## Thursday, June 29, 2023

### Time Transformation: UNIZOR.COM - Relativity 4 All - Einstein View

Notes to a video lecture on UNIZOR.COM

Time Transformation

We have introduced a concept of time dilation in the lecture Time Measuring and, arranging a thought experiment with a light clock, came up with a formula for the perceived time duration of our experiment by an observer outside of a reference frame where this experiment is conducted in terms of the proper time duration measured inside a frame of this experiment.

The formula for time dilation was
Tβ = γ·tβ
where
α is some inertial frame with Cartesian coordinates {X,Y,Z}, time T and α-observer in it;
β is another inertial frame with Cartesian coordinates {x,y,z}, time t and β-observer in it;
initially at time T=t=0 α-frame and β-frame coincide;
β-frame is moving with constant speed v relative to α-frame along the X-axis of α, preserving parallelism of corresponding axes;
Tβ is a perceived time duration of β-experiment by α-observer;
tβ is the proper time duration of β-experiment as observed by β-observer moving together with β-frame;
γ = 1/1−(v²/c²);
c is the speed of light in vacuum - a universal constant, the same in all inertial reference frames.

We have also derived the formulas of transformation of Cartesian coordinates and time from one inertial reference frame to another in Lorentz Transformation lecture.
The final form of Lorentz transformation from inertial α-frame coordinates {X,T} to inertial β-frame coordinates {x,t}, when β-frame is uniformly moving with speed v along α's X-axis, in the Special Theory of Relativity is:
x =
 X − v·T √1−(v/c)²
t =
 T − v·X/c² √1−(v/c)²

If β-frame moves with speed v relative to α-frame along X-axis, we can consider α-frame as moving with speed −v relative to β-frame.
Then the inverse transformation would be
X =
 x + v·t √1−(v/c)²
T =
 t + v·x/c² √1−(v/c)²

Armed with the above Lorentz Transformation formulas, we can attempt to derive the formula for time dilation more mathematically.
Assume some experiment is conducted at point x=x0 in β-frame.
The beginning of this experiment is at time tbeg and it ends at time tend, so the duration of an experiment is
Δt = tend − tbeg

The α-observer sees this experiment begins at point Xbeg and time Tbeg, ending at point Xend and time Tend.

According to Lorentz Transformation, the coordinates of β-experiment for α-observer are
Xbeg =
 x0 + v·tbeg √1−(v/c)²
Tbeg =
 tbeg + v·x0/c² √1−(v/c)²

Xend =
 x0 + v·tend √1−(v/c)²
Tend =
 tend + v·x0/c² √1−(v/c)²

Now we can calculate the duration of our experiment as perceived by α-observer.
ΔT = Tend − Tbeg =
=
 tend−tbeg √1−(v/c)²
=
Δt·γ
This is exactly the same formula obtained in lecture Time Measuring through a thought experiment with a light clock.

Incidentally, while in β-frame our experiment was conducted at a single point x0 on x-axis, an observer in α-frame sees it at moving along X-axis for a distance
ΔX = Xend − Xbeg =
=
 v·(tend − tbeg) √1−(v/c)²
= v·
Δt·γ

## Monday, June 26, 2023

### Lorentz Transformation: UNIZOR.COM - Relativity 4 All - Einstein View

Notes to a video lecture on UNIZOR.COM

Lorentz Transformation

As in the previous lecture about time dilation, consider two inertial reference frames:
frame α with Cartesian coordinates X, Y, Z and time T
and frame β with Cartesian coordinates x, y, z and time t.

So, all capital coordinates (X, Y, Z) and time (T) will be related to α-frame, events and experiments occurring in it will be referred to as α-events and α-experiments, and observer at rest in it will be referred to as α-observer.
Similarly, all lower case coordinates (x, y, z) and time (t) will be related to β-frame, β-events, β-experiments and β-observer.

Assume that at time T=t=0 both frames coincide, that is their origins are at the same point in space and their corresponding axes overlap, but frame β is moving relative to α along the X-axis with constant speed v, so the velocity vector of β-frame in α-coordinates has components {v,0,0}.

Since the movement is relative, we can say at the same time that frame α is moving relative to β along its x-axis with speed −v, so the velocity vector of α-frame in β-coordinates has components {−v,0,0}.

