one more integration

[grav]

How would one go about integrating something like (x^2 + y^2) sqrt(dx^2 + dy^2)?

[tdvance]

Try converting to Polar Coordinates.

[Incomplete]

How would one go about integrating something like (x^2 + y^2) sqrt(dx^2 + dy^2)?

My guess is your sqrt(dx^2 + dy^2) is there because you read somewhere that the root of the metric is the correct measure for integrations. If so you missed something: it's not the root of the metric, it's the root of the DETERMINANT of the metric, which is simply dx dy in that case.

[grav]

Try converting to Polar Coordinates.

Ironically, that's actually what I was trying to steer away from. I was originally trying to work through the Schwarzschild metric using polar coordinates, ds^2=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r) - r^2(dtheta^2 + sin(theta)^2 dphi^2), which I was asking about in this thread, but I had trouble visualizing the angles to work through that way, but have since found another where I can use the Cartesian coordinates, which I'm more familiar with.

[robross]

How would one go about integrating something like (x^2 + y^2) sqrt(dx^2 + dy^2)?

Are you familiar with the Math Help forum?

http://www.mathhelpforum.com/math-help/

It is to math what BAUT is to, uhm, whatever BAUT is

But seriously, it's a fantastic companion site to this one, and there are people there who can answer the most arcane abstract math questions.

Rob

[grav]

My guess is your sqrt(dx^2 + dy^2) is there because you read somewhere that the root of the metric is the correct measure for integrations. If so you missed something: it's not the root of the metric, it's the root of the DETERMINANT of the metric, which is simply dx dy in that case.

Thanks. I'm not sure what that means, though, except that I know I need to find dx dy in terms of the partial derivatives, which would be (x^2 + y^2) sqrt(dx^2 + dy^2) = (x^2 + y^2) dx sqrt(1 + (dy / dx)^2), which then works out to (x^2 + y^2) dx sqrt(1 + (x / y)^2) as far as I can tell from 0 to y, leaving factors of y and only dx left to work through, which would be okay since I'm finding for a curve from (x1 , 0) to (x2, y), but I want to be sure it can be used that way. For the surface of a sphere, for example, all of the x's, y's, and z's completely fall out, but that is because we have r^2 = x^2 + y^2 + z^2 and r is a constant, so I can find the curve in terms of r. In this case, we also have r^2 = x^2 + y^2 + z^2, but the curve is going from one value of r to another, and z = 0, so I didn't include dz, in which case the partial derivative would be dy / dx = -x / sqrt(r^2 - x^2) for the sphere, but since r also varies in this one, I figure finding it in terms of dy / dx = -x / y is just as good, but the y's remain in the equation even after dy drops out in that case. Is that correct?

[grav]

Are you familiar with the Math Help forum?

http://www.mathhelpforum.com/math-help/

It is to math what BAUT is to, uhm, whatever BAUT is

But seriously, it's a fantastic companion site to this one, and there are people there who can answer the most arcane abstract math questions.

Rob

Thanks. A question like this can usually be answered straight away here if I can understand the replies. I'll have to look through their threads, though. Maybe I can learn some more math that way. BAUT is the only forum I have been on since I joined. Sometimes I forget the internet consists of more than just BAUT.

[grav]

My next to last reply might be a little confusing, so I'll just show what I'm trying to do directly. I am using Eddington's isotropic metric using Cartesian coordinates in terms of r1 in place of the Schwarzschild metric which uses polar coordinates, because I can visualize and work with it better. Eddington's metric works out to dt = Int [(1 + r_s / (4 r1))^3 / (1 - r_s / (4 r1))] sqrt[dx^2 + dy^2 + zx^2] / c when finding for dt, where x^2 + y^2 + z^2 = r1^2. I am only finding for it in two dimensions, so z = 0 and I'm just using dt = Int [(1 + r_s / (4 r1))^3 / (1 - r_s / (4 r1))] sqrt[dx^2 + dy^2] / c = Int [(1 + r_s / (4 r1))^3 / (1 - r_s / (4 r1))] dx sqrt[1 + (dy / dx)^2] / c, and get dy / dx = -x / sqrt(r1^2 - x^2), which I have found before for the surface of a sphere, where r is constant, but since r1 is not constant here, I figure I might as well use y, for dy / dx = -x / y and r1 = sqrt(x^2 + y^2). That gives dt = Int dx [(1 + r_s / (4 sqrt(x^2 + y^2)))^3 / (1 - r_s / (4 sqrt(x^2 + y^2)))] sqrt[1 + x^2 / y^2] / c from 0 to y, although that still leaves terms of y, which might be okay, I'm not sure, and then just finish it in terms of dx, using the value of a constant y associated with each end value of x used for that after finding the integration for dx, which is again 0 with x1 and y with x2. Is that right so far?

