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This is my first attempt to tackle something I wish I had when I was doing my physics degree: some step-by-step solutions to problems that explained how each piece was done in a painfully clear way (sometimes I was pretty thick and missed important points that are maybe obvious to me now, looking back). I’ve purchased some “problems with solutions” books, but most of them assume that I have a much deeper and broader knowledge that I apparently have, so I have not gotten a lot of use out of them. As a result, another thing I realized I could have benefited from is a more integrated approach, where whatever tools that are needed are explained as the solution is presented. I’ve been told all my life that the best way to learn something is to try to teach it to someone else, and so this is part of my attempt to become more agile and comfortable with the ideas and tools needed to do modern physics. If you don’t understand something, please ask, because I haven’t explained it well enough and others will surely be confused as well. If it works out, I am going to be calling this series Phelonius’ Pedantic Solutions — hopefully more in the “excessively concerned with formalism, accuracy, and precision” rather than the “one who makes an ostentatious and arrogant show of learning” sense [1].
Things covered:
- intro to relativistic energy equation (including Lorentz transformation)
- vector versus scalar quantities and notation
- Taylor expansion as a way of representing arbitrary functions as an infinite series
- single-variable differentiation (including sum, power, and exponentiation rules)
Pretty much everyone knows Einstein’s famous declaration that \(E = mc^{2}\), which reads: energy is equal to mass, \(m\), times the speed of light, \(c\), squared. This could be written more explicitly as \(E = m\times c^{2}\) or \(E = m\cdot c^{2}\) to indicate the multiplication between the \(m\) and the \(c^{2}\); but when no operator is present between variables, multiplication is generally implied in these sorts of simple equations (for more advanced stuff, this is not always true as other operators can be expected, but the context usually makes it clear what to expect). This revolutionary idea meant that the notion in classical physics that matter can neither be created nor destroyed was incorrect (energy, on the other hand, is conserved so far as we have been able to measure). However, that equation is just an approximation at the low velocities we experience in everyday life (velocity, \(\vec{v}\) is just speed in a particular direction, it is a vector quantity). But when things get fast in the universe, Einstein’s theory of Special Relativity tells us the actual equation is \(E = \gamma mc^{2}\), where \(\gamma = 1/\sqrt{1 – \vec{v}^{2}/c^{2}} =\) \(1/\sqrt{1 – v^{2}/c^{2}}\) (which is the Lorentz transformation, a mathematical relationship that was developed before Einstein started his work on relativity… n.b. there will be more on the disappearing vector notation arrow shortly). Notice that as \(\vec{v} \rightarrow 0\) (as the velocity approaches zero), \(v^{2}/c^{2} \rightarrow 0\), so \(\gamma \rightarrow 1/\sqrt{1 – 0} = 1/1 = 1\), which means at low velocities, \(E \approx mc^{2}\) (\(E\) is approximately equal to \(mc^{2}\)). There is a lot more to relativity than the Lorentz transformation, but this is enough to set up the problem outlined below.
Two things are important to notice about \(\gamma\): \(v^{2}\) can never be as large as \(c^{2}\), and as \(v^{2} \rightarrow c^{2}\), \(E\) will increase without bound. The first is because if \(v^{2} = c^{2}\), then \(v^{2}/c^{2} = c^{2}/c^{2} = 1\), which means that \(\gamma = 1/\sqrt{1 – v^{2}/c^{2}} = 1/\sqrt{1 – 1} = 1/0\) which is a no-no in mathematics (anything divided by zero is an undefined infinity)! This implies that, in our universe, if anything with mass approaches the speed of light, \(\gamma\) grows without bound to infinity, which means that the energy grows without bound to infinity, which is the second observation. Because of this \(1 – v^{2}/c^{2}\) relationship, objects with mass can travel close to the speed of light but can never go faster because if it reached the speed of light it would have to somehow get past these infinities. You might well ask, well what about light? It travels at the speed of light doesn’t it? The answer is that because it travels at the speed of light, \(c\), it must have zero mass (\(m = 0\))! However, since light has energy, this obviously isn’t the whole story, but more on that some other time…
So… the relativistic energy equation is:
\begin{equation}
E = \gamma mc^{2} = \frac{mc^{2}}{\sqrt{1\, – v^{2}/c^{2}}}\label{emc2}\tag{1}
\end{equation}
Because I don’t want to introduce ideas without at least some explanation, here is a little aside about velocity. Velocity is a vector and is usually written \(\vec{v}\), where the arrow over the letter tells us it’s a vector rather than some other kind of value. A vector has both a magnitude and a direction. Looking at velocity, it’s magnitude is its speed (say, 60km/h) and it’s direction is exactly what you think (say, heading north-east down a hill). What other kinds of values are there? Well, there are all kinds in math, but one of the main other kinds used in physics at this level is called a scalar. A scalar is just a number with no direction. For instance, speed is a scalar, and so is temperature (it’s never, say, 20.6°C west… it’s just 20.6°C). So then, why am I writing \(v^{2}\) instead of \(\vec{v}^{2}\)? Without getting into the grit of it (which I will do at some later time presumably), a vector times itself is a scalar: \(\vec{v}^{2} = \vec{v}\cdot\vec{v} = v^{2} =\) speed\(^{2}\), where the dot in \(\vec{v}\cdot\vec{v}\) is just a kind of multiplication called a dot product, or scalar product since the answer is always a scalar. A scalar times a scalar is also always a scalar, e.g. 2 · 3 = 6, never something like 2 · 3 = 6 north. Any vector times a scalar is a vector: if you’re heading north at 20km/h and you double your speed, you will be heading north at 40km/h. This would be written: 2 · 20km/h north = 40km/h north (the magnitude of the vector, speed in this case, changes but not its direction). Because vectors are more complicated than scalars, they can sometimes[2] have two kinds of multiplication, and the other kind is called a cross product, or vector product because the answer is a vector (this operation is used only for 3-dimensional vectors in undergraduate physics, but it can be used in other ways in more advanced mathematics). Since the cross product is written \(\vec{v}\times\vec{v}\), it can get confusing with the × symbol used for grade-school math. The general approach is to use no symbol or the dot for the multiplication of scalars (e.g. \(mc^{2}\) or \(m\cdot c^{2}\)), or scalars times vectors (e.g. \(2\vec{v}\) or \(2\cdot\vec{v}\)); the dot symbol when multiplying vectors and expecting a scalar result (e.g. \(\vec{v}\cdot\vec{v} = v^{2}\)); and the cross product symbol should only be used for vector products when the result is expected to be another vector (e.g. \(\vec{u}\times\vec{v} = \vec{w}\)).
