# B Time derivatives in GR and SR

1. Feb 17, 2017

### davidge

Since in GR and SR the basis vectors are generally orthogonal, how can we take derivatives of position with respect to time? For example, the current four-vector is $$J^{\alpha} = \sum_n e_{n} \frac{\partial x^{\alpha}}{\partial t} \delta^{3}(x - x_{n})$$ where n labels the n-th particle. In this case the derivative will generally not be zero, so how can time be orthogonal to $x^{i}$?

2. Feb 17, 2017

### Ibix

The basis vectors of Euclidean space are orthogonal. How can we write dy/dx?

You're differentiating the coordinate values, not the basis vectors, along some path with respect to some parameter that indexes events along that path. Time (either proper or coordinate) is a convenient parameter for time-like worldlines.

3. Feb 17, 2017

### Staff: Mentor

Hmm, this quantity doesn't seem right. It doesn't look like a four vector at all.

The orthogonality of the basis vectors doesn't have much to do with taking derivatives of the a world line. Such a derivative is simply a tangent vector to a worldline, which is clearly a valid vector regardless of the basis.

4. Feb 18, 2017

### vanhees71

In special relativity the current is (nearly) correct
$$j^{\mu}(x) = \sum_n e_n \frac{\mathrm{d} x_n^{\mu}}{\mathrm{d} t} \delta^{(3)}(\vec{x}-\vec{x}_n(t)).$$
It can be written in manifestly covariant form by introducing an arbitrary world-line parameter $\lambda$,
$$j_{\mu}(x) = \sum_n e_n \int \mathrm{d} \lambda \frac{\mathrm{d} x_n^{\mu}}{\mathrm{d} \lambda} \delta^{(4)}(x-x_n(\lambda)).$$
In GR you must be a bit more careful since here $\mathrm{d}^4 x$ is of course not invariant, but $\mathrm{d}^4 x \sqrt{-g}$ is, where $g=\det(g_{\mu \nu})$. Thus the covariant $\delta$ distribution is $\frac{1}{\sqrt{-g}} \delta^{(4)}(x-x_n(\lambda))$, and that's why the current density is given by
$$j^{\mu}(x) = \sum_n e_n \int \mathrm{d} \lambda \frac{1}{\sqrt{-g}} \frac{\mathrm{d} x_n^{\mu}}{\mathrm{d} \lambda} \delta^{(4)}(x-x_n(\lambda)).$$
For massive particles you can use each particle's proper time as world-line parameter.

5. Mar 7, 2017

### davidge

So why in many textbooks we are told that $\partial x^{\mu} / \partial x^{\nu} = \delta^\mu{}_{\nu}$(1)? For example, substituting $x$ and $y$ in (1) we would get $\partial x / \partial y = \delta^x{}_{y}$, meaning that $x$ does not vary with $y$ and vice-versa.

@vanhees71 thank you for explaining it to me.

Please see the example I give above to @Ibix.

6. Mar 7, 2017

### Staff: Mentor

You should not substitute in x and y like that. $\mu$ and $\nu$ are indexes that range over all the coordinates.

So if you are thinking of 2D Cartesian coordinates then $\mu$ is not just x or y but is an index which ranges over both. If you like to think in terms of matrix notation then
$$\frac{\partial x^{\mu}}{\partial y^{\nu}} = \begin{pmatrix} \partial x/\partial x & \partial x/\partial y \\ \partial y/\partial x & \partial y/\partial y \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \delta^{\mu}_{\nu}$$

7. Mar 7, 2017

### davidge

Indeed. I think you misunderstood my notation.

Yea. But the problem persists, namely that $\partial x^\mu / \partial x^\nu = \delta^\mu {}_{\nu}$ indicates that the basis vectors are orthogonal. Ibix says in post #2 that this derivative is in fact a derivative of the coordinate values and not the basis vectors itself, but still we have the problem that e.g. $\partial x^\mu / \partial \tau = \delta^\mu{}_{\tau}$ and this will be zero for $\mu \neq \tau$, where $\tau$ can be an arbitrary parameter or the proper time.

I hope you understand that I'm using the same symbol to label indices and variables as well.

8. Mar 7, 2017

### Staff: Mentor

No. $\tau$ is not a coordinate and you can't substitute it into coordinate equations as an index.

9. Mar 7, 2017

### davidge

Oh, I thought this was the reason for the derivative not to vanish. Thanks for confirming it to me.

