# I Relativistic mass increase (simple or not)

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1. Aug 8, 2017

### sha1000

Hi, I have yet another question in the field of the special relativity theory.

I always thought that the derivation of the relativistic mass is trivial. But I discovered that there is some complexity due to the transverse and longitudinal relativistic mass derivation.

I saw a thread where people talked about different aproaches which can demonstrate that the transverse relation must be used for relativistic mass calculation.

For me it always seemed normal to only consider the transverse relation (in y direction). Since we have the length contraction phenomena in x direction which "compensate" the difference between transverse and longitudinal 2 way light travel.

I'm not an expert and maybe my vision is too simplistic but i would like to know if this arguments are sufficient for demonstration of transverse choice.

2. Aug 8, 2017

3. Aug 8, 2017

### Staff: Mentor

This is one of the main reasons to avoid the concept of relativistic mass entirely. The preferred concept is the invariant mass, and when you see the unqualified term "mass" in the modern professional literature it almost always refers to the invariant mass.

4. Aug 8, 2017

### pervect

Staff Emeritus

Relativistic mass seems to be very popular with the lay audience, but is mostly ignored in the modern treatments of relativity.

Relativistic mass is equivalent to energy, so as a matter of semantics people use "energy' when they want to talk about what was formerly called 'relativistic mass', and use invariant mass for mass.

But usually the confusion arises about the issues related to relativistic forces.

The modern formulation is to write $p = \gamma m v$ where p is momentum, $\beta = v/c$ and $\gamma = 1/\sqrt{1-\beta^2}$

Then if we define the 3-force F = dp/dt, i.e that force is the rate of change of momentum with time, we can write $F = d(\gamma m v) / dt$. That's all there is to it, though it's helpful to apply the chain rule from calculus to the resulting expression.

I've gone through this spiel before, and met resistance or incomprehension. My best guess at the difficulty is that it does require that one remember what a derivative is from elementary calculus. This isn't terribly demanding, really. But if one is remembering F=ma by rote, and has forgotten all the supporting calculus, I suppose I can see how it could seem difficult. It's less clear what to do about the seeming difficulty, I suppose my attitude is that those serious enough about it will learn calculus, and the less serious will do the best they can.

A further complication arises in that the 3-force F isn't really the best approach to learning relativistic dynamics. But while I think the point is relevant, it starts to digress from my best estimate of what the thread is about.

5. Aug 8, 2017

6. Aug 8, 2017

### Mister T

It's all in what you prefer. It has nothing to do with the physics. Originally transverse mass (which is equal to $\gamma m$) and longitudinal mass ($\gamma^3 m$) were used in an attempt to relate the force exerted on a particle to its acceleration. Particles were thought to have a different mass depending on the direction of the applied force, that is, directed parallel (longitudinal) to the motion or perpendicular (transverse) to the motion; and of course on how fast the particle is moving. The concept of relativistic mass was also introduced and it's equal to the transverse mass. Once all of this got sorted out it became clear that it's not the mass that's changing, but rather that the behavior is due to the geometry of spacetime. Unfortunately the concept of relativistic mass stuck around through most of the 20th century, even though the high energy physicists and engineers who deal with high speed particles every day rarely if ever used it.

In the 1990's we saw it begin to disappear from introductory physics textbooks and is now considered an antiquated concept. There is only one mass used, $m$, and it's the ordinary mass.

7. Aug 8, 2017

### SiennaTheGr8

One subtlety: relativistic mass was often defined as $E / c^2$, whereas transverse mass was typically defined as $\gamma m$. For a particle with non-zero [rest] mass, these are equivalent. But for a particle with zero [rest] mass, they are not.

(Semantics, really.)

8. Aug 8, 2017

### Mister T

Or history! Here's a possible example of the way the terminology may have been used. There may be others.

Remember that while Einstein was formulating GR he first predicted that light would be bent by a certain amount when it grazed the sun, but then later changed his prediction to be double his first prediction. It could very well be that during those few years in the early part of the 20th century at least some physicists were thinking of photons with a transverse mass equal to $E/c^2$ to explain the deflection.

