Uncovering the Mystery of Mass: What is it?

[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 $\Delta t^2$?

--
lightarrow

 

[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.