# Talk:Acceleration

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## Uniform acceleration

I don't understand the equation in the article, is it possibly wrong?

${\displaystyle \mathbf {s} =\mathbf {u} t+{{1} \over {2}}\mathbf {a} t^{2}=\mathbf {u} t+{{1} \over {2}}{{v} \over {t}}\mathbf {t} ^{2}={(\mathbf {u} +{{1} \over {2}}\mathbf {v} )t}}$
You have substituted a = v/t, but v is not a t. The final speed v is given by v = u + a t. Hope this helps. - DVdm (talk) 17:22, 12 September 2012 (UTC)

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 (talkcontribs) 18:48, 15 January 2016 (UTC)

## "Planar decomposition" of acceleration into tangential and normal component

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)

## Decomposition of acceleration into tangential and normal component is the same for space curves as well

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)

## Correction of some minor errors and of a common misconception related to planar-spatial curves

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)

## Name Change

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)

I think the current article (Acceleration) can safely remain as it is now, in a more general, non-relativistic context, in which Coordinate acceleration already is redirected to Proper acceleration, as it should. - DVdm (talk) 16:47, 9 January 2013 (UTC)
That I guess makes sense for the article title, but I think coordinate acceleration should definitely redirect to acceleration and not proper acceleration. KingSupernova (talk) 13:56, 10 January 2013 (UTC)
I'm not so sure about that, as coordinate acceleration is only used in the context of non-Galilean relativity, where it is opposed to proper acceleration. Strictly speaking, perhaps we should have an article relativistic acceleration to which both coordinate and proper would redirect, but I don't think we'll find consensus for that.

What say others? - DVdm (talk) 14:15, 10 January 2013 (UTC)

Since most of the page on acceleration is really talking about proper acceleration and neither page really make the differece clear, I think the best thing would be to merge both pages into one, under acceleration, with different sections for the different kinds. However this would probably cause the page to be too long and complicated, so what might be simpler would be to clarify acceleration, making clear the differences between the two, and remnoving a lot of the information about proper acceleration and giving links to proper acceleration instead. Some other people's thoughts would be helpful here. KingSupernova (talk) 15:28, 11 January 2013 (UTC)
No way. The current name is what most people looking for an article on acceleration would expect to see. Dger (talk) 19:00, 11 January 2013 (UTC)
Well it doesn't really matter as the page they searched for would redirect to the right one. KingSupernova (talk) 03:19, 14 January 2013 (UTC)

## What is acceleration?

In any differentiable manifold ${\displaystyle Q}$ (even in the Newtonian spacetime), in order to define acceleration, a connection ${\displaystyle \Gamma }$ (i.e., a covariant derivative ${\displaystyle \nabla }$) 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 ${\displaystyle Q}$.Mgvongoeden (talk) 12:41, 22 April 2015 (UTC)

Notes
1. ^ G. Giachetta, "Jet Methods in Nonholonomic Mechanics", Journal of Mathematical Physics, 33, pag. 1652, (1992).
Do you have some kind of proposal to improve the article? I mean, see wp:talk page guidelines. - DVdm (talk) 18:11, 20 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."

"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?

## Where to place this?

Have been reverted 2 times. Where to place it? ${\displaystyle \mathbf {a} =\lim _{{\Delta t}\to 0}{\frac {\Delta \mathbf {v} }{\Delta t}}={\frac {d\mathbf {v} }{dt}}=v{\frac {d\mathbf {v} }{dx}}}$
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)

