I Easiest possible way to derive the Lorentz transformation

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The discussion centers on the simplest methods to derive the Lorentz transformation equations, emphasizing both mathematical simplicity and expediency. Participants suggest various approaches, including starting with generic transformation forms and applying the two postulates of special relativity, as well as using hyperbolic functions for a more elegant derivation. Bondi's method, which employs radar measurements and basic algebra, is highlighted as an accessible approach, though some argue it may not align with the thread's focus on elegance. The conversation also touches on the importance of space-time diagrams and their role in understanding relativity, with a consensus that these tools are underutilized in education. Overall, the thread explores multiple perspectives on deriving the Lorentz transformations effectively.
  • #31
(It all looks nicer with boldface instead of arrows for vectors, but it wasn't working for Greek letters.)
 
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  • #33
Okay, one more. Lorentz transformation of the electric and magnetic fields ##\vec E## and ##\vec B##:

##\vec E^\prime = \cosh{\phi} \, \vec E + \sinh{\phi} \, (\hat \phi \times \vec B) - 2 \sinh^2 \dfrac{\phi}{2} \, (\hat \phi \cdot \vec E) \hat \phi##

##\vec B^\prime = \cosh{\phi} \, \vec B - \sinh{\phi} \, (\hat \phi \times \vec E) - 2 \sinh^2 \dfrac{\phi}{2} \, (\hat \phi \cdot \vec B) \hat \phi##
 
  • #34
SiennaTheGr8 said:
(It all looks nicer with boldface instead of arrows for vectors, but it wasn't working for Greek letters.)

It's all good stuff. But, I see that as a (beautiful) distillation of the subject. It's so well distilled that I would question its accessibility to someone new to SR. It's not unlike Hartle's summary of SR in his GR book. It's slick and elegant to the point that it almost makes a mockery of how difficult SR was to learn in the first place! And, I'm not sure I could ever have learned SR (with no previous knowledge) from Hartle's book.

Even the concept of a four-vector (for me) took a while to digest. Although I'm self taught in physics (not in maths) and seem to be able to pick up things directly from a textbook, I've noticed that I can't cope very well with more than one new idea at a time.

When I started learning SR it was the first serious study I'd done in 30 years. SR was completely new to me and any maths and classical physics were very rusty. I learned from Helliwell's book (which I thought at the time was excellent, and still do). It took me about six weeks to nail the basic concepts (up to the Lorentz Transformation) and about three months to get through the whole book (spacetime, energy-momentum, four vectors, particle collisions, some kinematics and a bit about gravitation).

I'm trying to imagine learning the subject starting with four-vectors and geometry. I wish I could say "yes, that would have been better", but I can't. I see your ideas as more what you do with a subject once you have mastered it. It's a bit like an expert piano player looking at someone doing their scales and saying, you don't need to struggle with that, just do this and rattling off a sonata! A poor analogy, perhaps, but there it is.
 
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  • #35
One more thing about rapidity (sorry!).

It's well-known that rapidities represent hyperbolic angles. What often goes unstated, though—perhaps because it's thought to follow trivially from the previous sentence—is that when Bob measures Alice's rapidity, he's measuring the hyperbolic angle between her four-velocity and his own.
 
  • #36
SiennaTheGr8 said:
One more thing about rapidity (sorry!).
Well, there is also the fact that you can use it to appease people who are unhappy with a finite/maximum speed of light ;)
 
  • #37
SiennaTheGr8 said:
One more thing about rapidity (sorry!).

It's well-known that rapidities represent hyperbolic angles. What often goes unstated, though—perhaps because it's thought to follow trivially from the previous sentence—is that when Bob measures Alice's rapidity, he's measuring the hyperbolic angle between her four-velocity and his own.

The big deal with rapidities is that they are additive (since Minkowski-arc-length along the "Minkowski-circle" [the hyperbola] is additive).
This is contrasted with velocities.
Note:
v_{AC}<br /> =\tanh\theta_{AC}<br /> =\tanh\left( \theta_{AB}+\theta_{BC}\right)<br /> =\frac{\tanh\theta_{AB}+\tanh\theta_{BC}}{1 + \tanh\theta_{AB}\tanh\theta_{BC}}<br /> =\frac{v_{AB}+v_{BC}}{1 + v_{AB}v_{BC}}<br />

In addition, \exp(\theta_{AC})=\sqrt{\frac{1+V_{ac}}{1-V_{ac}}}, the Doppler factor [which is an eigenvalue of the Lorentz boost].
It's worth noting that there are Galilean-rapidities, as well.
It's the Galilean-spacelike-arc-length along the "Galilean-circle" [the vertical line on a position-vs-time graph].
The analogue of the "tangent function of the Galilean-rapidity" in Galilean spacetime geometry is the Galilean-rapidity itself.
Because of this, Galilean-velocities are also additive (since Galilean-rapidity is additive).
This situation of the additivity-of-Galilean-velocities is one of those prejudices making up our common sense that makes relativity hard to understand.
 
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  • #38
robphy said:
The big deal with rapidities is that they are additive (since Minkowski-arc-length along the "Minkowski-circle" [the hyperbola] is additive).
This is contrasted with velocities.

Yes, and what I'm trying to emphasize is that in addition to the mathematical elegance this gives you (additivity is nice!), there's a conceptual elegance. Even if you aren't drawing a diagram, rapidity has a simple geometric interpretation (angle between an object's four-velocity and your own).
 
  • #39
The other "big deal with rapidities" is that differences/changes in them are invariant under a collinear boost (or in the 1+1D case), though you need to allow for signed rapidities for this to work in all cases.

More generally, under a boost in the ##x##-direction, it's differences/changes in the quantity ##\tanh^{-1} \beta_x## that are invariant. This is sometimes called the "longitudinal rapidity" (or just "rapidity" in the context of particle physics). Note that this quantity is not the ##x##-component of the rapidity vector ##\vec \phi = \phi \hat \beta##.
 

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