From: ram@zedat.fu-berlin.de   
      
   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!   
      
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