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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?

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  • 1
    $\begingroup$ Oh snap, that's a delta my bad. Anyways, 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. $\endgroup$ – JMac Aug 2 '17 at 19:58
  • $\begingroup$ 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. $\endgroup$ – UniqueWorldline Aug 2 '17 at 20:27
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  1. 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.

  2. 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:

  1. A.H.J. De Ruiter, C.J. Damaren & J.R. Forbes, Spacecraft Dynamics and Control: An Introduction; Section 1.4, p. 32.
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  • $\begingroup$ 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: $\frac{\delta \vec{H}}{\delta t} + \vec{r} \times m\vec{v} = \vec{\tau}$ $\endgroup$ – UniqueWorldline Aug 2 '17 at 21:35
  • $\begingroup$ Which page in Ginsburg & Genin? $\endgroup$ – Qmechanic Aug 3 '17 at 8:08
  • $\begingroup$ Page 829 equation 25 $\endgroup$ – UniqueWorldline Aug 8 '17 at 5:57
  • $\begingroup$ Wiley (1977) only has 558 pages. $\endgroup$ – Qmechanic Aug 8 '17 at 7:58
  • $\begingroup$ The equation is $\sum\vec{M_{A}} = \vec{\dot{H_A}} = \frac{\delta\vec{\dot{H_A}}}{\delta t} + \vec{\omega} \times \vec{H_A}$ $\endgroup$ – UniqueWorldline Aug 8 '17 at 13:29

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