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I don't understand the equation in the article, is it possibly wrong?
I recommend addition of a second plot: Displacement vs. Time – a Parabola. It will reduce confusion. Also, because free-fall acceleration is described, it will make the point clear that the Linear v vs. t corresponds with a parabolic Displacement vs. Time. — Preceding unsigned comment added by SirHolo ( talk • contribs) 18:48, 15 January 2016 (UTC)
The section of the article on tangential and centripetal acceleration contains the statement that "the acceleration of a particle moving on a curved path on a planar surface can be written using the chain rule of differentiation... etc" (immediately before the reference number 4).
This claim that such decomposition of acceleration is valid only for planar curves is further supported by the statement "Extension of this approach to three-dimensional space curves that cannot be contained on a planar surface leads to the Frenet–Serret formulas", which appears at the end of the section.
Would the authors care to explain why they believe that decomposition into tangential and normal acceleration cannot be done in the same way for "space curves"? I see nothing in the derivation presented that would support such a claim.-- Ilevanat ( talk) 23:33, 17 October 2012 (UTC)
Having established that the author of the article section under disscusion will not answer my question (see User talk:Brews ohare), and due to my determination not to edit work of others without consent, I can only here inform the interested readers that the tangential and normal acceleration formulas in the article are equally valid for planar and spatial curves. See, for example, the following link: http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/velacc/velacc.html
-- Ilevanat ( talk) 22:49, 25 October 2012 (UTC)
Allowing for the possibility that the previous author is forbidden even to discuss the physics-related articles, I have finally decided to make corrections described above.-- Ilevanat ( talk) 01:16, 26 November 2012 (UTC)
I suggest that the name of this page be changed to Coordinate acceleration so as to contrast it with Proper acceleration. KingSupernova ( talk) 15:29, 9 January 2013 (UTC)
What say others? - DVdm ( talk) 14:15, 10 January 2013 (UTC)
In any differentiable manifold (even in the Newtonian spacetime), in order to define acceleration, a connection (i.e., a covariant derivative ) must be given. Newton uses inertial frames because in his mathematical apparatus there is no concept of affine space, connection, parallelization (cf. teleparallelism ). Lagrange accelerations [1] and Lagrange forces are not vectors but their difference yes. Hence, Euler-Lagrange equations are tensorial. In Newton gravity theory, gravitational forces are vectors, in Albert Einstein GR gravity theory gravitational-inertial forces are Lagrange forces. In other words, the concept of acceleration is a covariant derivative concept, an additional structure on . Mgvongoeden ( talk) 12:41, 22 April 2015 (UTC)
I'm experiencing the same dichotomy in the definition of acceleration, but in a different sense... a person (I won't give his real name unless mods request it... I bet you didn't know 'Gandalf61' denies the fundamental physical laws, eh? That brings into question everything he's edited here...) who's been attempting to claim 2LoT can be violated at the quantum scale (for the record, 2LoT is even more rigorously observed at the quantum scale than macroscopically) has been backed into a deeper and deeper corner, to such an extent that we're now arguing over whether acceleration is a rank-1 tensor or a rank-2 tensor. He claims all derivatives of position (velocity, acceleration, jerk, snap, crackle, pop, lock, drop) are rank-1 tensors because apparently a temporal derivative is a 'special' kind of derivative that doesn't have to comport with differential calculus. He's edited this page to remove all references to the tensor rank of acceleration. I've reverted his changes.
As evidence that a derivative increases resultant tensor rank:
http://openaccess.thecvf.com/content_cvpr_2018/papers/Kim_High-Order_Tensor_Regularization_CVPR_2018_paper.pdf "In general, taking a derivative of a tensor increases its order by one: The derivative of function f is a vector, a first-order tensor. Similarly, the derivative of a second order tensor h is a third order tensor ∇gh."
/info/en/?search=Tensor_derivative_(continuum_mechanics)#Gradient_of_a_tensor_field
"Derivatives of scalar valued functions of vectors Let f(v) be a real valued function of the vector v. Then the derivative of f(v) with respect to v (or at v) is the vector defined through its dot product..."
"Derivatives of vector valued functions of vectors Let f(v) be a vector valued function of the vector v. Then the derivative of f(v) with respect to v (or at v) is the second order tensor defined through its dot product..."
"Derivatives of tensor valued functions of second-order tensors Let F(S) be a second order tensor valued function of the second order tensor S. Then the derivative of F(S) with respect to S (or at S) in the direction T is the fourth order tensor..."
"The gradient of a tensor field of order n is a tensor field of order n+1."
In other words, the derivative of a scalar is a vector, the derivative of a vector is a rank 2 tensor, and the 2nd derivative of a rank 2 tensor is a rank 4 tensor, because the gradient of a tensor field of order n is a tensor field of order n+1.
Of course, the gradient is dual to the derivative, one cannot have a derivative without a gradient, nor can one have a gradient without a derivative.
Gandalf61 stated in his change of the web page that "acceleration is derivative of velocity wrt time, *not* a gradient of a vector field". That is wrong. http://clas.sa.ucsb.edu/staff/alex/VCFAQ/vectorFields/vectorFields.htm The technical definition of a vector field is a map from R^3 to R^3 What this means is we can assign a 3 dimensional vector to every point in R^3. We can think of the vector field as an ordered set of 3 functions: F = (f1(x,y,z), f2(x,y,z),f3(x,y,z)). Here the functions f1, f2 and f3 are ordinary scalar functions of x, y, and z.