We still rely on two main principles:
(a) Principle of Relativity that states that all the Physics laws must be the same, if expressed quantitatively, using corresponding coordinates and time, in all inertial reference frames.
(b) Speed of light is constant and the same in all inertial reference frames in empty space, we will use a symbol c for this universal constant.

Our current task is purely mathematical - express the transformation of space and time coordinates from one inertial reference frame to another that moves relative to the first, preserving the integrity of the above principles.

At this point we would like to remind that this type of task is not unusual in Math.
One of the elementary tasks in coordinate geometry was to transform the Cartesian coordinates if XY-axes are rotated by some angle around the origin.
In this case the transformation from coordinate system {X,Y} to {x,y}, where axes are rotated by angle φ around origin is
x = X·cos(φ) − Y·sin(φ)
y = X·sin(φ) + Y·cos(φ)

Consider now our case of the transformation of space and time coordinates between two inertial reference frames moving relatively to each other as described above:
α-frame {X,T} and
β-frame {x,t} that uniformly moves relatively to α-frame along the X-axis with constant speed v, preserving the parallelism of corresponding axes, provided at time T=t=0 both reference frames coincide.

Since the direction of β-frame movement is along X-axis, coordinates y and z will be the same as, correspondingly, Y and Z. This simplifies our task, reducing it to transformation of only two coordinates - X in space and T in time into x and t.

We will look for a simple linear transformation from {X,T} system to {x,t} system of the form
x = p·X + q·T
t = r·X + s·T
where p, q, r and s are four unknown coefficients of transformation, which we are going to determine.

We should not add any constants into above transformations since coordinates (X=0,T=0) should transform into (x=0,t=0).

Equation 1
Since β-frame moves along X-axis of α-frame with speed v, its origin of space coordinate (point x=0) must at any moment of α-time T be on a distance v·T from the origin of coordinates of a stationary α-frame.
Hence, if X=v·T then x=0 for any T.
From this and the first transformation equation
x = p·X + q·T
we derive:
0 = p·v·T + q·T or
0 = (p·v + q)·T.

Since this equality is true for any time moment T,
p·v + q = 0
and, unconditionally,

q = −p·v

This is the first equation for our unknown coefficients.

Equation 2
Consider a ray of light issued at time T=t=0 from the point of coinciding origins of both reference frames in the direction of positive coordinates X and x.
Since the speed of light c is the same in both systems {X,T} and {x,t}, according to Principle of Relativity, an equation of the motion of the front of the light wave in the α-frame must be X=c·T and in the β-frame it is x=c·t.
Therefore, if X=c·T, then x=c·t.
Put X=c·T into both equations of transformation of coordinates.
We get
x = p·c·T + q·T and
t = r·c·T + s·T
Substitute these expressions into x=c·t:
p·c·T + q·T = r·c²·T + s·c·T.
Reduce by T,

p·c + q = r·c² + s·c

This is the second equation for unknown coefficients.

Equation 3
Consider a ray of light issued at time T=t=0 from the point of coinciding origins of both reference frames in the direction of negative coordinates X and x.
Repeat the logic of a previous paragraph for the light moving now in the opposite direction with a speed −c.
Therefore, if X=−c·T, then x=−c·t.
Put X=−c·T into both equations of transformation of coordinates.
We get
x = −p·c·T + q·T
t = −r·c·T + s·T
Substitute these expressions into x=−c·t:
−p·c·T + q·T = r·c²·T − s·c·T.
Reduce by T,

−p·c + q = r·c² − s·c

This is the third equation for unknown coefficients.

System of Equations
So, this is the system of three linear equations for four unknown coefficients of transformation p, q, r, s:
(a) q = −p·v
(b) p·c + q = r·c² + s·c
(c) −p·c + q = r·c² − s·c
Usually, we need four equations for four unknown, but additional considerations will help to resolve this situation.

Solution
From (b) and (c), adding and subtracting these equations, we get:
2q = 2r·c², therefore q = r·c²
2p·c = 2s·c, therefore p = s

Now the system of equations is
q = −p·v
q = r·c²
p = s

Let's express q, r and s in terms of only one unknown p and known constants c and v.
q = −p·v
r = q/c² = −p·v/c²
s = p

The original system of equations now looks like
x = p·X − p·v·T
t = −p·(v/c²)·X + p·T

In a matrix form the transformation from α-frame {X,T} to β-frame {x,t} looks like
 x t
 =
 p −p·v −p·v/c² p
 ·
 X T

