Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Arthur Geddes on 25/04/2016 19:42:03
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Could someone provide a practical description of tensors of first & second order, etc.? Trying to get this relativity thing (A.R., Absolute Relativity is looking good) & i'm hitting a roadblock with my limited understanding of differential linear algebra.
Thanks in advance...
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Could someone provide a practical description of tensors of first & second order, etc.? Trying to get this relativity thing (A.R., Absolute Relativity is looking good) & i'm hitting a roadblock with my limited understanding of differential linear algebra.
Thanks in advance...
A tensor is typically defined in two equivalent ways. One way is what I refer to as the geometric definition. The second way is what I refer to as the analytic definition. I created two webpages to explain these in detail for each viewpoint.
Geometric definition: A tensor is defined as a multi-linear map from vectors and 1-forms to the set of real numbers. See:
www.newenglandphysics.org/physics_world/math_phy/05_tensors_via_geometric/tensor_via_geometric.htm
Analytic definition: An equivalent way of defining a tensor, one commonly used in relativity, is a set of numbers which transform from one system of coordinates to another by a certain transformation rule. See:
http://www.newenglandphysics.org/physics_world/math_phy/04_tensors_via_analytic/tensors_via_analytic.htm
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A point worth keeping in mind is that a distinction is to be made between a mathematical value and a function whose domain is a set of such mathematical values and whose range is another set of such (or perhaps a different type of) mathematical values, the function expressing the mapping between one and the other. We may thus speak of the velocity of an object as a vector, and we may loosely speak of the windspeed as a vector, but to be semantically precise, the latter should be spoken of as a vector function of position because it does not necessarily have a single value, but a whole set of values varying from place to place. In some equations, this distinction is assumed and may not be immediately obvious to the uninitiated. The letter V could be used for either, but the significance is not the same. To be more precise, the expression for the windspeed case would be V(x, y, z) , or you could equivalently write V(r) with r being position expressed vectorially with respect to some reference point.
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Another way to look a tensors is as follows: Consider an object having a specific location and simple numerical value. That is a scalar. Now imagine that a second like object is placed in its near vicinity but with the opposite value. The difference between them multilied by the displacement is a dipole moment, and has not only a magnitude but also a direction, and in 3 dimensional space would be expressed by 3 components. Assuming linear mathematics, this would be a vector. Vectors can of course be combined by the usual procedures to give other vectors, that likewise each have (in this case) 3 components and are equivalent to a magnitude times a displacement. Now imagine you have a vector and then displaced from it another like vector except of opposite sign. This configuration is now associated with the underlying original value, the magnitude and direction of the first displacement, and the magnitude and direction of the second displacement. This object is vector-like but is more complex than a vector, and sums of such objects cannot necessarily be expressed as a simple replica of the exact style of object that one such object would be based only on the two-displacement model. (Although the similarity of this to a quadrupole moment is obvious, it is not correct to state that the structure is a quadrupole because one needs to distinguish between the case of starting out with a vector in the x direction followed by displacement in the y direction for the opposite vector, as distinguishted between starting out with the vector in the y direction and then displacing for the opposite vector in the x direction. These two procedures are not in general equivalent, and to have a true quadrupole, as in an electromagnetic sense, it is necessary that both processes be present simultaneously, something that is not true of all processes). In general, in 3 dimensional space, an object of this type require 9 components for its expression and admits of additional complexities, specifically, that tensors (of the second rank, which is what we are talking about here) can be expressed as combinations of (using ordinary geometrical language such as one would use in things such as mechanical engineering) symmetric, shear, and antisymmetric forms, or some combination of these. An alternative expression is in terms of dyads, none of which fit these descriptions but all of which are aligned along the coordinate axes of interest (here assumed rectangular) and can be combined to form the aforesaid forms. An example of a symmetric tensor is the surface of a lens, the gradient of whose surface constitutes an (effectively linear) vector field, so that the value of that gradient (which is a vector) as a function of the distance from the center (which is also a vector) is simply a numerical constant times the distance vector. This is a special kind of symmetric tensor, known as a unitary tensor, and is represented by a matrix having 0 as all off-diagonal values and a constant for the diagonal values. The multiplication whereby this tensor would actuate upon the position vector to give the gradient vector is simply to do the matrix multiplication of the diagonal-only matrix by the vector, which, of course, in this case simply multiplies each component of the position vector by the corresponding diagonal element in the matrix, resulting in simply a constant multiplication everywhere. If the lens is astigmatic, having different curvatures on perpendicular axes, the vector field varies from place to place in both magnitude and