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

Qmechanic
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John McAndrew
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  • This is partially addressed by http://physics.stackexchange.com/q/1368 – David Z Feb 05 '11 at 06:26
  • @David, I don't even see the word scalar mentioned in that post. – John McAndrew Feb 05 '11 at 13:25
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    Vague, confusing and attracting discussions. Closed. –  Feb 05 '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 http://en.wikipedia.org/wiki/Scalar_%28physics%29 . 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 05 '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 McAndrew Feb 05 '11 at 16:44
  • Maybe He didn't like a comment of me somewhere? – Georg Feb 05 '11 at 17:21
  • @Georg, I don't think it's you, since you gave a relevant answer people can vote on. It's probably because of the comment I made earlier on the lines of " My question has a -1, someone has deleted their answer in embarrassement, wrong answers are being marked up - great!", together with the text of the message asking for a discussion. I've deleted these, and so I think it should be reopened as a question that people can learn from. The comments from Marek and Lubos are particularly enlightening. – John McAndrew Feb 05 '11 at 19:23
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    This is a real question. I teach physics. My students have to answer this very question on tests every quarter. It is one they find difficult because kinetic energy comes from velocity and velocity is not a scalar. I would like mbq to reconsider the vote to close. – Carl Brannen Feb 05 '11 at 19:56
  • @John (2nd comment): sure, the word "scalar" doesn't appear in that post, but it does ask how KE transforms under boosts, and that is precisely what you need to figure out to decide whether a quantity is a (Lorentz) scalar. – David Z Feb 05 '11 at 21:58
  • @anna v: I think you have to think of the definition of a scaler. A scaler is a quantity that does not change with rotation and translation and not with "coordinate transformation" as you have written. A different coordinate system may indeed give different K.E. but that does not make it a vector in the 3D Euclidean space in non-relativistic case. You can define a quantity scaler or vector in non-relativistic cases on the basis of $SO(3)$ rotations only. –  Feb 06 '11 at 03:56

3 Answers3

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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.

Gerben
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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 05 '11 at 07:48
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    +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 05 '11 at 09:43
  • Thanks both Mark and Marek :) You both are absolutely right :) –  Feb 05 '11 at 09:46
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    @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 05 '11 at 10:07
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    @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. –  Feb 05 '11 at 14:11
  • @Marek, I would say that it doesn't depend upon the context at all. A 3-vector is a 4-vector for v = 0 where t = t'. – John McAndrew Feb 06 '11 at 15:34
  • @John: sorry, what? 3-vector is never a 4-vector. First transforms under SO(3), second under SO(1,3), so they are completely different objects. You can reduce the group to some subgroup but that is precisely what context means ;) – Marek Feb 06 '11 at 16:14
  • @Marek yes I can see you're right. But, would you agree that if a Galilean 4-vector was used instead of a 3-vector, there wouldn't be this problem of context? – John McAndrew Feb 07 '11 at 05:44
  • @sb1 Kinetic energy isn't a scalar in Relativistic mechanics so how can it gradually becomes a scalar as v -> 0 in Newton mechanics? – John McAndrew Feb 07 '11 at 05:56
  • @John McVirgo: In Newtonian context, the relevant group is $SO(3)$ and K.E. doesn't change under this group action. You can't use a Galilean 4-vector since there is no such thing. Why? see my earlier comment to Gerben. "how can it gradually becomes a scalar as v -> 0 in Newton mechanics?" It never becomes. You can always use relativity for small $v$ as well. So in general context it is a component of a 4-vector. Only when we talk about Newtonian mechanics, disregarding SR, we can treat it as a scaler. Hope, you got the point. –  Feb 08 '11 at 10:21
  • @sb1, a Galilean 4-vector is to a relativistic 4-vector what the Galilean transformations are to the LTs - cases for v<<c. To suggest there is no such thing as a Galilean 4-vector is to suggest there is no such thing as Galilean transformation, which is wrong. For v << c it works, try it. Also, there are two different definitions of scalar; one under so(3), the other under so(1,3) and the problem is deciding when to use one and not the other which everone can agree on. If there is ambiguity, then state which one. And that's the point. – John McAndrew Feb 09 '11 at 14:18
  • @John McVirgo: apologies, but you are almost hopelessly confused. You are so very ignorant about basic things, I have no intension to get involved into any debate with you any more. –  Feb 09 '11 at 14:42
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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.

Mr X
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    You can have boosts in Newtonian mechanics, too. Kinetic energy is not invariant wrt boosts. – Mark Eichenlaub Feb 05 '11 at 04:57
  • Boosts are not symmetries of Newtonian mechanics; it does not make much sense to boost a Newtonian system. – Mr X Feb 05 '11 at 05:04
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    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 05 '11 at 06:57
  • 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 05 '11 at 07:18
  • Kinetic energy isn't really a component of a four-vector. Total energy is. – Mark Eichenlaub Feb 05 '11 at 07:48
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    -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 05 '11 at 09:38
  • @Marek, well, yeah, but the OP probably doesn't know about groups, and all the groups we deal with in physics can be realized as matrix groups, which the group action can be realized in as a change of coordinates. And what I described is only really diffeomorphism invariance if there's a nontrivial dependence on position, in which case my statement is that scalars transform as rank zero tensors, and vectors transform with one jacobian. And at any rate, kinetic energy is a scalar under SO(3) action realized by matrix multiplication which is clearly what I was thinking of in my first paragraph – Mr X Feb 06 '11 at 05:07
  • @jeremy, it would have be clearer to me if you had said "a scalar is a quantity whose value doesn't change under a coordinate transformation" rather than something that doesn't transform. Also, it's the total energy rather than kinetic energy that transforms as the component of a four-vector. – John McAndrew Feb 06 '11 at 15:12
  • @john, that's pretty much the same as what I said, but it can be confusing to think about the "value" of things under coordinate transforms like that... And yes, it is the total energy, but for the simple case of a single object with nothing interesting going on, which is most likely what the OP was thinking, it's a reasonable answer (and considering his question was only one sentence, erring on the side of less sophistication seems like the thing to do). – Mr X Feb 08 '11 at 06:58