Distribution (Dirac&standard) formulations of f=ma how do they go again?

[James MC]

This is not well. First, the formula you assume in invalid, because the gamma factor has to be differentiated too. The correct equation is

$$
\mathbf F = \sum_i m_i \frac{d}{dt}(\gamma_i \mathbf v_i).
$$

Huh? But I see this all the time. Take the proof in the wikipedia article. Do you think there is something wrong with the wikipedia example? Surely I only need to separate things into the part parallel to the velocity and the part perpendicular to it. My example just used the perpendicular component for simplicity.

Second, the same acceleration of the parts is very special situation in both nonrel. and rel. mechanics, which almost never happens. You would derive the additivity only for such special situation.

I don't think that's a special situation because the acceleration of a composite is not well defined if its parts have distinct accelerations. And if an object's acceleration is not well defined, then neither can its mass be, because mass is defined in terms of it. Of course, any such composite may still have a mass in the following sense: it resists acceleration given applied forces when its acceleration is well defined. That means that for any composite, we can define its mass by considering counterfactual scenarios in which its components have identical accelerations, and then running my argument. So there is no composite that my argument does not apply to.

Yes, this was my point. I think that the derivation in rel. mechanics will be necessarily different and more difficult than in nonrel. mechanics, because in relativity, the interactions between the parts are not instantaneous anymore and this has consequence on the mass and motion of the composite.

I am afraid that you won't be able to find explanation which will be more satisfactory to you and correct, because physics is not so tight axiomatic system like mathematics. In my view, the additivity of mass in nonrel. physics is primarily a generalization of daily experience with weighing, and this later was vindicated in the form of motion of COM as implied by Newton's laws. In this way, I consider it very satisfactory and I would not attempt to derive it from anything more simpler. Because, what is more simple?

But if you think there is something to discover, keep trying, who knows, perhaps it is possible.

I guess I'll have to wait and see what you think of the wikipedia proof, that derives a form of the law that separates out gamma from acceleration.

 

[Jano L.]

I think you take very special situation:

1) accelerations are all the same
2) they are all in the direction of velocities

From this it follows that all particles have velocities in the same direction. This is just too special for composite body.

But if you insist on this situation, take the particular one in which all particles have the same velocity. Your formula gives inertial (relativistic) mass, while the original question is how the rest masses of the particles form the rest mass of the composite. In this case, the rest mass is just sum of rest masses

$$
M = \sum_i m_i.
$$

EDIT: Disregard, I misunderstood. I'll write the answer in while.

 

[Jano L.]

OK, you take this situation:

1) accelerations are all the same
2) they are all perpendicular to velocities

From this it follows that all velocities lie in a plane. This is the only case when the textbook procedure works in rel. mechanics, but you in fact derive inertial(relativistic) mass.

But the basic question is how the rest masses compose to form rest mass of the composite. If all velocities are the same, the rest mass is just sum of individual masses. Your formula gives something different.

 

[James MC]

This is a good point, but I think I in fact derive both the composite relativistic mass and the composite invariant mass.

Recall the reason why I rejected the energy/momentum based explanation of composite invariant mass. That's because the "M" that was derived, had no obvious connection to our concept of mass. I recall you agreed with me on this point.

The concept of mass is best characterised as "that property that explains the object's disposition to resist changes in motion given applied forces". According to the equation I appeal to, relativistic mass best fits this definition for elementary particles, because m*gamma is the proportionality constant between force and acceleration. And as you rightly note, I end up inferring the composite relativistic mass, because I show that the sum of the relativistic masses of the parts is the proportionality constant between composite force and composite acceleration.

However, when we are intentionally using theoretical vocabulary to pick out the most natural, fundamental properties of nature, we must add a clause to our definitions if our definitions are to capture our actual use of the vocabulary. There is therefore another notion of 'mass' that is defined as "the most natural/fundamental property that is involved in explaining resistance to changes in motion given applied forces". This definition does not pick out relativistic mass because relativistic mass is a frame variant (and hence non-natural/fundamental) property. For elementary particles, what therefore fits this definition is the property that plays the most substantive role in explaining the inertia of the particle, that is frame invariant. This is the invariant mass, which is equal to the relativistic mass when momentum = 0 (I think that's right). In that case, when we infer the relativistic mass of the composite, we simply infer its invariant mass by moving to the frame in which the composite has zero total momentum, and seeing what the real invariant mass is.

