B I have a conceptual question about relative mass

[Victorifrj]

If mass increases with velocity(v), can I say velocity is a quantity dependent on (v)?

m=Ymo, m>mo

so why can't we say that the mass (matter) increases and yes what increases is the energy to make your speed keep increasing?

 

[Helios]

can I say velocity is a quantity dependent on (v)

I dare say ANY quantity depends on itself. Relative mass is an archaic concept these days.

 

[PeterDonis]

If mass increases with velocity(v)

Relativistic mass does, but relativistic mass is an outdated concept which is not used in modern relativity. We just use energy when that is the right concept to use.

Invariant mass does not increase with velocity.

can I say velocity is a quantity dependent on (v)?

What would this even mean?

 

[Victorifrj]

I dare say ANY quantity depends on itself. Relative mass is an archaic concept these days.

I was reading a book now that confused me, it's from 2005. It has a chapter on relativistic mass, and it makes a joke of what it would be like if we calculated someone else's mass if they were moving close to the speed of light, with that we would notice a small difference in the mass of this person. So the mass that's in this equation would be the quantity that opposes motion, the inertial mass?

 

[Victorifrj]

Relativistic mass does, but relativistic mass is an outdated concept which is not used in modern relativity. We just use energy when that is the right concept to use.

Invariant mass does not increase with velocity.What would this even mean?

In the book I'm reading he says: "We can write in aspects of classical physics defining a quantity called relativistic mass..." I understand that it is an archaic term, but this term in this equation doesn't mean anything?

 

[Victorifrj]

Relativistic mass does, but relativistic mass is an outdated concept which is not used in modern relativity. We just use energy when that is the right concept to use.

Invariant mass does not increase with velocity.What would this even mean?

I wrote wrong: can I say Relativistic mass a quantity dependent on (v)?

 

[Grasshopper]

I dare say ANY quantity depends on itself. Relative mass is an archaic concept these days.

How does one prove this though? I’m mostly joking, but I did try to think it through, and I can’t imagine any proof of the reflexive property that at some point doesn’t say something like “f(x) = f(x) for all x ∈ X, therefore the relation xRx is reflexive”. Seems kind of circular.

 

[PeterDonis]

In the book I'm reading

What book?

 

[PeterDonis]

I can’t imagine any proof of the reflexive property that at some point doesn’t say something like “f(x) = f(x) for all x ∈ X, therefore the relation xRx is reflexive”.

But for the velocity $v$ you aren't even saying that; you're not saying $f(v) = f(v)$, you're just saying $v = v$. Which is already a tautology. No proof required.

 

[PeterDonis]

the mass that's in this equation would be the quantity that opposes motion, the inertial mass?

Only if we restrict to one spatial dimension, so the motion and the forces are all in the same direction.

As soon as you have motion and forces pointing in different directions, relativistic mass as inertia no longer works because "inertia" is no longer a single number; an object's response to a given force depends on the direction of the force relative to the object's motion. Look up "transverse mass" and "longitudinal mass" if you want to see the tangle that resulted before all this got straightened out by abandoning relativistic mass as a concept.

 

[Ibix]

Only if we restrict to one spatial dimension, so the motion and the forces are all in the same direction.

I'd say it's not even that simple since even in one dimension $p=m_rv$ but $F=\gamma^2 m_ra$, where $m_r$ is the relativistic mass. Is $m_r$ really the inertial mass if $F\neq m_ra$? I suppose you could make the case either way, but this is another example of the definitional tangle you mentioned.

OP: The fundamental problem here is trying to take concepts from Newtonian physics, which is an approximation to a more general theory, and trying to make them fit the more general theory. Sometimes it works, sometimes it doesn't. There being a simple relationship between mass and inertia is one case where it doesn't.

 

[DrStupid]

I understand that it is an archaic term, but this term in this equation doesn't mean anything?

Of course it means something but it is outdated. It is just about OK in special relativity but using it in general relativity would be like going with horse and knight's armour into a modern battle.

 

[PeterDonis]

It is just about OK in special relativity

Only if you restrict to motion in one spatial dimension, as I said in post #10 (and even then there are issues, as @Ibix pointed out in post #11).

 

[PeterDonis]

even in one dimension $p=m_rv$ but $F=\gamma^2 m_rv$, where $m_r$ is the relativistic mass.

Actually it's $F = \gamma^3 m a$ for motion in one spatial dimension ($a$, not $v$). But yes, this is a good point, the relation $F = m_r a$, where $m_r = \gamma m$, only works for transverse forces, not longitudinal forces.

 

[Ibix]

($a$, not $v$)

Thank you - corrected above.

 

[DrStupid]

Only if you restrict to motion in one spatial dimension

That depends on how you do it and what you start with. But we don't need to discuss that in detail. It is sufficent to say that relativistic mass is not useful and should be avoided.

 

[vanhees71]

It's really a pain to work with relativistic masses. There's no need for it. It's only obscuring the issue. It's best to work with Lorentz-covariant quantities. Then the equation of motion simply reads
$$m \frac{\mathrm{d}^2 x^{\mu}}{\mathrm{d} \tau^2}=K^{\mu}.$$
Here $m$ is simply the scalar invariant mass of the particle.

The four-force of course must fulfill the constraint
$$K^{\mu} \frac{\mathrm{d} x_{\mu}}{\mathrm{d} \tau}=0.$$
An example is the electromagnetic force which simply reads
$$K^{\mu}=\frac{q}{c} F^{\mu \nu} \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}.$$

One of the equations is redundant because of the constraint,
$$\eta_{\mu \nu} \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau} \frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau} = c^2.$$

 

[Herman Trivilino]

so why can't we say that the mass (matter) increases and yes what increases is the energy to make your speed keep increasing?

