A fun cube root simplification problem.

July 14, 2024 math and physics play No comments , , , , , ,

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I saw a thumbnail of a cube root simplification problem on youtube, and tried it myself before watching the video. I ended up needing two hints from the video to solve the problem.  The problem was to simplify
\begin{equation}\label{eqn:cuberootsimplify:20}
x = \lr{ \sqrt{5} – 2 }^{1/3}.
\end{equation}
My guess was that the solution was of the form
\begin{equation}\label{eqn:cuberootsimplify:40}
x = a \sqrt{5} + b,
\end{equation}
where \(a,b\) are rational numbers. I say that because, if we cube that expression for \(x\) we get
\begin{equation}\label{eqn:cuberootsimplify:60}
x^3 = a^3 5 \sqrt{5} + 15 a^2 b + 3 \sqrt{5} a b^2 + b^3,
\end{equation}
so if we can find rational solutions to the system
\begin{equation}\label{eqn:cuberootsimplify:80}
\begin{aligned}
\sqrt{5} \lr{ 5 a^3 + 3 a b^2 } &= \sqrt{5} \\
15 a^2 b + b^3 &= -2.
\end{aligned}
\end{equation}
My problem now was that this doesn’t look like it’s particularly easy to solve. Mathematica can do it easily, as shown in fig. 1.

fig. 1. Mathematica simultaneous rational cubic reduction.

But if I wanted to cheat, I can just ask Mathematica to simplify the expression, as in fig. 2

fig. 2. Direct Mathematica simplification.

So, back to the drawing board. One thing that we can notice is that the expression in the cube root, looks like it could be recast in terms of a difference of squares
\begin{equation}\label{eqn:cuberootsimplify:100}
\sqrt{5} – 2 = \sqrt{5} – \sqrt{4}.
\end{equation}
Let’s let \( a = \sqrt{5}, b = \sqrt{4} \), so that
\begin{equation}\label{eqn:cuberootsimplify:120}
\begin{aligned}
\sqrt{5} – \sqrt{4} &=
a – b \\
&= \frac{a^2 – b^2}{a + b} \\
&= \frac{5 – 4}{\sqrt{5} + \sqrt{4} }.
\end{aligned}
\end{equation}
This shows that we have a sort of “conjugate” relationship for this difference
\begin{equation}\label{eqn:cuberootsimplify:140}
\sqrt{5} – 2 = \inv{\sqrt{5} + 2}.
\end{equation}
Surely this can be exploited somehow in the simplification process. I was stumped at this point, and didn’t see where to go with this, so I cheated a different way (not using Mathematica this time) and looked at the video to see where he went with it. Sure enough, he used these related pairs, and let
\begin{equation}\label{eqn:cuberootsimplify:160}
\begin{aligned}
x &= \lr{ \sqrt{5} – 2 }^{1/3} \\
y &= \lr{ \sqrt{5} + 2 }^{1/3}.
\end{aligned}
\end{equation}
Without looking further, let’s see what we can do with these. Clearly, we’d like to cube them, so that we seek solutions to
\begin{equation}\label{eqn:cuberootsimplify:180}
\begin{aligned}
x^3 &= \sqrt{5} – 2 \\
y^3 &= \sqrt{5} + 2.
\end{aligned}
\end{equation}
Sums and differences look like they would be interesting
\begin{equation}\label{eqn:cuberootsimplify:200}
\begin{aligned}
x^3 + y^3 &= 2 \sqrt{5} \\
y^3 – x^3 &= 4.
\end{aligned}
\end{equation}
We’ve also seen that
\begin{equation}\label{eqn:cuberootsimplify:220}
x y = 1,
\end{equation}
so just like the initial guess problem, we are left with having to solve two simulateous cubics, but this time the cubics are simpler, and we have a constraint condition that should be helpful.
My next guess was to form the cubes of \( x \pm y \), and use our constraint equation \( x y = 1 \) to simplify that. We find
\begin{equation}\label{eqn:cuberootsimplify:240}
\begin{aligned}
\lr{ x + y }^3
&= x^3 + 3 x^2 y + 3 x y^2 + y^3 \\
&= 2 \sqrt{5} + 3 \lr{ x + y} x y \\
&= 2 \sqrt{5} + 3 \lr{ x + y },
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:cuberootsimplify:260}
\begin{aligned}
\lr{ y – x }^3
&= y^3 – 3 y^2 x + 3 y x^2 – x^3 \\
&= 4 – 3 \lr{ y – x } x y \\
&= 4 – 3 \lr{ y – x }.
\end{aligned}
\end{equation}
We can now let \( u = x + y, v = y – x \), and have a pair of independent equations to solve
\begin{equation}\label{eqn:cuberootsimplify:280}
\begin{aligned}
u^3 &= 2 \sqrt{5} + 3 u \\
v^3 &= 4 – 3 v.
\end{aligned}
\end{equation}
However, we still have cubic equations to solve, neither of which look particularly fun to reduce. I went around in circles from here and didn’t make much headway, and eventually went back to the video to see what he did. He ended up with an equivalent to my equation for \( v \) above, but I actually got there much more directly (my \( v \) was his \( -u \), so the exact steps he used differed.) His basic technique was to note that \( 4 = 3 + 1 \) so he looked for factors with \( 3 \) and \( 1 \) terms. In my case, that is equivalent to the observation that \( v = 1 \) is a root to the cubic in \( v \). So, we want to factor out \( v – 1 \) from
\begin{equation}\label{eqn:cuberootsimplify:300}
v^3 + 3 v – 4 = 0,
\end{equation}
Long dividing this by \( v -1 \) gives
\begin{equation}\label{eqn:cuberootsimplify:320}
\lr{ v – 1 } \lr{ v^2 + v + 4 } = 0.
\end{equation}
Completing the square for the quadratic factor gives
\begin{equation}\label{eqn:cuberootsimplify:340}
\lr{v + \inv{2} }^2 = -4 – \inv{4},
\end{equation}
which has only complex solutions (and we want a positive real solution.) Equating the remaining factor to zero, and reminding ourselves about our \( x y \) constraint, we are now left with
\begin{equation}\label{eqn:cuberootsimplify:360}
\begin{aligned}
v = y – x &= 1,
x y &= 1.
\end{aligned}
\end{equation}
Solving both for \( y \) gives
\begin{equation}\label{eqn:cuberootsimplify:380}
y = x + 1 = \inv{x},
\end{equation}
or
\begin{equation}\label{eqn:cuberootsimplify:400}
x^2 + x = 1,
\end{equation}
or
\begin{equation}\label{eqn:cuberootsimplify:420}
\lr{ x + \inv{2} }^2 = 1 – \inv{4} = \frac{5}{4}.
\end{equation}
We are left with two possible solutions for \( x \)
\begin{equation}\label{eqn:cuberootsimplify:440}
x = -\inv{2} \pm \frac{\sqrt{5}}{2},
\end{equation}
and we can now discard the negative solution, and find
\begin{equation}\label{eqn:cuberootsimplify:460}
x = \frac{ \sqrt{5} – 1 }{2},
\end{equation}
matching the answer that we’d found with the Mathematica cheat earlier.

