Derivation of the Equation for Relativistic Mass

nakurusil

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.

bernhard.rothenstein

Thanks. Please let me know in which way does the derivation I have presented leads back to Tolman as not involving conservation of momentum and energy?

Fredrik

This looks interesting, but I don't understand what you're doing here. In particular, what is the definition of the function g, and how did you obtain identity (1)?

bernhard.rothenstein

thanks. g(u) and g(u') stand for the gamma factor in the inertial reference frames I and I. We obtain the relativistic identity by expressing g(u) as a function of u' via the addition law of relativistic velocities (see W.G.V Rosser "Classical electromagnetism via relativity" London Butterworth 1968) pp 165-173)
For more details please have a critical look at
arXiv.org > physics > physics/0605203

Date: Tue, 23 May 2006 22:28:50 GMT (259kb)
Relativistic dynamics without conservation laws

Subj-class: Physics Education

We show that relativistic dynamics can be approached without using conservation laws (conservation of momentum, of energy and of the centre of mass). Our approach avoids collisions that are not easy to teach without mnemonic aids. The derivations are based on the principle of relativity and on its direct consequence, the addition law of relativistic velocities.
Full-text: PDF only
I would highly appreciate your experience with it.

nakurusil

I showed you how (1) is obtained in my first post about Tolman's solution (the one that you keep trying to re-explain to me). <<mentor snip>> You get it from the fact that The Lorentz transforms satisfy the cndition:

L(u)*L(v)=L(w) where w=(u+v)/(1+uv/c^2))

where L(v)=gamma(v)*|1............-v|
...............................|-v/c^2......1|

Try it, you might even be able to calculate it all by yourself.

nakurusil

You follow the same steps.
You use a hypothetical collision experiment
You use the conservation of momentum equation as "seen" from two different frames (you call them I and I', right?).
This is the formalism employed by Tolman about 100 years ago.

Fredrik

I haven't had time to look at this yet. Maybe tomorrow.

bernhard.rothenstein

With all respect, I think that in the derivation I poposed "collision", "conservation laws" are not mentioned and so it has nothing in common with Tolman but it has with the 100 years old special relativity.

nakurusil

Yes, I remember your paper now. We've talked about it.

NanakiXIII

It's been a while, I left this thread because of the now fortunately deleted bickering. However, having once again picked up this subject, no matter how useless, I've stumbled upon a problem. I refer to this post:

http://www.physicsforums.com/showpos...3&postcount=21

It states:

[tex]m(u_1)u_1+m(u_2)u_2 = (m(u_1)+m(u_2))V[/tex]

However, since after the collision the speeds are no longer [tex]u_1[/tex] and [tex]u_1[/tex], how can you say that the relativistic mass of the "new" object is simply [tex]m(u_1)+m(u_2)[/tex]?

NanakiXIII

Though he only had time for a quick glance, my physics teacher agreed that this looked incorrect.

He also suggested I should use the following equation to derive the equation for relativistic mass (I was too lazy to LaTeX, so here's a link):

http://hyperphysics.phy-astr.gsu.edu...grel/emcpc.gif

I'm not sure how to go about that, though. If anyone has any comment on why that which I think to be false is true or any suggestion on how to go about this using the referred equation, I'd much appreciate it.

nakurusil

Interesting, it appears that you found a weakness in Tolman's derivation. Since Tolman surmises that the two masses will have zero speed in S' (in order to move together with speed V wrt S after their collision) you would expect :

[tex]m(u_1)u_1+m(u_2)u_2 = (m(0)+m(0))V[/tex]

However,I think that Tolman must have assumed that mass cannot vary non-continously (from [tex]m_1(u_1)[/tex] to [tex]m_1(0)[/tex]), thus justifying his use of :

[tex]m(u_1)u_1+m(u_2)u_2 = (m(u_1)+m(u_2))V[/tex]

Granted, this is a very weak argument, so the best thing is to dump the blasted "relativistic mass" altogether, as I mentioned in the opening statement. The whole darned thing was introduced in order to reconcile the relativistic momentum/energy:

[tex]p=\gamma m(0)v[/tex] (1)
[tex]E=\gamma m(0)c^2[/tex]

with the Newtonian counterpart:


[tex]p=mv[/tex] (2)

So the best thing is to tell your teacher that your proof is you grouped together [tex]\gamma[/tex] and proper mass m(0) into [tex]\gamma m(0)[/tex] and you assigned that quantity to m

NanakiXIII

But since [tex]u_1[/tex] and [tex]u_2[/tex] are the speeds in S, why are you using the speed in S' after the collision?

Why should it vary non-continuously? For the relativistic mass to change non-continuously, wouldn't the velocity have to as well? I don't see why it would, it simply changes due to the collision, quite continuously, right?

nakurusil

I see, correction:

[tex]m_1(u_1)u_1+m_2(u_2)u_2=(m_1(V)+m_2(V))V[/tex]

Doesn't change anything, the derivation is still flawed.


Because the formula above would imply that m_1(u_1) before collision becomes m_1(V) after collision and this would entail a discontinuity. This is why Tolman must be silently assuming that m_1 before after collision are exactly the same. In fact , he writes :


[tex]m_1u_1+m_2u_2=(m_1+m_2)V[/tex]

So, Tolman must be assuming [tex]m_1(u_1)[/tex] before and after the collision in his formula. Otherwise, the derivation falls apart.

Conclusion: use the derivation based on relativistic momentum I gave you twice.

NanakiXIII

This I don't understand. Changing [tex]u_1[/tex] to [tex]V[/tex] doesn't break continuity, does it? There is just an acceleration. The collision isn't an instantaneous event. So if the velocity changes continuously, why wouldn't the relativistic mass?

I'm probably overlooking it or something, but I'll risk looking stupid and ask: what derivation are you talking about?

nakurusil

You got things backwards, if [tex]m_1(u_1)[/tex] before collision (LHS) is different from [tex]m_1(V)[/tex] after collision (RHS), Tolman's derivation falls apart. So, the only out for his derivation is that [tex]m_1[/tex] before and after collision are the same. THOUGH the [tex]m_1[/tex] speed has jumped from [tex]u_1[/tex] to V


http://www.physicsforums.com/showpos...3&postcount=46
Look at the bottom.

country boy

Try starting with:
ds^2 = (cdt)^2 - dx^2

Divide through by dt^2 and rearrange:
c^2 = v^2 + (ds/dt)^2

Multiply by m^2 to get p^2:
(E/c)^2 = p^2 + (m*ds/dt)^2

Make the last term a constant:
m*ds/dt = Eo/c

And arrive at:
mc = Eo*dt/ds

Which from the 2nd eqn above is:
mc^2 = Eo/[1-(v/c)^2] = E

The Lorentz transformations come from the invariance of ds. The mass behaves just so that the rest energy is invariant.