Some common vector fields in physics: Electric Fields
Magnetic Fields
Gravitational Fields
Wind Velocity <---
Fluid Velocity <---
https://en.wikiversity.org/wiki/Acceleration_field "Acceleration field is a two-component vector field..."
If Gandalf61 wishes to sabotage Wikipedia pages in support of his denial of the fundamental physical laws and his kooky take on reality, then his entire history of edits should be reviewed to ensure they actually comport with reality.
To RealOldOne2... the person doing the editing of this page to remove any reference to tensor rank is none other than our CFACT friend, Dave Burton, with whom you've had several encounters.
Ah, I see now... the new user 'RealOldOne2' has hijacked a well-known 'nym... it's none other than 'Gandalf61' aka Dave Burton, attempting yet again to sabotage a Wikipedia page to comport with his kooky fundamental-physical-law denying take on reality. Can an administrator take this nutter 'Gandalf61' aka 'RealOldOne2' down, please?
Have been reverted 2 times. Where to place it?
Moreover, I feel that using "t" is time interval instead of just time would be better. If we are here to improve the pedia, why not be clear that nobody in audience has any difficulty?
117.248.121.6 (
talk) 11:36, 16 May 2015 (UTC)
Would it be helpful or just complicate things to have a table of identities for uniform acceleration? I know I keep finding myself on this page for them:
so we have time as a function of v and a, time as a function of distance traveled and velocities:
and we have the acceleration required to change speed in a given time and the acceleration required to go a distance from a given velocity in a given time:
also, distance traveled given velocities and a:
Others? At least they are now here for my future reference. :-) —Ben FrantzDale ( talk) 12:03, 16 October 2015 (UTC)
The equation:
seems wrong, since the last part does not contain . The equation seems to mixing up displacement and position. See http://www.open.edu/openlearn/science-maths-technology/science/physics-and-astronomy/describing-motion-along-line/content-section-1.6.1#ueqn001-050 — Preceding unsigned comment added by MLópez-Ibáñez ( talk • contribs) 18:33, 27 February 2016 (UTC)
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Can the category of this vital article be category?-- Dthomsen8 ( talk) 19:39, 28 September 2019 (UTC)
I have reverted repeated revisions by anonymous contributor
71.135.47.16 (
talk ·
contribs) which claim that acceleration is not a vector/rank-1 tensor but is instead a rank-2 tensor. This seems to be based on the mistaken assumption that acceleration is the gradient of velocity and so should have a tensor rank that is one greater than that of velocity. Instead, of course, acceleration is the derivative of velocity with respect to time and so has the same rank as velocity, which in turn has the same rank as position displacement i.e. all three quantities are vectors/rank 1 tensors.
Gandalf61 (
talk) 08:56, 17 March 2020 (UTC)
What mess? The only mess is that people are led to believe that we live in a 3-D Euclidean space, but the real world is 4-D Minkowski space, where time is just another coordinate and behaves just as the other coordinates behave. Thus the acceleration tensor is rank-2, as even your *own pages* concede. https://en.wikiversity.org/wiki/Acceleration_tensor https://en.wikiversity.org/wiki/Gravitational_tensor http://www.tapir.caltech.edu/~chirata/ph236/lec07.pdf /info/en/?search=Einstein_tensor http://sergf.ru/aten.htm http://zen.uta.edu/me5312/03.pdf#page=3
You've confused the real-world 4-D Minkowski space (with coordinates of t,x,y,z) with the 'toy model' 3-D Euclidean space (with coordinates of x,y,z)... the 3-D 'toy model' is only used to simplify the calculations for a problem which is localized... it does not properly address time except in an instantaneous sense, thus it cannot reflect reality fully.
As it turns out, there is a close relationship between the dual basis and the vector derivative operator (usually denoted nabla, ∇). If dual basis vectors are written e^a (and ordinary basis vectors e_a), then we tend to say ∇ = ∑_a e^a ∂/ ∂x^a. Nabla, ∇, is defined in terms of dual vectors, not ordinary vectors... that's the key. There are typically no dual vectors in 3-D space because the tangent vectors to the coordinate axes form the basis of the vector space, and they are often set in Euclidean space as orthogonal. Which is why in Euclidean space the dual basis is generally the same as the tangent basis.
Contravariant components are expressed with respect to ordinary basis vectors. Covariant components are expressed with respect to dual basis vectors. Usually in 3-D Euclidean space, that's not an issue, since the dual and the ordinary bases are the same, and the contravariant and covariant components of a vector are equal.
F = ma
Force is a dual vector, its one-form adds one covariant component to the index of force, escalating its rank.. So the left-hand side of the equation is rank 2. On the right-hand side, m is scalar, and a rank 2 tensor multiplied with a scalar is a rank 2 tensor. Thus the equation balances in terms of tensor rank.
So the question really is... are we talking about space in the 4-D real world, or space in a 3-D 'toy model'?
