensor does not require structures that are generally associated with a geometric theory such as an inner product structure or metric structure.

 

[WannabeNewton]

Sorry but you have all the information you need to answer these questions yourself. I can no longer stand the pain and I'm not into repeating myself for no apparent reason. If you reject the definition then that's not my problem.

Your reference was an age old dover book on differential geometry. Tensors are not defined like that anymore. If you want to talk mathematics you must use modern definitions. I didn't reject the definition, the mathematical community did.

 

[WannabeNewton]

By the way, for anyone interested, the most general modern definition of a tensor can be found in e.g. chapter 14 of Roman's "Advanced Linear Algebra" or chapter 13 of Lang's "Linear Algebra".

 

[Dale]

I'm sorry my good man but I don't see a blue point.

Look harder, it is there. You need to click on the thumbnail to make it full-size.

I assume that you understand that when I said that setting up a coordinate system is identical to settiing up a basis that it can't be taken to mean that a set of basis vectors is a coordinate system, right?

If they are "identically the same thing" (your words) then given a specified set of basis vectors you should be able to determine the coordinates.

Regarding your question, I assume you meant to place a point in that diagram somewhere to represent the position vector which is a displacement vector from a point chosen as the origin?

Nope, I didn't identify any point as the origin. That is one specific problem with your claim that makes it wrong. Setting up a system of basis vectors in no way identifies an origin.

 

[Phy_Man]

Fair enough, but the modern definition of a tensor does not require structures that are generally associated with a geometric theory such as an inner product structure or metric structure.

Since I don't see a question mark I won't respond to this. I have no idea what that's supposed to mean anyway (not that I want to). The modern definition of a tensor is a multilinear map from vectors and 1-forms to reals. The definition of geometrical object is consistent with that fact. As I said, I won't discuss definitions anymore. I'm too old and too tired for such useless conversations. If people don't like the way things are defined I'd wish they'd simply say so and stop wasting time. Thanks.

 

[TrickyDicky]

Ohanian and Ruffini are wrong on this point ("no choice but to set up a coordinate system"). You can set up a set of basis vectors without setting up a coordinate system.

If I may, the set of basis you were thinking of happened to be ordered? There might be some confusion with your statements if you don't specify this. Ohanian and Ruffini are clearly referring to ordered bases.

Components are then contractions with one of the basis vectors, which is still coordinate-free. So just because you are dealing with components doesn't mean that you are dealing with coordinates. Their conclusion ("no choice ...") doesn't follow from their argument ("cannot ultimately avoid ... components"). Coordinates always imply a unique vector field called the coordinate basis, but a basis does not imply a unique coordinate system.

See above. Also, Ohanian and Ruffini are not talking about a "unique" coordinate system in the quoted paragraph but about coordinate systems in general.

 

[WannabeNewton]

DaleSpam's point was that you don't need a coordinate system to talk about the components of tensors. That only applies to the components of a tensor as represented in a coordinate basis. The problem is that there are GR books which use the word "component" as if it unequivocally meant "coordinate basis components". Wald does this too in fact, in the beginning of ch4.

 

[Samshorn]

You can set up a set of basis vectors without setting up a coordinate system... So just because you are dealing with components doesn't mean that you are dealing with coordinates... a basis does not imply a unique coordinate system.

No one is saying it does. Remember, this discussion began with the claim that by using the "coordinate-free" approach we can dispense with "reference frames and coordinate systems and Lorentz transformations". Since a reference frame is an equivalence class of coordinate systems that all share the same measures of spatial distances, temporal intervals, speeds, angles, etc., (and also to sidestep the ambiguous aspects of the word "frame", and also since it is the natural contra-distinction to "coordinate-free") we've been sometimes referring to reference frames informally as coordinate systems - but not with the intent of suggesting uniqueness, which would be absurd. The coordinate system obviously only needs to be specified up to the point of determining all measures of distances, times, speeds, angles, etc. Which basically means we need to specify the frame - or a basis if you prefer.

Note that those saying we can dispense with frames are also saying that components have no physical meanings - not just that coordinates have no physical meanings. This is a fairly standard notion of what the coordinate-free approach entails. Any time you resort to indices on your tensors, and actually quantify the components of a tensor, you are diverging from the coordinate-free precepts, by their own admission, because even they recognize that choosing a basis is tantamount to establishing a (equivalence class of) coordinate system. For example, D'Inverno has this to say

"There are two distinct approaches to the teaching of tensors: the abstract or index-free (coordinate-free) approach and the conventional approach based on indices... The disadvantage [of the coordinate-free approach] is that when one wants to do a real calculation with tensors, as one frequently needs to, then recourse has to be made to indices."

You see, the very use of indices (or quantifying components) is understood to be a transgression against the coordinate-free (index-free) approach. And, again, in order to do actual calculations, this is what we must do. And from the foundational standpoint (which is what this thread is about), since the comparison with observation unavoidably involves this kind of actual calculation, we can't dispense with (equivalence classes of) coordinate systems, or, if you prefer, frames, or basis, or however you prefer to think about it.

 

[robphy]

For example, D'Inverno has this to say

"There are two distinct approaches to the teaching of tensors: the abstract or index-free (coordinate-free) approach and the conventional approach based on indices... The disadvantage [of the coordinate-free approach] is that when one wants to do a real calculation with tensors, as one frequently needs to, then recourse has to be made to indices."

