The Michelson-Morley experiment

[arbol]

Let S' be an x'y'-coordinate system. Let the x'-axis of S' coincide with the x-axis of an xy-coordinate system S, and let the y'-axis of S' be parallel to the y-axis of S. Let S' move along the x-axis of S with velocity v in the direction of increasing x, and let the origin of S' coincide with the origin of S at t = t' = 0s.

Let a ray of light depart from x'0 = 0m at the time t'0 = 0s towards x' = x'1 and reach x' = x'1 at the time t' = t'1, and let it be reflected back at x' = x'1 and reach x' = x'0 at the time t' = t'2. Let x'1 - x'0 = |x'0 - x'1| = L. Then with respect to the moving system S',

t'1 - t'0 = L/c, and

t'2 - t'1 = L/c = t'1 - t'0, and with respect to the stationary system S,

t1 - t0 = L + v*(t'1 - t'0)/(c + v) = t'1 - t'0, and

t2 - t1 = L - v*(t'2 - t'1)/(c - v) = t'2 - t'1.

Now let the ray of light depart from y'0 = 0m at the time t'0 = 0s towards y' = y'1 and reach y' = y'1 at the time t' = t'1, and let it be reflected back at y' = y'1 and reach y' = y'0 at the time t' = t'2. Let y'1 - y'0 = |y'0 - y'1| = L. Then with respect to the moving system S',

t'1 - t'0 = L/c, and

t'2 - t'1 = L/c = t'1 - t'0, and with respect to the stationary system S,

t1 - t0 = sqrt(sq(L) + sq(v*(t'1 - t'0)))/sqrt(sq(c) + sq(v)) = t'1 - t'0, and

t2 - t1 = sqrt(sq(L) + sq(v*(t'2 - t'1)))/sqrt(sq(c) + sq(v)) = t'2 - t'1.


The result of the Michelson-Morley experiment confirmed that

t2 - t0 = (L + v*(t'1 - t'0))/(c + v) + (L - v*(t'2 - t'1))/(c - v) = 2*(c*L - sq(v)*(t'2 - t'0/2))/(sq(c) - sq(v)) = sqrt(sq(L) + sq(v*(t'1 - t'0)))/sqrt(sq(c) + sq(v)) + sqrt(sq(L) + sq(v*(t'2 - t'1)))/sqrt(sq(c) + sq(v)) = 2*sqrt(sq(L) + sq(v*(t'2 - t'0/2)))/sqrt(sq(c) + sq(v)).

The result of the experiment is described as negative because it was expected to confirmed that

T1 = 2*L*c/(sq(c) - sq(v)) was not equal T2 = 2*sqrt(sq(L) + sq(v*(T2/2)))/c.

Moreover, the result of the Michelson-Morley experiment is actually independent of the value of v. Thus it did not reflect at all that the velocity v of the moving system S' with respect to the stationary system S is 0m/s, where S' is to be taken as the Earth and S as the assumed ether.

 

[arbol]

The result of the Michelson-Morley experiment confirmed that

t2 - t0 = (L + v*(t'1 - t'0))/(c + v) + (L - v*(t'2 - t'1))/(c - v) = 2*(c*L - sq(v)*(t'2 - t'0/2))/(sq(c) - sq(v)) = sqrt(sq(L) + sq(v*(t'1 - t'0)))/sqrt(sq(c) + sq(v)) + sqrt(sq(L) + sq(v*(t'2 - t'1)))/sqrt(sq(c) + sq(v)) = 2*sqrt(sq(L) + sq(v*(t'2 - t'0/2)))/sqrt(sq(c) + sq(v)).

The result of the experiment is described as negative because it was expected to confirmed that

T1 = 2*L*c/(sq(c) - sq(v)) was not equal T2 = 2*sqrt(sq(L) + sq(v*(T2/2)))/c

.

I made some mistakes in the quote above. I apologize for the confusion. The quote above ought to be written as follows:

The result of the Michelson-Morley experiment confirmed that

t2 - t0 = (L + v*(t'1 - t'0))/(c + v) + (L - v*(t'2 - t'1))/(c - v) = 2*(c*L - sq(v)*(t'2 - t'0)/2))/(sq(c) - sq(v)) = sqrt(sq(L) + sq(v*(t'1 - t'0)))/sqrt(sq(c) + sq(v)) + sqrt(sq(L) + sq(v*(t'2 - t'1)))/sqrt(sq(c) + sq(v)) = 2*sqrt(sq(L) + sq(v*(t'2 - t'0/2)))/sqrt(sq(c) + sq(v)).