All to say that while it’s perfectly valid to put the arrow over the velocity squared, \(\vec{v}^{2}\), it doesn’t really matter, so people just leave it off, \(v^{2}\). It also shows explicitly that \(v^{2}\) is a number (a scalar, which is just the speed of something squared), just like the speed of light squared (another scalar), so that one divided by the other is just another number (and not a vector or something else) that can be subtracted from the 1 in the square root (you can’t subtract a vector from a scalar, for instance). One last comment on vectors before moving on, you may see the following notation: \(|\vec{v}| = v\). This is a shorthand way of saying: just give me the magnitude of the vector. So \(|\vec{v}|\) is just the speed component of the velocity vector (the direction is stripped off). It is sometimes also called the length of the vector.
Now to the main reason for this post! In Griffith’s Introduction to Elementary Particles, Second Edition (ISBN 978-3-527-40601-2) p. 99, they take the relativistic energy equation (\ref{emc2}) above and “expand the radical in a Taylor series” to get:
\begin{equation}
E = mc^{2} \left(1 + \frac{1}{2}\frac{v^{2}}{c^{2}} + \frac{3}{8}\frac{v^{4}}{c^{4}} + \cdots\right) \label{gexp1}\tag{2}
\end{equation}
so, multiplying it out (each term by \(mc^{2}\)) and simplifying, it gives:
\begin{align}
E & = 1\cdot mc^{2} + \frac{1}{2}\cdot mc^{2}\cdot\frac{v^{2}}{c^{2}} + \frac{3}{8}\cdot mc^{2}\cdot\frac{v^{4}}{c^{4}} + \cdots \\
& = mc^{2} + \frac{mv^{2}c^{2}}{2c^{2}} + \frac{3mv^{4}c^{2}}{8c^{4}} + \cdots\label{gexp2}\tag{3} \\
& = mc^{2} + \frac{1}{2}mv^{2} + \frac{3}{8}m\frac{v^{4}}{c^{2}} + \cdots
\end{align}
which is really interesting because, looking at the last line of equation (\ref{gexp2}), if the velocity is zero (i.e. the mass is at rest relative to the observer), then all but the first term on the right disappear (they are all zero) because they have a velocity term in them, and we get \(E = mc^{2}\), which is the rest mass of an object, as we would hope (physicists measure mass in terms of energy because of this equation since it’s easy to go back and forth between mass and energy)! More subtly, the second term is the classical (non-relativistic) kinetic energy (or energy of motion) of the object, which is very cool! The third and later terms (the higher order terms) in the series are then relativistic corrections to the energy equation, and when \(|\vec{v}|\) is much less than \(c\), then those higher order terms hardly contribute anything because of the higher powers of the velocity vs. the higher powers of the speed of light.
As a quick aside on exponentiation, just in case you’re rusty, when multiplying an exponentiated value by that same value to some other exponent, add the exponents together: e.g. \(a^{4}\cdot a^{3} = a^{4 + 3} = a^{7}\). When dividing exponentiated values, subtract the exponents: from above, e.g. \(c^{2} / c^{4} = c^{2 – 4} = c^{-2}\). Anything to a negative power is just 1 over the same value to its positive power: e.g. \(c^{-2} = 1 / c^{2}\), so \(c^{2} / c^{4} = 1 / c^{2}\) which is what we see above. When exponentiating an exponentiated value, multiply the exponents: e.g. \((d^{2})^{4} = d^{2\cdot 4} = d^{8}\) or \((q^{4})^{1/2} = q^{4\cdot 1/2} = q^{4 / 2} = q^{2}\). The values under the exponents must be the same to do any of the above operations (e.g. \(2^{2}\cdot3^{2}\neq6^{4}\) or some such)!
To demonstrate what was said about the contributions of higher order terms to the energy equation above, let’s assume that something is travelling at one tenth the speed of light, or to put it mathematically: \(|\vec{v}| = 1/10\cdot c = 0.1\,c\). For the second term, the ratio \(v^{2}/c^{2} =\) \((0.1\,c)^{2}/c^{2} =\) \(0.01\,c^{2}/c^{2} =\) \(0.01\), which is what \(1/2\cdot mc^{2}\) is multiplied by (in the top line of equation (\ref{gexp2}) above). For the third term, the ratio \(v^{4}/c^{4} =\) \((0.1\,c)^{4}/c^{4} =\) \(0.0001\,c^{4}/c^{4} =\) \(0.0001\), which is what \(3/8\cdot mc^{2}\) is multiplied by. If we were to keep going, the ratio \(v^{6}/c^{6} =\) \((0.1\,c)^{6}/c^{6} =\) \(0.000001\,c^{6}/c^{6} =\) \(0.000001\), and so on, each term getting 1000 times smaller than the one before it. There is a reason why early scientists didn’t notice the relativistic effects of moving objects: they just didn’t have accurate enough measuring tools! These days, smartphones have relativistic correction algorithms (using both Special and General Relativity) built into them so they can make sense of the GPS signals they receive for time and location from the fast-moving satellites (Special Relativity) orbiting in Earth’s gravitational field (General Relativity), both of which affect time and space as measured by the satellites. But more on that another time!
And now to the rub… Taylor series, aka Taylor expansions… I’ve always had trouble with them because I am not so good with the nuts and bolts of algebra and calculus (I make stupid mistakes and mess up things and have to go back and fix them). Most real physicists can look at “simple” equations like the relativistic energy equation and do the expansion in their heads (I am still a larval physicist as I just completed my undergraduate degree in 2017… I’m a bit late to the game, yes). I need a lot more practice, so when I was reading Griffith’s chapter on relativistic kinematics (the study of moving things) and ran across the phrase (paraphrased here, actually), “oh, you just need to expand this to see the implications”, and I had no intuitive sense of how to go from the before picture to the after picture, I realized it was time to sit down and do a refresher… thus this blog post.
The Taylor series is an infinite series that allows for any arbitrary “well behaved” (this has a mathematical definition that is pretty close to the way it sounds in plain English) mathematical function to be expressed as an infinite sum of terms. In practice, because of the diminishing effects of higher order terms in the series, most physicists stop after a few terms (which was done in Griffith’s) because there is no way to measure better than that level of precision in the real world, so it could be just wasted computational effort to go farther. Fyi, an equation that gives one answer on the left side of the equation, like \(E\), for every input value, like \(v\), is called a function. The equation for the Taylor series is written out:
\begin{equation}
f(x) = \sum_{n = 0}^{\infty}\frac{f^{(n)}(x_0)}{n!}(x\, – x_0)^{n}\label{taylor}\tag{4}
\end{equation}
which is quite a mouthful to say the least, but it’s a pretty straightforward equation to unpack if one can avoid panicking at the sight of it (which I certainly did when I was starting out).
