Derivation of the Equation for Relativistic Mass

[bernhard.rothenstein]

mass relativity

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?

 

[Fredrik]

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)?

 

[bernhard.rothenstein]

mass relativity

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.

 

[nakurusil]

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.

 

[nakurusil]

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 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]

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.

 

[bernhard.rothenstein]

mass relativity

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.:rofl:

 

[nakurusil]

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.:rofl:

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:

https://www.physicsforums.com/showpost.php?p=1227743&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 $$m(u_1)+m(u_2)$$?

 

[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/hbase/relativ/imgrel/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]

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:

https://www.physicsforums.com/showpost.php?p=1227743&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

 

[NanakiXIII]

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?

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?

 

[nakurusil]

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.

 

[NanakiXIII]

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?

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?

 

[nakurusil]

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?

https://www.physicsforums.com/showpost.php?p=1238273&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.

 

[bernhard.rothenstein]

a simple and convincing way to m=gm(0)

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.

I suggest to have a look at
Leo Karlov, "Paul Kard and Lorentz-free special relativity." Phys.Educ. 24,
165-168 (1969).
The derivation Kard proposes is based on a scenario which involves a body absorbin a photon, using conservation of momentum and mass, velocity dependent mass and the relation m=p/c between the mass m of the photon and its momentum p.

 

[rbj]

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.

you can do a thought experiment with two identical balls of identical mass moving toward each other at identical speeds (along the y-axis), striking each other in a perfectly elactic collision and bouncing back. let's call the top ball, "A" and the bottom ball, "B". now, if there is no "x" velocity, all this makes sense, the two balls having equal mass and equal speeds, then have equal momentum and ball A bounces back up (+y direction) with the same speed it had before and ball B bounces back down similarly.

now imagine that this same experiment is done but ball A is moving along the x-axis direction with a constant velocity of $v$. the y velocity is the same as before in the frame of reference of ball A. ball B is not moving along the x-axis direction but still has the previous y velocity and they collide at the origin. after the collision ball A is moving up, as before (but also to the right with velocity $v$) and ball B is moving down. now, for observers, one traveling with ball A and the other hanging around with ball B, we set this up so that both observers sense the y-axis velocity of the ball in their reference frame as the same as the other observer sees for their own ball.

now, because of time-dilation, the y velocity of ball A, as observed by an observer hanging around with ball B must be slower than the velocity that the "moving" observer measures for ball A by a factor of:

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

but for the y-axis momentums to be the same, then the mass of ball A, as observed by the "stationary" observer $m$ must be increased by the same factor (from what the "moving" observer sees as the mass of ball A $m_0$) or

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

and that is where the increased inertial mass comes from.

now a legitimate question would be "Why would the y-axis momentums of the two balls have to be the same? Why can't the masses of the two balls remain the same resulting in a decreased y-axis momentum for ball A (as observed by the "stationary" observer) and less than the y-momentum of ball B?"

the answer to that is then, after the collision, both balls would tend to be moving upward since ball B had more y-axis momentum than A from the "stationary" observer's POV. so what's wrong with that? well, instead of hanging out with the ball B observer, now let's hang out with the ball A observer who, "moving" at a constant velocity has just as legitimate perspective as does observer B. so from observer A's POV, he is stationary and it is ball B moving to the left at velocity $v$. so, if the y-axis momentums were not equal in magnitude, from observer B's perspective, they would be be moving up together after the collision, but from oberver A's perspective (which is just as legit as B's) they would be moving down after the collision. that is contradictory, so they must have the same y-axis momentum, whether you are hanging out with ball A or with ball B.

but since observer A sees ball B as having less y-axis velocity than ball A (due to time-dilation) observer A must see ball B as having larger mass so that the y-axis momentum of ball A is the same as the y-axis momentum of ball B. likewize since observer B sees ball A as having less y-axis velocity than ball B (due to time-dilation) observer B must see ball A as having larger mass so that the y-axis momentum of ball B is the same as the y-axis momentum of ball A.

 

[nakurusil]

I found a very nice treatment here. It seems devoid of the problem that plagues Tolman's solution. In addition, it gives a very nice derivation of relativistic momentum/energy from base principles.

 

[NanakiXIII]

Thanks, nakurusil, that site is quite perfect for me. However, I've stumbled upon one problem. I don't understand how they get from

http://upload.wikimedia.org/math/9/1/9/919ed04c37014cd574c512ec1333f658.png

to

http://upload.wikimedia.org/math/1/2/b/12b6bc3d9ac1a3bde8d9fa85d6afb8ed.png

There seems to be part of one of the interlying equations missing, and I just generally don't see how the did it. I tried getting there myself, but to no avail. Could anyone explain these steps in some detail?rjb, your way of handling the problem seems similar to the one on the site nakurusil posted and helped me understand it somewhat better, thanks.

