topics

• Maxwell's eqns
• Lorentz transfm

Relativity: Lorentz Transformation

A derivation

The postulates of special relativity lead directly to the Lorentz transformation on the co-ordinates of space and time, which governs how an object moving away from an observer is viewed. The faster the object is moving away, especially the closer its velocity is to that of light, the more marked are two effects (ignoring the distance between observer and object): the spatial dimensions of the object appear contracted and time goes more slowly.

Relativity is concerned with events. An event takes place at a particular instant of time and a particular point in space. It is specified by a time co-ordinate and three spatial co-ordinates (assuming three-dimensional space). The co-ordinates are assigned within a frame of reference. Frames of reference can have any orientation (translational, reflective and rotational) and motion relative to each other, each having their own co-ordinates for the same event.

Special relativity is restricted to inertial frames, in which Newton’s first law applies. Any frame of reference that moves with constant velocity relative to an inertial frame is also an inertial frame.

Consider an inertial frame F with co-ordinates (t, x, y, z) and another inertial frame F’ with co-ordinates (t’, x’, y’, z’), where the spatial axes are parallel and the origin of F’ moves with constant velocity v = (v, 0, 0) along the x-axis of F. Further, the unique event, defined by the co-incidence of the origins of F and F’, happens at time t = 0 in F and time t’ = 0 in F’. By satisfying these two requirements, the two frames are said to be in standard configuration. Such a configuration simplifies the derivation of results, but suitably applied does not limit their generality.

What is the relationship between the F co-ordinates of an event and the F’ co-ordinates of the same event? According to assumptions obtaining before special relativity was developed, where time and lengths are absolute:

x’ = x – vt
y’ = y
z’ = z
t’ = t

However, different results are obtained for x’ and t’ if the postulates of special relativity are taken into account. These postulates are:

1. Invariance of the speed of light in a vacuum.
2. Principle of relativity: no experiment can distinguish one inertial frame from another: the laws of physics have the same form in all inertial frames.

The relationship, under special relativity, between the two sets of four frame co-ordinates, is collectively known as the Lorentz transformation. Einstein provided a "simple derivation" of this transformation as an appendix to his book, Relativity, which he had written for a lay readership. The derivation is reproduced here, albeit shortened by summarising the text, but all steps in the calculation have been retained. I have in fact added some additional steps to show how certain intermediate results are obtained. The calculation involved is straightforward; it is the physical insights from special relativity which make it interesting.

