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Let's imagine that we have a parametric function f1(x(t),y(t),z(t)) and an analytic one f2(x,y,z) and we have to integrate their product over some volume dx dy dz.

So what are analytical tools for it?

Thanks!

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- #1

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Let's imagine that we have a parametric function f1(x(t),y(t),z(t)) and an analytic one f2(x,y,z) and we have to integrate their product over some volume dx dy dz.

So what are analytical tools for it?

Thanks!

- #2

Stephen Tashi

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I think you should give a specific example.

- #3

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Thank you for your response.

I think you should give a specific example.

I mean that function f1 represents a curve that can be defined only as parametric. And when i want to calculate an integral f1*f2 over dxdydz i need to turn all my functions into either parametric or into ordinary view as far as i understand. So the question is how can i do it ?

An example can be given as follows:

2d:

f1 = {

x = r1*cos(t)

y = r2*sin(t).

}

f2 = x^2+y^2;

Integral[f1*f2,{x,-1,1}{y,-1,1}];

- #4

Stephen Tashi

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The integral of the usual sort of curve over an area or volume will be zero because the usual sort of curve has zero area and zero volume.

It would make sense to ask about the integral of the product of two parameterized curves: [itex] \int_{t_0}^{t_1} { f_1(x(t),y(t)) f2(x(t),y(t))} dt.[/itex].

It would make sense to ask about a "line integral". Are you asking about line integrals?

It would make sense to ask about the integral of the product of two parameterized curves: [itex] \int_{t_0}^{t_1} { f_1(x(t),y(t)) f2(x(t),y(t))} dt.[/itex].

It would make sense to ask about a "line integral". Are you asking about line integrals?

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- #5

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To avoid "zero volume" problem let's put f1 into delta-function.The integral of the usual sort of curve over an area or volume will be zero because the usual sort of curve has zero area and zero volume.

It would make sense to ask about the integral of the product of two parameterized curves: [itex] \int_{t_0}^{t_1} { f_1(x(t),y(t)) f2(x(t),y(t))} dt.[/itex].

It would make sense to ask about a "line integral". Are you asking about line integrals?

Something like this : DiracDelta[x-x(t)] DiracDelta[y-y(t)] f2 . So after using properties of DiracDelta, integration will be done easily. That's right?

- #6

Stephen Tashi

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Using Dirac Delta functions doesn't define a mathematical problem.To avoid "zero volume" problem let's put f1 into delta-function.

Something like this : DiracDelta[x-x(t)] DiracDelta[y-y(t)] f2 . So after using properties of DiracDelta, integration will be done easily. That's right?

For example let [itex]f_2(x,yz) [/itex] to be the constant function [itex] f_2(x,y) = 1 [/itex]. Let [itex] f_1(x(t),y(t)) [/itex] be the line [itex] f_1(x(t),y(t)) = (t,0) [/itex].

Are you asking about whether the notation [itex] \int_0^1 \int_0^1 \delta(x-x(t)) \delta(y-y(t)) (1)\ dy\ dx [/itex] has a mathematical definition? It doesn't. The role of [itex] t [/itex] is unclear.

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that's the question, then. parameter t is the only way to define the curve. so what can we do here? we either have to find an analytical way to define the curve without "t" or to find some math tool that modifies the expression that takes into account "t" somehow , i don't know. that's what i'm looking for.Using Dirac Delta functions doesn't define a mathematical problem.

For example let [itex]f_2(x,yz) [/itex] to be the constant function [itex] f_2(x,y) = 1 [/itex]. Let [itex] f_1(x(t),y(t)) [/itex] be the line [itex] f_1(x(t),y(t)) = (t,0) [/itex].

Are you asking about whether the notation [itex] \int_0^1 \int_0^1 \delta(x-x(t)) \delta(y-y(t)) (1)\ dy\ dx [/itex] has a mathematical definition? It doesn't. The role of [itex] t [/itex] is unclear.

- #8

Stephen Tashi

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The most likely scenario is that you have incorrectly formulated a situation as a mathematical problem.. What situation do you think you are solving by writing down an expression the requires integrating a curve over a volume? For example, if this is supposed to solve a problem in physics, what is the statement of that problem?that's the question, then. parameter t is the only way to define the curve. so what can we do here? we either have to find an analytical way to define the curve without "t" or to find some math tool that modifies the expression that takes into account "t" somehow , i don't know. that's what i'm looking for.

- #9

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the problem is to calculate spectral-angular density ("sad") of radiation produced by electron that moves through spiral curve along it's axis.The most likely scenario is that you have incorrectly formulated a situation as a mathematical problem.. What situation do you think you are solving by writing down an expression the requires integrating a curve over a volume? For example, if this is supposed to solve a problem in physics, what is the statement of that problem?

so basic expression for "sad" is :

So after this problem of the integration appears:

- #10

Stephen Tashi

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radiation density - at large distances, field energy that contains withing a solid angle - dOmega - , frequency range - domega - , with towards wave vector k. Here "radiation" is diffraction radiation, produced by induced currents that caused by a point charge's field.

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Stephen Tashi

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Yes. The thing is that we want to see "integrated over time" field energy , so it will be constant in time withing the solid angle.

- #14

Stephen Tashi

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average radiation density = [itex] \int_{T}\ [\ \int_{(X,Y,Z)} f(x(t),y(t),z(t)) \ dx\ dy\ dz)\ ]\ dt [/itex]

The integral [itex] \int_{(X,Y,Z)} f(x(t),y(t),z(t))\ dx\ ,dy\ dz [/itex] is computed regarding [itex] t [/itex] as a constant. This is not the integral of a function along a line. It is the integral of a time varying function defined everywhere within a volume.

