Derivation of the Equation for Relativistic Mass

bernhard.rothenstein

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

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 [itex]v[/itex]. 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 [itex]v[/itex]) 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:

[tex] \sqrt{1 - \frac{v^2}{c^2}} [/tex]

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

[tex] \frac{m}{m_0} = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} [/tex]

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 [itex]v[/itex]. 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...ec1333f658.png

to

http://upload.wikimedia.org/math/1/2...85d6afb8ed.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

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

Sunset

Hi!

There is no derivation of this formula because it's simply the abreviation for the expression [tex] \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}} [/tex] which you CALL
m. Mass of a particle [tex]m_0[/tex] is a parameter of your theory , which doesn't change under Lorentz-transformation. The factor [tex]\gamma[/tex] only appears in formulas - by accident - in the combination with [tex]m_0[/tex]. That's why people speak of [tex]m_0[/tex] as "rest-mass", but the relation above is nothing but introducing an abreviation.
[tex]m_0[/tex] is a Lorentz-scalar, m is not [tex]m_0 '[/tex] 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

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

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

[tex]m = \frac{m_0}{\sqrt{1-\frac{c^2}{v^2}}} [/tex]

or this

[tex]m = m_0 \sqrt{1-\frac{c^2}{v^2}} [/tex]

or this

[tex]m = m_0 \frac{v^2}{c^2} [/tex]

???

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 [itex]v[/itex], it becomes momentum. why not use those? why is this one

[tex] p = \frac{m_0 v}{\sqrt{1-\frac{v^2}{c^2}}} [/tex]

the correct one for momentum as perceived by an observer that observes the mass [itex]m_0[/itex] whizzing by at velocity [itex]v[/itex]?

yeah, right.

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

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. [itex] m = (h \nu)/c^2 [/itex]) 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,

The relativistic momentum [tex] p_r = \frac{m_0 v}{\sqrt{1-\frac{v^2}{c^2}}} [/tex] 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 [tex]p_r[/tex] in such a way that your relativistic theory is covariant and reproduces classical mechanics for low v. There might be indeed other choices for [tex]p_r[/tex] which fullfill those two conditions - its simply the question if they describe reality.
[tex] p_r = \frac{m_0 v}{\sqrt{1-\frac{v^2}{c^2}}} [/tex]
is the easiest choice which garantees conservation of [tex]p_r[/tex] (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 [tex]p_r '[/tex] , too! The thing is you have a NEW physical quantity, which you should call "relativistic momentum".

Best regards, 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

but

[tex] p = \frac{m_0 v}{\sqrt{1-\frac{v^2}{c^2}}} [/tex]

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

[tex] m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}} [/tex]

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

You get:

[tex]u_x_R*v^2/c^2-2v+u_x_R=0[/tex]

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

jtbell

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

[tex]ax^2 + bx + c = 0[/tex]

then

[tex]x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

NanakiXIII

Alright, I think I got it. In nakurusil's solution, is the part between (brackets) multiplied by [tex]c^2[/tex] or [tex]u_xR[/tex]? 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

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

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?

[tex]v=c^2/u_x_R(1-\sqrt(1-u_x_R^2/c^2)[/tex]=[tex]v=(1-\sqrt(1-u_x_R^2/c^2)*c^2/u_x_R[/tex]


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