# Conservation of energy and momentum

1. Oct 8, 2014

### azerbajdzan

Hello,

Suppose that an object of rest mass m0 travels to the right with speed v1.
A photon of frequency f also travels to the right and hits the object. The photon is fully absorbed by the object and then the object travels to the right with the (unknown) speed v2 after collision.
Now, the total energy should be same before collision as after collision.
Also the total momentum should be same before collision as after collision.
I expressed these two conservation laws by two equations:
$$\frac{m_0 c^2}{\sqrt{1-\left(\frac{v_1}{c}\right){}^2}}+h f=\frac{m_0 c^2}{\sqrt{1-\left(\frac{v_2}{c}\right){}^2}}$$
$$\frac{m_0 v_1}{\sqrt{1-\left(\frac{v_1}{c}\right){}^2}}+\frac{h f}{c}=\frac{m_0 v_2}{\sqrt{1-\left(\frac{v_2}{c}\right){}^2}}$$
But since there is only one unknown variable - the speed v2 of the object after collision - the two equations cannot hold both.

So to compute the speed v2 either I choose first equation or second equation, but the two computed values of v2 would be different for each equation.

The question is, what am I doing wrong?

2. Oct 8, 2014

### A.T.

Total energy, but not kinetic energy, since it's an inelastic collision.

3. Oct 8, 2014

### PAllen

Well, the OP's first equation is for total energy. The key is that rest mass after absorbing the photon is larger. So you have m1 on the right side of both equations. So now you have two equations in two unknowns (m1, v2). It will all work fine, then.

4. Oct 8, 2014

### Orodruin

Staff Emeritus
To emphasize the point that was already made by PAllen: If you absorb the photon, then the mass has to increase. If it is a collision with an object of a definite rest mass, such as an electron or other elementary particle, this process cannot occur for exactly the reasons you stated. I don't know what your level is, but showing this is typically covered in introductory university courses on relativity.

5. Oct 8, 2014

### azerbajdzan

I know that absorbing a photon increases the mass, but I was not sure which mass - the rest mass or relativistic mass (mass at non-zero speed)...
But yes, it now makes more sense when I replace m0 with m1 on the right side of both equations.

Now another example... The photon reflects exactly the opposite direction, i.e. after collision it travels to the left with frequency f2.
Now the two equations become:
$$\frac{m_0 c^2}{\sqrt{1-\left(\frac{v_1}{c}\right){}^2}}+h f_1=\frac{m_1 c^2}{\sqrt{1-\left(\frac{v_2}{c}\right){}^2}}+h f_2$$
$$\frac{m_0 v_1}{\sqrt{1-\left(\frac{v_1}{c}\right){}^2}}+\frac{h f_1}{c}=\frac{m_1 v_2}{\sqrt{1-\left(\frac{v_2}{c}\right){}^2}}-\frac{h f_2}{c}$$
All quantities on the left are known, i.e. all quantities before collision.
After collision we have 3 new unknown quantities: m1, v2, f2, but only two equation. So it looks like the system is not determined because of one free parameter.
What am I missing now? Is there some third equation that we need to use to determine all the variables?

6. Oct 8, 2014

### azerbajdzan

So free electron (not inside atom) cannot absorb photon without emitting some other photon? Is it this you are trying to say?

7. Oct 8, 2014

### Orodruin

Staff Emeritus
Exactly. The free electron cannot simply absorb (or emit) a photon. In order for both energy and momentum to be conserved, there needs to be another photon involved. Note that this additional photon a priori can be a virtual photon taken from the background field near a nucleus or similar - this process is known as bremsstrahlung (from German, meaning essentially "braking radiation") and transfers part of the incoming momentum to the nucleus.

8. Oct 8, 2014

### PAllen

What you are really talking about here is (effectively) absorbing a photon of one energy and emitting one in a particular direction (opposite incoming) of another energy. The limitations on this process are all quantum mechanical - what energy transitions are available in the body. It is not a kinematic problem. Given some value for the outgoing photon energy, you can then solve for the kinematic variables of the massive body.

At the kinematic level, there is a free parameter. Even classically, this free parameter relates to the detailed physics of the bodies - if the collision is not elastic, what vibrational modes are available, what are the deformation possibilities, etc.

9. Oct 8, 2014

### azerbajdzan

Is it possible in real life, that the energies of incident and reflected photons are the same (f1=f2) if reflected photon went exactly the opposite direction to the direction of incident photon? ...it would mean that the rest mass of the body after collision have been decreased compared to rest mass before collision. Can it be even f2>f1?

10. Oct 8, 2014

### PAllen

If the ingoing an outgoing photon have the same energy in the original reference frame, then they don't have the same energy in the rest frame of the body after collision. In this frame, there is thus no mystery to the body's decrease in rest mass.

It is certainly possible for f2>f1 in some frame, e.g. the initial one. This could happen if, e.g. the absorbed photon sent a bound electron from a state above ground to higher, and the emission was due to higher state all the way to ground.