Can we deal with relativistic mass once and for all?

[Orodruin]

Ok, got it. So there is no relativistic notion of inertia in your conception.

No, it is a derivation of the mass-energy equivalence that tells you that, in the Newtonian limit, what we know as inertia from classical mechanics corresponds to the rest energy. This is a deep result. In classical mechanics there is no a priori link between Newtonian inertia and energy, but in relativity there is. This is what allows you to extract part of the mass of a system and convert it to electric energy.

 

[A.T.]

...so is there any general proof that inertial mass doesn't increase with velocity?

I gave you a counter example in post #50, where the "inertial mass" doesn't increase with velocity.

 

[vanhees71]

In another thread I demonstrated that relativistic mass is the same quantity as the mass as used in non-relativistic physics:

https://www.physicsforums.com/threads/relativistic-mass-still-a-no-no.892981/page-2#post-5620010

And relativistic mass is not mass in the modern sense of the word.
That's not gravitational mass but mass. The source of gravity in GR is the stress-energy-tensor. The fact, that this tensor may depend on mass only in special cases desn't turn mass into geravitational mass.

Well, there are many writings about these issues in popular-science websites and other non-peer-reviewed sources that are not up to date or even incorrect.

Again: To get the correct relation between Newtonian and special-relativistic mechanics one must not (I emphasize one must not!) use relativistic mass. The reason is simple: In Newtonian physics mass is frame independent, i.e., a scalar under Galileo transformations. If you use relativistic mass, the Newtonian limit, which you get by expanding in powers of $v/c$, and there you see that the various relativistic masses depend on the speed of the particle at order $\mathcal{O}(v^2/c^2)$, which is the same order you have to take for kinetic energy to get the correct non-relativistic limit.

The correct way to guess a relativistic equation of motion from a given Newtonian one (you cannot strictly derive it, because SR mechanics is more general than Newtonian mechanics), you have to go to an instantaneous inertial restframe of the particle, where Newtonian mechanics is supposed to hold as an approximation, and then write down the Newtonian equation of motion in terms of covariant quantities. Then you have a good guess for an SR-mechanics equation of motion.

For the kinematic quantities it's clear that to get a covariant expression you have to extend the position vector $\vec{x}$ by the time-poisition four-vector $(x^{\mu})$. The time in the intantaneous rest frame defines the proper time of the particle. It's given in covariant form by
$$\mathrm{d} \tau=\frac{1}{c} \mathrm{d} s=\frac{1}{c} \sqrt{\mathrm{d} x^{\mu} \mathrm{d} x^{\nu} \eta_{\mu \nu}}.$$
Thus the velocity and acceleration in Newton's Law is most conveniently extended to the corresponding four-velocity and four-acceleration
$$c u^{\mu}=\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}, \quad a^{\mu} = \frac{\mathrm{d}^2 x^{\mu}}{\mathrm{d} \tau^2}.$$
The simplified equation of motion for a point particle in Newtonian physics $\vec{F}=m \vec{a}$ is thus generalized to
$$m a^{\mu}=K^{\mu},$$
where $m$ is the invariant mass of the particle, and it's the same mass as in Newtonian physics, i.e., a scalar (in Newtonian physics under Galileo transformations in SR under Poincare transformations).

Of course also in SR there are only 3 independent degrees of freedom. That's because by definition
$$u_{\mu} u^{\mu}=1 \, \Rightarrow u_{\mu} a^{\mu}=0.$$
The Minkowski four-force $K^{\mu}$ thus must fulfill the constraint
$$K_{\mu} u^{\mu}=0.$$
It implies that both $a^{\mu}$ and $K^{\mu}$ are space-like four-vectors.

