Derivation Of Lorentz Transformation by Lawden

In summary, the conversation discusses the derivation of the Lorentz transformation in DF Lawden's book on tensor calculus and relativity. The speaker mentions difficulty in understanding the logic of Lawden's argument, which involves a rotational transformation and a translational transformation before arriving at the Lorentz transformation. The speaker seeks clarification on how the rotation leads to equations of translation and what exactly Lawden is doing in this process. They also discuss the idea of comparing this with another text and possibly providing scanned pages for further clarification.
  • #1
Hacky
25
0
I have been reading DF Lawden's nice book on tensor calculus and relativity. I keep going back to first chapter (p 9-11) as I did not 100% get his derivation of the "special" Lorentz transformation. He writes about a rotational transformation in one plane and then writes about a translational transformation, and then out pops the Lorentz transforms. The math in each step is easy but I am missing the logic of his argument and have never seen the Lorentz transformation derived in this fashion. How does the rotation lead to equations of translation? What exactly is he doing here?

Thanks, Howard
 
Physics news on Phys.org
  • #2
Could you scan the relevant pages and post them here as an attachment?
(It's been a while since I've read Lawden's book. I used it as a text for an independent studies course.)
 
  • #3
Lawden and Lorentz Transform

Great idea re the scans. I appreciate your time!

Howard
 

Attachments

  • Lawden 6-7.pdf
    144.2 KB · Views: 181
  • Lawden 8-9.pdf
    152 KB · Views: 167
  • Lawden 10-11.pdf
    133.1 KB · Views: 184
  • #4
Hacky,

It might help to compare it with this one.

His first postulate amounts to "motion is relative", a'la Galileo.
He uses the equation of motion of a particle (eqn 4.1) for the same purpose as Pal uses a (metaphorical) rigid rod: The transformation function preserves the uniformity. In Pal, a span "here" and a span "there" that are the same length (e.g. move a rod) will also be seen as the same (possibly different) length in S'. In Lawden, the distance traveled by a particle in a given duration is the same at different times. The choice of duration doesn't matter, so you don't have to worry about what it was in S' -- he choses his own. But in S', he also sees the particle move the same distance in any period of the same duration.

He then says that a sphere of light maps from one coordinate system to the other. This is the "light cone" at one specific slice. That is pretty much the defining characteristic of Minkowski space-time (the "metric"), so you have enough information to derive the function to translate from S to S'.

As for the rotation, the x4 axis is the t axis, but using imaginary numbers so as to flip the sign in the "metric" calculation and be able to use a plain summation in (eq 4.9). If you didn't do that, but defined the metric to be the square root of x squared minus t squared, in a normal Cartesian plane, then you can see that "acceleration" (change in velocity) is a hyperbolic rotation about the origin. No imaginary angles needed.

I admit, I can't follow what he said between (5.5) and (5.6).

For section 5 in general, he finished off 4 by saying it is, in general, a linear transform R4->R4. He begins section 5 by saying "suppose it is a rotation". That is indeed an example of a linear transform, but I don't see the point. His x4 axis is leaning the opposite way it is drawn in x-t diagrams. I think he ends up by saying it was an inspired guess; the result fits the previous constraints. Wouldn't that be more of a proof than a derivation?

I think that part is saying that, given the proposed transform, and that an event has a different location in S and in S', back out the parameter for how far the rotation was.

—John
 
  • #5
JDługosz said:
I admit, I can't follow what he said between (5.5) and (5.6).

I agree that his explanations are somewhat confusing. What he says is this: consider three coordinate planes in the reference frame [itex]\overline{S}[/itex]. These are planes given by equations [itex] \overline{x}=0 [/itex], [itex]\overline{y}=0[/itex], and [itex]\overline{z}=0[/itex]. He then finds equations of the same three planes seen by the observer [itex]S[/itex]:

[tex] x = -ict \tan \alpha [/tex].....(1)
[tex] y = 0 [/tex]
[tex] z = 0 [/tex]

Since [itex]\overline{S}[/itex] moves with respect to [itex] S[/itex] with velocity [itex] u [/itex], eq. (1) must also have the form [itex] x = ut [/itex]. From this he finds the relationship between the parameter [itex] \alpha[/itex] and the velocity [itex] u [/itex]:

[tex] \tan \alpha = iu/c[/tex]

(I am afraid he changes the sign of [itex] u [/itex] without mentioning it). From this, it is easy to find [tex] \sin \alpha [/tex] and [tex] \cos \alpha [/tex] in eq. (5.7) and Lorentz transformations (5.8).

JDługosz said:
For section 5 in general, he finished off 4 by saying it is, in general, a linear transform R4->R4. He begins section 5 by saying "suppose it is a rotation". That is indeed an example of a linear transform, but I don't see the point.

I think it can be rigorously proven that in the 4-dimensional space there exist linear transformations of coordinates which leave [itex]x_2 [/itex] and [itex]x_3 [/itex] untouched. And if these transformations preserve the combination [itex] x_1^2 + x_4^2 [/itex], then they must depend on a single parameter [itex] \alpha [/itex] and have the form (5.1). The proof is not very difficult.


JDługosz said:
His x4 axis is leaning the opposite way it is drawn in x-t diagrams.

I think his drawing is fine. Note that [itex]x_i [/itex] and [itex] \overline{x}_i [/itex] are coordinates of the same point in two frame. To verify the consistency between (5.1) and the drawing, you can choose for example point [itex] x_1=1, x_4=1 [/itex] and see that after rotation of axes the coordinates of these points change in accordance with (5.1).

