# Transformation of solid angle

1. Mar 17, 2015

### ShayanJ

I'm trying to figure out how the element of solid angle transforms under a transformation between two inertial frames moving with velocity v w.r.t. each other under an arbitrary direction. But I should say I disappointed myself! Anyway, some books which contain a brief discussion on this(which doesn't include an actual derivation!), simply state that $d\varphi'=d\varphi$, but this is not at all trivial to me! What should I do?
In my desperate actions toward the above goal, I managed to get myself confused about something else too! How curvilinear coordinates transform under Lorentz transformation?
I'll appreciate any suggestion about the above problems.
Thanks

2. Mar 17, 2015

### BiGyElLoWhAt

Does psi represent an angle element? Or does it represent the measure of the whole angle?

3. Mar 17, 2015

### ShayanJ

If you mean $\varphi$, then its the last coordinate in the spherical coordinate system($(r,\theta,\varphi)$).
Then the solid angle element becomes $d \Omega=\sin \theta d\varphi d\theta$.

4. Mar 17, 2015

### Orodruin

Staff Emeritus
The angle $\varphi$ represents a rotation in the plane perpendicular to the motion, thus it is given by the $y$ and $z$ coordinates (in standard configuration) which do not change. Therefore $\varphi = \varphi'$ and also $d\varphi = d\varphi'$.

5. Mar 17, 2015

### ShayanJ

But this is an assumption about the relative motion. It doesn't work if we consider an arbitrary direction!

6. Mar 17, 2015

### Orodruin

Staff Emeritus
You can always chose your coordinates in such a way that this is the case (as long as you satisfy yourself with rotation free Lorentz transformations). Just define $\theta$ as the angle between the direction of the transform and the differential solid angle.

7. Mar 17, 2015

### ShayanJ

Thanks.
So what about the other question? How can we transform curvilinear coordinates between inertial frames moving uniformly w.r.t. each other?

8. Mar 17, 2015

### Orodruin

Staff Emeritus
Usually when you deal with this kind of problems, only one event is of real importance (i.e., the decay or interaction). The $\varphi$ angle is unchanged and you can typically compute the $\theta$ transformation using the abberation formula.

9. Mar 17, 2015

### Roy_1981

"The angle φ represents a rotation in the plane perpendicular to the motion, thus it is given by the y and z coordinates (in standard configuration) which do not change. Therefore φ=φ′ and also dφ=dφ′."

This is true for transformation properties of the longitude, φ, which NOT a solid angle. The solid angle, usually denoted by Ω actually does involve directions parallel to the direction of relative velocity. In arbitrary dimensions, say n-space (plus time), the solid angle is defined in terms of cartesian components by the differential form,

dΩ= ∑i=1,n [(-1)(i-1) xi dx1 ∧...dxi-1 ∧ dxi+1....dxn] /rn.

You can check for test cases for n=2 and 3 that this general (n-1)-form expression indeed reduces to familiar planar polar angle element, dθ and solid angle element, dΩ= sinθ dθ ∧ dφ respectively. When cast In this cartersian form, the solid angle element does not appear to me to be Lorentz invariant due to "dt ∧ dz or dt∧dy" terms. I checked explicitly by plugging Lorentz transformation for n=2 case.

10. Mar 17, 2015

### PAllen

If you really want to play with explicit transform between spherical coords in one inertial frame to same in another, you can get them (in ugly form) by a straightforward process. Starting from the Lorentz transform between Cartesian coordinates. Then, substitute standard expressions of r',θ',φ' for for x',y',z' on the left; and the same for r,θ,φ for x,y,z on the right. Then simplify per φ'=φ. With just trivial rearrangement you end up with (for example, depends on conventions chosen):

φ' = φ
tanθ' = rsinθ/ (γ(rcosθ-vt))
r'sinθ' = rsinθ
t' = γ (t- vrcosθ)

I am sure someone, somewhere, has derived a much more clever way of writing these. This is just a quick scribble to show the ideas. Never, in my life, have I had occasion to use such an explicit set of relations, so I had no idea what they might look like before writing these down.

11. Mar 18, 2015

### ShayanJ

Its clear now, thanks guys!

12. Mar 18, 2015

### Orodruin

Staff Emeritus
I never claimed anything else ...

13. Mar 18, 2015

### Roy_1981

"I never claimed anything else ... "

In that case, my apologies...