Normalization of Four velocity

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SUMMARY

The normalization of four velocity, denoted as u, results in a scalar product of -1, which is a fundamental property of four-vectors in the context of special relativity. This outcome is contingent upon the choice of units, specifically when the speed of light is set to one. The sign of the scalar product can vary based on the metric signature used, either ##ds^2=-dt^2+dx^2+dy^2+dz^2## or ##ds^2=dt^2-dx^2-dy^2-dz^2##. Consistency in the chosen metric is crucial for accurate calculations.

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The scalar product of four velocity u with u gives -1. This is said as normalization of four vector.
But how does the scalar product yield -1 and what's normalization of four vector.
Immediate help required:oldconfused::oldconfused::oldconfused:
 
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Write down the velocity four-vector using coordinates in which the object in question is at rest, then calculate its scalar product with itself. What do you get?

Be aware that the magnitude of the result will depend on the units you use - it only comes out to be one if you use units in which the speed of light is one.

If you get the wrong sign, that will be because you're using a different sign convention for the metric: you can write ##ds^2=-dt^2+dx^2+dy^2+dz^2## or ##ds^2=dt^2-dx^2-dy^2-dz^2##; they both work as long as you're consistent and you'll see them both used.

If none of what I just said makes sense, you're missing some prerequisites - google for "four-vector" to learn about four-vectors.
 

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