The v dv/dx expression, referring to a derivative w.r.t. position, just does not seem to fit in the text where the topic is derivative w.r.t. time, as here and here. It is of course true, but is it useful? Do you have a source where it is actually used? If so, we could perhaps include it—and the source—in a separate subsection. - DVdm (talk) 11:50, 16 May 2015 (UTC)
Re your remark about time interval: usually the variable t is called time, whereas an explicit time interval is represented by Δt. - DVdm (talk) 12:04, 16 May 2015 (UTC)
@DVdm: t is called time, but in the formula, we put the time interval as the value. And, it always remains to the reader that hoe s/he remembers it. I feel the explicit correct/ wrong can't be decided. And that formula: My sir had taught me. I too thought it had no use. But a month later (3rd of May, 2015) in AIPMT, the question was find acceleration. v(x) = k (x^[-2n]) and there I found the use of this formula's use for the very first time. So, shall I include it?
117.248.121.6 (talk) 12:55, 16 May 2015 (UTC)
No, without a source that establishes its notability, let's not include it. - DVdm (talk) 12:58, 16 May 2015 (UTC)
By the way, this was a good find. However, I relinked time to the relevant article Time in physics. - DVdm (talk) 13:26, 16 May 2015 (UTC)

## Identities?

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:

${\displaystyle v=v_{0}+at}$
${\displaystyle s=v_{0}t+{\frac {1}{2}}at^{2}={\frac {v_{0}+v}{2}}t}$
${\displaystyle |v|^{2}=|v_{0}|^{2}+2\,a\cdot s}$

so we have time as a function of v and a, time as a function of distance traveled and velocities:

${\displaystyle t={\frac {v-v_{0}}{a}}={\frac {2s}{v_{0}+v}}}$

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:

${\displaystyle a={\frac {v-v_{0}}{t}}=2{\frac {s-v_{0}t}{t^{2}}}}$

also, distance traveled given velocities and a:

${\displaystyle s={\frac {v^{2}-v_{0}^{2}}{2a}}}$

Others? At least they are now here for my future reference. :-) —Ben FrantzDale (talk) 12:03, 16 October 2015 (UTC)

Could be helpful, provided (1) all the variables are carefully described, and provided (2) we work in one dimension (negative/positive "magnitudes" only, and vectors a, v and v0 parallel), in which case the third equation becomes
${\displaystyle v^{2}=v_{0}^{2}+2as}$
Otherwise it would get too complicated. And of course, as we don't want the readers to have to go verify for themselves, provided (3) we include a source, which should be easy to find. - DVdm (talk) 13:07, 16 October 2015 (UTC)

## Wrong equation in Uniform acceleration?

The equation:

${\displaystyle \mathbf {s} (t)=\mathbf {s} _{0}+\mathbf {v} _{0}t+{\tfrac {1}{2}}\mathbf {a} t^{2}={\frac {\mathbf {v} _{0}+\mathbf {v} (t)}{2}}t}$

seems wrong, since the last part does not contain ${\displaystyle \mathbf {s} _{0}}$. 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 (talkcontribs) 18:33, 27 February 2016 (UTC)

You are right and I've fixed it, -- 20:44, 27 February 2016 (UTC)
I think it is still wrong to talk about initial displacement. Displacement at time t is the difference between position at time t and initial position. Thus initial displacement has to be zero by definition. MLópez-Ibáñez (talk) 00:18, 28 February 2016 (UTC)
I've added words to clarify that, in this case, s is displacement from the origin, not displacement from the initial position. -- 00:36, 28 February 2016 (UTC)

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## category=category

Can the category of this vital article be category?--Dthomsen8 (talk) 19:39, 28 September 2019 (UTC)

## Acceleration is a vector

The following discussion is closed. Please do not modify it. Subsequent comments should be made in a new section. A summary of the conclusions reached follows.
Discussion closed. Reliable sources needed. - DVdm (talk) 11:15, 24 March 2020 (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)