Now, in a rest reference frame, the time coordinate is the proper time, right? Does it mean that we cannot evaluate a calculation using the time as measured in a rest reference frame (e.g. the frame moving along with a particle)?

10. Mar 7, 2017

### Staff: Mentor

Proper time for an object at rest in the frame, if the object at rest is in free fall and we are talking about an inertial frame in SR or a local inertial frame in GR, yes. Otherwise this will not be true in general.

11. Mar 7, 2017

### Staff: Mentor

No. That does not indicate orthogonality. Two vectors $a^{\mu}$ and $b^{\mu}$ are orthogonal if $g_{\mu\nu}a^{\mu}b^{\nu}=0$ which is not the same as the condition you wrote.

Sure, but that is more a matter of notational laziness than anything else. You can always write the basis vectors explicitly if you wish.

Hmm. That expression doesn't make sense to me. $\delta^{\mu}_{\nu}$ is a tensor with two indices. In your expression $\mu$ is an index, but $\tau$ is a scalar.

Edit: oh, I see @PeterDonis pointed that out already

12. Mar 7, 2017

### Staff: Mentor

I disagree with this statement. Proper time is an invariant quantity (a scalar field defined only on the worldline of the particle). The time coordinate is a frame variant quantity defined on the entire chart. They are not the same, even in the particle's rest frame.

For one thing, proper time does not define any notion of simultaneity, but coordinate time does.

That said, on the worldline of the particle they are equal. Sometimes this is expressed by saying that the coordinate time is "adapted" to the proper time of the particle.

13. Mar 7, 2017

### Staff: Mentor

Only for the special case I stated before, of an inertial frame that is the rest frame (globally in SR, locally in GR) of an object in free fall.

For a simple counterexample in the more general case, Rindler coordinate time is not equal to proper time along the worldlines of objects at rest in Rindler coordinates (except for one particular object at a particular Rindler $x$ coordinate).

14. Mar 7, 2017

### Staff: Mentor

Well, you can have other coordinates which are adapted to other objects. Rindler coordinates are an example, and Schwarzschild coordinates can be adapted to some non inertial objects.

15. Mar 8, 2017

### haushofer

My 2 cents: because the 4 coordinates $x^{\nu}$ describe directions which are, per definition (!), linear independent.

However, if you switch from one coordinate system $x^{\nu}$ to another one, $y^{\mu}( x^{\nu})$, y and x are per definition dependent on each other (even more; the transformation should be invertible)! The $x^{\nu}$ are still lineair independent, and the $y^{\mu}$ too. This means that
$$\frac{\partial x^{\mu}}{ \partial x^{\nu}} = \frac{\partial y^{\mu} }{ \partial y^{\nu}} = \delta^\mu{}_{\nu} \,.$$

Your $\tau$ is a coordinate on a wordline, not in spacetime (also called the target space). That's why you only need one; a line is 1-dimensional.

16. Mar 8, 2017

### Staff: Mentor

Yes, but coordinate time will not, in general, be the same as proper time for those objects. See the second paragraph of the post of mine that you quoted. Schwarzschild spacetime is an even better example, since in that case coordinate time is not equal to proper time for any observer that is static (at rest in the coordinates) at finite $r$.

17. Mar 8, 2017

### Staff: Mentor

I agree. I think people should never say that coordinate time is the same as proper time.

18. Mar 8, 2017

### davidge

@haushofer I think you misunderstood my notation. In my post #5, $x$ and $y$ are two of the coordinates of a point in a same chart, say point $P$. Then $P$ has coordinates $(x,y,...)$. Now, what you pointed out is for the case where ${x} = (x^0,x^1,...)$ are the coordinates of the point $P$ in some chart and ${y} = (y^0,y^1,...)$ are its coordinates in another chart. Btw, is it correct to say that what we call two reference frames in physics, is in math language roughly just two different coordinate charts?
Ah, ok.

Oh, I see. Thanks.
Why is $\delta^\mu{}_{\nu}$ a tensor? Does it transforms like a tensor?

19. Mar 8, 2017

### davidge

I was using $x$ and $y$ for the coordinates of a point, following the suggestion of @PeterDonis on a previous thread. He told me something like: "Don't make things so hard for yourself".
In that thread, I was using abstract notation, $x^0,x^1$ instead of $x,y$.

20. Mar 8, 2017

### Staff: Mentor

It depends on how precise you want to be.