I know, you thought I was a 'kook' spouting nonsense. I assure you, every bit of information above was drawn directly from physics texts. The concept flies over the heads of many, such as Dave Burton (aka gandalf61, aka looksquirrel101, aka (fake) RealOldOne2)... but in reality, he's the kook... he's been, over the past half-year that I've been in contention with him, twisting science in furtherance of his contention that continual 2LoT violations can cause catastrophic atmospheric warming... considering that a macroscopic 2LoT violation has never been empirically observed, and considering that 2LoT is even more rigorously observed at the quantum scale, I'm sure you can see the problem inherent in his claim. Did I mention that he uses multiple socks to gain 'consensus points' and hijacks nyms? RealOldOne2 is one of the people he regularly clashes with... poor Dave always loses those battles, hence Dave hijacks the nym given the chance. — Preceding unsigned comment added by 71.135.33.190 ( talk) 10:11, 19 March 2020 (UTC)
You tried this denial tactic with your looksquirrel101 sock, as well, Dave... but you forgot that you'd been previously outed on that sock (and let us not forget your braggadocio that you 'teach physics'... certainly not in any professional setting, considering your fundamental misunderstandings). Your wording above is the exact wording you've used on CFACT (to wit: "acceleration is the derivative of velocity with respect to time and so has the same rank as velocity"... because you don't understand that the gradient is dual to its derivative (one cannot have a derivative without a gradient, nor can one have a gradient without a derivative). When we state we are 'taking the derivative', we generally are taking the dot product of the gradient and the tensor being operated upon because the gradient is a measurable or calculable quantity, whereas the derivative isn't always), where you've demonstrated your inability to grasp simple physics concepts '80-some-odd' times. If you like, I can provide the list of concepts you have a fundamental misunderstanding of, to include links to your own words. That's not an 'ad hominem attack', that's reality... and reality can be harsh. Truth is not an ad hominem. Stop playing games with science in furtherance of your climate ideology. 2LoT is never violated (macroscopically or quantumly), individual atoms and molecules have a measurable temperature, the Equipartition Theorem applies to individual atoms and molecules, entropy applies to individual atoms and molecules, time reversibility doesn't occur except under very limited CPT symmetry conditions, Kirchhoff's Law of Thermal Radiation is a ratio not an equality, radiation pressure exists... all (and so much more) of which you've denied and/or twisted in furtherance of your claim that continual 2LoT violations can cause catastrophic atmospheric warming. Every time you're proven wrong, you backpedal, move the goal posts, and come up with another bastardization of science to support your contention. This last go-round was especially damaging to you, so you're reaching out to other sites for support... and hijacking nyms and using multiple socks to do so. — Preceding unsigned comment added by 71.135.34.23 ( talk) 00:41, 20 March 2020 (UTC)
From your own web pages: /info/en/?search=General_relativity "While general relativity replaces the scalar gravitational potential of classical physics by a **symmetric rank-two tensor**..."
A gravitational field is an accelerational field, is it not? The Equivalence Principle asserts that in free-fall the effect of gravity is totally abolished and general relativity reduces to special relativity, as in the inertial state. In other words, one cannot distinguish, in an inertial frame, any difference between acceleration and gravitational effects.
Same for special relativity... whereas general relativity uses a pseudo-Riemannian space (Lorentzian space with a metric tensor consisting of a nondegenerate symmetric bilinear form on the tangent space at each point), special relativity uses a pseudo-Euclidean Minkowski manifold... but even there, acceleration is a rank-2 tensor: https://en.wikiversity.org/wiki/Acceleration_tensor "u_uv is the antisymmetric tensor of **rank 2**..."
So again, the question is... are we talking about real-world 4-D space (Lorentzian or Minkowskian), or a 3-D 'toy model' ala classical physics used as an approximation to make the calculations easier? I posit that we should be treating the physical quantities as they are represented in the real world, with a caveat worded to alert readers to the fact that classical physics is an antiquated theory which only approximates reality, and as such, tensor rank is not accurately handled when compared to more modern theories, as explained in my prior comment above about dual vectors and nabla.
A rank 1 tensor (a vector quantity) can be described in terms of a uniform displacement per unit time (the magnitude of its velocity in the direction of motion). A body undergoing acceleration, however, cannot be described merely in terms of a uniform displacement per unit time, because the displacement per unit time is changing per unit time. If a body undergoing acceleration could be described in that manner, that would imply that it would fall at the same displacement per unit time in a gravitational field. There is a gradient (and thus its dual, the derivative), and thus a higher tensor.
Tensor rank is analogous to the number of interacting vector fields at each point in space-time. A gradient, which is dual to it derivative, implies there is another 'interacting vector field' of a higher order than the one currently being considered. For a tensor field of order n, the gradient of that tensor field *is* a tensor field of order n+1. — Preceding unsigned comment added by 71.135.34.23 ( talk) 05:36, 20 March 2020 (UTC)
Gedankenexperiment: Consider a large sealed box falling freely in a gravitational field (and thus accelerating) from a great height, inside the box is air filled with dust... the dust converges laterally because particles are following converging trajectories toward the center of the planet. The dust stretches out vertically because particles nearer the planet will accelerate faster. Two vectors are necessary to describe what is happening.
Two vector fields... three of the six independent components of the acceleration tensor associated with the components of the acceleration field strength S (longitudinal component, which causes the dust to stretch out vertically), and the other three with the components of the acceleration solenoidal vector N (transverse component, which causes the dust to converge laterally). That can't happen with a vector (rank 1 tensor).
The tensor rank is defined in a frame-independent manner as the number of tensors and 1-forms that it accepts as input... in this case, the tensor accepts two vector fields, ∴ acceleration is a rank 2 tensor. — Preceding unsigned comment added by 71.135.34.23 ( talk) 09:24, 20 March 2020 (UTC)
Here you go: https://i.imgur.com/pkXo2KE.png https://www.youtube.com/watch?v=yx0oql3LIiU Prof. Grinfeld, referring to acceleration: "What can you say about this variant? It's a tensor." That means rank 2.
