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# 15.15: Derivatives

• • Contributed by Jeremy Tatum
• Emeritus Professor (Physics & Astronomy) at University of Victoria

We’ll pause here and establish a few derivatives just for reference and in case we need them later.

We recall that the Lorentz relations are

$x=\gamma(x'+\nu t') \label{15.15.1}$

and

$t=\gamma\left( t' + \frac{\beta x'}{c}\right) \label{15.15.2}$

From these we immediately find that

$\left( \frac{\partial x}{\partial x'}\right)_{t'}=\gamma;\quad\left( \frac{\partial x}{\partial t'}\right)_{x'}=\gamma\nu;\quad\left( \frac{\partial t}{\partial x'}\right)_{t'}=\frac{\beta\gamma}{c};\quad\left( \frac{\partial t}{\partial t'}\right)_{x'}=\gamma. \label{15.15.3a,b,c,d}\tag{15.15.3a,b,c,d}$

We shall need these in future sections.

Caution: It is not impossible to make a mistake with some of these derivatives if one allows one’s attention to wander. For example, one might suppose that, since $$\frac{\partial x}{\partial x'}=\gamma$$ then “obviously” $$\frac{\partial x'}{\partial x}=\frac{1}{\gamma}$$ - and indeed this is correct if $$t'$$ is being held constant. However, we have to be sure that this is really what we want. The difficulty is likely to arise if, when writing a partial derivative, we neglect to specify what variables are being held constant, and no great harm would be done by insisting that these always be specified when writing a partial derivative. If you want the inverses rather than the reciprocals of Equations $$\ref{15.15.3a,b,c,d}$$ the rule, as ever, is: Interchange the primed and unprimed symbols and change the sign of $$\nu$$ or $$\beta$$. For example, the reciprocal of $$\left( \frac{\partial x}{\partial x'} \right)_{t'}$$ is $$\left( \frac{\partial x'}{\partial x} \right)_{t'}$$, while its inverse is $$\left( \frac{\partial x'}{\partial x} \right)_{t}$$. For completeness, and reference, then, I write down all the possibilities:

$\left( \frac{\partial x'}{\partial x}\right)_{t'}=\frac{1}{\gamma};\quad\left( \frac{\partial t'}{\partial x}\right)_{x'}=\frac{1}{\gamma\nu};\quad\left( \frac{\partial x'}{\partial t}\right)_{t'}=\frac{c}{\beta\gamma};\quad\left( \frac{\partial t'}{\partial t}\right)_{x'}=\frac{1}{\gamma}. \label{15.15.3e,f,g,h}\tag{5.15.3e,f,g,h}$

$\left( \frac{\partial x'}{\partial x}\right)_{t}=\gamma;\quad\left( \frac{\partial x'}{\partial t}\right)_{x}=-\gamma\nu;\quad\left( \frac{\partial t'}{\partial x}\right)_{t}=-\frac{\beta\gamma}{c};\quad\left( \frac{\partial t'}{\partial t}\right)_{x}=\gamma. \label{15.15.3i,j,k,l}\tag{15.15.3i,j,k,l}$

$\left( \frac{\partial x}{\partial x'}\right)_{t}=\frac{1}{\gamma};\quad\left( \frac{\partial t}{\partial x'}\right)_{x}=-\frac{1}{\gamma\nu};\quad\left( \frac{\partial x}{\partial t'}\right)_{t}=-\frac{c}{\beta\gamma};\quad\left( \frac{\partial t}{\partial t'}\right)_{x}=\frac{1}{\gamma}. \label{15.15.3m,n,o,p}\tag{15.15.3m,n,o,p}$

Now let’s suppose that $$\psi = \psi(x,t)$$ where $$x$$ and $$t$$ are in turn functions (Equations $$\ref{15.15.1}$$ and $$\ref{15.15.2}$$) of $$x'$$ and $$t'$$. Then

$\frac{\partial \psi}{\partial x'}=\frac{\partial x}{\partial x'}\frac{\partial \psi}{\partial t}+\frac{\partial t}{\partial x'}\frac{\partial \psi}{\partial t}=\gamma\frac{\partial \psi}{\partial x}+\frac{\beta\gamma}{c}\frac{\partial\psi}{\partial t} \label{15.15.4}$

and

$\frac{\partial \psi}{\partial t'}=\frac{\partial x}{\partial t'}\frac{\partial \psi}{\partial x}+\frac{\partial t}{\partial t'}\frac{\partial \psi}{\partial t}=\gamma\nu\frac{\partial \psi}{\partial x}+\gamma\frac{\partial\psi}{\partial t}. \label{15.15.5}$

The reader will doubtless notice that I have here ignored my own advice and I have not indicated which variables are to be held constant. It would be worth spending a moment here thinking about this.

We can write Equations $$\ref{15.15.4}$$ and $$\ref{15.15.5}$$ as equivalent operators:

$\frac{\partial}{\partial x'}=\gamma\left(\frac{\partial}{\partial x}+\frac{\beta}{c}\frac{\partial}{\partial t}\right) \label{15.15.6}$

and

$\frac{\partial}{\partial t'}=\gamma\left(\nu\frac{\partial}{\partial x}+\frac{\partial}{\partial t}\right). \label{15.15.7}$

We can also, if we wish, find the second derivatives. Thus

$\frac{\partial^{2}\psi}{\partial x'^{2}}=\gamma^{2}\left(\frac{\partial^{2}}{\partial x^{2}}+\frac{2\beta}{c}\frac{\partial^{2}}{\partial x\partial t}+\frac{\beta^{2}}{c^{2}}\frac{\partial^{2}}{\partial t^{2}}\right). \label{15.15.9}$

In a similar manner we obtain

$\frac{\partial^{2}}{\partial x'\partial t'}=\gamma^{2}\left(\nu\frac{\partial^{2}}{\partial x^{2}}+(1+\beta^{2})\frac{\partial^{2}}{\partial x\partial t}+\frac{\beta}{c}\frac{\partial^{2}}{\partial t^{2}}\right) \label{15.15.10}$

and

$\frac{\partial^{2}}{\partial t'^{2}}=\gamma^{2}\left(\nu^{2}\frac{\partial^{2}}{\partial x^{2}}+2\nu\frac{\partial^{2}}{\partial x\partial t}+\frac{\partial^{2}}{\partial t^{2}}\right). \label{15.15.11}$

The inverses of all of these relations are to be found by interchanging the primed and unprimed coordinates and changing the signs of $$\nu$$ and $$\beta$$.