9. Aug 9, 2017

### vanhees71

Einstein's first calculation was not done with the fully developed theory, and that's why he got a wrong result. Using the non-relativistic approximation to describe the gravitational field gives only the $g_{00}$ component in terms of the Newtonian gravitational potential,
$$g_{00}=1+2 \phi(\vec{x})/c^2$$
but not the corresponding radial component of the Schwarzschild pseudometric. That's why you get only half of the deflection of light in this naive non-relativistic approximation.

All this has nothing to do with the very misleading idea of a socalled "relativistic mass", and if you read Einstein's addendum to the famous paper of 1905, where he develops the energy-mass relationship, you'll see that he got it right from the very beginning (not very surprisingly given his really amazing physics intuition).

In modern terms you can summarize the definition of mass (as far as classical, i.e., non-quantum physics) is concerned as follows: You start defining the total energy and momentum of a many-body system such that these quantities form the components of a relativistic four-vector, the four-momentum vector, $p=(E/c,\vec{p})$. Then either $p^2>0$ or $p^2=0$. If $p^2>0$ you can always find an inertial frame of reference, where the total momentum vanishes. You just have to boost with a velocity of $c \vec{p}/E$. In the center-of-mass frame then $E=m c^2$, where $m$ is the invariant mass, and this is a Lorentz scalar. It can be expressed in any inertial frame with help of the invariant formula
$$p^2=E^2/c^2-\vec{p}^2=m^2 c^2.$$
If $p^2=0$, then there is no "rest frame", but you can use the above formula to define the invariant mass of such an object as $m=0$. Note that within classical physics that's a bit strange, but sometimes you can use it to describe "classical photons", which however is as bad a picture as it can be.

The full beauty of the definition of the fundamental physical quantities is revealed in quantum theory, which of course is based on the fundamental symmetries, which can be taken as the prime outcomes of empirical and theoretical research in about 400 years of physics.

First of all you start an analysis of the symmetries of space and time or, the better notion in the relativistic case, the spacetime manifold. In case of Special Relativity the space-time manifold is a fixed affine pseudo-Euclidean manifold with a fundamental form of signature (1,3) (I'm a west-coast conventionalist; of course the east-coast convention with the signature (3,1) is completely equivalent, it's just a matter of convention).

In classical physics the relevant symmetry group is the proper orthochronous Poincare group, i.e., the semidirect product of spacetime translations and proper orthochronous Lorentz transformations. To construct possible Hilbert spaces for the description of special-relativistic systems one should thus find the unitary ray representations of the proper orthochronous Poincare group to be able to formulate a quantum theory compatible with the mathematical structure of the spacetime manifold. It turns out (Wigner 1939) that all that happens is that the proper orthochronous Lorentz group is substituted by its covering group, $\mathrm{SL}(2,\mathbb{C})$, and all irreducible unitary ray representations can be lifted to unitary representations. In contradistinction to the Galilei group no non-trivial central charges occur, and that's why mass in relativistic physics turns out to be a Casimir operator of the Poincare group, related to energy and momentum (defined as the generators of spacetime translations) as in the classical case.

To make the long and somewhat math-loaden discussion short: Mass is a scalar and nothing but a scalar in relativistic physics!

Last edited: Aug 9, 2017
10. Aug 9, 2017

### DrStupid

Why should they do that? The deflection is independent from the mass.

11. Aug 9, 2017

### nitsuj

Great post!

I didn't know of that path of reasoning attempted between f=ma to the indexed one...but imo it's too bad nature didn't play out the more intuitive way, how cool to follow this "directional mass" around from one object, bumped into the next.

Still don't make sense once ya posit an invariant speed and isotropic space......not that "Oh it's just the geometry" is intuitive lol There is a lot to go over once c is determined to be invariant.

12. Aug 10, 2017

### sha1000

Thank you all for your help and kind responses.