Uncovering the Mystery of Mass: What is it?

[DrStupid]

It isn't really necessary.

Of course it is necessary to conclude

This shows that the impulse, $I=\int Fdt$ on Ball A is the same for all inertial observers. And since time is the same for all inertial observers, the time averaged force is the same. So let $I = \int Fdt = F_{avg}\Delta t$.

With my modified laws of motion above this is not the case. As I already mentioned all affected laws of nature would (of course) need to be adjusted accordingly. This also applies to Hooke's law. It would turn into

$F = k \cdot \Delta x \cdot \left( {1 + \frac{{v^2 }}{{c^2 }}} \right)$

resulting in the frame-dependent Impulse

$\int {F\left( t \right) \cdot dt} = k \cdot \int {\Delta x\left( t \right) \cdot \left[ {1 + \frac{{v\left( t \right)^2 }}{{c^2 }}} \right] \cdot dt}$

But the modified laws of motion and the corresponding Hooke's law would still be the same in all frames of reference and the resulting equations of motion are identical with the good old classical mechanics. Thus your assumption about force and compression of the spring does not result from Galilean relativity and/or the first law. You must explain where it comes from.

 

[Andrew Mason]

Of course it is necessary to concludeWith my modified laws of motion above this is not the case. As I already mentioned all affected laws of nature would (of course) need to be adjusted accordingly. This also applies to Hooke's law. It would turn into

$F = k \cdot \Delta x \cdot \left( {1 + \frac{{v^2 }}{{c^2 }}} \right)$

resulting in the frame-dependent Impulse

$\int {F\left( t \right) \cdot dt} = k \cdot \int {\Delta x\left( t \right) \cdot \left[ {1 + \frac{{v\left( t \right)^2 }}{{c^2 }}} \right] \cdot dt}$

Ok. But you are making some kind of fundamental distinction between the observed interactions based not on what is observed but only on the velocity of the frame of reference. So not all frames of reference are equivalent. I am just saying that without defining force in any way, the same interaction applied successively to the same body causes constant acceleration, and the same interaction applied to any number of identical bodies simultaneously results in the same change in motion to each. So any combination of these successive and simultaneous interactions has the result that $Impulse = \int Fdt \propto \int dp = \Delta p$ which is just Newton's second law.

But the modified laws of motion and the corresponding Hooke's law would still be the same in all frames of reference and the resulting equations of motion are identical with the good old classical mechanics. Thus your assumption about force and compression of the spring does not result from Galilean relativity and/or the first law. You must explain where it comes from.

It comes from symmetry and the equivalence of inertial frames of reference. There is no basis for having different effects of the same spring compressed the same distance applied the same way to the same mass but in different inertial frames.

 

[DrStupid]

But you are making some kind of fundamental distinction between the observed interactions based not on what is observed but only on the velocity of the frame of reference.

The equations of the natural laws result from observations. Why else should I need to adjust them to the modified laws of motion? This is necessary to keep the results in agreement with experimental observations.

So not all frames of reference are equivalent.

That wouldn't even follow from a simple distinction on the basis of velocity. The kinetic energy of a body with constant mass also depends on it's velocity only but that doesn't mean that there are preferred inertial frames.

So any combination of these successive and simultaneous interactions has the result that $Impulse = \int Fdt \propto \int dp = \Delta p$ which is just Newton's second law.

You still need to explain how this results from Galilean relativity and the first law.

There is no basis for having different effects of the same spring compressed the same distance applied the same way to the same mass but in different inertial frames.

I'm not talking about different effects but about different forces.

 

[Andrew Mason]

The equations of the natural laws result from observations. Why else should I need to adjust them to the modified laws of motion? This is necessary to keep the results in agreement with experimental observations.

