I "relativistic mass" still a no-no?

[vanhees71]

The SI unit for the "amount of substance" is not kilogram, which is the unit for mass, but mol.

 

[harrylin]

You need to be more specific. Where in these references is the term "quantity of matter" explicitly defined? I'm not looking for your own personal interpretation of what that term means. I'm looking for some explicit definition of it in an acceptable source.

It's not difficult to find Definition no.1 of the Principia - https://en.wikisource.org/wiki/The_Mathematical_Principles_of_Natural_Philosophy_(1846)/Definitions

 

[DrStupid]

You need to be more specific. Where in these references is the term "quantity of matter" explicitly defined? I'm not looking for your own personal interpretation of what that term means. I'm looking for some explicit definition of it in an acceptable source.

According to http://cudl.lib.cam.ac.uk/view/PR-ADV-B-00039-00001/25 quantity of matter $q$ (I don’t use the symbol $m$ in the following derivation in order to avoid confusions with mass) is the product of volume and density:

q = V \cdot \rho

That answers your question for the explicit definition. However, that doesn’t help you to determine whether q depends on velocity or not and if yes how. Without additional information would be at best an unknown function $q \left( q_0 , v \right)$ of the quantity of matter $q_0$ of a body at rest and its velocity $v$. But fortunately there are additional conditions which allow to derive this function:

According to http://cudl.lib.cam.ac.uk/view/PR-ADV-B-00039-00001/26 the momentum is the product of quantity of matter and velocity

p\left( {q_0 ,v} \right) = q\left( {q_0 ,v} \right) \cdot v

In the explanatory note Newton also defined that the momentum of a body is the sum of the momentums of its parts. That includes that the quantity of matter must be additive at least for the special case that all parts have the same velocity

q\left( {q_0 ,v} \right) = \sum\limits_i {q\left( {q_{0,i} ,v} \right)} = \sum\limits_j {q\left( {q_{0,j} ,v} \right)}

That results in

q\left( {q_0 ,v} \right) = q_0 \cdot f\left( v \right)

with

f\left( 0 \right) = 1

The next condition is isotropy. It requires that quantity of matter must be independent from direction. That includes

f\left( { - v} \right) = f\left( v \right)

Now let's continue with the laws of motion. According to the http://cudl.lib.cam.ac.uk/view/PR-ADV-B-00039-00001/46 force is defined as

F := \dot p

With the definition of momentum and the properties of quantity of matter we know so far this means

F = q_0 \cdot f \cdot a + q_0 \cdot \dot f \cdot v = q_0 \cdot f \cdot a + q_0 \cdot \left( {f' \cdot a} \right) \cdot v

In order to keep it simple I will limit the following calculations to the one-dimensional case. With

K\left( v \right) = f\left( v \right) + f'\left( v \right) \cdot v

the equation for force can be simplified to

F\left( v \right) = K\left( v \right) \cdot q_0 \cdot a

with

K\left( 0 \right) = 0
K\left( { - v} \right) = K\left( v \right)

According to the http://cudl.lib.cam.ac.uk/view/PR-ADV-B-00039-00001/49 the forces between two interacting bodies with the quantities of matter $q_1$ and $q_2$ (in their own rest frames) and the velocities v1 and v2 add to zero:

F_1 + F_2 = K\left( {v_1 } \right) \cdot q_1 \cdot a_1 + K\left( {v_2 } \right) \cdot q_2 \cdot a_2 = 0

The principle of relativity requires that everything mentioned above (including the function f) must be identical in all frames of reference. That means

F'_1 + F'_2 = K\left( {v'_1 } \right) \cdot q_1 \cdot a'_1 + K\left( {v'_2 } \right) \cdot q_2 \cdot a'_2 = 0

and therefore

- \frac{{q_1 }}{{q_2 }} = \frac{{K\left( {v_2 } \right) \cdot a_2 }}{{K\left( {v_1 } \right) \cdot a_1 }} = \frac{{K\left( {v'_2 } \right) \cdot a'_2 }}{{K\left( {v'_1 } \right) \cdot a'_1 }}

This is where the transformation comes into play. Let me do the calculation for classical mechanics first:

Galilean transformation results in

v' = v - u
a' = a

and therefore

\frac{{K\left( {v_2 } \right)}}{{K\left( {v_1 } \right)}} = \frac{{K\left( {v_2 - u} \right)}}{{K\left( {v_1 - u} \right)}}

This applies to all cases including

v_1 = 0
v_2 = u = v

That gives

K\left( v \right) = 1

and therefore

q\left( {q_0 ,v} \right) = q_0
p = q_0 \cdot v

That's the expected result for classical mechanics.Now let's go to special relativity:

