Four-vectors and conservation laws in relativity

This lecture was delivered to the final year honors class of 3 year science degree students on 21 November 2017 as part of the Classical Dynamics paper.

The web version will always have some more refinements, add-ons, content expansion, clarifications etc that are not to be found in the actual lecture I delivered. Because talking to the hello world is always relaxing and has as much time available as is comforting.

In this lecture we will discuss some of the important tools of relativistic mechanics. We will discuss the idea of proper-time, 4-velocity, 4-acceleration, 4-momentum, 4-force and related conservation law of the 4-momentum.

A. proper-time

The proper time is the time interval in the rest-frame of any event. The proper time is related to time-interval in other inertial frame by: $\boxed{d\tau = \frac{1}{\gamma}dt}$ where $\gamma > 1$ always.

Gamma ($\gamma$) is the Lorentz factor or Lorentz boost factor directly related to the speed of an object in speed-of-light units, i.e. beta ($\beta$).

$\boxed{\gamma = \frac{1}{\sqrt{1-v^2/c^2}}}$.

Hence proper-time is the “smallest possible time interval” for an object in motion in a frame of reference, among all possible inertial frames of reference and it occurs in the rest frame. Mathematically thus:

$\boxed{d\tau < dt}$.

Proper-time is necessary to define other basic quantities in theory of relativity if we are to preserve their basic meaning in terms of the non-relativistic mechanics definitions. Soon we will see this in the discussion to follow.

B. four velocity

Four velocity of a particle is the rate of change of 4-displacement $x^{\mu}$ wrt the proper time $\tau$.

So, four velocity $\eta^{\mu}$ is defined through the following equation, $\eta^{\mu} = \frac{dx^{\mu}}{d\tau}$, where 4-displacement $x^{\mu}$ is the position vector — or space-time interval, in the Minkowski  space — akin to the difference of two 3-dimensional vectors, in coordinate space, this time with 4 coordinates rather than 3.

The proper-time interval $d\tau$ is a Lorentz invariant i.e. when we move between arbitrary inertial frames of reference, governed by the Lorentz factor $\beta$ or $\gamma$ this interval retains its value — because it retains its form. Thus its equally good to recognize that Lorentz invariant quantities are scalars or constants under Lorentz transformation.

Any variable which would retain its form under Lorentz transformation are said to be Lorentz invariant quantities.

Since $x^{\mu}$ transforms like a 4-vector under Lorentz transformation, the 4-velocity $\eta^{\mu}$ as defined above, also transforms  like a 4-vector. Since proper-time $\tau \, (or \, d\tau)$ is a scalar, it does not dictate terms to the Lorentz transformation nor does it take them from the constraints of the transformation.

When we will learn more about quantities known as tensors we will see its not true for tensors. 4-vectors and tensors are quantities which transform in accordance with what the Lorentz transformation dictates them to.

Thus we can write: $\eta^{\mu} = \frac{dt}{d\tau}\frac{dx^{\mu}}{dt} = \gamma \frac{dx^{\mu}}{dt}$ as $\gamma = \frac{dt}{d\tau}=(1-v^2/c^2)^{-\frac{1}{2}}$.

So, $\eta^{\mu} =\gamma \frac{dx^{\mu}}{dt}= \gamma \frac{d}{dt}(x^0, \vec{x}) =\gamma \frac{d}{dt}(ct, \vec{x})= \gamma (c, \vec{v})$.

So, $\boxed{\eta^{\mu} =(\gamma c,\gamma \vec{v})}$.

The zeroth component of four-velocity: $\boxed{\eta^0 =\gamma c=\frac{c}{\sqrt{1-v^2/c^2}}}$.

The spatial components of four-velocity: $\boxed{\eta^k =\frac{v^k}{\sqrt{1-v^2/c^2}}}$, with $k=1,\,2,\,3$.

$v^k$ are the k-th component of $\vec{v}$, along $x^1, x^2, x^3$ — or $x, \,y, \,z$ axes.

So, $\boxed{\vec{\eta} =\gamma \vec{v}}$, $\boxed{\eta^{\mu} =\gamma (c, \vec{v})=(\eta^0=\gamma c,\vec{\eta}=\gamma \vec{v})}$.

Square of four-velocity — or dot product of the 4-vector with self is given as: $\eta^2 = \eta^{\mu} \eta_{\mu}=\gamma^2(c^2-v^2)=c^2 \gamma^2 (1-v^2/c^2)$.