And of course, in your *own pages*... /info/en/?search=Time_derivative "The components of a vector U expressed this way transform as a contravariant **tensor**..." Excuse me... but I do believe my oppositional interlocutor claimed that taking the time derivative of a vector quantity of rank n returned a vector quantity of rank n and thus all derivatives (velocity, jerk, snap, crackle, pop, lock, drop) were rank 1 tensors (vector quantities)... he'd be correct, *if one disregarded all the lower-rank gradients used to get to that derivative in the first place*. A tensor is a vector of vectors, after all. So simplistically, all tensors of all ranks are "vectors" (in that they have magnitude and direction). But only simplistically. — Preceding unsigned comment added by 71.135.34.23 ( talk) 08:20, 21 March 2020 (UTC)
Of course it is a tensor, just like every other vector is tensor (a rank-1 tensor). Furthermore, the YouTube video you pointed to does not mention that acceleration (or the second derivative) is a rank-2 tensor, it also just mentions that it is a tensor. On the contrary, he uses only one index, which points to the fact that it is just a rank-1 tensor.
So let's just ask a basic question: Let be vectors and let be the covariant derivative. What kind of object is to you? (If you want to use local coordinates .) Nmdwolf ( talk) 12:36, 22 March 2020 (UTC)
No I didn't mean the gradient, as I said, I was talking about the covariant derivative (since you insist on using general spacetime): . So once again, now that I clarified my notation, what is the rank of the object if you know that and are rank-1 tensors (i.e. vectors)?
Nmdwolf ( talk) 09:36, 23 March 2020 (UTC)
Can we put an end to this nonsense already? I'd say the Feynman Lectures are a definitive source. [2] He explicitly states that the acceleration vector is the time derivative of the velocity vector. Case closed. RealOldOne2 ( talk) 20:54, 23 March 2020 (UTC)
https://books.google.com/books?id=IyJhCHAryuUC&pg=PA180 Einstein's General Theory of Relativity: With Modern Applications in Cosmology Øyvind Grøn, Sigbjorn Hervik Springer Science & Business Media, Aug 24, 2007 ISBN 0387692002, 9780387692005
"An alternative way to see this is if we keep in mind eq. (1.33) which tells us how acceleration of gravity can be deduced in Newton's theory: acceleration of gravity is generated by the gradient of the potential Φ. Recall from chapter 6 that for a free particle instantaneously at rest, the acceleration is given by = Γ, see eq. (6.112); hence, the acceleration of gravity can be represented by the Christoffel symbols in an appropriate frame. According to eq.(1.33), we therefore seek a tensor which contains first derivatives of the Christoffel symbols.
The Christoffel symbols contain first derivatives of the metric, and the Riemann curvature tensor contains first derivatives of the Christoffel symbols. Hence, the Riemann curvature tensor contains the right order of derivatives to represent the left hand side of Poisson's equation. We have already argued that the right hand side is proportional to the energy-momentum tensor T_uv. This is a symmetric tensor of rank 2. The first natural choice is therefore to consider the Ricci tensor, which is obtained by contracting the Riemann tensor once. Therefore Einstein initially tried: .
He discovered, however, that this is not quite satisfactory; the Ricci tensor is in general not divergence-free. With the help of Marcel Grossman, Einstein then discovered that the combination - (1/2) is divergence-free. This is the Einstein tensor, see eq. (7.71), and is the simplest divergence-free combination of the Ricci tensor. The Einstein tensor has all the required properties: it is a divergence-free symmetric tensor of rank two."
The Christoffel symbol contains the first derivative of the metric. The Einstein tensor contains the first derivative of the Christoffel symbol, making it the 2nd derivative of the metric. Note the double-dot over the x... that denotes the 2nd derivative. Note the text blurb "This is a symmetric tensor of rank 2.".
You'll note this reference explicitly states that the acceleration tensor is rank 2. 71.135.43.236 ( talk) 05:15, 25 March 2020 (UTC)
.
If the right-hand side of the equation (proportional to the energy-momentum tensor) is a rank 2 tensor, the left-hand side of the equation (acceleration due to gravity) is a rank 2 tensor.
I've got more published sources, if you need them... ALL of them stating that the acceleration tensor is rank 2 in 4-D Minkowskian or Lorentzian space. Your claim that acceleration is rank 1 is for 3-D space, but 3-D space is an approximation of 4-D space. 71.135.43.236 ( talk) 08:44, 25 March 2020 (UTC)
This article was the subject of a Wiki Education Foundation-supported course assignment, between 7 September 2022 and 24 December 2022. Further details are available on the course page. Student editor(s): Zaustin801 ( article contribs).
— Assignment last updated by Janyahmercedes ( talk) 02:52, 26 January 2023 (UTC)
This article was the subject of a Wiki Education Foundation-supported course assignment, between 13 February 2023 and 12 June 2023. Further details are available on the course page. Student editor(s): JiachengGeng ( article contribs). Peer reviewers: Kbalagna, MRoger886.
— Assignment last updated by Lzepeda12 ( talk) 21:07, 26 April 2023 (UTC)
I guess it is important to explain the difference between "decelerating" and "negative acceleration", like this: "The term 'decelerating' means only that the acceleration vector points opposite to the velocity vector. Is is not specified whether the velocity vector of the car points in the positive or negative direction. Therefore, it is not possible to know whether the acceleration is positive or negative." Cutnell, J. D., Johnson, K. W., Young, D., Stadler, S. (2021). Physics. Reino Unido: Wiley. p. 40. https://www.google.com.br/books/edition/Physics/tH49EAAAQBAJ?hl=pt-BR&gbpv=1&dq=deceleration%20negative%20physics&pg=PA40&printsec=frontcover It is ok? Dalmas64 ( talk) 13:39, 30 March 2023 (UTC)
I think it should say just the velocity. I don't see why it would say change in velocity. 2.36.106.75 ( talk) 18:05, 13 April 2024 (UTC)
This
level-4 vital article is rated C-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | |||||||||||||||||||||
|
Text and/or other creative content from this version of Uniform acceleration was copied or moved into Acceleration with this edit on 5 December 2010. The former page's history now serves to provide attribution for that content in the latter page, and it must not be deleted as long as the latter page exists. |
|
|
I don't understand the equation in the article, is it possibly wrong?