Seeing the effort required to simplify this makes me impressed once again with Mathematica. I wonder what algorithm it uses to do the simplification?

Quantum Man, Richard Feynman’s Life in Science, by Lawrence Krauss.

July 13, 2024 Reviews No comments , , ,

I just finished “Quantum Man”, Richard Feynman’s Life in Science, by Lawrence Krauss.
 
I finished Gleick’s Genius recently.  When I finished that I thought I’d read all the Feynman biographies and autobiographies.  However, I heard Krauss mention his Feynman bio on his origins podcast, a book that was not just a book on Feynman, but on Feynman’s science.  I’m very pleased that the Toronto Public Library is finally operational again after the hacking fiasco last year, and was able to get this book to read with only a couple days wait.
A large part of this book isn’t a biography of Feynman, but the story of the creation of quantum electrodynamics.  From a technical (physics) perspective, I know little bits of that story, certainly not as much as I would like.  This book described a number of the key problems and solutions that were dealt with in the evolution of this science.  Should I ever find the time to study more of the physics of QED, which I’d like to do, having read this story will provide some useful context.
There are a lot of tricky details of the theory that are explained in this book in an accessible way.  I thought the description of positrons as electrons moving backwards in time was especially well done, possibly because it was set in the context of forward and backward time processes in classical electrodynamics.  It seems clear that forward and backwards propagators are really the interesting character hiding behind the curtain here (but the book didn’t include any of the mathematics that would confirm that.)  In a number of places, Krauss did a remarkable job of describing complex and abstract mathematical ideas in a way that was accessible.

A book on Feynman can’t just be about his physics, but also about the man.  There were lots of examples of that here too, with lots to keep the reader laughing and amused.  Here’s a characteristic quote from the book that I found particularly funny:

one of the items brought up in the divorce proceedings was telling.  She reported, “He begins working calculus problems in his head as soon as he awakens.  He did calculus while driving his car, while sitting in the living room and while lying in bed at night”
… all in all, this was a highly enjoyable book.

A kind of fun high school physics collision problem, generalized slightly.

June 15, 2024 math and physics play , , , ,

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fig. 1. The collision problem.

Karl’s studying for his grade 12 physics final, and I picked out some problems from his text [1] for him to work on. Here’s one, fig. 1, that he made a numerical error with.

I solved this two ways, the first was quick and dirty using Mathematica, so he could check his answer against a number, and then while he was working on it, I also tried it on paper. I found the specific numeric values annoying to work with, so tackled the slightly more general problem of an object of mass \( m_1 \) colliding with an object of mass \( m_2 \) initially at rest, and determined the final velocities of both.

If we want to solve this, we start with a plain old conservation of energy relationship, with initial potential energy, equal to pre-collision kinetic energy
\begin{equation}\label{eqn:collisionproblem:20}
m_1 g h = \inv{2} m_1 v^2,
\end{equation}
where for this problem \( h = 3 – 3 \cos(\pi/3) = 1.5 \,\textrm{m} \), and \( m_1 = 4 \,\textrm{kg} \). This gives us big ball’s pre-collision velocity
\begin{equation}\label{eqn:collisionproblem:40}
v = \sqrt{2 g h}.
\end{equation}

