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

[Orodruin]

If you already know what you're doing

I think this is the crucial point. The text is not written as an argument for this kind of person. The typical target audience of the text will consider $F=ma$ as the defining relation of inertial mass and that simply does not generalise to relativity. This point is made in the text where I specifically discuss this discrepancy with the relation to the time derivative of momentum. You cannot have F=ma and p=mv at the same time if m depends on v.

I just find this argument

That's not a problem of relativistic mass. Invariant mass isn't better either.

a bit off the mark. It is rather clear that it is a problem if you expect to recover classical results, as pointed out in #30. That invariant mass suffers from the same issue just provides a straw man argument. Once we have established that the concept has issues, we can start the consideration of what the more appropriate quantity to consider should be.

 

[SiennaTheGr8]

I think this is the crucial point. The text is not written as an argument for this kind of person. The typical target audience of the text will consider $F=ma$ as the defining relation of inertial mass and that simply does not generalise to relativity. This point is made in the text where I specifically discuss this discrepancy with the relation to the time derivative of momentum. You cannot have F=ma and p=mv at the same time if m depends on v.

Yes, and to be clear, my comment was meant as a general one on relativistic mass, not as a response to the Insights article.

 

[DrStupid]

in general, there simply is no multiplicative factor (let alone a constant of proportionality) that tells you how an object will accelerate under the influence of a given 3-force.

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.

The typical target audience of the text will consider $F=ma$ as the defining relation of inertial mass and that simply does not generalise to relativity.

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.

 

[Orodruin]

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.

You have taken the quote out of context. I believe the [...] makes it pretty clear that what is discussed in the sentence refers to classical mechanics. So yes, it should be a reaffirmation of the reader’s assumed knowledge of Newton-2.

 

[Orodruin]

Of course there is an M(v) with F=M·a, but it is almost ever different from relativistic mass.

Actually, this is false. 3-force and 3-acceleration are not necessarily parallel in relativity.

 

[DrStupid]

I believe the [...] makes it pretty clear that what is discussed in the sentence refers to classical mechanics.

That's what I'm talking about. This sentence reads like a confirmation for F=m·a as the defining relation of inertial mass in classical mechanics. This is simply not the case. It should be pointed out that this equation is valid for constant inertial mass only - no matter if you take it as a definition or derive it from F=dp/dt. Whith this knowledge it would be obvoius that is must not be used with a variable mass.

 

[DrStupid]

Actually, this is false. 3-force and 3-acceleration are not necessarily parallel in relativity.

M = \left( {I_3 + \frac{{v \cdot v^T }}{{c^2 - v^2 }}} \right) \cdot \frac{m}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}

edit: typo corrected

 

[Orodruin]

M = \left( {I_3 + \frac{{v \cdot v^T }}{{c^2 - v^2 }}} \right) \cdot \frac{m}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}

edit: typo corrected

Where M is obviously a matrix and not a number. Your previous post seemingly suggests that it is a number. "Scalar" in relativity is typically reserved for frame-independent scalar quantities.

That's what I'm talking about. This sentence reads like a confirmation for F=m·a as the defining relation of inertial mass in classical mechanics. This is simply not the case.

I believe you are missing the mark. At that point it would be detrimental to the Insight to start talking about how Newton-2 looks for variable mass systems in classical mechanics (which would be another Insight altogether). @Mister T has it figured out in #30. The point is that laymen try to use relativistic mass in expressions such as F = ma and K = mv^2/2 or think that a moving object's inner structure changes.

 

[DrStupid]

Your previous post seemingly suggests that it is a number.

Can you explain what exactly let you think that (in order to avoid it the next time).

"Scalar" in relativity is typically reserved for frame-independent scalar quantities.

Would my statement be wrong for lorentz scalars?

The point is that laymen try to use relativistic mass in expressions such as F = ma and K = mv^2/2 or think that a moving object's inner structure changes.

And such misconceptions shouldn't be suported by using by using them without immediate correction.

 

[Orodruin]

Can you explain what exactly let you think that (in order to avoid it the next time).

You had been using the word "scalar" with the meaning of a Lorentz scalar in a previous post and in the context of relativistic mass not being a Lorentz scalar. This made your post appear to simply state that $M$ was not a Lorentz scalar, but still a number.

And such misconceptions shouldn't be suported by using by using them without immediate correction.

I strongly disagree. With this mentality you would get absolutely nowhere in high-school physics. It is clearly within the lies-to-children category.

 

[SiennaTheGr8]

Incidentally, I think we lose something linguistically by reserving "scalar" for Lorentz invariants. I find "number" a poor substitute for describing the type of quantity that, say, total energy is—it would be easy to infer that a quantity that is a "number" doesn't have a unit. (And anyway, total energy is invariant under spatial rotation, so it is a scalar in that context, isn't it?)

 

[DrStupid]

You had been using the word "scalar" with the meaning of a Lorentz scalar in a previous post and in the context of relativistic mass not being a Lorentz scalar. This made your post appear to simply state that $M$ was not a Lorentz scalar, but still a number.

I still do not see why my post suggests that M is a number - even if you read "scalar" as lorentz scalar. Nevermind, it was just a misunderstanding that has been cleared.