The first thing to examine is that the left hand side says that any function of a variable, let’s say \(x\) (but it doesn’t matter what variable is used), can be expressed as the stuff on the right hand side of the equation. This is an extremely powerful tool for understanding the behaviour of equations! The next bit to tackle is the \(x_0\) on the right hand side: the Taylor expansion is said to be “around that point”, so it is important to choose the right point to be looking at. If possible, it is usually preferable to select a point where the function is symmetric on either side of it, for instance, \(x_0 = 0\) for a sine or cosine function that stretches off to infinity in either direction. If you choose this point wrong, then you will need to work harder to get an accurate representation of your function using a Taylor series (it can still be done, but you won’t be happy with the mess that results). Note that sometimes you might want to do a Taylor expansion around a point at some extreme of the function (e.g. where \(x_0 = \infty\) or where the function starts to behave badly), but most expansions I’ve seen tend to be around \(x_0 = 0\), which is what we’ll use here, and I’ll explain why.
If you read back, you’ll see that the relativistic energy equation “blows up” (goes to infinity) when the speed of the mass approaches the speed of light (we say it mathematically as “\(E \rightarrow \infty\) as \(|\vec{v}| \rightarrow c\)”). This means that the equation is well behaved on the interval of \(|\vec{v}| < c\) (i.e. when \(-c < v < +c\)), but is badly behaved at \(|\vec{v}|\approx c\) (when the magnitude of the velocity, i.e. speed, is approximately the speed of light... it is implied that it is less than the speed of light though because of the laws of physics don't allow going past the speed of light). Note that \(|\vec{v}| \geq 0\) since, for example, if you're driving backward at 5km/h (call it –5km/h), you’re going the same speed as if you were driving forward at 5km/h (which would be +5km/h), it’s just the direction that is different, and the \(|\vec{v}|\) operation strips the direction off the vector (which in this simple case would just be the sign). It is not a coincidence that the vertical bars used are the same as what is used to take the absolute value of a scalar (e.g. \(|3| = 3\) and \(|-3| = 3\))! So, the relation \(|\vec{v}| < c\) says that the speed of the mass must be greater than \(-c\), but less than \(c\) (remember, if \(v = c\), we get \(1/0\) and unmanageable infinities). Since we square everything, it doesn't make any difference to the value of the energy whether our velocity is in any particular direction, so it is logical in this case to hope that the simplest (and hopefully most intuitively meaningful) expansion will be around the point \(|\vec{v}| = 0\) for our equation because it is in the middle of the interval where \(\vec{v}\) gives well-behaved values for \(E\) (since \((+c) + (-c) = c - c = 0\)).
Which brings us to the last point before simplifying the Taylor series formula as much as possible for this application (always a good thing to do before starting the calculation): it doesn’t matter what variable you choose, it’s just a label. If your equation is a function of variable \(x\), then the given Taylor series equation is used “as is”. But if your variable is, as in this case, \(\vec{v}\), then you just change everywhere there’s an \(x\) to a \(\vec{v}\) (including changing \(x_0\) to \(\vec{v}_0\))! If your variable was \(\odot\), you’d just change your equation to read:
\[f(\odot) = \sum_{n = 0}^{\infty}\frac{f^{(n)}(\odot_0)}{n!}(\odot\, – \odot_0)^{n}\]
Just fill in the blank. So, in the case of expanding the relativistic energy equation, with the point chosen to expand around (i.e. \(\vec{v}_{0} = 0\)), we are “just” going to be solving:
\begin{align}
E = f(\vec{v}) & = \sum_{n = 0}^{\infty}\frac{f^{(n)}(\vec{v}_0)}{n!}(\vec{v}\, – \vec{v}_0)^{n} \\
& = \sum_{n = 0}^{\infty}\frac{f^{(n)}(0)}{n!}(\vec{v}\, – 0)^{n} = \sum_{n = 0}^{\infty}\frac{f^{(n)}(0)}{n!}\vec{v}^{\,n}\label{staylor}\tag{5}
\end{align}
which is already a lot simpler than what we started with, and it looks a little closer to where we’re going. The only thing that seems odd is the vector symbol \(\vec{v}\) in this equation versus the non-vector symbol in the relativistic energy equation (\ref{emc2}). Here, I’m being ultra-pedantic because I was always bothered when my physics profs would play fast and loose with notation. I could very well write the energy equation as follows to make it match:
\[ E = \frac{mc^{2}}{\sqrt{1\, – \vec{v}^{2}/c^{2}}} \]
since \(v^{2} = \vec{v}^{2}\), as explained above; however, until it’s squared, I really think the vector notation should be preserved. Why? Well, besides being picky (I would say “precise”), it suggests there might be a way of reducing the number of calculations we have to do by half! If you notice the \(\vec{v}^{\,n}\) term in our simplified Taylor series equation, it implies that for every even value of \(n\), we will get a scalar; but for every odd value of \(n\), we will get a vector (for example, \(\vec{v}^{2} = v^2\) is a scalar, but \(\vec{v}^{3} = \vec{v}^{2}\cdot\vec{v} = v^2\cdot \vec{v}\) is a scalar times a vector, which is a vector). Since energy is a scalar, the function \(f(\vec{v})\) will be a scalar for all values of \(n\), which means \(f^{(n)}(0)\) should also generate a scalar for all values of \(n\) (the term \(f^{(n)}(0)\) is the nth derivative of the function \(f(\vec{v})\) evaluated at 0, which will be explained in detail below… suffice it to say taking the derivative of a scalar function can’t magically turn it into a vector). The only way for this to work out is if \(f^{(n)}(0) = 0\) for all odd values of \(n\) so we never end up with a vector in any of our terms (recall when \(n\) is odd, \(\vec{v}^{\,n}\) is a vector, but we’re not expecting any vectors in the expansion, so we would need to multiply those terms by 0 to get rid of them). This is precisely what happens and which I’ll demonstrate explicitly below. So if you were doing this kind of problem on your own, and could think it through, you could make the above argument and then calculate only the values where \(n\) is even… effort saved (and you look smart because you thought things out, something I often had trouble doing until it was too late in the calculation)!
