# The concept of inertial mass from Newton to Einstein

1. Mar 7, 2013

### James MC

The most common definition of mass, at least in the Newtonian context, is in terms of a measure of inertia: The mass of an object is a measure of, and gives rise to, its resistance to changes in motion.

F=MA presumably quantifies this idea of inertial mass.

I'm wondering whether any property (at all) in special relativity exactly fits this definition, and therefore deserves the label "inertial mass".

In relativity, F=MA no longer holds, because the relationship between F and A depends on the orientation of F with respect to V. When F and V are perpendicular, F=$\gamma$MA, and when F and V are parallel, F=$\gamma^3$MA.

Consider a space ship firing its constant-thrust engine (constant F in the direction of V); even though F and M are constant, A continually descreases such that v never reaches C. (Incidently, if you understand how merely cubing gamma achieves this, please say.) Hence, M cannot be a measure of acceleration resistance.

What, if anything, is the inertial mass, in SR? Here is one answer: the best one can say is that there is an "inertial mass" associated with the F that is perpendicular to the V, which is equal to $\gamma$M, and there is an "inertial mass" associated with the F that is perpendicular to the V, which is equal to $\gamma^3$M.
Is this the only way to make sense of inertial mass in SR?

2. Mar 8, 2013

### crissyb1988

When we consider a space ship accelerating where v is close to c the mass of the object DOES NOT stay constant. $m = \gamma m_0$ . $m_0$ is the rest mass of the object and that stays constant. $\gamma$ is the gamma factor and it depends on v. m is the mass of the object in a "stationary, non-accelerating frame" . SO the decreasing of $a$ is due to the increase in relative mass. So the "inertial mass" is $m = \gamma m_0$. And i think you will need to use tensors to describe force if im not mistaken (someone correct me here) as you must factor in the link between space and time..

Hope that makes sense

EDIT: tried using latex code for the first time why isnt it working?
EDIT: Ok its working :)

Last edited: Mar 8, 2013
3. Mar 8, 2013

### James MC

The now standard formulation of SR does not introduce two masses, as you have, but only introduces a single notion of mass that refers to an invariant quantity that is constant in the situation I described.

Anyway, my original question can be asked without even using the word 'mass': in SR, what property (if any), is a measure of, and gives rise to, an object's resistance to changes in motion?

The answer may well have something to do with a move from the equations I mentioned (e.g. where gamma is cubed) to different equations, perhaps involving tensors or some such, but I'm not sure.

Thoughts anyone?

4. Mar 8, 2013

### A.T.

It's the quantity formerly known as "relativistic mass" that crissyb1988 mentioned.

5. Mar 8, 2013

### James MC

But why gamma m rather than gamma cubed m?

6. Mar 8, 2013

7. Mar 8, 2013

### elfmotat

If you define inertial mass as "the coefficient relating momentum to velocity", then in SR it's the relativistic mass $\gamma m$. This is a useful definition because $F=dp/dt$ still holds true in relativity when momentum is defined as $p=\gamma mv$. If you instead define inertial mass as "the coefficient relating force to acceleration" then it's actually a matrix. If a particle is moving in the x-direction then:

$$F_i=\sum_{j}M_{ij}a_j$$

where:

$$M_{ij}=\begin{bmatrix}\gamma^3 m & 0 & 0\\ 0 & \gamma m & 0\\ 0 & 0 & \gamma m\end{bmatrix}$$

8. Mar 9, 2013

### James MC

Thanks for the link, the original poster's discussion of the baseball example was very helpful.
However, I can't help but notice that post #4 from your link, conflicts with what you say in post #4 of this thread! In post #4 of this thread you state that the inertial mass is the relativistic mass. But in post #4 of your link ZikZak says "The concept of "Relativistic Mass" is a terrible idea because, as you have discovered, it is misleading and wrong, because it is not the inertial mass."

The key point seems to be (i) the value of the "inertial mass" of an object is higher in the direction of its velocity; whereas (ii) the value of the "relativisitic mass" of an object is not higher in the direction of its velocity; therefore (iii) the inertial mass is not the relativistic mass.

Are you thinking of inertial mass, not in terms of the way I defined it, but in terms of Elfmotat's alternative definition?

9. Mar 9, 2013

### James MC

That's a very useful point to make! If one party to the debate says relativistic mass = inertial mass, while the other party says relativistic mass ≠ inertial mass, yet both parties agree on all the empirical facts, then most likely, the two parties are assuming different definitions of inertia.

However, the alternative definition of inertial mass you provide, which enables one to say that inertial mass = relativistic mass, is curious. The fundamental idea behind inertial mass is that the harder an object is to push around in space, the more inertial mass it has. Any proposed definition of inertial mass must connect to this basic idea. Two questions:
(i) How exactly does your alternative definition connect to this idea?
(ii) How does it connect to this idea without ending up applying to a matrix quantity?

I imagine the answer to (i) appeals to the fact that defining inertial mass as "the coefficient relating force to rate of change of velocity" is somewhat restrictive, because our definition should leave open the possibility that the inertial mass changes with the velocity, as the force changes?

I don't know what the answer to (ii) could be. How can you avoid the need to cube gamma for a particular direction, simply by differentiating both gamma and rest mass, along with velocity, in the force equation? It's true that acceleration is explicitly gone from the equation, but we are still differentiating velocity...

This is really helpful! One question though: if I've understood the subscripts, then $a_j$ is a row vector, where each element corresponds to the acceleration in each of three spatial dimensions? But in that case, why require inertial mass to be a matrix? Why not just eliminate the zeros, and let it be a row (or column) vector too?

10. Mar 9, 2013

### A.T.

I was thinking in terms of momentum. But I agree that the term "relativistic mass" can be misleading for the reasons stated by others.

11. Mar 9, 2013

### Staff: Mentor

Excellent reference. I don't really understand why you are asking about relativistic mass when you have read this reference.

Mass. $F^{\mu}=m \; dU^{\mu}/d\tau$

Here m is the invariant mass, F is the four-force, and U is the four-velocity. There is no need whatsoever to introduce relativistic mass, a vector mass, nor any other strange mass concept.

http://en.wikipedia.org/wiki/Four-force
http://en.wikipedia.org/wiki/Invariant_mass
http://en.wikipedia.org/wiki/Four-velocity

Last edited: Mar 9, 2013
12. Mar 9, 2013

### Staff: Mentor

And when F and V are at some angle, neither parallel nor perpendicular, things get more complicated, because then F and A are not in the same direction. See the section "What is the relativistic version of F = ma?" in

http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html

However, I would argue along with DaleSpam that the true relativistic version of F = ma must use four-vectors instead of three-vectors.

13. Mar 9, 2013

### DrStupid

p=m·v and F=dp/dt quantifies this idea and it works both in classical mechanics and in special relativity. However, this mass is no longer used in physics.

14. Mar 9, 2013

### elfmotat

You could equally well say that "the fundamental idea behind inertial mass is how hard it is to change an object's momentum."

You can't relate three-force to three-acceleration without invoking longitudinal and transverse mass, so it has to be a matrix.

Because the inner product of two vectors is a scalar, not a vector. The object M has to act on the vector a in such a way that the result is another vector. So it has to either be a scalar (which doesn't work because of the above) or a matrix.

I agree completely. Working in terms of three-force and three-acceleration is notoriously sloppy.

15. Mar 9, 2013

### DrStupid

This would work in special cases only (if the velocity is parallel to an axis). The general formula is

$F = \frac{E}{{c^2 }} \cdot \left( {I + \frac{{v \cdot v^T }}{{c^2 - v^2 }}} \right) \cdot a$

(where I is the identity matrix)