It is clearly within the lies-to-children category.

F=m·a as defining relation for inertial mass actually is in that category. That shouldn't be mentioned, unless in order to correct this misconception.

 

[Orodruin]

F=m·a as defining relation for inertial mass actually is in that category.

I don't understand this answer. This is exactly what I was saying in my post.

That shouldn't be mentioned, unless in order to correct this misconception.

I disagree again. The concept of lies-to-children is a crucial idea in education wherein we get people a bit closer to the right idea than they were before. If you wish to clear that up, fine, but that is not the goal of the text and in this context counter-productive. The goal of the text is to clarify a different lie that in my experience is responsible for many more severe misunderstandings. For some time when this text was written, every other thread in the relativity forum was filled with misconceptions based on relativistic mass and trying to apply it in statements such as F = ma. You will not find the same type of misconceptions regarding the definition of mass in classical mechanics. Partially because you could just as well define it through F = ma and the variable mass case would look slightly different, but still contain the same physics. (I am not saying it is a good idea, just that you could do it.)

 

[Orodruin]

Incidentally, I think we lose something linguistically by reserving "scalar" for Lorentz invariants. I find "number" a poor substitute for describing the type of quantity that, say, total energy is—it would be easy to infer that a quantity that is a "number" doesn't have a unit. (And anyway, total energy is invariant under spatial rotation, so it is a scalar in that context, isn't it?)

This is not an issue only for scalars but for many other mathematical concepts as well. Just take such a thing as a "vector". In relativity we generally assume that it means a 4-vector or, when restricting ourselves to rotations, 3-vectors. However, a "vector" generally is just an element of a vector space, which would also include tensors or any other type of elements of vector spaces (such as function spaces etc). If we would wish to be precise, we would say that energy is a scalar under rotations, but not a Lorentz scalar.

 

[DrStupid]

The concept of lies-to-children is a crucial idea in education wherein we get people a bit closer to the right idea than they were before.

F=dp/dt instead of F=m·a gets them as close to the right idea as currently possible. The results are in agreement with all known independently reproducible experimental observations and nothing else decides between right or wrong in physics. Of course the use of relativistic mas is not the optimal choice in relativity, but that doen't mean it is wrong.

 

[Orodruin]

The results are in agreement with all known independently reproducible experimental observations and nothing else decides between right or wrong in physics.

So would the F = ma definition be if used it appropriately with the appropriate inclusion of terms in the variable-mass case on the force side (i.e., $\dot m \vec v_{\rm rel}$). I agree that the other option is cleaner, but for the purposes of the text, the distinction is not really relevant as that is not the definition that leads to severe misunderstandings of the basic theory - relativistic mass does just that.

I would agree with your comments if the text was longer and intended for university level students, but the thing is that it is not and you have to adapt your level to the target audience (and its typical attention span) even if that means not being as precise as you would like. The text is a very short argumentation for why physicists generally avoid talking about relativistic mass, intended for people who come to PF with some of the strong misconceptions that the concept induces.

 

[Dale]

And relativistic mass is the proportionality factor between velocity and momentum, which is by definition the iniertial mass in classical mechancis

I disagree with this. From classical mechanics we expect $\vec{f}=m\vec{a}$ and $\vec{p}=m\vec{v}$ and $\vec{f}=d\vec{p}/dt$ to all hold. The question is how to generalize this. Using relativistic mass in place of $m$ does not generalize all of them (neither with three vectors nor with four vectors). Using invariant mass does (with four vectors).

 

[PAllen]

To my mind, the Newtonian notion of inertia was tied to F=mA, with m constant. It is precisely the feature that this m and the m in universal gravitation were the same that led to notions of universal free fall that Newton was familiar with. The m in momentum has little to do with this. Further, Newton never defined m via momentum. He defined it via quantity of matter, which is not a very useful modern definition.

 

[DrStupid]

I disagree with this.

Newton's definition II says "Quantitas motus est mensura ejusdem orta ex Velocitate et quantitate Materiæ conjunctim." That means p:=m·v. If you disagree with this definition than you are outside classical mechanics.

From classical mechanics we expect $\vec{f}=m\vec{a}$ and $\vec{p}=m\vec{v}$ and $\vec{f}=d\vec{p}/dt$ to all hold.

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.

To my mind, the Newtonian notion of inertia was tied to F=mA, with m constant.

The Newtonean notation was F=dp/dt. That results in F=m·a if m is constant. But m was not defined to be constant in classical mechanics (in contrast to current meaning of the term "mass"). Such a definition wasn't required because the velocity dependence of m was already given by other basic requirement, including the transformation. Galilean transformation results in a constant mass for closed systems. Replacing Galilean transformation by Lorentz transformation turns m into the relativistic mass.

 

[Orodruin]

If you disagree with this definition than you are outside classical mechanics.

Newton's formulation with his three laws is not the only formulation of classical mechanics. Nor is Principia a Bible that cannot be questioned or improved upon. Imagine a similar claim in quantum mechanics; that you have to adhere to the Schrödinger picture or you are not doing QM.

 

[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.