Differentiating Vectors and their Magnitudes in Physics

[garrettm]

I've taken up to Calc. III (vector/multi-variable calculus) but have not had any classes that used intensive calculus for a few semesters. However, I'm now in a quantum physics class (its a glorified title really--it's more or less "modern physics") and am looking to see if I am correct in my understanding on how to approach differentiating vector and scalar quantities used in physics problems.

First let's start with some basic definitions.

1) If s is the position of an object, then its change in position with respect to time is its speed u:
$$ u\equiv \frac{ds}{dt} $$
2) If u is the speed of an object, then its change in speed with respect to time is its acceleration a:
$$ a\equiv \frac{du}{dt} = \frac{d}{dt} (\frac{ds}{dt}) = \frac{d^2s}{dt^2} $$
Now, if we differentiate speed squared we get:
$$ \frac{d}{dt} u^2 = \frac{d}{dt} [\frac{ds}{dt}]^2 $$

Treating this as a composition of functions, so we can use the chain rule, we see that

$$ \frac{d}{dt} [ \frac{ds}{dt} ]^2 = \frac{d}{dt} [t^2 \circ \frac{ds}{dt}] = (2 \frac{ds}{dt}) \frac{d}{dt} (\frac{du}{dt}) = (2 \frac{ds}{dt}) (\frac{d^2 u}{dt^2}) = 2ua $$

The same result can be had assuming u is a variable separate from t and using implicit differentiation on u, which really is just the chain rule again but without working with the strict definition of u.

For example:
$$ \frac{d}{dt} u^2 = \frac{d}{dt} [t^2 \circ u] = (2u)(u') = 2ua $$
So my question, would it be safe to use implicit differentiation when differentiating vectors and their magnitudes? Or should I stick to using the strict definitions?

Finally, is it best to work with vectors broken out into the product of their magnitudes and directional unit vectors?

Where, k arrow is the vector, k is the magnitude, and k hat is the directional unit vector:
$$ \vec{k} = k \hat{k}$$
For instance, force on a particle traveling at relativistic speeds is given by:
$$ \vec{F}\equiv \frac{d\vec{P}}{dt} = \frac{d}{dt} (\gamma m \vec{u}) $$
Using the product rule we get that
$$ \frac{d}{dt} (\gamma m \vec{u}) = \frac{d\gamma}{dt}m\vec{u} + \frac{dm}{dt}\gamma\vec{u} + \frac{d\vec{u}}{dt}\gamma m $$
Assuming rest mass is constant and simplifying the change in velocity WRT/ time to acceleration we get
$$ \vec{F} = \frac{d\gamma}{dt}m\vec{u} + \gamma m \vec{a} $$
Solving for gamma prime we see, using composition of functions, that:
$$ \frac{d\gamma}{dt} = \frac{d}{dt}(1 - \frac{u^2}{c^2})^{-1/2} = \frac{d}{dt} [t^{-1/2} \circ (1 - \frac{t^2}{c^2}) \circ u] $$
With aid of the power rule and the chain rule we get
$$ \frac{d}{dt} [t^{-1/2} \circ (1 - \frac{t^2}{c^2}) \circ u] = (-\frac{1}{2})(1 - \frac{u^2}{c^2})^{-3/2}(0 - \frac{2u}{c^2})(u') = \gamma^3 \frac{u}{c^2} a $$
Plugging this result back into the original equation for force we have that
$$ \vec{F} = (\gamma^3 \frac{u}{c^2} a)m\vec{u} + \gamma m \vec{a} $$
Now decomposing the velocity vector u into its magnitude and direction we see that
$$ (\gamma^3 \frac{u}{c^2} a)m(u \hat{u}) = \gamma^3 \frac{u^2}{c^2}ma\hat{u} = \gamma^3 \beta^2 ma\hat{u} $$

Now if the force is parallel to the velocity, and we know that acceleration is always parallel to the force, then the acceleration of the particle is parallel to the velocity. Thus the directional unit vector associated with the velocity can be paired with the magnitude of the acceleration. Thus:

$$ \vec{F} = \gamma^3 \beta^2 ma\hat{u} + \gamma m \vec{a} = \gamma^3 \beta^2 m\vec{a} + \gamma m \vec{a} $$

Which, is only true for the given condition that the force is parallel to the velocity.

So my second question, am I justified in the above assumptions? Can I, with certain conditions, pair directional unit vectors with magnitudes of other vectors? And is using this method of decomposing the vector into its component magnitudes and directional unit vectors helpful in general?

 

[jvicens]

Hello,

Thank you for your detailed explanation and questions. First, I would like to commend you for taking a proactive approach to understanding the use of calculus in physics problems. It is important to have a strong foundation in calculus in order to fully grasp the concepts of physics.

To answer your first question, yes, it is safe to use implicit differentiation when differentiating vectors and their magnitudes. In fact, implicit differentiation is often the preferred method in physics problems because it allows for a more general approach and can handle more complex situations.

As for your second question, yes, you can pair directional unit vectors with magnitudes of other vectors, as long as they are in the same direction. This is a common practice in physics and is helpful in simplifying calculations. However, it is important to note that this method may not always be applicable, especially in situations where the vectors are not parallel.

In general, decomposing vectors into their component magnitudes and directional unit vectors can be helpful in simplifying calculations and understanding the physical meaning of the problem. However, it is important to also understand the vector as a whole and not just its components. So, it is best to use this method when it is appropriate and relevant to the problem at hand.

I hope this helps clarify your understanding of using calculus in physics problems. Keep up the good work and keep asking questions! Best of luck in your studies.