Of course also energy and momentum easily are derived in this way. The three-momentum in Newtonian physics is $\vec{p}=m \vec{v}$. This suggests to extend this definition to the relativistic four-momentum
$$p^{\mu}=m \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}=m c u^{\mu}.$$
That indeed $p^0=E/c$ is an expression for the kinetic energy (up to an additive constant which is physically irrelevant) can be seen by expanding in powers of $v/c=|\vec{u}|/u^0$. Note that $\vec{v}=\mathrm{d} \vec{x}/\mathrm{d} t$ are not (!) spatial components of a four-vector because one takes a derivative of space components of the four-vector $x^{\mu}$ with respect to the coordinate time $t$. Of course we can use, for a moment, such non-invariant quantities to get the non-relativistic limit by expanding in powers of $v/c$. For energy we have
$$E=mc^2 u^0=m c^2 \frac{\mathrm{d} t}{\mathrm{d} \tau}=m c^2 \frac{1}{\mathrm{d} \tau/\mathrm{d} t}=m c^2 \frac{1}{\sqrt{1-\vec{v}^2/c^2}}=m c^2 \gamma.$$
Now expanding in powers of $v/c$
$$E=m c^2 \left (1+\frac{v^2}{2 c^2} +\mathcal{O}(v^4/c^4) \right )=m c^2 + \frac{m}{2} \vec{v}^2 + \mathcal{O}(v^4/c^4).$$
This shows that in the non-relativistic limit we get, up to the constant rest-energy $E_0=m c^2$ the Newtonian kinetic energy. To include the rest energy in the energy of a relativistic particle is convenient, because as we've just shown, then you get a Lorentz-four vector and make energy and momentum co-variant quantities as they are components of the energy-momentum four-vector
$$(p^{\mu})=\begin{pmatrix} E/c \\ \vec{p} \end{pmatrix}.$$
The mass is, by construction, a scalar obeying the "on-shell condition"
$$p_{\mu} p^{\mu}=m^2 c^2.$$
For a derivation from the point of view of Noether's theorem, see my SR FAQ article:
https://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[DrStupid]

Again: To get the correct relation between Newtonian and special-relativistic mechanics one must not (I emphasize one must not!) use relativistic mass. The reason is simple: In Newtonian physics mass is frame independent, i.e., a scalar under Galileo transformations.

I can't follow your argumentation. Yes, mass as used in classical mechanics is frame-independent under Galilean transformation but frame-dependent under Lorentz transformation. But why is that a reason not to use it to get the correct relation between Newtonian and special-relativistic mechanics? I would even say that getting a relation between Newtonian and SR mechanics is today the only remaining reason to use this property. Within classical mechanics it is always equal to mass and within special relativity it not efficient due to its frame-dependence but it allows a smooth transtition between classical and SR mechanics.

 

[Orodruin]

But why is that a reason not to use it to get the correct relation between Newtonian and special-relativistic mechanics?

Because mass, as used in classical mechanics, does not work the same way in relativity. In classical mechanics, mass is used in many different relations. Depending on which relation you are looking at, you will get different "correct relations".

 

[DrStupid]

Because mass, as used in classical mechanics, does not work the same way in relativity. In classical mechanics, mass is used in many different relations. Depending on which relation you are looking at, you will get different "correct relations".

Do you have examples for such relations which are not based on Galilean transformation.

 

[Orodruin]

Do you have examples for such relations which are not based on Galilean transformation.

This is completely irrelevant. You are not trying to check classical mechanics for internal consistency, you are trying to generalise the concept of mass in classical mechanics to relativity.

 

[PAllen]

I can't follow your argumentation. Yes, mass as used in classical mechanics is frame-independent under Galilean transformation but frame-dependent under Lorentz transformation. But why is that a reason not to use it to get the correct relation between Newtonian and special-relativistic mechanics? I would even say that getting a relation between Newtonian and SR mechanics is today the only remaining reason to use this property. Within classical mechanics it is always equal to mass and within special relativity it not efficient due to its frame-dependence but it allows a smooth transtition between classical and SR mechanics.

