I see a lot of reluctance to use relativity, although it is the best we have for the representation of the macroscopic universe.
So I will try to explain in simple terms why, starting from a basic observation, the constancy of the speed of light for all observers, we easily arrive at special relativity and the Lorentz transforms.
Let us take a point (x,y,z) in space. From the origin (0,0,0), let's send a light signal toward this point. The distance to travel is :
d = √(x²+y²+z²). To simplify let us write d²=x²+y²+z².
The light covers this distance d at speed c in a time t, we will have d=ct so c²t² = x²+y²+z² or
x²+y²+z²-c²t² = 0.
Let us suppose that for another observer, an observer being either a human or a measuring device or an electrical charge feeling effects... the point is seen in (x',y',z') relatively to (0,0,0).
As the speed of light does not depend on the observer, it will see: c²t'² = x'²+y'²+z'² or
x'²+y'²+z'²-c²t'² = 0.
Therefore
(1) x²+y²+z²-c²t² = x'²+y'²+z'²-c²t'². s = √(x²+y²+z²-c²t²) is called a space-time interval, i.e. a "distance" between 2 points in a 4D space, and its value
is the same for all observers.
In a 4D space, we note (x,y,z,-ct) the coordinate of an event, which is the spatial coordinates of a point to which we add -ct, the temporal coordinate, made homogeneous with the others since in this form, multiplied by c, it is in fact like a fourth spatial coordinate, this changing only the form because c is a constant.
An event is to 4D space what a position is to 3D space, and the space-time interval s, "distance" between 2 events, the same for everyone, is to 4D space what a distance between 2 positions is to 3D space.
If you have understood this invariance of the space-time interval in a 4D universe, you know everything about special relativity, the rest being only logical consequences..
One of these consequences are the Lorentz transforms, allowing to express in one observer, what another sees. So now that everything is clear and straightforward, let's check it.
Let us consider only the x-axis, so y=z=0, with a fixed observer at the origin (0,0), and another one moving on this axis at the speed v.
Then according to
(1) x²-c²t² = x'²-c²t'² where (x,t) is seen by the fixed observer and (x',t') by the moving observer. The mobile observer moving at the speed v, the fixed observer will see it at the distance x=vt, and x'=0 because x' is the position of the mobile observer seen by itself.
It thus remains: x²-c²t² = -c²t'² => (vt)²-c²t² = -c²t'² => v²/c².t² - t² = - t'² => t² = t'²/(1-v²/c²) =>
t = t'/√(1-v²/c²)We find the Lorentz transform for the time dilation. Indeed v/c < 1, so 1/√(1-v²/c²) > 1 so t > t' : the time which passes for the fixed observer is longer than the one to which the mobile observer is subjected.
I let you do the same calculation exercise for the contraction of lengths.
These questions concern the space and time coordinates, i.e. the geometry.
Everything we express depends on them, electric and magnetic fields, potentials, fluxes, the movement of charges... and energy!
If the reference frame is not implicit and you do not specify it, you are wrong.
If you believe that a moving electric charge will see the same fields as you, you are wrong.
If you mix quantities measured from different reference frames, you are wrong.
If you neglect relativistic effects without justifying it, you are wrong...
Relativity is just common sense drawing inevitable conclusions from the facts. It applies everywhere. We are really in a 4D space.The 19th century ideas of absolute time are no longer valid, those who still have them must absolutely update their brains!
Author
Topic: Electromagnetism and relativity (Read 10897 times)
