- #1
trelek2
- 88
- 0
Force is equal to change in potential (r is a point in space suppose integrating between points r1 to r2 in time t1 to t2, i did not indicate this but it is not relevant for my quenstion:
[tex]mr'' =F(r) = -\delta V[/tex]
after multiplying both sides by dr it is obvious RHS will give potential difference, but LHS needs to be transformed to show corresponding kinetic energy difference:
[tex]mr'' \cdot dr = m \int_{}^{} r'' \cdot r' \cdot dt = m \int_{}^{} \frac{d}{dt}r' ^{2}dt [/tex]
:(
The problem is there should be a 1/2 term there as well. I don't know how to integrate this LHS. Can you please help- show me clearly step by stem what is done and where this 1/2 comes in...
[tex]mr'' =F(r) = -\delta V[/tex]
after multiplying both sides by dr it is obvious RHS will give potential difference, but LHS needs to be transformed to show corresponding kinetic energy difference:
[tex]mr'' \cdot dr = m \int_{}^{} r'' \cdot r' \cdot dt = m \int_{}^{} \frac{d}{dt}r' ^{2}dt [/tex]
:(
The problem is there should be a 1/2 term there as well. I don't know how to integrate this LHS. Can you please help- show me clearly step by stem what is done and where this 1/2 comes in...