With c=1 (to make the formulas less ugly) Lorentz transformation results in

v' = \frac{{v - u}}{{1 - u \cdot v}}
a' = a \cdot \left( {\frac{{\sqrt {1 - u^2 } }}{{1 - u \cdot v}}} \right)^3

and therefore

\frac{{K\left( {\frac{{v_1 - u}}{{1 - u \cdot v_1 }}} \right)}}{{K\left( {v_1 } \right) \cdot \left( {1 - u \cdot v_1 } \right)^3 }} = \frac{{K\left( {\frac{{v_2 - u}}{{1 - u \cdot v_2 }}} \right)}}{{K\left( {v_2 } \right) \cdot \left( {1 - u \cdot v_2 } \right)^3 }}

With the special case

v_1 = 0
v_2 = u = v

this turns into

K\left( v \right)^2 = \left( {1 - v^2 } \right)^{ - 3}

The resulting differential equation

f' = \frac{1}{v}\left( {\sqrt {1 - v^2 } ^{ - 3} - f} \right)

has only one physical solution:

f = \sqrt {1 - v^2 } ^{ - 1}

That means

q\left( {q_0 ,v} \right) = \frac{{q_0 }}{{\sqrt {1 - v^2 } }}
p = \frac{{q_0 \cdot v}}{{\sqrt {1 - v^2 } }}

This is the expected result for special relativity.

 

[DrStupid]

Are you using that as a definition of quantity of matter, or as a definition of mass?

I use it as a definition of quantity of matter (see above).

The $\mathrm{SI}$ unit of mass is the kilogram.

And so is the unit of quantity of matter.

It's the amount of substance that is equivalent to the quantity of matter, not the mass.

No, it isn’t. You can easily see that in case of your example with the heated gold brick. If the it moves with constant velocity during heating, its momentum and therefore its quantity of matter will be increased (see definition 2) whereas the amount of substance remains unchanged. Therefore amount of substance cannot be equivalent to quantity of matter. However, quantity of matter can also not be equivalent to mass because changing the speed of the brick at constant mass also changes its quantity of matter (see my calculation above).

 

[jbriggs444]

No, it isn’t. You can easily see that in case of your example with the heated gold brick. If the it moves with constant velocity during heating, its momentum and therefore its quantity of matter will be increased

So, for you, "quantity of matter" is the the m in $p=m \gamma v$. And indeed, if you heat up a gold bar (carefully, using flames that move at the same velocity as the gold bar and leave no soot), it will gain momentum.

 

[DrStupid]

So, for you, "quantity of matter" is the the m in $p=m \gamma v$.

No, according to definition 2 quantity of matter is the $m \gamma$ in $p=m \gamma v$. Of course that makes no difference in classical mechanics due to

\mathop {\lim }\limits_{c \to \infty } \frac{m}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }} = m

 

[Herman Trivilino]

I use it as a definition of quantity of matter (see above).

What then is your definition of mass?

 

[Herman Trivilino]

The SI unit for the "amount of substance" is not kilogram, which is the unit for mass, but mol.

Right. Since we measure mass in kilograms and amount of substance in moles we don't measure amount of substance in kilograms.

I guess I don't understand your point.

 

[DrStupid]

What then is your definition of mass?

Mass can be defined in many different ways. One of the most popular definitions is based on the Minkowski norm of the four-momentum:

m = \frac{{\sqrt {E^2 - p^2 c^2 } }}{{c^2 }}

In order to keep the link to classical mechanics it can also be defined as the quantity of matter (aka relativistic mass) in the rest frame of the system.

 

[PeterDonis]

That answers your question for the explicit definition.

Yes, but now we need an explicit definition of "volume" and "density". Are those explicitly defined in the Principia?

And since both of those things behave differently in relativity than they do in Newtonian mechanics, I don't see how any of this argues against the point I have been making.

 

[Herman Trivilino]

In order to keep the link to classical mechanics it can also be defined as the quantity of matter.

Then your original claim that the two are equal is, as I already told you, a tautology.

 

[DrStupid]

Yes, but now we need an explicit definition of "volume" and "density".

For which purpose?

I don't see how any of this argues against the point I have been making.

You need to be more specific. Which point are you talking about?

 

[DrStupid]

Then your original claim that the two are equal is, as I already told you, a tautology.

Learn to quote correctly.

 

[PeterDonis]

For which purpose?

Because without such explicit definitions, your definition of "quantity of matter" is not explicit either. It just pushes back the implicitness one step, so to speak.

Which point are you talking about?

See post #5 of this thread.

 

[Herman Trivilino]

I don't define "quantity of matter" to be the mass

[mass] can also be defined as the quantity of matter

When one term is used to define another, then a claim that the two terms are equivalent is a mere tautology, devoid of any meaning.

When we're told the quantity of matter contained in a body equals $\gamma m$ it implies that the quantity of matter contained in that body depends on the relative motion of an observer. Hence it changes according to the observer's speed relative to it.

When we're told the quantity of matter contained in a body is $m$ it implies that the quantity of matter depends on the energies of the body's constituents relative to its rest frame. Hence it changes when those energies change.

 

[harrylin]

[..] When we're told the quantity of matter contained in a body equals $\gamma m$ it implies that the quantity of matter contained in that body depends on the relative motion of an observer. Hence it changes according to the observer's speed relative to it.[..]