The terms "reference frame" and "coordinate chart" are commonly used as synonyms, yes. But "reference frame", strictly speaking, means something different: it means what is also called a "frame field", which is an assignment of four mutually orthogonal unit vectors, one timelike and three spacelike, to every event in some open region of spacetime. The set of vectors at each event represent, heuristically, a clock and three mutually orthogonal rulers used by a (hypothetical) observer at that event to make measurements. This is basically the GR version of the set of measuring rods and clocks that was used by Einstein and others in the early days of SR to describe what they meant by an "inertial frame".

The reason the two terms are often used synonymously is basically a holdover from SR, where, because there are global inertial frames and global inertial coordinate charts in which the frame vectors at every event are also the coordinate basis vectors at that event, there is an exact correspondence between the two concepts. But I think it's a good idea to unlearn this habit, because in GR there is in general no such correspondence: for many, if not most, frame fields of interest in GR, there is no coordinate chart in which the frame vectors at every event are also the coordinate basis vectors at that event. So I think it's a good idea to keep the concepts separate in your mind, and to use the proper term for whichever one you actually mean in a given case.

21. Mar 8, 2017

### Staff: Mentor

Yes, it is a legitimate tensor and transforms like a tensor. It also has the same numerical value in every coordinate system.

22. Mar 9, 2017

### vanhees71

Physicists are sometimes sloppy in their language. ${\delta^{\mu}}_{\nu}$ are tensor components with respect to a basis. In GR tensors themselves are invariant under general coordinate transformations.

In GR one often uses holonomous coordinate bases, and then components with upper indices transform like the differentials of the coordinates ("contravariant"), i.e.,
$$\mathrm{d} \tilde{x}^{\mu} = \mathrm{d} x^{\nu} \frac{\partial \tilde{x}^{\mu}}{\partial x^{\nu}},$$
while those with lower indices like $\partial_{\mu}$ (covariant), i.e.,
$$\partial_{\mu}'=\frac{\partial}{\partial \tilde{x}^{\mu}}=\frac{\partial x^{\nu}}{\partial \tilde{x}^{\mu}} \partial_{\nu}.$$
From this you have
$${\tilde{\delta}^{\mu}}_{\nu} = \frac{\partial \tilde{x}^{\mu}}{\partial x^{\rho}} \frac{\partial x^{\sigma}}{\partial \tilde{x}^{\nu}} {\delta^{\rho}}_{\sigma} = \frac{\partial \tilde{x}^{\mu}}{\partial x^{\rho}} \frac{\partial x^{\rho}}{\partial \tilde{x}^{\nu}} = \frac{\partial \tilde{x}^{\mu}}{\partial \tilde{x}^{\nu}}={\delta^{\mu}}_{\nu},$$
i.e., the tensor components don't change. The Kronecker-$\delta$ thus are invariant tensorcomponents under general coordinate transformations.

23. Mar 9, 2017

### stevendaryl

Staff Emeritus
There is a more abstract way to see that $\delta_\mu^\nu$ is a tensor (or rather, components of a tensor):

A $(1,1)$ tensor is by definition a linear function that takes a vector and returns a vector (see the note below*). The very simplest such function is the identity function: $I(v) = v$. If you work out the components of the tensor $I$, you find that $I^\mu_\nu = \delta^\mu_\nu$.

*Note: A $(1,1)$ tensor can be characterized in a number of different, equivalent ways:
1. A linear function that takes one vector and returns a vector: $A(v) = A^\mu_\nu v^\nu e_\mu$
2. A linear function that takes one covector and returns a covector: $A(B) = A^\mu_\nu B_\mu e^\nu$
3. A function, linear in each argument, that takes one vector and one covector and returns a scalar. $A(v,B) = A^\mu_\nu v^\nu B_\mu$

24. Mar 9, 2017

### davidge

So... we often use the notation $\partial / \partial x^\mu$ for a coordinate basis vector defined at a point $x$. What would be a notation for the frame vectors? Or would it be meaningless to use a notation for frame vectors?

Thanks @vanhees71 for showing how it transforms.
@stevendaryl Interesting way to see it

25. Mar 9, 2017

### Staff: Mentor

If you have chosen a coordinate chart, you can write the frame vectors in terms of the coordinate basis vectors (i.e., by giving their components in the coordinate basis), just as for any vectors.

In coordinate-free terms, frame vectors are sometimes written as letters with indices, for example $e_\mu$ (note the lower index), where $\mu$ runs over the four spacetime indices. But this notation can get confusing since the index $\mu$ in this case does not refer to a component of a vector, but to a vector from the set of four vectors.