Relativistic Momentum of a photon

In summary, an elementary particle with mass M completely absorbs a photon, resulting in a new mass of 1.01M. The energy of the incoming photon can be calculated using the equations p(photon) = E/c and p(i) = p(f). The energy of the photon is greater than 0.01Mc^2 because the mass change of the particle is due to its relativistic mass, not its rest mass.
  • #1
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1.
An elementary particle of mass M completely absorbs a photon, after which its mass is 1.01M. (a) What was the energy of the incoming photon? (b) Why is that energy greater than 0.01Mc2?




Homework Equations


p (photon) = E/c
p (particle) = γmv
p(i) = p(f)
Ek = γmc^2 -mc^2

The Attempt at a Solution


I really have no idea where to start without having knowledge of the velocity of the particle after the collision
 
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  • #2
Just use v as velocity. Momentum conservation will give you that velocity during the calculations.
 
  • #3
jacksonb62 said:

Homework Statement



An elementary particle of mass M completely absorbs a photon, after which its mass is 1.01M. (a) What was the energy of the incoming photon? (b) Why is that energy greater than 0.01Mc2?


Homework Equations


p (photon) = E/c
p (particle) = γmv
p(i) = p(f)
Ek = γmc^2 -mc^2


The Attempt at a Solution


I really have no idea where to start without having knowledge of the velocity of the particle after the collision
Was that the complete problem statement? I don't think the situation described is physically possible.
 
  • #4
It is possible, it is the reverse process of a gamma decay.
 
  • #5
But that's with a nucleus, which has a substructure. I guess it seems a little odd to me to say that an elementary particle changes its mass by absorbing a photon unless it's referring to its relativistic mass.
 
  • #6
It becomes a different elementary particle. There are no suitable elementary particles for that process (and it would probably need weak higher-order processes), but I don't think we have to worry about that.
"Mass" has to be rest mass, otherwise the second question does not make sense.
Elementary just guarantees we don't have to worry about rotations and so on.
 

1. What is the equation for calculating the relativistic momentum of a photon?

The equation for calculating the relativistic momentum of a photon is p = h/λ, where p is the momentum, h is Planck's constant, and λ is the wavelength of the photon.

2. How does the relativistic momentum of a photon differ from classical momentum?

The relativistic momentum of a photon takes into account the photon's energy and mass, whereas classical momentum only considers mass. This is due to the fact that photons have no rest mass, but they do have energy and momentum.

3. Can the relativistic momentum of a photon be negative?

No, the relativistic momentum of a photon cannot be negative. Since photons are massless particles, their momentum is always positive and is directly proportional to their energy.

4. How does the relativistic momentum of a photon change with increasing speed?

The relativistic momentum of a photon does not change with increasing speed, as photons always travel at the speed of light, which is a constant. However, their energy and frequency do change with increasing speed, according to the equation E = hf, where E is energy, h is Planck's constant, and f is frequency.

5. What is the significance of the relativistic momentum of a photon?

The relativistic momentum of a photon is significant because it helps us understand the behavior of light and other electromagnetic radiation. It also plays a crucial role in theories such as special relativity and quantum mechanics.

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