The last tricky bit of the right side of the equation is \(f^{(n)}(0)\), which states: take the nth derivative of the function, then evaluate it when \(\vec{v} = 0\). Taking the derivative of a function is figuring out the rate of change of that function for all values of the variable, \(\vec{v}\) in this case. When there is only one variable, it is taking the full derivative (which is what to do here); and when there are multiple variables, taking the derivative with respect to one variable is a partial derivative. The process is called differentiation (in contrast to its inverse operation, integration… note: differentiation is way easier than integration usually). There are only a couple of rules to know in order to expand the relativistic energy function. These are simple “plug and chug” techniques that are purely rules-based (I say simple because the rule is easy, even if applying it and not messing up might not be easy). One of the most common (and I would argue, flexible) ways to write “take the derivative of a function with respect to a specified variable” is the Leibniz notation [3] — let’s go with \(E\) which is a function of \(\vec{v}\), so \(E = f(\vec{v})\) (which literally just says “\(E\) is a function of \(\vec{v}\)):
\[\frac{d}{d\vec{v}}E = \frac{d}{d\vec{v}}f(\vec{v}) = \frac{d}{d\vec{v}}\left(\frac{mc^{2}}{\sqrt{1 – \vec{v}^{2}/c^{2}}}\right)\]
where I have used \(\vec{v}^{2}\) instead of \(v^{2}\) just to be crystal clear that the variable is \(\vec{v}\) (a vector) even though \(\vec{v}^{2} = v^{2}\) (which begs the question, why not just use \(\vec{v}^{2}\) everywhere, but the answer is that in any textbook or reference it will be \(v^{2}\) and you just have to know that there’s a vector underneath that becomes a scalar when squared). The \(d/d\vec{v}\) just says “take the full derivative of whatever comes next, with respect to the variable \(\vec{v}\)”. It can be written more compactly with the “whatever comes next” on top (\(dE/d\vec{v}\))… it’s just like multiplication in that regard, but differentiation is not multiplication, but rather something called an operator, which may have different rules (the differentiation operator doesn’t behave the same way as multiplication in what it does: for instance, multiplication is commutative, so \(a\cdot b = b\cdot a\), but \(d/d\vec{v}\cdot E \neq E\cdot d/d\vec{v}\)… the differentiation operator operates on what is to the right of it only).
If the function can be separated into chunks that are added or subtracted (which is just adding a chunk that is multiplied by \(-1\)), then each chunk is differentiated separately (the results should then be simplified if possible). e.g. If there is a function where \(g\), \(h\), \(j\), and \(k\) are constants:
\begin{align}
f(x) & = g + hx + jx^{2} – kx^{3} \\
& = gx^{0} + hx^{1} + jx^{2} – kx^{3}
\end{align}
(this is a polynomial of order 3, since that’s the highest exponent of its variable), then:
\[\frac{d}{dx}f(x) = \left(\frac{d}{dx}(gx^{0})\right) + \left(\frac{d}{dx}(hx^{1})\right) + \left(\frac{d}{dx}(jx^{2})\right) – \left(\frac{d}{dx}(kx^{3})\right)\]
Note that anything to the power of \(1\), e.g. \(x^{1}\), is just that same thing back, e.g. \(x\), and that anything to the power of zero equals \(1\), e.g. \(x^{0} = 1\), and yes, even zero to the power of zero equals \(1\), e.g. \(0^{0} = 1\). Don’t let it hurt your brain too much, it certainly hurts mine! Writing the exponents on \(x\) explicitly like this will help in a couple of steps.
The next thing is that any constants multiplying the variable can be taken to the left of the differentiation operator: \(d/dx(ax^{2}) = d/dx(a\cdot x^{2}) = a\cdot d/dx(x^{2})\). Using the same equation as above:
\[\frac{d}{dx}f(x) = \left(g\frac{d}{dx}(x^{0})\right) + \left(h\frac{d}{dx}(x^{1})\right) + \left(j\frac{d}{dx}(x^{2})\right) – \left(k\frac{d}{dx}(x^{3})\right)\]
Constants can get pretty complicated, so this often simplifies things a lot. The next thing is to actually take the derivative, and this has two simple rules that get used quite a lot (that can get messy as we shall see later). The first is called the product rule and says that if you have two functions multiplying each other that are functions of the same variable, e.g. \(g(y)\) and \(h(y)\) where \(f(y) = g(y)\cdot h(y)\), then apply the following expansion:
\[\frac{d}{dy}f(y) = \frac{d}{dy}\left[g(y)\cdot h(y)\right] = g(y)\cdot \left(\frac{d}{dy}h(y)\right) + \left(\frac{d}{dy}g(y)\right)\cdot h(y)\label{prodrule}\tag{6}\]
When the sub-chunks that are being differentiated also end up as two functions of the same variable multiplied together, this can go deep and get quite complicated, but the rule is still simple even though there is potentially a lot of “bookkeeping”. I will give an example below, after the second rule.
The second rule is to take the exponent of the chunk being differentiated and put it out front, subtract one from the exponent (this is the next term in the result, all of which are multiplied together), and then take the derivative of what’s below the exponentiation:
\[\frac{d}{d\oplus}f(\oplus)^{m} = m\cdot f(\oplus)^{m – 1}\cdot \frac{d}{d\oplus}f(\oplus)\]
Going back to the polynomial above:
\begin{align}
\frac{d}{dx}f(x) & = \left(g\cdot 0 \cdot x^{0 – 1}\frac{d}{dx}(x)\right) + \left(h\cdot 1\cdot x^{1 – 1}\frac{d}{dx}(x)\right) \\
& + \left(j\cdot 2\cdot x^{2 – 1}\frac{d}{dx}(x)\right) – \left(k\cdot 3\cdot x^{3 – 1}\frac{d}{dx}(x)\right)
\end{align}
so,
\begin{align}
\frac{d}{dx}f(x) & = g\cdot 0\cdot x^{-1}\cdot 1 + h\cdot 1\cdot x^{0}\cdot 1 + j\cdot 2\cdot x^{1}\cdot 1\, – k\cdot 3\cdot x^{2}\cdot 1 \\
& = 0 + hx^{0} + 2jx^{1} – 3kx^{2} = h + 2jx – 3kx^{2}
\end{align}
where, following the same rules:
\[\frac{d}{dx}(x) = \frac{d}{dx}(x^{1}) = 1\cdot x^{0}\cdot\frac{d}{dx}(x) = 1\cdot 1\cdot\frac{d}{dx}(x) = \frac{d}{dx}(x) = \cdots = 1\]
where we get an infinitely deep recursion that just keeps generating \(1\) to multiply the previous terms by, so by observation we can see that \(d/dx(x) = 1\). Another way of looking at it is just to write it \(dx/dx = 1\), which seems to make intuitive sense (any value divided by itself, as long as it’s not zero, is equal to \(1\)). This kind of thing supposedly makes a mathematician’s skin crawl, but in physics we treat differential operators as symbols that we can multiply and divide like any other symbol (as long as one remembers that the operators are not commutative and it matters what side you come in from, but that’s another topic for another day). Yet another way of looking at it is if \(f(x) = x\), then if \(x = 0\), \(f(x) = 0\); if \(x = 1\), \(f(x) = 1\); if \(x = 2\), \(f(x) = 2\); and so on. So, the rate of change (which is what taking the derivative measures) is: for every \(1\) the variable \(x\) increases, then \(f(x)\) increases by \(1\), so the rate of change is \(1\) for all \(x\), and therefore \(d/dx(x) = 1\).