For the collision part of the problem, we have energy and momentum balance equations
\begin{equation}\label{eqn:collisionproblem:60}
\begin{aligned}
\inv{2} m_1 v^2 &= \inv{2} m_1 v_1^2 + \inv{2} m_2 v_2^2 \\
m_1 v &= m_1 v_1 + m_2 v_2.
\end{aligned}
\end{equation}
Clearly, the ratio of masses is more interesting than the masses themselves, so let’s write
\begin{equation}\label{eqn:collisionproblem:80}
\mu = \frac{m_1}{m_2}.
\end{equation}
For the specific problem at hand, this is a value of \( \mu = 2 \), but let’s not plug that in now, instead writing
\begin{equation}\label{eqn:collisionproblem:100}
\begin{aligned}
\mu v^2 &= \mu v_1^2 + v_2^2 \\
\mu \lr{ v – v_1 } &= v_2,
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:collisionproblem:120}
v^2 = v_1^2 + \mu \lr{ v – v_1 }^2,
\end{equation}
or
\begin{equation}\label{eqn:collisionproblem:140}
v_1^2 \lr{ 1 + \mu } – 2 \mu v v_1 = v^2 \lr{ 1 – \mu }.
\end{equation}
Completing the square gives

\begin{equation}\label{eqn:collisionproblem:160}
\lr{ v_1 – \frac{\mu}{1 + \mu} v }^2 = \frac{\mu^2}{(1 + \mu)^2} v^2 + v^2 \frac{ 1 – \mu }{1 + \mu},
\end{equation}
or
\begin{equation}\label{eqn:collisionproblem:180}
\begin{aligned}
\frac{v_1}{v}
&= \frac{\mu}{1 + \mu} \pm \inv{1 + \mu} \sqrt{ \mu^2 + 1 – \mu^2 } \\
&= \frac{\mu \pm 1}{1 + \mu}.
\end{aligned}
\end{equation}
Our second velocity, relative to the initial, is
\begin{equation}\label{eqn:collisionproblem:200}
\begin{aligned}
\frac{v_2}{v}
&= \mu \lr{ 1 – \frac{v_1}{v} } \\
&= \mu \lr{ 1 – \frac{\mu \pm 1}{1 + \mu} } \\
&= \mu \frac{ 1 + \mu – \mu \mp 1 }{1 + \mu} \\
&= \mu \frac{ 1 \mp 1 }{1 + \mu}.
\end{aligned}
\end{equation}

The post collision velocities are
\begin{equation}\label{eqn:collisionproblem:220}
\begin{aligned}
v_1 &= \frac{\mu \pm 1}{1 + \mu} v \\
v_2 &= \mu v \frac{ 1 \mp 1 }{1 + \mu},
\end{aligned}
\end{equation}
but we see the equations describe one scenario that doesn’t make sense physically, because the positive case, describes the first mass teleporting through and past the second mass, and continuing merrily on its way with its initial velocity. That means that our final solution is
\begin{equation}\label{eqn:collisionproblem:240}
\begin{aligned}
v_1 &= \frac{\mu – 1}{1 + \mu} v \\
v_2 &= 2 \frac{ \mu }{1 + \mu} v,
\end{aligned}
\end{equation}
For the original problem, that is \( v_1 = 2 v / 3 \) and \( v_2 = 4 v /3 \), where \( v = \sqrt{ 2(9.8) 1.5 } \,\textrm{m/s} \).

For the post-collision heights part of the question, we have
\begin{equation}\label{eqn:collisionproblem:260}
\begin{aligned}
\inv{2} m_1 \lr{ \frac{2 v}{3} }^2 &= m_1 g h_1 \\
\inv{2} m_2 \lr{ \frac{4 v}{3} }^2 &= m_2 g h_1,
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:collisionproblem:280}
\begin{aligned}
h_1 &= \frac{2}{9} \frac{v^2}{g} = \frac{4}{9} h \\
h_2 &= \frac{8}{9} \frac{v^2}{g} = \frac{16}{9} h,
\end{aligned}
\end{equation}
where \( h = 1.5 \,\textrm{m} \).

The original question doesn’t ask for the second, or Nth, collision. That would be a bit more fun to try.

References

[1] Bruni, Dick, Speijer, and Stewart. Physics 12, University Preparation. Nelson, 2012.

More on time derivatives of integrals.

June 9, 2024 math and physics play , , , , , , , , , , , , ,

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Motivation.

I was asked about geometric algebra equivalents for a couple identities found in [1], one for line integrals
\begin{equation}\label{eqn:more_feynmans_trick:20}
\ddt{} \int_{C(t)} \Bf \cdot d\Bx =
\int_{C(t)} \lr{
\PD{t}{\Bf} + \spacegrad \lr{ \Bv \cdot \Bf } – \Bv \cross \lr{ \spacegrad \cross \Bf }
}
\cdot d\Bx,
\end{equation}
and one for area integrals
\begin{equation}\label{eqn:more_feynmans_trick:40}
\ddt{} \int_{S(t)} \Bf \cdot d\BA =
\int_{S(t)} \lr{
\PD{t}{\Bf} + \Bv \lr{ \spacegrad \cdot \Bf } – \spacegrad \cross \lr{ \Bv \cross \Bf }
}
\cdot d\BA.
\end{equation}

Both of these look questionable at first glance, because neither has boundary term. However, they can be transformed with Stokes theorem to
\begin{equation}\label{eqn:more_feynmans_trick:60}
\ddt{} \int_{C(t)} \Bf \cdot d\Bx
=
\int_{C(t)} \lr{
\PD{t}{\Bf} – \Bv \cross \lr{ \spacegrad \cross \Bf }
}
\cdot d\Bx
+
\evalbar{\Bv \cdot \Bf }{\Delta C},
\end{equation}
and
\begin{equation}\label{eqn:more_feynmans_trick:80}
\ddt{} \int_{S(t)} \Bf \cdot d\BA =
\int_{S(t)} \lr{
\PD{t}{\Bf} + \Bv \lr{ \spacegrad \cdot \Bf }
}
\cdot d\BA
–
\oint_{\partial S(t)} \lr{ \Bv \cross \Bf } \cdot d\Bx.
\end{equation}
The area integral derivative is now seen to be a variation of one of the special cases of the Leibniz integral rule, see for example [2]. The author admits that the line integral relationship is not well used, and doesn’t show up in the wikipedia page.

