Derivation of the Equation for Relativistic Mass

[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.

 

[Roman]

hello,

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

 

[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.

 

[cristo]

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 $m=m_0\gamma(u)$.

 

[bernhard.rothenstein]

mass in special relativity

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.

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 which 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.

 

[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.

 

[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 ...

 

[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.

 

[jtbell]

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 $m = \gamma m_0$.

 

[bernhard.rothenstein]

relativistic dynamics

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 $m = \gamma m_0$.

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

 

[Doc Al]

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.

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

 

[bernhard.rothenstein]

mass momentum

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

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

 

[robphy]

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

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.

 

[nakurusil]

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.

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.

 

[quantum123]

He used conservation of momentum for deriving eqn 5?

 

[nakurusil]

He used conservation of momentum for deriving eqn 5?

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

 

[Doc Al]

For me, I prefer an approach that uses the velocity-composition formula to analyze a general collision.

Exactly. That's what I was thinking of.

 

[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.

 

[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?

 

[nakurusil]

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.

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

 

[Fredrik]

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

I had to think for more than an hour to make sense of this calculation, so I'll try to explain it to those who gave up after running into the same problems.

S' is the frame where the speeds are u' and -u'. S is an arbitrary frame for the moment. We will make a specific choice later.

V = the velocity of S' in S
u1 = the velocity of object 1 in S
u2 = the velocity of object 2 in S

I'm using units such that c=1.

The addition of velocities formula yields (2) and (3), except that there's a typo in (3). The LHS should say u2.

$$u_1=\frac{u'+V}{1+u'V}$$

$$u_2=\frac{-u'+V}{1-u'V}$$

Now suppose that the two objects will stick to each other. Let's choose their masses to be the same. Then the speed of the two objects in S' will be 0. This means that the speed in S' will be V.

Now suppose that relativistic conservation of momentum in frame S can be expressed as

$$m(u_1)u_1+m(u_2)u_2 = (m(u_1)+m(u_2))V$$

where this "relativistic mass" is assumed to be a function of velocity. m(0) is the rest mass. This is (1) expressed in a more transparent way. Note that this wouldn't hold if we hadn't already assumed that the two rest masses are the same.

Our goal is to determine the form of the function m.

First solve (1) for m(u1)/m(u2). The result is

$$\frac{m(u_1)}{m(u_2)}=\frac{V-u_2}{u_1-V}$$

Then insert the results (2) and (3) (mind the typo though) into this, and simplify. The result is (4).

$$\frac{m(u_1)}{m(u_2)}=\frac{1+u'V}{1-u'V}$$

Now choose the frame S so that u2=0. This turns (3) into u'=V. Let's write u instead of u1, and $m_0$ instead of m(0). Equation (4) now takes the form

[tex]m(u)=m_0\cdot\frac{1+V^2}{1-V^2}=m_0\cdot\frac{1+V^2}{1-V^2}[/itex]

We want the result as a function of u, so we use the result u'=V in (2), and solve for V. The result is

$$V=\frac{1+\sqrt{1-u^2}}{u}$$

It should be easy from here, but for some reason I don't see how to do the simplification. Anyone else feel like finishing this one off?

Edit: I got it.

The result u'=V turns (2) into

$$u=\frac{2V}{1+V^2}$$

I found a way to use this.

$$\frac{1+V^2}{1-V^2}=\frac{1+V^2}{1+V^2-2V^2}=\frac{1}{1-\frac{2V^2}{1+V^2}}=\frac{1}{1-uV}=\frac{1}{1-(1+\sqrt{1-u^2})}=\frac{1}{\sqrt{1-u^2}}=\gamma$$

 

[nakurusil]

I had to think for more than an hour to make sense of this calculation, so I'll try to explain it to those who gave up after running into the same problems.

S' is the frame where the speeds are u' and -u'. S is an arbitrary frame for the moment. We will make a specific choice later.

V = the velocity of S' in S
u1 = the velocity of object 1 in S
u2 = the velocity of object 2 in S

I'm using units such that c=1.

The addition of velocities formula yields (2) and (3), except that there's a typo in (3). The LHS should say u2.

$$u_1=\frac{u'+V}{1+u'V}$$

$$u_2=\frac{-u'+V}{1-u'V}$$

Now suppose that the two objects will stick to each other. Let's choose their masses to be the same. Then the speed of the two objects in S' will be 0. This means that the speed in S' will be V.

Now suppose that relativistic conservation of momentum in frame S can be expressed as

$$m(u_1)u_1+m(u_2)u_2 = (m(u_1)+m(u_2))V$$

where this "relativistic mass" is assumed to be a function of velocity. m(0) is the rest mass. This is (1) expressed in a more transparent way. Note that this wouldn't hold if we hadn't already assumed that the two rest masses are the same.

