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In some literatures, the Hamilton's principle for conservative systems is introduced by this equation: $$\delta \int_{t_1}^{t_2}(T-V) ~\mathrm{d}t~=~0$$

In some others, this principle is introduces as follow:

$$\int_{t_1}^{t_2}\delta(T-V) ~\mathrm{d}t~=~0$$

What is the difference between two equations? Are those expressions the same?

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Note that the general expression of Hamilton's principle is: $\delta S=0$. –  jinawee Jun 18 at 21:20
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From the variational calculus point of view, we can get the first expression. I mean if we use the Principle of least action the first equation will appear. "The path taken by the system between times t1 and t2 is the one for which the action is stationary (no change) to first order." Why we could take the $ \delta $ operator into the integral? What is the physical meaning of the second equation? –  Mas ooD Jun 18 at 21:34
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I just wanted to point out the general expression. I suggest you read Lanczos The Variational Principles of Mechanics, which is quite amusing (which gives the same answer as the given below). –  jinawee Jun 18 at 21:41
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1 Answer 1

up vote 5 down vote accepted

They're the same, because integration is linear:

$$\int _{t_1}^{t_2} \left( f(t) - g(t) \right) \,dt = \int_{t_1}^{t_2}f(t)\,dt - \int_{t_1}^{t_2}g(t)\,dt$$

Addendum:

Consider two paths. Let the system trajectory in the first path be denoted $X_1(t)$ and the trajectory for the second path be denoted $X_2(t)$. Let the Lagrangian be denoted $L$. Then the action for the first trajectory is $$S_1 = \int_{t_1}^{t_2}L(X_1(t))\,dt$$ and the action for the second trajectory is $$S_2 = \int_{t_1}^{t_2}L(X_2(t))\,dt.$$

The difference in the action between the two paths is \begin{eqnarray} \Delta S \equiv S_1 - S_2 &=& \int_{t_1}^{t_2}L(X_1(t))\,dt - \int_{t_1}^{t_2}L(X_2(t))\,dt \\ &=& \int_{t_1}^{t_2} \left[ L(X_1(t))-L(X_2(t)) \right]\,dt \\ &=& \int_{t_1}^{t_2}\Delta L(t)\,dt \end{eqnarray} In the last equation we wrote $\Delta L$ to denote $L(X_1) - L(X_2)$. This is not quite what we're going for: we want to get $\delta L$ inside the integral, where $\delta L$ is the variation of the Lagrangian. The variation is the derivative of a function with respect to something. When you think about the definition of a derivative, you have something like $$\frac{df}{dx}(x_0)\equiv \lim_{h\rightarrow 0}\frac{f(x_0+h)-f(x_0)}{h}.$$ So really, taking the $\delta$ inside the integral requires the linearity property we already demonstrated and the ability to move the limit inside the integral. This is ok in almost every case you will ever encounter. To prove why you can move the limit inside the integral you have to do some analysis which I really don't want to recall and type here. You can find that sort of thing in analysis books.

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I'm Not Convinced yet. –  Mas ooD Jun 18 at 21:40
    
@MasooD: I added an explicit demonstration. Note that it's just using exactly the linearity property of the integral. –  DanielSank Jun 18 at 21:42
    
@DanielSank Although the spirit of this is correct, it's not quite this simple. Recall that $\delta$ denotes a variation, not simply a change. In other words, it is a first order change in some parameter, say $\epsilon$, when one deforms a given path $q$ via a one-parameter family of paths $q + \epsilon\eta$. The validity of taking the $\delta$ inside then comes down to a question of when one can take derivatives inside of integrals, which one can under certain assumptions. –  joshphysics Jun 19 at 9:20
    
@joshphysics: The cases in which you can't take the derivative inside integrals almost never show up in any physics problem, so I didn't want to bother about that. Also, while $\delta$ is indeed a variation, the reason it (And the derivative) can be put inside the integral follows from exactly the linearity property already discussed. That said, I didn't explicitly explain this, so I'll edit my answer. –  DanielSank Jun 19 at 15:27
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