My end goal will be to evaluate the derivative of a general multivector line integral
\begin{equation}\label{eqn:more_feynmans_trick:100}
\ddt{} \int_{C(t)} F d\Bx G,
\end{equation}
and area integral
\begin{equation}\label{eqn:more_feynmans_trick:120}
\ddt{} \int_{S(t)} F d^2\Bx G.
\end{equation}
We’ve derived that line integral result in a different fashion previously, but it’s interesting to see a different approach. Perhaps this approach will lend itself nicely to non-scalar integrands?

Prerequisites.

Definition 1.1: Convective derivative.

The convective derivative,
of \( \phi(t, \Bx(t)) \) is defined as
\begin{equation*}
\frac{D \phi}{D t} = \lim_{\Delta t \rightarrow 0} \frac{ \phi(t + \Delta t, \Bx + \Delta t \Bv) – \phi(t, \Bx)}{\Delta t},
\end{equation*}
where \( \Bv = d\Bx/dt \).

Theorem 1.1: Convective derivative.

The convective derivative operator may be written
\begin{equation*}
\frac{D}{D t} = \PD{t}{} + \Bv \cdot \spacegrad.
\end{equation*}

Start proof:

Let’s write
\begin{equation}\label{eqn:more_feynmans_trick:140}
\begin{aligned}
v_0 &= 1 \\
u_0 &= t + v_0 h \\
u_k &= x_k + v_k h, k \in [1,3] \\
\end{aligned}
\end{equation}

The limit, if it exists, must equal the sum of the individual limits
\begin{equation}\label{eqn:more_feynmans_trick:160}
\frac{D \phi}{D t} = \sum_{\alpha = 0}^3 \lim_{\Delta t \rightarrow 0} \frac{ \phi(u_\alpha + v_\alpha h) – \phi(t, Bx)}{h},
\end{equation}
but that is just a sum of derivitives, which can be evaluated by chain rule
\begin{equation}\label{eqn:more_feynmans_trick:180}
\begin{aligned}
\frac{D \phi}{D t}
&= \sum_{\alpha = 0}^{3} \evalbar{ \PD{u_\alpha}{\phi(u_\alpha)} \PD{h}{u_\alpha} }{h = 0} \\
&= \PD{t}{\phi} + \sum_{k = 1}^3 v_k \PD{x_k}{\phi} \\
&= \lr{ \PD{t}{} + \Bv \cdot \spacegrad } \phi.
\end{aligned}
\end{equation}

End proof.

Definition 1.2: Hestenes overdot notation.

We may use a dot or a tick with a derivative operator, to designate the scope of that operator, allowing it to operate bidirectionally, or in a restricted fashion, holding specific multivector elements constant. This is called the Hestenes overdot notation.Illustrating by example, with multivectors \( F, G \), and allowing the gradient to act bidirectionally, we have
\begin{equation*}
\begin{aligned}
F \spacegrad G
&=
\dot{F} \dot{\spacegrad} G
+
F \dot{\spacegrad} \dot{G} \\
&=
\sum_i \lr{ \partial_i F } \Be_i G + \sum_i F \Be_i \lr{ \partial_i G }.
\end{aligned}
\end{equation*}
The last step is a precise statement of the meaning of the overdot notation, showing that we hold the position of the vector elements of the gradient constant, while the (scalar) partials are allowed to commute, acting on the designated elements.

We will need one additional identity

Lemma 1.1: Gradient of dot product (one constant vector.)

Given vectors \( \Ba, \Bb \) the gradient of their dot product is given by
\begin{equation*}
\spacegrad \lr{ \Ba \cdot \Bb }
= \lr{ \Bb \cdot \spacegrad } \Ba – \Bb \cdot \lr{ \spacegrad \wedge \Ba }
+ \lr{ \Ba \cdot \spacegrad } \Bb – \Ba \cdot \lr{ \spacegrad \wedge \Bb }.
\end{equation*}
If \( \Bb \) is constant, this reduces to
\begin{equation*}
\spacegrad \lr{ \Ba \cdot \Bb }
=
\dot{\spacegrad} \lr{ \dot{\Ba} \cdot \Bb }
= \lr{ \Bb \cdot \spacegrad } \Ba – \Bb \cdot \lr{ \spacegrad \wedge \Ba }.
\end{equation*}

Start proof:

The \( \Bb \) constant case is trivial to prove. We use \( \Ba \cdot \lr{ \Bb \wedge \Bc } = \lr{ \Ba \cdot \Bb} \Bc – \Bb \lr{ \Ba \cdot \Bc } \), and simply expand the vector, curl dot product
\begin{equation}\label{eqn:more_feynmans_trick:200}
\Bb \cdot \lr{ \spacegrad \wedge \Ba }
=
\Bb \cdot \lr{ \dot{\spacegrad} \wedge \dot{\Ba} }
= \lr{ \Bb \cdot \dot{\spacegrad} } \dot{\Ba} – \dot{\spacegrad} \lr{ \dot{\Ba} \cdot \Bb }. \end{equation}
Rearrangement proves that \( \Bb \) constant identity. The more general statement follows from a chain rule evaluation of the gradient, holding each vector constant in turn
\begin{equation}\label{eqn:more_feynmans_trick:320}
\spacegrad \lr{ \Ba \cdot \Bb }
=
\dot{\spacegrad} \lr{ \dot{\Ba} \cdot \Bb }
+
\dot{\spacegrad} \lr{ \dot{\Bb} \cdot \Ba }.
\end{equation}

End proof.

