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Is it correct to say that kinetic energy is a scalar?

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closed as not a real question by mbq Feb 5 '11 at 13:41

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

This is partially addressed by – David Z Feb 5 '11 at 6:26
@David, I don't even see the word scalar mentioned in that post. – John McVirgo Feb 5 '11 at 13:25
Vague, confusing and attracting discussions. Closed. – mbq Feb 5 '11 at 13:42
@JohnMcVirgo Maybe you have to think of the definition of a scalar? A scalar is a quantity that does not change with coordinate transformations, see . As the kinetic energy is a function of velocity, a different coordinate system will give a different value, particularly if it is moving. So no kinetic energy is not a scalar. It is part of the total energy which total energy is the fourth component of a four vector. – anna v Feb 5 '11 at 15:30
@mbq, the confusion and vagueness doesn't lie with the question, it lies with the differing views of what kinetic energy is. Yes, it is attracting discussion. – John McVirgo Feb 5 '11 at 16:44

Just to add some explicitness to the above answers: take an isolated particle at rest; it's KE is zero. Now switch to a reference frame with relative velocity $\beta$ wrt the particle. In this frame, it's KE is $$KE' = E - mc^2 = (\gamma - 1)mc^2 = \frac{1}{2}mv^2 + O(\beta^4).$$ We see that $KE' \neq KE,$ thus it's not a scalar.

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A scalar is something that doesn't transform under coordinate transformations. A vector is something that transforms "like a vector," in other words, its coordinate transforms are realized as a local multiplication by a linear operator (e.g., a matrix).

In Newtonian mechanics, it's simple to see that kinetic energy, being proportional to the square of a vector (a length) doesn't change under the allowable coordinate transformations (rotations and translations) in Newtonian mechanics, so it is a scalar.

In relativity, we mix up space and time coordinates with coordinate transformations, so there, it transforms as the time component of a 4-vector. This is because energy comes from symmetries associated to time translation. Symmetries associated with space translations are associated with conservation of momentum. Because coordinate transforms in relativity mix up space and time coordinates, we have a single 4-vector whose components are (energy, momentum in x direction, momentum in y direction, momentum in z direction) instead of separate notions of energy and momentum.

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You can have boosts in Newtonian mechanics, too. Kinetic energy is not invariant wrt boosts. – Mark Eichenlaub Feb 5 '11 at 4:57
Boosts are not symmetries of Newtonian mechanics; it does not make much sense to boost a Newtonian system. – Mr X Feb 5 '11 at 5:04
Jeremy, of course that boosts are symmetries of Newtonian mechanics - the Galilean non-relativistic boosts, $x=x+\Delta vt$, parameterized by $v$. They're the simple non-relativistic limits of the Lorentz boosts. Energy transforms nontrivially under them. But it doesn't form a simple linear representation with the momenta etc. in this case. One has to deal with the representations of the non-simple Galilean group which is a semidirect product. Terminologically, in non-relativistic theories, we usually use the word "scalar, vector" with respect to the $SO(3)$ rotations only. – Luboš Motl Feb 5 '11 at 6:57
Oh, yes, of course nonrelativistic boosts are symmetries of Newtonian mechanics! I was just not being clear about what I was saying ;). – Mr X Feb 5 '11 at 7:18
-1 Here's why: 1) "A scalar is something that doesn't transform under coordinate transformations." no. Scalar is something that is invariant with respect to some group action. What you described is a diffeomorphism invariance. 2) "KE is scalar" -> again, this depends on the group. For rotations + translations true, but surely not for the total Galilean group and it is this group that's important here. In particular Sir Isaac Newton with his first law wouldn't agree ;) – Marek Feb 5 '11 at 9:38

I would say it depends on the context. In an Euclidean 3 space it is a scaler (provided its total energy is kinetic) for non relativistic case. In 4D relativistic case, it is a component of a 4 vector.

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I'd modify that because total energy is a component of a four-vector, not kinetic energy. – Mark Eichenlaub Feb 5 '11 at 7:48
+1 for (correctly) stating that it depends on the context. Nevertheless, this isn't a complete answer because there is even more context to be chosen. Namely, whether one takes just the usual Euclidean space ${\mathbb E}^3$, or also includes time to obtain ${\mathbb E}^3 \times {\mathbb R}$. In the latter case the group of symmetries is larger (it includes boosts) and the KE is no longer a scalar under this action. – Marek Feb 5 '11 at 9:43
Thanks both Mark and Marek :) You both are absolutely right :) – user1355 Feb 5 '11 at 9:46
@sb1: Actually, Marek's response seems to render yours meaningless; if you don't include time, then there is no notion of velocity/KE at all; you can't really speak of classical mechanics in that case. In any system with a time axis, you have boosts, thus the KE is not invariant. – Gerben Feb 5 '11 at 10:07
@Gerben: sorry, but the above comment is nonsense. In Newtonian mechanics, time can not be mixed with space in a globally consistent way to form a 4 dimensional continuum. Kinetic energy is invariant under rotation and translation in the 3 space. Time plays an external and universal role unlike relativity. So the statement that K.E. is a scaler in Newtonian mechanics makes a perfect sense. – user1355 Feb 5 '11 at 14:11

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