Minkowski metric, scalar product, why the minus sign?

[peter46464]

In Schutz's A First Course in General Relativity (second edition, page 45, in the context of special relativity) he gives the scalar product of four basis vectors in a frame as follows:

$$\vec{e}_{0}\cdot\vec{e}_{0}=-1,$$
$$\vec{e}_{1}\cdot\vec{e}_{1}=\vec{e}_{2}\cdot\vec{e}_{2}=\vec{e}_{3}\cdot\vec{e}_{3}=1,$$
with all other permutations equalling zero. On page 36 he gives the components of ##\vec{e}_{0}## as ##\left(1,0,0,0\right)##. Why then is the scalar product of ##\vec{e}_{0}## with itself equal to ##-1## and not ##+1##? In my innocence I thought you found the scalar product by simply multiplying together the components. What am I missing?

 

[Orodruin]

This is not a regular scalar product. Scalar products in Minkowski space are different, the metric is not positive definite but has signature (1,3). In other words, Minkowski space is pseudo-Riemannian. This is the major difference to Euclidean space and the sign difference is essential for the scalar product to be invariant under Lorentz boosts instead of regular rotations.

 

[Dale]

I thought you found the scalar product by simply multiplying together the components

If the components are ##(t,x,y,z)## then the scalar product is ##-t^2+x^2+y^2+z^2##. The timelike component is distinguished from the spacelike components by exactly the fact that you noticed.

 

[vanhees71]

It's not a scalar product at all, because it's not positive definite. Physicists call it a scalar product since it has all algebraic rules in common with a scalar product; it's only lacking the positive definiteness. It's called a symmetric bilinear form. Given an arbitrary basis ##\vec{e}_{\mu}##, a symmetric bilinear form is a function ##V \times V \rightarrow \mathbb{R}##, ##\vec{x},\vec{y} \mapsto \vec{x} \cdot \vec{y}##, where ##V## is a real vector space, is completely characterized by a matrix
$$g_{\mu \nu} = \vec{e}_{\mu} \cdot \vec{e}_{\nu}.$$
By assumption it obeys ##\vec{x} \cdot \vec{y} = \vec{y} \cdot \vec{x}## (that's, why it's called a symmetric bilinear form). One can show there exists always a basis ##\vec{e}_1,\ldots,\vec{e}_d## (##d##: dimension of the vector space) such that ##(g_{\mu \nu})=\mathrm{diag}(1,1,\ldots,-1,-1,\ldots,0,0,\ldots)## (with of course ##d## entries in total).

A bilinear form is called non-degenerate, if there are no zeroes in the above formula, and a symmetric non-degenerate bilinear form is called a "fundamental form" of the vector space. It's characterized by the number of ones in the above diagonal matrix (then the remaining entries are of course just ##-1##'s). A Lorentz bilinear form (or Lorentz product) has (in your convention) 3 ones and 1 ##-1## in the diagonal matrix, and that's precisely what you need for special relativity: It's a 4D real vector space with a fundamental bilinear form with the signature (3,1) (giving the number of ##+1##'s and ##-1##'s in the diagonal matrix).

Why does one need now precisely this form? That's, because of Einstein's postulate about the invariance of the speed of light in a vacuum under "boosts", where boosts are such transformations of the spatial coordinates and the time which describe the transformation from one inertial reference frame to another, without rotating the spatial axis, i.e., the one is simply moving with a constant velocity against another.

To figure out the transformations, describing such a boost was done by Einstein in 1905 in a quite cumbersome way (but you should try to understand it, because it's very physical compared to the one I summarize here). This derivation is much simplifight by the ingeneous insight by Minkowski of 1908, that one can describe special relativistic spacetime as a vector space (when fixing just an arbitrary spacetime point as an "origin") with exactly a Lorentzian fundamental bilinear form! Choosing an inertial frame then means that it's exactly one of the preferred bases which make the scalar product look simple, i.e.,
$$\vec{e}_{\mu} \cdot \vec{e}_{\nu} = \eta_{\mu \nu} = \mathrm{diag}(1,1,1,-1).$$
The vector components with respect to such a basis are then
$$(\boldsymbol{x},c t),$$
where ##\boldsymbol{x}## is the position vector of some point and ##t## the time an observer measures with an ideal clock who is at rest relative to this coordinate system.

Now think about a flashlight, located in the origin of the coordinate system, which is switched on and off again after a short while. Then a shperical light-pulse will spread out, which is located at the spherical shell given by the equation ##\boldsymbol{x}^2=c^2 t^2## (with the usual 3D Euclidean scalar product for the spatial 3D vector) since after a time ##t## the pulse, traveling with the speed of light in a vacuum ##c##, has reached a distance of ##c t## from the origin. This we can of course write in terms of the Minkowski product as
$$\vec{x} \cdot \vec{x}=\boldsymbol{x}^2-c^2 t^2=0.$$
You can now characterize the Lorentz boosts as such linear transformations (or the corresponding change of the four basis vectors) that don't rotate the spatial axes against each other, for which the Minkowski product reads precisely the same. This socalled "covariant" notation makes life in relativity much easier than any other notation.

