Inconsistency in formulas for relativistic energy

Click For Summary

Discussion Overview

The discussion revolves around the apparent inconsistencies in formulas related to relativistic energy, specifically the relationships between total energy, momentum, and mass for particles, including massless particles. Participants explore the implications of these formulas in various contexts, including particle decay and the behavior of systems of particles.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants express confusion regarding the implications of different energy and momentum formulas, particularly in the context of massless particles and their energy contributions.
  • It is noted that formulas 1 and 2 do not apply to massless particles, leading to discussions about the definitions of energy and momentum in relativity.
  • One participant claims that formula 3 is the correct one for all cases, while formulas 1 and 2 are only valid when mass is non-zero.
  • Another participant suggests that there are multiple derivations of the formulas, including classical arguments and those based on electromagnetic theory, which lead to the same conclusions despite different starting points.
  • Concerns are raised about the implications of using the momentum term in formula 3 for systems of particles, particularly regarding the total energy and kinetic energy of colliding particles.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the interpretations of the formulas and their applications, particularly regarding massless particles and the treatment of systems of particles. Multiple competing views remain on how to reconcile the formulas with physical scenarios.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about mass, velocity, and the definitions of energy and momentum in different contexts. The treatment of massless particles and the implications for systems of particles are particularly noted as areas of complexity.

Nikitin
Messages
734
Reaction score
27
I'm very noob at this and am a bit confused:

Formula 1: [itex]E_T = \gamma \cdot m c^2[/itex]
Formula 2: [itex]p = \gamma m v[/itex]
Formula 3: [itex]E_T^2 = (pc)^2 + (mc^2)^2[/itex]

Formula 3 says a particle of negligible mass can have energy, but isn't this in contradiction to
formula 1? Unless maybe the velocity of the particle is c, such that ##\gamma## becomes infinite too?

So if I get an assignment where I'm supposed to neglect the mass of a moving particle, I must either neglect its total energy or set its velocity to c in my calculations?

For instance, let's say I need to calculate on myon decay: ##\pi_+ \rightarrow \mu_+ + \nu##. Do I have to set the velocity of the myon-neutrino to c? But why can't I just neglect the energy of the neutrino?
 
Physics news on Phys.org
Formulas 1 and 2 don't hold for massless particles.

Also, you have to properly define all the properties.
In Newtonian physics rest mass and velocity are basic properties and energy and momentum are derived from them.
In relativity it's the opposite. Energy and momentum are basic properties and rest mass and velocity are derived from them. We avoid division by zero this way.
 
As haael said, formula 3 is the correct one. Formulas 1 and 2 can be derived from formula 3 in the special case that mass is non-zero.
 
Uhmmmmm, in my book formula 3 was derived FROM formula 1 and 2, so I assumed formula 3 was built upon formula 1 and 2.

But OK.. so formula 3 is the one which is correct for all cases, while 1 and 2 are only correct when the particle has non-neglible/non-zero mass? Is there an easy-to-understand derivation for formula 3 lying around somewhere on the web?
 
In units where c=1, the four-momentum is given by ##\mathbf{P}=(E,\mathbf{p})##. The Minkowski norm of this quantity is then ##|\mathbf{P}|^2=E^2-\mathbf{p}^2\equiv m^2##. Rearrange and you get your equation 3.
 
Last edited:
Nikitin said:
Uhmmmmm, in my book formula 3 was derived FROM formula 1 and 2, so I assumed formula 3 was built upon formula 1 and 2.

But OK.. so formula 3 is the one which is correct for all cases, while 1 and 2 are only correct when the particle has non-neglible/non-zero mass? Is there an easy-to-understand derivation for formula 3 lying around somewhere on the web?

1 and 2 hold for anything except an exactly massless particle. For a massless particle, gamma is undefined, so any formula using it is useless for a massless particle.
 
Nikitin said:
For instance, let's say I need to calculate on myon decay: ##\pi_+ \rightarrow \mu_+ + \nu##.

You can solve for all the momenta and energies in any two-body decay like this one, using only energy and momentum conservation, and your "Formula 3" above (applied separately to each particle). This is most easily done in the rest frame of the initial particle (the pion in your example).

After you have the energies and momenta, if you want the speeds of the decay products, you can get them most simply using

$$\beta = \frac{v}{c} = \frac{pc}{E}$$
 
Last edited:
Nikitin said:
Uhmmmmm, in my book formula 3 was derived FROM formula 1 and 2, so I assumed formula 3 was built upon formula 1 and 2.

But OK.. so formula 3 is the one which is correct for all cases, while 1 and 2 are only correct when the particle has non-neglible/non-zero mass? Is there an easy-to-understand derivation for formula 3 lying around somewhere on the web?

There are several different arguments leading to formulas 1-3 that all have the same conclusion, even though their starting points are very different. There is a mathematical argument that it's the only possibility, which is intuitively unsatisfying, perhaps. In addition, there is a purely classical argument (from Maxwell's equations plus the Lorentz force law) that electromagnetic waves carry energy and momentum that are related by [itex]E = pc[/itex] (which is the special case of 3 that applies when the mass is zero). Then there are several different heuristic arguments that massive particles have to satisfy equations 1&2, which imply equation 3. One derivation (I think this is original with Einstein) considers a massive box with a pulse of light bouncing back and forth between the two walls. Another derivation starts with the assumptions that (1) Energy is proportional to mass, and is conserved in collisions, (2) momentum is proportional to mass, and is in the same direction as the velocity, and is conserved in collisions, (3) in the nonrelativistic limit, kinetic energy and momentum approach their Newtonian values, [itex]\frac{1}{2} m v^2[/itex] and [itex]m v[/itex].
 
  • Like
Likes   Reactions: 1 person
Thanks for all the replies, but I have one more question:

For a system of particles, is the momentum term in formula 3 the sum of the momentums, right? Well, doesn't that bring forward contradictions?

For example: Let's say you have two electrons heading towards each-other at equal speeds (seen from inertial frame). Then the momentum for the system as a whole is zero, and thus total energy is just equal to 2 times the rest energy of an electron? But this is ridiculous, as this would mean the system contains zero kinetic energy...
 
  • #10
Nikitin said:
Thanks for all the replies, but I have one more question:

For a system of particles, is the momentum term in formula 3 the sum of the momentums, right? Well, doesn't that bring forward contradictions?

For example: Let's say you have two electrons heading towards each-other at equal speeds (seen from inertial frame). Then the momentum for the system as a whole is zero, and thus total energy is just equal to 2 times the rest energy of an electron? But this is ridiculous, as this would mean the system contains zero kinetic energy...

For a system of particles you add the 4-momenta, the quantity (E,p) in units of c=1. Thus, for two oppositely moving electrons you get (2E,0) as the 4-momentum of the system. Then, m^2= E^2 - p^2 tells that the invariant mass of the system is equal to the total energy of the two electrons, with no momentum for the system. Effectively, the systems can be treated as one object with no momentum and rest mass determined by total kinetic energy and rest mass of the electrons.

The key point is that the 'rest mass' of a system is not simply the sum of the rest mass of its components.
 
  • Like
Likes   Reactions: 1 person

Similar threads

  • · Replies 15 ·
Replies
15
Views
1K
  • · Replies 55 ·
2
Replies
55
Views
6K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 82 ·
3
Replies
82
Views
7K