Exactly. All I am doing is saying that if two interactions are identical, they are identical in all inertial frames of reference under Galilean Relativity. If they produced a different change in motion of the same unit body in the same unit time period depending on the inertial reference frame in which it is observed, the premise of Galilean Relativity is violated. You are using a different concept of force. So let me ask you: what is the basis for saying that a unit mass on a uniformly moving rail car subjected to the same interaction as an identical unit mass sitting on a stationary rail car, is subjected to a different change in motion?

That wouldn't even follow from a simple distinction on the basis of velocity. The kinetic energy of a body with constant mass also depends on it's velocity only but that doesn't mean that there are preferred inertial frames.

But kinetic energy being constant in all reference frames is not a law of motion. Kinetic energy of a body is defined by motion relative to an arbitrary frame of reference. Change in motion of a body is always with respect to a particular inertial frame of reference of the body prior to the change.

You still need to explain how this results from Galilean relativity and the first law.

I have done that in my post #64

I'm not talking about different effects but about different forces.

All I am saying is that since the same interaction between the same bodies produces the same changes in motion in all reference frames, there is no basis for distinguishing between the forces that apply during those identical interactions. Force is defined in the first law as a physical phenomenon that causes a body to change its motion. I am not sure why you want to say that the same phenomenon that has the same effect on identical bodies in all reference frames should be called a different "force".

AM

 

[DrStupid]

All I am doing is saying that if two interactions are identical, they are identical in all inertial frames of reference under Galilean Relativity.

No, that's not all you are saying. You also say that the corresponding forces are identical and that this is not just the result of a definition but a consequence of Galilean relativity and the first law.

So let me ask you: what is the basis for saying that a unit mass on a uniformly moving rail car subjected to the same interaction as an identical unit mass sitting on a stationary rail car, is subjected to a different change in motion?

I don't know why somebody should say something like that.

Kinetic energy of a body is defined by motion relative to an arbitrary frame of reference.

And with my modification of the second law force is also defined by motion relative to an arbitrary frame of reference (among other things). If kinetic energy does not define a preferred frame of reference why should that be a problem in case of my definition of force?

I have done that in my post #64

No, you didn't because you didn't explain why "a force, F, at a given time is represented by the compression distance". Your argumentation is broken without this explanation.

All I am saying is that since the same interaction between the same bodies produces the same changes in motion in all reference frames, there is no basis for distinguishing between the forces that apply during those identical interactions.

No, you are saying that F=m·a follows from Galilean relativity and the first law. That implies that distinguishing between the forces that apply during those identical interactions is not only unmotivated but forbidden.

I am not sure why you want to say that the same phenomenon that has the same effect on identical bodies in all reference frames should be called a different "force".

Just to disprove your claim that Galilean relativity and the first law results in F=m·a.

 

[lightarrow]

Yes, but not in classical mechanics. Newton's definition of mass as "quantity of matter" applies to classical mechanics.

Did the OP ask "what is mass" or "what Newton thought mass is"?
Actually with "quantity of matter" we usually mean "number of moles", nowadays.

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[DrStupid]

Actually with "quantity of matter" we usually mean "number of moles", nowadays.

Isn't that amount of substance?

 

[lightarrow]

Isn't that amount of substance?

Yes, I have just controlled, you're right, I remembered wrong.

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[lightarrow]

Anyway, saying "mass is the quantity of matter" doesn't answer much to the question "what is mass?", unless someone defines "matter" in terms of something measurable.

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[Andrew Mason]

No, that's not all you are saying. You also say that the corresponding forces are identical and that this is not just the result of a definition but a consequence of Galilean relativity and the first law.

Well, one does have to have a concept of what a force is. And I say that this concept of force comes from the first law: the physical phenomenon that causes a body to change its motion.

So now we want to compare forces. So in IFR A, the same physical interaction applied to each of two identical unit bodies at the same time for the same duration would result in the same change of velocity. If that was not the case, then after the interaction ended (at the same time for both unit bodies), the bodies would define two different inertial frames of reference in which either: 1. the laws of motion were not the same for each body, or 2. time and space were measured differently for each body.

