0
$\begingroup$

The scalar product in special relativity is given by

\begin{equation} V \cdot W = V^{\mu} g_{ \mu \nu} W^{\nu} \end{equation} and the components of the vectors $V^{\mu}$ and $W^{\nu}$. With the metric

\begin{equation} \eta_{ \mu \nu} = \mathrm{diag} (-1,1,1,1) \end{equation} if we using Cartesian coordinates.

a) First of all, is this always valid? If we change coordinates and $g \neq \eta$ is it still valid?

Then we can calculate that scalar product given the components of $V$ and $W$ and the metric. If we say, for example, that

\begin{equation} V = \begin{pmatrix} 1 \\ 0 \\ 2 \\ 1 \\ \end{pmatrix} \end{equation}

b) What is the meaning of this expression? Is this vector implicitly meaning a vector starting from the origin of the coordinates, assuming Cartesian coordinates and with components in the $\hat{t}$, $\hat{x}$ etc directions the one specified above? So when taking the scalar product in $(1)$ we take it in the origin of the coordinates and with the Minkowski metric?

Then we can have a change of coordinates and therefore the metric changes as well. For example spherical coordinates where the metric is

\begin{equation} g_{\mu \nu} = \mathrm{diag} (-1, 1, r^2, r^2 \sin \theta) \end{equation} The components of our vectors should change according to

\begin{equation} V^{\prime \mu}=\frac{\partial x^{\mu}}{\partial x^{\nu}} V^{\nu} \end{equation} So the numbers that we obtain as components of the vectors are now saying how to build the vector with these "new" axes $\hat{t}$, $\hat{r}$, $\hat{\theta}$ and $\hat{\phi}$ so giving the components of the vector in these new axes. If we base the coordinate transformation on the vector $V$ than it will have $0$ component in direction $\theta$ and only in $\hat{r}$ but $W$ will have both.

c) Now we want to calculate the scalar product as in $(1)$. We will have to use $r=0$ in the metric right? And we expect to get the same result as we obtained with Cartesian coordinates.

d) If we now have two vectors in a different point instead one in the origin and $W$ in the point $x_P$ where we can give the coordinates of the point in Cartesian coordinates for example. How can we perform the dot product? The vectors do not have the same $r$ anymore.

e) In Cartesian coordinates we can still do it I think. We just transport $W$ to the origin and we get the product. Is it the same in polar coordinates?

f) If $V$ is now in a point $x_Q$. We can transport $W$ in $x_Q$ and do the product, but in polar this product depends on $r$ and if instead we transport $V$ in $x_P$ we have a different $r$ and then a different result. Where is the mistake in this reasoning?

$\endgroup$
2
$\begingroup$

a) Yes, this equation is valid even when we are in a coordinate system in which the metric is not the Minkowski metric. In fact, since the inner product of two vectors is a scalar (and thus, a coordinate invariant quantity) you will get exactly the same valued in a different coordinate system.


b) You have to be careful here, Minkowski space can be treated as $\Bbb R^4$ equipped with an inner product (making it a vector space), but if you want to talk about vectors at different points in Minkowski space you need to treat it as a manifold. On a manifold each point has a tangent space which has vector space structure which allows us to justifiably talk about vectors that aren't "at the origin" or "at the same point".

Even here we again have to be careful, because we are not necessarily allowed to compare vectors in different tangent spaces if we are working in, say, spherical coordinates (or if our spacetime has curvature).


c) The scalar product is invariant under coordinate transformations.


d), e), f) These issues again relate to the fact that we are talking about Minkowski space as a manifold, not just as a vector space (for which the concept of having two vectors at two different points is not defined). Also as I have said above you cannot always compare vectors living in different tangent spaces with a manifold alone, this requires extra structure called a connection. See also the more familiar notion in physics of a covariant derivative.

