For massive particles, force translates into acceleration which again is a change of the velocity vector in direction and/or magnitude.

For a photon, the velocity magnitude cannot change, only the direction. It curves towards the source of gravitaion, which, I think, can be compared to a force acting at some angle to the velocity direction. If it is not a right angle, a massive particle would increase its velocity's magnitude. We can get the velocity vector by integation of the acceleration. The photon instead, I assume, changes its frequency. Can the combined velocity/frequency pair be derived also by simple integration? Of what?


1 Answer 1


After some investigation it turns out that my question is a bit of a convoluted way to ask for the momentum, which contains the information as follows. The momentum-energy relation

$$ E(t)^2/c^2 - \vec{p}(t)^2 = m_0^2$$

with $\vec{p}$ the momentum, $E$ the engery and $m_0$ the invariant rest mass, simplifies to

$$ \vec{p}(t)^2 = E(t)^2/c^2 $$

for photons, because they have zero rest mass. Another way to write this down is to say that the magnitude of the momentum is

$$ |\vec{p}(t)| = E(t)/c $$

On the other hand, the momentum points into the same direction as the velocity $\vec{v}(t)$ with its magnitude $c$, so we also have

$$ \vec{p}(t) = E(t)/c \cdot \vec{v}(t)/c ,$$

where the first factor describes the magnitude and the second factor, a unit vector, describes the direction.

The time derivative of the momentum is force $\vec{F}(t)$, so that we get

$$ \vec{F}(t) = 1/c^2 \left(\vec{a}(t) E(t) + \vec{v}(t) E'(t)\right) ,$$

where $\vec{a}(t)$ is the acceleration, the time derivative of $\vec{v}(t)$. Now, since $|\vec{v}(t)|=c$ is constant, the acceleration can not point into the direction of the velocity, not even to a small part, which means that velocity and acceleration must always be perpendicular to each other. By normalizing to their respective magnitudes, we get two orthogonal unit vectors nicely combining to the force as follows:

$$ \vec{F}(t) = 1/c \left( \frac{\vec{a}(t)}{|\vec{a}(t)|}\frac{|\vec{a}(t)| E(t)}{c} + \frac{\vec{v}(t)}{c}E'(t) \right) .$$

And here I have the two parts I was looking for. When the force points along the velocity, the factor of the perpendicular acceleration must be zero. Since $E(t)$ is not zero, it must be $|\vec{a}(t)|$ which is zero. The momentum then only changes per a change in energy, $E'(t)$, which is a change in frequency. If, however, the force is perpendicular to the velocity, $E'(t)$ must be zero and the force is $\vec{F}(t) = \frac{E(t)}{c^2}\vec{a}(t)$, resembling the classical $F=m\cdot a$, per $m=E/c^2$. For force a orientation between these two extremes, the photon partly changes it path and partly changes its energy.


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