# Can relativistic mass be a useful concept

1. Dec 30, 2013

Can "relativistic mass" be a useful concept

Many basic texts on relativity refer to the mass variation equation:
m= gamma mo
[gamma = Lorentz factor, mo= invariant(rest) mass]

It seems that there are some good reasons including,for example, problems with definitions of mass, why the equation is not favoured by a majority of physicists the alternative equation involving E squared, p squared and so on being preferred. What I am trying to find out is the following:

How widely is the equation (m= gamma mo) still favoured and is it considered to be a useful equation? To me it seems to be a useful and simple equation for calculating the kinetic energy of a relativistic particle.

Thank you.

2. Dec 30, 2013

Staff Emeritus
It is not a useful concept. It allows people to generate wrong equations quickly.

If you want a simple equation that is actually correct, why not E = γm?

3. Dec 30, 2013

### Bill_K

You don't have to use E2 = p2c2 + m2c4. To get the total energy, just use E = γmc2. Just don't factor this as E = (γm)c2 and pretend that γm is the mass of something.

Although they're related by a factor c2, energy and mass are conceptually different. Mass (inertial mass) relates force to acceleration, and in Special Relativity the relationship becomes more complicated. One needs to distinguish two kinds of relativistic mass, "transverse mass" and "longitudinal mass" depending on whether the force is parallel or perpendicular to the velocity. This is why we don't use relativistic mass!

4. Dec 30, 2013

Thank you. It seems you are showing me how to find the total energy (in mass units). If so I think I'm already familiar with that but my specific comment was about finding, not the total energy but the kinetic energy. As I understand it at this moment the E in the equation stands for the sum of the invariant mass plus the kinetic energy(expressed in mass units). If so the equation can still be written as
m=gamma mo (where m=mo + kinetic energy)
Therefore KE= gamma mo-mo

Is it not so then that the out of favour so called "mass variation equation" can still be succesfully used to calculate K.E.?

5. Dec 30, 2013

Thank you.Your initial advice seems to be the same as that Vandium 50 except that you are using energy units.As I stated above I am interested in finding the KE not the total energy.

6. Dec 30, 2013

### Bill_K

That's correct, KE = (γ-1)mc2. Although in relativity, kinetic energy is not so terribly useful either.

7. Dec 30, 2013

Thank you. I understand that Lev Okun is a major opponent of the equation. If I get some time and if it's not too gruelling I will take a look at his objections.

8. Dec 30, 2013

### PAllen

Going back to Vanadium50's point, the issue is what value is relativistic mass as a guide to correct calculations. It would be nice if you could generate correct relativistic equations using relativistic mass, but this just doesn't work:

KE = 1/2 m v^2 is wrong with relativistic mass
F= m A is wrong

Also, it impedes understanding of 4-vectors in SR, from which it is seen that the gamma factor is 'tied' to the 4-velocity, not the mass. You have, in 4-vector terms:

P = m U ; m is rest mass, U is 4 velocity; this encompasses total energy and momentum.
Its norm is invariant mass = rest mass for one particle; and the norm leads to the
E^2 formula.

F = dP/d$\tau$

Note also that 4-velocity is naturally defined as derivative of coordinate position by proper time, making it a unit vector.

Last edited: Dec 30, 2013
9. Dec 30, 2013

Thanks but I seem to be getting some contradictory advice and I would be grateful for clarification. I understand that there are reasons why the concept of relative mass may not work for some (perhaps most) situations but my enquiry is specifically (and initially) about the use of the equation to calculate the kinetic energy of relativistic particles.
I understand also that the term "relativistic mass" may be misleading and requires careful definition. As I stated above the m in the equation stands for the invariant (rest) mass plus the kinetic energy {m=mo+KE....units must be consistent}.Is this not the case?
Also, the equation for KE reduces to 1/2 Mv^2 as speeds reduce to non relativistic values.
In a nutshell can I use the equation to calculate the KE of say an electron moving at 99% of c?

10. Dec 30, 2013

### PAllen

To me, that makes it a useless tautology. Since total energy is, by definition, rest energy + KE, then, if you define 'relativistic mass' as total energy / c^2, then (relativistic mass - rest mass) * c^2 gives you back KE. The modern view is simpler: you have a direct formula for total energy; KE = total energy - rest energy. No need to introduce an otherwise useless mass. You only need the formula in terms of momentum if you are given momentum. Otherwise you simply use that total energy is rest energy times gamma.

On the other hand, if you try to use relativistic mass to compute anything based on Newtonian formulas, you are led into errors.

Note again, that as soon as you use 4-vectors, you see that the gamma is attached to the velocity anyway, not to the mass.

11. Dec 30, 2013

### vela

Staff Emeritus
Kinetic energy isn't equal to $m-m_0$ as you've defined $m$ and $m_0$. The units don't work out. It's equal to
$$KE = (m-m_0)c^2 = \gamma m_0 c^2 - m_0 c^2 = (\gamma-1)m_0c^2.$$ But $E=\gamma m_0 c^2$ is just the total energy of the object, and $E_0 = m_0c^2$ is its rest energy. So really you just have $KE = E - E_0$. There's no compelling reason to artificially isolate the quantity $m=\gamma m_0$.

To put it another way, the only difference between $E=\gamma m_0 c^2$ and the relativistic mass $m = \gamma m_0$ is a factor of $c^2$. So why not just use $E$, which is a useful quantity, instead of $m$, which isn't and which leads to misconceptions?

12. Dec 30, 2013

Staff Emeritus
As you discovered later, it is not the kinetic energy. It's the total energy.

