Re: The momentum - a cotangent vector?

ram@zedat.fu-berlin.de (Stefan Ram) schrieb oder zitierte:

How can I see that (given that q' is a tangential vector)
p is a cotangential vector?

  Here's a little calculation I whipped up in the realm of good
  old classical mechanics, no relativity involved.

  I'm starting with the well-known formula

E = 1/2 m v^2

  Using Cartan's calculus, from this, I come up with:

dE = m v dv + 1/2 v^2 dm.

  And since dm = 0 (I assume the mass is constant):

dE = p dv.

  Now let's write out the implied scalar product as "*":

dE = p * dv.

  This "p *" is now a covector acting like a linear function, mapping
  changes in velocity (a vector) to changes in energy (a scalar).

  BTW, we also can derive the "other" relationship dE = v dp!

  Writing "1/2 m v^2" as "1/2 m v v", we can see that

E = 1/2 p v

  , so,

dE = 1/2 p dv + 1/2 v dp

  . But since we already had established that dE is "p dv" for a
  constant mass m, "1/2 p dv" must be "1/2 dE", so that,

dE = 1/2 dE + 1/2 v dp.

  Subtracting "1/2 dE" on both sides gives:

1/2 dE = 1/2 v dp,

  and multiplication by 2,

dE = v dp.

  So, dE is both "v dp" and "p dv" when the mass m is constant!