It seems to me that your whole argument rests on an assumption you add to Newton's definitions which I think is not in the spirit of Newon's writing. The assumption you add is that Newton's definition of momentum is also the definition of mass, and you ignore Newton's actual definition of mass. I claim the best correspondence between Newtonian and SR mechanics is achieved by formalizing and amending Newton's actual definition of mass, preserving its crucial properties - frame inedpendence, change only via flow of something, and resistance to change of motion. Definition of momentum and kinetic energy then get amended such that they approach the Newtonian definitions for slow speeds. Definition of force is preserved exactly. You insist yours is the only valid correspondence, while I claim it is a possible correspondence, but not the most faithful one.

 

[DrStupid]

This is completely irrelevant. You are not trying to check classical mechanics for internal consistency, you are trying to generalise the concept of mass in classical mechanics to relativity.

I didn’t ask you for internal consistencies in classical mechanics but for an example for relations that result in different "correct relations". As that was your objection you should be more interested in supporting it than me.

The assumption you add is that Newton's definition of momentum is also the definition of mass, and you ignore Newton's actual definition of mass.

No, I just take it for granted that the concept of mass used in the definition of momentum is consistent with this definition.

I claim the best correspondence between Newtonian and SR mechanics is achieved by formalizing and amending Newton's actual definition of mass, preserving its crucial properties - frame inedpendence, change only via flow of something, and resistance to change of motion.

Which definition are you talking about? Newton’s definition of mass doesn’t say anything about its frame-dependence or -independence.

Definition of momentum and kinetic energy then get amended such that they approach the Newtonian definitions for slow speeds.

And that’s the problem. By forcing the mass as used in classical mechanics to remain frame-independent in SR, everything where mass is involved doesn’t work anymore and needs to be amended. How is that a better correspondence than leaving as many definitions unchanged as possible?

You insist yours is the only valid correspondence, while I claim it is a possible correspondence, but not the most faithful one.

I don’t insist more in my opinion than everybody else. Would you blame vanhees71 in the same way for advancing his opinion?

 

[Orodruin]

I didn’t ask you for internal consistencies in classical mechanics but for an example for relations that result in different "correct relations". As that was your objection you should be more interested in supporting it than me.

You obviously cannot talk about things derived from Galilean transformations to discuss the generalisation to SR since SR instead have Lorentz transformations. You get things such as the inertia in different directions being different, as discussed in this thread. You simply cannot replace m with relativistic mass in F=ma as you seemingly want to do (that would qualify as a correct relation in classical mechanics).

 

[Orodruin]

And that’s the problem. By forcing the mass as used in classical mechanics to remain frame-independent in SR, everything where mass is involved doesn’t work anymore and needs to be amended. How is that a better correspondence than leaving as many definitions unchanged as possible?

This in my opinion shows a fundamental lack of understanding of how SR works and how it corresponds to the Newtonian limit. The only reason you need to ”amend” anything is your insistence on using 3-vectors rather than 4-vectors, which is the natural thing to use in SR.

 

[stevendaryl]

And that’s the problem. By forcing the mass as used in classical mechanics to remain frame-independent in SR, everything where mass is involved doesn’t work anymore and needs to be amended. How is that a better correspondence than leaving as many definitions unchanged as possible?

Well, as I said in another thread, you can retcon Newtonian physics so that the concept of mass, and the dynamics in terms of mass are almost exactly unchanged in going from Newton's physics to Einstein's physics (Special Relativity, not General Relativity).

1. Associated with every (slower-than-light) object is a monotonically increasing affine parameter $s$.
2. Also associated with every object is a scalar value, $m$. $m$ is invariant under coordinate changes.
3. The path of the object through spacetime can then be characterized by a 4-velocity $V$ with components in an inertial Cartesian coordinate system given by: $(\frac{dx}{ds}, \frac{dy}{ds}, \frac{dz}{ds}, \frac{dt}{ds})$. $V$ is a vector under coordinate changes.
4. Combining the scalar and the 4-velocity gives a 4-momentum: $P = m V$. The 4-momenta of a collection of objects is additive, and is conserved in collisions (at least if we ignore long-range forces).
5. Each object obeys Newton's second law of motion (both relativistically and nonrelativistically): $F = m \frac{dV}{ds}$, where $F$ is a 4-vector quantity under changes of coordinates.

