## How to get this equation - Mechanics by Landau and Lifshitz

In page 28 of Mechanics by Landau and Lifshitz, there is the following equation.

$$\int_0^\alpha \frac{T(E) dE}{\sqrt{\alpha-E}}=\sqrt{2m}\int_0^\alpha \int_0^E \left[ \frac{dx_2}{dU}-\frac{dx_1}{dU}\right] \frac{dU dE}{\sqrt{[(\alpha-E)(E-U)]}}$$

Then, by changing the order of integration, it is converted to,

$$\int_0^\alpha \frac{T(E) dE}{\sqrt{\alpha-E}}=\sqrt{2m}\int_0^\alpha \left[ \frac{dx_2}{dU}-\frac{dx_1}{dU}\right] dU \int_U^\alpha \frac{dE}{\sqrt{[(\alpha-E)(E-U)]}}$$

I don't really understand how this happens. Could someone please break it down for me? Thanks.

An image of the page is attached.
Attached Thumbnails

 PhysOrg.com physics news on PhysOrg.com >> Is there an invisible tug-of-war behind bad hearts and power outages?>> Penetrating the quantum nature of magnetism>> Rethinking the universe: Groundbreaking theory proposed in 1997 suggests a 'multiverse'
 Recognitions: Homework Help Do you understand how to change the order of integration? http://mathinsight.org/double_integr...ation_examples http://www.youtube.com/watch?v=NETmfwOAKpQ

Recognitions:
Homework Help
 Quote by omoplata Then, by changing the order of integration, it is converted to, $$\int_0^\alpha \frac{T(E) dE}{\sqrt{\alpha-E}}=\sqrt{2m}\int_0^\alpha \left[ \frac{dx_2}{dU}-\frac{dx_1}{dU}\right] dU \int_U^\alpha \frac{dE}{\sqrt{[(\alpha-E)(E-U)]}}$$
I don't like the way they've written the last line. It makes it look as though the integrals have been decoupled. Clearer would be:
$$\sqrt{2m}\int_{U=0}^\alpha \left(\left[ \frac{dx_2}{dU}-\frac{dx_1}{dU}\right] dU \int_U^\alpha \frac{dE}{\sqrt{[(\alpha-E)(E-U)]}}\right)$$
For the change of order, think about the total set of values of U and E being integrated over. (Perhaps think of it as a region in the U-E plane.) Both range from 0 to α, but with the condition that U < E (a triangle in the plane). If you let E range second, that becomes "U = 0 to E for each E, while E = 0 to α". If you swap the order then it's "E = U to α for each U, while U = 0 to α"

## How to get this equation - Mechanics by Landau and Lifshitz

The square bracketed term doesn't contain any dependence on E, so it can be brought outside the inner integral.

I don't think the order has actually been changed, it's just a simplification, in the same way that you might simplify d(2x^2)/dx to 2(d(x^2)/dx), because 2 is not a function of x.

Recognitions:
Homework Help
 Quote by MikeyW The square bracketed term doesn't contain any dependence on E, so it can be brought outside the inner integral.
You're overlooking the presence of E in the range for the (original) inner integral. This makes the inner integral a function of E.
 Ah- that makes sense! Note to self- don't second guess Landau and Lifshitz.
 Thanks for the links and the explanations. But I still don't get one small part. So what they mean by saying "$\alpha$ is a parameter" is that it is a constant, right? So if I draw the $E = \alpha$ line in the $U - E$ plane it will be a straight line parallel to the $U$ axis? Similarly, the $U = \alpha$ line will be straight line parallel to the $E$ axis? Assuming that, I get this from the original equation, $$\int_0^\alpha \frac{T(E) dE}{\sqrt{\alpha-E}}=\sqrt{2m}\int_0^\alpha \left[ \frac{dx_2}{dU}-\frac{dx_1}{dU}\right] \int_U^\alpha \frac{dE dU}{\sqrt{[(\alpha-E)(E-U)]}}$$ How did they move the $dU$ to the left of the second integration sign to get the final answer? Doesn't that decouple the integrals? The integration with respect to $E$ is still a function of $U$, so it cannot be taken out of the integration with respect to $U$.

Recognitions:
Homework Help
 Quote by omoplata How did they move the $dU$ to the left of the second integration sign to get the final answer? Doesn't that decouple the integrals? The integration with respect to $E$ is still a function of $U$, so it cannot be taken out of the integration with respect to $U$.
That's what I was complaining about in my earlier post. The way it's written in the book, you could misread it as having decoupled the integrals. What they mean is:
$$\sqrt{2m}\int_{U=0}^\alpha \left(\left[ \frac{dx_2}{dU}-\frac{dx_1}{dU}\right] dU \int_U^\alpha \frac{dE}{\sqrt{[(\alpha-E)(E-U)]}}\right)$$
Here I've deliberately made only the minimal change of putting in parentheses, but it says the same as this, which is even clearer:
$$\sqrt{2m}\int_{U=0}^\alpha \left(\left[ \frac{dx_2}{dU}-\frac{dx_1}{dU}\right] \int_U^\alpha \frac{dE}{\sqrt{[(\alpha-E)(E-U)]}}dU\right)$$
 Oh OK. I was always under the impression that when you use integration as an operation the operand always has to be between the $\int$ and the $dU$. But when it's written down like this, $$\int_0^\alpha \frac{T(E) dE}{\sqrt{\alpha-E}}=\sqrt{2m}\int_0^\alpha \left[ \frac{dx_2}{dU}-\frac{dx_1}{dU}\right] dU \int_U^\alpha \frac{dE}{\sqrt{[(\alpha-E)(E-U)]}}$$ how do you know that the term $\frac{1}{\sqrt{[(\alpha-E)(E-U)]}}$ is to be integrated with respect to $U$? Because there is a $U$ in it?

Recognitions:
Homework Help
 Quote by omoplata Oh OK. I was always under the impression that when you use integration as an operation the operand always has to be between the $\int$ and the $dU$.
 But when it's written down like this, $$\int_0^\alpha \frac{T(E) dE}{\sqrt{\alpha-E}}=\sqrt{2m}\int_0^\alpha \left[ \frac{dx_2}{dU}-\frac{dx_1}{dU}\right] dU \int_U^\alpha \frac{dE}{\sqrt{[(\alpha-E)(E-U)]}}$$ how do you know that the term $\frac{1}{\sqrt{[(\alpha-E)(E-U)]}}$ is to be integrated with respect to $U$? Because there is a $U$ in it?
3. But, it is very common to use the same symbol inside and outside an integral, e.g. $$y = x + \int_0^xx^2.dx$$This really means $$y = x + \int_{t=0}^xt^2.dt$$The use of x both inside and outside is a 'pun'. Since this means you cannot rely on (2), it's better to make the expression clear with suitable use of parentheses.