# Special Relativity/Dynamics

## Introduction

The way that the velocity of a particle can differ between observers who are moving relative to each other means that momentum needs to be redefined as a result of relativity theory.

The illustration below shows a typical collision of two particles. In the right hand frame the collision is observed from the viewpoint of someone moving at the same velocity as one of the particles, in the left hand frame it is observed by someone moving at a velocity that is intermediate between those of the particles.

If momentum is redefined then all the variables such as force (rate of change of momentum), energy etc. will become redefined and relativity will lead to an entirely new physics. The new physics has an effect at the ordinary level of experience through the relation ${\displaystyle K=\gamma mc^{2}-mc^{2}\,}$ whereby it is the tiny deviations in gamma from unity that are expressed as everyday kinetic energy so that the whole of physics is related to "relativistic" reasoning rather than Newton's empirical ideas.

## Momentum

In physics momentum is conserved within a closed system, the law of conservation of momentum applies. Consider the special case of identical particles colliding symmetrically as illustrated below:

The momentum change by the red ball is:

${\displaystyle 2m\mathbf {u_{yR}} }$

The momentum change by the blue ball is:

${\displaystyle -2m\mathbf {u_{yB}} }$

The situation is symmetrical so the Newtonian conservation of momentum law is demonstrated:

${\displaystyle 2m\mathbf {u_{yR}} =2m\mathbf {u_{yB}} }$

Notice that this result depends upon the y components of the velocities being equal, that is, ${\displaystyle \mathbf {u_{yR}} =\mathbf {u_{yB}} }$.

The relativistic case is rather different. The collision is illustrated below, the left hand frame shows the collision as it appears for one observer and the right hand frame shows exactly the same collision as it appears for another observer moving at the same velocity as the blue ball:

The configuration shown above has been simplified because one frame contains a stationary blue ball (ie: ${\displaystyle u_{xB}=0}$) and the velocities are chosen so that the vertical velocity of the red ball is exactly reversed after the collision ie:${\displaystyle u_{yR}^{'}=u_{yB}^{'}}$. Both frames show exactly the same event, it is only the observers who differ between frames. The relativistic velocity transformations between frames is:

${\displaystyle u_{yR}^{'}={\frac {u_{yR}{\sqrt {1-v^{2}/c^{2}}}}{1-u_{xR}v/c^{2}}}}$

${\displaystyle u_{yB}^{'}={\frac {u_{yB}{\sqrt {1-v^{2}/c^{2}}}}{1-u_{xB}v/c^{2}}}=u_{yB}{\sqrt {1-v^{2}/c^{2}}}}$ given that ${\displaystyle u_{xB}=0\,}$.

Suppose that the y components are equal in one frame, in Newtonian physics they will also be equal in the other frame. However, in relativity, if the y components are equal in one frame they are not necessarily equal in the other frame (time dilation is not directional so perpendicular velocities differ between the observers). For instance if ${\displaystyle u_{yR}^{'}=u_{yB}^{'}}$ then:

${\displaystyle u_{yB}={\frac {u_{yR}}{1-u_{xR}v/c^{2}}}}$

So if ${\displaystyle u_{yR}^{'}=u_{yB}^{'}}$ then in this case ${\displaystyle u_{yR}\neq u_{yB}}$.

If the mass were constant between collisions and between frames then although ${\displaystyle 2mu_{yR}^{'}=2mu_{yB}^{'}}$ it is found that:

${\displaystyle 2mu_{yR}\neq 2mu_{yB}}$

So momentum defined as mass times velocity is not conserved in a collision when the collision is described in frames moving relative to each other. Notice that the discrepancy is very small if ${\displaystyle u_{xR}}$ and ${\displaystyle v}$ are small.

To preserve the principle of momentum conservation in all inertial reference frames, the definition of momentum has to be changed. The new definition must reduce to the Newtonian expression when objects move at speeds much smaller than the speed of light, so as to recover the Newtonian formulas.

The velocities in the y direction are related by the following equation when the observer is travelling at the same velocity as the blue ball ie: when ${\displaystyle u_{xB}=0\,}$:

${\displaystyle u_{yB}={\frac {u_{yR}}{1-u_{xR}v/c^{2}}}}$

If we write ${\displaystyle m_{B}}$ for the mass of the blue ball) and ${\displaystyle m_{R}}$ for the mass of the red ball as observed from the frame of the blue ball then, if the principle of relativity applies:

${\displaystyle 2m_{R}u_{yR}=2m_{B}u_{yB}\,}$

So:

${\displaystyle m_{R}=m_{B}{\frac {u_{yB}}{u_{yR}}}}$

But:

${\displaystyle u_{yB}={\frac {u_{yR}}{1-u_{xR}v/c^{2}}}}$

Therefore:

${\displaystyle m_{R}={\frac {m_{B}}{1-u_{xR}v/c^{2}}}}$

This means that, if the principle of relativity is to apply then the mass must change by the amount shown in the equation above for the conservation of momentum law to be true.