These facts hold for both Newton's and Einstein's physics, at least if we only consider short-range forces. The biggest difference is the affine parameter $s$. In nonrelativistic physics, $ds = dt$. In relativistic physics, $ds = \sqrt{dt^2 - \frac{1}{c^2} (dx^2 + dy^2 + dz^2)}$. Because the relation between $s$ and $t$ is so trivial in Newtonian physics, the 4th component of 4-vectors is pretty boring:

• $V^t = 1$
• $F^t = 0$
• $P^t = m$

So Newtonian physics is essentially 3-dimensional, rather than 4-dimensional, although the 4th components are necessary in order for velocity, momentum, and force to be vectors under coordinate changes involving time.

 

[DrStupid]

You obviously cannot talk about things derived from Galilean transformations to discuss the generalisation to SR since SR instead have Lorentz transformations.

That’s why I asked you for relations which are not based on Galilean transformation.

You simply cannot replace m with relativistic mass in F=ma as you seemingly want to do (that would qualify as a correct relation in classical mechanics).

No, I don’t want to do something like that. This is an example for a relation that is based on Galilean transformation because it only holds for constant mass and the mass (as used in classical mechanics) of a closed system is constant under Galilean transformation but not under Lorentz transformation. The original relation F=dp/dt works with both transformations.

 

[DrStupid]

This in my opinion shows a fundamental lack of understanding of how SR works and how it corresponds to the Newtonian limit. The only reason you need to ”amend” anything is your insistence on using 3-vectors rather than 4-vectors, which is the natural thing to use in SR.

SR is not a matter of 3- or 4-vectors. They are just different formalisms to describe the same physics.

 

[DrStupid]

So Newtonian physics is essentially 3-dimensional, rather than 4-dimensional, although the 4th components are necessary in order for velocity, momentum, and force to be vectors under coordinate changes involving time.

That's why Newton's definition of momentum is p=m·v and not P=m·V. The equations may look similar but they are actually different and have different physical meanings. Replacing one by the other has concequences. One of them is a different concept of mass.

 

[stevendaryl]

SR is not a matter of 3- or 4-vectors. They are just different formalisms to describe the same physics.

I guess I would agree that using 3-vectors or 4-vectors amounts to different formalisms for expressing the same physics. But that applies to both Newtonian and SR physics. You can describe Newtonian physics using 4-vectors (as I did above) and you can describe Special Relativity using 3-vectors.

 

[stevendaryl]

That's why Newton's definition of momentum is p=m·v and not P=m·V. The equations may look similar but they are actually different and have different physical meanings. Replacing one by the other has concequences. One of them is a different concept of mass.

If you are going to say that SR can be described in a 3-vector or 4-vector formalism, you should be able to describe Newtonian physics in a 3-vector or 4-vector formalism. The physics is not changed. What I think is nice about the 4-vector formalism is that

• conservation of mass is a consequence of conservation of momentum in Newtonian physics
• 4-velocity, 4-momenta and 4-force are true vectors (while 3-velocity, 3-momenta and 3-force are only vectors under changes of pure spatial coordinates, not Galilean or Lorentz transformations)
• mass is a true scalar in the 4-vector formulation, both in SR and Newtonian physics.

 

[DrStupid]

If you are going to say that SR can be described in a 3-vector or 4-vector formalism, you should be able to describe Newtonian physics in a 3-vector or 4-vector formalism.

How does that contribute to the topic?

 

[stevendaryl]

How does that contribute to the topic?

I think it shows the relationship between mass in SR and mass in Newtonian physics most clearly. In the 4-vector approach, it's not necessary to have a velocity-dependent mass in order for $F = m \frac{dV}{ds}$ to hold.