direction, so that in general the gradient is neither simply proportional to the distance from the center nor is it necessarily parallel to that displacement. It would be expressed by a matrix whose non-diagonal elements are still 0 but whose diagonal elements are unequal. That is of course if the coordinate system used is aligned with the principle axes of the lens. If they are not, then we will end up with something having nonzero off-diagonal components but they will be symmetrically balanced across the diagonal (hence the term symmetric tensor). In general, symmetric tensors can be associated with ellipsoidal or hyperboloidal configurations , whether in 2, 3, or more dimensions, and as such will be characterized by principle axes, and when these axes are aligned with the coordinate system being used, will result in a matrix expression having only diagonal values, the off-diagonals being zero. If the values are all of one sign, then the configuration will be ellipsoidal; otherwise hyperboloidal. (If some of the diagonal elements are zero, you get a configuration with cylindrical characteristics). Then we have the antisymmetric tensor. That is characterized by, when aligned with the coordinate axes, of a matrix expression that has zero for all diagonal positions, and in the off-diagonal positions, values on opposite sides of the diagonal are of opposite sign. An example of this is the velocity distribution of a spinning object. The dyadic tensor, mentioned earlier, is one that is aligned strictly on one pair of coordinate axes, so that it is like a vector along the x direction, and another like but opposite vector displace from it along the y direction, but antiparallel to the original. For some purposes these are the easiest and simplest to work with, and whenever you have a matrix expression for the tensor, each value in the matrix actually corresponds to one such dyad. A physical example of a pure dyadic tensor is the stress present in a beam when one face of it is being pulled in a direction parallel to its length while the other face is being pulled in the opposite direction.
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A point worth keeping in mind is that a distinction is to be made between a mathematical value and a function whose domain is a set of such mathematical values and whose range is another set of such (or perhaps a different type of) mathematical values, the function expressing the mapping between one and the other.
What does this have to do with the definition of tensors? This appears to be off topic to me. I
Suggestion: If you had any point that was on topic then you should have mentioned that vectors are examples of tensors, i.e. a vector is a tensor of rank 1. It'd be a good idea to keep this in mind for the future. Otherwise people might take this as an opportunity to drag the subject way off topic.
We may thus speak of the velocity of an object as a vector, and we may loosely speak of the windspeed as a vector, but to be semantically precise, the latter should be spoken of as a vector function of position because it does not necessarily have a single value, but a whole set of values varying from place to place.
The velocity of a particle is a vector whose value depends on the position of the particle along its trajectory where the position is in turn a function of time. I.e. V = V(r(t)). In this case the velocity is a function of time whereas wind-speed is a vector field which depends on both position and time, i.e. V = V(r, t).
Note: A vector is a tensor of rank 1 and a scalar is a tensor of rank 0. The term scalar as its used in physics and tensor analysis is a real valued function, possibly of position, whose value remains unchanged upon a change in coordinates.
More on vectors since they're examples of tensors: There are two kinds of vectors (1) free vectors and (2) bound vectors. The former refers to vectors with fixed initial and terminal point whereas the later only the magnitude and direction completely determine the vector. For details please see: https://en.wikipedia.org/wiki/Euclidean_vector
An example of a second rank tensor is the Newtonian tidal force tensor. See: http://www.newenglandphysics.org/physics_world/mech/07_tidal_force_tensor/tidal_force_tensor.htm
This tensor determines the forces on an object in a non-uniform gravitational field.
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A point worth keeping in mind is that a distinction is to be made between a mathematical value and a function whose domain is a set of such mathematical values and whose range is another set of such (or perhaps a different type of) mathematical values, the function expressing the mapping between one and the other.
What does this have to do with the definition of tensors?
The point of the thread is to further understanding of gravity as understood by G.R.. Atomic-S' reply is inline with my train of thought; i was gonna mention. Gravity is a 'acceleration field/spacetime shape resolution' issue; i think we're gonna get to "metric" & variance issues soon, if this goes well.
I'm eyeing an interpretation of me on a non-spinning planet in the middle of nowhere; in spherical coordinates & considering also the G field from a massless point in the vicinity.
Anyone wanna jump into that?
I have two questions wrt equivalence: if i'm accelerating at 9.8 m/s2, why am i not going the speed of light? I'm old enough, to be sure. And; if i'm experiencing a force but *not* accelerating; am i traveling in a circle in some sense?
& remember, i'm on a non-spinning (apparently earth-sized) planet in free space.
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An example of a second rank tensor is the Newtonian tidal force tensor. See: http://www.newenglandphysics.org/physics_world/mech/07_tidal_force_tensor/tidal_force_tensor.htm
This tensor determines the forces on an object in a non-uniform gravitational field.
So, the tidal force tensor describes the local spacetime (cuz it's about the 'metric?' or just space?) relative to a massless test point in free fall? .. Kinda by definition cuz relativity just says this is transposable to other reference frames (i.e. a tensor?)