Position isn't really a vector! (Displacement in Euclidean space maybe, but not position).TR 09:25, 17 March 2020 (UTC)
Agreed. My bad. Fixed above. Gandalf61 (talk) 11:23, 17 March 2020 (UTC)
Huh. That's a new one for me — I've seen people get physics wrong in all sorts of ways, but "acceleration is a rank-2 tensor" was not among them until today. XOR'easter (talk) 16:42, 17 March 2020 (UTC)
To the IP, then the answer is simple show some reliable sourcing for this. I don't care what .16 says, I don't care what Gandalf says, I care what's out in published WP:RS already. We've had three hundred years to describe acceleration, a hundred to do so in terms of tensors. If acceleration wants to be described as a tensor of rank 2, then that's either out there already reliably, or it's not for us to add it, it's not for the IP to add it. Yours, kookily Andy Dingley (talk) 17:04, 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 https://wiki.alquds.edu/?query=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)

Please stop making unfounded allegations of sockpuppetry. I have no idea who Dave Burton is, but I resent and refute the accusation that I am his (or anyone else's) sockpuppet. My lengthy Wikipedia history (starting in November 2003) will bear this out. As Andy Dingley (talk · contribs) says above, the way to make your case that acceleration is a rank 2 tensor is to provide reliable sources for that claim. Your ad hominem attacks on myself and other editors have no place in Wikipedia. Gandalf61 (talk) 13:24, 19 March 2020 (UTC)

From your own web pages: https://wiki.alquds.edu/?query=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*... https://wiki.alquds.edu/?query=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.

You didn't watch the video (https://www.youtube.com/watch?v=yx0oql3LIiU), then, Dave, nor did you even look at (understand?) the graphic above (https://i.imgur.com/pkXo2KE.png). In the video, he explicitly refers to the entirety of the right-hand side of the equation as a tensor. In scientific circles, scalars and vectors are referred to as such, and tensors >= rank 2 referred to as tensors... a rank 2 tensor is referred to as simply a tensor, whereas higher-rank tensors generally include their rank when context is not apparent. You'll further note he breaks down the expressions on the right-hand side of the equation: "In any case, this {referring to the left-hand expression} is not a tensor, and this {referring to the Christoffel derivative} is not a tensor, but as a combination {referring to the entirety of the right-hand side of the equation} it's a tensor." And if the right-hand side of the equation in toto is a rank-2 tensor, that means the left-hand side (the symbol for acceleration) is the same rank. The Einstein summation convention means that an index is counted if it is repeated, yes? Z_i is factored out, m is substituted for j and the Christoffel symbol being a non-tensor function leaves indices of k and m... but didn't you state above "he uses only one index" (your words)? Having run out of words and scientific concepts to redefine, are you now redefining *numbers*, Dave?
Now you may look at that equation in that video and claim, "But the covariant and contravariant indices contract, leaving only i! It's a rank 1 tensor!"... not so. As I stated immediately above, the Christoffel symbol is a non-tensor function (in this case, in which m and k are repeated)... and one can only contract tensor rank using opposing covariant and contravariant indices of *tensors*... which is why I wrote prior that the acceleration tensor rank is irreducible. It is, and it remains, rank 2.
As Professor Grinfeld states in the video, dV^i/dt is not a tensor, and the Christoffel symbol is not a tensor. V^k and V^m are tensors... two indices. Rank 2 tensor.
To reiterate, the 'toy model' of 3-D Euclidean space doesn't properly address tensor rank escalation 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. Thus there are generally no dual vectors (the basis of the nabla operator) in 3-D Euclidean space, thus people who use the 'toy model' of 3-D Euclidean space as an approximation of 4-D reality trip themselves up by claiming all derivatives are rank 1 tensors.
https://i.imgur.com/BPC1kL5.png — Preceding unsigned comment added by 71.135.38.240 (talk) 08:44, 22 March 2020 (UTC)

So let's just ask a basic question: Let ${\displaystyle X,Y}$ be vectors and let ${\displaystyle \nabla }$ be the covariant derivative. What kind of object is ${\displaystyle \nabla _{X}Y}$ to you? (If you want to use local coordinates ${\displaystyle X^{i}\nabla _{i}Y^{j}}$.)Nmdwolf (talk) 12:36, 22 March 2020 (UTC)