Then it is just a matter of equating this concept of "interaction" (ie. that which causes a body to change its motion) to the concept of force by virtue of the first law. So we define a standard interaction of some kind and call it a unit of force.

Now we translate to IFR B which is moving at velocity v relative to IFR A. In the reference frame of IFR B, we can say that based on the equivalence of IFRs (a premise of Galilean Relativity), one unit of force applied to each of two identical unit bodies at the same time for the same duration would result in the same change of velocity and that this change in velocity would be the same change of velocity as observed in IFR A. If that was not the case, we could easily distinguish between IFRs.

If you disagree, please explain why this would not be the case yet Galilean Relativity would not be violated.

I don't know why somebody should say something like that.

So you would appear to agree that the same change in motion results from the same interaction. All I am doing is assigning the concept of "average force" to the interaction for its duration, because it is an interaction that results in a change in motion. What is your concept of force?

And with my modification of the second law force is also defined by motion relative to an arbitrary frame of reference (among other things). If kinetic energy does not define a preferred frame of reference why should that be a problem in case of my definition of force?

Because the laws of motion have to be the same in all IFRs and the first law of motion says that a force is something that causes a change in motion of a body that was, before being subjected to the force, either at rest or in uniform motion. You appear to want to define force as something else. All I am saying is that if all IFRs are equivalent, then the same "force" for the same duration will cause the same change in motion to the same body in all IFRs.

Using your suggestion, $F = \dot{p}(1-v^2/c^2)$, a force, F, applied for a duration $\Delta t$ in IFR A (v=0) to a unit mass would result in a change of velocity of $\Delta v$ (i.e. $F\Delta t = \Delta v$) but in IFR B (v=v_B) $F\Delta t = \Delta v(1-v_B^2/c^2) < \Delta v$

No, you didn't because you didn't explain why "a force, F, at a given time is represented by the compression distance". Your argumentation is broken without this explanation.

You don't need the spring. You just need the collision. The spring just allows you to see the collision dynamics better. If two identical bodies subjected to the same spring with the same compression for the same duration had different changes in motion depending only the IFR they were in, then the premise of Galilean relativity would not hold.

No, you are saying that F=m·a follows from Galilean relativity and the first law.

I am really saying that Galilean Relativity and Newton's Second Law, together with the underlying premises that time, mass and space are absolute, are equivalent. The second law leads to the equivalence of all IFRs. The equivalence of all IFRs leads to the second law.

That implies that distinguishing between the forces that apply during those identical interactions is not only unmotivated but forbidden.

I have lost you there. If the interactions are identical, and this is an interaction involving forces, then what basis is there for saying that the forces would be different?

AM

 

[Andrew Mason]

Anyway, saying "mass is the quantity of matter" doesn't answer much to the question "what is mass?", unless someone defines "matter" in terms of something measurable.

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Newton conceived that all matter was comprised of the same fundamental units of matter. That was a guess. But it was an amazingly accurate guess. We now know that the difference in mass of two bodies of matter is very closely approximated by the difference in the number of nucleons each contain. So just imagine the "quantity of matter" to be the "number of nucleons".

AM

 

[lightarrow]

Newton conceived that all matter was comprised of the same fundamental units of matter. That was a guess. But it was an amazingly accurate guess. We now know that the difference in mass of two bodies of matter is very closely approximated by the difference in the number of nucleons each contain. So just imagine the "quantity of matter" to be the "number of nucleons".

So he was quite right even about it, interesting. But, aren't you a Newton's descending? You seem to praise him a lot

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[DrStupid]

So we define a standard interaction of some kind and call it a unit of force.

In that case F=ma for constant mass would not result from Galilean relativity and the first law but from Galilean relativity, the first law and the definition above. But if you already admit that an additional definition is necessary, why not simply define F=dp/dt instead? You wouldn't need additional derivations and it would also work for systems with variable mass.

All I am doing is assigning the concept of "average force" to the interaction for its duration, because it is an interaction that results in a change in motion.

