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I Differentiating Vectors and their Magnitudes in Physics

  1. Oct 15, 2016 #1
    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?
     
  2. jcsd
  3. Oct 20, 2016 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
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