# Differentials? Can you elaborate?

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1. Oct 29, 2014

### BiGyElLoWhAt

Disclaimer: This isn't a homework assignment, so maybe it shouldn't be in the homework forums. If you feel it should be located elsewhere, feel free to move it, but the template doesn't really apply to this question so...

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So I'm reading Fundamentals of Modern Physics by Eisberg, and I just got through all the Lorentz transforms for space-time coordinates.

In order to do a Lorentz transform on the velocity between reference frames, he differientated (somehow) the spacetime lorentz transforms for the ST coords.

The velocity is presumed to be exclusively in the x direction between both sets of coordinates.

He started with

$x'(x,t) = \frac{1}{\sqrt{1-\beta ^2}}(x-vt)$
and differentiated to get
$dx'(x,t)=\frac{1}{\sqrt{1-\beta ^2}}(dx-vdt)$

I have a couple questions about the legitimacy of this.

1) I think this is called the total differential (?) is this correct? It appears to take the form $\frac{\partial}{\partial x}x' *dx + \frac{\partial}{\partial t}x' *dt$

2) Assuming this is 100% legitimate (I don't necessarily doubt that it is), when is this method applicable? What are the implications of this method?

I understand that x' is a function of x and t, (it's actually denoted simply x', not x'(x,t) in the book), so the velocity should be dependent on both differentials, but I'm not really seeing the logic behind differentiating this way.

Any help would be greatly appreciated.
Thanks

2. Oct 29, 2014

### Fredrik

Staff Emeritus
If $f:\mathbb R^2\to\mathbb R$, then you can define $df:\mathbb R^4\to\mathbb R$ by
$$df(x,y,h,k)=D_1f(x,y) h+D_2 f(x,y)k,$$ for all x,y,h,k. If you denote the numbers h and k by dx and dy respectively, and simplify the notation for the left-hand side to just df, the above can be written
$$df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy.$$ You're dealing with four variables x,t,x',t', that satisfy two constraints, the Lorentz transformation equations. These equations ensure that the values of the primed variables are fixed by a choice of values for the unprimed variables. In particular, we can define a function f by
$$f(x,t)=\gamma(x-vt)$$ for all x,t. It's confusing but common to use the notation x' instead of f. So x' now denotes two different things, both a real number and a function.
$$dx'=df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial t}dt =\frac{\partial x'}{\partial x}dx+\frac{\partial x'}{\partial t}dt.$$ Why do it this way? I have no idea.

3. Oct 29, 2014

### BiGyElLoWhAt

I'm just hoping to get a little more insight into the why.

4. Oct 29, 2014

### Ray Vickson

You can avoid "differentials", by using finite differences and the basic definition of derivative. For a space-time increment $(\Delta x, \Delta t)$ in the first frame, you get an increment $(\Delta x', \Delta t')$ in the second frame, and these are related by
$$\Delta x' = \gamma\,(\Delta x - v \Delta t)\\ \Delta t' = \gamma \, (\Delta t - \frac{v}{c^2} \Delta x )$$
Here, $x = x(t)$ is on some trajectory. Now use the definitions
$$u_x = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t}\\ u_x' = \lim_{\Delta t' \to 0} \frac{\Delta x'}{\Delta t'} = \lim_{\Delta t' \to 0} \frac{\Delta x - v \Delta t}{\Delta t - (v/c^2) \Delta x}\\ = \lim_{\Delta t' \to 0} \frac{(u_x-v) \Delta t}{(1-(u_x v/c^2)) \Delta t}$$
and note that $\Delta t' \to 0$ if and only if $\Delta t \to 0$.