I assume you're referring to the gradient at Y with respect to X.
∇xY is a tensor whose entities ∂x_i_Y are just standard derivatives of multi-variable function Y with respect to the real variable x_i.
If x=(x_1,x_2,…,x_n) then ∇xY=(∂x_1Y,∂x_2Y,…,∂x_nY) where x_i are variables.
In vector analysis, the gradient of a tensor of rank n *is* a tensor of rank n+1.
Remember, the gradient and its derivative are dual (transpose) to (of) each other. Tensor rank is invariant under transposition.
This is why the gradient of (affine space) scalar position is a rank 1 tensor (velocity). Now apply that to higher tensors, same concept.
The nabla operator acts as a vector-calculus equivalent of a spatial derivative.
As explained here:
https://i.imgur.com/Qd6EuNi.png
Here's a neat visualization of the concept:
Sorry, I don't know how to make superscript / subscript properly here, so I've denoted them with carat and underline. Is there a guide to the notation used?
From *your* *own* *pages*:
https://wiki.alquds.edu/?query=Tensor_derivative_(continuum_mechanics)
Start reading at: "Since the basis vectors do not vary in a Cartesian coordinate system we have the following relations for the gradients of a scalar field phi, a vector field v, and a second-order tensor field S"... note the indices for each resultant. The resultant of the gradient of a scalar is a vector, the resultant of the gradient of a vector is a rank 2 tensor, the resultant of a rank 2 tensor is a rank 3 tensor.
Remember, the gradient and its derivative are dual (transpose) to (of) each other. Tensor rank is invariant under transposition. — Preceding unsigned comment added by 71.135.39.130 (talk) 02:17, 23 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): ${\displaystyle \nabla _{i}Y^{j}=\partial _{i}Y^{j}-\Gamma _{ik}^{j}Y^{k}}$. So once again, now that I clarified my notation, what is the rank of the object ${\displaystyle X^{i}\nabla _{i}Y^{j}}$ if you know that ${\displaystyle X^{i}}$ and ${\displaystyle Y^{j}}$ are rank-1 tensors (i.e. vectors)?

For the math formatting, use < math>< /math> (without the spaces). Inside these tags you can use ordinary LaTeX commands.

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)