[grav]

Okay, so is it not possible to perform an integration in two dimensions using this metric? It can be found in one dimension where y=0 and z=0, because that only leaves x=r1, so integrated along just r1 for that. I could find it in two dimensions if r1 were constant, but it's not, so I need dy or dr1, and I have tried several ways of extracting that and it's not happening. I could also perform an integration for this if I knew something about the equation for the curve, and the light would usually only take one specific curved path between two endpoints, but the metric itself doesn't appear to supply that information either in and of itself. Similarly, if I knew how the gravity affected the path of light, I could possibly use that, but although the metric contains r_s as a portion of its gravitational effect, basically the only thing the metric really supplies is the instantaneous speed at some point in the field, and that's all. For instance, if we start with just v_r1 = c (1 - r_s / (4 r1)) / (1 + r_s / (4 r1)), which is simply the instantaneous speed of the photon at r1 in the field, then we can rearrange to find

v_r1 = c (1 - r_s / (4 r1)) / (1 + r_s / (4 r1))^3, where

v_r1 = sqrt(vx_r1^2 + vy_r1^2 + vz_r1^2) = sqrt[(dx / dt)^2 + dy / dt)^2 + (dz / dt)^2], so

sqrt[(dx / dt)^2 + (dy / dt)^2 + (dz / dt)^2]= c (1 - r_s / (4 r1)) / (1 + r_s / (4 r1))^3

sqrt[dx^2 + dy^2 + dz^2] = c dt (1 - r_s / (4 r1)) / (1 + r_s / (4 r1))^3

dx^2 + dy^2 + dz^2 = (c dt)^2 (1 - r_s / (4 r1))^2 / (1 + r_s / (4 r1))^6

ds^2 = 0 = - (c dt)^2 (1 - r_s / (4 r1))^2 / (1 + r_s / (4 r1))^2 + (1 + r_s / (4 r1))^4 [dx^2 + dy^2 + dz^2]

which gives us Eddington's metric, although I am not sure what to multiply by in that last step to gain ds^2 directly, but it matches how the metric reads when multiplied by (1 + r_s / (4 r1))^4. So there is apparently nothing that is given in terms of dy or dr1, the curve of the path of the light, or the effect of gravity other than that of the instantaneous speed at position r1.

[mugaliens]

I can use the Cartesian coordinates, which I'm more familiar with.

No, you can't, as it doesn't work with Cartesian coordinates as they result in indeterminate solutions.

Take tdvance's advice - convert to polar coordinates.

[grav]

No, you can't, as it doesn't work with Cartesian coordinates as they result in indeterminate solutions.

Take tdvance's advice - convert to polar coordinates.

Oh, okay, thanks. Is that the same thing Incomplete was saying also? I can't tell. So it is determinate using dr and d0 but not dx and dy? That's strange. So that brings me back to the other thread about the polar coordinates then. I guess I'll just have to learn them if I want to get this.

[Incomplete]

Oh, okay, thanks. Is that the same thing Incomplete was saying also? I can't tell. So it is determinate using dr and d0 but not dx and dy? That's strange. So that brings me back to the other thread about the polar coordinates then. I guess I'll just have to learn them if I want to get this.

Hi grav, no, that wasn't what I was saying. Anything you can do in polar coordinates you can also do in Cartesian, but sometimes it's more complicated.

I'm not sure what you're trying to do here, though. In one of your posts you mention "finding for" dt. Are you trying to find a null geodesic? If so, the only simple ones are radial. In general a geodesic that starts moving in (say) the x direction will bend, so the path will be in (say) the x-y plane. But to find it you need a little more than the basic calculus you're trying to apply.

