Converting Newtonian momentum into relativistic momentum

[PainterGuy]

Hi,

I was trying to derive relativistic momentum equation using classical momentum equation but it didn't work. Could you please help me? Thank you!

Where am I wrong? Or, is not possible, in any way, to derive relativistic momentum using Newtonian momentum equation? Thanks!

 

[PAllen]

You are using the wrong approach to go from Newtonian to relativistic momentum. There are several ways to do it correctly. One is to solve for a function of m and v that, if conserved in one frame, is necessarily conserved in another frame, via Lorentz transform. You would pick some generic collision to do this.

A more direct way is to use 4 vectors. An equation of 4 vectors that holds in one frame necessarily holds in all. There is 4 vector corresponding to velocity, called 4-velocity. Assuming mass is a scalar quantity, then if momentum defined as mass times 4 velocity is conserved in one frame, it is necessarily conserved in all.

Your method makes an erroneous assumption about what feature of Newtonian momentum you should try to preserve. The key things to preserve are that the momentum of two bricks that move at the same speed is twice that of one brick, that faster means more momentum, and that conservation of momentum holds in all frames.

 

[PainterGuy]

Thank you!

Does the same reasoning go for the case when relativistic kinetic energy is derived from classical kinetic energy?

Actually I tried to do it as shown below but, as is obvious, got the wrong result.

 

[Ibix]

The Newtonian formulae are approximations to the relativistic ones. You can't deduce accurate expressions from approximations any more than you can deduce my exact age from a statement that I'm in my forties.

The best you can do is note that relativistic equations are based on four vectors and tensors and guess a four vector generalisation of a Newtonian expression, as @PAllen suggests.

Incidentally, even if the whole approach weren't mistaken, your algebra's wrong in your last calculation. Correct algebra would yield $\frac 12mv^2\left(1-\frac{v^2}{c^2}\right)^{3/2}$. Also you are using $d$ and $t$ for both the coordinate and proper lengths and times, which is just confusing.

 

[vanhees71]

The irony is that the first thing you have to do is to abandon the idea of a relativistic mass and just use the definition of the momentum as in Newtonian physics to get to the right relativistic momentum. The important point is that you have to use the proper time, i.e., the time a clock measures moving with the particle. That's also true in Newtonian mechanics, only that there time is absolute and there is no difference in the universal, absolute coordinate time and proper time.

In relativistic theory the connection between proper-time increments and coordinate-time increments (in coordinates with respect to an inertial reference frame) is
$$\mathrm{d} \tau=\mathrm{d} t \sqrt{1-\vec{v}^2/c^2}.$$
Whith this you get the correct relativistic momentum as
$$\vec{p}=m \mathrm{d}_{\tau} \vec{x} = m (\mathrm{d} t/\mathrm{d} \tau) \mathrm{d}_t \vec{x} = m \gamma \vec{v},$$
where $\gamma=1/\sqrt{1-\vec{v}^2/c^2}$ is the famous Lorentz factor and $m$ is the invariant mass, related to the energy of the particle in its rest frame $E_0=m c^2$, which is a Lorentz scalar.

It's also immediately clear that the corresponding four-vector is
$$p^{\mu}=m \mathrm{d}_{\tau} x^{\mu} = m \gamma \begin{pmatrix}c \\ \vec{v} \end{pmatrix},$$
and the temporal component turns out to be the relativistic energy of the particle, including the rest energy $E_0$, i.e.,
$$E=c p^0=E_0 + E_{\text{kin}} = m c^2 \gamma.$$
For $|\vec{v}/c| \ll 1$, i.e., in Newtonian physics
$$\gamma \simeq 1+\vec{v}^2/(c^2)$$
and thus
$$E \simeq m c^2 + \frac{m}{2} \vec{v}^2 \; \Rightarrow \; E_{\text{kin}} \simeq \frac{m}{2} \vec{v}^2,$$
as it should be in the Newtonian approximation.

 

[DrStupid]

The irony is that the first thing you have to do is to abandon the idea of a relativistic mass and just use the definition of the momentum as in Newtonian physics to get to the right relativistic momentum. The important point is that you have to use the proper time, i.e., the time a clock measures moving with the particle.

You can do it but you don't have to. If you start with Newton's definition of the momentum and use coordinate time you get the relativistic mass (and therefore the relativistic momentum) as a result.

 

[Nugatory]

You can do it but you don't have to. If you start with Newton's definition of the momentum and use coordinate time you get the relativistic mass (and therefore the relativistic momentum) as a result.

What do you mean by "Newton's definition of the momentum"? The $p(t)$ that makes $F=\frac{\mathrm d p}{\mathrm d t}$ work?

 

[DrStupid]

What do you mean by "Newton's definition of the momentum"? The $p(t)$ that makes $F=\frac{\mathrm d p}{\mathrm d t}$ work?

I mean p:=m·v with the m that makes F=dp/dt work.