Feynman also states in that lecture: "All quantities that have a direction, like a step in space, are called vectors." He's using the generic term, as in "has a magnitude and direction", he's not referring to the quantity as relates to its tensor rank. Nowhere in that Feynman lecture do the words "rank", "order" (as relates to tensor rank), "index" or "indices" occur. But this isn't the first time you've misconstrued quotes... remember you claiming that Rudolf Clausius (who wrote in his Clausius Statement of 2LoT: "Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time.") claimed that 2LoT could be violated? Or that Feynman claimed that 2LoT could be violated?
In vector analysis, the gradient of a tensor of rank n *is* a tensor of rank n+1.
The nabla operator acts as a vector-calculus equivalent of a spatial derivative.
The gradient and its derivative are dual (transpose) to (of) each other. Tensor rank is invariant under transposition.
The derivative and the gradient are essentially the same thing, transposed.
From *your* *own* *pages*:
https://wiki.alquds.edu/?query=Tensor_derivative_(continuum_mechanics)
Let f(v) be a real valued function of the vector v. Then the ${\displaystyle {\boldsymbol {derivative}}}$ of f(v) with respect to v (or at v) is the vector defined through its dot product....
IOW, the derivative of a rank 0 tensor is a rank 1 tensor.
Let f(v) be a vector valued function of the vector v. Then the ${\displaystyle {\boldsymbol {derivative}}}$ of f(v) with respect to v (or at v) is the second order tensor defined through its dot product...
IOW, the derivative of a rank 1 tensor is a rank 2 tensor.
Let ${\displaystyle f({\boldsymbol {S}})}$ be a real valued function of the second order tensor ${\displaystyle {\boldsymbol {S}}}$. Then the ${\displaystyle {\boldsymbol {derivative}}}$ of ${\displaystyle f({\boldsymbol {S}})}$ with respect to ${\displaystyle {\boldsymbol {S}}}$ (or at ${\displaystyle {\boldsymbol {S}}}$) in the direction ${\displaystyle {\boldsymbol {T}}}$ is the second order tensor...
IOW, the 2nd derivative of a rank 0 tensor is a rank 2 tensor.
Let ${\displaystyle {\boldsymbol {F}}({\boldsymbol {S}})}$ be a second order tensor valued function of the second order tensor ${\displaystyle {\boldsymbol {S}}}$. Then the ${\displaystyle {\boldsymbol {derivative}}}$ of ${\displaystyle {\boldsymbol {F}}({\boldsymbol {S}})}$ with respect to ${\displaystyle {\boldsymbol {S}}}$ (or at ${\displaystyle {\boldsymbol {S}}}$) in the direction ${\displaystyle {\boldsymbol {T}}}$ is the fourth order tensor
IOW, the 2nd derivative of a rank 2 tensor is a rank 4 tensor.
Only ${\displaystyle {\boldsymbol {after}}}$ one takes the dot product of the derivative ${\displaystyle {\boldsymbol {with}}}$ ${\displaystyle {\boldsymbol {another}}}$ ${\displaystyle {\boldsymbol {quantity}}}$ (for instance, another completely unrelated vector quantity) does the resultant rank reduce by n-1. But the gradient of a tensor of rank n *is* a tensor of rank n+1, and thus its derivative is as well, because tensor rank is invariant under transposition. Acceleration being a derivative, that means it's rank 2. You're not taking the dot product of the 2nd derivative of affine space position with another quantity, it's a quantity that stands on its own.
So we start at rank 0 (affine space position). Its gradient is rank 0+1 (rank 1), thus its derivative (velocity) is rank 1 ${\displaystyle {\boldsymbol {(does}}}$ ${\displaystyle {\boldsymbol {anyone}}}$ ${\displaystyle {\boldsymbol {argue}}}$ ${\displaystyle {\boldsymbol {that}}}$ ${\displaystyle {\boldsymbol {velocity}}}$ ${\displaystyle {\boldsymbol {is}}}$ ${\displaystyle {\boldsymbol {*not*}}}$ ${\displaystyle {\boldsymbol {a}}}$ ${\displaystyle {\boldsymbol {rank}}}$ ${\displaystyle {\boldsymbol {1}}}$ ${\displaystyle {\boldsymbol {tensor?}}}$). Velocity's gradient is rank 1+1 (rank 2), thus its derivative (acceleration) is rank 2.
Now given that a derivative is dual (transpose) to (of) its gradient (and vice versa), and given that tensor rank is invariant under transposition, that means that the derivative and the gradient have the same tensor rank.
Derivatives are only rank 1 tensors (as many believe) if you consider them in the simple sense (ie: if you disregard all the gradients used to get to the gradient being considered). But given that the gradient and its derivative are dual (transpose) to (of) each other, that's not a valid assumption.
I've stated prior that the gradient and its derivative are dual (transpose) to (of) each other.
The outer product u ${\displaystyle \otimes }$ v is equivalent to a matrix multiplication u·vᵀ. <-- The ᵀ means 'transpose'.
Note the middle dot. When we say we are 'taking the derivative', we generally take the dot product of the gradient and the derivative immediately below it.