Can you compute it as:

average radiation density = [itex] \int_{(X,Y,Z)} \ [ \ \int_{T} f(x(t),y(t),z(t)) dt\ ]\ dx\ dy\ dz [/itex]

?

- #15

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i guess it's not an option because [itex] \ \int_{T} f(x(t),y(t),z(t)) dt\ [/itex] does not depend on any x,y,z ( x(t) = r1*cos(t), y(t) = ... , z(t) = ... ) so integration over x,y,z in [itex] \int_{(X,Y,Z)} dx\ dy\ dz [/itex] will give us V.

average radiation density = [itex] \int_{T}\ [\ \int_{(X,Y,Z)} f(x(t),y(t),z(t)) \ dx\ dy\ dz)\ ]\ dt [/itex]

The integral [itex] \int_{(X,Y,Z)} f(x(t),y(t),z(t))\ dx\ ,dy\ dz [/itex] is computed regarding [itex] t [/itex] as a constant. This is not the integral of a function along a line. It is the integral of a time varying function defined everywhere within a volume.

Can you compute it as:

average radiation density = [itex] \int_{(X,Y,Z)} \ [ \ \int_{T} f(x(t),y(t),z(t)) dt\ ]\ dx\ dy\ dz [/itex]

?

- #16

Stephen Tashi

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If we are consider a fixed point in space [itex] (x,y,z)[/itex], the E&M field at that point depends on the distance between the [itex] (x,y,z) [/itex] and the position of the particle [itex] (x_p(t),y_p(t),z_p(t))[/itex] The function we need to integrate can be expressed as [itex] f( x - x_p(t), y - y_p(t), z - z_p(t) ) [/itex] soi guess it's not an option because [itex] \ \int_{T} f(x(t),y(t),z(t)) dt\ [/itex] does not depend on any x,y,z ( x(t) = r1*cos(t), y(t) = ... , z(t) = ... ) so integration over x,y,z in [itex] \int_{(X,Y,Z)} dx\ dy\ dz [/itex] will give us V.

[itex]\int_T f(x - x_p(t),y - y_p(t), z - z_p(t)) dt = g(x,y,z) [/itex] This is defined for each [itex] (x,y,z) [/itex] in a volume.

- #17

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i don'e get it. Parameter t has only one definition - parametric parameter of the curve, spiral curve. Has it anything to do with the position of the particle [itex] (x_p(t),y_p(t),z_p(t))[/itex] ? we have to take into account parametric parameter in the integration over the volume. or maybe i'm wrong?If we are consider a fixed point in space [itex] (x,y,z)[/itex], the E&M field at that point depends on the distance between the [itex] (x,y,z) [/itex] and the position of the particle [itex] (x_p(t),y_p(t),z_p(t))[/itex] The function we need to integrate can be expressed as [itex] f( x - x_p(t), y - y_p(t), z - z_p(t) ) [/itex] so

[itex]\int_T f(x - x_p(t),y - y_p(t), z - z_p(t)) dt = g(x,y,z) [/itex] This is defined for each [itex] (x,y,z) [/itex] in a volume.

- #18

Stephen Tashi

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That is correct. let [itex] t = [/itex] time.i don'e get it. Parameter t has only one definition - parametric parameter of the curve, spiral curve.

Let (x,y,z) be a point in the volume. A point on the path of the electron is [itex] (x_p(t),y_p(t),z_p(t)) [/itex].Has it anything to do with the position of the particle [itex] (x_p(t),y_p(t),z_p(t))[/itex] ? we have to take into account parametric parameter in the integration over the volume. or maybe i'm wrong?

At each point [itex] (x,y,z) [/itex] the the volume, the radiation field depends on the distance between [itex] (x,y,z) [/itex] and the electron. The radiation field is a function of

[itex] \frac{1}{r^2} = \frac{1}{(x - x_p(t))^2 + (y - y_p(t))^2 + (z - z_p(t))^2} [/itex]. So the field and field density is a function F(x,y,z,t).

[itex] \int_T F(x,y,z,t) dt = G(x,y,z) [/itex] There is no variable [itex] t [/itex] in G(x,y,z). We can integrate G(x,y,z) over the volume.

- #19

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i believe that is not exactly right, because there is also dependence on the path that the electron has covered and this dependence takes into account the spiral curve and it's shape, because it is the source of the radiation we consider.That is correct. let [itex] t = [/itex] time.

Let (x,y,z) be a point in the volume. A point on the path of the electron is [itex] (x_p(t),y_p(t),z_p(t)) [/itex].

At each point [itex] (x,y,z) [/itex] the the volume, the radiation field depends on the distance between [itex] (x,y,z) [/itex] and the electron. The radiation field is a function of

[itex] \frac{1}{r^2} = \frac{1}{(x - x_p(t))^2 + (y - y_p(t))^2 + (z - z_p(t))^2} [/itex]. So the field and field density is a function F(x,y,z,t).

[itex] \int_T F(x,y,z,t) dt = G(x,y,z) [/itex] There is no variable [itex] t [/itex] in G(x,y,z). We can integrate G(x,y,z) over the volume.

- #20

Stephen Tashi

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The dependence on the path the electron has covered is implemented when we do the integration with respect to [itex] t [/itex]. Of course you have to know the correct function to integrate with respect to t.i believe that is not exactly right, because there is also dependence on the path that the electron has covered.

- #21

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that's right, but i don't have it because field itself is the electron field and radiation field. And to find the last one i have to solve Maxwell's equation first and then do the next step whatever it is. The thing is that to find spectral-angular density i don't have to know radiation field itself, only coordinate dependence.The dependence on the path the electron has covered is implemented when we do the integration with respect to [itex] t [/itex]. Of course you have to know the correct function to integrate with respect to t.

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