Note that taking the derivative of a constant is always zero. E.g. for a function of \(x\), a constant, say \(a\), can be thought of as \(f(x) = a = a\cdot 1 = a\cdot x^{0}\), so \(df(x)/dx = d/dx(a\cdot x^{0}) = a\cdot 0\cdot x^{-1}\cdot dx/dx = 0\). If you think about it conceptually, a derivative determines the rate of change of what it is operating on, and since a constant never changes, it’s rate of change is always \(0\). Knowing this can speed things up a bit in doing differentiation. The second thing to note is that taking the derivative of an un-exponentiated variable just gives back \(1\), so if a term is a constant times a variable to the power of \(1\), then the result of taking the derivative is just the constant, e.g. \(d/dx(ax) = a\).
Putting the two rules together, let’s say we have a function:
\[f(\bowtie) = (2 – \bowtie^{2})(2\bowtie – 5)^{3}\]
One possibility of differentiating this is to expand the second part (multiply rightmost part by itself three times) and then multiply that by the first part. Then we could take the derivative of each of the resultant pieces and then simplify as best we could. This would be a long and tedious process to say the least. But using the various rules covered, we can go nearly straight to the answer. There are two parts of the equation that are multiplied together, so let’s declare that \(g(\bowtie) = (2 – \bowtie^{2}) = (2 – \bowtie^{2})^{1}\) and \(h(\bowtie) = (2\bowtie – 5)^{3}\). Using the product rule (\ref{prodrule}):
\[\frac{d}{d\bowtie}f(\bowtie) = (2 – \bowtie^{2})^{1}\cdot \left(\frac{d}{d\bowtie}(2\bowtie – 5)^{3}\right) + \left(\frac{d}{d\bowtie}(2 – \bowtie^{2})^{1}\right)\cdot (2\bowtie – 5)^{3}\]
Starting with the derivative of \(g(\bowtie)\):
\[\frac{d}{d\bowtie}(2 – \bowtie^{2})^{1} = 1\cdot (2 – \bowtie^{2})^{1 – 1}\cdot\frac{d}{d\bowtie}(2 – \bowtie^{2}) = 1\cdot 1\cdot\frac{d}{d\bowtie}(2 – \bowtie^{2})\]
\[= \frac{d}{d\bowtie}(2) – \frac{d}{d\bowtie}(\bowtie^{2}) = 0 – 2\cdot \bowtie^{1}\cdot 1 = -2\bowtie\]
In general, for an un-exponentiated term, and recognizing that one of the subterms is a simple exponentiated variable, we can jump pretty much right to the answer (using the distributive properly of the operator):
\[\frac{d}{d\bowtie}(2 – \bowtie^{2}) = \frac{d}{d\bowtie}(2)\,- \frac{d}{d\bowtie}(\bowtie^{2}) = 0 – 2\bowtie = -2\bowtie\]
Next, tackling the derivative of \(h(\bowtie)\):
\[\frac{d}{d\bowtie}(2\bowtie – 5)^{3} = 3\cdot (2\bowtie – 5)^{2}\cdot \frac{d}{d\bowtie}(2\bowtie – 5) \]
\[= 3\cdot (2\bowtie – 5)^{2}\cdot \left(\frac{d}{d\bowtie}(2\bowtie) – \frac{d}{d\bowtie}(5)\right) = 3\cdot (2\bowtie – 5)^{2}\cdot (2 – 0) = 6\cdot (2\bowtie – 5)^{2}\]
So putting it all together:
\[\frac{d}{d\bowtie}f(\bowtie) = (2 – \bowtie^{2})\cdot 6\cdot (2\bowtie – 5)^{2} + (-2\bowtie)\cdot (2\bowtie – 5)^{3}\]
And rearranging (simplifying would take a lot of multiplication, which might be necessary depending on the problem):
\[\frac{d}{d\bowtie}f(\bowtie) = 6 (2 – \bowtie^{2})(2\bowtie – 5)^{2} – 2\bowtie (2\bowtie – 5)^{3}\]
There’s one more piece of machinery needed before proceeding, and that is to make sure that everything is exponentiated rather than having some construct we don’t know how to handle, like a square root symbol or variables in the denominator of a fraction. So we can write:
\[\gamma = \frac{1}{\sqrt{1 – v^{2}/c^{2}}} = (1 – v^{2}/c^{2})^{-1/2}\]
Anything to a negative power is the reciprocal of that thing (e.g. \(x^{-1} = 1/x\) and \(z^{-2} = 1/z^{2}\)). A square root of anything is that thing to the power of one half. You can see this by remembering that exponents add when two of the same thing are multiplied together, so \(x^a\cdot x^b = x^{(a + b)}\). If you multiply a square root of something by that same square root of something, you’re supposed to get the something as the answer, e.g.
\[\sqrt{3\odot – 2\oplus}\cdot \sqrt{3\odot – 2\oplus} = 3\odot – 2\oplus\]
similarly
\[(3\odot – 2\oplus)^{1/2}\cdot (3\odot – 2\oplus)^{1/2} = (3\odot – 2\oplus)^{(1/2 + 1/2)} = (3\odot – 2\oplus)^{1} = 3\odot – 2\oplus\]
so it makes sense that a square root of something is just that thing raised to the power of 1/2. Similarly, the cube root of something is that thing raised to the power of 1/3, and a square root raised to the power of 3 is to the power of 3/2 (e.g. \((\sqrt{3\odot – 2\oplus})^{3} =\) \(((3\odot – 2\oplus)^{1/2})^{3} =\) \((3\odot – 2\oplus)^{3/2}\) because \((a^{n})^{m} = a^{(n\cdot m)}\)).
And finally back to the relativistic energy equation (remember that thing?). The \(f^{(n)}(\vec{v})\) term says to take the derivative of the function \(f(\vec{v})\), but do it \(n\) times. This can be written this way as well:
\[f^{(n)}(\vec{v}) = \frac{d^{n}}{d\vec{v}^{n}}f(\vec{v})\]
So, as an example, if \(n = 4\):
\[f^{(4)}(\vec{v}) = \left(\frac{d}{d\vec{v}}\left(\frac{d}{d\vec{v}}\left(\frac{d}{d\vec{v}}\left(\frac{d}{d\vec{v}}f(\vec{v})\right)\right)\right)\right)\]
Which says, take the derivative of \(f(\vec{v})\) (with respect to \(\vec{v}\)), then take the derivative of the result of the first derivative, then take the derivative of the result of the second derivative, then take the derivative of the result of the third derivative. The statement \(f^{(4)}(0)\) says to take those four derivatives (with respect to \(\vec{v}\) is implied since that’s the variable in \(f(\vec{v})\)), and only then substitute \(\vec{v} = 0\) into that final result.