Time derivative of a line integral of a vector field.

We now have all our tools assembled, and can proceed to evaluate the derivative of the line integral. We want to show that

Theorem 1.2:

Given a path parameterized by \( \Bx(\lambda) \), where \( d\Bx = (\PDi{\lambda}{\Bx}) d\lambda \), with points along a \( C(t) \) moving through space at a velocity \( \Bv(\Bx(\lambda)) \), and a vector function \( \Bf = \Bf(t, \Bx(\lambda)) \),
\begin{equation*}
\ddt{} \int_{C(t)} \Bf \cdot d\Bx =
\int_{C(t)} \lr{
\PD{t}{\Bf} + \spacegrad \lr{ \Bf \cdot \Bv } + \Bv \cdot \lr{ \spacegrad \wedge \Bf}
} \cdot d\Bx
\end{equation*}

Start proof:

I’m going to avoid thinking about the rigorous details, like any requirements for curve continuity and smoothness. We will however, specify that the end points are given by \( [\lambda_1, \lambda_2] \). Expanding out the parameterization, we seek to evaluate
\begin{equation}\label{eqn:more_feynmans_trick:240}
\int_{C(t)} \Bf \cdot d\Bx
=
\int_{\lambda_1}^{\lambda_2} \Bf(t, \Bx(\lambda) ) \cdot \frac{\partial \Bx}{\partial \lambda} d\lambda.
\end{equation}
The parametric form nicely moves all the boundary time dependence into the integrand, allowing us to write
\begin{equation}\label{eqn:more_feynmans_trick:260}
\begin{aligned}
\ddt{} \int_{C(t)} \Bf \cdot d\Bx
&=
\lim_{\Delta t \rightarrow 0}
\inv{\Delta t}
\int_{\lambda_1}^{\lambda_2}
\lr{ \Bf(t + \Delta t, \Bx(\lambda) + \Delta t \Bv(\Bx(\lambda) ) \cdot \frac{\partial}{\partial \lambda} \lr{ \Bx + \Delta t \Bv(\Bx(\lambda)) } – \Bf(t, \Bx(\lambda)) \cdot \frac{\partial \Bx}{\partial \lambda} } d\lambda \\
&=
\lim_{\Delta t \rightarrow 0}
\inv{\Delta t}
\int_{\lambda_1}^{\lambda_2}
\lr{ \Bf(t + \Delta t, \Bx(\lambda) + \Delta t \Bv(\Bx(\lambda) ) – \Bf(t, \Bx)} \cdot \frac{\partial \Bx}{\partial \lambda} d\lambda \\
&\quad+
\lim_{\Delta t \rightarrow 0}
\int_{\lambda_1}^{\lambda_2}
\Bf(t + \Delta t, \Bx(\lambda) + \Delta t \Bv(\Bx(\lambda) )) \cdot \PD{\lambda}{}\Bv(\Bx(\lambda)) d\lambda \\
&=
\int_{\lambda_1}^{\lambda_2}
\frac{D \Bf}{Dt} \cdot \frac{\partial \Bx}{\partial \lambda} d\lambda +
\lim_{\Delta t \rightarrow 0}
\int_{\lambda_1}^{\lambda_2}
\Bf(t + \Delta t, \Bx(\lambda) + \Delta t \Bv(\Bx(\lambda) \cdot \frac{\partial}{\partial \lambda} \Bv(\Bx(\lambda)) d\lambda \\
&=
\int_{\lambda_1}^{\lambda_2}
\lr{ \PD{t}{\Bf} + \lr{ \Bv \cdot \spacegrad } \Bf } \cdot \frac{\partial \Bx}{\partial \lambda} d\lambda
+
\int_{\lambda_1}^{\lambda_2}
\Bf \cdot \frac{\partial \Bv}{\partial \lambda} d\lambda
\end{aligned}
\end{equation}
At this point, we have a \( d\Bx \) in the first integrand, and a \( d\Bv \) in the second. We can expand the second integrand, evaluating the derivative using chain rule to find
\begin{equation}\label{eqn:more_feynmans_trick:280}
\begin{aligned}
\Bf \cdot \PD{\lambda}{\Bv}
&=
\sum_i \Bf \cdot \PD{x_i}{\Bv} \PD{\lambda}{x_i} \\
&=
\sum_{i,j} f_j \PD{x_i}{v_j} \PD{\lambda}{x_i} \\
&=
\sum_{j} f_j \lr{ \spacegrad v_j } \cdot \PD{\lambda}{\Bx} \\
&=
\sum_{j} \lr{ \dot{\spacegrad} f_j \dot{v_j} } \cdot \PD{\lambda}{\Bx} \\
&=
\dot{\spacegrad} \lr{ \Bf \cdot \dot{\Bv} } \cdot \PD{\lambda}{\Bx}.
\end{aligned}
\end{equation}
Substitution gives
\begin{equation}\label{eqn:more_feynmans_trick:300}
\begin{aligned}
\ddt{} \int_{C(t)} \Bf \cdot d\Bx
&=
\int_{C(t)}
\lr{ \PD{t}{\Bf} + \lr{ \Bv \cdot \spacegrad } \Bf + \dot{\spacegrad} \lr{ \Bf \cdot \dot{\Bv} } } \cdot \frac{\partial \Bx}{\partial \lambda} d\lambda \\
&=
\int_{C(t)}
\lr{ \PD{t}{\Bf}
+ \spacegrad \lr{ \Bf \cdot \Bv }
+ \lr{ \Bv \cdot \spacegrad } \Bf
– \dot{\spacegrad} \lr{ \dot{\Bf} \cdot \Bv }
} \cdot d\Bx \\
&=
\int_{C(t)}
\lr{ \PD{t}{\Bf}
+ \spacegrad \lr{ \Bf \cdot \Bv }
+ \Bv \cdot \lr{ \spacegrad \wedge \Bf }
} \cdot d\Bx,
\end{aligned}
\end{equation}
where the last simplification utilizes lemma 1.1.

