How to determine real inertia without 4 vectors?

In summary, relativistic mass equation is out of favor among most theoretical physicists, but it may still have some uses. People still like it, but an example is one of the reasons why most don't.
  • #1
Dadface
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I know the reasons why the relativistic mass equation (M= gamma Mo) is out of favour with what seems to be a majority of theoretical physicists but I am unsure about one particular thing:

Suppose we knew the mass and velocity of a relativistic particle and wanted to calculate its KE.
Isn't use of the relativistic mass equation, along with (M-M0)c squared, the quickest and easiest way to get to the answer? In other words although some may consider the equation to be a historical curiosity, does it still have its uses?

I have considerd this question before but am still not sure about the answer.

Thanks
 
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  • #2
Dadface said:
I know the reasons why the relativistic mass equation (M= gamma Mo) is out of favour with what seems to be a majority of theoretical physicists but I am unsure about one particular thing:

Suppose we knew the mass and velocity of a relativistic particle and wanted to calculate its KE.
Isn't use of the relativistic mass equation, along with (M-M0)c squared, the quickest and easiest way to get to the answer? In other words although some may consider the equation to be a historical curiosity, does it still have its uses?

I have considerd this question before but am still not sure about the answer.

Thanks

A few people still like it, but your example is one of the reasons most don't like it. The original value was to replace the m in p=mv. For KE one would want, by the same philosophy, to replace the m in 1/2 mv^2, but that doesn't work with relativistic mass. If you you are going to use a relativistic formula anyway (yours is s relativistic formula, not Newtonian), why not consider KE = E - E0 = mc^2(γ-1).

For me, it all comes back to the 4-vector notion, in which gamma doesn't go with the mass at all, it is part of the 4-velocity. 4-Momentum is then mU, and energy it the timelike component of this in the basis of the measuring instrument.
 
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  • #3
Dadface said:
Suppose we knew the mass and velocity of a relativistic particle and wanted to calculate its KE.
Isn't use of the relativistic mass equation, along with (M-M0)c squared, the quickest and easiest way to get to the answer? In other words although some may consider the equation to be a historical curiosity, does it still have its uses?
Kinetic energy itself is a leftover concept from Newtonian mechanics. In relativity the quantities you deal with are the covariant quantities, and there's hardly ever an occasion when you'd want or need to calculate KE.
 
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  • #4
Thank you PAllen but the equation you presented is the same as the one I presented. The only difference is I used Mo instead of M. I know that the concept of rest mass is out of favour and I know why but I used Mo in reference to the original equation. I'm quite happy to use M or m in place of Mo but when I use your equation I am just using the original equation in a different format and without the confusion of definitions of mass.
 
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  • #5
Bill_K said:
Kinetic energy itself is a leftover concept from Newtonian mechanics. In relativity the quantities you deal with are the covariant quantities, and there's hardly ever an occasion when you'd want or need to calculate KE.

"Hardly ever" is not the same as never. There are occasions when I might want to calculate KE and that's why I asked the question.
 
  • #6
Dadface said:
In other words although some may consider the equation to be a historical curiosity, does it still have its uses?

There are some problems for which some people find relativistic mass to be useful. For example, many years ago I had to calculate the trajectory of relativistic electrons passing between two parallel charged plates. Applying ##F=m_{r}a## seemed as easy as anything else, although I wouldn't argue with anyone who prefers to do the calculation some other way.

However, problems of this sort are few and far between, and the overwhelming majority of people studying relativity do so not because they need to solve a very specific class of problems, but because they want to understand relativity. If understanding relativity is the goal, than spending a lot of time learning about relativistic mass, then even more time learning about its limitations, is counterproductive.
 
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  • #7
Dadface said:
Thank you PAllen but the equation you presented is the same as the one I presented. The only difference is I used Mo instead of M. I know that the concept of rest mass is out of favour and I know why but I used Mo in reference to the original equation. I'm quite happy to use M or m in place of Mo but when I use your equation I am just using the original equation in a different format and without the confusion of definitions of mass.

In mine, only one mass is used. There is no concept or relativistic mass in my formula. The gamma has nothing to do with the mass.
 
