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## Derivation of the Equation for Relativistic Mass

Quote by bernhard.rothenstein 					 				 				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.

Quote by 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.
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?

Mentor

Quote by bernhard.rothenstein 					 				 				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 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)?

Quote by 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)?
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.

Quote by 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)?
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.

Quote by 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?
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.

Mentor

Quote by bernhard.rothenstein 					 				 				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...
I haven't had time to look at this yet. Maybe tomorrow.

Quote by 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.
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.

Quote by 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.

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:        $$m(u_1)u_1+m(u_2)u_2 = (m(u_1)+m(u_2))V$$  However, since after the collision the speeds are no longer $$u_1$$ and $$u_1$$, how can you say that the relativistic mass of the "new" object is simply $$m(u_1)+m(u_2)$$?

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.

Quote by 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:        $$m(u_1)u_1+m(u_2)u_2 = (m(u_1)+m(u_2))V$$  However, since after the collision the speeds are no longer $$u_1$$ and $$u_1$$, how can you say that the relativistic mass of the "new" object is simply $$m(u_1)+m(u_2)$$?
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 :

$$m(u_1)u_1+m(u_2)u_2 = (m(0)+m(0))V$$

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

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

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:

$$p=\gamma m(0)v$$ (1)
$$E=\gamma m(0)c^2$$

with the Newtonian counterpart:

$$p=mv$$ (2)

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

Quote by 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 :  $$m(u_1)u_1+m(u_2)u_2 = (m(0)+m(0))V$$
But since $$u_1$$ and $$u_2$$ are the speeds in S, why are you using the speed in S' after the collision?

Quote by nakurusil 					 				 				However,I think that Tolman must have assumed that mass cannot vary non-continously (from $$m_1(u_1)$$ to $$m_1(0)$$), thus justifying his use of :
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?

Quote by NanakiXIII 					 				 				But since $$u_1$$ and $$u_2$$ are the speeds in S, why are you using the speed in S' after the collision?
I see, correction:

$$m_1(u_1)u_1+m_2(u_2)u_2=(m_1(V)+m_2(V))V$$

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

Why should it vary non-continuously?
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 :

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

So, Tolman must be assuming $$m_1(u_1)$$ 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.

Quote by nakurusil 					 				 				 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 I don't understand. Changing $$u_1$$ to $$V$$ 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?

Quote by nakurusil 					 				 				Conclusion: use the derivation based on relativistic momentum I gave you twice.
I'm probably overlooking it or something, but I'll risk looking stupid and ask: what derivation are you talking about?

Quote by NanakiXIII 					 				 				This I don't understand. Changing $$u_1$$ to $$V$$ 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?
You got things backwards, if $$m_1(u_1)$$ before collision (LHS) is different from $$m_1(V)$$ after collision (RHS), Tolman's derivation falls apart. So, the only out for his derivation is that $$m_1$$ before and after collision are the same. THOUGH the $$m_1$$ speed has jumped from $$u_1$$ to V

I'm probably overlooking it or something, but I'll risk looking stupid and ask: what derivation are you talking about?
http://www.physicsforums.com/showpos...3&postcount=46
Look at the bottom.

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.

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