- #36

aliens123

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If this is the case, then why isn't the Lorentz transformation given in this link:PeterDonis said:The correct statements of what the Lorentz transformation does are:

$$(E_G)_\mu \mapsto (E_G)'_\mu$$

$$(E_H)_\mu \mapsto (E_H)'_\mu$$

And, for completeness:

$$F_{\mu \nu} \mapsto F'_{\mu \nu}$$

$$(v_G)^\nu \mapsto (v_G)'^\nu$$

$$(v_H)^\nu \mapsto (v_H)'^\nu$$

In other words: a Lorentz transformation doesn't changewhich object you are dealing with. It just transforms the components of that object from one frame to another. But going from ##E_G## to ##E_H##changes which object you are dealing with. That is simply a different thing from a Lorentz transformation.

https://webhome.phy.duke.edu/~rgb/Class/phy319/phy319/node136.html

Related by a "simple" Lorentz transformation? That is, ##(E_G)_\mu \mapsto (E_G)'_\mu##

Is just

$$(E_G)_\mu = (E_G)'_\alpha \Lambda_{\mu}^{\ \ \alpha}$$

But this is not what the Lorentz transformation given in the link is. Or, in other words, what should we call

$$F_{\mu 0} \mapsto F'_{\mu 0}$$

This is what I was calling the "Lorentz transformation." But this is equivalent to

$$(E_G)_\mu \mapsto (E_H)'_\mu$$

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