I recommend addition of a second plot: Displacement vs. Time – a Parabola. It will reduce confusion. Also, because free-fall acceleration is described, it will make the point clear that the Linear v vs. t corresponds with a parabolic Displacement vs. Time. — Preceding unsigned comment added by SirHolo ( talk • contribs) 18:48, 15 January 2016 (UTC)
The section of the article on tangential and centripetal acceleration contains the statement that "the acceleration of a particle moving on a curved path on a planar surface can be written using the chain rule of differentiation... etc" (immediately before the reference number 4).
This claim that such decomposition of acceleration is valid only for planar curves is further supported by the statement "Extension of this approach to three-dimensional space curves that cannot be contained on a planar surface leads to the Frenet–Serret formulas", which appears at the end of the section.
Would the authors care to explain why they believe that decomposition into tangential and normal acceleration cannot be done in the same way for "space curves"? I see nothing in the derivation presented that would support such a claim.-- Ilevanat ( talk) 23:33, 17 October 2012 (UTC)
Having established that the author of the article section under disscusion will not answer my question (see User talk:Brews ohare), and due to my determination not to edit work of others without consent, I can only here inform the interested readers that the tangential and normal acceleration formulas in the article are equally valid for planar and spatial curves. See, for example, the following link: http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/velacc/velacc.html
-- Ilevanat ( talk) 22:49, 25 October 2012 (UTC)
Allowing for the possibility that the previous author is forbidden even to discuss the physics-related articles, I have finally decided to make corrections described above.-- Ilevanat ( talk) 01:16, 26 November 2012 (UTC)
I suggest that the name of this page be changed to Coordinate acceleration so as to contrast it with Proper acceleration. KingSupernova ( talk) 15:29, 9 January 2013 (UTC)
What say others? - DVdm ( talk) 14:15, 10 January 2013 (UTC)
In any differentiable manifold (even in the Newtonian spacetime), in order to define acceleration, a connection (i.e., a covariant derivative ) must be given. Newton uses inertial frames because in his mathematical apparatus there is no concept of affine space, connection, parallelization (cf. teleparallelism ). Lagrange accelerations [1] and Lagrange forces are not vectors but their difference yes. Hence, Euler-Lagrange equations are tensorial. In Newton gravity theory, gravitational forces are vectors, in Albert Einstein GR gravity theory gravitational-inertial forces are Lagrange forces. In other words, the concept of acceleration is a covariant derivative concept, an additional structure on . Mgvongoeden ( talk) 12:41, 22 April 2015 (UTC)
I'm experiencing the same dichotomy in the definition of acceleration, but in a different sense... a person (I won't give his real name unless mods request it... I bet you didn't know 'Gandalf61' denies the fundamental physical laws, eh? That brings into question everything he's edited here...) who's been attempting to claim 2LoT can be violated at the quantum scale (for the record, 2LoT is even more rigorously observed at the quantum scale than macroscopically) has been backed into a deeper and deeper corner, to such an extent that we're now arguing over whether acceleration is a rank-1 tensor or a rank-2 tensor. He claims all derivatives of position (velocity, acceleration, jerk, snap, crackle, pop, lock, drop) are rank-1 tensors because apparently a temporal derivative is a 'special' kind of derivative that doesn't have to comport with differential calculus. He's edited this page to remove all references to the tensor rank of acceleration. I've reverted his changes.
As evidence that a derivative increases resultant tensor rank:
http://openaccess.thecvf.com/content_cvpr_2018/papers/Kim_High-Order_Tensor_Regularization_CVPR_2018_paper.pdf "In general, taking a derivative of a tensor increases its order by one: The derivative of function f is a vector, a first-order tensor. Similarly, the derivative of a second order tensor h is a third order tensor ∇gh."
/info/en/?search=Tensor_derivative_(continuum_mechanics)#Gradient_of_a_tensor_field
"Derivatives of scalar valued functions of vectors Let f(v) be a real valued function of the vector v. Then the derivative of f(v) with respect to v (or at v) is the vector defined through its dot product..."
"Derivatives of vector valued functions of vectors Let f(v) be a vector valued function of the vector v. Then the derivative of f(v) with respect to v (or at v) is the second order tensor defined through its dot product..."
"Derivatives of tensor valued functions of second-order tensors Let F(S) be a second order tensor valued function of the second order tensor S. Then the derivative of F(S) with respect to S (or at S) in the direction T is the fourth order tensor..."
"The gradient of a tensor field of order n is a tensor field of order n+1."
In other words, the derivative of a scalar is a vector, the derivative of a vector is a rank 2 tensor, and the 2nd derivative of a rank 2 tensor is a rank 4 tensor, because the gradient of a tensor field of order n is a tensor field of order n+1.
Of course, the gradient is dual to the derivative, one cannot have a derivative without a gradient, nor can one have a gradient without a derivative.