  • #8
Dadface said:
I know the reasons why the relativistic mass equation (M= gamma Mo) is out of favour with what seems to be a majority of theoretical physicists but I am unsure about one particular thing:

Suppose we knew the mass and velocity of a relativistic particle and wanted to calculate its KE.
Isn't use of the relativistic mass equation, along with (M-M0)c squared, the quickest and easiest way to get to the answer? In other words although some may consider the equation to be a historical curiosity, does it still have its uses?

I have considerd this question before but am still not sure about the answer.

Thanks

You can just as easily use the energy equation, and subtract the rest energy.
 
  • #9
PAllen said:
In mine, only one mass is used. There is no concept or relativistic mass in my formula. The gamma has nothing to do with the mass.

Yes I agree but in the original equation,the one I presented, there is only one mass namely Mo. The only difference is one of symbols. I know that generally speaking relativistic mass is considered to be outdated and misleading and I am familiar with the reasons for that. I like to think of M (as in the original M=gamma Mo) as being equal to the mass plus the kinetic energy expressed in mass units.
 
  • #10
pervect said:
You can just as easily use the energy equation, and subtract the rest energy.
Yes you can express it mass units or energy units.
 
  • #11
Nugatory said:
There are some problems for which some people find relativistic mass to be useful. For example, many years ago I had to calculate the trajectory of relativistic electrons passing between two parallel charged plates. Applying ##F=m_{r}a## seemed as easy as anything else, although I wouldn't argue with anyone who prefers to do the calculation some other way.

However, problems of this sort are few and far between, and the overwhelming majority of people studying relativity do so not because they need to solve a very specific class of problems, but because they want to understand relativity. If understanding relativity is the goal, than spending a lot of time learning about relativistic mass, then even more time learning about its limitations, is counterproductive.

Thanks but the problems are probably not as rare as you think. The old mass variation equation is still taught in A level physics courses in UK schools (eg AQA physics A "turning points in physics" topic). Until and if the examiners modify the specification teachers still have to go along with it. Admittedly the relativity taught is at a very basic level.
 
  • #12
Apart from a few things that need clarifying with regards to things like symbols used I think I am in broad agreement with everybody here.But there are still a couple of problems:

!.What should A level students be told?
2.The second point is about terminology. There are three terms in the equation when expressed in energy units KE = E-Eo
What should each of these three terms be called? KE is called kinetic energy but what do we call E and Eo. It might be tempting to call E total energy and Eo rest energy but those terms imply the existence of total mass and rest mass.
 
  • #13
Dadface said:
Apart from a few things that need clarifying with regards to things like symbols used I think I am in broad agreement with everybody here.But there are still a couple of problems:

!.What should A level students be told?
2.The second point is about terminology. There are three terms in the equation when expressed in energy units KE = E-Eo
What should each of these three terms be called? KE is called kinetic energy but what do we call E and Eo. It might be tempting to call E total energy and Eo rest energy but those terms imply the existence of total mass and rest mass.

Energy is energy (a frame variant quantity); mass is mass, an invariant. The full equation to focus on is:

E^2 = (mc^2)^2 + p^2 c^2

Then you are not tempted to convert total energy to total mass, making a frame dependent notion of mass.
 
  • #14
PAllen said:
Energy is energy (a frame variant quantity); mass is mass, an invariant. The full equation to focus on is:

E^2 = (mc^2)^2 + p^2 c^2

Then you are not tempted to convert total energy to total mass, making a frame dependent notion of mass.

That's good for me but it's probably too heavy going for A level students.
 
  • #15
Dadface said:
The old mass variation equation is still taught in A level physics courses in UK schools (eg AQA physics A "turning points in physics" topic). Until and if the examiners modify the specification teachers still have to go along with it. Admittedly the relativity taught is at a very basic level.
http://filestore.aqa.org.uk/subjects/AQA-2450-W-TRB-OGTP.PDF also tells us that "time runs slower when you are moving." I guess in comparison, their use of relativistic mass is a minor point.
 