Symmetrically, the motion of system {X,T} relative to system {x,t} is a uniform motion with speed −v along x-axis.
That means that expression of {X,T} coordinates in terms of {x,t} should look similar to above with the only difference of the sign of the speed v
X = p·x + p·v·t
T = p·(v/c²)·x + p·t

In a matrix form the transformation from β-frame {x,t} to α-frame {X,T} looks like
 X T
 =
 p p·v p·v/c² p
 ·
 x t

Since the latter matrix of transformation is an inverse of the one that represents transformation from α-frame {X,T} to β-frame {x,t}, the product of these two matrices must produce a unit matrix.
 1 0 0 1

Therefore, multiplying two matrices of transformation, we should have satisfied the following equations for all four elements of the unit matrix
p² − p²·v²/c² = 1
p²·v − p²·v = 0
−p²·v²/c² + p²·v²/c² = 0
−p²·v²/c² + p² = 1

The first equation is the same as the fourth and produce the value for coefficient p
p²·(1 − v²/c²) = 1 and
p = 1/√1−v²/c² .

The second and the third equations above are identities that we can disregard.

Consequently, the values for other coefficients of transformation of coordinates are
q = −p·v = −v/√1−v²/c²
r = −p·v/c² = (−v/c²)/√1−v²/c²
s = p = 1/√1−v²/c²

Traditionally, factor v/c is replaced with Greek letter β, which results in formulas:
x = (1/√1−β² )·X +
+ (−v/√
1−β² )·T
t = ((−v/c²)/√
1−β² )·X +
+ (1/√
1−β² )·T

One more simplification is usually done by introducing Lorentz factor γ equaled to 1/√1−β² :
x = γX − γv·T = γ(X − v·T)
t =−γv·X/c²+γT = γ(T−v·X/c²)

The final form of Lorentz transformation from inertial α-frame coordinates {X,T} to inertial β-frame coordinates {x,t}, when β-frame is uniformly moving with speed v along α's X-axis, in the Special Theory of Relativity is:
x =
 X − v·T √1−(v/c)²
t =
 T − v·X/c² √1−(v/c)²
where c is the speed of light in empty space - the same constant for all inertial reference frames.

The final transformation looks exactly like in Einstein's article On the Electrodynamics of Moving Bodies (1905), but the derivation is mathematically rigorous based on two postulates - Principle of Relativity and the universality of the constant speed of light in all inertial reference frames in vacuum.

## Saturday, June 17, 2023

### Time Measuring: UNIZOR.COM - Relativity 4 All - Einstein View

Notes to a video lecture on UNIZOR.COM

Time Measuring

We are planning to conduct a few thought experiments. For this we need some simple clock to measure time between events in different inertial reference frames.

Consider the following clock construction.
Take a rigid rod, its edges are points A and B.
At edge point A we install the source of light and a sensor of light.
At edge point B we install a mirror that reflects a beam of light back to its origin.

We define the unit of time as the time from a moment we send a beam of light from point A to point B till we receive back a reflected beam back at point A.
This will be our light clock that we can use to measure any time interval.

As confirmed by numerous experiments and follows from the Maxwell equations, the speed of light c is exactly the same in any inertial reference frame in empty space.
That's what makes this clock perfect for time measurement.

According to the Principle of Relativity, which we accept as an axiom, any observer in one inertial reference frame, who measured a unit of time by this light clock, can go to another inertial reference frame, take this clock with him, and the interval of time measured by this clock there would be exactly the same.

If some process, like boiling a pot of water or growing corn, in one inertial frame takes certain time, according to this light clock, exactly the same process would take exactly the same amount of time in any other inertial reference frame using the same clock locally to that frame.

Now consider two inertial reference frames:
frame α with Cartesian coordinates X, Y, Z and time T
and frame β with Cartesian coordinates x, y, z and time t.

So, all capital coordinates (X, Y, Z) and time (T) will be related to α-frame, events occurring in it (α-events) and observer at rest in it (α-observer).
Similarly, all lower case coordinates (x, y, z) and time (t) will be related to β-frame, events occurring in it (β-events) and observer at rest in it (β-observer).

Assume that at time T=t=0 both frames coincide, that is their origins are at the same point in space and their axes overlap, but frame β is moving relative to α along the X-axis with speed v, so the velocity vector of β-frame in α-frame coordinates has components {v,0,0}.

Since the movement is relative, we can say at the same time that frame α is moving relative to β along the X-axis with speed −v, so the velocity vector of α-frame in β-frame coordinates has components {−v,0,0}.