Gandalf61 stated in his change of the web page that "acceleration is derivative of velocity wrt time, *not* a gradient of a vector field". That is wrong. http://clas.sa.ucsb.edu/staff/alex/VCFAQ/vectorFields/vectorFields.htm The technical definition of a vector field is a map from R^3 to R^3 What this means is we can assign a 3 dimensional vector to every point in R^3. We can think of the vector field as an ordered set of 3 functions: F = (f1(x,y,z), f2(x,y,z),f3(x,y,z)). Here the functions f1, f2 and f3 are ordinary scalar functions of x, y, and z.
Some common vector fields in physics: Electric Fields
Magnetic Fields
Gravitational Fields
Wind Velocity <---
Fluid Velocity <---
https://en.wikiversity.org/wiki/Acceleration_field "Acceleration field is a two-component vector field..."
If Gandalf61 wishes to sabotage Wikipedia pages in support of his denial of the fundamental physical laws and his kooky take on reality, then his entire history of edits should be reviewed to ensure they actually comport with reality.
To RealOldOne2... the person doing the editing of this page to remove any reference to tensor rank is none other than our CFACT friend, Dave Burton, with whom you've had several encounters.
Ah, I see now... the new user 'RealOldOne2' has hijacked a well-known 'nym... it's none other than 'Gandalf61' aka Dave Burton, attempting yet again to sabotage a Wikipedia page to comport with his kooky fundamental-physical-law denying take on reality. Can an administrator take this nutter 'Gandalf61' aka 'RealOldOne2' down, please?
Have been reverted 2 times. Where to place it?
Moreover, I feel that using "t" is time interval instead of just time would be better. If we are here to improve the pedia, why not be clear that nobody in audience has any difficulty?
117.248.121.6 (
talk) 11:36, 16 May 2015 (UTC)
Would it be helpful or just complicate things to have a table of identities for uniform acceleration? I know I keep finding myself on this page for them:
so we have time as a function of v and a, time as a function of distance traveled and velocities:
and we have the acceleration required to change speed in a given time and the acceleration required to go a distance from a given velocity in a given time:
also, distance traveled given velocities and a:
Others? At least they are now here for my future reference. :-) —Ben FrantzDale ( talk) 12:03, 16 October 2015 (UTC)
The equation:
seems wrong, since the last part does not contain . The equation seems to mixing up displacement and position. See http://www.open.edu/openlearn/science-maths-technology/science/physics-and-astronomy/describing-motion-along-line/content-section-1.6.1#ueqn001-050 — Preceding unsigned comment added by MLópez-Ibáñez ( talk • contribs) 18:33, 27 February 2016 (UTC)
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Can the category of this vital article be category?-- Dthomsen8 ( talk) 19:39, 28 September 2019 (UTC)
I have reverted repeated revisions by anonymous contributor
71.135.47.16 (
talk ·
contribs) which claim that acceleration is not a vector/rank-1 tensor but is instead a rank-2 tensor. This seems to be based on the mistaken assumption that acceleration is the gradient of velocity and so should have a tensor rank that is one greater than that of velocity. Instead, of course, acceleration is the derivative of velocity with respect to time and so has the same rank as velocity, which in turn has the same rank as position displacement i.e. all three quantities are vectors/rank 1 tensors.
Gandalf61 (
talk) 08:56, 17 March 2020 (UTC)
What mess? The only mess is that people are led to believe that we live in a 3-D Euclidean space, but the real world is 4-D Minkowski space, where time is just another coordinate and behaves just as the other coordinates behave. Thus the acceleration tensor is rank-2, as even your *own pages* concede. https://en.wikiversity.org/wiki/Acceleration_tensor https://en.wikiversity.org/wiki/Gravitational_tensor http://www.tapir.caltech.edu/~chirata/ph236/lec07.pdf /info/en/?search=Einstein_tensor http://sergf.ru/aten.htm http://zen.uta.edu/me5312/03.pdf#page=3
You've confused the real-world 4-D Minkowski space (with coordinates of t,x,y,z) with the 'toy model' 3-D Euclidean space (with coordinates of x,y,z)... the 3-D 'toy model' is only used to simplify the calculations for a problem which is localized... it does not properly address time except in an instantaneous sense, thus it cannot reflect reality fully.
As it turns out, there is a close relationship between the dual basis and the vector derivative operator (usually denoted nabla, ∇). If dual basis vectors are written e^a (and ordinary basis vectors e_a), then we tend to say ∇ = ∑_a e^a ∂/ ∂x^a. Nabla, ∇, is defined in terms of dual vectors, not ordinary vectors... that's the key. There are typically no dual vectors in 3-D space because the tangent vectors to the coordinate axes form the basis of the vector space, and they are often set in Euclidean space as orthogonal. Which is why in Euclidean space the dual basis is generally the same as the tangent basis.
Contravariant components are expressed with respect to ordinary basis vectors. Covariant components are expressed with respect to dual basis vectors. Usually in 3-D Euclidean space, that's not an issue, since the dual and the ordinary bases are the same, and the contravariant and covariant components of a vector are equal.
F = ma
Force is a dual vector, its one-form adds one covariant component to the index of force, escalating its rank.. So the left-hand side of the equation is rank 2. On the right-hand side, m is scalar, and a rank 2 tensor multiplied with a scalar is a rank 2 tensor. Thus the equation balances in terms of tensor rank.
So the question really is... are we talking about space in the 4-D real world, or space in a 3-D 'toy model'?