For more on this (but with the other convention of the sign of the Minkowski product), see

http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[martinbn]

In Schutz's A First Course in General Relativity (second edition, page 45, in the context of special relativity) he gives the scalar product of four basis vectors in a frame as follows:

$$\vec{e}_{0}\cdot\vec{e}_{0}=-1,$$
$$\vec{e}_{1}\cdot\vec{e}_{1}=\vec{e}_{2}\cdot\vec{e}_{2}=\vec{e}_{3}\cdot\vec{e}_{3}=1,$$
with all other permutations equalling zero. On page 36 he gives the components of ##\vec{e}_{0}## as ##\left(1,0,0,0\right)##. Why then is the scalar product of ##\vec{e}_{0}## with itself equal to ##-1## and not ##+1##? In my innocence I thought you found the scalar product by simply multiplying together the components. What am I missing?

On the same page, just before the example, he gives the definition of the scalar product that he uses. Here it is:

We define the scalar product of ##\vec{A}## and ##\vec{B}## to be ##\vec A \cdot \vec B = −A_0B_0 + A_1B_1 + A_2B_2 + A_3B_3## (2.26)
​

So, if ##\vec A = \vec e_0=\left(1, 0, 0, 0\right) ## and ##\vec B = \vec e_0=\left(1, 0, 0, 0\right) ## what do you get?

 

[Nugatory]

It's not a scalar product at all, because it's not positive definite. Physicists call it a scalar product since it has all algebraic rules in common with a scalar product;

Vanhees is absolutely correct here, both that it's not a scalar product and that physicists call it that anyway. In practical terms this means that everyone else in this thread is going to keep on calling it a scalar product.

 

[peter46464]

So, as I understand the answers given here, ##\vec{e}_{0}\cdot\vec{e}_{0}=-1## is not a scalar product but a symmetric bilinear form. And the definition of the metric $$g_{\mu\nu}=\vec{e}_{\mu}\cdot\vec{e}_{\nu}$$involves not scalar products but a bunch of (please excuse the technical language) symmetric bilinear forms. But what is the rule that allows me to get from ##\vec{e}_{0}=\left(1,0,0,0\right)## to ##\vec{e}_{0}\cdot\vec{e}_{0}=-1##? In layman's language, would it be fair to say that the rule for finding this mysterious symmetric bilinear form is (1) multiply together the two components of ##\vec{e}_{0}=\left(1,0,0,0\right)##, and then (2) multiply by the negative coefficient of the Minkowski metric?

 

[martinbn]

Did you not see my post, and more importantly did you not see the definition on page 45 of the book!?

 

[Orodruin]

So, as I understand the answers given here, ⃗e0⋅⃗e0=−1e→0⋅e→0=−1\vec{e}_{0}\cdot\vec{e}_{0}=-1 is not a scalar product but a symmetric bilinear form. And the definition of the metric

gμν=⃗eμ⋅⃗eνgμν=e→μ⋅e→ν​

g_{\mu\nu}=\vec{e}_{\mu}\cdot\vec{e}_{\nu}involves not scalar products but a bunch of (please excuse the technical language) symmetric bilinear forms. But what is the rule that allows me to get from

The metric (or more correctly if we are being precise, the pseudo-metric) is the bilinear form - there is only one. ##\vec e_0\cdot \vec e_0 = -1## is not a bilinear form, it is the value of the bilinear form when plugging in ##\vec e_0## in both arguments.

 

[peter46464]

martinbn - yes, I saw your post and the definition. I assumed, when you said "what do you get?" that that was a genuine question, ie that you didn't know the answer and were as puzzled as I am. Whoops, please don't take offence! A definition isn't really an answer though, is it? Am I correct in thinking that in order to find the symmetric bilinear form you need to multiply the components together along with a symmetric matrix, aka the Minkowski metric? At my high school maths level, that's a rule I can understand. Thanks.

 

[Orodruin]

Am I correct in thinking that in order to find the symmetric bilinear form you need to multiply the components together along with a symmetric matrix, aka the Minkowski metric?

No, the symmetric bilinear form is the metric. What physicists call "scalar product between two 4-vectors" is the bilinear form acting on those 4-vectors and we denote it ##\vec v \cdot \vec w##.

 

[peter46464]

So when Schutz talks about the scalar product of two vectors he actually means a symmetric bilinear form (aka the Minkowski metric) acting on those two vectors?

 

[Orodruin]

Yes. Or, as Nugatory noted, what essentially every physicist will refer to as "scalar product" (or "inner product") although it really isn't one.

 

[peter46464]

We got there in the end. Thanks ever so much everyone for your patience. This has been bothering me for years, so it's a relief to get it cleared up. Even in MTW's Gravitation and Wald's General Relativity there's no reference to symmetric bilinear forms.

 

[Orodruin]

We got there in the end. Thanks ever so much everyone for your patience. This has been bothering me for years, so it's a relief to get it cleared up. Even in MTW's Gravitation and Wald's General Relativity there's no reference to symmetric bilinear forms.

There is not really any reason to call it a symmetric bilinear form in GR and just any symmetric bilinear form will not do (it needs to be non-degenerate). It works pretty well just calling it "metric" (or in more precise texts "pseudo-metric"). However, the metric is a particular symmetric type (0,2) tensor so ...