# Tag Info

6

It depends on how the quantity in question transforms. Almost always, densities in the form of "stuff per unit volume" and generally the "stuff" (like a charge) is a scalar (a number of things - number of elementary charges), but the volume it is contained in is observer dependent, owing to the Lorentz contraction. Therefore the density is ...

0

Why do we need coordinate-free description in the first place? In physics, we have coordinate-free description in the first place. Consider your example: When I am walking from school to my house The description of this consist of you and the school (building) departing from each other, you and your house meeting each other, and, filling in ...

1

A vector is a scalar with direction. So Time can be a vector, but what it means depends on the context. In 1D it has only 2 directions, positive and negative with zero being positive. In 2D it can be an angle between ÷/-Pi radians. And so on. Time can be a single dimension attached to the familiar 3 Euclidian spacial dimensions and in this case it is ...

1

Mathematicians don't talk about rotations or isometries or whatnot because they already know if they're talking about a scalar or something else; a physicist has to determine whether a physical quantity has the properties of a scalar or a vector or something else. The easiest way to do that is to look at transformation laws--to devise some experiment (real ...

1

When talking about scalars, mathematicians usually use your definition, that is, something which doesn't vary with coordinate changes. (Basically, that there's some mapping to the actual points in space, in which the scalar is well defined) When physicists talk about scalars, we usually refer to Lorentz scalars, which requires two things: Invariance under ...

5

This is a great question, with three great answers, and here I am, a bit late to the party. There are two crucial things which the above answers don't seem to address, so I am going to try to give a really simple explanation at those levels. Locality / Manifolds I'm going to come close to giving you one technical definition of a manifold, following ...

0

$$I \equiv \int \mathcal{L}_\lambda d\lambda, \ \ \ \text{for} \ \ \mathcal{L}_\lambda \equiv\sqrt{g_{\mu\nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} } \\ ds = \mathcal{L}_\lambda d\lambda \\ \frac{d}{d \lambda} \left(\frac{dx}{ds}\right) = \frac{d}{d \lambda} \left(\frac{1}{\mathcal{L}_\lambda} \frac{dx}{d\lambda }\right) = - ... 8 Why do we need coordinate free in the first place? Let me tell you about a related experience I have had teaching students. When I ask them to define the scalar product then the vast majority will write down something along the lines of$$ \vec{v}\cdot \vec{w} = v_x w_x + v_y w_y + v_z w_z \qquad (1) $$i.e. a coordinate-based description. However, ... 0 From what I know the easiest way to do this is to go back to the action from which you derive the momentum$$ I = m \int \sqrt{g_{\mu\nu} dx^\mu dx^\nu} = m \int \sqrt{g_{\mu\nu} \frac{dx^\mu}{d \lambda} \frac{dx^\nu}{d \lambda}} d\lambda  Now taking $\lambda'= \lambda'(\lambda)$, thus \$d\lambda'= \frac{d \lambda'}{d\lambda} d\lambda \ \to \ d\lambda = ...

5

Let me first tell you that what you read is very vague. In GR the laws of physics are assumed to be independent of the observer. An observer is represented by the a reference frame the observer uses to measure physical phenomena. There are a set of coordinate transformations that relates observables for different observers. Say for example the speed of a ...

19

That's a very good question. While it may seem "natural" that the world is ordered like a vector space (it is the order that we are accustomed to!), it's indeed a completely unnatural requirement for physics that is supposed to be built on local laws only. Why should there be a perfect long range order of space, at all? Why would space extend from here to ...

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