 

[DrStupid]

I think it shows the relationship between mass in SR and mass in Newtonian physics most clearly.

Can you please demonstrate that?

 

[PeroK]

One thing is clear:

Q. Can we deal with relativistic mass once and for all?

A. Apparently not!

 

[stevendaryl]

Can you please demonstrate that?

I thought I did. In the 4-vector formalism:

1. 4-velocity, 4-momentum, 4-force are true vectors under all coordinate changes (while 3-velocity, 3-momentum and 3-force are only vectors under a change of spatial coordinates)
2. Mass is a true scalar under coordinate changes.
3. Momentum is just the product of the mass and the 4-velocity.
4. Force and velocity are related by Newton's law: $F = m \frac{dV}{ds}$

In the 4-vector formalism, all 4 statements are true without change in both Newtonian physics and Special Relativity.

 

[stevendaryl]

I thought I did. In the 4-vector formalism:

1. 4-velocity, 4-momentum, 4-force are true vectors under all coordinate changes (while 3-velocity, 3-momentum and 3-force are only vectors under a change of spatial coordinates)
2. Mass is a true scalar under coordinate changes.
3. Momentum is just the product of the mass and the 4-velocity.
4. Force and velocity are related by Newton's law: $F = m \frac{dV}{ds}$

In the 4-vector formalism, all 4 statements are true without change in both Newtonian physics and Special Relativity.

It's sort of interesting that in the 4-vector formalism of Newtonian physics, there are two different places where mass appears:

1. It's a scale factor to convert a 4-velocity into a 4-momentum.
2. It's also the 4th component of the 4-momentum.

In SR, #1 corresponds to invariant mass, and #2 corresponds to "relativistic mass". So I guess the 4-vector formalism gives support to both claims:

• Invariant mass in SR corresponds to Newtonian mass (as the scalar multiple of 4-velocity to produce 4-momentum)
• Relativistic mass in SR corresponds to Newtonian mass (as the 4th component of the 4-momentum)

 

[DrStupid]

I thought I did.

I still don't see it.

Mass is a true scalar under coordinate changes.

Yes, that's true for mass but where is the relation to mass as used in Newtonian physics?

Momentum is just the product of the mass and the 4-velocity.

No, momentum is defined the product of mass (as used in Newtonian physics) and velocity. What you mean is 4-momentum.

Force and velocity are related by Newton's law: $F = m \frac{dV}{ds}$

That's true if m is mass. But i still do not see the the relationship between mass in SR and mass in Newtonian physics.

 

[stevendaryl]

I still don't see it.

Yes, that's true for mass but where is the relation to mass as used in Newtonian physics?

It's the same: In both Newtonian physics and SR, mass is what you multiply velocity by to get momentum. Mass is what you multiply acceleration by to get force.

No, momentum is defined the product of mass (as used in classical mecatnis) and velocity. What you mean is 4-momentum.

3-momentum is just the first 3 components of 4-momentum.

That's true if m is mass. But i still do not see the the relationship between mass in SR and mass in Newtonian physics.

They are the SAME! Mass is the scalar quantity that you multiply velocity by to get momentum.

 

[DrStupid]

It's the same: In both Newtonian physics and SR, mass is what you multiply velocity by to get momentum. Mass is what you multiply acceleration by to get force.

That's only true for closed systems in classical mechanics. In SR mass is what you multiply 4-velocity by to get 4-momentum and what you multiply with 4-acceleration to get 4-force.

3-momentum is just the first 3 components of 4-momentum.

That's just another way to say that they are different.

 

[stevendaryl]

That's only true for closed systems in classical mechanics. In SR mass is what you multiply 4-velocity by to get 4-momentum and what you multiply with 4-acceleration to get 4-force.

It's true for both Newtonian physics and SR, it's just that nobody bothered with the 4th component of 4-velocity or 4-momentum in Newtonian physics since it's so boring.

That's just another way to say that they are different.