Our goal is to determine the form of the function m.

First solve (1) for m(u1)/m(u2). The result is

$$\frac{m(u_1)}{m(u_2)}=\frac{V-u_2}{u_1-V}$$

Then insert the results (2) and (3) (mind the typo though) into this, and simplify. The result is (4).

$$\frac{m(u_1)}{m(u_2)}=\frac{1+u'V}{1-u'V}$$

So far so good.

Now choose the frame S so that $$u_2=0$$.

I am not sure you can do that, S has already been chosen such that it moves with speed $$V$$ wrt S', remember. You just added another condition. I think you need to continue to use Tolman's reasoning:

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

 

[Fredrik]

S has already been chosen such that it moves with speed $$V$$ wrt S', remember.

This is actually the detail the caused me the most problems when I was trying to understand the derivation you posted, but I think I got it right. S is not fixed by the choice that the velocity of S' in S is V, as long as we haven't specified the value of either u1, u2 or V.

By the way, you also put u2=0 near the end, so if my approach has this problem, then so does Tolman's.

 

[bernhard.rothenstein]

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.

Start with the identity (Rosser)
g(u)=g(V)g(u')(1+u'V/cc) (1)
Multiply both its sides with m (invariant rest mass) in order to obtain
mg(u)=g(V)mg(u')(1+u'V/cc). (2)
Find out names for
mg(u), mg(u') and mg(u')u'
probably in accordance with theirs physical dimensions. In order to avoid criticism on this Forum avoid the name relativistic mass for mg(u) and mg(u) using instead E=mc^2g(u) and E'=mc^2g(u') calling them in accordance with theirs physical dimensions relativistic energy in I and I' respectively using for p=Eu/c^2 and p'=E'u'/c^2 the names of relativistic energy. Consider a simple collision from I and I' in order to convince yourself that they lead to results in accordance with conservation of momentum and energy.
Is there more to say?
sine ira et studio

 

[nakurusil]

This is actually the detail the caused me the most problems when I was trying to understand the derivation you posted, but I think I got it right. S is not fixed by the choice that the velocity of S' in S is V, as long as we haven't specified the value of either u1, u2 or V.

By the way, you also put u2=0 near the end, so if my approach has this problem, then so does Tolman's.

Ummm, no. Tolman deduces the condition for arbitrary $$u_1$$ and
$$u_2$$ and that makes it perfectly legal to put $$u_2=0$$

 

[Fredrik]

But u1 and u2 are not arbitrary once we have fixed the value of V. These velocities have to transform to u' and -u' respectively, so they are fixed by that. (If the frame S isn't arbitrary in the first calculations, then u1 and u2 aren't arbitrary either). I still say that u2 is arbitrary in the first calculations (because S is arbitrary), and that the choice u2=0 is what finally fixes the value of V (and makes S one specific frame rather than an arbitrary frame).

By the way, I have updated #21 with the part of the calculation that I wasn't able to figure out yesterday, so it's complete now.

 

[nakurusil]

But u1 and u2 are not arbitrary once we have fixed the value of V. These velocities have to transform to u' and -u' respectively, so they are fixed by that. (If the frame S isn't arbitrary in the first calculations, then u1 and u2 aren't arbitrary either). I still say that u2 is arbitrary in the first calculations (because S is arbitrary), and that the choice u2=0 is what finally fixes the value of V (and makes S one specific frame rather than an arbitrary frame).

By the way, I have updated #21 with the part of the calculation that I wasn't able to figure out yesterday, so it's complete now.

I understand your confusion. Tolman's derivation is indeed a very ugly one but it is correct. Let me try again:

Tolman derives the formula

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

The formula holds for ARBITRARY $$u_1$$ and $$u_2$$. So we can now forget how the formula was derived, set $$u_2=0$$ and get:

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

Does this make sense now?

 

[Fredrik]

No, it doesn't hold for arbitrary u1 and u2. It holds for an arbitrary choice of frame S. We choose a specific S by setting u2=0. I'm not confused about this at all.

 

[nakurusil]

No, it doesn't hold for arbitrary u1 and u2. It holds for an arbitrary choice of frame S. We choose a specific S by setting u2=0. I'm not confused about this at all.

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

$$u_2=(-u'+V)/(1-u'V/c^2)$$ (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)

$$u'$$ has been taken arbitrary (this makes both $$u_1$$ and $$u_2$$ arbitrary) , therefore on can easily make $$u'=V$$ (or $$V=u'$$) thus making $$u_2=0$$ and $$u_1=2V/(1+V^2/c^2)=v$$ then (5) becomes:

$$m_1=m(v)=m(0)/\sqrt(1-v^2/c^2)=m_0/\sqrt(1-v^2/c^2)$$ (6)

 

[Fredrik]

I'm not sure what you were hoping to accomplish by just reposting stuff that you've said before, but after some thinking I've come to the conclusion that your approach works too. I still think it's a very strange approach though. In my opinion, my approach makes a lot more sense.

