One-forms in differentiable manifolds and differentials in calculus

[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.

 

[lavinia]

Not sure whar you are asking.
Here are a few thoughts.

- dt and dx are the differentials of functions. They are differentials of the coordinate functions of Minkowski space.

- An arbitrary 1 form is not the differential of a function and so is not df for any f. This may be true globally as in dtheta in the plane minus the origin or locally as well. A 1 form is locally a differential if it is closed.

- When one thinks of df as a quantity, rather than a linear function on a vector, one thinks of a small increment in a measurement of some physical quantity. This increment is well approximated by the inner product of the gradient of f with the underlying parameter increments if they are small enough. So df may be thought of as the change in f for small changes in the parameters.

- Along any curve, a 1 form may in fact be thought of as the differential of a function. But in general, this function will be different in different directions.-

 

[victorvmotti]

I see the difference, what I am asking is that are there examples that unlike the above one, if we do not care about the difference and treat a one-form like a differential in our calculations then we will get incorrect results?

 

[lavinia]

I see the difference, what I am asking is that are there examples that unlike the above one, if we do not care about the difference and treat a one-form like a differential in our calculations then we will get incorrect results?

If a form is not a differential how would you treat it as one?

If you treat an arbitrary 1 form as a differential then you will get the interesting result that the integral of any 1 form over any closed path is zero.

 

[victorvmotti]

It's not about how, but given the practical aim of a shortcut calculations, apparent in the example above, I was wondering when and in which examples such a shortcut will fail and will not match the outcome of the rigorous way.

 

[lavinia]

It's not about how, but given the practical aim of a shortcut calculations, apparent in the example above, I was wondering when and in which examples such a shortcut will fail and will not match the outcome of the rigorous way.

If you are careful you can always think of differetials of functions as small increments.

 

[victorvmotti]

Are one-forms, as operators, commutative? Can we say $$dxdy=dydx$$?

 

[lavinia]

Are one-forms, as operators, commutative? Can we say $$dxdy=dydx$$?

They are anticommutative if you take multiplcation to mean the wedge product.

But if if you take the product to be multiplication of increments then they do.

The produlct dxdy of increments, though, is not a form.

 

[victorvmotti]

I think that I have found a counterexample where you cannot overlook the difference between a one-form and a differential.

Consider a classical field theory.

When applying the least action I see that a term is considered total derivative.

We say that
$$\int \partial_\mu (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi) d^4x= \int d(\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)= (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)$$
And then because the variation at the spatial infinity vanishes this terms is equal to zero.

I do not get the calculation from $$\partial_\mu (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi) d^4x=\frac {\partial (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)}{\partial x^\mu} dtdxdydz$$ to
$$d(\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)=\frac {\partial (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)}{\partial t}dt+\frac {\partial (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)}{\partial x}dx+\frac {\partial (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)}{\partial y}dy+\frac {\partial (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)}{\partial z}dz$$
$$\neq \frac {\partial (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)}{\partial x^\mu} dtdxdydz$$
Can you expand this to fill the gap for me.

Also, why we require "spatial infinity" here, isn't it also true that $$\delta \phi$$ in the $$\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi$$ vanishes at any two endpoints of the events path, but why we require infinity here?

 

[lavinia]

I think that I have found a counterexample where you cannot overlook the difference between a one-form and a differential.

Consider a classical field theory.

When applying the least action I see that a term is considered total derivative.

We say that
$$\int \partial_\mu (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi) d^4x= \int d(\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)= (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)$$
And then because the variation at the spatial infinity vanishes this terms is equal to zero.

I do not get the calculation from $$\partial_\mu (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi) d^4x=\frac {\partial (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)}{\partial x^\mu} dtdxdydz$$ to
$$d(\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)=\frac {\partial (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)}{\partial t}dt+\frac {\partial (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)}{\partial x}dx+\frac {\partial (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)}{\partial y}dy+\frac {\partial (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)}{\partial z}dz$$
$$\neq \frac {\partial (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)}{\partial x^\mu} dtdxdydz$$
Can you expand this to fill the gap for me.

Also, why we require "spatial infinity" here, isn't it also true that $$\delta \phi$$ in the $$\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi$$ vanishes at any two endpoints of the events path, but why we require infinity here?

Sorry but I don't know what you are talking about here. Can you explain the terms in detail?

 

[ShayanJ]

Sorry but I don't know what you are talking about here. Can you explain the terms in detail?

Although I answered his question in its own thread, he's hijacking this one!

 

[victorvmotti]

Given the Lagrangian density of a field, the action is

$$S=\int \mathcal L(\phi, \partial_\mu \phi) d^4x $$

Now if we apply the Euler-Lagrange equation of motion via the principle of the least action.

$$\delta S=\int [\frac {\partial \mathcal L}{\partial \phi}\delta \phi+\frac {\partial \mathcal L}{\partial(\partial_\mu \phi)}\delta (\partial_\mu \phi] d^4x$$

You see the above is simplified to $$\delta S=\int \partial_\mu (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi) d^4x$$

The integrand is an "exact form" over a volume, which, by Stokes' theorem, reduces to the boundary values of a 3-form.

But if we treat the $$\mathrm{d}^4x$$ as the product of differentials $$d^4x=dtdxdydz$$ you will run into problems as shown above.

So this is an example in which you cannot overlook the difference between a one-form and a differential.

 

[lavinia]

Given the Lagrangian density of a field, the action is

$$S=\int \mathcal L(\phi, \partial_\mu \phi) d^4x $$

Now if we apply the Euler-Lagrange equation of motion via the principle of the least action.

$$\delta S=\int [\frac {\partial \mathcal L}{\partial \phi}\delta \phi+\frac {\partial \mathcal L}{\partial(\partial_\mu \phi)}\delta (\partial_\mu \phi] d^4x$$

You see the above is simplified to $$\delta S=\int \partial_\mu (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi) d^4x$$

The integrand is an "exact form" over a volume, which, by Stokes' theorem, reduces to the boundary values of a 3-form.

But if we treat the $$\mathrm{d}^4x$$ as the product of differentials $$d^4x=dtdxdydz$$ you will run into problems as shown above.

So this is an example in which you cannot overlook the difference between a one-form and a differential.

No. There is not a problem as I said before. It looks like you took a differential when you should have just taken partial derivatives.