Inconsistency in formulas for relativistic energy

In summary, there are three formulas discussed in the conversation: Formula 1, Formula 2, and Formula 3. Formula 1 and 2 are only correct for particles with non-negligible/non-zero mass, while Formula 3 is the correct one for all cases. Formula 3 states that a particle of negligible mass can still have energy, and is derived from the Minkowski norm of the four-momentum. For a system of particles, the momentum term in Formula 3 is the sum of the individual momentums. The system can be treated as one object with no momentum and a determined rest mass.
  • #1
Nikitin
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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?
 
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  • #2
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.
 
  • #3
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.
 
  • #4
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?
 
  • #5
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.
 
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  • #6
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.
 
  • #7
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

$$\beta = \frac{v}{c} = \frac{pc}{E}$$
 
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  • #8
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].
 
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  • #9
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.
 
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1. What is the concept of inconsistency in formulas for relativistic energy?

Inconsistency in formulas for relativistic energy refers to the fact that there are multiple equations used to calculate the energy of a relativistic object, and these equations can sometimes give different results. This can be confusing and has led to some debate and controversy in the scientific community.

2. Why are there different formulas for calculating relativistic energy?

The different formulas for calculating relativistic energy arise from different ways of approaching the problem and different assumptions made about the system being studied. For example, some equations may only be applicable to objects moving in a straight line, while others may take into account acceleration and curved paths.

3. Which formula should be used to calculate relativistic energy?

There is no one "correct" formula for calculating relativistic energy. The appropriate formula to use depends on the specific situation and what information is known about the object's motion. Scientists often use the formula that best fits the parameters of their experiment or observation.

4. How do scientists deal with inconsistency in formulas for relativistic energy?

Scientists are aware of the potential for inconsistency in formulas for relativistic energy and take this into account when conducting experiments or making calculations. They carefully select the most appropriate formula for their specific situation and may also use multiple equations to compare results and verify their findings.

5. Is inconsistency in formulas for relativistic energy a problem for the theory of relativity?

No, inconsistency in formulas for relativistic energy does not invalidate the theory of relativity. The theory has been extensively tested and confirmed by numerous experiments and observations. Any discrepancies in formulas are simply a result of different approaches and assumptions, and do not undermine the overall validity of the theory.

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