Given two vectors:
u = (u_1, u_2, ..., u_m)
v = (v_1, v_2, ..., v_n)
Their outer product u ${\displaystyle \otimes }$ v is defined as the m x n matrix A obtained by multiplying each element of u by each element of v.
(Since I don't know how to make a long square bracket, I'll use a column of square brackets to denote a single long square bracket.)
u ${\displaystyle \otimes }$ v = A =
[u_1v_1 u_1v_2 ... u_1v_n]
[u_2v_1 u_2vv_2 ... u_2v_n]
[u_mv_1 u_mv_2 ... u_mv_n]
${\displaystyle {\boldsymbol {This}}}$ ${\displaystyle {\boldsymbol {increases}}}$ ${\displaystyle {\boldsymbol {resultant}}}$ ${\displaystyle {\boldsymbol {tensor}}}$ ${\displaystyle {\boldsymbol {rank}}}$. 71.135.40.131 (talk) 09:22, 24 March 2020 (UTC)
Please sign all your talk page messages with four tildes (~~~~) — See Help:Using talk pages. Thanks.
Anon, you need a solid wp:reliable source to support your proposal. As Wikipedia is not a reliable source for itself—see wp:circular, please do not mention "*our* *own* *pages*" again. That is useless. Unless you can produce reliable, academic sources that expicitly say that acceleration is a rank-2 tensor, this discussion must be closed—see wp:Talk page guidelines: "Talk pages are not a place for editors to argue their personal point of view about a controversial issue. They are a place to discuss how the points of view of reliable sources should be included in the article, so that the end result is neutral. The best way to present a case is to find properly referenced material." If you fail to produce such a source, and continue to discuss the subject here, then you are disrupting this talk page. I have put a first level formal warning on your currernt IP user talk page User talk:71.135.40.131. Thank you. - DVdm (talk) 08:23, 24 March 2020 (UTC)
Should I go back and quad-tilde all my prior comments?
Well what do you consider to be a "reliable" source? That which is taught in colleges?
University of Texas, Arlington - http://zen.uta.edu/me5312/03.pdf#page=3
California Institute of Technology - http://www.tapir.caltech.edu/~chirata/ph236/lec07.pdf#page=3
Stanford University - http://math.stanford.edu/~jmadnick/R3.pdf
I'm not sure how to get the Wiki-formatted list of references on the Wkipedia page I referenced (the list is blank when I edit the subsection to grab the Wiki-formatted text), so I'll just list them:
J. C. Simo and T. J. R. Hughes, 1998, Computational Inelasticity, Springer
J. E. Marsden and T. J. R. Hughes, 2000, Mathematical Foundations of Elasticity, Dover.
Ogden, R. W., 2000, Nonlinear Elastic Deformations, Dover.
http://homepages.engineering.auckland.ac.nz/~pkel015/SolidMechanicsBooks/Part_III/Chapter_1_Vectors_Tensors/Vectors_Tensors_14_Tensor_Calculus.pdf
Hjelmstad, Keith (2004). Fundamentals of Structural Mechanics. Springer Science & Business Media. p. 45. ISBN 9780387233307. 71.135.40.131 (talk) 09:07, 24 March 2020 (UTC)
Please sign all your talk page messages with four tildes (~~~~) — See Help:Using talk pages.
These 03.pdf, lec07.pdf and Vectors_Tensors_14_Tensor_Calculus.pdf are non-published wp:primary sources, and they don't even say that acceleration is a rank-2 tensor. Deducing that it "must be", would be wp:original research based on improper sources, a double no-no for Wikipedia. An established text book, or an often quoted journal article would be a reliable source, and given the opposition you seem to be facing here, I gather that, in order to achieve wp:consensus on this, you would need at least two of them. See also wp:secondary sources.
I don't have access to the other sources. Do any of the text books explicitly say that acceleration is a rank-2 tensor? - DVdm (talk) 09:11, 24 March 2020 (UTC)
The mathematics emphatically state it. If you can't read the maths, I can't help you. 71.135.40.131 (talk) 09:17, 24 March 2020 (UTC)
We all can read mathematics, but alas, Wikipedia policy forbids wp:original research. If the literature doesn't explictly mention that acceleration is a rank 2 tensor, then it is insufficiently important for Wikipedia to mention it—see wp:UNDUE.
So, I'm afraid that if you do not or cannot provide sources that explicitly say it, we can't help you. Cheers. - DVdm (talk) 09:54, 24 March 2020 (UTC)
This anonymous poster is confusing the gradient with a simple time derivative. Not a single one of his or her sources ever states that the time derivative of a vector is a second rank tensor. I have Tom Hughes' and Ray Ogden's books. They never state that the time derivative of a vector is a second rank tensor. RealOldOne2 (talk) 11:06, 24 March 2020 (UTC)
Let's close this discussion per wp:talk page guidelines. - DVdm (talk) 11:15, 24 March 2020 (UTC)
The discussion above is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