No, that's not all you are doing. You also claimed that F=ma results from Galilean relativity and the first law. But with your statement above you seem to distance yourself from this claim and start to realize that it is in fact the result of an useful but nevertheless arbitrary definition.

What is your concept of force?

I usually use Newton's concept of force as defined by his second and third law of motion. But that does not mean that this is the only possible definition. It is just the definition that has been finally established.

Because the laws of motion have to be the same in all IFRs and the first law of motion says that a force is something that causes a change in motion of a body that was, before being subjected to the force, either at rest or in uniform motion.

As this applies to the modified laws of motion above I still do not see the problem.

All I am saying is that if all IFRs are equivalent, then the same "force" for the same duration will cause the same change in motion to the same body in all IFRs.

It is not sufficient to say that. You need to justify this claim.

You don't need the spring. You just need the collision.

Than try to finish your argumentation without the spring.

I am really saying that Galilean Relativity and Newton's Second Law, together with the underlying premises that time, mass and space are absolute, are equivalent.

That's actually not all you are saying.

If the interactions are identical, and this is an interaction involving forces, then what basis is there for saying that the forces would be different?

That's the wrong question. The correct question is: What prevents me from saying it? F=ma cannot result from Galilean relativity and the first law if I can simply define something else (no matter why) without violating these conditions.

 

[DrStupid]

So he was quite right even about it, interesting. But, aren't you a Newton's descending? You seem to praise him a lot

The modern concept of atoms and elementary particles dates back to Robert Boyle's idea of basic particles from 1661. As Newton was not only a Physicist but also an Alchemist it is very likely that he was aware of this idea when he wrote the Principia in 1686. It would still be a guess but not necessarily Newton's guess.

 

[Andrew Mason]

In that case F=ma for constant mass would not result from Galilean relativity and the first law but from Galilean relativity, the first law and the definition above. But if you already admit that an additional definition is necessary, why not simply define F=dp/dt instead? You wouldn't need additional derivations and it would also work for systems with variable mass.

No, that's not all you are doing. You also claimed that F=ma results from Galilean relativity and the first law. But with your statement above you seem to distance yourself from this claim and start to realize that it is in fact the result of an useful but nevertheless arbitrary definition

I usually use Newton's concept of force as defined by his second and third law of motion. But that does not mean that this is the only possible definition. It is just the definition that has been finally established.

It is not arbitrary to assign the concept of force to dp/dt. One could say that f=dp/dt is a natural definition because it corresponds to reality.If a body is being pulled by a spring with a constant extension then $\Delta p \propto \Delta t$. I think we agree that this is a true statement within the limits of classical mechanics.

We could do a bunch of experiments and observe that $\Delta p \propto \Delta t$ is constant for any constant spring extension and any mass of body. From this we could deduce that the experiment would give the same result in all inertial frames of reference.

OR: (which is what I am saying) we could start with the a priori rule that all the laws of motion will be the same in all inertial frames of reference and predict that $\Delta p \propto \Delta t$ must be true (i.e. without actually observing it) for any spring extension and any mass of body. If this was not the case then we could use the spring to distinguish between different inertial frames.

AM

 

[lightarrow]

It is not arbitrary to assign the concept of force to dp/dt. One could say that f=dp/dt is a natural definition because it corresponds to reality.

If a body is being pulled by a spring with a constant extension then $\Delta p \propto \Delta t$. I think we agree that this is a true statement within the limits of classical mechanics.

We could do a bunch of experiments and observe that $\Delta p \propto \Delta t$ is constant for any constant spring extension and any mass of body. From this we could deduce that the experiment would give the same result in all inertial frames of reference.

OR: (which is what I am saying) we could start with the a priori rule that all the laws of motion will be the same in all inertial frames of reference and predict that $\Delta p \propto \Delta t$ must be true (i.e. without actually observing it) for any spring extension and any mass of body. If this was not the case then we could use the spring to distinguish between different inertial frames.

