Author: vasco [ 25 Jan 2010, 09:44 ] Post subject: Re: 7.1 Contour Integrals - Help needed! billroot wrote:What is the line from a to b meant to represent? Why not a straight line? I tried substituting simple complex values for points a and b, doing the integration, applying the limits and I always get the same value for the integral regardless of whether I go straight from a to b or via some other artbitary points, chosen to go different ways round the origin. When I look more carefully at fig 7.3. I notice that Prof. Penrose has not drawn a circle round the origin from b which is what I would expect, the last contour in fig 7.3(d) has point b well inside it. What am I missing here?Without an understanding of this I won't understand the solution to 7.1 or what comes after.Quote:What is the line from a to b meant to represent? The line from to is meant to represent the path of integration. In the theory of integration for real variables you always integrate along the real x-axis, but as soon as you go into the complex domain you have a dilemma: how to go from to ?Quote:Why not a straight line?Of course you could draw a straight line but Penrose has avoided that because it doesn't look general enough - it would be a particular path from to , and he's trying to talk generally here.In diagram 7.3(b) - top right - instead of going directly to from the path goes round the origin once to get to .Now, you could say, "but point in diagram (b) is the same point as point in diagram (a)".This is correct in purely geometric terms, but when you are doing integration, you are moving around the curve in small increments and is changing continuously as you go round the origin.So, if you think of the initial value of as , then after you've gone round the origin once and back to the same point the value of is now .That means that whatever the value was for the integral in diagram (a), let's say , then it's value in diagram (b) is , because the value of has increased by .In diagram (c), Penrose shows the integration continuing from back to . The value of this integral from to must be the negative of the original integral in diagram (a) since the path of integration has been reversed, and so its value must be . So the total integral round a closed contour enclosing the origin from and back to must be . This is true whatever values we choose for and . (Added by vasco on 25th January 2010 at 17:22 GMT).Quote:I notice that Prof. Penrose has not drawn a circle round the origin from b which is what I would expect, the last contour in fig 7.3(d) has point b well inside it. What am I missing here?The point here is that and represent two completely arbitrary points in the complex plane and therefore the squid-shaped contour in diagram (c) might as well be any shape you like and pass through any points you like, as long as it encloses the origin. You can continuously deform the contour into any shape you like without crossing over the origin and the integral will still be equal to .I hope this helps. If you need any further clarification then please post again.Vasco
 Author: billroot [ 25 Jan 2010, 18:07 ] Post subject: Re: 7.1 Contour Integrals - Help needed! Thank you Vasco. Your reply was just what I needed.So an integral from any point returning to that point will equal zero unless it goes round the origin to form a closed loop of any shape then the integral will equal where is the number of times the origin is looped.It's kind of obvious really and stems from the fact that the argument of the integral is . My Road to Reality is going to take me some time!
 Author: vasco [ 25 Jan 2010, 19:39 ] Post subject: Re: 7.1 Contour Integrals - Help needed! Quote:So an integral from any point returning to that point will equal zero unless it goes round the origin to form a closed loop of any shape then the integral will equal where is the number of times the origin is looped.You've got it! Well done! This only applies to the integral of course. Other powers of integrate to zero even if the path of integration goes round the origin.VascoPS I'm only a couple of chapters ahead of you, so keep going and you might catch me up!