Consider identical light clocks in both α- and β-frames arranged between local in each frame points A{0,0,0} and B{0,0,R}, where R is any real number. The first point is an origin of a corresponding frame and the second point is on the Z-axis on a distance R from the origin.

Since the distance between these points in each frame is R, the interval of time between emitting a ray of light from the origin and receiving a reflected ray back at the origin (with the total distance covered by light ray equaled to 2R and the speed of light c) equals in each frame for a local observer at rest in each corresponding frame to
Tα = tβ = 2R/c

Besides this local observation from within each frame, consider how the same time interval is evaluated by an observer in the other reference frame.

The interval of time for a ray of light covering in α-frame the distance from the origin of coordinates to a mirror and back can be analyzed by an observer at rest in β-frame (tα) and an interval of time for a ray of light moving in β-frame along the analogous path can be analyzed by an observer at rest in α-frame (Tβ).
These remote observations are not necessarily produce the same measurements of time interval as observed locally.

Let's see now how an observer in α-frame {X,Y,Z,T} evaluates the time interval of a round trip of a ray of light in the β-frame {x,y,z,t} moving with speed v along α's X-axis.

Assume, at some moment, when β's origin is at α coordinate {X0,0,0}, the light was emitted by the β light clock from its origin.
From a β-observer viewpoint the ray emitted by a light clock from β point {0,0,0} goes vertically along its Z-axis, reaches the point {0,0,R}, is reflected back and received at the β's origin after tβ=2R/c interval of time.

However, an observer at α who analyses the events in β-frame thinks differently.
Since β-frame is moving relatively to α-frame, from an α-observer viewpoint the ray that should reach the mirror in β-frame and reflect back should go along trajectory as on the following picture

The α-observer concludes that, while the ray moves from β's origin at point X0 in α-frame coordinates towards the mirror Q, the mirror moves further to the right in α-frame with speed v, and, while the reflected ray moves back to the β's origin, this origin moves even more to the right to point X1 in α-frame coordinates.

Let's calculate the time Tβ that should pass on α-observer's clock during the time the ray in β-frame moves to a mirror and back (index β signifies the frame where an event takes place, in this case we are talking about β-event).

The distance from X0 to X1 is the distance β-frame moves during time Tβ on α-observer's clock with uniform speed v.
During the same time Tβ the light with uniform speed c covers the distance from β-frame's origin at point X0 in α-frame up to the mirror at point Q and back to β-frame's origin that by that time has moved with uniform speed v to point X1 in α-frame.

Therefore,
X0X1 = v·Tβ
X0Q + QX1 = c·Tβ
PQ = R
X0P = PX1 = ½·v·Tβ
X0Q = QX1 = ½·c·Tβ
R² + (½·v·Tβ)² = (½·c·Tβ
4R² = −(v·Tβ)² + (c·Tβ

But the β-observer's clock would show that this same process took tβ local time, which is the time light covers the distance from point β{0,0,0} straight up to point β{0,0,R} and back, covering the distance of 2R with speed c.
Therefore,
2R = c·tβ

Comparing the above expressions, we come with the following
4R² = −(v·Tβ)² + (c·Tβ)² =
= (c·tβ

From this we can derive the relationship between time measurement Tβ of a β-event in α-frame by a remote α-observer and time measurement tβ of the same β-event in β-frame by local β-observer.
−(v·Tβ)² + (c·Tβ)² = (c·tβ

From this follows
(1−(v²/c²))·Tβ² = tβ²

Traditionally, physicists use symbol γ to express this relationships between the timing in both reference frames
γ = 1/1−(v²/c²)

Then the above relationship takes the form of
Tβ = γ·tβ

As you see, a remote evaluation of a time interval of an event is always greater than local.
This property of time is symmetrical.
The α-event takes longer, if evaluated from the β-frame than locally in the α-frame.
tα = γ·Tα
In some way it's similar to a situation when two people look at each other from a distance. Each person sees the other smaller than that person really is.

This relationship between measuring a time interval of certain event by a local observer in some inertial frame and the time of the same event measured by an observer in another inertial frame moving relative to the first one is called time dilation.

TIME IS NOT ABSOLUTE

## Thursday, June 15, 2023

### Galilean NON-Invariance: UNIZOR.COM - Relativity 4 All - Galilean View

Notes to a video lecture on UNIZOR.COM

Galilean NON-Invariance

Clouds on the Horizon

At the end of the 19th century physicists were generally of a positive opinion about the state of Physics.