So the proof infers both composite relativistic mass and composite invariant mass (as it presumably should). I also think it illuminates why two notions of 'mass' arose in the first place.

 

[Jano L.]

In that case, when we infer the relativistic mass of the composite, we simply infer its invariant mass by moving to the frame in which the composite has zero total momentum, and seeing what the real invariant mass is.

How do you "see what the real invariant mass is"? Your procedure of finding inertial mass depends on the assumption that accelerations are perpendicular to velocities. If this is so in the original reference frame, it need not be so in the frame where total momentum is zero. Imagine swarm of electrons with equal velocities perpendicular to a uniform magnetic field. In this frame, forces and accelerations are perpendicular to velocities, but in the frame of the electrons, this is no longer true, since the speed of the particles change.

Try to think of a more general procedure, where velocities and accelerations can be arbitrary.

 

[James MC]

How do you "see what the real invariant mass is"?

By looking to the microphysical resources that the explanation appeals to and finding what fits the definition of 'real mass'.

By definition, we use 'mass' to refer to the property that plays the inertial role (that is, the property that (best) explains object's resistance to changes in motion given applied forces). We find that "relativistic mass" plays the inertial role for particles.

The frame-variant (and hence unreal) nature of "relativisitic mass" provokes us to introduce a notion of 'mass' that picks out the real property of an object that is responsible for inertia - 'real mass'.

For particles, we find this property by moving to the frame in which v=o (or $\gamma$=1). This is motivated by the need to cancel out the frame-variant (unreal) contributions to inertia, to lock on to the frame-invariant (real) contribution.

Once we have inferred the composite relativistic mass, we know that we have inferred both the real and unreal contributions to the composite's inertia. We also know the procedure by which we lock on to the real aspect - it's the procedure invoked by the microphysics. We just apply that same procedure to the composite.

Your procedure of finding inertial mass depends on the assumption that accelerations are perpendicular to velocities. If this is so in the original reference frame, it need not be so in the frame where total momentum is zero. Imagine swarm of electrons with equal velocities perpendicular to a uniform magnetic field. In this frame, forces and accelerations are perpendicular to velocities, but in the frame of the electrons, this is no longer true, since the speed of the particles change.

I don't understand why you think my procedure depends on the assumption that accelerations are perpendicular to velocities. But let's assume that it does. Even so, I don't see why it matters that the assumption doesn't hold in the frame in which total momentum is zero.

We start with a frame in which forces and accelerations are perpendicular to velocities, and we infer composite relativistic mass. We then infer composite invariant mass just by a procedure that allows us to distinguish the real component responsible for the inertia, from the unreal component that was so apparent in the original frame. Why should it matter that this procedure yields a frame in which the original assumption does not hold? I think it doesn't matter because once we move to the centre of momentum frame, we don't need to reapply the original procedure (swapping the sum to the rhs), because by that stage we are done: we've reached our explanandum already.

Try to think of a more general procedure, where velocities and accelerations can be arbitrary.

I suspect there might be two reasons why you think my procedure is not completely general.

One is that you think my procedure depends on the assumption that accelerations are perpendicular to velocities. But this doesn't seem right: for simplicity I demonstrated the procedure using the equation for when the relationship is perpendicular. But I could have also used the equation for when the relationship is parallel (i.e. when gamma is cubed). And I could have also used the general equation that sums these contributions: the general equation I referenced from wiki. I don't see that I need to make any such assumptions if I run the proof using that general equation.

Another reason might be that you want to see the proof apply to composites whose parts have arbitrary accelerations. But my proof in fact applies to all of them! Take a composite whose parts have distinct accelerations. If we want to know the composite's mass then we need to determine the extent to which it is disposed to resist acceleration given a force. We can only do that when the composite has an acceleration. But it only has an acceleration when the acceleration of its parts are identical. In that case, take the target composite and revise the accelerations of its parts until the composite has an acceleration and then apply my procedure to it. That's how my argument applies to all possible composites.

Have I understood why you think my procedure is not completely general?

 

[Jano L.]

But I could have also used the equation for when the relationship is parallel (i.e. when gamma is cubed). And I could have also used the general equation that sums these contributions: the general equation I referenced from wiki.

Can you please show how? I doubt that what you suggest will work, since the acceleration cannot be pulled in front of the expression in general.

If we want to know the composite's mass then we need to determine the extent to which it is disposed to resist acceleration given a force. We can only do that when the composite has an acceleration. But it only has an acceleration when the acceleration of its parts are identical.