You claim that u', u1 and u2 are all arbitrary, but that V is fixed. If V is fixed, then (3) shows that once you choose a specific value of u2, this also fixes the value of u', and (2) shows that that fixes the value of u1. So if things are as you say, then u1 is no longer a variable.

This means that it's far from obvious that the final result can be interpreted as a function of u. All we have so far is a result that's valid for that particular value of u. What saves the day is that a different choice of V in the beginning would yield a different u in the final result. So by pretending that we're repeating the calculations for an infinite number of frames S(V), one for each V in the interval (0,1), we can see that it does in fact make sense to interpret the final result as a function.

 

[nakurusil]

I'm not sure what you were hoping to accomplish by just reposting stuff that you've said before, but after some thinking I've come to the conclusion that your approach works too. I still think it's a very strange approach though. In my opinion, my approach makes a lot more sense.

Of course

You claim that u', u1 and u2 are all arbitrary,

Why don't you read the previous post (more carefully):

$$u'$$ has been taken arbitrary (this makes both $$u_1$$ and $$u_2$$ arbitrary) , therefore one can easily make $$u'=V$$ thus making $$u_2=0$$ and $$u_1=2V/(1+V^2/c^2)=v$$ then (5) becomes:

$$m_1=m(v)=m(0)/\sqrt(1-v^2/c^2)=m_0/\sqrt(1-v^2/c^2)$$

but that V is fixed. If V is fixed, then (3) shows that once you choose a specific value of u2,

...but this is not what I am doing, why don't you read above?<rest snipped as irrelevant since it is based on a basic misunderstanding>

 

[Fredrik]

You're not choosing u2=0?! What you're saying in the last post is that you're choosing u'=0. Fine. You can do that, but that's equivalent to choosing u2=0. It's the same thing.

(2) and (3) is a system of equations with four "variables": u1, u2, u' and V. If you fix two of them, then the other two are completely determined by (2) and (3). You fixed V in your definition of S, so when you choose a specific value of any of the remaining three variables, the other two will be fixed as well. That's why choosing u'=V is the same thing as choosing u2=0.

 

[nakurusil]

You're not choosing u2=0?! What you're saying in the last post is that you're choosing u'=0. Fine. You can do that, but that's equivalent to choosing u2=0. It's the same thing.

(2) and (3) is a system of equations with four "variables": u1, u2, u' and V. If you fix two of them, then the other two are completely determined by (2) and (3). You fixed V in your definition of S, so when you choose a specific value of any of the remaining three variables, the other two will be fixed as well. That's why choosing u'=V is the same thing as choosing u2=0.

This is why math was invented, in order to make sure that there is no ambiguity.

$$u'$$ has been taken arbitrary (this makes both $$u_1$$ and $$u_2$$ arbitrary) , therefore one can easily make $$u'=V$$ thus making $$u_2=0$$ and $$u_1=2V/(1+V^2/c^2)=v$$ then (5) becomes:

$$m_1=m(v)=m(0)/\sqrt(1-v^2/c^2)=m_0/\sqrt(1-v^2/c^2)$$
So, whenever you have problems you can read Tolman , pages 43,44 in the reference. If you still have problems with his solution, I suggest you take it up with the editor since Tolman is long dead.

 

[Fredrik]

I don't have problems with Tolman or his solution. I found two consistent interpretations of the partial solution you posted. One of them was based on stuff that you had said but later denied. I posted both interpretations in this thread.

I do have a problem with you though. A lot of the things you're saying are wrong. In this thread your statements aren't wrong in the absolute sense, but they are contradicting each other. Simply reposting the same thing again won't change that.

My part in this discussion ends here. <<mentor snip>>

 

[nakurusil]

Start with the identity (Rosser)
g(u)=g(V)g(u')(1+u'V/cc) (1)
Multiply both its sides with m (invariant rest mass) in order to obtain
mg(u)=g(V)mg(u')(1+u'V/cc). (2)
Find out names for
mg(u), mg(u') and mg(u')u'
probably in accordance with theirs physical dimensions. In order to avoid criticism on this Forum avoid the name relativistic mass for mg(u) and mg(u) using instead E=mc^2g(u) and E'=mc^2g(u') calling them in accordance with theirs physical dimensions relativistic energy in I and I' respectively using for p=Eu/c^2 and p'=E'u'/c^2 the names of relativistic energy. Consider a simple collision from I and I' in order to convince yourself that they lead to results in accordance with conservation of momentum and energy.
Is there more to say?
sine ira et studio

This seems to lead back to Tolman's derivation mentioned above. Same equations , you would need to add the momentum conservation in the collision as you mention.