Changing to polar coordinates in an exponential

• physichu
In summary, the conversation discusses an integral involving momentum and energy, specifically in the context of Peskin & Schroeder P. 27. The first expression is a simplified version of the integral, where the author mentions the need for justification of spherical symmetry. The second expression uses the equation E = p^2/2m to simplify further. The third expression introduces an additional exponential and denominator, but the origin of these terms is not clear.
physichu
Hello :)
I don't get this integral (Peskin & Schroeder P. 27 )

##\int {{{{d^3}p} \over {{{\left( {2\pi } \right)}^3}}}{1 \over {{E_{\bf{p}}}}}{e^{i{\bf{p}} \cdot {\bf{r}}}}} = {{2\pi } \over {{{\left( {2\pi } \right)}^3}}}\int\limits_0^\infty {dp{{{p^2}} \over {2{E_{\bf{p}}}}}{{{e^{ipr}} - {e^{ - ipr}}} \over {ipr}}} ##.

Actualy I don't get the next stage iethr :(

## = {{ - i} \over {2{{\left( {2\pi } \right)}^2}r}}\int\limits_{ - \infty }^\infty {dp{{p{e^{ipr}}} \over {\sqrt {{p^2} + {m^2}} }}} ##.

Thanx in advence :)

Between the first two, it appears as if they have integrated over the angular coordinates in a spherical coordinate sytem, leaving only the radial integral. There needs to be some justification about spherical symmetry to simply yield a factor of 2 pi by integrating over the angular coordinates.

Between the second and third expressions, they seem to be using E = p^2/2m and simplifying.

where do the other exponential and the ##ipr## denominator come from?

Last edited by a moderator:

1. How do you convert from Cartesian coordinates to polar coordinates?

To convert from Cartesian coordinates (x,y) to polar coordinates (r,θ), you can use the following equations:
r = √(x² + y²)
θ = tan⁻¹(y/x)
Where r represents the distance from the origin (0,0) to the point and θ represents the angle between the positive x-axis and the line connecting the origin to the point.

2. What is the relationship between exponential functions and polar coordinates?

Exponential functions can be represented in polar coordinates using the formula r = a^θ, where a is the base of the exponential function. This formula describes a spiral that increases or decreases in radius as θ increases or decreases, depending on the value of a.

3. Can you use polar coordinates to graph exponential functions?

Yes, polar coordinates can be used to graph exponential functions. The polar equation r = a^θ can be graphed on a polar coordinate system, with r representing the distance from the origin and θ representing the angle. This graph will show a spiral shape that increases or decreases in radius as θ increases or decreases, depending on the value of a.

4. How do you determine the domain of an exponential function in polar coordinates?

The domain of an exponential function in polar coordinates is determined by the range of values for θ. Typically, the domain of an exponential function in polar coordinates is all real numbers except for the values of θ that would cause the function to be undefined, such as when θ = 0 for an exponential function with a negative base.

5. What are the advantages of using polar coordinates to represent exponential functions?

Using polar coordinates to represent exponential functions allows for a more intuitive and visual understanding of the function. The spiral shape of the graph can show the behavior of the function as θ increases or decreases, and can also make it easier to identify key features such as the asymptote and range. Additionally, polar coordinates can be useful for solving problems involving complex numbers and for graphing functions in three dimensions.

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