[grav]

I'm not sure what you're trying to do here, though. In one of your posts you mention "finding for" dt. Are you trying to find a null geodesic? If so, the only simple ones are radial. In general a geodesic that starts moving in (say) the x direction will bend, so the path will be in (say) the x-y plane. But to find it you need a little more than the basic calculus you're trying to apply.

Right, that's what I'm trying to do. I took Eddington's metric, ds^2 = - (c dt)^2 (1 - r_s / (4 r1))^2 / (1 + r_s / (4 r1))^2 + (1 + r_s / (4 r1))^4 [dx^2 + dy^2 + dz^2], dropping the dz since it is not necessary here, and rearranged it to find the time for light to travel along a curved path from (x1, y1) to (x2, y2) with t = Int dt = Int [(1 + r_s / (4 r1))^3 / (1 - r_s / (4 r1))] sqrt[dx^2 + dy^2] / c. Finding in one dimension radially is simple, like you said, because dr = dx in that case and the rest is all zero. For two dimensions, I can draw out the dx with sqrt[dx^2 + dy^2] = dx sqrt[1 + (dy / dx)^2], but from there I'm stuck.

[Incomplete]

Right, that's what I'm trying to do. I took Eddington's metric, ds^2 = - (c dt)^2 (1 - r_s / (4 r1))^2 / (1 + r_s / (4 r1))^2 + (1 + r_s / (4 r1))^4 [dx^2 + dy^2 + dz^2], dropping the dz since it is not necessary here, and rearranged it to find the time for light to travel along a curved path from (x1, y1) to (x2, y2) with dt = [(1 + r_s / (4 r1))^3 / (1 - r_s / (4 r1))] sqrt[dx^2 + dy^2] / c. Finding in one dimension radially is simple, like you said, because dr = dx in that case and the rest is all zero. For two dimensions, I can draw out the dx with sqrt[dx^2 + dy^2] = dx sqrt[1 + (dy / dx)^2], but from there I'm stuck.

You can't proceed from that, because that equation alone doesn't contain enough information. One way to understand what it says is to divide both sides by dt. Then it's an equation that relates the coordinate velocity in x to the coordinate velocity in y, but there are many solutions (as there should be, if you think about it: there are many possible null geodesics in the x-y plane of this metric).

I could teach you a nice trick for how to obtain three 2nd order differential equations for t, x, and y in terms of an affine parameter, and where to go from there. But I'm not sure your maths level is high enough. At this point I'd suggest reading a book on GR, or taking a course.

[grav]

You can't proceed from that, because that equation alone doesn't contain enough information.

Ah, good. So it's not just me. I had come to the conclusion that there was not enough information to derive the integration in two dimensions, as per post #9, since there is nothing that describes the equation for the specific curved path the light must follow or how the gravity affects the light other than the instantaneous speed at some position in the field. Is there something else that can be used in conjuction with the metric to gain a solution?

[Incomplete]

Ah, good. So it's not just me. I had come to the conclusion that there was not enough information to derive the integration in two dimensions, as per post #9, since there is nothing that describes the equation for the specific curved path the light must follow or how the gravity affects the light other than the instantaneous speed at some position in the field. Is there something else that can be used in conjuction with the metric to gain a solution?

The metric IS enough, but you have to know what to do with it. You can't just naively integrate that differential.

[grav]

The metric IS enough, but you have to know what to do with it. You can't just naively integrate that differential.

Now I'm more confused, if that's possible. Okay, so are you saying that the metric contains enough information, but the integration itself the way I found it doesn't directly? What else do I do?

[Incomplete]

Now I'm more confused, if that's possible. Okay, so are you saying that the metric contains enough information, but the integration itself the way I found it doesn't directly? What else do I do?

Solve the geodesic equation.

You can read about it in any book on GR or Riemannian geometry.

[grav]

Solve the geodesic equation.

You can read about it in any book on GR or Riemannian geometry.

Ouch. That's a whole 'nother deal. Well, I have a couple of books which deal with what you are describing that I wasn't able to read through before, "Black Holes - An Introduction" by Derek Raine and Edwin Thomas, and "The Road to Reality" by Roger Penrose I can use as a reference. Maybe I've gained enough familiarity with the metric so far that I can try to start reading through those again. Thanks very much for your help.