Newtonian vs. Relativistic momentum

[Davephaelon]

Perhaps naively I assumed that for a relativistic particle the product of mass times velocity would be the same in both Newtonian and Einsteinian mechanics. The simplistic thinking was the mass increase in relativistic dynamics would balance out the non-real superluminal velocity of a particle in Newtonian dynamics. I've assumed that my entire life.

So I went to this online calculator below which I presumed gives the final velocity of a particle (electron) accelerated through a particular voltage the user can select. Much to my surprise the expected equality of momentum for Newtonian and Relativistic mechanics wasn't true. I selected "2000", or 2 million volts in the box. In the relativistic case the electron's momentum then comes out to 131.37 x 10^-23, while for the Newtonian case it comes out to 76.4 x 10^-23, approaching half as much.

I'm frankly confused. My intuition was clearly wrong. Something is evidently not being taken into account.

https://www.ou.edu/research/electron/bmz5364/calc-kv.html

 

[Nugatory]

Stuff like this is the reason why relativistic mass is seldom used these days. The momentum is $\gamma{m_0}v$ so the Newtonian relationship holds if you consider $\gamma{m_0}$ to be the "mass", but the kinetic energy is not $\gamma{m_0}v^2/2$... and you were providing an energy, not a momentum, when you supplied a voltage difference.

 

[Dale]

The simplistic thinking was the mass increase in relativistic dynamics

That is probably the source of your confusion. Relativistic mass is a concept that has been largely discarded for decades. It pretty much only shows up now in pop-sci media.

 

[Ibix]

Perhaps naively I assumed that for a relativistic particle the product of mass times velocity would be the same in both Newtonian and Einsteinian mechanics.

It is. It's just that mass times velocity isn't momentum in relativistic physics.

 

[sweet springs]

The momentum is γm0vγm0v\gamma{m_0}v so the Newtonian relationship holds if you consider γm0γm0\gamma{m_0} to be the "mass", but the kinetic energy is not γm0v2/2γm0v2/2\gamma{m_0}v^2/2... and you were providing an energy, not a momentum, when you supplied a voltage difference.

Yea. I would add $\gamma m v$ should be interpreted NOT as relativistic mass $\gamma m$ times velocity v, but as mass times "relativistic velocity" $\gamma v$.

 

[Davephaelon]

Thank you everyone. I just found an online calculator that incorporates gamma for relativistic particles. It's at one of my favorite sites. Completely forgot to look there: http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/relmom.html

 

[Vanadium 50]

I would add γmv\gamma m v should be interpreted NOT as relativistic mass γm\gamma m times velocity v, but as mass times "relativistic velocity" γv\gamma v.

Why is that helpful?

 

[sweet springs]

$\gamma \vec{v}$ is space components of 4-velocity in normal way of SR. Product $\gamma m$ is not a good combination to be interpreted now.

 

[Nugatory]

$\gamma \vec{v}$ is space components of 4-velocity in normal way of SR. Product $\gamma m$ is not a good combination to be interpreted now.

I am not seeing how that makes any sense. If we use coordinates in which an object is at rest, its three-velocity is zero; if we use coordinates in which it is moving at speed $v$ then its three-velocity is $v$. What value of $v$ do you use to calculate a value of $\gamma$?

 

[sweet springs]

If we use coordinates in which an object is at rest, its three-velocity is zero

$v=0$ then $\gamma=1$

if we use coordinates in which it is moving at speed vvv then its three-velocity is vvv.

then $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$.
For both the cases contravariant components of 4-velocity are $$(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}},\frac{\vec{v}}{\sqrt{1-\frac{v^2}{c^2}}})$$
Multiplying m, contravariant components of 4-momentum are
$$(m\frac{1}{\sqrt{1-\frac{v^2}{c^2}}},m\frac{\vec{v}}{\sqrt{1-\frac{v^2}{c^2}}})$$

$m\gamma v$ should be interpreted as $m(\gamma v)$ rather than $(m\gamma)v$.

 

[jartsa]

Perhaps naively I assumed that for a relativistic particle the product of mass times velocity would be the same in both Newtonian and Einsteinian mechanics. The simplistic thinking was the mass increase in relativistic dynamics would balance out the non-real superluminal velocity of a particle in Newtonian dynamics. I've assumed that my entire life.

Well yes, if a Newtonian particle and an Einsteinian particle are pushed by the same force for the same time, then they both gain the same momentum.

