# Derivation of the Equation for Relativistic Mass

1. Jan 27, 2007

### NanakiXIII

Throughout pages on the internet I've seen the following relationship between rest mass and relativistic mass:

$$m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$

However, I have been utterly unable to find any sort of derivation or explanation of this formula, other than such explanations as "nearing the speed of light, an object's mass nears infinity and thus.. etc.". Where did this equation come from? Does it somehow follow from the Lorentz transformations? Any helpful insights would be much appreciated.

2. Jan 27, 2007

### Roman

hello,

this equation comes from lorentz transformations. just look in any book about special relativity

3. Jan 27, 2007

### cesiumfrog

One possible starting point is $E=mc^2$, $E^2=(m_0c^2)^2+(pc)^2$, so I suppose you could get it from the symmetries of minkowskii space (and the equivalence of relativistic mass and energy). Alternatively, you might want to consider from first-principles a constantly accelerated box of moving particles or photons, and show that the accelerating force on the box (hence the inertial mass of the system) depends on the internal momentum (relativistic mass) of the contents.

4. Jan 27, 2007

### cristo

Staff Emeritus
One way to see how the relativistic mass arises, is to consider the particle's four momentum P=m0U, where $$\bold{U}=\left(\frac{dx}{d\tau},\frac{dy}{d\tau},\frac{dz}{d\tau},\frac{dct}{d\tau}\right)$$, the particle's four velocity. So, $$\bold{P}=m_0\bold{U}=m_0\gamma(u)(\bold{u},c)=(m\bold{u},mc)=:(\bold{p},mc)$$ where u and p are the particle's three velocity and "relativistic momentum", respectively, and m is the relativistic mass defined as $m=m_0\gamma(u)$.

Last edited: Jan 27, 2007
5. Jan 27, 2007

### bernhard.rothenstein

mass in special relativity

IMHO you can consider the equation which relates proper mass and relativistic mass (horribile dictu) as an experimental fact obtained by Kaufmann, Bucherer and probably many others. American Journal of Physics offers its derivation via thought experiments, whereas others derive it considering an intercation between a photon and a tardion, and taking into account conservation of momentum and energy. I am fond of such derivations wich are simple and involve subtle physical thinking. If you are interested in the papers I mention, please let me know.
The best things a physicist can offer to another one are information and constructive criticism in the spirit of IMHO.

6. Jan 28, 2007

### NanakiXIII

Roman: The only book about Special Relativity I have at my disposal at the moment is Relativity by Albert Einstein. It doesn't mention anything about the topic. If you know of any sources on the internet that contain the same information, I'd like to hear of them.

cesiumfrog: I'm actually interested in understanding this equation because it is at the basis of a derivation for E=mc^2 I found.

In general, my preference lies with a derivation starting from the Lorentz transformations, not in the least because my knowledge and understanding of physics and mathematics is limited. cristo's post, for example - and you may think me lazy for not doing my homework - is incomprehensible to me.

bernhard.rothenstein: If those papers are understandable for someone with little knowledge of the matter, in this case a mere high school student, then yes, I might be interested. If you could provide me with some more information on these papers, I'd much appreciate it.

So thank you all for your replies. If anyone knows, however, of a derivation of this equation using the Lorentz transformations and preferably no other assumptions that are specifically within the theory of Special Relativity, if such a derivation exists at all, that would be the greatest help.

7. Jan 28, 2007

### Gib Z

It would be hard to derive that with no assumptions from Special Relativity, since one of the axioms is that C is constant ...

8. Jan 28, 2007

### NanakiXIII

I didn't say no assumptions. I said preferably as few as possible besides the Lorentz transformations, which account for the constancy of the speed of light.

9. Jan 28, 2007

### Staff: Mentor

Some introductory textbooks analyze glancing collisions between objects, using time dilation and length contraction, to derive that momentum should equal $m_0 \gamma \vec v$ in order to preserve conservation of momentum. That is, they show that $m_0 \vec v$ is not conserved, but $m_0 \gamma \vec v$ is (where $\gamma = 1 / \sqrt {1 - v^2 / c^2}$) and use this as justification for redefining momentum accordingly.

From that, one can take an additional step and say that we can preserve the classical formula $p = mv$ by defining $m = \gamma m_0$.