$\endgroup$
4
  • $\begingroup$ But since the Minkowski space is a vector space itself this allows us to make an identification of vectors in tangent spaces at points (events) with vectors (points so events) in Minkowski space, right? $\endgroup$
    – TheoPhy
    Nov 16 '20 at 13:01
  • $\begingroup$ More or less, yes. The tangent spaces to the Minkowski space manifold carry the usual vector space structure that is introduced as "Minkowski space". $\endgroup$
    – Charlie
    Nov 16 '20 at 13:13
  • $\begingroup$ Is then the statement in (b) correct? $\endgroup$
    – TheoPhy
    Nov 16 '20 at 13:44
  • $\begingroup$ I'm not actually sure what you're getting at in (b). You have written down the transformation law for a rank-1 contravariant tensor correctly though (except for the prime on the left hand side which shouldn't be there). $\endgroup$
    – Charlie
    Nov 16 '20 at 13:55
1
$\begingroup$

As Charlie explained, one way to treat Minkowski spacetime is as manifold, which is what one does when he computes metric components as function of coordinates or when one uses the transformation formula $$V'^\mu=\frac{\partial x'^\mu}{\partial x^\nu}V^\nu.$$

Another way is to treat it as affine space, which is what you are trying to do in your question. Affine space is a set together with accompanying vector space and operation that takes point, vector and produces another point: $$P_1+\vec{v}=P_2$$ This is how vectors are usually introduced in high school, as arrows from one point to another.

The change of coordinates does not play a role here however. The vector is fully specified by points at its tip and at its tail. Nowhere do the coordinates come in this abstract definition.

The problem comes when one starts to treat points using coordinates. The operation in cartesian coordinates is simple enough. If the vector has certain components in basis associated with given cartesian coordinates, then the new point is simply given by adding components of initial point with components of the vector. In curvilinear coordinates however, the rule is not so simple and straightforward.

First of all, there is no vector basis associated with these new coordinates. On an intuitive level, this makes sense as curvilinear coordinates change their direction and "speed" and thus you cannot associate only one arrow with them. So to write down the component version of the equation $P_1+\vec{v}=P_2$, you do not even know how to decide which basis vectors to use to get components of the vector $\vec{v}$. The affine space simply tells you to forget curvilinear coordinates and work with cartesian coordinates as these coordinates are adapted to structure of affine space.

The manifold viewpoint gets around this problem by demanding that the formula $P_1+\vec{v}=P_2$ can be used only in an infinitesimal neighborhood of the point $P_1$ and scales the vector $\vec{v}$ down by some scalar $\epsilon$. I.e. in manifold formulation each point has accompanying vector space, which we can use to get point at the tip of the vector: $P_1+\epsilon\vec{v}=P_2$, for ininitesimal value of $\epsilon$.

As Charlie was explaining, in full generality manifold does not enforce the ability to compare two vectors at different points. The vector spaces at different points are simply fully independent.

Minkowski spacetime however is known to have structure of affine space. So there exists one canonical isomorphism between all the vector spaces at different points. Simply put, all the vector spaces can be identified. Using curvilinear coordinates can then be interpreted as demand that at different points you are using decomposition to different basis.

b) What is the meaning of this expression? Is this vector implicitly meaning a vector starting from the origin of the coordinates, assuming Cartesian coordinates and with components in the t^, x^ etc directions the one specified above? So when taking the scalar product in (1) we take it in the origin of the coordinates and with the Minkowski metric?

As I said, in Minkowski spacetime there are two interpretations. One is, that this are components of vector living at some point as decomposed to some basis at that point. The basis is most probably given by "direction and speed" of coordinate curves at the point. This is fully manifold-view point. Another intepretation, that makes use of affine structure of Minkowski spacetime tells you, that these are components of the vector decomposed w.r.t to basis defined at the point P. The difference between the two interpretation is not actually very big. In one interpretation both the basis and vector live at the point $P$, in the other interpretation, they live in the "one true" vector space but the decomposition happens according to the point in question.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.