Here you neglect the ordinary meaning of the word mass. According to your above idea, rotating body does not have mass, while physics assigns mass to a body irrespective of its motion.

 

[James MC]

Can you please show how? I doubt that what you suggest will work, since the acceleration cannot be pulled in front of the expression in general.

Sure; if one is thinking of the equation for total force in the following terms:
$$
\mathbf F_T = \sum_i (\gamma^{3}_i m_i \mathbf a_i^{||} + \gamma_i m_i \mathbf a_i^{τ}),
$$
...then it might be hard to see how one can take the acceleration outside the scope of the sum. But notice that this is equivalent to:
$$
\mathbf F_T = \sum_i (\gamma^{3}_i m_i \mathbf a_i^{||}) + \sum_i(\gamma_i m_i \mathbf a_i^{τ}),
$$
Now we can run the argument in full generality: We want composite mass. Mass is disposition to resist acceleration given force. So we focus on the only situation in which composite has acceleration: when its parts have the same acceleration. Then we have:
$$
\mathbf F_T = \mathbf a_C^{||}\sum_i (\gamma^{3}_i m_i) + \mathbf a_C^{τ}\sum_i(\gamma_i m_i),
$$
We derive composite force as total force of parts:
$$
\mathbf F_C = \mathbf a_C^{||}\sum_i (\gamma^{3}_i m_i) + \mathbf a_C^{τ}\sum_i(\gamma_i m_i),
$$
From this we derive:
(i) $\sum_i (\gamma^{3}_i m_i) + \sum_i (\gamma_i m_i)$ plays the composite's inertial role, and is "composite mass".
(ii) This composite mass is not fully real (due to frame invariance) motivating the introduction of "real mass", a term defined to apply to the real aspect.
(iii) Procedure for isolating real aspect of composite mass = procedure for isolating real aspect of particle mass: setting v=0, $\gamma$=1, i.e. finding the rest frame. Hence, composite (real) mass = sum of relativistic masses of parts (i.e. sum of sum of longitudinal and transverse masses of the parts) as defined in frame where $\mathbf p$=0.

Here you neglect the ordinary meaning of the word mass. According to your above idea, rotating body does not have mass, while physics assigns mass to a body irrespective of its motion.

I don't think that follows from my idea. The ordinary meaning of the word 'mass' treats mass as a disposition, and in particular a disposition to resist acceleration when it has an acceleration. You've chosen a situation where the disposition cannot manifest due to no well defined acceleration. But situations in which a disposition cannot be made manifest are not situations in which the disposition does not exist. Hence, the rotating body has a mass because it has a disposition (always) to resist acceleration. It still has this disposition in situations when it has no acceleration. I always have the disposition to eat breakfast in the morning - I have that disposition even when it is not morning.

Notice that a version of your argument could be used against the textbook argument for classical additivity: "according to the textbook idea, body with no force on it does not have mass, while physics assigns mass to a body irrespective of whether it is a closed system". Textbook response: we consider situations in which the object actually has a force on it, and simply apply that result to the case of a closed system.

Thus, the rotating body has inertial mass in virtue of how it behaves when we try to accelerate the body. Am I making sense here?

 

[Jano L.]

From this we derive:
(i) ∑i(γ3imi)+∑i(γimi) plays the composite's inertial role, and is "composite mass".

This is wrong. There is no sense in adding these two masses to find "total mass". It would give you twice the correct mass if the body is at rest. In the frame of thought of pre-relativistic physics there is no such thing as one total mass, and this was the reason for introduction of those two separate masses.

After the discovery of relativity, the force is thought to be defined as
$$
\mathbf F = \frac{d\mathbf p}{dt}
$$

where

$$
\mathbf p = \gamma m_0\mathbf v.~~~(*)
$$

Today, the quantity $\gamma m_0$ is called the inertial mass, which is what you want to derive. Its value is independent of the direction of the force, and it is not sum of the above two masses.

The ordinary meaning of the word 'mass' treats mass as a disposition, and in particular a disposition to resist acceleration when it has an acceleration.

OK, but composite bodies can be assigned acceleration even if their parts do not have the same acceleration. It suffices that the system remains compact. Then some COM can be defined and used to find $m_0$ according to (*).

 

[James MC]

OK, but composite bodies can be assigned acceleration even if their parts do not have the same acceleration. It suffices that the system remains compact. Then some COM can be defined and used to find $m_0$ according to (*).