And if in frame1 and in frame2 you would observe, e.g., that $\Delta p \propto \Delta t^2$, what would you deduce?

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[DrStupid]

It is not arbitrary to assign the concept of force to dp/dt. One could say that f=dp/dt is a natural definition because it corresponds to reality.

My modified definition corresponds to reality as well.

OR: (which is what I am saying) [...]

No, that is not what you are saying. You made a claim about force and not just about the principle of relativity.

 

[Andrew Mason]

And if in frame1 and in frame2 you would observe, e.g., that $\Delta p \propto \Delta t^2$, what would you deduce?

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One could conclude that the premises of Galilean relativity are incorrect.

This is because the application of the spring for the period $2\Delta t$ to a body at rest in IFR A is equivalent to:

1. applying the spring in IFR A for $\Delta t$ at which time the body is moving at velocity v_A, and then
2. re-applying the spring for $\Delta t$, in which case the body starts the second application at rest in IFR A', (which is moving at v_A relative to IFR A) and ends up moving at velocity v_A relative to IFR A'

If $\Delta p \propto \Delta t^2$, the total $\Delta p = 4v_A$ if the application of the spring is viewed entirely in IFR A. But if you look at it as two successive applications of the spring for the same total duration the final velocity relative to A' would be $v_A$, so relative to A it would be $2v_A$ i.e. $\Delta p = 2v_A$ in A'.

AM

 

[Andrew Mason]

My modified definition corresponds to reality as well.

So explain to me how $\Delta p \propto \Delta t$ if $\Delta p = F\Delta t/(1-v^2/c^2)$.

You would have to say that force magically varies as $(1-v^2/c^2)$ , despite there being a complete lack of physical change to the thing that is causing the force. That sounds a bit like the attempts by Fitzgerald and Lorentz to preserve the concept of ether while explaining the results of the Michelson-Morley experiment.

No, that is not what you are saying. You made a claim about force and not just about the principle of relativity.

I see that you are reluctant to concede a natural concept of force. I thought there was an obvious way to think of a force but you can define force however you wish. So we won't even refer to force. We can just refer to springs applied to bodies while maintaining constant extensions or compressions.

I will just say that for applications of springs to bodies while maintaining constant extensions the relationship of $\Delta p \propto \Delta t$ can be predicted from Galilean Relativity. I will not go so far as to say that f = dp/dt can be determined because that depends on what you mean by f.

AM

 

[Neon]

Mass is energy. Most of it comes from the interactions between gluons and quarks IN subatomic particles.The rest comes from matter interacting with the higgs field.
This is off topic but good to know
Massless particles CAN ONLY travel at light speed and can NEVER slow down.We(have mass) can move at any speed we want, but not the speed of light.

 

[lightarrow]

One could conclude that the premises of Galilean relativity are incorrect.

This is because the application of the spring for the period $2\Delta t$ to a body at rest in IFR A is equivalent to:

1. applying the spring in IFR A for $\Delta t$ at which time the body is moving at velocity v_A, and then
2. re-applying the spring for $\Delta t$, in which case the body starts the second application in IFR A', moving at v_A relative to IFR A

If $\Delta p \propto \Delta t^2$, the total $\Delta p = 4\Delta v_A$ if the application of the spring is viewed entirely in IFR A; but if you look at it as two successive applications of the spring for the same total duration the change would only be $2\Delta v_A$ viewed entirely in IFR A'.

AM

Viewed "entirely in A' " you have $2\Delta t$ and so $\Delta p \propto 4\Delta t^2$. But maybe I don't have grasped your reasoning: if you substitute $\Delta p$ with $\Delta s$ (space interval) you would conclude that $\Delta s$ cannot be proportional to $\Delta t^2$ ?
P.S. why you wrote $\Delta v_A$ instead of $\Delta t^2$?

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[Andrew Mason]

Viewed "entirely in A' " you have $2\Delta t$ and so $\Delta p \propto 4\Delta t^2$.

