[back][up]

4.4 Further Examples for Series and Limits

4.4.1 Newtonian Limit of Relativistic Energy

According to Einstein, the total energy of a particle of rest mass {m}_{0} and velocity v is

\begin{eqnarray} E = {{m}_{0}{c}^{2}\over \sqrt{1 −{ \left ({v\over c}\right )}^{2}}},& & %&(4.38) \\ \end{eqnarray}

where c is the speed of light on vacuum. We would like to find an approximation of this formula for small velocities v ≪ c, in order to compare to Newton’s expression for the kinetic energy, {E}_{kin} = (1∕2){m}_{0}{v}^{2}. Defining β := v∕c, we recognise that our mathematical task is to (x := {β}^{2})

Expand f(x) = {1\over \sqrt{1−x}} around x = 0:

We write f(x) = {(1 − x)}^{−1∕2} and

\begin{eqnarray} f(0)& =& 1 %& \\ f'(0)& =&{ \left .(−1)(−1∕2){(1 − x)}^{−3∕2}\right |}_{ x=0} = 1∕2 %& \\ f''(0)& =&{ \left .(−1)(−1∕2)(−1)(−3∕2){(1 − x)}^{−3∕2}\right |}_{ x=0} = 3∕4%& \\ ...& & %&(4.39) \\ \end{eqnarray}

With our

Taylor expansion of f(x) around x = 0,

\begin{eqnarray} f(x) = {f(x = 0)\over 0!} + {f'(x = 0)\over 1!} x + {f''(x = 0)\over 2!} {x}^{2} + ... ={ \mathop{∑ }}_{n=0}^{∞}{{f}^{(n)}(x = 0)\over n!} {x}^{n},& & %&(4.40) \\ \end{eqnarray}

we find the first terms as

\begin{eqnarray} f(x) = 1 + {1\over 2}x + {3\over 8}{x}^{2} + ...& & %&(4.41) \\ \end{eqnarray}

(note that 2! = 2). Therefore, with x = {β}^{2} = {(v∕c)}^{2}, we obtain

\begin{eqnarray} E = {m}_{0}{c}^{2}\left [1 + {1\over 2}{\left ({v\over c}\right )}^{2} + {3\over 8}{\left ({v\over c}\right )}^{4} + O{\left ({v\over c}\right )}^{6}\right ].& & %&(4.42) \\ \end{eqnarray}

Here, we introduced the O–symbol (speak ‘order of’), i.e. O{(x)}^{6} means ‘terms of order {x}^{6} or higher powers’ like {x}^{7}, {x}^{8} etc. This is a convenient way to express that in a Taylor expansion with the first few terms written down as above, there are higher order terms to follow that one does not care to write down explicitely here. These higher order terms in fact become smaller and smaller for |x| < 1.

We can take use of the O–symbol to write

\begin{eqnarray} E = {m}_{0}{c}^{2} + {1\over 2}{m}_{0}{v}^{2} + O{\left ({v\over c}\right )}^{4}.& & %&(4.43) \\ \end{eqnarray}

This shows that the first term in the total energy is a velocity–independent rest energy of the particle, and the second term is the lowest order approximation to its kinetic energy. The relativistic correction to the kinetic energy if of order {(v∕c)}^{4}, i.e. very small for velocities small compared to the speed of light.

4.4.2 Limits

Expressions of the type ‘0∕0’

Often we have to discuss and sketch functions like

\begin{eqnarray} f(x) = {\mathop{sin}\nolimits (x)\over x} & & %&(4.44) \\ \end{eqnarray}

with a seemingly ill-defined behaviour at x = 0. Direct substitution gives 0∕0, which is indeed not well defined. However, a closer look shows that we can make sense of the function even at x = 0: we expand \mathop{sin}\nolimits (x) by its Taylor series near x = 0, i.e.

\begin{eqnarray} f(x) = {\mathop{sin}\nolimits (x)\over x} = {x −{{x}^{3}\over 3!} + O({x}^{5})\over x} = 1 −{{x}^{2}\over 3!} + O({x}^{4}).& & %&(4.45) \\ \end{eqnarray}

This means, that if x approaches x = 0 we have the finite value f(x = 0) = 1, i.e.

\begin{eqnarray} {\mathop{lim}}_{x→0}f(x) = 1.& & %&(4.46) \\ \end{eqnarray}

The deviation from f(x = 0) = 1 close to x = 0 is described by the second term, − {x}^{2}∕6, i.e. a quadratic decrease of the function for small values of x.

[back][up]