The particle velocities were chosen so that ${\displaystyle u_{yR}^{'}=u_{yB}^{'}}$. ${\displaystyle v}$ was selected so that ${\displaystyle v=u_{xR}^{'}}$. This allows ${\displaystyle v}$ to be expressed in terms of ${\displaystyle u_{xR}\,}$:

${\displaystyle u_{xR}^{'}={\frac {u_{xR}-v}{1-u_{xR}v/c^{2}}}=v}$

and hence:

${\displaystyle v={\frac {c^{2}}{u_{xR}}}(1-{\sqrt {1-u_{xR}^{2}/c^{2}}})}$

So substituting for ${\displaystyle v}$ in ${\displaystyle m_{R}={\frac {m_{B}}{1-u_{xR}v/c^{2}}}}$:

${\displaystyle m_{R}={\frac {m_{B}}{\sqrt {1-u_{xR}^{2}/c^{2}}}}}$

The blue ball is at rest so its mass is sometimes known as its rest mass, and is given the symbol ${\displaystyle m}$. As the balls were identical at the start of the boost the mass of the red ball is the mass that a blue ball would have if it were in motion relative to an observer; this mass is sometimes known as the relativistic mass symbolised by ${\displaystyle M}$. These terms are now infrequently used in modern physics, as will be explained at the end of this section. The discussion given above was related to the relative motions of the blue and red balls, as a result ${\displaystyle u_{xR}}$ corresponds to the speed of the moving ball relative to an observer who is stationary with respect to the blue ball. These considerations mean that the relativistic mass is given by:

${\displaystyle M={\frac {m}{\sqrt {1-u^{2}/c^{2}}}}}$

The relativistic momentum is given by the product of the relativistic mass and the velocity ${\displaystyle \mathbf {p} =M\mathbf {u} }$.

The overall expression for momentum in terms of rest mass is:

${\displaystyle \mathbf {p} ={\frac {m\mathbf {u} }{\sqrt {1-u^{2}/c^{2}}}}}$

and the components of the momentum are:

${\displaystyle p_{x}={\frac {mu_{x}}{\sqrt {1-u^{2}/c^{2}}}}}$

${\displaystyle p_{y}={\frac {mu_{y}}{\sqrt {1-u^{2}/c^{2}}}}}$

${\displaystyle p_{z}={\frac {mu_{z}}{\sqrt {1-u^{2}/c^{2}}}}}$

So the components of the momentum depend upon the appropriate velocity component and the speed.

Since the factor with the square root is cumbersome to write, the following abbreviation is often used, called the Lorentz gamma factor:

${\displaystyle \gamma ={\frac {1}{\sqrt {1-u^{2}/c^{2}}}}}$

The expression for the momentum then reads ${\displaystyle \mathbf {p} =m\gamma \mathbf {u} }$.

It can be seen from the discussion above that we can write the momentum of an object moving with velocity ${\displaystyle \mathbf {u} }$ as the product of a function ${\displaystyle M(u)}$ of the speed ${\displaystyle u}$ and the velocity ${\displaystyle \mathbf {u} }$:

${\displaystyle M(u)\mathbf {u} }$

The function ${\displaystyle M(u)}$ must reduce to the object's mass ${\displaystyle m}$ at small speeds, in particular when the object is at rest ${\displaystyle M(0)=m}$.

There is a debate about the usage of the term "mass" in relativity theory. If inertial mass is defined in terms of momentum then it does indeed vary as ${\displaystyle M=\gamma m}$ for a single particle that has rest mass, furthermore, as will be shown below the energy of a particle that has a rest mass is given by ${\displaystyle E=Mc^{2}}$. Prior to the debate about nomenclature the function ${\displaystyle M(u)}$, or the relation ${\displaystyle M=\gamma m}$, used to be called 'relativistic mass', and its value in the frame of the particle was referred to as the 'rest mass' or 'invariant mass'. The relativistic mass, ${\displaystyle M=\gamma m}$, would increase with velocity. Both terms are now largely obsolete: the 'rest mass' is today simply called the mass, and the 'relativistic mass' is often no longer used since, as will be seen in the discussion of energy below, it is identical to the energy but for the units.