The observer in A' sees the second application of the spring begin while the body is at rest in A'. The second application of the spring lasts for $\Delta t$. At that point the velocity of the body relative to A' will be v_A. If it were not, the application of the spring to the body for $\Delta t$ in A would result a change of velocity of v_A but in A' the result would be different: Galilean relativity would be violated. So the velocity of the body relative to A will be the velocity of the body relative to A' + the velocity of A' relative to A = 2v_A.

But maybe I don't have grasped your reasoning: if you substitute $\Delta p$ with $\Delta s$ (space interval) you would conclude that $\Delta s$ cannot be proportional to $\Delta t^2$ ?

Distance and time translate differently between frames. You have to apply the Galilean transformation for distance: x = x'+vt. For time, t = t'.

AM

 

[DrStupid]

So explain to me how $\Delta p \propto \Delta t$ if $\Delta p = F\Delta t/(1-v^2/c^2)$.

Simply by

$\frac{F}{{1 + \frac{{v^2 }}{{c^2 }}}} = const.$

You would have to say that force magically varies as $(1-v^2/c^2)$

It results from the modified definition of force. There is no magic involved.
But if you can't distinguish it from magic and if Arthur C. Clarke's thesis about magic and technology can be generalized to theories, then it would suggest that this theory is sufficiently advanced.

despite there being a complete lack of physical change to the thing that is causing the force.

The corresponding change of kinetic energy also varies without physical change to the thing that is causing it.

That sounds a bit like the attempts by Fitzgerald and Lorentz to preserve the concept of ether while explaining the results of the Michelson-Morley experiment.

And that sounds like stratagem 32 of Schopenhauer's Art of Being Right.

I see that you are reluctant to concede a natural concept of force. I thought there was an obvious way to think of a force but you can define force however you wish.

Does that mean you finally got it?

So we won't even refer to force.

There would be no physics left if we won't refer to definitions.

 

[Andrew Mason]

Simply by

$\frac{F}{{1 + \frac{{v^2 }}{{c^2 }}}} = const.$
The corresponding change of kinetic energy also varies without physical change to the thing that is causing it.

Under the first law, nothing is "causing" uniform velocity of a body to occur, so nothing is "causing" a body to maintain its kinetic energy. But something is required to cause a change in motion: ie. a force if you want to use Newton's terminology. If the change in kinetic energy is caused by a spring with constant extension applied to the body, the rate at which work is being done in maintaining that spring extension has to keep increasing. So there is a physical change occurring - it is just not evident by just looking at the spring. Rather it is in what is causing that spring to maintain its constant extension.

Does that mean you finally got it?

I was trying to explain something which I think helps students understand Newton's laws. Many students wonder why F=ma. If you look at it from the premise of Galilean Relativity it is a relationship that is required if we think of force in the natural way that Newton did. No one is trying to make anyone feel stupid for not getting it. It would be antithetical to all the reasons I post here to ask someone if they had "finally got it". So I will just politely ignore your question.

There would be no physics left if we won't refer to definitions.

I suppose. I am just explaining why f=ma it is not an arbitrary definition.

AM

 

[Andrew Mason]

And that sounds like stratagem 32 of Schopenhauer's Art of Being Right,

There was nothing odious about the Fitzgerald-Lorentz contraction. It explained why the ether could not be detected. It just did not offer an explanation why measuring sticks would contract like that as speed of the light source increased. Einstein discovered the reason.

If you are going to suggest a new definition of a physical phenomenon (force) whose value depends on 1 + v^2/c^2, without being able to explain why, you should be prepared to deal with the same kind of questions that I expect Fitzgerald and Lorentz had to deal with.

AM

 

[DrStupid]

I am just explaining why f=ma it is not an arbitrary definition.

No, you are claiming it and instead of an explanation you repeat this claim again and again with just another wording. Now I leave this discussion because it will lead to nothing.