 

[nakurusil]

Thanks, nakurusil, that site is quite perfect for me. However, I've stumbled upon one problem. I don't understand how they get from

http://upload.wikimedia.org/math/9/1/9/919ed04c37014cd574c512ec1333f658.png

to

http://upload.wikimedia.org/math/1/2/b/12b6bc3d9ac1a3bde8d9fa85d6afb8ed.png

There seems to be part of one of the interlying equations missing, and I just generally don't see how the did it. I tried getting there myself, but to no avail. Could anyone explain these steps in some detail?

Hmm, I looked, it requires solving a simple equation degree 2 in v followed by an easy series of substitutions. Try again.

 

[Sunset]

Hi!

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.

There is no derivation of this formula because it's simply the abreviation for the expression $$\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$ which you CALL
m. Mass of a particle $$m_0$$ is a parameter of your theory , which doesn't change under Lorentz-transformation. The factor $$\gamma$$ only appears in formulas - by accident - in the combination with $$m_0$$. That's why people speak of $$m_0$$ as "rest-mass", but the relation above is nothing but introducing an abreviation.
$$m_0$$ is a Lorentz-scalar, m is not $$m_0 '$$ i.e. it is NOT the mass measured by a moving guy. There exists only a act of measurement for particles at rest! You can't put a moving particle on a balance. That what you call mass is always for particles at rest - the prefix rest is unnecessary!

Best regards Martin

 

[NanakiXIII]

Hmm, I looked, it requires solving a simple equation degree 2 in v followed by an easy series of substitutions. Try again.

Simple to you, perhaps. Either I am to use techniques I'm not familiar with, or I'm just not seeing it. The variable v is in there three times, I've been trying but I can't seem to get it to one side the the equal sign.

By "degree 2 in v", you mean I should square the equation? I looked into that but still see no obvious solution.

 

[rbj]

There is no derivation of this formula because it's simply the abreviation for the expression $$\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$ which you CALL m.

okay, so why don't we just CALL m this:

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

or this

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

or this

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

?

there's a lot of formulae (that are dimensionally correct) that we could pull out of our butt and say it's the inertial mass so that when you multiply it by $v$, it becomes momentum. why not use those? why is this one

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

the correct one for momentum as perceived by an observer that observes the mass $m_0$ whizzing by at velocity $v$?

Mass of a particle $$m_0$$ is a parameter of your theory , which doesn't change under Lorentz-transformation. The factor $$\gamma$$ only appears in formulas - by accident - in the combination with $$m_0$$.

yeah, right.

That's why people speak of $$m_0$$ as "rest-mass", but the relation above is nothing but introducing an abreviation.
$$m_0$$ is a Lorentz-scalar, m is not $$m_0 '$$ i.e. it is NOT the mass measured by a moving guy. There exists only a act of measurement for particles at rest! You can't put a moving particle on a balance.

but you can see how it behaves in a collision with a particle that you can possibly put on a scale.

That what you call mass is always for particles at rest - the prefix rest is unnecessary!

that's an opinion, not an undisputed fact.

i know this is controversial, but i think more confusion happens when speaking of photons ("massless" vs. $m = (h \nu)/c^2$) and such. even the new proposed definition of the kilogram uses the term "at rest" and if they adopt it, i'll bet those two words survive in the final definition.

 

[Sunset]

Hi rbj,

why is this one

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

the correct one for momentum as perceived by an observer that observes the mass $m_0$ whizzing by at velocity $v$.

The relativistic momentum $$p_r = \frac{m_0 v}{\sqrt{1-\frac{v^2}{c^2}}}$$ is defined in that way, because you want to have momentum-conservation.
Let me explain:
When you step from classical mechanics (Newtons axioms) to relativistic mechanics you need a new axiom-system. You have to define $$p_r$$ in such a way that your relativistic theory is covariant and reproduces classical mechanics for low v. There might be indeed other choices for $$p_r$$ which fullfill those two conditions - its simply the question if they describe reality.
$$p_r = \frac{m_0 v}{\sqrt{1-\frac{v^2}{c^2}}}$$
is the easiest choice which garantees conservation of $$p_r$$ (e.g. Nolting, a german text-book, discusses this). Indeed it doesn't mean there are other possible choices, then you could describe physics with this $$p_r '$$ , too! The thing is you have a NEW physical quantity, which you should call "relativistic momentum".

Martin

 

[Sunset]

Let me add:

I only wanted to say that there exists a "access to" SR where the formula m=... simply is a abreviation. I admit, for understanding some textbooks this might not be a helpfull remark... I can only commend Noltings textbooks - I don't know if they have been translated to english.

 

[country boy]

rjb, I liked your thought experiment. And Sunset, a valid reply. It made me realize that y in your setup is similar to ds in mine (above). As far as special relativity is concerned, directions transverse to the motion (y, z) are just as invariant as the proper interval s. But in my example, the tranverse momentum (along s) is invariant because it is proportional to the rest energy. Relativity is also designed to make that true.