Newton's mechanics and theory of gravitation, Coulomb's laws of electrostatic were pretty much in good shape.

The properties of electromagnetic field were well described by Maxwell equations.

The theory of aether as a medium for electromagnetic waves, while not being absolutely satisfactory for physicists, was the only one available to explain the observable wave character of light and its constant speed that depends only on medium's characteristics of electric permittivity and magnetic permeability.

All was not bad... except a couple of "small" things.

First, the experiment of Michelson-Morley in 1887 attempted to detect aether and showed negative result, as was confirmed by many other experiments afterwards.

The aether was supposed to have too many contradictory properties to satisfy all the experimental data, and physicists became more and more skeptical about this theory.

Second, the Newtonian laws of mechanics an gravitation, as well as Coulomb's law of electrostatics showed the invariance relative to Galilean transformation of coordinates between two inertial frames, but the laws related to electromagnetism and, in particular, Lorentz force and Maxwell equations showed that these laws and equations are not invariant, if we use Galilean transformations.
Their expression in one inertial reference frame, transformed into another frame using the Galilean transformation, took a different form than in the original frame.

An important details about the laws that are invariant relative to Galilean transformation is that they are dependent either on a distance between two points, like in case of gravity or electrostatic forces, which is invariant under Galilean transformation as demonstrated in the lecture Metric Invariance of this chapter, or on acceleration, like in case of Newton's laws of mechanics, which is the second derivative of the coordinates as linear functions of time in Galilean transformation, and the second derivative nullifies the dependency on the velocity of one reference frame relative to another.

At the same time, electromagnetic properties and forces depend on the velocity of charged objects (Lorentz force), which is the first derivative of coordinates by time, and the first derivative leaves the velocity as a participant in the equations, thus making the laws non-invariant to Galilean transformation.

Indeed,
if x=X−v·t/
then dx/dt=dX/dt−v,
which shows clear dependency on velocity of a moving frame relatively to a stationary, and velocity participates in Maxwell equations.

Aether as Light Medium

Wave properties of light were known and well accepted by physicists at the end of 19th century.
It was also assumed that light waves need a medium to carry them analogously to sound waves that need a medium, like air, to propagate.
The medium that carries light was called aether (or ether).

A consequence from Maxwell equations was that the waves of light have a fixed speed of propagation that depends only on the medium's electromagnetic properties, the electric permittivity ε and magnetic permeability μ.

But, if the aether is a physical substance, its movement relative to the source of light should affect the speed of light propagation similarly to air wind affecting the propagation of sound waves, which are faster along the direction of a wind and slower against it.

Many experiments intended towards detection some difference in light propagation related to movement of the Earth through aether ended up with negative results. For some of them physicists came up with explanations why the results were negative, but others were just hanging as unexplainable in the framework of the aether theory.

The most famous experiment to detect aether wind was that of Michelson and Morley in 1887.
If aether exists and our planet is moving in it like a submarine in the ocean, the light should have different speed if measured along two perpendicular to each other directions.
The experiment convincingly showed that there is no difference in the speed of light.

Magnetism of Electric Current

The following thought experiment demonstrates that from the pre-relativistic viewpoint, when space and time are understood in absolute classic Galilean sense, the behavior of a moving charge in constant uniform electromagnetic field depends on the reference frame from which it is observed.
This contradicts the Relativity Principle, which is considered one of the major cornerstones of Physics.

Consider an electrically neutral straight line wire in a vacuum. Since it's electrically neutral, there is no electrostatic field around it and the number of electrons equals to the number of protons.

Let's connect it to a battery; electric current will start running, and electrons will start moving in one direction, consumed by the positive pole of a battery and emitted from its negative pole.

During this process the number of electrons consumed by a positive pole per unit of time will be equal to the number of electrons emitted from the negative pole, which preserves the electrostatic neutrality of a wire as a whole.
Therefore, there is no electrostatic field around the wire.

As we know, there is a magnetic field around an electric current produced by moving electrons inside a wire. Its field lines form circles around a wire. This field does not change in time, as long as the battery maintains the constant voltage and the electric current is not changed, all its characteristics are constant.