End proof.

Since \( \Ba \cdot \lr{ \Bb \wedge \Bc } = -\Ba \cross \lr{ \Bb \cross \Bc } \), observe that we have also recovered \ref{eqn:more_feynmans_trick:20}.

Time derivative of a line integral of a bivector field.

For a bivector line integral, we have

Theorem 1.3:

Given a path parameterized by \( \Bx(\lambda) \), where \( d\Bx = (\PDi{\lambda}{\Bx}) d\lambda \), with points along a \( C(t) \) moving through space at a velocity \( \Bv(\Bx(\lambda)) \), and a bivector function \( B = B(t, \Bx(\lambda)) \),
\begin{equation*}
\ddt{} \int_{C(t)} B \cdot d\Bx =
\int_{C(t)}
\PD{t}{B} \cdot d\Bx + \lr{ d\Bx \cdot \spacegrad } \lr{ B \cdot \Bv } + \lr{ \lr{ \Bv \wedge d\Bx } \cdot \spacegrad } \cdot B.
\end{equation*}

Start proof:

Skipping the steps that follow our previous proceedure exactly, we have
\begin{equation}\label{eqn:more_feynmans_trick:340}
\ddt{} \int_{C(t)} B \cdot d\Bx =
\int_{C(t)}
\PD{t}{B} \cdot d\Bx + \lr{ \Bv \cdot \spacegrad } B \cdot d\Bx + B \cdot d\Bv.
\end{equation}
Since
\begin{equation}\label{eqn:more_feynmans_trick:360}
\begin{aligned}
B \cdot d\Bv
&= B \cdot \PD{\lambda}{\Bv} d\lambda \\
&= B \cdot \PD{x_i}{\Bv} \PD{\lambda}{x_i} d\lambda \\
&= B \cdot \lr{ \lr{ d\Bx \cdot \spacegrad } \Bv },
\end{aligned}
\end{equation}
we have
\begin{equation}\label{eqn:more_feynmans_trick:380}
\ddt{} \int_{C(t)} B \cdot d\Bx
=
\int_{C(t)}
\PD{t}{B} \cdot d\Bx + \lr{ \Bv \cdot \spacegrad } B \cdot d\Bx + B \cdot \lr{ \lr{ d\Bx \cdot \spacegrad } \Bv } \\
\end{equation}
Let’s reduce the two last terms in this integrand
\begin{equation}\label{eqn:more_feynmans_trick:400}
\begin{aligned}
\lr{ \Bv \cdot \spacegrad } B \cdot d\Bx + B \cdot \lr{ \lr{ d\Bx \cdot \spacegrad } \Bv }
&=
\lr{ \Bv \cdot \spacegrad } B \cdot d\Bx –
\lr{ d\Bx \cdot \dot{\spacegrad} } \lr{ \dot{\Bv} \cdot B } \\
&=
\lr{ \Bv \cdot \spacegrad } B \cdot d\Bx
– \lr{ d\Bx \cdot \spacegrad} \lr{ \Bv \cdot B }
+ \lr{ d\Bx \cdot \dot{\spacegrad} } \lr{ \Bv \cdot \dot{B} } \\
&=
\lr{ d\Bx \cdot \spacegrad} \lr{ B \cdot \Bv }
+ \lr{ \Bv \cdot \dot{\spacegrad} } \dot{B} \cdot d\Bx
+ \lr{ d\Bx \cdot \dot{\spacegrad} } \lr{ \Bv \cdot \dot{B} } \\
&=
\lr{ d\Bx \cdot \spacegrad} \lr{ B \cdot \Bv }
+ \lr{ \Bv \lr{ d\Bx \cdot \spacegrad } – d\Bx \lr{ \Bv \cdot \spacegrad } } \cdot B \\
&=
\lr{ d\Bx \cdot \spacegrad} \lr{ B \cdot \Bv }
+ \lr{ \lr{ \Bv \wedge d\Bx } \cdot \spacegrad } \cdot B.
\end{aligned}
\end{equation}
Back substitution finishes the job.

End proof.

Time derivative of a multivector line integral.

Theorem 1.4: Time derivative of multivector line integral.