I don't think this is quite right. At best, such "compact" composites can be assigned a vague-acceleration $a_v$ or a pragmatic acceleration $a_p$, but not a real acceleration $a$. The mass of an object is not defined in terms of a disposition to resist $a_v$ (as this would only yield a "vague mass"), nor is it defined in terms of a disposition to resist $a_p$ (as this would only yield a "pragmatic mass"). So situations in which a composite has no acceleration, but only an $a_v$ or an $a_c$ are not situations that the derivation of composite mass needs to cater for, it seems to me.

My procedure for deriving (Newtonian) mass additivity assumes mass is defined solely in terms of a disposition to resist (precise) acceleration and that (precise) composite acceleration only occurs when the composite's parts have identical accelerations. Yet my procedure yields masses for all composite objects no matter the acceleration of their parts (recall the counterfactual procedure). Now, since one knows the masses and forces of all composites by the procedure, one can then ask: is there a point in the vicinity of the composite that can be attributed with a composite's mass and force such that tracking its evolution enables one to track the composite's evolution using f=ma? This is how COM is arrived at. Hence, my procedure not only explains mass additivity, but justifies the introduction of COM.

This is wrong. There is no sense in adding these two masses to find "total mass". It would give you twice the correct mass if the body is at rest. In the frame of thought of pre-relativistic physics there is no such thing as one total mass, and this was the reason for introduction of those two separate masses.

After the discovery of relativity, the force is thought to be defined as
$$
\mathbf F = \frac{d\mathbf p}{dt}
$$

where

$$
\mathbf p = \gamma m_0\mathbf v.~~~(*)
$$

Today, the quantity $\gamma m_0$ is called the inertial mass, which is what you want to derive. Its value is independent of the direction of the force, and it is not sum of the above two masses.

I don't understand. In SR, the expression relating force and acceleration for a point-mass m moving in the x direction is:
$$
\mathbf F_x = m\gamma^3\mathbf a_x
$$
$$
\mathbf F_y = m\gamma\mathbf a_y
$$
$$
\mathbf F_z = m\gamma\mathbf a_z
$$
I take it this can be rewritten as:
$$
\mathbf F_{x+y+z} = m\gamma^3\mathbf a_x + m\gamma\mathbf a_{y+z}
$$
...such that this sum yields the total force. I take it that because length contraction is in the direction of motion (here, x), there is an inertial mass in the direction of movement and a different inertial mass in any direction transverse to this motion. That is, acceleration would not be strictly proportional to force, but would vary by direction relative to the particle's motion. (Somehow cubing the x-direction gamma, captures all that!)

Now, you refer to these equations as "pre-relativistic". What does that mean? Are you saying they are false? I thought they were both true (in SR) and relativistic in the sense that the cube captures length contraction - a relativisitic phenomenon. Then you say "Today, the quantity $\gamma m_0$ is called the inertial mass". Here's my question: if "inertial mass" just means "disposition to resist acceleration given applied forces" then how can you say that $\gamma m_0$ is the inertial mass if you don't think it can enter into any equation as the proportionality constant between force and acceleration?

 

[Jano L.]

Longitudinal and transverse mass were introduced before the discovery of relativity, and are motivated by the equation F=ma. They are alright in relativity. It is just that I do not see any reason for mentioning them here, since they do not help in any way to derive composite mass.

In relativity one can work with equation $F = \frac{d}{dt}(m \mathbf v)$, where $m$ is the inertial mass, which is much simpler concept. As long as some center of mass is adopted, its acceleration is unambiguous. Then there is no problem with it.

Your requirement that mass of a body refers to situation when all parts of this body have the same acceleration is unreasonable and devoid of practical use. According to standard views, bodies almost never move in such a way.

I think that what you are trying to describe is not the common concept of mass, but instead your own concept. It is OK to have eccentric views on anything, but the rules of PF do not permit us to discuss personal theories here, so if you want to continue just with your personal concept, I think I'll finish.

 

[MikeGomez]

it only has an acceleration when the acceleration of its parts are identical.

That’s not true. In fact, almost nothing ever does accelerate with the acceleration of its parts being identical.

only rigid bodies can be said to have mass in the first place!

Seriously? That seems to be a fantastic discovery of yours which will certainly make headlines around the world.

Here is the conclusion I have come to: (inertial) mass is defined in terms of the "most natural" property responsible for resistance to changes in motion given applied forces.