That doesn't make any sense to me. If multiplying mass by 4-velocity produces 4-momentum, and 3-velocity is the spatial components of 4-velocity, then it follows that multiplying 3-velocity by mass produces 3-momentum. Why would you say that it's a different concept of mass?

 

[DrStupid]

It's true for both Newtonian physics and SR, it's just that nobody bothered with the 4th component of 4-velocity or 4-momentum in Newtonian physics since it's so boring.

In SR 4-momentum is

$P = m \cdot V$

but momentum is

$p = \frac{{m \cdot v}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}$

and not

$p = m \cdot v$

If multiplying mass by 4-velocity produces 4-momentum, and 3-velocity is the spatial components of 4-velocity, then it follows that multiplying 3-velocity by mass produces 3-momentum.

Obviously not (see above).

 

[stevendaryl]

In SR 4-momentum is

$P = m \cdot V$

but momentum is

$p = \frac{{m \cdot v}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}$

Yes, but the spatial components of the 4-momentum $P$ are equal to the components of the 3-momentum $p$. So I don't understand what point you are making.

Newtonian physics doesn't make the distinction between clock time $s$ and coordinate time $t$ because the two are trivially related: $ds = dt$. But formulating Newtonian physics in terms of $s$ produces a theory where the generalization to SR is a lot clearer. Almost everything is the same in the two theories---the only difference is a different notion of how $s$ is related to coordinate time $t$.

It is a fact of SR that clock time $s$ is not equal to coordinate time $t$, so you have to introduce that change, regardless. But once you're introduced that change, there is no additional change needed to go from $m$ to $\gamma m$. The $\gamma$ can be seen as just the factor $\frac{dt}{ds}$.

 

[PeterDonis]

and 3-velocity is the spatial components of 4-velocity

Which it isn't. The spatial components of the 4-velocity are the 3-velocity times $\gamma$, not the 3-velocity itself.

 

[PeterDonis]

The $\gamma$ can be seen as just the factor $\frac{dt}{ds}$.

Yes, but that doesn't change the definition of "3-velocity". It changes the relationship between 3-velocity and 4-velocity. When you measure the 3-velocity of an object in SR, you still measure $\mathbf{v}$, not $\gamma \mathbf{v}$.

 

[stevendaryl]

Which it isn't. The spatial components of the 4-velocity are the 3-velocity times $\gamma$, not the 3-velocity itself.

I think you missed my "retcon" of Newtonian kinematics in terms of clock time s, rather than coordinate time t. In retrospect, everything could have been formulated in a 4-vector formalism. The reason it wasn't was, first, because nobody thought of it, and second, because the 4-the component of most vectors is pretty boring in Newtonian physics: $V^t = \frac{dt}{ds} = 1$, $P^t = m V^t = m$.

 

[PeterDonis]

I think you missed my "retcon" of Newtonian kinematics in terms of clock time, rather than coordinate time

No, I didn't miss it. I'm just pointing out that your retcon doesn't change the meaning of the term "3-velocity". It formulates Newtonian mechanics in terms of "4-velocity", but that just means, as I said, that the relationship between 4-velocity and 3-velocity is different in Newtonian mechanics and SR. That shouldn't be surprising, since Newtonian mechanics and SR are different theories that make different predictions, so there will be no way to "retcon" one so that everything "matches" the other; there will always be a difference somewhere, it's just a matter of preference for where you want to put it.

 

[DrStupid]

In retrospect, everything could have been formulated in a 4-vector formalism.

It could but it doesn't. You can't simply replace Newton's p=m·v by P=m·V and expect that it is still the same definition.

 

[stevendaryl]

No, I didn't miss it. I'm just pointing out that your retcon doesn't change the meaning of the term "3-velocity". It formulates Newtonian mechanics in terms of "4-velocity", but that just means, as I said, that the relationship between 4-velocity and 3-velocity is different in Newtonian mechanics and SR.

But I would trace that difference to the fact that the relationship between clock time (or proper time) and coordinate time is more complicated in SR.