 

[rbj]

The relativistic momentum $$p_r = \frac{m_0 v}{\sqrt{1-\frac{v^2}{c^2}}}$$ is defined in that way, because you want to have momentum-conservation.
Let me explain:
When you step from classical mechanics (Newtons axioms) to relativistic mechanics you need a new axiom-system.

but

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

is not either of the axioms. it is a result.

the axioms are:

1. Observation of physical phenomena by more than one inertial observer must result in agreement between the observers as to the nature of reality. Or, the nature of the universe must not change for an observer if their inertial state changes. Or, every physical theory should look the same mathematically to every inertial observer. Or, the laws of the universe are the same regardless of inertial frame of reference.

2. (invariance of c) Light is always propagated in empty space with a definite velocity c that is independent of the state of motion of the emitting body, and all observers observe the speed of light in vacuo to be the same value, c.

and the original poster was asking how to get to

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

from those axioms. i skipped over deriving time-dilation (since he/she didn't ask about that and i assume knew about it) but from these first principles and time-dilation (which also comes from the same two first principles). it's a derived result, not an axiom.

 

[nakurusil]

Simple to you, perhaps. Either I am to use techniques I'm not familiar with, or I'm just not seeing it. The variable v is in there three times, I've been trying but I can't seem to get it to one side the the equal sign.

By "degree 2 in v", you mean I should square the equation? I looked into that but still see no obvious solution.

You get:

$$u_x_R*v^2/c^2-2v+u_x_R=0$$

with the obvious solution : $$v=c^2/u_x_R(1-\sqrt(1-u_x_R^2/c^2)$$

 

[jtbell]

By "degree 2 in v", you mean I should square the equation?

Do you remember the formula for solving a quadratic equation? If

$$ax^2 + bx + c = 0$$

then

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

 

[NanakiXIII]

Alright, I think I got it. In nakurusil's solution, is the part between (brackets) multiplied by $$c^2$$ or $$u_xR$$? In your equation it looks like it's multiplied by the denominator, while as I calculated, it's multiplied by the numerator. Did I do something wrong or am I just reading the equation wrong?

Also, to get the solution, the abc-formula with the minus-sign was used and not with the plus-sign (I apologize for my lack of a better description so early in the morning). Why?

 

[bernhard.rothenstein]

Hi rbj,



The relativistic momentum $$p_r = \frac{m_0 v}{\sqrt{1-\frac{v^2}{c^2}}}$$ is defined in that way, because you want to have momentum-conservation.
Let me explain:
When you step from classical mechanics (Newtons axioms) to relativistic mechanics you need a new axiom-system. You have to define $$p_r$$ in such a way that your relativistic theory is covariant and reproduces classical mechanics for low v. There might be indeed other choices for $$p_r$$ which fullfill those two conditions - its simply the question if they describe reality.
$$p_r = \frac{m_0 v}{\sqrt{1-\frac{v^2}{c^2}}}$$
is the easiest choice which garantees conservation of $$p_r$$ (e.g. Nolting, a german text-book, discusses this). Indeed it doesn't mean there are other possible choices, then you could describe physics with this $$p_r '$$ , too! The thing is you have a NEW physical quantity, which you should call "relativistic momentum".

Martin

Is it wrong to say that when you steo from classical mechanics to relativistic mechanics then there should use the addition law of relativistic velocities? So
p=mu (1)
p'=m'u' (2)
p/m=p'/m'(u/u') (3).
Expressing the right side of (3) as a function of physical quantities measured in I' via the addition law of relativistic velocities the way is paved for the transformation of momentum, mass(energy).

 

[nakurusil]

Alright, I think I got it. In nakurusil's solution, is the part between (brackets) multiplied by $$c^2$$ or $$u_xR$$? In your equation it looks like it's multiplied by the denominator, while as I calculated, it's multiplied by the numerator. Did I do something wrong or am I just reading the equation wrong?

Come on, I gave you the equation, for sure you can solve it. How can you get involved with physics when your math is so weak?

$$v=c^2/u_x_R(1-\sqrt(1-u_x_R^2/c^2)$$=$$v=(1-\sqrt(1-u_x_R^2/c^2)*c^2/u_x_R$$

Also, to get the solution, the abc-formula with the minus-sign was used and not with the plus-sign (I apologize for my lack of a better description so early in the morning). Why?

For the obvious reason that the "+" solution would produce a v larger than c.

 

[NanakiXIII]

Actually, I found your method of writing the equation rather ambiguous. But let's not turn this into another name-calling match, nakurusil.

I'll thank you for your help and leave it at that.

 

[nakurusil]

Actually, I found your method of writing the equation rather ambiguous. But let's not turn this into another name-calling match, nakurusil.

I'll thank you for your help and leave it at that.

Good, I am sorry. I got frustrated with your math. The important thing is that you got a good solution.