It might help if you didn’t think of it that way. You are associating an active attribute to mass, and in reality it is passive, at least in the context of this discussion.

Mass doesn’t “resist” anything.

In fact, it’s a pretty convenient property of mass that it does exactly as we request of it in the form of response to applied forces.

Let’s say (from space) I throw a rock, meaning I apply a force to it, meaning I gave it some momentum.

The rock will take the momentum I have given it, and use it as an instruction to travel in that direction forever.

No resistance! It obeys what is asked of it.

If I wish it to stop, I command it to do so with an applied equal force in the opposite direct. It obeys my command and remain motionless forever (until acted on by another force).

If I wish it to go faster instead of stopping I apply more force in the direction it is heading. It always does what I wish. Never any resistance.

So forget about inertia. The concept to me seems to do more harm than good. Think only of momentum and you will be fine.

 

[James MC]

Your requirement that mass of a body refers to situation when all parts of this body have the same acceleration is unreasonable and devoid of practical use. According to standard views, bodies almost never move in such a way.

That is not my requirement. I have stated on several occasions that Newtonian composites always have mass no matter their arrangement. In fact, in my very last post I said "my procedure yields masses for all composite objects no matter the acceleration of their parts".

My requirement is just this: if you want to know the exact inertial mass of a Newtonian composite, using the method of applying forces and measuring resulting accelerations, then you require an exact acceleration for the composite i.e. exact acceleration of parts.

That is not an unreasonable requirement, it's just how things are. As regards practical use, it is useful because it makes sense of what our real life approximate measurements of composite mass, are approximating in the first place! (It's perhaps worth reiterating: exact COM acceleration is no help here for the usual reasons - can't know what COM is without first knowing composite mass.)

As equivocation upon three separate notions of acceleration is causing mess here, I reintroduce subscripts. I'll change "v" (for vague) into "a" (for approximate), and introduce "e" for exact (so that new posters don't take my claims out of context). Thus, a composite has an $a_e$ when its parts have identical accelerations, it has an $a_a$ when its parts are compact enough to be approximated as having some exact acceleration value, and it has a pragmatic acceleration $a_p$ when there is some way of tracking the composite by applying Newton's laws to some point and pretending (for pragmatic purposes) that the acceleration of that point is the acceleration of the composite.

As regards practical use, we can determine the mass of a composite by applying forces and measuring resulting $a_a$'s. We can then say that we know the mass of the composite with a degree of uncertainty that is proportional to the degree to which the resulting $a_a$'s approximate the acceleration of a rigid body. (Of course our degree of uncertainly will also be a function of many other things too.) So as regards practical use, it gives a rationale for the entire enterprise of determining composite Newtonian mass using applied forces.

I think that what you are trying to describe is not the common concept of mass, but instead your own concept. It is OK to have eccentric views on anything, but the rules of PF do not permit us to discuss personal theories here, so if you want to continue just with your personal concept, I think I'll finish.

As I'm getting a lot out of this discussion, that would be a shame, but you should certainly feel free to finish at any stage. But to say that I'm not describing the common concept of mass is quite extraodinary. All I'm doing is taking the completely standard definition, being absolutely clear about the definition's reference to 'acceleration' with a technique called "disambiguation", and then working out the consequences. As Eugene Hecht writes in his paper On Defining Mass:

"A sophisticated definition appearing in countless textbooks and classrooms is based on the idea of inertia. The mass of an object is a measure of, and gives rise to, its resistance to changes in motion;
F=ma, which stands on a rich experimental history, presumably quantifies the traditional idea of “inertia”.” (2011: 40)

Longitudinal and transverse mass were introduced before the discovery of relativity, and are motivated by the equation F=ma. They are alright in relativity. It is just that I do not see any reason for mentioning them here, since they do not help in any way to derive composite mass.

In relativity one can work with equation $F = \frac{d}{dt}(m \mathbf v)$, where $m$ is the inertial mass, which is much simpler concept. As long as some center of mass is adopted, its acceleration is unambiguous. Then there is no problem with it.

