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I currently study special relativity and some authors write stuff like: $$ r^\mu = \left(ct, \vec x\right) $$

This is awful since $\mathsf r$ is a vector, and $r^\mu$ ist just a single component of that vector.

Now I am wondering whether something I write occasionally is similarly wrong: As far as I know, I can define a function $f$ like $f\colon x \mapsto x^2$ or $f(x) = x^2$. Sometimes I just write $f = x^2$. Would that be incorrect in a strict sense?

Did
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Martin Ueding
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    Yes, unless it refers to the associated polynomial $x^2 \in K[x]$. – Siméon Jan 15 '13 at 09:22
  • @StefanHansen: I think you meant “ambigious“ when you wrote “unambiguous”. – Martin Ueding Jan 15 '13 at 09:51
  • I updated the question: What happens to functions of other functions that have the same variable? – Martin Ueding Jan 15 '13 at 09:57
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    it does not make sense to edit your post and add an additional question after a lot of people hafe written ansers adn comments. Now it is completely unclear to what questions the statements refer to. I think you should rollback your edit and pose another question. – miracle173 Jan 15 '13 at 10:05
  • It is certainly quite common in physics to omit function arguments; but note that the two notations you give for kinetic energy are not equivalent: E.g. for a rocket, the complete expression would be $E(t)=\frac{1}{2}m(t)v^2(t)$ since mass is not constant. – Ansgar Esztermann Jan 15 '13 at 10:07
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    The notion $f=x^2$ is ambiguous. It could refer to both the polynomial $x\mapsto x^2$ and the constant function $y\mapsto x^2$. I would recommend that you use one of the two notions you mention yourself. – Stefan Hansen Jan 15 '13 at 10:12

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Yes, strictly speaking it would mean that $f$ is the square of whatever $x$ is. So if $x$ was a real number then $f$ would be the square of that number.

Johan
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Yes, that would be incorrect. If $f$ is - say - a continous function from $\mathbb{R}$ to $\mathbb{R}$. Then $f$ would be an element of $C(\mathbb{R},\mathbb{R})$ and $x$ as well as $x^2$ would be an element of $\mathbb{R}$. It makes no sense to identify them. $f$ is a function, $f(x)$ is the value of this function at a point $x$, namely $x^2$.

mjb
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  • Not quite - $x$ is not some value, but instead a variable. In that case, $x^2$ can be seen as an element of $\mathbb{R}[x]$, the ring of polynomials, which is isomorphic to a subset of the ring of continuous functions. – akkkk Jan 15 '13 at 10:30
  • @akkkk This is just another way to view it. Yet, if $f: \mathbb{R} \to \mathbb{R}$ one can (for $x \in \mathbb{R}$) still view $f(x)$ as an element of $\mathbb{R}$. – mjb Jan 15 '13 at 14:05
  • So why is your view correct? Under my view, the notation is correct (albeit confusing, since I am leaving away all the isomorphisms), but under your view, it is not. – akkkk Jan 15 '13 at 14:30
  • @akkkk I'm not sure I understand what you mean. As Ju'x pointed out in his comment to the original question: for a function $f : \mathbb{R} \to \mathbb{R}$ it is correct to write $f=x^2 \in \mathbb{R}[x]$, but not $f(x)=x^2 \in \mathbb{R}[x]$, since clearly $f(x) \in \mathbb{R}$ for $x \in \mathbb{R}$, no? – mjb Jan 15 '13 at 16:47
  • if, by $x$, the /variable/ is meant, then I do not see how that would be incorrect. You are right however, that (in words) "plugging in a value" produces a new "value". – akkkk Jan 15 '13 at 17:00
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The notation $r^\mu = \left(ct, \vec x\right)$ is not awful. This notation is defined in relativity. Upper indices denote contravariant components.

Džuris
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  • The problem I see with it that $r$ is the four-vector and $r^\mu$ is one of its components. It does not make sense to me to refer to a single component $\mu$ while specifying a tuple. It is kind of like writing $\vec a = a_i$, I think. – Martin Ueding Jan 16 '13 at 15:31