The summation sign \(\sum\) with its indices of \(n = 0 \rightarrow \infty\), says to do everything to its right once for each index, \(n\), and add them all together. Yes, that indicates that we need to do this an infinite number of times… but we definitely stop after only a few as stated earlier. So the first four terms (\(n = 0, 1, 2, 3\)) of the Taylor series for the relativistic energy equation are:
\[E = f(\vec{v}) \approx \sum_{n = 0}^{3}\frac{f^{(n)}(0)}{n!}\vec{v}^{n} = \frac{f^{(0)}(0)}{0!}\vec{v}^{0} + \frac{f^{(1)}(0)}{1!}\vec{v}^{1} + \frac{f^{(2)}(0)}{2!}\vec{v}^{2} + \frac{f^{(3)}(0)}{3!}\vec{v}^{3}\]
Note that since we’re only doing the first few terms, that it’s an approximation (but close enough for Griffith’s). Also, in case you don’t know what \(n!\) is, it’s just the product of all the integers from \(1\) up to that integer (e.g. \(4! = 1\cdot 2\cdot 3\cdot 4 = 24\)), with the exception that \(0! = 1\) and factorials can be done for positive \(n\) only. So:
\[E = f(\vec{v}) \approx \sum_{n = 0}^{3}\frac{f^{(n)}(0)}{n!}\vec{v}^{n} = f^{(0)}(0) + f^{(1)}(0)\cdot\vec{v} + \frac{f^{(2)}(0)}{2}\cdot v^{2} + \frac{f^{(3)}(0)}{6}\cdot\vec{v}^{3}\]
This reads as “The energy is a function of the velocity vector \(\vec{v}\) and is approximately equal to the sum of, from \(n = 0\) to \(n = 3\), the \(n\)th derivative of the energy function evaluated at \(\vec{v} = 0\) divided by \(n\)-factorial multiplied by the velocity vector to the \(n\)th power, which equals the zeroth derivative of the energy function evaluated at \(\vec{v} = 0\), plus the first derivative of the energy function evaluated at \(\vec{v} = 0\) multiplied by the velocity (a vector), plus the second derivative of the energy function evaluated at \(\vec{v} = 0\) divided by \(2\) and multiplied by the velocity squared (a scalar), plus the third derivative of the energy function evaluated at \(\vec{v} = 0\) divided by \(6\) and multiplied by the velocity cubed (a vector)”.
It should be noted that in many textbooks and papers since only the first few derivatives are ever used, there is a shorthand to use the prime symbol to indicate how many times to differentiate. e.g. \(f^{(1)}(x)\equiv f^{\prime}(x)\), \(f^{(2)}(x)\equiv f^{\prime\prime}(x)\), \(f^{(3)}(x)\equiv f^{\prime\prime\prime}(x)\), \(\ldots\). Also note that \(f^{(0)}(x)\equiv f(x)\) (i.e. the function with no differentiation).
Now we’re finally ready to tackle generating the Taylor series for the relativistic energy equation in Griffith’s, starting with the \(n = 0\) term:
\[f^{(0)}(0) = \left. f(\vec{v})\right|_{\vec{v} = 0} \equiv \left. f(\vec{v})\right|_{0}\]
where the bar on the right reads “evaluate at \(\vec{v} = 0\)” (in either of the equivalent cases… in the event that there is only one variable, it’s customary to drop the explicit mention of the variable unless there is a reason why it’s needed for clarity). So, remembering that \(f^{(0)}(\bowtie) = f(\bowtie)\):
\[f^{(0)}(0) = \left. f(\vec{v})\right|_{0} = \left. mc^{2}\cdot (1 – v^{2}/c^{2})^{-1/2}\right|_{0}\]
\[= mc^{2}\cdot (1 – 0^{2}/c^{2})^{-1/2} = mc^{2}\cdot (1 – 0)^{-1/2} = mc^{2}\]
Yay, we have the first term in the series that Griffith’s specified! It was just a substitution into the equation, so it wasn’t a lot of work. For the second term (which needs the first derivative), there’s a bit more to do, so it makes sense to start with only part of it, and for that part to be the derivative (also, because of what I mentioned earlier about only even values of \(n\) having non-zero values, I’m expecting this to be zero).
\[f^{(1)}(0) = \left. f^{(1)}(\vec{v})\right|_{0}\]
\[= \left.\frac{d}{d\vec{v}}f^{(0)}(\vec{v})\right|_{0} = \left.\frac{d}{d\vec{v}}(mc^{2}\cdot (1 – v^{2}/c^{2})^{-1/2})\right|_{0} = \left. mc^{2}\frac{d}{d\vec{v}}(1 – v^{2}/c^{2})^{-1/2}\right|_{0}\]
With the constant \(mc^{2}\) outside of the derivative, there’s only one term, so looking at just the derivative part:
\[\frac{d}{d\vec{v}}(1 – v^{2}/c^{2})^{-1/2} = \left(-\frac{1}{2}\right)\cdot\left((1 – v^{2}/c^{2})^{((-1/2) – 1)}\right)\cdot\left(\frac{d}{d\vec{v}}(1 – v^{2}/c^{2})\right)\]
\[= \left(-\frac{1}{2}\right)\cdot\left((1 – v^{2}/c^{2})^{-3/2}\right)\cdot\left(\frac{d}{d\vec{v}}(1) – \frac{d}{d\vec{v}}\left(\frac{v^{2}}{c^{2}}\right)\right) \]
\[= -\frac{1}{2}\cdot(1 – v^{2}/c^{2})^{-3/2}\cdot\left(0 – \frac{2\vec{v}}{c^{2}}\right) = \frac{\vec{v}}{c^{2}}\cdot(1 – v^{2}/c^{2})^{-3/2}\]
(since \((-1/2)\cdot(-2) = 1\), and we need to put the vector symbol back on \(\vec{v}\) where it is not squared). And then if we evaluate it at \(\vec{v} = 0\):
\[\left.\frac{d}{d\vec{v}}(1 – \vec{v}^{2}/c^{2})^{-1/2}\right|_{0} = \left.\frac{\vec{v}}{c^{2}}\cdot (1 – v^{2}/c^{2})^{-3/2}\right|_{0} = 0\cdot (1 – 0^{2}/c^{2})^{-3/2} = 0\cdot 1 = 0\]
As expected, the term for \(n = 1\) is zero. Remember though, that the constant \(mc^{2}\) was not included in the calculation of the derivative, so properly:
\[f^{(1)}(\vec{v}) = mc^{2}\frac{d}{d\vec{v}}(1 – v^{2}/c^{2})^{-1/2} = mc^{2}\cdot \frac{\vec{v}}{c^{2}}\cdot(1 – v^{2}/c^{2})^{-3/2} = m\vec{v}\cdot(1 – v^{2}/c^{2})^{-3/2}\]
This is important because we need this value to compute the next derivative.
