How to determine real inertia without 4 vectors?

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Discussion Overview

The discussion revolves around the use of the relativistic mass equation (M = γM₀) in calculating kinetic energy (KE) for relativistic particles. Participants explore the relevance of this equation in modern physics, particularly in the context of teaching and understanding relativity, while addressing the implications of using relativistic mass versus rest mass.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification
  • Homework-related

Main Points Raised

  • Some participants question whether the relativistic mass equation is still useful for calculating KE, despite its declining favor among physicists.
  • Others argue that using the energy equation (KE = E - E₀) is more appropriate and avoids confusion associated with relativistic mass.
  • A few participants mention that there are specific problems where relativistic mass can be useful, such as calculating trajectories of relativistic electrons.
  • Concerns are raised about the educational implications of teaching relativistic mass in A level physics, with some suggesting it may still be relevant in certain contexts.
  • There is a discussion about the terminology used in energy equations, particularly regarding the terms E and E₀, and how they should be defined without implying outdated concepts of mass.

Areas of Agreement / Disagreement

Participants express a mix of agreement and disagreement regarding the utility of the relativistic mass equation. While some acknowledge its historical significance and occasional usefulness, others advocate for a focus on covariant quantities and express skepticism about its relevance in modern physics education.

Contextual Notes

Participants note that the discussion is influenced by the historical context of the relativistic mass concept and its implications for teaching. There are unresolved questions about the best terminology to use for energy terms in relativistic equations.

Who May Find This Useful

This discussion may be of interest to educators, students in physics courses, and those exploring the foundations of relativity and its implications for teaching and understanding modern physics concepts.

  • #31
PAllen said:
the second wrong in relativity

In the context if physics "wrong" means disproved by experiments. Which experiments do you mean?

PAllen said:
there was no m and m0 in Newtonian physics

In classical mechanics m and mo are identical. This changes if Galilei transformation is replaced by Lorentz transformation. I can show you the corresponding derivations if you are really interested.

PAllen said:
To me, the ratio of 4-force to 4-acceleration is the exact application of Newton's definition in a relativistic context.

Newton never published such a definition. He defined p=m·v and F=dp/dt and these formulas were intended for use with 3-vectors only. Using them with 4-vectors results in new formulas. Of course that doesn't mean that this new formulas are wrong or useless (in fact they are very useful in SR) but they are different from Newtons original formulas and so is the resulting new definition of mass.
 
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  • #32
DrStupid said:
PAllen said:
To me, the ratio of 4-force to 4-acceleration is the exact application of Newton's definition in a relativistic context.
Newton never published such a definition. He defined p=m·v and F=dp/dt and these formulas were intended for use with 3-vectors only.
From Newton's Principia,

Law II. The alteration of motion is ever proportional to the motive force impress'd; and is made in the direction of the right line in which the motive force is impress'd.
"Alteration of motion" is what we today would call acceleration, dv/dt. In other words, he said F = ma. Newton made no mention of momentum.
 
  • #33
PAllen said:
Hmm. In my mind, the only form of inertial mass in SR is invariant mass, which includes KE only to the extent it becomes part of the invariant mass of the system.

You mean like in the 4-vector analogue of "F=ma" as in the posts above?
 
  • #34
atyy said:
You mean like in the 4-vector analogue of "F=ma" as in the posts above?

Yes.
 
  • #35
PAllen said:
Yes.

How do you get from there to the Einstein equation G=T?

The ADM mass and Komar mass seem closer to the notion of gravitational mass as invariant mass. But the stress energy tensor seems closer to the notion of gravitational mass as energy.
 
  • #36
atyy said:
How do you get from there to the Einstein equation G=T?

The ADM mass and Komar mass seem closer to the notion of gravitational mass as invariant mass. But the stress energy tensor seems closer to the notion of gravitational mass as energy.

I don't know. I was only speaking to what plays a role of inertia in SR, not an approach for motivating stress energy tensor. If you don't use 4 vectors, how do you conclude which is the real inertia: longitudinal relativistic mass or transverse?

Maybe get to T via flows of 4-momentum, which brings energy and spatial momentum in together.
 
  • #37
Bill_K said:
"Alteration of motion" is what we today would call acceleration, dv/dt.

That's simply wrong.

From Newton's Principia,

Newton said:
Definition II: The quantity of motion is the measure of the same, arising from the velocity and the quantity of matter conjunctly.

or in the original:

Newton said:
Def. II: Quantitas motus est mensura ejusdem orta ex velocitate et quantitate materiae conjunctim.

Obviously Newton used the term "velocitate" for velocity. Therefore acceleration would have been "alteration of velocity" or "mutationem velocitate" in original. But he wrote "mutationem motus" and the quantitative definition of "motus" is given above. As "quantitate materiae" is Newtons term for mass, definition 2 means

motion = velocity * mass

That's what we today call momentum.
 
  • #38
PAllen said:
If you don't use 4 vectors, how do you conclude which is the real inertia: longitudinal relativistic mass or transverse?

If "real inertia" means M in F=M·a than it is

M = \frac{{m_0 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }} \cdot \left( {I + \frac{{v \cdot v^T }}{{c^2 - v^2 }}} \right)

Longitudinal and transversal mass are the eigenvalues of M. Of course that is not what Newton called mass (or anybody else today).
 

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