So for an electron being accelerated by an electric field we can write:

Newtonian change of mass of the particle * Newtonian change of velocity of the particle = Einsteinian change of relativistic mass of the particle * Einsteinian change of velocity of the particle

Correction: That above formula is wrong. The correct formula is ... something else. It's left as an exercise to the reader to find it... Because I might still get it wrong.Now if we compare two particles, one of which is Newtonian while the other is Einsteinian, being accelerated between two electrodes, both particles starting from zero speed at the same time, then as long as both particles are between the plates we can write:

momentum of the Newtonian particle = momentum of the Einsteinian particle

 

[PeterDonis]

if a Newtonian particle and an Einsteinian particle are pushed by the same force for the same time, then they both gain the same momentum

Only if you interpret "time" to mean "coordinate time in the inertial frame in which the particle started out at rest" for the Einsteinian case.

Newtonian change of mass of the particle * Newtonian change of velocity of the particle = Einsteinian change of relativistic mass of the particle * Einsteinian change of relativistic velocity of the particle

This is wrong, because there is no "Newtonian change of mass of the particle" (the mass of the particle is constant in Newtonian physics), and the change in momentum is not the change in mass times the change in velocity in either case.

 

[Ibix]

I think we need to be a bit careful with our language here. There's no such thing as a "Newtonian particle". It's perfectly legitimate to compare the Newtonian prediction for what will happen to the relativistic one, but the reality always matches the relativistic prediction.

 

[jartsa]

This is wrong, because there is no "Newtonian change of mass of the particle" (the mass of the particle is constant in Newtonian physics), and the change in momentum is not the change in mass times the change in velocity in either case.

Oh. "Newtonian mass change factor" would make a bit more sense there.

But I need to rethink the whole equation. It wasn't quite as simple as I thought.

Hmm. Maybe I could say that Newtonian momentum change factor is Newtonian mass change factor multiplied by Newtonian velocity change factor, and Einsteinian momentum change factor is Einsteinian relativistic mass change factor multiplied by Einsteinian velocity change factor.

 

[PAllen]

I am not seeing how that makes any sense. If we use coordinates in which an object is at rest, its three-velocity is zero; if we use coordinates in which it is moving at speed $v$ then its three-velocity is $v$. What value of $v$ do you use to calculate a value of $\gamma$?

It is just the spatial components of 4 velocity, which is certainly a useful 3 vector quantity. Equivalantly, it is proper time derivative of coordinates, using proper time of the particle. Some authors call it celerity. It leads directly to the notion that 3 force (when no mass/energy flow is involved) is mass times time derivative of celerity. Finally, it captures the notion that since momentum has no upper limit, and mass is invariant, momentum needs to be defined in terms of an unbounded function of velocity (since velocity is limited to c).

 

[PeterDonis]

"Newtonian mass change factor" would make a bit more sense there.

Huh? There's no such thing: in Newtonian physics the mass of the particle is constant.

 

[jartsa]

Huh? There's no such thing: in Newtonian physics the mass of the particle is constant.

Well then the "mass change factor" is one in Newtonian physics.

In Einsteinian physics "mass change factor" is the same number as "gamma", and momenum is $m_{relativistic} v$, if we use the old fashioned relativistic mass to calculate the momentum.

 

[PeterDonis]

the "mass change factor" is one in Newtonian physics.

In Einsteinian physics "mass change factor" is the same number as "gamma",

Hm, ok. But I still don't see how your proposed equations make things any clearer.

 

[jartsa]

Hm, ok. But I still don't see how your proposed equations make things any clearer.

Well I have to agree, it doesn't make anything any clearer.

 

[vanhees71]

That is probably the source of your confusion. Relativistic mass is a concept that has been largely discarded for decades. It pretty much only shows up now in pop-sci media.

It has been discarded by Einstein very quickly after his second additional paper of 1905. Unfortunately the damage by introducing the concept of relativistic mass(es) has been done in 1905 and it stuck with a large part of the physics community until today. The problem is that of course for Einstein, Planck, and von Laue this sloppy early notions were no problem, because they were ingenious enough to have the right physics intuition to avoid mistakes. For us normal mortals sloppy notions are a great danger getting confused. Relativistic mass is a relatively harmless sin. It becomes very messy with the thermodynamical quantities like temperature, chemical potential, etc. Even just last week or so a new paper appeared in Annals of Physics getting it confusing again, falling back to the sins of Planck. To say it very clearly, today all these quantities are scalars, defined and to be measured in the (local) rest frame of the heat bath...