Derivation of the Equation for Relativistic Mass

[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.

 

[pmb_phy]

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.

I gave this to you in your PM but I think its okay if I post it now. Since I had a feeling that someone would want to follow this derivation someday I placed it on my web page here.

http://www.geocities.com/physics_world/sr/inertial_mass.htm

Now others can read it and ask any questions they might have about it. The Lorent transformation manifests itself through the velocity transformation relations.

Pete

 

[jhigbie]

Yes NanakiXIII, you are right. The literature usually skips over this derivation and the four-vector approach, while rigorous, usually leaves me cold. And, it is not the one that Einstein used.

I think, but I am not sure, that he used the idea of bouncing elastic balls, you know, one thrown from a moving train and the other one on the stationary platform.

But in the meantime, if you are comfortable with the idea of a Taylor's Series, you can expand "gamma" as a power series in "beta" and if you *start* with the idea that m/m₀ = gamma = 1 + (1/2)beta² + ... (higher powers) and this leads to the expression for kinetic energy:

(m-m₀)c² = (1/2)m₀v² + ... (higher terms),

So, it then becomes reasonable that (m-m₀)c² is in fact the kinetic energy and this of course makes us feel that this is probably the correct expression for relativistic mass. It does have the right form.

Einstein's approach to SR was quite novel. He *assumed* that a relativity principle applied not only to mechanical phenomena, but electromagnetic as well ... and that was a great leap of faith. It turns out that he was quite right, but he was the only one at the time to venture such an idea.

 

[stevmg]

By special relativity, we have time dilation and space contraction by this "gamma." Time is increased by division of the reference time by this gamma and distance is reduced by multiplication by this gamma. For f = ma, because the translation distance is shortened by this factor, the acceleration, compared to the inertial reference is shortened, but force is constant, so the mass must be increased to make up for the lost acceleration.

The derivation of this gamma is done by Einstein in his relativity book. It is an algebraic derivation with no calculus.

Once we know that relativistic mass changes with velocity, we show that a loss in velocity results in a loss in mass and pure energy being releasedf (delta m x c^2)

remember [(1 - v^2/c^2)]^-1/2 = (closely) 1 + (1/2)(v^2/c^2)

relativistic mass - inertial mass = inertial mass + (1/2)inertial mass x loss in velocity^2/c^2) - inertial mass = (1/2)inertial mass x loss in velocity^2/c^2)
delta mass x c^w = 1/2inertail mass x loss in velocity^2 = "mc^2" qed

Again, this requires that relativistic mass is increased from inertial mass by division by this gamma and it is the law of ther conservation of momentum or energy which requires mass to increase when linear acceleration id reduced by space contraction.

 

[PeterDonis]

Einstein's approach to SR was quite novel. He *assumed* that a relativity principle applied not only to mechanical phenomena, but electromagnetic as well ... and that was a great leap of faith. It turns out that he was quite right, but he was the only one at the time to venture such an idea.

PMFJI, but this isn't true. Lorentz derived the Lorentz transformation well before Einstein did, as did Larmor. Their derivations were specifically intended to find the transformation laws for the electromagnetic field that would be compatible with Maxwell's Equations. See the section on "Historical Developments" in the "[URL page on Maxwell's Equations[/URL]. What Einstein did was to assume that *mechanical* phenomena were governed by the Lorentz transformation, and work out the consequences; the consequences for electromagnetism were already well known.

 

[PeterDonis]

I think, but I am not sure, that he used the idea of bouncing elastic balls, you know, one thrown from a moving train and the other one on the stationary platform.

In the version of this thought experiment that's in his popular book on relativity, _Relativity: A Clear Explanation That Anyone Can Understand_, he used light flashes, emitted from each end of a moving train. I haven't been able to find an online reference that contains his description of the scenario.

 

[stevmg]

The flashing lights, etc. are in his book "Relativity"

 

[stevmg]

Remember that assumptions of constant c and the preservation of laws of physics in all inertial time frames leads to a mathematical "proof" of the Lorentz transforms. They are equivalent statements.

Many college textbooks actually derive the "relatvistic mass" equation by itself based on the Lorentz transforms.

 

[jesusfreak]

Energy-Momentum Relation:

E² = (m₀c²)² + (pc)²

Substitute mc² for E (energy-mass equivilence)

Substitute mv for p (relation between mass, velocity, and momentum)

it should look like this:

(mc²)² = (m₀c²)² + (mvc)²

Subtract (mvc)² over..

(mc²)² - (mvc)² = (m₀c²)²

m²c⁴ - m²v²c² = m₀²c⁴

Factor out (mc²) on the left..

(mc²) (c² - v²) = (m₀c²)²

Factor out a c² from the second set of parenthesis above...

(mc²) (c²) (1 - v² / c²) = m₀²c⁴

Take the square root of both sides...

mc² * √(1 - v² / c²) = m₀c²

"c²"s cancel out and then divide the √(1 - v² / c²) over..

m = m₀ / √(1 - v² / c²) [relativistic mass-invariant mass relation]

m: relativistic mass
m₀: rest mass (invariant mass)
v: velocity
c: the speed limit of the universe
p: momentum
E: energy

 

[gpeach]

pa₀+pb₀=pa₁+pb₁

γ(va₀).ma.va₀+γ(vb₀).mb.vb₀=γ(va₁).ma.va₁+γ(vb₁).mb.vb₁

This is the general expression for the relativistic momentum conservation with two particles, where 'ma' and 'mb' are the rest masses of the particles.
To understand the relativistic momentum definition, we need to see why this expression is valid for moving systems, upon the multiplication of the factor γ(v). So it is important to bear in mind the relation between time and proper time, as we can deduce from the basic experiment of a light ray in a moving mirror:

Δt²-Δx²/c²=Δζ² →

Δζ/Δt=√(1-v²/c²)=γ(v) →

γ(v)/Δt=1/Δζ (proper time)

Now, using the speed "diferential definition" (Δx/Δt), and applying the relation whith proper time above to the expression of momentum conservation along the 'x' coordinate, we get:

ma.Δxa₀/Δζa+mb.Δxb₀/Δζb=ma.Δxa₁/Δζa+mb.Δxb₁/Δζb

Applying the Lorentz transformation Δx=γ(V)(Δx'-V.Δt'), we obtain (canceling the commun term γ(V)):

ma.Δx'a₀/Δζa-ma.Δx'a₀V.Δt'/Δζa+mb.Δx'b₀/Δζb-mb.Δx'b₀V.Δt'/Δζb=ma.Δx'a₁/Δζa-ma.Δx'a₁V.Δt'/Δζa+mb.Δx'b₁/Δζb-mb.Δx'b₁V.Δt'/Δζb

Applying again the relation between proper time and time we have:

γ(v'a₀).ma.v'a₀+γ(v'b₀).mb.v'b₀-γ(v'a₀).ma.V-γ(v'b₀).mb.V=γ(v'a₁).ma.v'a₁+γ(v'b₁).mb.v'b₁-γ(v'a₁).ma.V-γ(v'b₁).mb.V

In this expression, if the equality

γ(v'a₀).ma+γ(v'b₀).mb=γ(v'a₁).ma+γ(v'b₁).mb

holds (this is the relatIvistic "condition" tor relativistic momentum conservation), then we have the momentum conservation:

γ(v'a₀).ma.v'a₀+γ(v'b₀).mb.v'b₀=γ(v'a₁).ma.v'a₁+γ(v'b₁).mb.v'b₁ →

p'a₀+p'b₀=p'a₁+p'b₁

With respect to the transversal components (x and y), the result is analogous and imediate...