Given a path parameterized by \( \Bx(\lambda) \), where \( d\Bx = (\PDi{\lambda}{\Bx}) d\lambda \), with points along a \( C(t) \) moving through space at a velocity \( \Bv(\Bx(\lambda)) \), and multivector functions \( M = M(t, \Bx(\lambda)), N = N(t, \Bx(\lambda)) \),
\begin{equation*}
\ddt{} \int_{C(t)} M d\Bx N =
\int_{C(t)}
\frac{D}{D t} M d\Bx N + M \lr{ \lr{ d\Bx \cdot \dot{\spacegrad} } \dot{\Bv} } N.
\end{equation*}

It is useful to write this out explicitly for clarity
\begin{equation}\label{eqn:more_feynmans_trick:420}
\ddt{} \int_{C(t)} M d\Bx N =
\int_{C(t)}
\PD{t}{M} d\Bx N + M d\Bx \PD{t}{N}
+ \dot{M} \lr{ \Bv \cdot \dot{\spacegrad} } N
+ M \lr{ \Bv \cdot \dot{\spacegrad} } \dot{N}
+ M \lr{ \lr{ d\Bx \cdot \dot{\spacegrad} } \dot{\Bv} } N.
\end{equation}

Proof is left to the reader, but follows the patterns above.

It’s not obvious whether there is a nice way to reduce this, as we did for the scalar valued line integral of a vector function, and the vector valued line integral of a bivector function. In particular, our vector and bivector results had \( \spacegrad \lr{ \Bf \cdot \Bv } \), and \( \spacegrad \lr{ B \cdot \Bv } \) terms respectively, which allows for the boundary term to be evaluated using Stokes’ theorem. Is such a manipulation possible here?

Coming later: surface integrals!

References

[1] Nicholas Kemmer. Vector Analysis: A physicist’s guide to the mathematics of fields in three dimensions. CUP Archive, 1977.

[2] Wikipedia contributors. Leibniz integral rule — Wikipedia, the free encyclopedia. https://en.wikipedia.org/w/index.php?title=Leibniz_integral_rule&oldid=1223666713, 2024. [Online; accessed 22-May-2024].

Multivector form of Leibniz integral theorem for line integrals.

June 2, 2024 math and physics play , , , , ,

[Click here for a PDF version of this post]

Goal.

Here we will explore the multivector form of the Leibniz integral theorem (aka. Feynman’s trick in one dimension), as discussed in [1].

Given a boundary \( \Omega(t) \) that varies in time, we seek to evaluate
\begin{equation}\label{eqn:LeibnizIntegralTheorem:20}
\ddt{} \int_{\Omega(t)} F d^p \Bx \lrpartial G.
\end{equation}
Recall that when the bounding volume is fixed, we have
\begin{equation}\label{eqn:LeibnizIntegralTheorem:40}
\int_{\Omega} F d^p \Bx \lrpartial G = \int_{\partial \Omega} F d^{p-1} \Bx G,
\end{equation}
and expect a few terms that are variations of the RHS if we take derivatives.

Simplest case: scalar function, one variable.

With
\begin{equation}\label{eqn:LeibnizIntegralTheorem:60}
A(t) = \int_{a(t)}^{b(t)} f(u, t) du,
\end{equation}
If we can find an antiderivative, such that
\begin{equation}\label{eqn:LeibnizIntegralTheorem:80}
\PD{u}{F(u,t)} = f(u, t),
\end{equation}
or
\begin{equation}\label{eqn:LeibnizIntegralTheorem:90}
F(u, t) = \int f(u, t) du.
\end{equation}
The integral is made trivial
\begin{equation}\label{eqn:LeibnizIntegralTheorem:100}
\begin{aligned}
A(t)
&=
\int_{a(t)}^{b(t)} f(u, t) du \\
&=
\int_{a(t)}^{b(t)} \PD{u}{F(u,t)} du \\
&= F( b(t), t ) – F( a(t), t ).
\end{aligned}
\end{equation}
Should we attempt to take derivatives, we have a contribution from the first parameter that is entirely dependent on the boundary, and a contribution from the second parameter that is entirely independent of the boundary. That is
\begin{equation}\label{eqn:LeibnizIntegralTheorem:120}
\begin{aligned}
\ddt{} \int_{a(t)}^{b(t)} f(u, t) du
&=
\PD{b}{ F } \PD{t}{b}
-\PD{a}{ F } \PD{t}{a}
+ \evalrange{\PD{t}{F(u, t)}}{u = a(t)}{b(t)} \\
&=
f(b(t), t) b'(t) –
f(a(t), t) a'(t)
+ \int_{a(t)}^{b(t)} \PD{t}{} f(u, t) du.
\end{aligned}
\end{equation}
In the second step, the antiderivative function \( F \) has been restated in it’s original integral form \ref{eqn:LeibnizIntegralTheorem:90}. We are able to take the derivative into the integral, since we first evaluate that derivative, independent of the boundary, and then evaluate the result at the respective end points of the boundary.

Next simplest case: Multivector line integral (perfect derivative.)

Given an \( N \) dimensional vector space, and a path parameterized by vector \( \Bx = \Bx(u) \). The line integral special case of the fundamental theorem of calculus is found by evaluating
\begin{equation}\label{eqn:LeibnizIntegralTheorem:140}
\int F(u) d\Bx \lrpartial G(u),
\end{equation}
where \( F, G \) are multivectors, and
\begin{equation}\label{eqn:LeibnizIntegralTheorem:160}
\begin{aligned}
d\Bx &= \PD{u}{\Bx} du = \Bx_u du \\
\lrpartial &= \Bx^u \stackrel{ \leftrightarrow }{\PD{u}{}},
\end{aligned}
\end{equation}
where \( \Bx_u \Bx^u = \Bx_u \cdot \Bx^u = 1 \).