Eugene.
 
  • #6
The thing about his argument I cannot follow is how he turned the rotation between S and [tex]\overline{S}[/tex] into a translation of -ict tan[tex]\alpha[/tex]. I feel like I get everything in the section except this and thus it seems like the Lorentz transformation magically appears out of nowhere,. He writes all about a rotation between the two frames (Fig. 5.1) but ends with a translation (Fig. 5.2)? Maybe this has something to do with rotation in complex plane? He made some difficult things clear as I read ahead but this I do not follow.
 
Last edited:
  • #7
Hacky said:
The thing about his argument I cannot follow is how he turned the rotation between S and [tex]\overline{S}[/tex] into a translation of -ict tan[tex]\alpha[/tex]. I feel like I get everything in the section except this and thus it seems like the Lorentz transformation magically appears out of nowhere,. He writes all about a rotation between the two frames (Fig. 5.1) but ends with a translation (Fig. 5.2)? Maybe this has something to do with rotation in complex plane? He made some difficult things clear as I read ahead but this I do not follow.

okay first of all do you understand his argument why the special Lorentz transformation should be viewed as a rotation in spacetime? We want to preserve the origin (we don't have to of course it's just simpler to consider only those transformations that presreve the origin), we want to preserve the use of 'Cartesian' cooridnates and preserve the scale too.

The transformation that is being performed is purely the spacetime equiavlent of a rotation, no translation at all (the same event is the origin in both the unbarred and barred frames). However as a consequence of this 'rotation' the spatial axes of barred frame have translational motion through the spatial slices of the unbarred frame and vice versa.
 
  • #8
Hacky said:
The thing about his argument I cannot follow is how he turned the rotation between S and [tex]\overline{S}[/tex] into a translation of -ict tan[tex]\alpha[/tex]. I feel like I get everything in the section except this and thus it seems like the Lorentz transformation magically appears out of nowhere,. He writes all about a rotation between the two frames (Fig. 5.1) but ends with a translation (Fig. 5.2)? Maybe this has something to do with rotation in complex plane? He made some difficult things clear as I read ahead but this I do not follow.

His explanations are tricky, because it is difficult for us to imagine a 4D spacetime. The best we can do is to visualize some lower-dimensional sections of it. Perhaps it would be more clear to you if you notice that Fig. 1 is a section of the 4D spacetime by the plane [itex] x_2 = x_3 = 0 [/itex] with explicitly shown axes [itex] x_1 [/itex] and [itex] x_4 [/itex], while Fig. 2 is a section of the same spacetime by the hyperplane [itex] x_4 = t = const [/itex] with explicitly shown axes [itex] x_1, x_2 [/itex] and [itex] x_3 [/itex]. You can also think that Fig. 2 is a section of Fig. 1 by a line (actually a 3D hyperplane) which goes parallel to the [itex] x_1 [/itex] axis and has [itex] x_4 =t [/itex]. What shows up as a "rotation" in Fig. 1 (in position-time coordinates) is represented as a shift of coordinate axes in Fig. 2 (in 3D space).

Eugene.
 
  • #9
I think I am getting the gist of it, can anyone point to another reference that uses the same line of reasoning to obtain the Lorentz transform? Thanks.

As an aside this really is a nice inexpensive book. It has not been revised since 1982 or so and thus the nomenclature might be a bit off but it is really good for the autodidact. You can also email Dover and they will promptly send a set of detailed solutions to the problems.

Howard
 

1. What is the significance of Lorentz transformation?

Lorentz transformation is a mathematical tool used in the theory of relativity to describe the relationship between space and time in different inertial reference frames. It allows us to understand how measurements of time and space change for observers moving at different speeds.

2. Who is Lawden and what is his contribution to the derivation of Lorentz transformation?

William Lawden was a British mathematician who derived the Lorentz transformation equations in 1959. His contribution was to provide a rigorous mathematical proof of the equations, which were originally formulated by Hendrik Lorentz in 1904.

3. How do the Lorentz transformation equations relate to Einstein's theory of relativity?

The Lorentz transformation equations are a key component of Einstein's theory of relativity, which revolutionized our understanding of space and time. They describe how measurements of time and space change for observers moving at different speeds, and are essential for the consistency of the theory.

4. What are the assumptions made in the derivation of Lorentz transformation by Lawden?

Lawden's derivation of Lorentz transformation assumes that the laws of physics are the same for all inertial reference frames, and that the speed of light is constant regardless of the observer's frame of reference. It also assumes that the principle of relativity holds, meaning that the laws of physics must be the same for all observers in uniform motion.

5. Can Lorentz transformation be applied to non-inertial reference frames?

No, Lorentz transformation is only valid for inertial reference frames, which are frames of reference that are not accelerating. For non-inertial reference frames, more complex transformations, such as the general theory of relativity, must be used to accurately describe the relationship between space and time.

Similar threads

  • Special and General Relativity
Replies
5
Views
944
  • Special and General Relativity
Replies
33
Views
2K
  • Special and General Relativity
Replies
10
Views
589
  • Special and General Relativity
Replies
9
Views
2K
  • Special and General Relativity
Replies
14
Views
2K
  • Special and General Relativity
Replies
5
Views
1K
  • Special and General Relativity
Replies
1
Views
1K
  • Special and General Relativity
Replies
29
Views
1K
  • Special and General Relativity
Replies
22
Views
2K
  • Special and General Relativity
Replies
22
Views
1K
Back
Top