Well, it was you that brought them up. I had an argument for Newtonian mass additivity ("the distributions argument") and you showed that it didn't work because it wrongly extends to the relativistic equation for transverse mass. I then realized a different approach ("the counterfactual argument") that not only proves Newtonian mass additivity, but enables us to derive both relativistic mass and rest mass, using that relativistic equation. You then pointed out that the 'mass' in the transverse mass equation only equals the mass-of-interest, in the specific situation in which (1) accelerations are all the same and (2) they are all perpendicular to velocities. So I tried to extend the argument by appeal to the equation for both longitudinal and transverse mass. You responded by saying that you cannot in general relate the 'm' in these equations to the 'm' that is used in modern relativisitic equations. And you appeared to suggest that there is simply no way of formulating a general equation in SR relating force and acceleration. That's where we are now.

Incidently, Hecht argues that because of this, we need to completely rethink our concept of mass, and give up on a definition in terms of inertia. Apparently, you don't agree, because you still appear to want to think of relativistic mass as inertial mass. So my question was: if mass (relativistic, rest or whatever) cannot, in SR, be quantified in an equation that relates it to force and acceleration, then why do you still think there is such a thing as inertial mass in SR?

 

[Jano L.]

My requirement is just this: if you want to know the exact inertial mass of a Newtonian composite, using the method of applying forces and measuring resulting accelerations, then you require an exact acceleration for the composite i.e. exact acceleration of parts.

That is the same I meant above.


In Newtonian mechanics, the motion of parts is of no consequence to the inertial mass, it is always additive. Your procedure will give you correct mass, but the restriction to equal accelerations is pointless.

In relativistic mechanics, the motion of the parts is of consequence to the inertial mass. In other words, the inertial mass is a function of the motion of the parts. You cannot find it if you tamper with the accelerations.

Well, it was you that brought them up. I had an argument for Newtonian mass additivity ("the distributions argument") and you showed that it didn't work because it wrongly extends to the relativistic equation for transverse mass.

I think you are mistaken. Please read my posts #26,#30 again. I did not introduce transverse mass, you did in post #29.

And you appeared to suggest that there is simply no way of formulating a general equation in SR relating force and acceleration. That's where we are now.

Incidently, Hecht argues that because of this, we need to completely rethink our concept of mass, and give up on a definition in terms of inertia. Apparently, you don't agree, because you still appear to want to think of relativistic mass as inertial mass. So my question was: if mass (relativistic, rest or whatever) cannot, in SR, be quantified in an equation that relates it to force and acceleration, then why do you still think there is such a thing as inertial mass in SR?

I do not know why, but you misunderstood me heavily. In fact, for a compact body, COM and its velocity $\mathbf v_c$ can be defined and based on the equation of its motion

$$
\sum_i \mathbf F_i = \frac{d}{dt}\big( m_C \mathbf v_C \big),
$$

where $\mathbf F_i$ are external forces, the inertial mass of the body can be defined. The procedure is just more complicated in relativity since the inter-particle forces do not cancel and have to be considered (they influence the motion of the COM).

The important point is that this mass depends on the motion of the parts, as expected from the Einstein relation $\mathscr{E}_C = m_Cc^2$. You cannot find it by making the parts move in the same way, because that is different situation which will lead to different mass.

 

[James MC]

Sorry for the delay, just finished moving house...

In Newtonian mechanics, the motion of parts is of no consequence to the inertial mass, it is always additive.

Yes, in NM inertial mass is not even partially determined by motions of parts. But that doesn't mean that motions of parts can't be used to obtain knowledge of inertial mass. It's important to keep the physics separate from the epistemology here. Physically, motion of parts is partially determined by mass, epistemically we can look to motions of parts (together with forces) to know the mass.

Your procedure will give you correct mass, but the restriction to equal accelerations is pointless.

The sentence fragment before the comma appears to contradict the sentence fragment after the comma: if the procedure that involves the "restriction" to equal accelerations gives the correct mass, then the restriction cannot be pointless. After all, getting the correct mass was the goal.

One objection one might have to my approach for explaining NM composite mass, is that my procedure doesn't rule out the possibility, that when accelerations of parts differ, mass additivity fails. But I'm not entirely sure if this is an objection you have in mind.

In relativistic mechanics, the motion of the parts is of consequence to the inertial mass. In other words, the inertial mass is a function of the motion of the parts. You cannot find it if you tamper with the accelerations.

I assume that by "inertial mass" you mean "relativistic mass". Then this is right: to find it you must also tamper with velocities.

I think you are mistaken. Please read my posts #26,#30 again. I did not introduce transverse mass, you did in post #29.

