Difference between 2 Langrangians, proof

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  • #1
fluidistic
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Homework Statement


Demonstrate that if the Lagrangians [tex]L(q, \dot q ,t)[/tex] and [tex]L'(q, \dot q , t)[/tex] who differ in a total derivative of a function f(q,t) give the same motion equations.
That is, [tex]L'=L +\frac{d}{dt}f(q,t)[/tex]

Homework Equations


Euler-Lagrange's equation.


The Attempt at a Solution


I tried to use Euler-Lagrange's equation to see if I could reach the same equations for L and L' but without any success. I've checked out in Landau & Lifgarbagez's book. Here is what it more or less says: "If we have [tex]L'=L+\frac{d}{dt}(2a \vec r \cdot \dot \vec r+a \dot r ^2 t)[/tex], we can omit the second term since it's a total derivative with respect to time."
So according to this book it's obvious while I have to prove it. But I don't understand why it's obvious nor why it's true.
 

Answers and Replies

  • #2
Dick
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It's obvious because the Euler-Lagrange equations come from the variation of the integral of L*dt. A total time derivative in L can only contribute a constant to the integral of L*dt. The variation of a constant is 0.
 
  • #3
fluidistic
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It's obvious because the Euler-Lagrange equations come from the variation of the integral of L*dt. A total time derivative in L can only contribute a constant to the integral of L*dt. The variation of a constant is 0.
I thank you for the reply, much appreciated.
I'm still not getting it. Ok for your 1st and 3rd sentence, but I don't see why the second is true.
If I understand you, you are implying that [tex]\int _{t_0}^{t_1} \frac{d}{dt} f(q,t) dt=\int _{t_0}^{t_1} d f(q,t)=K[/tex], a constant. Also the integral of d f(q,t) doesn't make sense to me. I'm missing something...
 
  • #4
gabbagabbahey
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If I understand you, you are implying that [tex]\int _{t_0}^{t_1} \frac{d}{dt} f(q,t) dt=\int _{t_0}^{t_1} d f(q,t)=K[/tex], a constant. Also the integral of d f(q,t) doesn't make sense to me. I'm missing something...

Just use the fundamental theorem of calculus...
 
  • #5
fluidistic
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Just use the fundamental theorem of calculus...
That's what I had done first on my draft, but I wasn't sure it was right. So it's worth [tex]f(q,t_2)-f(q, t_1)[/tex]. But I'm not sure q depends on t. If it does then we indeed reach a constant, but if it doesn't, I don't see it.
Also, I'm not sure I can use the FTC since I have an integral of a differential of a function, not a differential of a variable. I mean I don't even have a "dt" or whatsoever.
 
  • #6
Dick
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q does depend on t. But in euler-lagrange you fix the endpoints. So q(t1) is fixed and q(t2) is fixed.
 
  • #7
fluidistic
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q does depend on t. But in euler-lagrange you fix the endpoints. So q(t1) is fixed and q(t2) is fixed.

Ok I see. If q is constant we're done and if q depends on t we're also done. It cannot depend on anything else I guess.
Now am I right in post #3? How can I apply the FTC? I don't have a differential of a variable.
 
  • #8
Dick
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You have integral of d/dt(f(q(t),t)*dt from t1 to t2. That's f(q(t2),t2)-f(q(t1),t1) by FTW as gabbagabbahey said. Everything is fixed at the endpoints. It's a constant.
 
  • #9
fluidistic
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You have integral of d/dt(f(q(t),t)*dt from t1 to t2. That's f(q(t2),t2)-f(q(t1),t1) by FTW as gabbagabbahey said. Everything is fixed at the endpoints. It's a constant.

Thanks I now get it. Those differentials! I missed that I shouldn't have simplified the dt's in the equation [tex]\int _{t_0}^{t_1} \frac{d}{dt} f(q,t) dt=\int _{t_0}^{t_1} d f(q,t)[/tex] and that I should have applied it (the FTC) in the left hand side. Sorry to both for being so slow minded. I'm happy I've understood at least.
Thanks once again.
 
  • #10
gabbagabbahey
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On a side note, you can also use the chain rule to show that

[tex]\frac{\partial L'}{\partial q}-\frac{d}{dt}\frac{\partial L'}{\partial \dot{q}}=\frac{\partial L}{\partial q}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}[/tex]
 

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