# Differentiation in Minkowski Spacetime

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1. Sep 5, 2014

### branch

Hello everybody,

I'm currently reading the book Special Relativity in General Frames by Gourgoulhon. There, Minkowski Spacetime is introduced as an affine space $\mathscr{E}$ over $\mathbb{R}$ with a bilinear form g on the underlying vector space E that is symmetric, nondegenerate an has signature (-,+,+,+) (there is actually more to complete the definition of Spacetime but it's not relevant here).

The author later defines (p.30 for those that have the book) that massive particles are represented by a piecewise twice continuously differentiable curve $\mathscr{L}$ of Minkwowski Spacetime such that any vector tanget to $\mathscr{L}$ is timelike. (It is not important to know what timelike is to understand the question.)

I wonder how exactly twice continously differentiable is understood in the context affine spaces, especially when the underlying vector space has no norm (g does not induce a norm).

Given a parametrization $\varphi$ of $\mathscr{L}$ he then defines the derivative vector of $\varphi$:
$$\forall \lambda \in \mathbb{R} \quad \vec{v}(\lambda):=\lim_{\epsilon \rightarrow 0} \frac{1}{\epsilon}(\varphi(\lambda + \epsilon) - \varphi(\lambda))$$

Here I basically have the same question: how do you define the limit on the vector space E ($(\varphi(\lambda + \epsilon) - \varphi(\lambda)) \in E$) if there is no norm?

I mean, the simplest thing is probably to simply choose an origin and a basis and then differentiate the coordinates but I wonder if there is a frame independent way to define it (especially since the author makes it seem this way).

Edit: I just realized this should probably be posted in the Topology and Analysis forum, if this is the case I apologize and ask a moderator to move the thread.

Last edited: Sep 5, 2014
2. Sep 5, 2014

### Incnis Mrsi

When one speaks about differentiation in a pseudo-Euclidean space, the standard topology (of ℝ4 in Minkowski case) is assumed. It is no different from Euclidean and affine. For limits, you need topology only, no metric.

3. Sep 5, 2014

### branch

Hi,

thanks for your answer. Not sure if I understood you correctly. But to define the "standard topology" on $\mathscr{E}$ I still have to choose an origin and a basis, right? So this would amount to the same thing like I said before:

4. Sep 5, 2014

### Fredrik

Staff Emeritus
I think the hard part is to understand why the difference in the numerator makes sense. An affine space can be defined as a triple $(V,S,\phi)$, where $\phi:V\times S\to S$ is required to satisfy a number of conditions. One of them says that $\phi(0,s)=s$ for all $s\in S$. Another says that for all $s,s'\in S$, there's a unique $v\in V$ such that $\phi(v,s)=s'$. (I don't remember if there are any more. I know it's a very short list of conditions). The difference s'-s is defined by saying that s'-s is equal to this particular v. So the difference of two elements of S is a vector, even though no addition operation is defined on S. If we define the notation v+s by $v+s=\phi(v,s)$, then the definition ensures that the following equivalence holds
$$v+s=s'\ \Leftrightarrow\ v=s'-s.$$ This definition of difference ensures that your $\displaystyle\frac{\varphi(\lambda+\varepsilon)-\varphi(\lambda)}{\varepsilon}$ makes sense.

Regarding the topology, you can use the left action $\phi$ and an arbitrary $s\in S$ to define the map $\phi_s:V\to S$ by $\phi_s(v)=\phi(v,s)$ for all $v\in S$. This identifies elements of V with elements of S. Then you can use any norm on V. There's a theorem that ensures that all norms on finite-dimensional spaces are equivalent in the sense that they lead to the same sets being called "open". This means that they all agree about limits.

I don't have a handy reference, but if you google for "norms are equivalent", I'm sure you will find something useful.

Last edited: Sep 5, 2014
5. Sep 5, 2014

### Incnis Mrsi

Standard topology is an affine invariant. Choose whichever origin and basis for coordinates, it won’t change. This topology is easy to get from coordinates, but mathematically, you do not need coordinates to define the topology at all.

If you have a function from real numbers to pseudo-Euclidean vectors (see the formula for derivative), componentwise continuity is the same thing as continuity in the standard topology. The same for curves (functions from real numbers to pseudo-Euclidean space itself). In this simple case you can check directly that affine transformations preserve componentwise continuity. More generally, it is topological structure that defines limits in a frame-independent way. You have not necessarily know how a topological structure is defined; just believe that this piece of mathematics has no flaw.

6. Sep 5, 2014

### branch

This was clear to me. Actually, the book does not use the notation where you write differences, but I thought to people who know affine spaces this notation is clear. Later I explicitely mentioned that $(\varphi(\lambda + \epsilon) - \varphi(\lambda)) \in E$, where E is the underlying vector space. Also I think in your definition of the domain of $\Phi$ there is one S too much.

This is something I did not have in mind, but it helps alot.

@Incnis Mrsi: That the topology is independent of the origin and basis chosen is clear to me. but it's a bit like proving a theorem with some arbitrary basis when there is a way to prove it without using a basis at all.
I wonder however how you do not need coordinates to define the topology at all. Maybe I have to research on the subject a bit myself.

Last edited: Sep 5, 2014
7. Sep 5, 2014

### Incnis Mrsi

Standard topology in invariant terms

Given an abstract vector or affine real space, as an option invented in a couple of minutes, you can rely on the subbase of all (open) sets UL, x0 := { x from V | L (x − x0) > 0 }, where L ∈ V′ and x0 ∈ V. For a vector space V, V′ is the dual space (for an affine space you have to use the dual to the corresponding vector space). Hence, you introduced the standard topology for finite-dimensional real spaces in completely invariant terms.

8. Sep 5, 2014

### branch

That's interesting and answers all of my questions so far. Thanks.

9. Sep 5, 2014

### Fredrik

Staff Emeritus
Yes, that was a typo. It should be $\phi:V\times S\to S$, not $\phi:V\times S\times S$. Domain $V\times S$. Codomain $S$. I have edited my post now.

If you want to know more about these functions, the key term to look up is "group action". (You don't need to learn more about them for your current purposes). This $\phi$ is a left group action on the set S. The group involved in this is the vector space V. (Its addition operation makes it a group).

Every vector space has a basis. Every basis B can be used to define a norm $\|\cdot\|_B$, which can be used to define a metric $d_B$, which can be used to define a topology $\tau_B$, which is actually independent of B. This is the topology you're using.

I don't know if this topology can be defined without mentioning bases, but it doesn't matter. You don't have to be given a basis, since every vector space has one. So you can just start by letting B be an arbitrary basis and proceed from there. The end result will be independent of B anyway.

10. Sep 5, 2014

### WWGD

Can you justify this by saying that translation is a homeomorphism?

11. Sep 6, 2014

### Incnis Mrsi

A translation is, and, importantly, any affine transformation (invertible, of course) is too. That’s the same thought.