- #1
victorvmotti
- 152
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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.
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.
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