Problem with a piece of calculation!

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  • #1
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I'm reading Gourgoulhon's Special Relativity in general frames. In chapter 11, which is about the principle of least action, he constructs the relativistic version of this principle but I have problem with one part of it.
At first, he introduces Lagrangian and then from the fact that the action should be independent of the parametrization of the world line, concludes that the Lagrangian should be a positive homogeneous function of degree 1 w.r.t. the components of 4-velocity. So, using Euler's theorem for homogeneous functions, he writes:
[itex]
\dot x^\alpha \frac{\partial L}{\partial \dot x^\alpha}=L
[/itex].
After this, he derives the Euler-Lagrange equations [itex] \frac{\partial L}{\partial x^\alpha}-\frac{d}{d \lambda}\frac{\partial L}{\partial \dot x ^\alpha}=0 [/itex].
Then, in a remark, he says that the fact that we have four Euler-Lagrange equations in the Relativistic case, which is one more than in the Newtonian case, doesn't mean that an extra degree of freedom is added to the system because the four equations aren't independent. I don't have a problem with this statement but the calculations he does afterwards(which are off-shell) doesn't seem clear to me and I can't reconstruct them!
[itex]
\dot x ^\alpha (\frac{\partial L}{\partial x^\alpha}-\frac{d}{d \lambda}\frac{\partial L}{\partial \dot x ^\alpha})=
\dot x^\alpha \frac{\partial L}{\partial x^\alpha}-\frac{d}{d\lambda}(\dot x^\alpha \frac{\partial L}{\partial \dot x^\alpha})+\frac{d \dot x ^\alpha}{d \lambda}\frac{\partial L}{\partial \dot x^\alpha}=\frac{dL}{d \lambda}-\frac{d}{d \lambda}(\dot x^\alpha \frac{\partial L}{\partial \dot x^\alpha})
[/itex]
I have problem with the 2nd equality. How is it done?
Thanks
 
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Answers and Replies

  • #2
ChrisVer
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[itex] \dot{x}^a \frac{\partial L}{\partial x^a}= \dot{x}^a \frac{d}{d \lambda} \frac{\partial L}{\partial \dot{x}^a}[/itex] (E-L equat)

[itex] \frac{d \dot{x}^a}{d\lambda} \frac{\partial L}{\partial \dot{x}^a}+ \dot{x}^a \frac{d}{d \lambda} \frac{\partial L}{\partial \dot{x}^a}= \frac{d}{d \lambda} (\dot{x}^a \frac{\partial L}{\partial \dot{x}^a})[/itex] the 1st and 3rd terms in your equation after replacing the 1st term with the E-L equation result... it's a total derivative of lambda...

[itex]\frac{d}{d \lambda} (\dot{x}^a \frac{\partial L}{\partial \dot{x}^a})= \frac{d}{d\lambda} (L)[/itex] (using your first equation for L)

Then there you got the result
 
  • #3
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Yeah, that works but as I said his calculations are of-shell. He says: "To make this explicit, let us evaluate the following expression, without assuming that (11.17) holds " where 11.17 are the E-L equations!
 
  • #4
ChrisVer
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Maybe then you have to look what the total derivative of L wrt to lambda is...assuming that it has no explicit dependence on it,...
 
  • #5
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Oh my god...Its just I was being too careless. Of course 1st and 3rd terms add up to the L's total derivative. Thanks man!
 

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