How does this relate to geodesics? & if i have rocket boots holding me stationary in that potential field, what's happening to that acceleration?
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if i have rocket boots holding me stationary in that potential field, what's happening to that acceleration?
Maintaining my potential energy... Hmmm.. potential energy release is entropic relief.... The more intense the field, the quicker the release... & time goes slower... is there a constant in there somewhere?
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To use your rocket boots you are simply countering the instantaneous velocity at your altitude with an input of constant energy. Thus you are creating an artificially engineered 'inertial' frame of reference. However, this is still an accelerating frame of reference.
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To use your rocket boots you are simply countering the instantaneous velocity at your altitude with an input of constant energy. Thus you are creating an artificially engineered 'inertial' frame of reference. However, this is still an accelerating frame of reference.
Yeah; i'm asking how that presents mathematically. If free fall is a straight line on a geodesic(?); stationary in a potential gradient must be a non-straight line in some respect... mustn't it?
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Yeah; i'm asking how that presents mathematically. If free fall is a straight line on a geodesic(?); stationary in a potential gradient must be a non-straight line in some respect... mustn't it?
I believe you are correct about that. The worldline of someone standing "stationary" on the Earth's surface is not straight in spacetime, because spacetime is bent.
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I have two questions wrt equivalence: if i'm accelerating at 9.8 m/s2, why am i not going the speed of light? I'm old enough, to be sure. And; if i'm experiencing a force but *not* accelerating; am i traveling in a circle in some sense?
I believe why you are still where you are rather than going at the speed of light is because while you are accelerating at 9.8 m/s2 away from the Earth's center (note that the force you feel is equivalent to that generated by an outward, not inward, acceleration), spacetime is accelerating (as defined by the geometry of its distortion as it manifests itself from one time to the next) inward and a counterbalancing rate.
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How does this relate to geodesics? & if i have rocket boots holding me stationary in that potential field, what's happening to that acceleration?
It is being eaten up by spacetime bending in the other direction.
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if i'm accelerating at 9.8 m/s2, why am i not going the speed of light? I'm old enough, to be sure.
If we apply some physics from early high school, you would expect your velocity to be:
v = at where:
v: velocity at time t
a: acceleration (9.8m/s2 in this example)
t: elapsed time
According to this formula, to reach the speed of light v=c=3x108 m/s would take t=3x108/9.8 = 3x107 seconds = 347 days, or just under a year.
However (and it's big however!), Einstein showed that you can never accelerate a massive object to reach the speed of light - no matter how hard you accelerate, or for how long.
These concepts normally come up at the end of high school or early university physics.
We don't notice it with our daily lives here on Earth, but when you accelerate objects close to the speed of light, strange things occur: time slows down, and mass increases, so it gets harder and harder to get closer to the speed of light. Oddly, this is not visible to the person being accelerated.
Such peculiarities are very familiar to physicists at the LHC, who try to get particles traveling as close as possible to the speed of light so they can do experiments at very high energies.
See: https://en.wikipedia.org/wiki/Speed_of_light#Upper_limit_on_speeds
So you may feel old, but you are not nearly old enough to reach the speed of light, when accelerating at 9.8m/s2!
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It is being eaten up by spacetime bending in the other direction.
This is what i intuit but; this is the first i've heard it from someone else.
Spacetime *is* the aether &, deviations from flat *is* acceleration field. But, no circle &, what about time? I intuit eating space & spewing time to conserve 4D volume; just cuz.
Woo hoo!
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However (and it's big however!), Einstein showed that you can never accelerate a massive object to reach the speed of light - no matter how hard you accelerate, or for how long.
I don't believe this is a relativistic case.
You're suggesting i'm radiating & less than a year old?
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The point of the thread is to further understanding of gravity as understood by G.R..
Oh! Okay. Thanks for mentioning that. Had you not then I'd never have known since nobody mentioned gravity until I did in the last post. I'm curious. Why didn't you explain that in your opening post? How were we to know that the point of this thread is as you just said? Please understand that I'm not trying to be snide whatsoever. I'm merely trying to understand if I missed something that was obvious to others and not to me.
Atomic-S' reply is inline with my train of thought; i was gonna mention.
I'm confused here my dear sir. Please explain how so? I only ask because I can't see how its related since its almost completely unrelated to tensor analysis. I asked Atomic-S What does this have to do with the definition of tensors? but he chose not to answer my question. Perhaps if you describe the train of thought that you mentioned above then I'll understand. Thanks.
Gravity is a 'acceleration field/spacetime shape resolution' issue; ..