  • #16
Dadface said:
Thanks but the problems are probably not as rare as you think. The old mass variation equation is still taught in A level physics courses in UK schools (eg AQA physics A "turning points in physics" topic). Until and if the examiners modify the specification teachers still have to go along with it. Admittedly the relativity taught is at a very basic level.

No one disputes that it's still taught, and therefore that you may need it just to pass some tests. The dispute is over whether it's good pedagogy to teach it - that is, SHOULD it still be taught, or is there a better way of using limited classroom time to communicate the basic principles of SR. Note that relativistic mass is NOT one of those basic principles; it's a very specific application of them in a specific class of problems.

In some ways this discussion reminds me of discussions of whether Latin should be taught in American high schools. There's no doubt that most native English speakers will derive some non-zero benefit from a year or so of diligent Latin study; but enough to justify displacing a year or so of English literature?
 
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  • #17
Bill_K said:
http://filestore.aqa.org.uk/subjects/AQA-2450-W-TRB-OGTP.PDF also tells us that "time runs slower when you are moving." I guess in comparison, their use of relativistic mass is a minor point.

But is there a conceptual connection to GR, since the principle of equivalence in Newtonian gravity is that inertial mass is gravitational mass? Since the inertial mass is energy in special relativity, that suggests that the energy should be the source of gravity in relativistic gravity. Then requiring that the equations be tensorial, the source should be either the stress-energy tensor or something like its trace.
 
  • #18
atyy said:
But is there a conceptual connection to GR, since the principle of equivalence in Newtonian gravity is that inertial mass is gravitational mass? Since the inertial mass is energy in special relativity, that suggests that the energy should be the source of gravity in relativistic gravity. Then requiring that the equations be tensorial, the source should be either the stress-energy tensor or something like its trace.

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.
 
  • #19
Bill_K said:
http://filestore.aqa.org.uk/subjects/AQA-2450-W-TRB-OGTP.PDF also tells us that "time runs slower when you are moving." I guess in comparison, their use of relativistic mass is a minor point.

I'm not sure where you are getting that from. The specification is very brief referring to proper time,time dilation the equation and muon decay.
 
  • #20
Dadface said:
I'm not sure where you are getting that from. The specification is very brief referring to proper time,time dilation the equation and muon decay.
That is literally what it says on Page 21:
C - Time Dilation and Proper Time said:
A consequence of the invariance of the speed of light in free space is that ’moving clocks run slow’ or time runs slower when you are moving
It then goes on to discuss the usual train example, but there is no hint given that the effect is symmetrical, which I think is the whole point of relativity.

Are we looking at the same http://filestore.aqa.org.uk/subjects/AQA-2450-W-TRB-OGTP.PDF?
 
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  • #21
Nugatory said:
No one disputes that it's still taught, and therefore that you may need it just to pass some tests. The dispute is over whether it's good pedagogy to teach it - that is, SHOULD it still be taught, or is there a better way of using limited classroom time to communicate the basic principles of SR. Note that relativistic mass is NOT one of those basic principles; it's a very specific application of them in a specific class of problems.

In some ways this discussion reminds me of discussions of whether Latin should be taught in American high schools. There's no doubt that most native English speakers will derive some non-zero benefit from a year or so of diligent Latin study; but enough to justify displacing a year or so of English literature?

As the syllabus stands at present it has to be taught. The problem is how do teachers do the extra that is necessary without taking up too much extra time and without jeopardising students chances of exam success.
Syllabus changes have to take into account the likely future educational paths of the student population and only a very small fraction of this population will study physics beyond A level.
 
  • #22
Bill_K said:
That is literally what it says on Page 21:

It then goes on to discuss the usual train example, but there is no hint given that the effect is symmetrical, which I think is the whole point of relativity.

Are we looking at the same http://filestore.aqa.org.uk/subjects/AQA-2450-W-TRB-OGTP.PDF?

We're looking at different things. I was referring to the AQA 2014 specifications. Page 33 for syllabus A and page 26 for syllabus B.
 
  • #23
PAllen said:
The original value was to replace the m in p=mv.