## Acceleration as Rank 2 tensor in accord with Einstein's General Relativity - published references included

The following discussion is closed. Please do not modify it. Subsequent comments should be made in a new section. A summary of the conclusions reached follows.
Discussion closed, wp:original research

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 ${\displaystyle {\ddot {x}}^{i}}$ = Γ${\displaystyle ^{i}}$${\displaystyle _{00}}$, 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: ${\displaystyle R_{u}v}$ ${\displaystyle \propto }$ ${\displaystyle T_{u}v}$.

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 ${\displaystyle R_{u}v}$ - (1/2) ${\displaystyle R_{g}}$${\displaystyle _{uv}}$ 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)

The source does not say that acceleration is a rank 2 tensor. Furthermore, see, for instance, [3], [4], [5]: Christoffel symbols are not tensor quantities.
If you have any further questions, you might go to the wp:Reference desk/Math. If you still want to persue this, see wp:Dispute resolution. I think you already tried the article talk page approach, and since you seem to be unable to establish a wp:consensus here, you'll have to take other steps—see some tips at wp:DP. Third level warning at talk User talk:71.135.43.236 - DVdm (talk) 07:51, 25 March 2020 (UTC)
The discussion above is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

.

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)

## Wiki Education assignment: PHY 381 History of Modern Physics

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— Assignment last updated by Janyahmercedes (talk) 02:52, 26 January 2023 (UTC)

## Wiki Education assignment: 4A Wikipedia assignment

This article is currently 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.

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## The difference between "decelerating" and "negative acceleration"

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)

Our article currently says: "The acceleration of the vehicle in its current direction of motion is called a linear (or tangential during circular motions) acceleration." A bit later it says: "If the speed of the vehicle decreases, this is an acceleration in the opposite direction and mathematically a negative, sometimes called deceleration." So, in this context, the velocity vector of the car points in the positive direction, and deceleration corresponds to negative acceleration. I don't see any problem with any of that. - DVdm (talk) 14:53, 30 March 2023 (UTC)
The article is correctly using this concept but there is no explanation of this difference. I think this would be a relevant explanation to include in the text. Dalmas64 (talk) 15:25, 30 March 2023 (UTC)
Only if we can find a source that explicitly makes the explanation of this difference. Otherwise it amounts to wp:original research. - DVdm (talk) 16:40, 30 March 2023 (UTC)
I guess that this is a clear explanation:
"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."
From a Wiley book:
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 2804:14C:BD80:878D:29C1:5E58:AC46:3A41 (talk) 16:46, 30 March 2023 (UTC)
That source says that it is impossible to know wheter the acceleration is positive or negative, when it is not specified whether the velocity vector of the car points in the positive or negative direction, i.o.w. when the context is not given. But in the article, the context is given. - DVdm (talk) 17:11, 30 March 2023 (UTC)
On Wiley's website, chapter 2 by this author is in completed:
https://catalogimages.wiley.com/images/db/pdf/0471713988.2.pdf
On page 28 there is an important warning:
"As you reason through a problem before attempting to solve it, be sure to interpret the terms '“decelerating' or 'deceleration' correctly, should they occur in the problem statement"
And it confirms that there is indeed a lot of confusion with these terms:
"These terms are the source of frequent confusion, and Conceptual Example 7 offers help in understanding them."
The end of this Example 7 ends by concluding:
"We see, then, that a decelerating object does not necessarily have a negative acceleration."
Note that the author has even used bold for the avoidance of doubt. Dalmas64 (talk) 18:01, 30 March 2023 (UTC)