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$$\begin{array}{l}\text{A light signal transmitted along the positive x-axis satisfies}\\ x=\mathit{ct}\text{or}\\ x-\mathit{ct}=0\text{(1)}\\ \\ \text{The same light signal relative to F' has}\text{the same velocity c}\text{, so}\\ x\text{'}=\mathit{ct}\text{'}\text{or}\\ x\text{'}-\mathit{ct}\text{'}=0\text{(2)}\\ \\ \text{Those space-time points (events) which satisfy (1) must also satisfy (2),}\\ \text{which is the case when, in general,}\\ x\text{'}-\mathit{ct}\text{'}=\lambda (x-\mathit{ct})\text{where}\lambda \text{is a constant}\text{(3)}\\ \text{since by (3), the disappearance of}x-\mathit{ct}\text{involves the disappearance of}\\ x\text{'}-\mathit{ct}\text{'}\\ \\ \text{Applying the same reasoning to a light signal transmitted along the negative}\\ \text{x-axis}\\ x\text{'}+\mathit{ct}\text{'}=\mu (x+\mathit{ct})\text{where}\mu \text{is a constant}\text{(4)}\\ \\ \text{(3) + (4) gives}2x\text{'}=(\lambda +\mu )x-(\lambda -\mu )\mathit{ct}\\ \text{(4) - (3) gives}2\mathit{ct}\text{'}=(\lambda +\mu )\mathit{ct}-(\lambda -\mu )x\\ \\ \text{Let}\gamma =\frac{\lambda +\mu }{2}\text{,}\kappa =\frac{\lambda -\mu }{2}\text{, then}\\ x\text{'}=\gamma x-\kappa \mathit{ct}\text{(5.1)}\\ \mathit{ct}\text{'}=\gamma \mathit{ct}-\kappa x\text{(5.2)}\\ \\ \text{At the origin of F',}x\text{'}=0\text{permanently, so here}x=\frac{\kappa c}{\gamma }t\text{from (5.1)}\\ \text{but}v\text{is the velocity with which the origin of F' is moving relative to F, so}\\ v=\frac{\kappa c}{\gamma }\text{(6)}\\ \text{The same value of}v\text{can be obtained from equations (5) for the velocity of}\\ \text{another point of F' relative to F, or for the velocity, in the negative direction, of}\\ \text{a point of F relative to F'. Thus}v\text{is the relative velocity of the two systems.}\\ \\ \text{By the principle of relativity, a measuring rod at rest and of unit length in F'}\\ \text{has a length in F equal to the length in F' of a measuring rod at rest and of}\\ \text{unit length in F.}\\ \\ \text{To see how points on the x'-axis appear viewed from F, take a 'snapshot' of F'}\\ \text{from F, that is, at a particular time t, for example when t = 0. Equation (5.1)}\\ \text{becomes}\\ x\text{'}=\gamma x\\ \\ \text{So, in the snapshot, two points on the x'-axis, separated by the distance}\\ \text{measured in F' as}\mathrm{\Delta }x\text{'}=1\text{, are separated by the distance measured in F as}\\ \mathrm{\Delta }x=\frac{1}{\gamma }\text{(7.1)}\\ \\ \text{Now, to view points on the x-axis from F', take a snapshot of F from F' at time}\\ t\text{'}=0\text{, using equations (5) after}t\text{has been eliminated}\\ \text{(5.1)}\times \gamma \text{gives}\gamma x\text{'}={\gamma }^{2}x-\gamma \kappa \mathit{ct}\\ \text{(5.2)}\times \kappa \text{gives}\kappa \mathit{ct}\text{'}=\gamma \kappa \mathit{ct}-{\kappa }^{2}x\\ \text{Adding these}\gamma x\text{'}+\kappa \mathit{ct}\text{'}=({\gamma }^2-{\kappa }^2)x\\ \text{When}t\text{'}=0\text{,}\gamma x\text{'}=({\gamma }^2-{\kappa }^2)x\\ x\text{'}=\frac{({\gamma }^2-{\kappa }^2)x}{\gamma }=\frac{\gamma ({\gamma }^2-{\kappa }^2)x}{{\gamma }^2}=\gamma \left(1-{\left(\frac{\kappa }{\gamma }\right)}^2\right)x=\\ \gamma \left(1-{\left(\frac{\kappa c}{\gamma }\right)}^2\frac{1}{c^2}\right)x=\gamma \left(1-\frac{v^2}{c^2}\right)x\\ \\ \text{So, in the snapshot, two points on the x-axis, separated by the distance}\\ \text{measured in F as}\mathrm{\Delta }x=1\text{, are separated by the distance measured in F' as}\\ \mathrm{\Delta }x\text{'}=\gamma \left(1-\frac{v^2}{c^2}\right)\text{(7.2)}\\ \\ \text{But, by the principle of relativity, the two snapshots must be identical, with}\\ \mathrm{\Delta }x\text{in (7.1) equal to}\mathrm{\Delta }x\text{'}\text{in (7.2), so}\\ \frac{1}{\gamma }=\gamma \left(1-\frac{v^2}{c^2}\right)\\ \gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\text{(8)}\\ \text{(8) is an important result. The value}\gamma \text{appears in expressions for length}\\ \text{contraction, time dilation and more.}\\ \text{From (6),}\kappa =\frac{\gamma v}{c}\text{so equation (5.1) becomes}\\ x\text{'}=\gamma x-\frac{\gamma \mathit{vc}}{c}t=\gamma (x-\mathit{vt})=\frac{x-\mathit{vt}}{\sqrt{1-v^2/c^2}}\text{(9.1)}\\ \text{and equation (5.2)}\\ \mathit{ct}\text{'}=\gamma \mathit{ct}-\frac{\gamma v}{c}x\\ t\text{'}=\gamma t-\frac{\gamma v}{c^2}x=\gamma (t-\frac{v}{c^2}x)=\frac{t-(v/c^2)x}{\sqrt{1-v^2/c^2}}\text{(9.2)}\\ \\ \text{The co-ordinate mappings in equations (9) are the Lorentz transformation for}\\ \text{events on the x-axis. It satisfies the condition}\\ x{\text{'}}^2-c^{2}t{\text{'}}^2=x^2-c^2{t}^2\text{(10)}\\ \text{(obtained by using equations (8) to calculate the right-hand side)}\\ \\ \text{To extend the result to include events not located on the x-axis, include}\\ y\text{'}=y\text{(11.1)}\\ z\text{'}=z\text{(11.2)}\\ \\ \text{To show that these results satisfy the postulates of special relativity, suppose}\\ \text{a light signal is sent from the origin of F at time}t=0\text{. It will be propagated}\\ \text{as given by}\\ r=\sqrt{x^2+y^2+z^2}=\mathit{ct}\\ \\ \text{or in squared form}\\ x^2+y^2+z^2-c^2{t}^2=0\text{(12.1)}\\ \text{The postulates of special relativity require that this same signal viewed from}\\ \text{F' is given by}\\ r\text{'}=\sqrt{x{\text{'}}^2+y{\text{'}}^2+z{\text{'}}^2}=\mathit{ct}\text{'}\\ \\ \text{or}\\ x{\text{'}}^2+y{\text{'}}^2+z{\text{'}}^2-c^{2}t{\text{'}}^2=0\text{(12.2)}\\ \\ \text{Given that the light signal which satisfies (11.1) must satisfy (11.2), this}\\ \text{requires, for some constant}\sigma \\ x{\text{'}}^2+y{\text{'}}^2+z{\text{'}}^2-c^{2}t{\text{'}}^2=\sigma (x^2+y^2+z^2-c^2{t}^2)\text{(13)}\\ \\ \text{but since equation (9) holds for points on the x-axis, it must be the case that}\\ \sigma =1\\ \\ \text{Hence the Lorentz transformation satisfies}\\ x{\text{'}}^2+y{\text{'}}^2+z{\text{'}}^2-c^{2}t{\text{'}}^2=x^2+y^2+z^2-c^2{t}^2\text{(14)}\\ \text{as the equation follows from (10) and (11), and (10) follows from (9)}\\ \end{array}$$

It remains to derive the Lorentz transformation for frames whose axes are arbitrarily orientated with respect to each other, where the direction of the relative velocity v between frames is arbitrary and constant (so that both frames are inertial). This can be done by combining the Lorentz transformation for the standard configuration with other purely spatial transformations.

The general Lorentz transformation again satisfies equation (13).

Note

The introductory paragraphs, apart from the first, are a partial summary of section 10.1.1 Essentials of special relativity in Book 2 of the level 3 Open University course SMT359 Electromagnetism.

Bibliography

Einstein, A. (2009 [1952]) Relativity, London, Folio Society, appendix I, p.141-7
Freake, S., 2006. ed SMT359 Book 2 Electromagnetism: Electromagnetic Fields, Milton Keynes, The Open University, Sec 10.1.1, 220-3.

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