Tensors & Lorentz Transform: Is There a Connection?

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SUMMARY

The discussion establishes a definitive connection between tensors and Lorentz transformations, emphasizing that the Lorentz interval remains invariant across different inertial frames. The mathematical representation of the Lorentz interval, expressed as ##\eta_{\alpha \beta} dx^\alpha dx^\beta##, illustrates that both frames maintain the same metric tensor components, ##\eta_{\alpha \beta}##. This invariance is crucial for understanding how tensors operate within the framework of special relativity, confirming that tensors serve as a fundamental tool for describing physical phenomena in varying coordinate systems.

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  • Understanding of tensors and their properties
  • Familiarity with Lorentz transformations in special relativity
  • Knowledge of metric tensors and their role in physics
  • Basic proficiency in mathematical notation used in physics
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  • Explore the implications of Lorentz invariance in relativistic physics
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Physicists, mathematicians, and students of theoretical physics who are interested in the relationship between tensors and the principles of special relativity.

geordief
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Are these two subjects closely related?

It seems a tensor can be invariant when viewed from any **co ordinate system and

The Lorentz Transformation seems to allow 2 moving co ordinate frames to agree on a space time intervals.

Is there some deep connection going on?

**=moving frames of reference?
 
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geordief said:
Are these two subjects closely related?
Yes. Because Lorentz interval should be invariant in two inertial frames, and can be written as ##\eta_{\alpha \beta} dx^\alpha dx^\beta##, then if in frame with coordinate system ##x## the interval is ##\eta_{\alpha \beta} dx^\alpha dx^\beta## and in frame with coord sys ##x'## it is ##\eta_{\sigma \kappa} dx'^\sigma dx'^\kappa## (notice that we have the same ##\eta## in the two frames because the lorentz interval should take the same form in the two inertial frames) then ##\eta_{\alpha \beta} dx^\alpha dx^\beta = \eta_{\sigma \kappa} dx'^\sigma dx'^\kappa##, from which one defines the components of the metric tensor as ##\eta_{\alpha \beta}## and the basis as ##dx^\alpha dx^\beta## in the "frame ##x##" and by noticing that ##x' = \Lambda x + b## one realizes that the components in ##x'##, in terms of the components in ##x##, are ##\eta_{\sigma \kappa} = \Lambda^\alpha{}_\sigma \Lambda^\beta{}_\kappa \eta_{\alpha \beta}##, but the tensor remains the same.
 

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