The original value was NOT to replace the m in p=mv. In his paper from 1905 Einstein replaced Galilei transformation by Lorentz transformation only. Everything else remained unchanged. In the result the former invariant m became frame dependent. Many physicist (including Einstein himself) continued to use it in SR. Later it has been antiquated in favor of the rest mass in order to get a frame independent property again. This is what the term "mass" is generally used for today.
 
  • #24
DrStupid said:
The original value was NOT to replace the m in p=mv. In his paper from 1905 Einstein replaced Galilei transformation by Lorentz transformation only. Everything else remained unchanged. In the result the former invariant m became frame dependent. Many physicist (including Einstein himself) continued to use it in SR. Later it has been antiquated in favor of the rest mass in order to get a frame independent property again. This is what the term "mass" is generally used for today.

Einstein abandoned it, and wrote about doing so, calling it a bad idea. Mass remains an invariant scalar in SR. For a system of particles, the invariant mass is the norm of the total 4-momentum. Einstein, in 1905, derived that momentum was mγv, from which he factored mγ as a mass factor. This was before any notion of 4-vector treatment was developed. Even before adopting 4-vectors, Einstein was writing that m versus m0 was a bad idea.

4-vector treatments leads to a different factoring: gamma comes form derivative of position by proper time. It is part of the tangent vector, nothing to do with the mass.
 
  • #25
PAllen said:
Mass remains an invariant scalar in SR.

Mass as defined by Newton (by def. 2 and lex 2 & 3) does not remain invariant in SR. It needs to be refined in order to make it invariant again.
 
  • #26
DrStupid said:
Mass as defined by Newton (by def. 2 and lex 2 & 3) does not remain invariant in SR. It needs to be refined in order to make it invariant again.

Mass doesn't need redefinition. It is Newtonian formulas that need redefinition. Using the natural re-definition in terms of 4-velocity, 4-acceleration, and 4-force, you get that the ratio of force to acceleration is the (invariant) mass.
 
  • #27
PAllen said:
Mass doesn't need redefinition. It is Newtonian formulas that need redefinition. Using the natural re-definition in terms of 4-velocity, 4-acceleration, and 4-force, you get that the ratio of force to acceleration is the (invariant) mass.

This actually is a redefinition of mass.
 
  • #28
DrStupid said:
This actually is a redefinition of mass.

Not as I and most physicists see it. Given that time and simultaneity are relative, and that velocity is direction in spacetime, there is no plausible redefinition of velocity other than the unit tangent vector. Similarly, 4-acceleration follows as the proper time derivative of this. No mass or dynamics in involved. Then, dynamics is introduced on this in the most natural way, with mass being invariant just as it is in Newtonian physics. Variable mass only results from trying to use a mix of relativistic and non-relativistic formulas.
 
  • #29
PAllen said:
Then, dynamics is introduced on this in the most natural way, with mass being invariant just as it is in Newtonian physics.

You are right, but this mass is not the mass as defined by Newton.

PAllen said:
Variable mass only results from trying to use a mix of relativistic and non-relativistic formulas.

Please explain what you mean with "relativistic and non-relativistic formulas". In addition to the particular transformation the required formulas are definition 2 and lex 2 and 3 of Newtons Principia. They are valid both in classical mechanics and SR.

Just for the case that you think p=mo·v/sqrt(1-v²/c²) is relativistic and p=m·v is non-relativistic: These formulas are identical but m and mo are different.
 
  • #30
DrStupid said:
You are right, but this mass is not the mass as defined by Newton.



Please explain what you mean with "relativistic and non-relativistic formulas". In addition to the particular transformation the required formulas are definition 2 and lex 2 and 3 of Newtons Principia. They are valid both in classical mechanics and SR.

Just for the case that you think p=mo·v/sqrt(1-v²/c²) is relativistic and p=m·v is non-relativistic: These formulas are identical but m and mo are different.

We will never agree, obviously. To me, these formulas are not identical, the second wrong in relativity, the first just has m (there was no m and m0 in Newtonian physics). To me, the ratio of 4-force to 4-acceleration is the exact application of Newton's definition in a relativistic context.
 
  • #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.
 
  • #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.
 

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