Find Null Paths in Differentiable Manifolds Using One-Forms

[victorvmotti]

Suppose that we have this metric and want to find null paths:

$$ds^2=-dt^2+dx^2$$

We can easily treat $$dt$$ and $$dx$$ "like" differentials in calculus and obtain for $$ds=0$$

$$dx=\pm dt \to x=\pm t$$

Now switch to the more abstract and rigorous one-forms in differentiable manifolds.

Here $$\mathrm{d}t (v)$$ is a one-form that takes a tangent vector from $$T_p$$ and returns a real number, $$\mathrm {d}t(v) \in \mathbb {R}$$.

The tangent vector to a curve $$x^{\mu}(\lambda)$$ in the basis $$\partial_\mu$$ is

$$v=\frac {dx^\mu}{d\lambda}\partial_\mu$$

Now apply the one-form to this vector

$$\mathrm {d}t(\frac {dx^\mu}{d\lambda}\partial_\mu)=\frac{dx^\mu}{d\lambda}\mathrm {d}t(\partial_\mu)$$
$$ =\frac{dx^\mu}{d\lambda}\frac {\partial t}{\partial x^\mu}$$
$$=\frac {dt}{d\lambda}$$

Now the above metric, in terms of one-forms read

$$0=-\mathrm {d}t^2(v,v)+\mathrm {d}x^2(v,v)=-\mathrm {d}t(v)\mathrm {d}t(v)+\mathrm{d}x(v)\mathrm {d}x(v)$$$$=-(\frac {dt}{d\lambda})^2+(\frac {dx}{d\lambda})^2$$

If we use the chain rule $$\frac {dx}{dt}=\frac {dx}{d\lambda}\frac {d\lambda}{dt}$$

We eventually obtain $$dx=\pm dt \to x=\pm t$$

The above is from Carroll's page 77. He reminds us that we should stick to the second more formal one-forms, that is not using differentials as in calculus, because in the first shortcut we have "sloppily" did not make the distinction between $$\mathrm {d}t^2(v)$$ and $$dt^2$$.

Now my question is that can someone please provide an example that unlike the above example, treating the one-form in differentiable manifolds as a differential in calculus will indeed produce incorrect results or conclusions.

 

[pervect]

Well, while I agree that it's good to conceptually remember that one-forms are operators (which operate on vectors) which return scalar values, I'm not aware of any difficulties if you happen to forget to be that formal and just manipulate the operators as if they were scalars, rather than operators.

The operators are linear, and associative. I suppose the only issue is if they commute, I haven't tried to think about this rigorously enough to come to any conclusion. Specifically we should ask if dx*dy = dy*dx, if they both operate on v^2.