The formula describing this magnetic field (to refresh your knowledge, see Magnetism of Electric Current in the Electromagnetism part of Physics 4 Teens course on UNIZOR.COM) is
B = μ0·I/(2πR)
where
B - intensity of a magnetic field (in tesla)
μ0 - permeability of vacuum (constant characteristic of vacuum)
I - constant electric current
R - distance from a wire to a point where magnetic field intensity is measured.

As we see from this formula, at any given point around this wire the intensity of the magnetic field is constant.

Assume that at some distance from a wire a positively charged particle starts moving on a parallel to a wire trajectory in the same direction and the same speed as electrons inside a wire.

This particle moves along a trajectory that perpendicularly intersects magnetic field lines around a wire and, therefore, will experience the Lorentz force that is directed perpendicularly to its velocity and to magnetic field lines.
In vector form this force is
F = q·vB
where
F - force acting on a particle
(in newtons N)
q - particle's charge
(in coulomb C)
v - linear speed of a particle
(in meters per second m/s)
B - intensity of a magnetic field
(in tesla T)

This Lorentz force will cause a particle to deviate from the straight line parallel to a wire and the distance from this particle to a wire will change.

Consider now the same arrangement from the point of view of an observer traveling with our particle.
In this reference frame a particle and electrons in the wire are standing still, since our observer moves in the same direction and with the same speed as they are.
Instead, our observer sees protons moving in the opposite direction than electrons in the previous case. That means that we still have the same electric current in the wire, the same magnetic field is produced by this electric current, and it has the same circular magnetic field lines.

The magnetic field seems the same, but a particle now is not moving in this reference frame, is NOT crossing magnetic field lines, which means there is no Lorentz force and the distance between a particle and a wire remains constant.

So, it looks like the laws of nature are different, if observed from two different inertial frames. In the first reference frame the particle's distance from the wire was changing as the time goes, while in the second reference frame the particle was at the same distance from a wire all the time. This contradicts the Principle of Relativity generally accepted as universal.

Maxwell Equations and Galilean Transformation

As it turns out, some Maxwell Equations ARE NOT INVARIANT relative to Galilean Transformation.

Without going into mathematical details of actually proving this non-invariance, it can be seen from the previous qualitative analysis of an electrically charged particle moving along a wire with electric current viewed from two inertial reference frames described above.

In one reference frame there is a Lorentz force acting on this particle, causing its deviation from a straight line trajectory and changing its distance from a wire, while in another reference frame there is no such force, a particle continues its movement along a trajectory parallel to a wire, preserving the distance between a particle and a wire.

## Tuesday, June 13, 2023

### Problem 3: UNIZOR.COM - Physics 4 Teens - Electromagnetism - Magnetism o...

Notes to a video lecture on http://www.unizor.com

Moving Charge - Problem 3

This problem is based on previous lectures about magnetic field around a straight wire with electric current running through it and the one generated by a moving electrically charged particle.

We would like to establish a connection between two formulas:

(a) the formula for a magnitude of a magnetic field intensity vector of a long straight wire with electric current running through it
B = μ·I/(2πR)
where
B is an intensity of a magnetic field at observation point
μ is permeability of media around a wire
I is an electric current running through a wire
R is a shortest distance of an observer from a wire
(all constants)

and

(b) the formula for a magnitude of a magnetic field intensity vector of the moving electrically charged particle
B = μ·q·v·sin(φ)/(4πr²)
where
B(t) is a variable intensity of a magnetic field at observation point
μ is permeability of media between a particle and an observer
q is an electric charge of a particle
v is a particle's speed
φ(t) is a variable angle between the direction of a movement of a particle and direction from an observer to a particle.
r(t) is a variable distance of an observer from a particle
(notice that φ, r and, therefore, B are variable in this case)

The problem is to derive formula (a) from formula (b) considering an electric current inside a wire of formula (a) as a flow of particles of formula (b).

We compare only expressions for magnitudes, as the direction of magnetic field intensity vectors is, obviously, the same - perpendicularly to both trajectory of electric charges and direction from an observer to a wire or to a moving particle.

Solution

To bring analytics and math to our problem, we assume that each infinitesimal part of a wire with electric current running through it is a moving electrically charged particle.
We will examine the magnetic field generated by this moving electrically charged particle and integrate all the individual magnetic field intensities generated by all such particles along an infinite wire.

We intentionally did not use the vector expression of magnetic field intensity with a vector product of velocity vector of a particle and a unit vector from an observer to a particle vu. As we stated, the direction of magnetic intensity vector in both cases (a) and (b) is, obviously, the same. So, using v·sin(φ) as a magnitude of vector vu simplifies the problem.