 

[PAllen]

I mean p:=m·v with the m that makes F=dp/dt work.

Derivations of SR kinematics I’ve seen (those that don’t start from 4 vectors) always derive relativistic momentum first, by study of a collision in two frames, requiring conservation in both. Then force is defined from momentum. Elementary arguments require that the functional form of momentum be mf(v)v. A wholly separate choice is whether you want to call the first two factors a mass of some type. The form of f(v) can be uniquely determined from the collision analysis in two frames.

 

[pervect]

I mean p:=m·v with the m that makes F=dp/dt work.

Well, in relativity , as others have noted, $p = m \cdot u$, where u is the 4 velocity, and m is of course the invariant mass. This makes $F = dp / d\tau$ work, where F is the 4-force, p is the 4-momentum, and $\tau$ is proper time.

So, with the 4-vector approaches, things are indeed quite similar to the Newtonian approach.

The proper velocity u, the 4 force F, the invaraint mass m, and $\tau$ are all geometric objects / tensors, objects that exist regardless of a choice of partciular frame of reference. This makes using these formulas much easier than approaches that use objects that have no meaning unless a frame is specified, like a time coordinate t, or a force, or a 3-momentum, which do not have the required coordinate independence. One is having to label the 3-objects with what coordinate system they are applicable in, and convert them from one coordinate system to the other, an be careful not to use the wrong object, rather than thinking of them as something that exists independent of the coordinates.

 

[vanhees71]

You can do it but you don't have to. If you start with Newton's definition of the momentum and use coordinate time you get the relativistic mass (and therefore the relativistic momentum) as a result.

Then you need a direction-dependent mass, which is a quantity with no clear mathematical structure concerning the spacetime model, i.e., it's not a tensor and thus behaves very complicated under Poincare transformations.

 

[vanhees71]

I mean p:=m·v with the m that makes F=dp/dt work.

The relativistic equation of motion should of course also be written in covariant form, making live easier. Again, it's as in Newtonian physics taking the time derivative as the derivative with respect to the proper time of the particle. This gives
$$\mathrm{d}_{\tau} p^{\mu}=K^{\mu},$$
where $K^{\mu}$ is a force four-vector. Contrary to the Newtonian case usually $K^{\mu}$ is a function of both $x^{\mu}$ and $p^{\mu}$. The reason is that, of course, there's a constraint, because the four comonents of $p^{\mu}$ are not independent of each other. This should be so, because there are only 3 independent degrees of freedom in configuration space, as in Newtonian mechanics.

The constraint is easy to find, because by the definition of $p^{\mu}$ you have
$$p_{\mu} p^{\mu}=m^2 c^2=\text{const}.$$
Taking the derivative wrt. proper time, leads to
$$\dot{p}_{\mu} p^{\mu}=0 \; \Rightarrow \; K^{\mu} p_{\mu}=0.$$
The dot means derivatives with respect to $\tau$.

One of the most simple examples is the electromagnetic force on a point charge in an external electromagnetic field. The field is described by the antisymmetric tensor $F_{\mu \nu}$, and the equation of motion reads
$$\dot{p}^{\mu}=\frac{q}{m} F^{\mu \nu} p_{\nu}.$$
You can of course also write this as a 2nd-order differential equation for $x^{\mu}$ using the definition $p^{\mu}=m \dot{x}^{\mu}=m c u^{\mu}$, which gives
$$m \ddot{x}^{\mu}=q F^{\mu \nu} \dot{x}_{\nu}.$$

 

[DrStupid]

Then you need a direction-dependent mass

No, you don't. The relativistic mass (that's what you get for m in p=m·v in SR) is independend from the direction.

 

[DrStupid]

Derivations of SR kinematics I’ve seen (those that don’t start from 4 vectors) always derive relativistic momentum first, by study of a collision in two frames, requiring conservation in both.

Here you can see another derivation.

 

[robphy]

... possible, in any way, to derive relativistic momentum using Newtonian momentum equation? Thanks!

...what feature of Newtonian momentum you should try to preserve.

You can't deduce accurate expressions from approximations any more than you can deduce my exact age from a statement that I'm in my forties.

Starting with the "classical limit",
one has to know some aspects of the [desired] result... or be extremely lucky.
Otherwise, one encounters all sorts of likely-unphysical generalizations of the classical idea.

What likely happens in any "derivation from the classical limit" is that
the relativistic result [together with its relativistic structures] is formulated to make the "classical limit" clearly appear as the limit as some parameter varies.
(See the Bronstein Cube in @George Jones 's post https://www.physicsforums.com/threads/teach-me-about-classical-mechanics-please.930704/post-5876358 and my followup comment for further background on it.)
Often the trick is to make the relativistic result look like the classical result in some way so that it is easier to undo the steps to produce the "derivation".

In my opinion, this hindsight often suggests the "better ways" to interpret the classical quantities, since this aspect is preserved in the full theory. Of course, this applies to more than special relativity... one has quantum mechanics, statistical mechanics, etc...