You see, the very use of indices (or quantifying components) is understood to be a transgression against the coordinate-free (index-free) approach.

"Coordinate-free" and "index-free" are not synonyms.
(Penrose's) Abstract Index Notation is coordinate-free, but uses indices to label slots of a tensor.
http://en.wikipedia.org/wiki/Abstract_index_notation
(abstract-index does NOT mean, e.g., "\mu stands for \{ t,x,y,z \}".)

Classical tensor calculus notation, however, uses indices which refer to particular choices of coordinates or bases. Here, it may be that "\mu stands for \{ t,x,y,z \}".

The abstract index notation tries to bridge
the component-based methods of classical tensor calculus
with the coordinate-free methods of modern mathematical treatments.
http://books.google.com/books?id=hQdh3SVgZ8MC&pg=PA56&dq="abstract+index"+penrose

 

[TrickyDicky]

DaleSpam's point was that you don't need a coordinate system to talk about the components of tensors.

I doubt that was Dalespam's point. Such triviality has nothing to do with the Ohanian and Rufinni quote he said was incorrect. You seem to be stuck with the algebraic aspect of tensors and maybe didn't notice that O&R mention: "the physicist who wishes to measure a tensor" suggesting they are referring to the geometric aspect of tensors of a geometric space (such as the Minkowski manifold if we talk about SR).

 

[WannabeNewton]

That's the whole point of a frame field; the integral curves of the frame field represent a family of ideal observers who each carry a measuring apparatus with which they make a complete set of measurements of the components of tensor fields on space-time. These measurements will be independent of whatever coordinate chart you use to represent the frame field.

 

[TrickyDicky]

That's the whole point of a frame field; the integral curves of the frame field represent a family of ideal observers who each carry a measuring apparatus with which they make a complete set of measurements of the components of tensor fields on space-time. These measurements will be independent of whatever coordinate chart you use to represent the frame field.

Mate, I suggest you read again Samshorn's posts in this thread and then consider the apparently innocent phrase " a family of ideal observers who each carry a measuring apparatus". You might find it iluminating.

 

[TrickyDicky]

WN I see you edited your initial post and I'm not disagreeing with your new last line at all.

 

[WannabeNewton]

I'm not disagreeing with you either. I'm just trying to describe the physical/geometric nature of frame fields. If I have a frame field $\{(e_{a})^{\mu}\}$, the integral curves of $(e_0)^{\mu}$ (which is chosen to be time-like by definition of a frame field) will be a congruence of time-like curves, at the least on some proper open subset of space-time. At each event, we think of the orthonormal basis $\{(e_a)^{\mu}|_p\}$ for $T_p M$ as representing three perpendicular meter sticks and a clock, carried by the observer represented by the unique time-like curve in the congruence intersecting that event. I refer again to section 13.1 of Wald.

Notice how I said before that "these measurements will be independent of whatever coordinate chart you use to represent the frame field" but you still need some coordinate chart to actually write down the frame field and make down to Earth computations. You can just be rest assured that it wouldn't have mattered which coordinate chart you chose.

EDIT: By the way, I'm not berating Ohanian's book or anything. I have the book myself (assuming we are talking about "Gravitation and Spacetime") and I absolutely treasure it, especially its devoted focus to experiment instead of theory.

 

[TrickyDicky]

I'm not disagreeing with you either.
Notice how I said before that "these measurements will be independent of whatever coordinate chart you use to represent the frame field" but you still need some coordinate chart to actually write down the frame field and make down to Earth computations. You can just be rest assured that it wouldn't have mattered which coordinate chart you chose.

Totally agreed.
But then I'm not sure what the disagreement regarding the O&R quote is about (or the PeterDonis-Samshorn exchange for that matter). Are you guys making some nuanced distinction between coordinate charts and coordinate systems? I thought they were the same thing.

 

[WannabeNewton]

I wasn't going against or with the quote in any way. I was just trying my best to possibly clear up the various ways in which tensors are used/described in GR texts because there is a noticeable lack of consistency amongst them (which is not to say it's anyone's fault of course).

 

[TrickyDicky]

I was just trying my best to possibly clear up the various ways in which tensors are used/described in GR.

I know, and I always find it informative. Even if the discussion in this case seemed more centered in SR.

 

[WannabeNewton]

I think even in SR, the fact that the states of the physical theory on Minkowski space-time can be taken to be tensor fields and that its isometry group is the proper Poincare group allows for an instructive use of the same type of formalism i.e. frame fields and the link between the measurements of components of tensor fields made using one frame field and another through special covariance under the proper Poincare transformations.

 

[Dale]

If I may, the set of basis you were thinking of happened to be ordered? There might be some confusion with your statements if you don't specify this. Ohanian and Ruffini are clearly referring to ordered bases.

Yes, I was assuming a standard tetrad, so smooth, orthonormal, and ordered. It still doesn't give you a coordinate system. See the picture I drew earlier. The basis vector field simply doesn't imply a coordinate system.

 

[TrickyDicky]

Yes, I was assuming a standard tetrad, so smooth, orthonormal, and ordered. It still doesn't give you a coordinate system. See the picture I drew earlier. The basis vector field simply doesn't imply a coordinate system.

I think we all agree your abstract example doesn't give you a unique coordinate system. But this misses the point of the O&R quote.