Difference between 2 Langrangians, proof

In summary: L'}{L}-\frac{d}{dt}\frac{L'}{\dot{L}}=0In summary, Euler-Lagrange's equations give the same motion equations as L'=L+\frac{d}{dt}f(q,t).
  • #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.
 
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  • #2
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
Dick said:
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
fluidistic said:
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
gabbagabbahey said:
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
q does depend on t. But in euler-lagrange you fix the endpoints. So q(t1) is fixed and q(t2) is fixed.
 
  • #7
Dick said:
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
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
Dick said:
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
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]
 

1. What is a Langrangian?

A Langrangian is a function that is used in classical mechanics to describe the dynamics of a physical system. It is defined as the difference between the kinetic and potential energy of the system.

2. What is the difference between two Langrangians?

The difference between two Langrangians is simply the difference in their mathematical expressions. Each Langrangian represents a different system or scenario, and comparing the two can provide insight into the similarities and differences between these systems.

3. How is the difference between two Langrangians calculated?

The difference between two Langrangians is calculated by subtracting one Langrangian from the other. This results in a new Langrangian that represents the difference between the two original ones.

4. Why is it important to prove the difference between two Langrangians?

Proving the difference between two Langrangians is important because it allows us to verify the accuracy and validity of our calculations. It also helps us understand the underlying principles and relationships between different physical systems.

5. What is the significance of the difference between two Langrangians in physics?

The difference between two Langrangians is significant in physics because it can provide insight into the dynamics of a physical system. By comparing and analyzing the differences, we can better understand the behavior and interactions of the system.

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