$\delta$ differential notation

Question

Various textbooks that I am currently consulting (including Spacecraft Dynamics and Control An Introduction - Anton H.J. De Ruiter | Christopher J. Damaren | James R. Forbes Section 1.4, page 32) use $\delta$, not $d$ or $\partial$ to express an infinitesimal quantity. In the context of the reference text above, the symbol is used as part of a general definition for the derivative of a vector. Specifically, given

$$\mathbf{\vec{r}}=\mathcal{\vec{F^{T}_{1}}}\mathbf{\vec{r_{1}}}$$

where

$$\mathcal{\vec{F^{T}_{1}}}$$

is a vectrix

$$[\mathbf{\vec{x_1}} \phantom{s} \mathbf{\vec{y_1}} \phantom{s} \mathbf{\vec{z_1}}].$$

The time derivative of the vector is defined as

$$\dot{\mathbf{\vec{r}}}\triangleq \lim_{\delta t\to 0}\frac{\delta\mathbf{\vec{r}}}{\delta t}.$$

In this context, what is the difference between $\frac{\delta\mathbf{\vec{r}}}{\delta t}$ and $\frac{d \mathbf{\vec{r}}}{d t}$ ?

Note about thermodynamic use of $\delta$: My understanding is that $\delta$ is used in thermodynamic equations to express path dependence of a scalar quantity such as heat or work. What does it mean in the context of an abstract physical vector?

Oh snap, that's a delta my bad. Anyways, https://math.stackexchange.com/q/317338/402453 that should explain it. The second answer gives a good physics answer. I've seen it in the context that answer talks about. — JMac, Aug 02 '17 at 19:58
I have updated the question to address your comments. Thank you @JMac for the link to the other question. However, I am afraid it did not answer my question as I am interested in the notation in the context of physical vectors, not thermodynamics quantities. — UniqueWorldline, Aug 02 '17 at 20:27

Qmechanic · Accepted Answer · 2017-08-06T13:21:57.120

6

Typically, $\frac{d}{d(\ldots)}$ is a (total) derivative, $\frac{\partial}{\partial(\ldots)}$ is a partial derivative, and $\frac{\delta}{\delta(\ldots)}$ is a functional/variational derivative. See also e.g. this & this Phys.SE posts and links therein.
In Ref. 1 the symbol $\frac{d\mathbf{\vec{r}}}{d t}$ denotes the derivative, while $\frac{\delta\mathbf{\vec{r}}}{\delta t}$ denotes the difference quotient. A more common notation for the latter is $\frac{\Delta\mathbf{\vec{r}}}{\Delta t}$.

References:

A.H.J. De Ruiter, C.J. Damaren & J.R. Forbes, Spacecraft Dynamics and Control: An Introduction; Section 1.4, p. 32.

edited Aug 06 '17 at 13:21

answered Aug 02 '17 at 21:18

Qmechanic

201,751

I agree that, in the context above, the difference quotient would make sense. However, I have seen in $Dynamics$ by Ginsburg and Genin (old textbook) an example of an angular momentum framework that used $\delta$, and when solving problems with that framework, it seemed $\frac{\delta \vec{r}}{\delta t}$ was functionally equivalent to a total derivative. I think the framework was: $\frac{\delta \vec{H}}{\delta t} + \vec{r} \times m\vec{v} = \vec{\tau}$ – UniqueWorldline Aug 02 '17 at 21:35
Which page in Ginsburg & Genin? – Qmechanic Aug 03 '17 at 08:08
Page 829 equation 25 – UniqueWorldline Aug 08 '17 at 05:57
Wiley (1977) only has 558 pages. – Qmechanic Aug 08 '17 at 07:58
The equation is $\sum\vec{M_{A}} = \vec{\dot{H_A}} = \frac{\delta\vec{\dot{H_A}}}{\delta t} + \vec{\omega} \times \vec{H_A}$ – UniqueWorldline Aug 08 '17 at 13:29
I think there might be two different parts to it because my copy starts at chapter 10 and page 381 – UniqueWorldline Aug 08 '17 at 13:31
Unfortunately, I don't have access to the book. – Qmechanic Aug 08 '17 at 13:41
My guess is that it is the variational derivative, but I don't really have a firm reason for such a statement. The more I looked into this notation, the muddier and more ambiguous it got. Thanks for your help anyway. – UniqueWorldline Aug 08 '17 at 20:38

$\delta$ differential notation

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