Insights What is relativistic mass and why it is not used much? - Comments

[Dale]

We get F=m·a in classical mechanics for closed systems. But that doesn't mean that we need to expect it outside classical mechanics as well.

I disagree. In the classical domain f=ma is experimentally validated. So any theory which would generalize classical mechanics must reduce to f=ma in the appropriate limit. We therefore have a very strong expectation that we will see something like it in relativity, with some suitable modifications for the generalization.

Here the most appropriate generalization is the use of four vectors, in which case the familiar formulas hold and clearly reduce to the Newtonian expressions in the appropriate limit. The mass then is the invariant mass.

 

[SiennaTheGr8]

As far as I can tell, it's only a matter of convention whether we define force as $\dot{\vec p}$ or $m \vec a$ in Newtonian mechanics. The only difference is what symbols and terminology we use in a variable-mass situation. Given that $\vec f \equiv \dot{\vec p}$ becomes necessary in special relativity, it is perhaps unfortunate that the $m \vec a$ definition is generally preferred in "university physics" courses—but I imagine it's easier to teach.

Another argument in favor of $\vec f \equiv \dot{\vec p}$ in Newtonian physics is aesthetic: isn't it nice to have a specific word and symbol for the time-derivative of momentum, a conserved vector of central importance?

On the other hand, that definition means that force isn't necessarily invariant under a Galilean boost (because of the frame-dependent $\dot{m} \vec v$ term), and the "proper force" concept becomes useful (by which I mean the force as measured in the forcee's instantaneous rest frame, always equal to $m \vec a$). An argument that this opens up a can of worms and should be avoided can be found here.

 

[PAllen]

Note, my point had nothing to do with the definition of force. I was discussing the notion of inertia, which was conceptually resistance to force via F=mA. It is irrelevant whether this is a derived relation or a definition.

 

[SiennaTheGr8]

On the other hand, that definition means that force isn't necessarily invariant under a Galilean boost (because of the frame-dependent $\dot{m} \vec v$ term), and the "proper force" concept becomes useful (by which I mean the force as measured in the forcee's instantaneous rest frame, always equal to $m \vec a$). An argument that this opens up a can of worms and should be avoided can be found here.

I might also argue that the authors of the linked article "stack the deck" against $\dot{\vec p}$ on page 2, where they introduce the following "apparent paradox":

If we consider the simple case of a variable mass, and we write Newton's second law as:

$\vec F = m \dfrac{\mathrm{d} \vec v}{\mathrm{d}t} + \vec v \dfrac{\mathrm{d}m}{\mathrm{d}t} \qquad \qquad (2)$

we can easily see that it violates the relativity principle under Galilean transformations. When $\vec F$ is zero, in particular, Equation (2) implies that the particle will remain at rest in a system where it is originally at rest, but it will be accelerated by the "force" $-\vec v \, \mathrm{d}m / \mathrm{d}t$ in a system where the particle moves with velocity $\vec v$!

To solve this apparent paradox ...

Of course, if you've defined force as $\dot{\vec p}$, then you've accepted that it isn't a Galilean invariant, and you know perfectly well that $\vec F$ cannot equal zero if the mass varies while the (non-zero) velocity doesn't. Seems to me that there's no "apparent paradox" here, unless you allow the $m \vec a$ definition to sneak in.

 

[DrStupid]

In the classical domain f=ma is experimentally validated.

For constant m.

So any theory which would generalize classical mechanics must reduce to f=ma in the appropriate limit.

For constant m. Extrapolations beyond the limits of experimental observations are not reliable. Therefore you must ot expect that the relation will hold in such cases.

We therefore have a very strong expectation that we will see something like it in relativity, with some suitable modifications for the generalization.

For constant m. There are conditions, where this is the case (e.g. during uniform circular motion or in good approximation for non-relativistic velocities) and in these cases F=m·a still remains vaid. But significantant changes of m are outside the experimental validated range of validity of F=m·a. Therefore it is no surprise that the relation doesn't hold in these cases.

with some suitable modifications for the generalization.

There are no modifications required except for the transformation. You just need to replace Galilean transformation by Lorentz transformation and than derive the resulting equations for force, kinetic energy etc.

Here the most appropriate generalization is the use of four vectors

That's just another formalism to describe the same physics.

 

[SiennaTheGr8]

Link to the Insights article appears to be broken (wasn't working yesterday, either).

 

[vanhees71]

At least not a scalar. Of course there is an M(v) with F=M·a, but it is almost ever different from relativistic mass.
Wouldn't it be a good idea to clear this misconception from the beginning? The sentence "[...] inertial mass, which determines the resistance to acceleration of an object ($\vec a = \vec F/m$)" reather sounds like a confirmation.

Can you demonstrate these very strange claims for the most simple relativistic force, i.e., the electromagnetic force on a classical charged particle in an external electromagnetic field (neglecting radiation reaction forces of course)? I doubt it very much.

The four-dimensionally covariant EoM is much simpler than such confusing constructs anyway. I simply reads
$$m \frac{\mathrm{d}^2 x^{\mu}}{\mathrm{d} \tau^2}=\frac{q}{c} F^{\mu \nu} \frac{\mathrm{d} x_{\nu}}{\mathrm{d} \tau},$$
where $\tau$ is the proper tine of the particle.

 

[Dale]

For constant m.

Yes, this is part of what I intended by "in the appropriate limit", but I certainly could have been explicit.

Extrapolations beyond the limits of experimental observations are not reliable

I am not talking about extrapolation.

That's just another formalism to describe the same physics.

Sure, but it is also a formalism which clearly preserves all of the expected relationships. That doesn't make it right or wrong, but it does contradict your arguments in post 27.

 

[sweet springs]

The physicists should not have used the word relativistic mass but relativistic velocity that is spatial components of now called four-velocity.
Not mass but velocity they should have attribute relativistic results.