Mass is not a measure of the amount of matter. Consider a ball of mass $m$ inside a box of much larger mass $M$. The ball is moving around inside the box. The mass of the system is $m+M$ plus the kinetic energy of the ball. If you instead say that the relativistic mass of the ball is $\gamma m$ then the mass of the system is $M+\gamma m$.

 

[DrStupid]

Mass is not a measure of the amount of matter.

That depends on the context. Newton's term for mass was "quantitas materiae".

Consider a ball of mass $m$ inside a box of much larger mass $M$. The ball is moving around inside the box. The mass of the system is $m+M$ plus the kinetic energy of the ball.

I do not think it is that easy. For the case that the box is at rest I get

$m_{system}^2 = \left( {m + M + \frac{{E_{kin} }}{{c^2 }}} \right)^2 - \left( {2 \cdot m + \frac{{E_{kin} }}{{c^2 }}} \right) \cdot \frac{{E_{kin} }}{{c^2 }} = \left( {m + M} \right)^2 + 2 \cdot M \cdot \frac{{E_{kin} }}{{c^2 }}$

 

[Victorifrj]

Actually it's $F = \gamma^3 m a$ for motion in one spatial dimension ($a$, not $v$). But yes, this is a good point, the relation $F = m_r a$, where $m_r = \gamma m$, only works for transverse forces, not longitudinal forces.

yes, that's what I mean. So I can say that the concept of relativistic mass is an arbitrary concept, where I make a simple equality using the gamma(Y). For this term came out of the moment P=mN(x), where N(X) is the velocity as invariant of lorents in one dimension, Nx=Yv(x). With that P=mYv(x), right?

 

[PeterDonis]

Newton's term for mass

Is irrelevant here since we are talking about relativity, not Newtonian mechanics. The definitions of both relativistic mass and invariant mass in relativity are well known, and those are the only concepts of mass that are relevant to this discussion. @Mister T is correct that neither of those concepts correspond to "quantity of matter".

 

[DrStupid]

Is irrelevant here since we are talking about relativity, not Newtonian mechanics.

And what is "amount of matter" in relativity? I know this term from Newtonian mechanics only.

 

[PeterDonis]

what is "amount of matter" in relativity?

There isn't a single concept in relativity that corresponds to this. Some references use something like "number of atoms" or "number of particles", but all of those proposals have issues. The simplest option is to just not use the concept at all; you can do relativity physics perfectly well without it.

 

[DrStupid]

Some references use something like "number of atoms" or "number of particles"

That is amount of substance.

 

[PeterDonis]

That is amount of substance.

Whatever. If you think that's somehow different from "amount of matter", that's up to you. Neither one matches up with either invariant mass or relativistic mass in relativity.

 

[vanhees71]

Is irrelevant here since we are talking about relativity, not Newtonian mechanics. The definitions of both relativistic mass and invariant mass in relativity are well known, and those are the only concepts of mass that are relevant to this discussion. @Mister T is correct that neither of those concepts correspond to "quantity of matter".

But one should keep in mind that what's "mass" in Newtonian physics is the "invariant mass" of special relativity (at least for "point particles"). After all invariant mass is the mass as measured in the (momentaneous) rest frame of the particle, and there the Newtonian approximation applies.

For sure, there's no way to interpret "relativistic mass" as a mass in the sense of Newtonian physics. This is just relativistic energy, where with "relativistic" I mean, choosing the additive constant of the energy such that it gets the time-component of the energy-momentum four-vector, $(E/c,\vec{p})$. For a classical point particle (or a "real" particle, i.e., an asymptotic free state, in relativistic QFT) then you have the on-shell condition $p_{\mu} p^{\mu}=m^2 c^2$ (with $m$ the invariant mass, which thus is consistently a scalar).

 

[PeterDonis]

what's "mass" in Newtonian physics is the "invariant mass" of special relativity (at least for "point particles").

Yes, for "point particles". But the qualifier is crucial. As soon as you have many-particle systems, this equivalence no longer holds, because Newtonian mass is additive and invariant mass is not.

 

[DrStupid]

For sure, there's no way to interpret "relativistic mass" as a mass in the sense of Newtonian physics.

That depends on the meaning of "mass in the sense of Newtonian physics". But you can't explain that without talking about Newtonian mechanics and we have been instructed not to do that in this thread. Thus, this doesn't seem to be the right place for such statements.

 

[pervect]

To be more specific on the point just raised, so-called relativistic mass is not a relativistically invarant quantity. Relativistic mass depends on the frame of reference as can be seen by considering the relativistic mass of an object in a frame in which the object is at rest, and then considering the relativistic mass of the same object in another frame, a frame where the object is moving. If the quantity were frame invariant, it would be the same in both frames, but it is obviously not.

Thus, we must conclude that the relativistic mass cannot meaningfully be interpreted as a relativistically invariant measure of the "amount of material", because of this frame dependence.

 

[DrStupid]

Thus, we must conclude that the relativistic mass cannot meaningfully be interpreted as a relativistically invariant measure of the "amount of material", because of this frame dependence.

Of course a frame dependent property cannot be interpreted as a relativistically invariant measure. But if we cannot discuss what "amount of material" is and wheather is must be relativistically invariant or not this will get us to nowhere. Better keep it out of the discussion.