 

[vanhees71]

No, you don't. The relativistic mass (that's what you get for m in p=m·v in SR) is independend from the direction.

No, you need transverse and longitudinal relativistic mass, if you want to use this mediocre quantities for some incomprehensible reason:

https://en.wikipedia.org/wiki/Mass_in_special_relativity#Transverse_and_longitudinal_mass

 

[vanhees71]

Starting with the "classical limit",
one has to know some aspects of the [desired] result... or be extremely lucky.
Otherwise, one encounters all sorts of likely-unphysical generalizations of the classical idea.

What likely happens in any "derivation from the classical limit" is that
the relativistic result [together with its relativistic structures] is formulated to make the "classical limit" clearly appear as the limit as some parameter varies.
(See the Bronstein Cube in @George Jones 's post https://www.physicsforums.com/threads/teach-me-about-classical-mechanics-please.930704/post-5876358 and my followup comment for further background on it.)
Often the trick is to make the relativistic result look like the classical result in some way so that it is easier to undo the steps to produce the "derivation".

In my opinion, this hindsight often suggests the "better ways" to interpret the classical quantities, since this aspect is preserved in the full theory. Of course, this applies to more than special relativity... one has quantum mechanics, statistical mechanics, etc...

I guess by "classical limit" you mean "non-relativistic limit". Then it's clear how to proceed. You go to a momentary restframe of the particle, use the non-relativistic laws there and transform back (using the Lorentz transformation of course) to the original (lab) frame. Then you get at least covariant laws. It's immediately clear that time derivatives in the momentaneous rest frame translate into derivatives with respect to precisely this quantity, which is written in a covariant way the proper time of the particle.

That's why in the definition of momentum (and necessarily also energy) you have the usual scalar (invariant) mass and the proper-time derivative. There's no need to introduce concepts not really fitting into the framework of Minkowski space.

 

[DrStupid]

No, you need transverse and longitudinal relativistic mass

Not with F=dp/dt.

 

[PAllen]

Here you can see another derivation.

Interesting. Books I’ve seen get momentum just from conservation of momentum as a requirement in two frames (using the Lorentz transform at this point), then define force from it, then use the work energy theorem in some form to get relativistic energy.

Of course, your use of Newton’s third law is equivalent to conservation of momentum, but requires that force be brought in before you get the formula for momentum.

 

[vanhees71]

Not with F=dp/dt.

Why are you insiting on $t$? It's just inconvenient not to work using not manifestly covariant quantities.

 

[Vanadium 50]

This thread has gone in a bad direction for sure.

First, the right thing to do is to start out with the correct expression and show that it reduces to the approximation in the region where the approximation is best. What is proposed is to start with the approximation and try and guess at what needs to be included to make it work right in general and try to show the guess was right. Not only is it the hard way, there is no guarantee this process will ever get to the right answer.

I see relativistic mass has reared its ugly head again. It was proposed in 1906 and abandoned in 1908: 112 years ago. Yet people - science writers and poseurs - still push it because "it's easy to understand". It's not. It's easy to misunderstand. I point out in passing that it only took 108 years to kill phlogistoin. Caloric even less.

People pushing relatvistic mass as anything but a long-disposed of theory should be treated exactly the same as if they were bringing back caloric and phlogistion.

 

[PeroK]

People pushing relatvistic mass as anything but a long-disposed of theory should be treated exactly the same as if they were bringing back caloric and phlogistion.

Caloric might be useful in a game of Scrabble.

 

[DrStupid]

Why are you insiting on $t$?

Why is the Earth a disc?

 

[Ibix]

Caloric might be useful in a game of Scrabble.

Phlogiston is good if someone's already played gist.

 

[PeterDonis]

you need transverse and longitudinal relativistic mass

You do if you want to relate the (frame-dependent) 3-force to the (frame-dependent) 3-acceleration. However, if all you want is to relate the (frame-dependent) 3-force to the (frame-dependent) coordinate time derivative of the 3-momentum, you don't need different transverse and longitudinal masses, because $d \vec{p} / dt$ includes the time derivative of $\gamma$ if we have $\vec{p} = m_{\text{rel}} \vec{v} = m_0 \gamma \vec{v}$, where $m_0$ is the rest mass. Taking that time derivative and then trying to write things in terms of the 3-acceleration $d \vec{v} / dt$ is where the different transverse and longitudinal coefficients come in.

Of course, as has already been said, it's much easier to just use the 4-vector formulation for all this and forget about relativistic mass altogether.

 

[PeroK]

Phlogiston is good if someone's already played gist.

That's fantasy Scrabble.

 

[DrStupid]

First, the right thing to do is to start out with the correct expression

The right thing to do a derivation is to start with the result? That's an interesting approach.

 

[Vanadium 50]

I said no such thing. You know I said no such thing. You're just here to stir the pot.

 

[PeterDonis]

The OP question has been answered, and the thread is degenerating into noise. Thread closed.