# Is there a more precise definition of $dS^2$?

In order to compute the interval between two events we perform the following calculation: $$dS^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$$ I am A little confused by this notation. Are these $dx^{\mu}$'s (a basis of) one forms? What binary operation is going on here? Should it be a tensor product, a wedge product or something else entirely? I think I may have read somewhere that these $dx^{\mu}$'s are infinitesimals of the coordinates, but could someone please make this a little more precise in the terminology of differential geometry?

So, should I be interpreting $dS^2$ as something to which you feed two vectors and get out a real number? I.e. the distance between two events V and U is $$g(U,V)=dS^2(U,V)\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=g_{\mu\nu}(dx^{\mu}\otimes dx^{\nu})(V,U)\\~~~~~~~~~~~~~~~~~~~~~~~~~~~=g_{\mu\nu}dx^{\mu}(V)\cdot dx^{\nu}(U)\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=g_{\mu\nu}dx^{\mu}(v^i\partial_i) dx^{\nu}(u^j\partial_j)\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=g_{\mu\nu}v^iu^jdx^{\mu}(\partial_i) dx^{\nu}(\partial_j)\\~~~~~~~~~~~~~~~~=g_{\mu\nu}v^iu^j\delta^{\mu}_{~~i} \delta^{\nu}_{~~j}\\~~~~~~~~=g_{\mu\nu}v^{\mu}u^{\nu}$$

Where all components of the above tensor's are with respect to the coordinate basis $\{\partial_{i}\}$ and its dual basis $\{dx^{i}\}$. But then I often see expressions which involve the integral of dS and the integral of the $dx^{\mu}$, which leads me to believe that somewhere we need wedge products and all that jazz. So my question is how can we make $dS^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$ more precise in the terminology of differential geometry, or is it perfectly fine how it is and I am just over thinking things?

• – Qmechanic Apr 27 '17 at 6:13

Informally, $ds^2$ is the square of the distance taken when one travels from $x$ to $x+dx$.

Formally, the $dx^\mu$-s are basis 1-forms and we can define the symmetric product as $dx^\mu dx^\nu=\frac{1}{2}(dx^\mu\otimes dx^\nu+dx^\nu\otimes dx^\mu)$, then we have $$ds^2\equiv g=g_{\mu\nu}dx^\mu dx^\nu.$$ This expression would be valid even if we used tensor products instead of symmetric products, but if a metric is written explicitly as, say $ds^2=-dt^2+2A(r)drd\vartheta+r^2(d\vartheta^2+\sin^2\vartheta d\varphi^2)$, then the commutativity of the differentials is usually assumed in the (informal) literature, which is only true for symmetric products. In particular, the above expression would contain the $A(r)dr\otimes d\vartheta +A(r)d\vartheta\otimes dr$ terms instead of the simplified $2A(r)dr d\vartheta$ term.

So, basically, the symbol $ds^2$ is a relic from old-school differential geometry, when mathematicians weren't so anal about rigour as they are now in the post-Bourbaki world.

In order to keep up with traditions, we still use $ds^2$, but if you want to employ modern DG, you should interpret $ds^2$ to be the same as $g$ (the metric tensor field).

Note: About integration, integrating $ds$ is also informal. There does not exist an 1-form that assigns a curve's length to every curve (at least not to my knowledge - EDIT: There does not exist one.).

What really happens is that given a smooth curve $\gamma$, you either define its length as $$L[\gamma]=\int_{t_0}^{t_1}\sqrt{g(\dot{\gamma}(t),\dot{\gamma}(t))}dt,$$ which you can show to be reparametrization-invariant (and when written out in components, looks like $\int ds$, if you manipulate "differentials" heuristically), OR, if $\gamma:I\rightarrow M$, where $I$ is an (open) interval, you take the induced metric tensor $\gamma^*g$ (induced in the one-dimensional manifold $I$), and you integrate the volume form induced by $\gamma^*g$ on $I$ (which will be an 1-form, since $I$ is one dimensional). This 1-form will have the local component appearance of $đs=\sqrt{g_{\mu\nu}\frac{dx^\mu}{dt}\frac{dx^\nu}{dt}}dt,$ but it should be noted that this is an 1-form on $I$ and not on $M$!

• Uldreth: "given a smooth curve $\gamma$ [...]" -- Why this restriction? How even to determine whether some subset of spacetime constitutes (the image of) a "smooth curve"? "either define its length as" $$L[\, \gamma\, ] :=\int_{t_0}^{t_1} {\rm d}t\,\sqrt{g[\,\dot\gamma[\, t\, ],\dot\gamma[\, t\, ]\, ]}$$ -- Why allow complex "length"? Why not $$L[\, \gamma\, ] := \int_{t_0}^{t_1} {\rm d}t\, \text{sgn}[\, g[\,\dot\gamma[\, t\, ],\dot\gamma[\, t\, ]\, ] \, ] \, \sqrt{\text{sgn}[\, g[\,\dot\gamma[\, t\, ],\dot\gamma[\, t\, ]\, ] \, ] \, g[\,\dot\gamma[\, t\, ],\dot\gamma[\, t\, ]\, ]}$$?? – user12262 Apr 28 '17 at 14:20
• @user12262 The restriction is unnecessary, however the curve needs to be piece-wise differentiable for this to make sense. It is customary in DG to assume everything is smooth though, this restriction was not the point of my post. Likewise, about "complex length", assume that we are in a Riemannian manifold, or the curve is spacelike, or I am automatically switching to proper time instead of proper length when the curve is timelike. – Bence Racskó Apr 28 '17 at 16:04
• Uldreth: "the curve needs to be piece-wise differentiable for this to make sense." -- Apparently; but that doesn't address the problem how to find out which subsets of spacetime are "curves", and specifically "differentiable"; or how to determine "spacetime topology". "It is customary in DG to assume [...]" -- Right. In Physics, in contrast, it is necessary to measure (quantities; such as Lorentzian distance), and therefore foremost: to consider and to determine how to measure. So: thanks for our response ... – user12262 Apr 29 '17 at 6:01