\[f^{(2)}(\vec{v}) = \frac{d}{d\vec{v}}f^{(1)}(\vec{v}) = \frac{d}{d\vec{v}}\left(m\vec{v}\cdot(1 – v^{2}/c^{2})^{-3/2}\right) = m\cdot\frac{d}{d\vec{v}}\left(\vec{v}\cdot(1 – v^{2}/c^{2})^{-3/2}\right)\]
Here, we need to use the product rule, where \(g(\vec{v}) = \vec{v}\) and \(h(\vec{v}) = (1 – v^{2}/c^{2})^{-3/2}\) (leaving out the constant \(m\) before the differential operator for the calculation):
\[\frac{d}{d\vec{v}}\left(\vec{v}\cdot(1 – v^{2}/c^{2})^{-3/2}\right) = \vec{v}\cdot\left(\frac{d}{d\vec{v}}(1 – v^{2}/c^{2})^{-3/2}\right) + \left(\frac{d}{d\vec{v}}(\vec{v})\right)\cdot (1 – v^{2}/c^{2})^{-3/2}\]
The second derivative is trivial, since:
\[\frac{d}{d\vec{v}}(\vec{v}) = \frac{d\vec{v}}{d\vec{v}} = 1\]
For the first one:
\[\frac{d}{d\vec{v}}(1 – v^{2}/c^{2})^{-3/2} = \left(-\frac{3}{2}\right)\cdot\left((1 – v^{2}/c^{2})^{((-3/2) – 1)}\right)\cdot\left(\frac{d}{d\vec{v}}(1 – v^{2}/c^{2})\right)\]
\[= -\frac{3}{2}\cdot(1 – v^{2}/c^{2})^{-5/2}\cdot\left(0 – \frac{2\vec{v}}{c^{2}}\right) = 3\cdot\frac{\vec{v}}{c^{2}}\cdot(1 – v^{2}/c^{2})^{-5/2}\]
Then putting it back together:
\[f^{(2)}(\vec{v}) = m\cdot\left[\vec{v}\cdot\left(3\cdot\frac{\vec{v}}{c^{2}}\cdot(1 – v^{2}/c^{2})^{-5/2}\right) + (1 – v^{2}/c^{2})^{-3/2}\right]\]
\[= 3m\cdot\frac{v^{2}}{c^{2}}\cdot(1 – v^{2}/c^{2})^{-5/2} + m\cdot(1 – v^{2}/c^{2})^{-3/2}\]
Evaluating it at \(\vec{v} = 0\):
\[f^{(2)}(0) = \left.f^{(2)}(\vec{v})\right|_{0} = 3m\cdot\frac{0}{c^{2}}\cdot(1 – 0)^{-5/2} + m\cdot(1 – 0)^{-3/2} = 0 + m = m\]
So the third term is:
\[\frac{f^{(2)}(0)}{2!}\vec{v}^{2} = \frac{m}{2}\vec{v}^{2} = \frac{1}{2}mv^{2}\]
which is what we were expecting for this term. So far so good! Two more to go: for \(n = 3\) and \(n = 4\), where we expect the former to be zero (but we still have to calculate it to be able to figure out the latter… and it’s always good to check to make sure no errors have been made along the way).
Hopefully everything is making some sort of sense by now, and the process really is just “plug and chug”. Next is the \(n = 3\) term:
\[f^{(3)}(\vec{v}) = \frac{d}{d\vec{v}}f^{(2)}(\vec{v}) = \frac{d}{d\vec{v}}\left[3m\cdot\frac{v^{2}}{c^{2}}\cdot(1 – v^{2}/c^{2})^{-5/2} + m\cdot(1 – v^{2}/c^{2})^{-3/2}\right]\]
\[= \frac{3m}{c^{2}}\cdot\left(\frac{d}{d\vec{v}}v^{2}\cdot(1 – v^{2}/c^{2})^{-5/2}\right) + m\cdot\left(\frac{d}{d\vec{v}}(1 – v^{2}/c^{2})^{-3/2}\right)\]
Looking at the second derivative, we’ve already done it above (this kind of repeating is not too uncommon):
\[\frac{d}{d\vec{v}}(1 – v^{2}/c^{2})^{-3/2} = 3\cdot\frac{\vec{v}}{c^{2}}\cdot(1 – v^{2}/c^{2})^{-5/2}\]
Then tackling the first one, we have already done something quite like it before:
\[\frac{d}{d\vec{v}}v^{2}\cdot(1 – v^{2}/c^{2})^{-5/2} = v^{2}\cdot\left(\frac{d}{d\vec{v}}(1 – v^{2}/c^{2})^{-5/2}\right) + \left(\frac{d}{d\vec{v}}(v^{2})\right)\cdot(1 – v^{2}/c^{2})^{-5/2}\]
\[= v^{2}\cdot\left(-\frac{5}{2}\right)\cdot\left((1 – v^{2}/c^{2})^{-7/2}\right)\cdot\left(-\frac{2\vec{v}}{c^{2}}\right) + 2\vec{v}\cdot(1 – v^{2}/c^{2})^{-5/2}\]
\[= 5\cdot\frac{\vec{v}^{3}}{c^{2}}\cdot(1 – v^{2}/c^{2})^{-7/2} + 2\vec{v}\cdot(1 – v^{2}/c^{2})^{-5/2}\]
So,
\[f^{(3)}(\vec{v}) = \frac{3m}{c^{2}}\cdot\left(5\cdot\frac{\vec{v}^{3}}{c^{2}}\cdot(1 – v^{2}/c^{2})^{-7/2} + 2\vec{v}\cdot(1 – v^{2}/c^{2})^{-5/2}\right) + m\cdot\left(3\cdot\frac{\vec{v}}{c^{2}}\cdot(1 – v^{2}/c^{2})^{-5/2}\right)\]
\[= \frac{15m}{c^{4}}\cdot\vec{v}^{3}\cdot(1 – v^{2}/c^{2})^{-7/2} + \frac{6m}{c^{2}}\cdot\vec{v}\cdot(1 – v^{2}/c^{2})^{-5/2} + \frac{3m}{c^{2}}\cdot\vec{v}\cdot(1 – v^{2}/c^{2})^{-5/2}\]