I thought you introduced the equation for transverse mass in post #24. That's just because you did not add the extra expression where gamma is cubed (I first mentioned the full expression in post #22).
In your post #26, you say "there is nothing restricting it to perpendicular force - why would you think that."
In my post #27 I responded by saying that to account for parallel force, gamma must be cubed. You did not respond to this in your post #28, so I assumed I made no mistake. In my post #29 I reintroduced your equations from your #24. In #30 you rightly note that my "reintroduction" of your #24 equation isn't identical because I didn't differentiate the gamma factor, and you state that my equation is invalid. But then in my #31 I reference wiki to show that my equation is not invalid, and that the gamma factor is not to be differentiated. Your next posts #32,33 did not respond to this, but just went with my equations. For this reason I thought you agreed that gamma is not to be differentiated.

So you're right that your original equation is different than the one I attributed to you, due to the difference in differentiating the gamma factor. But are you now saying that (i) the equation that does not differentiate gamma is correct and (ii) the equation that does differentiate gamma is also correct? And are you trying to say that (i) applies to transverse mass while (ii) applies to invariant mass in general? I don't understand how you can move from the equation from transverse mass to a general equation for (invariant?) mass simply by differentiating gamma.

I do not know why, but you misunderstood me heavily.

It's because you are asserting that there is such a thing as inertial mass in relativity, but you are defending this assertion by appeal to an equation that does not obviously quantify inertia. I need you to say why your equation is relevant to inertia, otherwise, there's no reason to think your equation concerns inertial mass. Wiki defines inertia in terms of resistance to a change in state of motion (persumably, given a force). An equation that quantifies this is an equation relating mass to (i) force and (ii) change in state of motion, i.e. acceleration. Your equation does not quantify that, it seems to me, because you differentiate the mass as well as the velocity. That is why I don't understand.

In fact, for a compact body, COM and its velocity $\mathbf v_c$ can be defined and based on the equation of its motion

$$
\sum_i \mathbf F_i = \frac{d}{dt}\big( m_C \mathbf v_C \big),
$$

where $\mathbf F_i$ are external forces, the inertial mass of the body can be defined.

What is the notion of inertia that you are working with that allow you to use this equation to define inertial mass?

 

[Jano L.]

I thought you introduced the equation for transverse mass in post #24. That's just because you did not add the extra expression where gamma is cubed (I first mentioned the full expression in post #22).

No extra expression should be added. The equation
$$
\mathbf F = \frac{d}{dt}(m \mathbf v)
$$


is the same as the equation

$$
\mathbf F = m_0\gamma^3 \mathbf a_{parallel} + m_0\gamma \mathbf a_{perpendicular},
$$

since $m = \gamma m_0$.

My point is that there is no reason in using the second form instead of the first, when defining inertial mass. This is defined well by the quantity $m$ in

$$
\mathbf F = \frac{d}{dt}(m \mathbf v)
$$
in terms of known force and velocity.

 

[James MC]

No extra expression should be added. The equation
$$
\mathbf F = \frac{d}{dt}(m \mathbf v)
$$
is the same as the equation
$$
\mathbf F = m_0\gamma^3 \mathbf a_{parallel} + m_0\gamma \mathbf a_{perpendicular},
$$
since $m = \gamma m_0$.

Since this makes quite a difference, I would quite like to understand why these equations are equivalent; but right now I'm not seeing it, and I'm not sure what I'm missing.

If you're right that both these equations quantify inertia, then it is essential that both equations capture the crucial phenomenon we have been discussing: if an object is traveling fast in the x direction, then applying a force in the x direction yields a much greater resistance to acceleration than applying a force in a direction perpendicular to the velocity. I can (sort of) see how the second equation captures this phenomenon: by distinguishing these directions and cubing gamma for the direction with which the object has more inertia. I cannot see how the first equation captures this phenomenon.

My point is that there is no reason in using the second form instead of the first, when defining inertial mass. This is defined well by the quantity $m$ in

$$
\mathbf F = \frac{d}{dt}(m \mathbf v)
$$
in terms of known force and velocity.

To quantify inertia this equation must entail the above phenomenon. Can you derive the direction-relative inertial behaviour from this equation? If not, I don't see how it quantifies inertia.

Incidently, if you can show that this equation can capture the phenomena it needs to capture to quantify inertia, then you've got a powerful objection to Eugene Hecht's recent publication, which uses the second equation and the phenomena I mention to argue that although mass and inertia are connected concepts, "there is no simple relationship that informs a straightforward definition of mass" (p40).