Gravity is the reason why objects, free of all forces other than gravity, fall at a rate which is independent of the mass of the object. That is precisely what Einstein meant when he spoke of gravity. What you referred to, i.e. spacetime shape, are actually tidal gradients in the gravitational field. In the language of general relativity, spacetime curvature.
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Yeah; i'm asking how that presents mathematically. If free fall is a straight line on a geodesic(?); stationary in a potential gradient must be a non-straight line in some respect... mustn't it?
What do you mean by a straight line on a geodesic? A geodesic is defined here: https://en.wikipedia.org/wiki/Geodesic
There are several equal ways to define geodesics. They are
1) The straightest possible line in a curved space.
2) A path of extremal length.
3) The path, which is identical to zero acceleration of a point.
I provide a concise definition of those in my webpage on geodesics. See:
http://www.newenglandphysics.org/physics_world/math_phy/03_geodesics/geodesics.htm
However if you're in a gravitational field then the spatial trajectory is not a straight line in space.
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I'm merely trying to understand if I missed something that was obvious to others and not to me.
From the original post: "Trying to get this relativity thing..."
Atomic-S' reply is inline with my train of thought; i was gonna mention.
I'm confused here my dear sir. Please explain how so? I only ask because I can't see how its related since its almost completely unrelated to tensor analysis.
It's complicated; don't worry about it.
I asked Atomic-S What does this have to do with the definition of tensors? but he chose not to answer my question. Perhaps if you describe the train of thought that you mentioned above then I'll understand. Thanks.
It has to do with a particular type of tensor in a particular application to describe the nature of "relatively straight lines."
Gravity is a 'acceleration field/spacetime shape resolution' issue; ..
...actually tidal gradients in the gravitational field. In the language of general relativity, spacetime curvature.
Right; different languages; one including relative shapes.
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if i'm accelerating at 9.8 m/s2, why am i not going the speed of light? I'm old enough, to be sure.
If we apply some physics from early high school, you would expect your velocity to be:
v = at where:
v: velocity at time t
a: acceleration (9.8m/s2 in this example)
t: elapsed time
According to this formula, to reach the speed of light v=c=3x108 m/s would take t=3x108/9.8 = 3x107 seconds = 347 days, or just under a year.
However (and it's big however!), Einstein showed that you can never accelerate a massive object to reach the speed of light - no matter how hard you accelerate, or for how long.
These concepts normally come up at the end of high school or early university physics.
We don't notice it with our daily lives here on Earth, but when you accelerate objects close to the speed of light, strange things occur: time slows down, and mass increases, so it gets harder and harder to get closer to the speed of light. Oddly, this is not visible to the person being accelerated.
Such peculiarities are very familiar to physicists at the LHC, who try to get particles traveling as close as possible to the speed of light so they can do experiments at very high energies.
See: https://en.wikipedia.org/wiki/Speed_of_light#Upper_limit_on_speeds
So you may feel old, but you are not nearly old enough to reach the speed of light, when accelerating at 9.8m/s2!
The equivalence of an accelerating frame of reference and a gravitational field is highlighted by Evans post. In less than one year of acceleration we reach a point equivalent to approaching the event horizon of a black hole where the object is approaching light speed. If we cannot reach light speed that begs the question can gravity force an object to reach light speed at the event horizon of a black hole.
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From the original post: "Trying to get this relativity thing..."
Ahhh! I understand now. For some reason I thought you only had special relativity (SR) in mind because people almost always study SR before they study general relativity (GR). Thanks for making that clear. :)
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can gravity force an object to reach light speed at the event horizon of a black hole?
For a "distant observer", an object can reach just over c/3 before it disappears inside the event horizon.
If you scroll down to the graphs and equations on the following web page, you will see that the answer is "yes and no", all depending on your frame of reference.
http://physics.stackexchange.com/questions/170502/will-an-object-always-fall-at-an-infinite-speed-in-a-black-hole
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can gravity force an object to reach light speed at the event horizon of a black hole?
For a "distant observer", an object can reach just over c/3 before it disappears inside the event horizon.
If you scroll down to the graphs and equations on the following web page, you will see that the answer is "yes and no", all depending on your frame of reference.
http://physics.stackexchange.com/questions/170502/will-an-object-always-fall-at-an-infinite-speed-in-a-black-hole
Quite a while back now I did produce a set of graphs much like those of John Rennie in the linked thread. I was never convinced by some of my own conclusions. I may revisit this with a new perspective.
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"Apparent horizon" compelled me to look up "Photon sphere:" https://en.wikipedia.org/wiki/Photon_sphere & i wonder if these are essentially the same thing to a falling observer...?
Note "ds=o (a light-like interval)", i.e. relative to the photon.