Four-vector problem


by Wox
Tags: four-velocity, special relativity
Wox
Wox is offline
#19
Mar13-12, 04:19 AM
P: 71
Depends on your definition of the dot product, but I see what you mean. But I don't see why [Ux/Ut,Uy/Ut,Uz/Ut] would correspond to a spatial velocity.
PAllen
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#20
Mar13-12, 08:08 AM
Sci Advisor
PF Gold
P: 4,863
Quote Quote by Wox View Post
Depends on your definition of the dot product, but I see what you mean. But I don't see why [Ux/Ut,Uy/Ut,Uz/Ut] would correspond to a spatial velocity.
In a metric space there is one definition of the dot product. The Euclidean one looks the way it does solely because the Euclidean metric is diag(1,1,1).

If U is some tangent vector, and x,y,z,t are unit vectors for some frame, then the dot product of U with such unit vectors expresses U in that frame basis. Then Ux/Ut gives the x speed (well, actually, x-speed/c , but that is just as good). Look at the tangent vector itself expressed in your starting coordinates (c-normed, canonic metric; works the same in any other convention):

U = gamma(c,u)

Ux = gamma * ux is not the x speed; but note Ux/Ut = ux/c. This feature will be true in any other basis. In particular, in an orthonormal basis with U itself taken as the time unit vector, you get spatial speed of zero - the particle has no spatial speed in its own basis.


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