## Force

Newton's second law states that the total force acting on a particle equals the rate of change of its momentum. The same form of Newton's second law holds in relativistic mechanics. The relativistic 3 force is given by:

${\displaystyle \mathbf {f} =d\mathbf {p} /dt}$

If the relativistic mass is used:

${\displaystyle {\frac {d\mathbf {p} }{dt}}={\frac {d(m\mathbf {u} )}{dt}}}$

By Leibniz's law where ${\displaystyle d(xy)=xdy+ydx}$:

${\displaystyle \mathbf {f} ={\frac {d\mathbf {p} }{dt}}=m{\frac {d\mathbf {u} }{dt}}+\mathbf {u} {\frac {dm}{dt}}}$

This equation for force will be used below to derive relativistic expressions for the energy of a particle in terms of the old concept of "relativistic mass".

The relativistic force can also be written in terms of acceleration. Newton's second law can be written in the familiar form

${\displaystyle \mathbf {F} =m\mathbf {a} }$

where ${\displaystyle \mathbf {a} =d\mathbf {v} /dt}$ is the acceleration.

here m is not the relativistic mass but is the invariant mass.

In relativistic mechanics, momentum is ${\displaystyle \mathbf {p} =m\gamma \mathbf {v} }$

again m being the invariant mass and the force is given by ${\displaystyle \mathbf {F} ={\frac {d\mathbf {p} }{dt}}=m{\frac {d(\gamma \mathbf {v} )}{dt}}}$

This form of force is used in the derivation of the expression for energy without relying on relativistic mass.

It will be seen in the second section of this book that Newton's second law in terms of acceleration is given by:

${\displaystyle \mathbf {F} =m\gamma (\mathbf {a} +{\frac {\gamma ^{2}v}{c^{2}}}{\frac {dv}{dt}}\mathbf {v} )}$

## Energy

The debate over the use of the concept "relativistic mass" means that modern physics courses may forbid the use of this in the derivation of energy. The newer derivation of energy without using relativistic mass is given in the first section and the older derivation using relativistic mass is given in the second section. The two derivations can be compared to gain insight into the debate about mass but a knowledge of 4 vectors is really required to discuss the problem in depth. In principle the first derivation is most mathematically correct because "relativistic mass" is given by: ${\displaystyle M={\frac {m}{\sqrt {1-u^{2}/c^{2}}}}}$ which involves the constants ${\displaystyle m}$ and ${\displaystyle c}$.

### Derivation of relativistic energy using the relativistic momentum

In the following, modern derivation, m means the invariant mass - what used to be called the "rest mass". Energy is defined as the work done in moving a body from one place to another. We will make use of the relativistic momentum ${\displaystyle p=\gamma mv}$. Energy is given from:

${\displaystyle dE=\mathbf {f} d\mathbf {x} }$

so, over the whole path:

${\displaystyle E=\int _{0}^{x}\mathbf {f} d\mathbf {x} }$

Kinetic energy (K) is the energy used to move a body from a velocity of 0 to a velocity ${\displaystyle \mathbf {u} }$. Restricting the motion to one dimension:

${\displaystyle K=\int _{u=0}^{u=u}\mathbf {f} dx}$

Using the relativistic 3 force:

${\displaystyle K=\int _{u=0}^{u=u}{\frac {d(m\gamma u)}{dt}}dx=\int _{u=0}^{u=u}m{\frac {d(\gamma u)}{dt}}dx=\int _{u=0}^{u=u}md(\gamma u){\frac {dx}{dt}}}$

substituting for ${\displaystyle d(\gamma u)}$ and using ${\displaystyle dx/dt=u}$:

${\displaystyle K=\int _{u=0}^{u=u}m(\gamma du+ud\gamma )u}$

Which gives:

${\displaystyle K=\int _{u=0}^{u=u}m(u\gamma du+u^{2}d\gamma )}$

The Lorentz factor ${\displaystyle \gamma }$ is given by:

${\displaystyle \gamma ={\frac {1}{\sqrt {1-u^{2}/c^{2}}}}}$

meaning that :

${\displaystyle d\gamma ={\frac {u}{c^{2}}}\gamma ^{3}du}$

${\displaystyle du={\frac {c^{2}}{u\gamma ^{3}}}d\gamma }$

So that

${\displaystyle K=\int _{\gamma =1}^{\gamma =\gamma }m(u\gamma {\frac {c^{2}}{u\gamma ^{3}}}d\gamma +u^{2}d\gamma )=\int _{\gamma =1}^{\gamma =\gamma }m({\frac {c^{2}}{\gamma ^{2}}}+u^{2})d\gamma =\int _{\gamma =1}^{\gamma =\gamma }mc^{2}d\gamma }$