So $\boxed{{\eta}^2 = c^2 }$. i.e. ${\eta}^2$ is a Lorentz invariant. Magnitudes of vectors as expected do not change when we relocate from one inertial frame to another. This is true about all sorts of 4-vectors, their magnitude are Lorentz invariant.

C. four acceleration

Four acceleration of a particle is defined as: $\omega^{\mu} = \frac{d\eta^{\mu}}{d\tau}=\frac{d^2x^{\mu}}{d\tau^2}$.

As in the case of 4-velocity — and 4-displacement, from where this property is defined, the 4-acceleration can easily be shown to be Lorentz invariant, see B, above.

Since 4-velocity is a Lorentz invariant, we can write: as in B, $\eta^{\mu} \eta_{\mu}=c^2$. Differentiating this wrt $\tau$,

$\frac{d\eta^{\mu}}{d\tau}\eta_{\mu} +\eta^{\mu} \frac{d\eta_{\mu}}{d\tau}=0$.

⇒ $\omega^{\mu}\eta_{\mu} +\eta^{\mu} \omega_{\mu}=0$.

But $A^{\mu}A_{\nu} =A_{\mu} A^{\nu}$.

⇒ $\omega^{\mu}\eta_{\mu} = \eta^{\mu} \omega_{\mu}$.

⇒ $2\omega^{\mu}\eta_{\mu}=0$.

or $\omega . \eta =0$.

Thus $\omega$ and $\eta$ are orthogonal.

Or in words, four velocity and four acceleration are mutually orthogonal 4-vectors.

D. four momentum

The four-momentum or momentum four-vector of a particle is defined as product of its “rest mass” with its “four-velocity”. i.e.

$\boxed{p^{\mu} = m_0 \eta^{\mu}}$.

Spatial components of four-momentum$p^k =m_0 \eta^k=m_0 \gamma v^k=mv^k = \frac{m_0 v^k}{\sqrt{1-v^2/c^2}}$ or $\boxed{\vec{p}=m_0 \gamma \vec{v}=m\vec{v} = \frac{m_0 \vec{v}}{\sqrt{1-v^2/c^2}}}$.

Here we note that: the relativistic mass — i.e. mass in any arbitrary inertial frame of reference of a particle, is related to the particle’s rest-mass, i.e. the particles mass in a vantage which has zero relative motion wrt the particle, and is an inertial vantage are related by the following simple relation.

$\boxed{m=m_0 \gamma = \frac{m_0}{\sqrt{1-v^2/c^2}}}$.

$m_0$ is the rest mass and $m$ is the relativistic mass.

Time component of four-momentum is$\boxed{p^0 =m_0 \eta^0= m_0\gamma c= mc = \frac{mc^2}{c}= \frac{E}{c}}$.

where $E=mc^2$ is relativistic energy. This is the wildly famous Einstein’s mass-energy relation. See how easily it was unleashed from its obscure existence through the relevant vector formalism, as we have been discussing.

Also note that this relation is valid only when both sides of the equation are valid, hence by manifest clarity its valid for matter particles only, ie. particles with mass — such as electrons. Its not valid directly for mass-less particles like photons, where another relevant equation is valid. But not this one.

Remember photons have energy but no mass. When I say directly I mean when exclusively photons are present this equation can’t be invoked. Give the photons some good friends it hangs out with like electrons or pions or muons, in that relationship this equation is valid, simply because conservation of mass-energy has to hold for matter particles.

I am being contumacious here in stressing this point as I have seen physicists making this simple mistake. Famous things are easily misused. Also note that in the expression for relativistic energy above, there is relativistic mass (m) and not rest-mass, with the  subscript zero.

That means there is a Lorentz factor gamma sitting which enforces some kind of consistency in whether you applied the equation correctly or not. Ask yourself this innocuous question, whats the Lorentz factor gamma for a single photon nonchalantly traversing the free space. You will see what I am arguing.

So lets promise ourselves to be applying caution, between rest-mass, relativistic mass, rest-energy, relativistic energy, matter particles (mass haves) and mass-less zombies known as photons.

In totality the 4-momentum is thusly spelled out as: $\boxed{p^{\mu} \equiv (\frac{E}{c},p^1, p^2, p^3)=(\frac{E}{c},\vec{p})}$.

Four-momentum is also known by the appellation “energy momentum four vector” as energy and momenta is what it stores in its stomach.

Square of four-momentum: $p^2 = p.p=p^{\mu}p_{\mu}=\frac{E^2}{c^2}-\vec{p}\hspace{2pt}^2$.

But $p^{\mu}p_{\mu}=(m_0 \eta^{\mu})(m_0 \eta_{\mu})= m_0^2 \eta^{\mu} \eta_{\mu} = m_0^2c^2$.