10. Jan 28, 2007

### bernhard.rothenstein

relativistic dynamics

Please quote places where using glancing collisions, time dilation and length contraction lead to derive the momentum. Thanks

11. Jan 28, 2007

### Staff: Mentor

Wouldn't such an analysis apply to any collision, not just glancing?

12. Jan 28, 2007

### bernhard.rothenstein

mass momentum

have a look at
American Journal of Physics -- September 1986 -- Volume 54, Issue 9, pp. 804-808

An alternate derivation of relativistic momentum
P. C. Peters
Department of Physics FM-15, University of Washington, Seattle, Washington 98195

(Received 5 August 1985; accepted 13 September 1985)

An alternate derivation of the expression for relativistic momentum is given which does not rely on the symmetric glancing collision first introduced by Lewis and Tolman in 1909 and used by most authors today. The collision in the alternate derivation involves a non-head-on elastic collision of one body with an identical one initially at rest, in which the two bodies after the collision move symmetrically with respect to the initial axis of the collision. Newtonian momentum is found not to be conserved in this collision and the expression for relativistic momentum emerges when momentum conservation is imposed. In addition, kinetic energy conservation can be verified in the collision. Alternatively, the collision can be used to derive the expression for relativistic kinetic energy without resorting to a work-energy calculation. Some consequences of a totally inelastic collision between these two bodies are also explored. ©1986 American Association of Physics Teachers

13. Jan 28, 2007

### robphy

If I recall correctly, the glancing collision is a simple case which facilitates a motivation of the relativistic 3-momentum. However, I am generally unhappy with that approach. For me, I prefer an approach that uses the velocity-composition formula to analyze a general collision.

14. Jan 28, 2007

### nakurusil

This is a preocupation with a useless notion (there are many threads that explain why "relativistic mass" is a waste of time, you can check them out in this forum).
Having said that, the type of "elementary" derivatin that you are looking for can be found in the many articles written by R.C.Tolman on the subject. For example, he uses collision thought experiments, see pages 43,44 in his book "Relativity, Thermodynamics and Cosmology". In this experiment he looks at a collision between two masses:

$$m_1u_1+m_2u_2=(m_1+m_2)V$$ (1)

from the point of two different frams S and S' in relative motion with speed V.

In frame S' the two masses are moving at speed +u' and -u' respectivelly.
Using the speed composition law, Tolman shows that :

$$u_1=(u'+V)/(1+u'V/c^2)$$ (2)

$$u_1=(-u'+V)/(1-u'V/c^2)$$ (3)

Substituting (2)(3) into (1) he gets:

$$m_1/m_2=(1+u'V/c^2)/(1-u'V/c^2)$$ (4)

Since he showed earlier that:

$$1+u'V/c^2=\sqrt(1-u'^2/c^2)*\sqrt(1-V^2/c^2)/\sqrt(1-u^2/c^2)$$

substituting in (4) he gets:

$$m_1/m_2=\sqrt(1-u_2^2/c^2)/\sqrt(1-u_1^2/c^2)$$ (5)

If one takes $$u_2=0$$ then $$m_2=m_0$$ (the "rest mass"), $$u_1=u$$ and (5) becomes:

$$m_1=m(u)=m_0/\sqrt(1-u^2/c^2)$$ (6)

Very ugly and useless.

15. Jan 28, 2007

### quantum123

He used conservation of momentum for deriving eqn 5?

16. Jan 28, 2007

### nakurusil

Yes, look at (1). Or check out the book if you prefer. Tolman wrote a series of papers on the subject.

17. Jan 28, 2007

### Staff: Mentor

Exactly. That's what I was thinking of.

18. Jan 28, 2007

### arcnets

I think it's more simple, but I can't do the nice formulae, sorry...
First, from the invariance of c for all observers, you get the equation
(ct)^2 - x^2 = invariant for all observers.
Next, you multiply with m^2 and divide by t^2:
(mc)^2 - (mx/t)^2 = invariant.
Now if one observer (0) is in rest frame, then x0/t0 = v0 = 0:
(m0 c)^2 = (mc)^2 - (mx/t)^2
Solve that for m and there you are.

19. Jan 28, 2007

### cesiumfrog

arcnets, i don't like the RHS of your second equation.. can you show why m^2 x invariant / t^2 should itself be invariant?

20. Jan 28, 2007

### nakurusil

This is not an acceptable derivation, it is a "sleigh of hand". Can you detect the "trick"?