I'm not sure what you try to argue there, but it's simply wrong, due to a misapplication of the laws of physics. Physically it's not the same if you accelerate or if that body accelerates. The velocity and kinetic energy of a fast particle cannot change due to your relative speed to it - that would be magical action at a distance. Such values are relative in that they depend on your choice of reference system, which does not mean that they can fluctuate as function of your velocity relative to it.

 

[jbriggs444]

Such values are relative in that they depend on your choice of reference system, which does not mean that they can fluctuate as function of your velocity relative to it.

If one chooses to use the term "observer" to refer to a reference system in which that observer is at rest, as it is clear that @Mister T does then the kinetic energy, velocity and momentum of an object change as a function of the observer's velocity relative to that object precisely because those things depend on the choice of reference system.

 

[Herman Trivilino]

Physically it's not the same if you accelerate or if that body accelerates.

Its history is not relevant, however physical it might have been. It makes no difference if at sometime in its past it accelerated.

 

[DrStupid]

Because without such explicit definitions, your definition of "quantity of matter" is not explicit either.

Of course not. I already told you in #26 that I use an implicite definition. What's your point?

See post #5 of this thread.

OK, let me see:

No, it isn't, and that's a big part of the problem. In Newtonian physics, the concept of "mass" conflated several different things that, in relativistic physics, turn out to be different. One of those things (roughly, "quantity of matter") turns out to correspond with $m$; another (roughly, "amount of inertia") turns out to correspond with $\gamma m$ (with some caveats, since the relationship between force and acceleration is direction-dependent in SR).

My calculation shows that "quantity of matter" also corresponds to $\gamma m$ and the resulting relationship between force and acceleration is also direction-dependent in SR.

And yet a third (gravitational mass) turns out to correspond with neither, since in GR the source of gravity is not "mass" but the stress-energy tensor.

That's not a problem of quantity of matter but of the law of gravitation.

 

[DrStupid]

When one term [quantity of matter] is used to define another [mass], then a claim that the two terms are equivalent is a mere tautology, devoid of any meaning.

I don't claim that the two terms are equivalent and I do not use one of them to define the other in the way that your truncated quote suggests. This is what I actually wrote:

And it [relativistic mass] is identical with Newton's quantity of matter.

I don't define "quantity of matter" to be the mass and that wouldn't be true.

In order to keep the link to classical mechanics it [mass] can also be defined as the quantity of matter (aka relativistic mass) in the rest frame of the system.

 

[jbriggs444]

My calculation shows that "quantity of matter" also corresponds to $\gamma m$

You cannot successfully calculate "quantity of matter" without having a definition for "quantity of matter". So... what is your definition for "quantity of matter"?

 

[DrStupid]

You cannot successfully calculate "quantity of matter" without having a definition for "quantity of matter".

I actually did it in #33.

 

[jbriggs444]

I actually did it in #33.

Without a definition for density, that's a little pointless, don't you think?

 

[PeterDonis]

I already told you in #26 that I use an implicite definition.

And then in #29 I asked you for an explicit one, and in #33 you gave a definition that you claimed satisfied my requirement for an explicit definition. But that's not the case unless you can also give an explicit definition of "volume" (Newton used the term "bulk") and "density". Which you haven't.

 

[DrStupid]

Without a definition for density, that's a little pointless, don't you think?

Did you even read #33?

 

[jbriggs444]

Did you even read #33?

Yes, I did. The relevant definition for "quantity of matter" was as the product of volume and density. The rest of the post went on without providing any definitions to ground either of those terms.

 

[DrStupid]

The relevant definition for "quantity of matter" was as the product of volume and density.

What makes this definition relevant? I didn't used it in my calculation.

 

[DrStupid]

And then in #29 I asked you for an explicit one, and in #33 you gave a definition that you claimed satisfied my requirement for an explicit definition. But that's not the case unless you can also give an explicit definition of "volume" (Newton used the term "bulk") and "density". Which you haven't.

I provided you with Newton's explicit definition for quantity of matter. It's not my problem if it doesn't satisfied your requirement. I don't need it.

 

[jbriggs444]

What makes this definition relevant? I didn't used it in my calculation.

It seems that you want to work backwards. You are taking momentum as primitive and using the assertion that $p=q \cdot v$ as the defining property for quantity of matter q.

That's fine, but if you are going to do that, it would be good to discard the other definition (or be clear that you are interpreting Newton to be defining density in terms of momentum).

 

[DrStupid]

That's fine, but if you are going to do that, it would be good to discard the other definition (or be clear that you are interpreting Newton to be defining density in terms of momentum).

1. Post #33 was an answer to PeterDonis who asked me for an explicite definition for quantity of matter. I did him this favour but explained why this definition is neither helpful nor required to derive the properties of quantity of matter. For this purpose I needed to put it toghether with my calculation into the same post.

2. There is no reason to discard definition 1. It might not helful for this topic but that doesn't make it wrong. I already wrote in #26 that we today use it to define density.