2
$\begingroup$

Suppose I have some photon with a 4-momentum $p_\mu$, with $\mu = t,r,\theta, \phi$.

There is also a point with coordinates $(x,y,z)$. The 3-vector $x^i$ describes the vector between the origin of the coordinate system and the point.

The photon is at the origin of this coordinate system, I want to find the angle that the photon makes with the vector $x^i$.

Now, if we were just dealing with normal vectors in normal 3-space, I could just take a dot product to find the angle.

How would I do it in this case? Specifically, I am having trouble understanding the components of $p_{\mu}$; e.g. can I think of $p_{\phi}$ as a $\phi$ angle?

$\endgroup$
2
$\begingroup$

The angle between photon momentum and a vector is only defined in three dimensions. It is given by $\theta = \arccos (\vec p \cdot \vec x)/ (|\vec p||\vec x|))$. It has different values when observed in different reference frames. In other words, it is not a relativistic invariant.

The inproduct of four momentum with some coordinate four-vector , $\varphi = \omega t - \vec p \cdot \vec x$, however, is invariant. It is the phase of the plane wave describing the photon.

$\endgroup$
2
$\begingroup$

I want to find the angle that the photon makes with the vector $x^i$.

As pointed out in my2cts's answer, the answer to this question is frame-dependent. This is referred to as aberration of light rays. You need to start by fixing the frame of reference in which you want an answer. The observer you have in mind (who is at rest in this frame) has some velocity vector $v^i$. Presumably your 3-vector $x^i$ is already expressed, or can be reexpressed, as a 4-vector that it is orthogonal to $v^i$, i.e., the observer would consider it to be purely spatial: $v^ix_i=0$. Now project out the part $q^i$ of the photon's momentum vector that is orthogonal to $v_i$, so that $v^iq_i=0$. Then the inner product $q^ix_i$ is the quantity that your observer would describe as the dot product of these two 3-vectors, and you can extract the angle in the usual way.

can I think of $p_{\phi}$ as a $\phi$ angle?

$p_\phi$ can't be interpreted as an angle, because it doesn't have the right units to be an angle. All of the manipulations described above (projecting out components, taking inner products) have a coordinate-independent meaning, so you don't need to worry about the interpretation of the components in order t get the right answer. But note that you will need to raise and lower indices using the metric, even though the spacetime you're talking about is presumably flat.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.