So, $\frac{E^2}{c^2}-\vec{p}\,^2=m_0^2c^2$ or $\boxed{E^2 = m_0^2c^4 + \vec{p}\,^2c^2}$.

The boxed equation is known as “relativistic energy-momentum relation”. If simpletons didn’t have their say this should have been the most famous equation, not the relativistic mass-energy equation of Einstein that we discussed above. This equation is more general in its scope.

Although there is a caveat here. Strictly speaking this equation in not well defined for a photon. Rest-masses are not defined for perpetually rest-less particles like photons whose relativistic mass is zero and whose rest-mass is undefined. But if you remove the mass term by a make-do approach what you see is whats valid for photons. The relation between energy and momentum.

Although this works, let me warn you the equation boxed above is not defined for photon. Its a special particle and you must respect that fact. Why removing mass works though has to do with rigorous calculations of the photon mechanics that we are not stepping onto here.

You can see all the things I am saying here regarding photon can be checked by calculating gamma. Yes, its infinity, as I had hinted earlier. Hence relativistic mass of photon (which is zero) does not come from a zero rest-mass by multiplying zero to infinity (gamma).

Thats the hint that photon’s don’t have zero rest-mass. Its undefined. Its a very important lesson in relativity theory that I have seen scores of great luminaries who have completely forgotten and sidestepped as they have grown up.

Although the energy and momentum 3-vector of a particle alter with Lorentz transformation — i.e. change of inertial frames of reference, the quantity $p^2 = \frac{E^2}{c^2}-\vec{p}\,^2$ is Lorentz invariant, thats because as we have found earlier its the square of a 4-vector.

Thus 4-momentum magnitude is a conserved quantity under Lorentz transformation. Note that this conservation statement is different from the statement of energy conservation. Its fairly easy and not uncommon to confuse between the two.

E. four force

Four force is defined as: $\boxed{K^{\mu} = \frac{dp^{\mu}}{d\tau}}$ where $p^{\mu}$ is the four-momentum. This four-force is also known as Minkowski force.

Lets see how favorably our formalism would be taken by Newton. Did Newton have reservations for Einstein. I know they didn’t meet, but time-travel anyone?

Newton’s law? Lets use the definition of four-momentum, four-velocity  etc that we have discussed.

$p^{\mu}=m_0 \eta^{\mu}$$\eta^{\mu}=\frac{dx^{\mu}}{d\tau}$.

So $\boxed{K^{\mu} = m_0 \frac{d\eta^{\mu}}{d\tau} = m_0 \omega ^{\mu}=m_0 \frac{d^2 x^{\mu}}{d\tau^2}}$ where $\omega ^{\mu}$ is the four acceleration, as we have discussed earlier.

We see that Einstein is in good books of Newton. Newton’s law is valid. Four-force is rest-mass times four-acceleration. Also there is a secondary form of Newton’s law, force as the time rate of change of momentum, the so called alternative form of second law of Newton. But we see a gamma (Lorentz factor) here. Gammas have the purported responsibility of bringing consistency in the frame work of theory of relativity.

$\boxed{K^{\mu} = \frac{dt}{d\tau} \frac{dp^{\mu}}{dt} = \gamma \frac{dp^{\mu}}{dt}}$.

Lets now discuss a little about the spatial components and the temporal components of the four-force.

Spatial component of four force. Its given by $K^i = \gamma \frac{dp^i}{dt}= \gamma F^i$ where the indices $i = 1,\, 2, \,3$ run over ordinary Cartesian coordinate axes, $x, \,y, \,z$. The i-th component of the 3-force is given by: $F^i =\frac{dp^i}{dt}$.

So, the spatial component of the four-force is given by $\vec{K}=\frac{\vec{F}}{\sqrt{1-v^2/c^2}}=\gamma \vec{F}$ in terms of the 3-dimensional force.

Similarly we have the time component.

Temporal component of four force$K^0 = \gamma \frac{dp^0}{dt}= \gamma \frac{d}{dt}(mc)=\frac{\gamma}{c}\frac{dE}{dt}$.

$\frac{dE}{dt}$ is the rate at which energy is gained by the particle, what we call the power in ordinary mechanics. So $\frac{dE}{dt}=\vec{F}.\vec{v}$ and the temporal component is given by $K^0 = \frac{\gamma}{c}\vec{F}\cdot\vec{v}$.

The full four-vector form of the four-force is then: $\boxed{K^{\mu} = (\frac{\gamma}{c} \vec{F}\cdot\vec{v}, \gamma \vec{F})}$.