Wikipedia:Reference desk/Archives/Mathematics/2012 July 19

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July 19

Number that isn't a scalar

As I understand it, a scalar is a number which doesn't depend on the coordinate system. For example, no matter what coordinate system you use, the temperature of an object remains the same, so temperature is a scalar.

Are there examples of numbers which aren't scalars? 65.92.7.148 (talk) 03:12, 19 July 2012 (UTC)

The x-component of a vector is not a scalar. Bo Jacoby (talk) 03:31, 19 July 2012 (UTC).

I think you're talking about the definition in physics. Scalar has another meaning in mathematics, see Scalar (mathematics) and Scalar multiplication — Fly by Night (talk) 03:44, 19 July 2012 (UTC)

Indeed. I think what the OP is really asking about is cases where the mathematical definition and the physics definition disagree. Bo's example is a good one. Speed is another example. --Tango (talk) 03:55, 19 July 2012 (UTC)

Also agree. When physicists say "scalar" they (often) mean a scalar field. So:

1. A temperature distribution is a scalar field that associates a scalar (in the mathematical sense) to every point in space.
2. As Bo says, the x-component of a vector field associates a scalar (in the mathematical sense) to every point in space but is not a scalar field (unless it means "x component relative to a given fixed co-ordinate system" in which case it is a scalar field, but not a very natural one).
3. On the other hand, x-component² + y-component² + z-component² of a vector field is a scalar field because it represents the square of the magnitude of the vector field at each point, and so is invariant under co-ordinate transformations.
4. To be really pedantic, you have to keep unit length the same in the previous example. If you allow transformations that change the unit length then the magnitude of a vector field is a scalar density or relative scalar. Gandalf61 (talk) 08:37, 19 July 2012 (UTC)

Scalars are quantities which do not depend on certain group of symmetry transforms (see Symmetry for mathematical context and Symmetry (physics) for physical). But the term "scalar" is confusing because there is not always clear from the context, which symmetry group is assumed. For example, a function corresponding to what physicists call a scalar field is a scalar corresponding to the space transforms (such as different coordinate systems), and hence is a scalar function from the PoV of differential geometry. But if that physical theory provides gauge symmetry, the function is not a scalar value (with respect to gauge symmetry). For example, complex-valued ψ from Ginzburg–Landau theory is a geometrical scalar, but it is not a gauge scalar – only its absolute value, a real number, is a scalar in both senses. Oppositely, the magnetic field F_1 2 = $\partial _{1}A_{2}-\partial _{2}A_{1}$ from the same theory (2-dimentional + 1 time) is a gauge scalar (F is a 2-form of curvature), but it is not a geometrical scalar (relatively to Lorentz transforms). So, the notion of "scalar" is not absolute. Incnis Mrsi (talk) 11:53, 19 July 2012 (UTC)

Extendible cardinals vs Reinhardt cardinals

How are they related? Or are they the same thing? Rich (talk) 08:08, 19 July 2012 (UTC)

rotating a sphere in higher dimensions

how many dimensions is this tesseract rotating in?

If you rotate a circular object such as a bicycle tire about its axis (the axis at its center and normal to it) a force will be exerted that will tend to make it want to expand outward equally in all directions in the plane normal to the axis of rotation. If you spin an elastic spherical object about an axis passing through its center it will expand outward at its equator in the plane normal to its center of rotation. Would it be mathematically possible to spin an expandable sphere in some higher dimension so that all points on its surface would move outward equally in three dimensions (like a balloon expanding) away from the point at its center, rather than just at its equator? If so, what would this rotation be about, (obviously not about a two-dimensional linear axis) and in what dimension would the rotation have to be? Thanks. μηδείς (talk) 19:57, 19 July 2012 (UTC)

The first thing I will say is that rotation in even and odd dimensional spaces behaves very differently. I think you'll need to consider the cases $\mathbb {R} ^{2n}$ and $\mathbb {R} ^{2n+1}$ separately. — Fly by Night (talk) 21:00, 19 July 2012 (UTC)

You can do this in four dimension. It's enough to cook up a rotation that moves each point on the sphere by the same amount. You can find such a rotation by representing rotations in four dimensions by left and right multiplication by unit quaternions. Sławomir Biały (talk) 21:18, 19 July 2012 (UTC)

Thanks. I have to say that I am not familiar with anything more than one year of high school euclidean geometry and can understand how a tesseract works from Sagan and Flatland and comprehend its rotation from this wonderful animation. But the meaning of the terms $\mathbb {R} ^{2n}$ and quaternion are entirely unfamiliar to me. I will read the link to quaternion. My assumption is that if you can rotate a sphere in two dimensions about a one-dimensional axis in a three dimensional space and have its equator expand in a plane, by analogy you can rotate a sphere in three dimensions about a two dimensional plane in a 4d hyperspace and have its surface expand in three dimensions. Is that right? If so, can someone help me visualize what it is to rotate about a plane? And would it actually be a plane, or just a circle that bisects the sphere? What articles should I read?μηδείς (talk) 21:39, 19 July 2012 (UTC)

You might look at Plane of rotation and Rotations in 4-dimensional Euclidean space. To answer your question in four dimensions an isoclinic rotation has the property you desire: every point on the 3D surface of a sphere in four dimensions rotates at the same speed with such a rotation. The tesseract in the animation is actually rotating isoclinally in 4D, though I don't know if that helps visualise it: both the shape of the tesseract and the projection make hard to see what's going on.

In higher even dimensions you get analogous rotations, I guess also called isoclinic, where every point on a hypersphere is rotating with the same speed. In odd dimensions you always have a non-rotating axis so at least two fixed points.--JohnBlackburne^words_deeds 21:53, 19 July 2012 (UTC)

You can visualize the "isoclinic" rotations in 4 dim easily with the aid of the Hopf fibration, but this sort of exceeds my ability to explain. Sławomir Biały (talk) 22:06, 19 July 2012 (UTC)

Ok, so here's how you can visualize a sphere in four dimensions. At each point of a spherical globe, place a circle (a "clock") flat against the globe (so the face is pointing outward from the center). The four dimensional sphere is then parametrized by picking a point on the sphere and a clock value at that point. Then simultaneous rotation of every clock by the same amount is an isoclinic rotation ("passage of time", if you like). Sławomir Biały (talk) 00:22, 20 July 2012 (UTC)

Isn't that $S^{2}\times S^{1}$? That's not the same as $S^{3}$ is it? (In the same way that $S^{1}\times S^{1}$ is a torus, not a sphere.) --Tango (talk) 17:04, 20 July 2012 (UTC)

No, what I have described is the unit tangent bundle of S^2, which is the 3-sphere. Note that you cannot smoothly orient all of the clocks, so there is no preferred global time. This is something of a small technical point that is likely to be of little interest to the OP though. Sławomir Biały (talk) 19:42, 20 July 2012 (UTC)

And I thought the OP was talking about rotating a regular 3D sphere in a 4 dimensional space $(S^{2}$ embedded in $\mathbb {R} ^{4})$. Even after re-reading, the wording is ambiguous... SemanticMantis (talk) 19:21, 20 July 2012 (UTC)

That's how I read it too. 86.179.1.131 (talk) 20:18, 20 July 2012 (UTC)