Alternatively, we can use the fact that:

${\displaystyle \gamma ^{2}c^{2}-\gamma ^{2}u^{2}=c^{2}\,}$

Differentiating:

${\displaystyle 2\gamma c^{2}d\gamma -\gamma ^{2}2udu-u^{2}2\gamma d\gamma =0\,}$

So, rearranging:

${\displaystyle \gamma udu+u^{2}d\gamma =c^{2}d\gamma \,}$

In which case:

${\displaystyle K=\int _{u=0}^{u=u}m(u\gamma du+u^{2}d\gamma )=\int _{u=0}^{u=u}mc^{2}d\gamma \,}$

As ${\displaystyle u}$ goes from 0 to ${\displaystyle u}$, the Lorentz factor ${\displaystyle \gamma }$ goes from 1 to ${\displaystyle \gamma }$, so:

${\displaystyle K=mc^{2}\int _{\gamma =1}^{\gamma =\gamma }d\gamma \,}$

and hence:

${\displaystyle K=\gamma mc^{2}-mc^{2}\,}$

The amount ${\displaystyle \gamma mc^{2}}$ is known as the total energy of the particle. The amount ${\displaystyle mc^{2}}$ is known as the rest energy of the particle. If the total energy of the particle is given the symbol ${\displaystyle E}$:

${\displaystyle E=\gamma mc^{2}=mc^{2}+K\,}$

So it can be seen that ${\displaystyle mc^{2}}$ is the energy of a mass that is stationary. This energy is known as mass energy.

The Newtonian approximation for kinetic energy can be derived by using the binomial theorem to expand ${\displaystyle \gamma =(1-u^{2}/c^{2})^{-{\frac {1}{2}}}}$.

The binomial expansion is:

${\displaystyle (a+x)^{n}=a^{n}+na^{n-1}x+{\frac {n(n-1)}{2!}}a^{n-2}x^{2}....}$

So expanding ${\displaystyle (1-u^{2}/c^{2})^{-{\frac {1}{2}}}}$:

${\displaystyle K={\frac {1}{2}}mu^{2}+{\frac {3mu^{4}}{8c^{2}}}+{\frac {5mu^{6}}{16c^{4}}}+...}$

So if ${\displaystyle u}$ is much less than ${\displaystyle c}$:

${\displaystyle K={\frac {1}{2}}mu^{2}}$

which is the Newtonian approximation for low velocities.

### Derivation of relativistic energy using the concept of relativistic mass

Energy is defined as the work done in moving a body from one place to another. Energy is given from:

${\displaystyle dE=\mathbf {F} d\mathbf {x} }$

so, over the whole path:

${\displaystyle E=\int _{0}^{x}\mathbf {F} d\mathbf {x} }$

Kinetic energy (K) is the energy used to move a body from a velocity of 0 to a velocity ${\displaystyle u}$. So:

${\displaystyle K=\int _{u=0}^{u=u}Fdx}$

Using the relativistic force:

${\displaystyle K=\int _{u=0}^{u=u}{\frac {d(Mu)}{dt}}dx}$

So:

${\displaystyle K=\int _{u=0}^{u=u}d(Mu){\frac {dx}{dt}}}$

substituting for ${\displaystyle d(Mu)}$ and using ${\displaystyle dx/dt=u}$:

${\displaystyle K=\int _{u=0}^{u=u}(Mdu+udM)u}$

Which gives:

${\displaystyle K=\int _{u=0}^{u=u}(Mudu+u^{2}dM)}$

The relativistic mass is given by:

${\displaystyle M={\frac {m}{\sqrt {1-u^{2}/c^{2}}}}}$

Which can be expanded as:

${\displaystyle M^{2}c^{2}-M^{2}u^{2}=m^{2}c^{2}}$

Differentiating:

${\displaystyle 2Mc^{2}dM-M^{2}2udu-u^{2}2MdM=0}$

So, rearranging:

${\displaystyle Mudu+u^{2}dM=c^{2}dM}$

In which case:

${\displaystyle K=\int _{u=0}^{u=u}(Mudu+u^{2}dM)}$

is simplified to:

${\displaystyle K=\int _{u=0}^{u=u}c^{2}dM}$

But the mass goes from ${\displaystyle m}$ to ${\displaystyle M}$ so:

${\displaystyle K=c^{2}\int _{M=m}^{M=M}dM)}$

and hence:

${\displaystyle K=Mc^{2}-mc^{2}}$

The amount ${\displaystyle Mc^{2}}$ is known as the total energy of the particle. The amount ${\displaystyle mc^{2}}$ is known as the rest energy of the particle. If the total energy of the particle is given the symbol ${\displaystyle E}$:

${\displaystyle E=mc^{2}+K}$

So it can be seen that ${\displaystyle mc^{2}}$ is the energy of a mass that is stationary. This energy is known as mass energy and is the origin of the famous formula ${\displaystyle E=mc^{2}}$ that is iconic of the nuclear age.