So you've already fallen into the trap and generated a wrong equation using relativistic mass. This is a good reason to hate the concept: it causes the inexperienced to think they understand things when they don't, and to make mistakes because of it. (And I am perpetually puzzled why these people end up defending it)

You want an equation for kinetic energy? T = (γ-1)m. It's no more complicated than what you wrote down, and has the advantage of being correct.

13. Dec 30, 2013

I don't see where the "useless tautology" comes in because you have used the same terminology referred to previously.Please take another look. You stated that:
Total energy = rest energy + KE. Agreed if the equation is to be expressed in terms of units such as Joules. But surely this is the same as Total mass =rest mass + KE if the equation including the KE is to be expressed in units such as Kg. It's often normal to express energy, including K.E. in units other than Joules.

14. Dec 30, 2013

### PAllen

It's a tautology because you've given a second name to total energy, which has no need of a second name. Whatever your units, you don't need a second name or a second formula, which leads to erroneous conclusions.

15. Dec 30, 2013

Thanks to all those who have replied.A general sticking point seems to be that people seem to dislike some of the symbols I referred to and the fact that I referred to mass rather than energy.The thing is that mass can be measured in energy units and energy can be measured in mass units.There are different units for energy and mass,for example the J, the Kg, the u and the eV.We would normally use the most convenient units for the job in hand. I agree with equations such as KE= gamma mo c^2 if KE is to be expressed in J but I also see that KE=mo if KE is to be expressed in Kg .I can also see that expressing it the second way can cause misunderstanding if care is not taken to define the terms and units used.
From now onwards I will try to use E instead of m and the Joule instead of the Kg. I am not defending the equation,I just wanted it confirmed that I can use it to calculate KE. It seems I can (with the preferred symbols etc of course)

Last edited: Dec 30, 2013
16. Dec 30, 2013

### vela

Staff Emeritus
It's not so much the units. You're perfectly free to use the convention where $c=1$. In that case, however, you already have that the total energy of a particle is given by $E=\gamma m_0$. What's the point then of using $m = \gamma m_0$? It's redundant. If you don't use $c=1$, the only difference between $E$ and $m$ quantitatively is a constant factor of $c^2$, and conceptually, $E$ is much more useful. The concept of relativistic mass, on the other hand, leads to errors and misconceptions. It's best to avoid it.

I guess my question to you is why bother defining $m=\gamma m_0$ so that you can write $K = m-m_0$ instead of just saying $K = E-m_0$?

17. Dec 30, 2013

### PAllen

And, in terms of conceptual consistency,

KE = E - E0

makes the most sense (independent of units, all the terms of the same kind).

18. Dec 30, 2013

E=gamma mo is the same as m=gamma mo if m is defined to be E. It's just that I am more used to writing m instead of E,something I will change when appropriate. KE=E-Eo looks neater.
Given the rest mass of an electron (Mo) and given its speed I would calculate its KE as being given by:
KE=Mo(gamma-1) if the KE is to be expressed in Kg
if it is to be expressed in the more usual Joules I would multiply by c squared.
It is KE that I am interested in and I dont see any need to refer to E in my calculation.

Really tired now and probably making mistakes.Night night.

19. Dec 31, 2013

### devang2

Relative mass is used to calculate momntum of particle in motion The equation which relates energy of particle in motion to the energy at rest and momentum is based on relative mass equation therefore it is very useful.

20. Dec 31, 2013

Staff Emeritus
Baloney.

The equation is p = mγv

There is no reason to lump it together as either p = (mγ)v (relativistic mass) or as p = m(γv) (I guess you could call it relativistic velocity, but nobody does that). In either case, you can only use this in this particular equation - and you can always pick one equation in which you can keep the form intact. We could have picked E = 1/2 mv2, and have gotten an entirely different relativistic mass definition.

The whole thing is based on the idea that an equation with three variables is too complicated. Which is just plain silly, given that there are plenty of such equations that appear even earlier in the standard physics treatment.

21. Dec 31, 2013

### PAllen

I consider that 4-vectors give a preferred way to factor such equations.

We have:

P = m U

the gamma is encoded in the 4-velocity, and this need not even be considered in a coordinate basis. And U is simply the unit tangent vector to the world line. Further, for simple dynamics, we have, generally:

F = dP/d$\tau$

and if m is constant,

F = m A, with A being 4-acceleration = dU/$\tau$

The 4-acceleration's norm is the proper acceleration - what is measured by an accelerometer.

Note that to ask about the energy or KE of a body, you must really describe the measuring device as well as the particle. Thus, if you posit U1 for the particle, and U2 for the device, then total energy as measured by this device is simply:

E = m U1 dot U2

and KE is E - m (with units of c=1).

22. Dec 31, 2013

### pervect

Staff Emeritus
Obviously we have a fan of relativistic mass here. It would be unfair to say that the concept is always useless. But the majority of the time it causes confusion, just visit any of the threads about "if you go too fast, do you turn into a black hole" - and try to explain why you do not turn into a black hole in the paradigm of "relativistic mass".

If you believe that you DO turn into a black hole when you move rapidly, you are most likely unwittingly already a victim of the shortcomings in the relativistic mass paradigm :-(.

The wording of the question seems to be begging for people to say positive things about relativistic mass. "Can it" be a useful concept? In the right circumstances, maybe. Does it turn out to be more useful than a hindernace? I would say no.

23. Dec 31, 2013

### Staff: Mentor

I think that is an excellent note on which to close this thread. There is no science content here, just semantics and opinion.

As pervect mentioned, of course "relativistic mass" is on occasion a useful concept, otherwise it would never have been coined in the first place. However, the modern scientific view is that it is typically far less useful than "invariant mass", so the unqualified term "mass" almost always refers to "invariant mass" rather than "relativistic mass".