Velocity Goes Into Itself.

talanum1

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We have $\gamma = 1/sqrt(1-v^2/c^2)$ and then: $\vector{v}=\gamma (\vector{e}_0c+\vector{u}(t))$. This has $\vector{v}$ ill defined: it occurs in it's own definition.
 
We have $\gamma = 1/sqrt(1-v^2/c^2)$ and then: $\vector{v}=\gamma (\vector{e}_0c+\vector{u}(t))$. This has $\vector{v}$ ill defined: it occurs in it's own definition.

How are you doing?
 
We have $\gamma = 1/sqrt(1-v^2/c^2)$ and then: $\vector{v}=\gamma (\vector{e}_0c+\vector{u}(t))$. This has $\vector{v}$ ill defined: it occurs in it's own definition.

ChatGPT:

It only looks like a circular definition because two different things are being called “velocity.” In relativity, there’s the ordinary 3-velocity u (how fast something moves through space) and the 4-velocity U (how it moves through spacetime — space and time together). The familiar Lorentz factor comes from the 3-speed:
(\gamma = 1/\sqrt{1 - |\mathbf{u}|^2/c^2}).
Once you have that, the 4-velocity is built from it as
(\mathbf{U} = \gamma(\mathbf{u})(c,\mathbf{e}_0 + \mathbf{u})).
If you accidentally use the same symbol for both, it looks like the velocity is hiding inside its own definition — but that’s just a labeling mix-up. Keep the two meanings separate, and the formula makes perfect sense.


(Think of it as the difference between how fast you move through space and how fast you move through time.)
 
@talanum1 , why is it that you usually choose not to first check with ChatGPT, you've done so before.... could have saved yourself the embarrassment
If it's off-track, try a couple of minutes fine tuning your questions / directives, thrown into a squashed single input if need be.

Anyway I'm done doing that.
 
@bwana @The_Librarian I would like to propose a suggestion , that all these threads could be in one single thread for the OP:
 
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