Evaluating the integral, we have
\begin{equation}\label{eqn:LeibnizIntegralTheorem:180}
\begin{aligned}
\int F(u) d\Bx \lrpartial G(u)
&=
\int F(u) \Bx_u du \Bx^u \stackrel{ \leftrightarrow }{\PD{u}{}} G(u) \\
&=
\int du \PD{u}{} \lr{ F(u) G(u) } \\
&=
F(u) G(u).
\end{aligned}
\end{equation}

If we allow \( F, G, \Bx \) to each have time dependence
\begin{equation}\label{eqn:LeibnizIntegralTheorem:200}
\begin{aligned}
F &= F(u, t) \\
G &= G(u, t) \\
\Bx &= \Bx(u, t),
\end{aligned}
\end{equation}
so we have
\begin{equation}\label{eqn:LeibnizIntegralTheorem:220}
\ddt{} \int_{u = a(t)}^{b(t)} F(u, t) d\Bx \lrpartial G(u, t)
=
\evalrange{ \ddt{u} \PD{u}{} \lr{ F(u, t) G(u, t) } }{u = a(t)}{b(t)}
+ \evalrange{\ddt{} \lr{ F(u, t) G(u, t) } }{u = a(t)}{b(t)}
.
\end{equation}

General multivector line integral.

Now suppose that we have a general multivector line integral
\begin{equation}\label{eqn:LeibnizIntegralTheorem:240}
A(t) = \int_{a(t)}^{b(t)} F(u, t) d\Bx G(u, t),
\end{equation}
where \( d\Bx = \Bx_u du \), \( \Bx_u = \partial \Bx(u, t)/\partial u \). Writing out the integrand explicitly, we have
\begin{equation}\label{eqn:LeibnizIntegralTheorem:260}
A(t) = \int_{a(t)}^{b(t)} du F(u, t) \Bx_u(u, t) G(u, t).
\end{equation}
Following our logic with the first scalar case, let
\begin{equation}\label{eqn:LeibnizIntegralTheorem:280}
\PD{u}{B(u, t)} = F(u, t) \Bx_u(u, t) G(u, t).
\end{equation}
We can now evaluate the derivative
\begin{equation}\label{eqn:LeibnizIntegralTheorem:300}
\ddt{A(t)} = \evalrange{ \ddt{u} \PD{u}{B} }{u = a(t)}{b(t)} + \evalrange{ \PD{t}{}B(u, t) }{u = a(t)}{b(t)}.
\end{equation}
Writing \ref{eqn:LeibnizIntegralTheorem:280} in integral form, we have
\begin{equation}\label{eqn:LeibnizIntegralTheorem:320}
B(u, t) = \int du F(u, t) \Bx_u(u, t) G(u, t),
\end{equation}
so
\begin{equation}\label{eqn:LeibnizIntegralTheorem:340}
\begin{aligned}
\ddt{A(t)}
&= \evalrange{ \ddt{u} \PD{u}{B} }{u = a(t)}{b(t)} +
\evalbar{ \PD{t’}{} \int_{a(t)}^{b(t)} du F(u, t’) d\Bx_u(u, t’) G(u, t’) }{t’ = t} \\
&= \evalrange{ \ddt{u} F(u, t) \Bx_u(u, t) G(u, t) }{u = a(t)}{b(t)} +
\int_{a(t)}^{b(t)} \PD{t}{} F(u, t) d\Bx(u, t) G(u, t),
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:LeibnizIntegralTheorem:360}
\ddt{} \int_{a(t)}^{b(t)} F(u, t) d\Bx(u, t) G(u, t)
= \evalrange{ F(u, t) \ddt{\Bx}(u, t) G(u, t) }{u = a(t)}{b(t)} +
\int_{a(t)}^{b(t)} \PD{t}{} F(u, t) d\Bx(u, t) G(u, t).
\end{equation}

This is perhaps clearer, if just written as:
\begin{equation}\label{eqn:LeibnizIntegralTheorem:380}
\ddt{} \int_{a(t)}^{b(t)} F d\Bx G
= \evalrange{ F \ddt{\Bx} G }{u = a(t)}{b(t)} +
\int_{a(t)}^{b(t)} \PD{t}{} F d\Bx G.
\end{equation}
As a check, it’s worth pointing out that we can recover the one dimensional result, writing \( \Bx = u \Be_1 \), \( f = F \Be_1^{-1} \), and \( G = 1 \), for
\begin{equation}\label{eqn:LeibnizIntegralTheorem:400}
\ddt{} \int_{a(t)}^{b(t)} f du
= \evalrange{ f(u) \ddt{u} }{u = a(t)}{b(t)} +
\int_{a(t)}^{b(t)} du \PD{t}{f}.
\end{equation}

Next steps.

I’ve tried a couple times on paper to do surface integral variations of this (allowing the surface to vary with time), and don’t think that I’ve gotten it right. Will try again (or perhaps just look it up and see what the result is supposed to look like, then see how that translates into the GC formalism.)

References

[1] Wikipedia contributors. Leibniz integral rule — Wikipedia, the free encyclopedia. https://en.wikipedia.org/w/index.php?title=Leibniz_integral_rule&oldid=1223666713, 2024. [Online; accessed 22-May-2024].