# Transfer of momentum in relativistic collisions

1. Jul 6, 2012

### Ookke

A thought experiment: X is a particle moving horizontally 0.8c to right, L is a light beam moving vertically to up, and O is an observer at rest. With precise timing in test arrangement, light beam hits X directly from below and gets absorbed by X (all beam's energy transfers to kinetic energy, for simplicity).

As the light beam has only upwards momentum in the reference frame of O, O would expect that the collision changes only the vertical velocity of X and the horizontal velocity remains unchanged.

However, in the reference frame of X, light beam doesn't approach directly from below, but diagonally from bottom-right. In collision, beam's momentum would split to horizontal and vertical component, therefore changing also the relative horizontal velocity between O and X, which doesn't change in the reference frame of O.

As the two reference frames seem to disagree (at least in my understanding), is there some absolute way to solve this? Thanks in advance.

2. Jul 6, 2012

### tiny-tim

Hi Ookke!
nooo … the mass of X increases when it absorbs the light,

and since the horizontal momentum stays the same, the horizontal velocity must decrease

3. Jul 7, 2012

### Ookke

That's right. I didn't see the mass increase at all. Thanks for the solution!

4. Jul 7, 2012

### collinsmark

tiny-tim is correct that the mass increases, but that's not the whole story.

In relativity, momentum is defined by

$$\vec p = \gamma m \vec v$$

where $\gamma$ is

$$\gamma \equiv \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$

When a particle is at rest, it's $\gamma$ is equal to 1. As the particle approaches the speed of light, its $\gamma$ becomes greater than 1.

In your thought experiment, not only does the mass of the particle increase, but $\gamma$ increases too. The final speed of the particle is greater than its initial speed (even though its velocity in the x-direction decreases). So there are two things going on.

Let's talk about the mass for a moment. The mass does increase, and by this I mean the particle's rest mass (invariant mass). That's because in your thought experiment, the collision is inelastic. The particle has gained internal energy (such as exciting an electron, maybe placing the nucleolus in an excited state, or whatever). This internal energy increases the particle's rest mass by $E = mc^2$.

But the particle also gains kinetic energy because its speed increases. So not all of the photon's energy goes into the particle's rest mass; some of it ends up as the particle's kinetic energy.

The total energy of a particle is given by $\gamma m c^2$. That includes both its rest mass energy and its kinetic energy.

The energy of a photon is given by $\frac{hc}{\lambda}$ where $h$ is Planck's constant and $\lambda$ is the photon's wavelength.

Taking into account the particle's increase in rest mass, both energy and momentum are conserved in this inelastic collision.

Energy final = Energy initial

Energy final = Initial energy of particle + Photon energy

$$\gamma_f m_f c^2 = \gamma_0 m_0 c^2 + \frac{hc}{\lambda}$$

Dividing by $c^2$,

$$\gamma_f m_f = \gamma_0 m_0 + \frac{h}{\lambda c}$$

Keep that above equation handy. We're going to use it a couple times below.

Let's move onto momentum. Not only is momentum conserved, it is conserved individually in each x- and y-component.

Let's start with the x-component, since that is what your original thought experiment was about. Consider the particle initially moving on the x-axis, and the photon on the y-axis.

$$\gamma_f m_f v_x = \gamma_0 m_0 v_{0x}$$

Now we substitute $\gamma_f m_f$ into that equation and solve for the final $v_x$

$$v_x \left( \gamma_0 m_0 + \frac{h}{\lambda c} \right)= \gamma_0 m_0 v_{0x}$$

$$v_x = v_{0x} \frac{\gamma_0 m_0 }{\gamma_0 m_0 + \frac{h}{\lambda c}}$$

Multiplying both the numerator and denominator by $c^2$ we get

$$v_x = v_{0x} \frac{\gamma_0 m_0 c^2}{\gamma_0 m_0 c^2 + \frac{hc}{\lambda}}$$

And interestingly, that is

$$v_x = v_{0x} \left( \frac{\mathrm{Initial, \ total \ energy \ of \ particle}}{\mathrm{Initial, \ total \ energy \ of \ particle \ + \ Photon \ energy}} \right)$$

So as you can see, the x-component of the particle's velocity is reduced after the collision.

------------------

Although not part of your original thought experiment, we can do the same thing for the y-component of momentum. The momentum of a photon is $\frac{h}{\lambda}$.

$$\gamma_f m_f v_y = \frac{h}{\lambda}$$

Making our substitution,

$$v_y \left( \gamma_0 m_0 + \frac{h}{\lambda c} \right)= \frac{h}{\lambda}$$

$$v_y = \frac{\frac{h}{\lambda}}{\gamma_0 m_0 + \frac{h}{\lambda c}}$$

Multiply numerator and denominator by $c^2$

$$v_y = c \frac{\frac{hc}{\lambda}}{\gamma_0 m_0 c^2 + \frac{hc}{\lambda}}$$

And interestingly, that is

$$v_y = c \left( \frac{\mathrm{\ Photon \ energy}}{\mathrm{Initial, \ total \ energy \ of \ particle \ + \ Photon \ energy}} \right)$$

Last edited: Jul 7, 2012
5. Jul 7, 2012

### collinsmark

Oh, and I didn't mention, but you seem to already know, in the reference frame of the particle, the photon does not come directly from the y-direction due to relativistic beaming like/related effects (http://en.wikipedia.org/wiki/Relativistic_beaming). (And don't forget about Doppler.)

So in the particle's frame of reference its mass increases (as tiny-tim points out), but everything still ends up in relativistic agreement with the other frame, since $m_f$ is the same in both cases. (And even in the particle's frame of reference, the speed of everything else increases thus $\gamma_f$ increases accordingly in that frame of reference too.)

Last edited: Jul 7, 2012
6. Jul 8, 2012

### Ookke

collinsmark, thanks for your work! It's fascinating to see how different frames can see things very differently, and still there is certain consistency behind all this, so that the frames will agree on the most important things like energy and momentum conservation. For me, syncing between different frames is the most interesting part of Relativity.

7. Jul 8, 2012

### jartsa

I changed some words in the above thought experiment, and derived this thought experiment:

A thought experiment number 2: X and Y are particles moving horizontally 0.8c, one to right, other to left. L and M are light beams moving vertically to up, and O is an observer at rest. With precise timing in test arrangement, light beams hit X and Y directly from below and get reflected by X and Y (all beam's momentum transfers to momentum of X and Y, for simplicity).

As the light beams have only upwards momentum in the reference frame of O, O would expect that the collision is an acceleration of the X-Y particle system in the vertical direction, and the horizontal velocities of X and Y are changed by the time dilation effect.

However, in the reference frame of X and Y, light beams don't approach directly from below, but diagonally from bottom-right. In collision, beam's momentum would split to horizontal and vertical component, therefore changing also the relative horizontal velocity between O and X and Y, which also changes in the reference frame of O.

As the two reference frames seem to agree, everything seems to be fine.