I know, you thought I was a 'kook' spouting nonsense. I assure you, every bit of information above was drawn directly from physics texts. The concept flies over the heads of many, such as Dave Burton (aka gandalf61, aka looksquirrel101, aka (fake) RealOldOne2)... but in reality, he's the kook... he's been, over the past half-year that I've been in contention with him, twisting science in furtherance of his contention that continual 2LoT violations can cause catastrophic atmospheric warming... considering that a macroscopic 2LoT violation has never been empirically observed, and considering that 2LoT is even more rigorously observed at the quantum scale, I'm sure you can see the problem inherent in his claim. Did I mention that he uses multiple socks to gain 'consensus points' and hijacks nyms? RealOldOne2 is one of the people he regularly clashes with... poor Dave always loses those battles, hence Dave hijacks the nym given the chance. — Preceding unsigned comment added by 71.135.33.190 ( talk) 10:11, 19 March 2020 (UTC)
You tried this denial tactic with your looksquirrel101 sock, as well, Dave... but you forgot that you'd been previously outed on that sock (and let us not forget your braggadocio that you 'teach physics'... certainly not in any professional setting, considering your fundamental misunderstandings). Your wording above is the exact wording you've used on CFACT (to wit: "acceleration is the derivative of velocity with respect to time and so has the same rank as velocity"... because you don't understand that the gradient is dual to its derivative (one cannot have a derivative without a gradient, nor can one have a gradient without a derivative). When we state we are 'taking the derivative', we generally are taking the dot product of the gradient and the tensor being operated upon because the gradient is a measurable or calculable quantity, whereas the derivative isn't always), where you've demonstrated your inability to grasp simple physics concepts '80-some-odd' times. If you like, I can provide the list of concepts you have a fundamental misunderstanding of, to include links to your own words. That's not an 'ad hominem attack', that's reality... and reality can be harsh. Truth is not an ad hominem. Stop playing games with science in furtherance of your climate ideology. 2LoT is never violated (macroscopically or quantumly), individual atoms and molecules have a measurable temperature, the Equipartition Theorem applies to individual atoms and molecules, entropy applies to individual atoms and molecules, time reversibility doesn't occur except under very limited CPT symmetry conditions, Kirchhoff's Law of Thermal Radiation is a ratio not an equality, radiation pressure exists... all (and so much more) of which you've denied and/or twisted in furtherance of your claim that continual 2LoT violations can cause catastrophic atmospheric warming. Every time you're proven wrong, you backpedal, move the goal posts, and come up with another bastardization of science to support your contention. This last go-round was especially damaging to you, so you're reaching out to other sites for support... and hijacking nyms and using multiple socks to do so. — Preceding unsigned comment added by 71.135.34.23 ( talk) 00:41, 20 March 2020 (UTC)
From your own web pages: /info/en/?search=General_relativity "While general relativity replaces the scalar gravitational potential of classical physics by a **symmetric rank-two tensor**..."
A gravitational field is an accelerational field, is it not? The Equivalence Principle asserts that in free-fall the effect of gravity is totally abolished and general relativity reduces to special relativity, as in the inertial state. In other words, one cannot distinguish, in an inertial frame, any difference between acceleration and gravitational effects.
Same for special relativity... whereas general relativity uses a pseudo-Riemannian space (Lorentzian space with a metric tensor consisting of a nondegenerate symmetric bilinear form on the tangent space at each point), special relativity uses a pseudo-Euclidean Minkowski manifold... but even there, acceleration is a rank-2 tensor: https://en.wikiversity.org/wiki/Acceleration_tensor "u_uv is the antisymmetric tensor of **rank 2**..."
So again, the question is... are we talking about real-world 4-D space (Lorentzian or Minkowskian), or a 3-D 'toy model' ala classical physics used as an approximation to make the calculations easier? I posit that we should be treating the physical quantities as they are represented in the real world, with a caveat worded to alert readers to the fact that classical physics is an antiquated theory which only approximates reality, and as such, tensor rank is not accurately handled when compared to more modern theories, as explained in my prior comment above about dual vectors and nabla.
A rank 1 tensor (a vector quantity) can be described in terms of a uniform displacement per unit time (the magnitude of its velocity in the direction of motion). A body undergoing acceleration, however, cannot be described merely in terms of a uniform displacement per unit time, because the displacement per unit time is changing per unit time. If a body undergoing acceleration could be described in that manner, that would imply that it would fall at the same displacement per unit time in a gravitational field. There is a gradient (and thus its dual, the derivative), and thus a higher tensor.
Tensor rank is analogous to the number of interacting vector fields at each point in space-time. A gradient, which is dual to it derivative, implies there is another 'interacting vector field' of a higher order than the one currently being considered. For a tensor field of order n, the gradient of that tensor field *is* a tensor field of order n+1. — Preceding unsigned comment added by 71.135.34.23 ( talk) 05:36, 20 March 2020 (UTC)
Gedankenexperiment: Consider a large sealed box falling freely in a gravitational field (and thus accelerating) from a great height, inside the box is air filled with dust... the dust converges laterally because particles are following converging trajectories toward the center of the planet. The dust stretches out vertically because particles nearer the planet will accelerate faster. Two vectors are necessary to describe what is happening.
Two vector fields... three of the six independent components of the acceleration tensor associated with the components of the acceleration field strength S (longitudinal component, which causes the dust to stretch out vertically), and the other three with the components of the acceleration solenoidal vector N (transverse component, which causes the dust to converge laterally). That can't happen with a vector (rank 1 tensor).
The tensor rank is defined in a frame-independent manner as the number of tensors and 1-forms that it accepts as input... in this case, the tensor accepts two vector fields, ∴ acceleration is a rank 2 tensor. — Preceding unsigned comment added by 71.135.34.23 ( talk) 09:24, 20 March 2020 (UTC)
Here you go: https://i.imgur.com/pkXo2KE.png https://www.youtube.com/watch?v=yx0oql3LIiU Prof. Grinfeld, referring to acceleration: "What can you say about this variant? It's a tensor." That means rank 2.