The most important part of a solution is to represent an electric current I as an infinite set of infinitesimal particles carrying infinitesimal electric charge and moving along a wire with constant speed.

Electric current inside a wire consists of electrons moving in some direction. Ideally, they all move with a uniform and constant speed v.
Assume, a linear density of electrons inside a wire is δ(coulomb per meter C/m).

Wire is infinite, so let's take some point on it as a zero coordinate and consider a piece of wire from coordinate x to x+dx.
This linear piece of wire of the length dx(m) contains dq=δ·dx(C) of electric charge.

Moving with speed v(m/s), this small charge will cover a distance from x to x+dx during the time dt=dx/v.

Charge dq moving along a wire during time dt constitute an electric current I=dq/dt.
Therefore,
I = dq/dt = (δ·dx)/(dx/v) = δ·v
Using this expression for electric current, formula (a) would be equivalent to
(aa) B = μ·δ·v/(2πR)

We will prove that formula (b) for a magnetic field of a moving particle will lead to the above formula (aa), if a current in a wire is viewed as a set of particles moving along a wire.

Let's use the formula (b) to calculate the intensity of a magnetic field generated by a particle carrying infinitesimal charge dq at position x of a wire and moving along a wire with speed v.
dB = μ·dq·v·sin(φ)/(4πr²)
Here we can replace r with R²+x².
Charge dq can be replaced with δ·dx.
Also we can replace sin(φ) with R/√R²+x².

Now the expression for infinitesimal magnetic field intensity generated by infinitesimal charge dq moving with speed v is as follows
dB = μ·δ·dx·v·R/[4π(R²+x²)3/2]

Integrating this expression by x from −∞ to +∞ produce the total impact of magnetic field intensity produced by an entire wire. It should be the same as in formula (aa). If it does, it would be an additional proof that Physics does not have inner contradictions in its laws.
Let's check it out.

First of all, both expressions, (aa) for a magnetic field of an entire wire and the above expression for a field produced by a charge dq, contain constant multipliers μ (permeability of media), δ (linear density of electric charge) and v (speed of particles inside a wire). They will, obviously cancel each other.
So, all we have to prove is that the integration of
dx·R/[4π(R²+x²)3/2]
along all values of x (positive and negative) results in 1/(2πR).

More simplifications can be obtained by considering that integration of an even function from −∞ to +∞ is equivalent to integration from 0 to +∞ times 2.
So, we have to prove that
[0,+∞]dx·R/[4π(R²+x²)3/2]
is equal to 1/(2πR).
Also, by multiplying both formula by 2πR, our task is reduced to proving that
[0,+∞]dx·R²/[(R²+x²)3/2] = 1
The last simplification is to substitute
y = x/R
x = y·R
dx = dy·R
Limits of integration from 0 to +∞ remain the same in this substitution.

The result is to prove that
[0,+∞]dy·R³/[(R²+y²·R²)3/2] = 1
Here R is, obviously, cancels out, and the only thing to prove is that
[0,+∞]dy/[(1+y²)3/2] = 1

To calculate the above integral, let's use the substitution
y = tan(z)
dy = dz/cos²(z)
The limits of integration for y from zero to +∞ will change from zero to π/2 for z.

Now our integral looks like
[0,π/2]dz/[cos²(z)·(1+tan²(z))3/2]

Since 1+tan²(z)=1/cos²(z), our integral equals to
[0,π/2]dz·cos(z) = sin(z)|[0,π/2] =
= sin(π/2)−sin(0) = 1

As we see, the equality has been proven.
That means that two different formulas mentioned above, (a) for a magnetic field of an electric current and (b) for a magnetic field of a moving charge, are completely in agreement with each other.

## Saturday, June 10, 2023

### Moving Charge: UNIZOR.COM - Physics4Teens - Electromagnetism - Magnetism...

Notes to a video lecture on http://www.unizor.com

Magnetism of
Moving Charged Particle

In the previous lecture, using some common sense consideration and logic supported by experiments, we came up with an expression for a magnetic field around a straight line wire with electric current running through it
B = μ0I/(2πR)
where
I is an electric current running through a wire in ampere (A),
R is a distance from a wire with an electric current in meter (m),
B is an intensity of a magnetic field in tesla (T),
μ0 is the permeability of free space in newtons per ampere squared (N/A²)

Let's use similar logic to evaluate an intensity of a magnetic field generated by a single moving electrically charged particle.