\[= \frac{15m}{c^{4}}\cdot\vec{v}^{3}\cdot(1 – v^{2}/c^{2})^{-7/2} + \frac{9m}{c^{2}}\cdot\vec{v}\cdot(1 – v^{2}/c^{2})^{-5/2}\]
Then evaluating at \(\vec{v} = 0\):
\[f^{(3)}(0) = \left.f^{(3)}(\vec{v})\right|_{0} = \frac{15m}{c^{4}}\cdot 0\cdot(1 – 0)^{-7/2} + \frac{9m}{c^{2}}\cdot0\cdot(1 – 0)^{-5/2} = 0\]
So, as expected, the fourth term (\(n = 3\)) is zero. Now on to the fifth and last term (\(n = 4\)) to duplicate the series for the relativistic energy equation in Griffith’s:
\[f^{(4)}(\vec{v}) = \frac{d}{d\vec{v}}f^{(3)}(\vec{v}) = \frac{d}{d\vec{v}}\left[\frac{15m}{c^{4}}\cdot\vec{v}^{3}\cdot(1 – v^{2}/c^{2})^{-7/2} + \frac{9m}{c^{2}}\cdot\vec{v}\cdot(1 – v^{2}/c^{2})^{-5/2}\right]\]
\[= \frac{15m}{c^{4}}\cdot\left(\frac{d}{d\vec{v}}\vec{v}^{3}\cdot(1 – v^{2}/c^{2})^{-7/2}\right) + \frac{9m}{c^{2}}\cdot\left(\frac{d}{d\vec{v}}\vec{v}\cdot(1 – v^{2}/c^{2})^{-5/2}\right)\]
Starting with the derivative in the leftmost term this time:
\[\frac{d}{d\vec{v}}\vec{v}^{3}\cdot(1 – v^{2}/c^{2})^{-7/2} = \vec{v}^{3}\cdot\frac{d}{d\vec{v}}\left((1 – v^{2}/c^{2})^{-7/2}\right) + \frac{d}{d\vec{v}}\left(\vec{v}^{3}\right)\cdot(1 – v^{2}/c^{2})^{-7/2}\]
\[= \vec{v}^{3}\cdot\left(-\frac{7}{2}\right)\cdot\left((1 – v^{2}/c^{2})^{-9/2}\right)\cdot\left(-\frac{2\vec{v}}{c^{2}}\right) + 3v^{2}\cdot(1 – v^{2}/c^{2})^{-7/2}\]
\[= 7\cdot\frac{v^{4}}{c^{2}}\cdot(1 – v^{2}/c^{2})^{-9/2} + 3v^{2}\cdot(1 – v^{2}/c^{2})^{-7/2}\]
Doing the rightmost derivative:
\[\frac{d}{d\vec{v}}\vec{v}\cdot(1 – v^{2}/c^{2})^{-5/2} = \vec{v}\cdot\frac{d}{d\vec{v}}\left((1 – v^{2}/c^{2})^{-5/2}\right) + \frac{d}{d\vec{v}}\left(\vec{v}\right)\cdot(1 – v^{2}/c^{2})^{-5/2}\]
\[= \vec{v}\cdot\left(-\frac{5}{2}\right)\cdot\left((1 – v^{2}/c^{2})^{-7/2}\right)\cdot\left(-\frac{2\vec{v}}{c^{2}}\right) + (1 – v^{2}/c^{2})^{-5/2}\]
\[= 5\cdot\frac{v^{2}}{c^{2}}\cdot(1 – v^{2}/c^{2})^{-7/2} + (1 – v^{2}/c^{2})^{-5/2}\]
Then,
\[f^{(4)}(\vec{v}) = \frac{15m}{c^{4}}\cdot\left(7\cdot\frac{v^{4}}{c^{2}}\cdot(1 – v^{2}/c^{2})^{-9/2} + 3v^{2}\cdot(1 – v^{2}/c^{2})^{-7/2}\right)\]
\[+ \frac{9m}{c^{2}}\cdot\left(5\cdot\frac{v^{2}}{c^{2}}\cdot(1 – v^{2}/c^{2})^{-7/2} + (1 – v^{2}/c^{2})^{-5/2}\right)\]
\[= \frac{105mv^{4}}{c^{6}}\cdot(1 – v^{2}/c^{2})^{-9/2} + \frac{45mv^{2}}{c^{4}}\cdot(1 – v^{2}/c^{2})^{-7/2}\]
\[+ \frac{45mv^{2}}{c^{4}}\cdot(1 – v^{2}/c^{2})^{-7/2} + \frac{9m}{c^{2}}\cdot(1 – v^{2}/c^{2})^{-5/2}\]
\[= \frac{105mv^{4}}{c^{6}}\cdot(1 – v^{2}/c^{2})^{-9/2} + \frac{90mv^{2}}{c^{4}}\cdot(1 – v^{2}/c^{2})^{-7/2} + \frac{9m}{c^{2}}\cdot(1 – v^{2}/c^{2})^{-5/2}\]
Then evaluating at \(\vec{v} = 0\):
\[f^{(4)}(0) = \left.f^{(4)}(\vec{v})\right|_{0} = 0\cdot(1 – 0)^{-9/2} + 0\cdot(1 – 0)^{-7/2} + \frac{9m}{c^{2}}\cdot(1 – 0)^{-5/2} = \frac{9m}{c^{2}}\]
So the fifth term is:
\[\frac{f^{(4)}(0)}{4!}\vec{v}^{4} = \frac{9m}{c^{2}}\cdot\frac{1}{24}\cdot v^{4} = \frac{3}{8} m \frac{v^{4}}{c^{2}}\]
And there it is. Terms where \(n\) was odd were zero (\(n = 1, 3\)), and were non-zero when it was even \(n = 0, 2, 4\). Putting the terms together:
\[E = f(\vec{v}) \approx \sum_{n = 0}^{4}\frac{f^{(n)}(0)}{n!}\vec{v}^{n} = mc^{2} + \frac{1}{2}mv^{2} + \frac{3}{8}m\frac{v^{4}}{c^{2}}\]
We only did the first few terms, so there is no ellipsis at the end, although it would still be fair to write:
\[E = \sum_{n = 0}^{\infty}\frac{f^{(n)}(0)}{n!}\vec{v}^{n} = mc^{2} + \frac{1}{2}mv^{2} + \frac{3}{8}m\frac{v^{4}}{c^{2}} + \cdots\]
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