And of course, in your *own pages*... /info/en/?search=Time_derivative "The components of a vector U expressed this way transform as a contravariant **tensor**..." Excuse me... but I do believe my oppositional interlocutor claimed that taking the time derivative of a vector quantity of rank n returned a vector quantity of rank n and thus all derivatives (velocity, jerk, snap, crackle, pop, lock, drop) were rank 1 tensors (vector quantities)... he'd be correct, *if one disregarded all the lower-rank gradients used to get to that derivative in the first place*. A tensor is a vector of vectors, after all. So simplistically, all tensors of all ranks are "vectors" (in that they have magnitude and direction). But only simplistically. — Preceding unsigned comment added by 71.135.34.23 ( talk) 08:20, 21 March 2020 (UTC)
Of course it is a tensor, just like every other vector is tensor (a rank-1 tensor). Furthermore, the YouTube video you pointed to does not mention that acceleration (or the second derivative) is a rank-2 tensor, it also just mentions that it is a tensor. On the contrary, he uses only one index, which points to the fact that it is just a rank-1 tensor.
So let's just ask a basic question: Let be vectors and let be the covariant derivative. What kind of object is to you? (If you want to use local coordinates .) Nmdwolf ( talk) 12:36, 22 March 2020 (UTC)
No I didn't mean the gradient, as I said, I was talking about the covariant derivative (since you insist on using general spacetime): . So once again, now that I clarified my notation, what is the rank of the object if you know that and are rank-1 tensors (i.e. vectors)?
Nmdwolf ( talk) 09:36, 23 March 2020 (UTC)
Can we put an end to this nonsense already? I'd say the Feynman Lectures are a definitive source. [2] He explicitly states that the acceleration vector is the time derivative of the velocity vector. Case closed. RealOldOne2 ( talk) 20:54, 23 March 2020 (UTC)
https://books.google.com/books?id=IyJhCHAryuUC&pg=PA180 Einstein's General Theory of Relativity: With Modern Applications in Cosmology Øyvind Grøn, Sigbjorn Hervik Springer Science & Business Media, Aug 24, 2007 ISBN 0387692002, 9780387692005
"An alternative way to see this is if we keep in mind eq. (1.33) which tells us how acceleration of gravity can be deduced in Newton's theory: acceleration of gravity is generated by the gradient of the potential Φ. Recall from chapter 6 that for a free particle instantaneously at rest, the acceleration is given by = Γ, see eq. (6.112); hence, the acceleration of gravity can be represented by the Christoffel symbols in an appropriate frame. According to eq.(1.33), we therefore seek a tensor which contains first derivatives of the Christoffel symbols.
The Christoffel symbols contain first derivatives of the metric, and the Riemann curvature tensor contains first derivatives of the Christoffel symbols. Hence, the Riemann curvature tensor contains the right order of derivatives to represent the left hand side of Poisson's equation. We have already argued that the right hand side is proportional to the energy-momentum tensor T_uv. This is a symmetric tensor of rank 2. The first natural choice is therefore to consider the Ricci tensor, which is obtained by contracting the Riemann tensor once. Therefore Einstein initially tried: .
He discovered, however, that this is not quite satisfactory; the Ricci tensor is in general not divergence-free. With the help of Marcel Grossman, Einstein then discovered that the combination - (1/2) is divergence-free. This is the Einstein tensor, see eq. (7.71), and is the simplest divergence-free combination of the Ricci tensor. The Einstein tensor has all the required properties: it is a divergence-free symmetric tensor of rank two."
The Christoffel symbol contains the first derivative of the metric. The Einstein tensor contains the first derivative of the Christoffel symbol, making it the 2nd derivative of the metric. Note the double-dot over the x... that denotes the 2nd derivative. Note the text blurb "This is a symmetric tensor of rank 2.".
You'll note this reference explicitly states that the acceleration tensor is rank 2. 71.135.43.236 ( talk) 05:15, 25 March 2020 (UTC)
.
If the right-hand side of the equation (proportional to the energy-momentum tensor) is a rank 2 tensor, the left-hand side of the equation (acceleration due to gravity) is a rank 2 tensor.
I've got more published sources, if you need them... ALL of them stating that the acceleration tensor is rank 2 in 4-D Minkowskian or Lorentzian space. Your claim that acceleration is rank 1 is for 3-D space, but 3-D space is an approximation of 4-D space. 71.135.43.236 ( talk) 08:44, 25 March 2020 (UTC)
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I guess it is important to explain the difference between "decelerating" and "negative acceleration", like this: "The term 'decelerating' means only that the acceleration vector points opposite to the velocity vector. Is is not specified whether the velocity vector of the car points in the positive or negative direction. Therefore, it is not possible to know whether the acceleration is positive or negative." Cutnell, J. D., Johnson, K. W., Young, D., Stadler, S. (2021). Physics. Reino Unido: Wiley. p. 40. https://www.google.com.br/books/edition/Physics/tH49EAAAQBAJ?hl=pt-BR&gbpv=1&dq=deceleration%20negative%20physics&pg=PA40&printsec=frontcover It is ok? Dalmas64 ( talk) 13:39, 30 March 2023 (UTC)
I think it should say just the velocity. I don't see why it would say change in velocity. 2.36.106.75 ( talk) 18:05, 13 April 2024 (UTC)