 

[Jano L.]

Since this makes quite a difference, I would quite like to understand why these equations are equivalent; but right now I'm not seeing it, and I'm not sure what I'm missing.

This is quite a simple issue that is explained in some books on special theory of relativity. For example, see the book by W. Rindler, Introduction to Special Relativity, sec. 35.

you've got a powerful objection to Eugene Hecht's recent publication, which uses the second equation and the phenomena I mention to argue that although mass and inertia are connected concepts, "there is no simple relationship that informs a straightforward definition of mass" (p40).

I do not think Hecht's writings will help you to understand the concept of mass. He is too immersed in fighting for recognition of his point of view on purely semantical issue.

I advise you to read textbooks on mechanics and relativity instead, and original papers. From your posts I think some reading on mechanics and relativity will be more helpful to you than long discussions here.

 

[James MC]

Thanks, the reference is useful.

My questions were:

Q(1) Why is finding the coefficient between momentum and velocity tantamount to quantifying what's crucial to our pre-theoretical concept of mass ("the more mass an object has the harder it is to push around in space)?

Q(2) Why are F=MA and F=dp/dt equivalent in Newtonian mechanics, but not in SR?

Q(3) How can it be that just by moving to the F=dp/dt formulation in SR, we no longer need to explicitly distinguish equations for force parallel to velocity and force perpendicular to velocity? Can the direction relative properties of inertia be derived from F=dp/dt?

I've so far come up with answers to Q(1) and Q(2):

A(1): Because if mass is constant then F=MA and F=dp/dt equivalent, so if mass is constant then finding the coefficient between momentum and velocity is equivalent to finding the coefficient between force and acceleration, and the coefficient between force and acceleration quantifies what's crucial to our pre-theoretical concept of mass. If mass is not constant then finding the coefficient between momentum and velocity is like a default plan B.

A(2): Rindler (p102) says that in SR, the equations are equivalent only if the rest mass is constant. But rest mass is often not constant in SR. For example, if two particles collide elastically, their rest masses during collision will vary. In such a case, the force on one of the particles cannot be the product of its (new!) mass and acceleration, because the force must also be a function of the change in mass. The second equation captures this by keeping mass within the time derivative.

But I still don't know how to answer (3), and Rindler does not appear to directly answer the question. There is some relevant discussion going from p102 to 103 but it is hard to follow. How could one possibly work out that inertia is relative to direction of velocity, from an equation that just states that the force on an object is equivalent to the time derivative of the product of that object's mass, velocity, and gamma factor?

 

[Jano L.]

Q(3) How can it be that just by moving to the F=dp/dt formulation in SR, we no longer need to explicitly distinguish equations for force parallel to velocity and force perpendicular to velocity?

F = dp/dt is just a shorthand form of three separate equations containing different component of force. Parallel and perpendicular components of force are just one among many equally valid ways to write these three equations.

Can the direction relative properties of inertia be derived from F=dp/dt?

What do you mean by "direction relative properties of inertia L"? If you mean by it the ratio of force and acceleration in the same direction, then yes, this follows from that equation.

 

[James MC]

F = dp/dt is just a shorthand form of three separate equations containing different component of force. Parallel and perpendicular components of force are just one among many equally valid ways to write these three equations.

What do you mean by "direction relative properties of inertia L"? If you mean by it the ratio of force and acceleration in the same direction, then yes, this follows from that equation.

That is what I mean. I just don't understand how it follows.

If F=dp/dt is just shorthand for three separate equations for each component of force, then F=dp/dt entails that there is nothing special about one particular component. [ii] But there is something special about the component that is parallel to the velocity: there is more inertia in that direction.

This argument against F=dp/dt contains only two premises, and [ii], and is clearly valid. So either or [ii] (or both) is false?

 

[Jano L.]

F - dp/dt give no component a distinct behaviour, but p = gamma m v does.

 

[MikeGomez]

That is what I mean. I just don't understand how it follows.

If F=dp/dt is just shorthand for three separate equations for each component of force, then F=dp/dt entails that there is nothing special about one particular component. [ii] But there is something special about the component that is parallel to the velocity: there is more inertia in that direction.

This argument against F=dp/dt contains only two premises, and [ii], and is clearly valid. So either or [ii] (or both) is false?

James, it gets worse. Align the force/change in momentum with the x-axis. Now the entire effect is in the direction of the x component, and ther is no effect in the y or z componnents!