Assume, an electrically charged particle carrying charge q coulombs (C) is uniformly moving in vacuum along a straight line trajectory r(t) with constant velocity vector
v = dr(t)/dt
Assume further that an observer is located at the origin of coordinates O, and his task is to measure the intensity of a magnetic field created by this moving particle.

Since a charged particle is moving, it creates a magnetic field around it.
The force lines of this magnetic field form circles around a straight line trajectory of a particle. Each circle lies in the plane perpendicular to a trajectory and centered on a point of this trajectory.

Vectors of magnetic field forces (intensity vectors B(t,x,y,z)) at any point in space around a moving particle will be tangent to these circles.

The magnitude of the intensity of a magnetic field depends on certain parameters. The following statements about this magnitude are theoretically sound and confirmed by experiments.

This magnitude should be
(a) proportional to an amount of charge q carried by a particle,
(b) proportional to a particle's speed v,
(c) inversely proportional to a square of a distance r from a moving particle,
(d) depends on the angle between a particle's velocity vector v and a direction from an observer to a moving particle r(t) that we will describe for now as ∠(r(t),v).

The dependency (a) and (b) are obvious and need no explanation.

Inverse square rule (c) occurs everywhere in other fields (gravitational as Newton's Law, electrostatic as Coulomb's Law).
We usually put 4πr² into denominator of expressions for a force, where r is a distance from the source of this force.
The reason for this is that the energy emitted by this source is evenly distributed in our three-dimensional world in all directions and, when it reaches an observer on a distance r, actually falls on a whole sphere of a radius r around the source with an area of this sphere, as we know, 4πr².

The dependence (d) of the magnetic field intensity on the angle between a particle's velocity vector and a direction from an observer to a moving particle needs more explanation and analysis.

Consider two cases presented on pictures below.
Here we present only two dimensions to make the explanation more obvious.

An observer is measuring the intensity of a magnetic field by analyzing a density of an amount of magnetic energy falling on some measuring gadget.

In both cases the distance from a source of magnetic field (a moving charge) to an observer in the same.
In the first case the observation line denoted by a unit vector u is perpendicular to a velocity vector v of a particle. In the second case there is an acute angle between these two vectors.

The amount of energy emitted by a source (the proper term is magnetic field flux) in the first case and falling on a segment AB is the same as an amount of energy falling on a longer segment AB in the second case, and the length of this segment AB is longer by a factor 1/sin(φ) than in the first case.
Therefore, the observed density of magnetic field flux, which we call field intensity, must be less in the second case than in the first by a factor sin(φ).

This situation is exactly analogous to an amount of heat per unit of area (heat density) from the sun, when it shines perpendicularly to the Earth surface around noon, and at some later (or earlier) time, when the sun is not in zenith. The heat density is at maximum, when the sun shines straight down onto Earth surface and is less in the morning or evening, when the sun is closer to horizon and shines at other than the right angle.

All the above considerations lead us to the following formula for a magnitude of magnetic field intensity B
B ∝ q·v·sin(φ)/(4πr²)

We can transform this formula to incorporate the direction of the vector of magnetic field intensity. It's supposed to be tangent to a circle that lies in the plane perpendicular to a particle's velocity, centered on this trajectory and goes through a point of observation.

All these considerations lead us to the following expression for intensity of a magnetic field
B ∝ q·vu/(4πr²)
The usage of a vector product of a particle's velocity v and a unit vector in the direction of observation u assures correct representation of both direction and magnitude of the vector of magnetic field intensity.

Sometimes, to avoid using a unit vector u, it's replaced with r/r, where r is a vector from an observation point to a moving particle and r is its magnitude.
If an observation point is at the origin of Cartesian coordinates, vector r(t) describes the position of an particle as a function of time.
The first derivative of this vector by time is a velocity vector v(t)=dr(t)/dt=r'(t)
Now the magnetic field intensity of a charged particle whose position is described by a vector r(t) at the observation point at the origin of coordinates can be expressed as
B(t) ∝ q·r'(t)r(t)/(4πr³(t))

Experiments show that the coefficient of proportionality in this formula is the same media permeability μ introduced in the previous lecture (μ0 in case of vacuum).
The final formula is
B(t) = μ·q·r'(t)r(t)/(4πr³(t))