The Newtonian approximation for kinetic energy can be derived by substituting the rest mass for the relativistic mass ie:

${\displaystyle M={\frac {m}{\sqrt {1-u^{2}/c^{2}}}}}$

and:

${\displaystyle K=Mc^{2}-mc^{2}}$

So:

${\displaystyle K={\frac {mc^{2}}{\sqrt {1-u^{2}/c^{2}}}}-mc^{2}}$

ie:

${\displaystyle K=mc^{2}((1-u^{2}/c^{2})^{-{\frac {1}{2}}}-1)}$

The binomial theorem can be used to expand ${\displaystyle (1-u^{2}/c^{2})^{-{\frac {1}{2}}}}$:

The binomial theorem is:

${\displaystyle (a+x)^{n}=a^{n}+na^{n-1}x+{\frac {n(n-1)}{2!}}a^{n-2}x^{2}....}$

So expanding ${\displaystyle (1-u^{2}/c^{2})^{-{\frac {1}{2}}}}$:

${\displaystyle K={\frac {1}{2}}mu^{2}+{\frac {3mu^{4}}{8c^{2}}}+{\frac {5mu^{6}}{16c^{4}}}+...}$

So if ${\displaystyle u}$ is much less than ${\displaystyle c}$:

${\displaystyle K={\frac {1}{2}}mu^{2}}$

Which is the Newtonian approximation for low velocities.

## Nuclear Energy

When protons and neutrons (nucleons) combine to form elements the combination of particles tends to be in a lower energy state than the free neutrons and protons. Iron has the lowest energy and elements above and below iron in the scale of atomic masses tend to have higher energies. This decrease in energy as neutrons and protons bind together is known as the binding energy. The atomic masses of elements are slightly different from that calculated from their constituent particles and this difference in mass energy, calculated from ${\displaystyle E=mc^{2}}$, is almost exactly equal to the binding energy.

The binding energy can be released by converting elements with higher masses per nucleon to those with lower masses per nucleon. This can be done by either splitting heavy elements such as uranium into lighter elements such as barium and krypton or by joining together light elements such as hydrogen into heavier elements such as deuterium. If atoms are split the process is known as nuclear fission and if atoms are joined the process is known as nuclear fusion. Atoms that are lighter than iron can be fused to release energy and those heavier than iron can be split to release energy.

When hydrogen and a neutron are combined to make deuterium the energy released can be calculated as follows:

The mass of a proton is 1.00731 amu, the mass of a neutron is 1.00867 amu and the mass of a deuterium nucleus is 2.0136 amu. The difference in mass between a deuterium nucleus and its components is 0.00238 amu. The energy of this mass difference is:

${\displaystyle E=mc^{2}=1.66\times 10^{-27}\times 0.00238\times (3\times 10^{8})^{2}}$

So the energy released is ${\displaystyle 3.57\times 10^{-13}}$ joules or about ${\displaystyle 2\times 10^{11}}$ joules per gram of protons (ionised hydrogen).

(Assuming 1 amu = ${\displaystyle 1.66\times 10^{-27}}$ Kg, Avogadro's number = ${\displaystyle 6\times 10^{23}}$ and the speed of light is ${\displaystyle 3\times 10^{8}}$ metres per second)

Present day nuclear reactors use a process called nuclear fission in which rods of uranium emit neutrons which combine with the uranium in the rod to produce uranium isotopes such as 236U which rapidly decay into smaller nuclei such as Barium and Krypton plus three neutrons which can cause further generation of 236U and further decay. The fact that each neutron can cause the generation of three more neutrons means that a self sustaining or chain reaction can occur. The generation of energy results from the equivalence of mass and energy; the decay products, barium and krypton have a lower mass than the original 236U, the missing mass being released as 177 MeV of radiation. The nuclear equation for the decay of 236U is written as follows:

${\displaystyle _{92}^{236}U\rightarrow _{56}^{144}Ba+_{36}^{89}Kr+3n+177MeV}$

Nuclear explosion

If a large amount of the uranium isotope 235U (the critical mass) is confined the chain reaction can get out of control and almost instantly release a large amount of energy. A device that confines a critical mass of uranium is known as an atomic bomb or A-bomb. A bomb based on the fusion of deuterium atoms is known as a thermonuclear bomb, hydrogen bomb or H-bomb.