The result of the experiment is described as negative because it was expected to confirmed that

T1 = 2*L*c/(sq(c) - sq(v)) was not equal to T2 = 2*sqrt(sq(L) + sq(v*(t'2-t'0)/2)))/c.

Here I elaborate further the subject of this post.

let t' = L/c.

If t1 = 2*(c*L-v^2*t')/(c^2 - v^2) = 2*t'*(c^2 - v^2)/(c^2 - v^2) = 2*t', and

t2 = 2*sqrt(L^2+(v*t')^2)/sqrt(c^2 + v^2) = 2*t'*sqrt(c^2 + v^2)/sqrt(c^2 + v^2) = 2*t', then

t1/t2 =1.

The following programs return the following values respectively:


def t():
L=input("L: ")
c=299792458.0
t=L/c
print t

L: 299792458.0
t: 1.0

def t():
L=input("L: ")
c=299792458.0
t=L/c
v=input("v: ")
t=(L+v*t)/(c+v)
print t

L: 299792458.0
v: 29800.0
t: 1.0

def t():
L=input("L: ")
c=299792458.0
t=L/c
v=input("v: ")
t=(L-v*t)/(c-v)
print t

L: 299792458.0
v: 29800.0
t: 1.0

def t():
L=input("L: ")
c=299792458.0
t=L/c
v=input("v: ")
t=(L**2.0+(v*t)**2.0)**0.5/(c**2.0+v**2.0)**0.5
print t

L; 299792458.0
v: 29800.0
t: 1.0

def t():
L=input("L: ")
c=299792458.0
t=L/c
v=input("v: ")
t=2*(L**2.0+(v*t)**2.0)**0.5/(c**2.0+v**2.0)**0.5
print t

L: 299792458.0
v: 29800.0
t: 2.0

def t():
L=input("L: ")
c=299792458.0
t=L/c
v=input("v: ")
t=2*(c*L-v**2.0*t)/(c**2.0-v**2.0)
print t

L; 299792458.0
v: 29800.0
t: 2.0

let t' = L/c.

If t1 = 2*c*L/(c^2 - v^2) = 2*c^2*t'/(c^2 - v^2), and

t2 = 2*sqrt(L^2 + (v*t')^2)/c = 2*t'sqrt(c^2 + v^2)/c, then

t1/t2 = c^3/((c^2 - v^2)*sqrt(c^2 + v^2)).

The following programs return the following values respectively:

def t():
L=input("L: ")
c=299792458.0
t=L/c
v=input("v: ")
t=2*c*L/(c**2.0-v**2.0)
print t

L: 299792458.0
v: 29800.0
t: 2.00000001976.

def t():
L=input("L: ")
c=299792458.0
t=L/c
v=input("v: ")
t=2*(L**2.0+(v*t)**2.0)**0.5/c
print t

L: 299792458.0
v: 29800.0
2.00000000988.

 

[dkv]

The experiment confirmed that there was no absolute reference frame.Do you agree or disagree?

 

[arbol]

The experiment confirmed that there was no absolute reference frame.Do you agree or disagree?

Let S' be an x'-coordinate system. Let the x'-axis of S' coincide with the x-axis of an x-coordinate system S, and let S' move along the x-axis of S with velocity v in the direction of increasing x, and let the origin of S' coincide with the origin of S at t = t' = 0s.

Let a ray of light depart from x'0 = 0m at the time t'0 = 0s towards x' = x'1 and reach x' = x'1 at the time t' = t'1, and let it be reflected at x' = x'1 back to x'0 = 0m, and reach x' = 0m at the time t' = t'2. Let x'1 - x'0 = |x'0 - x'1| = L. Then with respect to the moving system S',

t'1 - t'0 = L/c, and

t'2 - t'1 = L/c = t'1 - t'0, and with respect to the stationary system S,

t1 - t0 = L + v*(t'1 - t'0)/(c + v), and

t2 - t1 = L - v*(t'2 - t'1)/(c - v), where by the Michelson-Morley experiment, it was established that

t1 - t0 = t2 - t1 = L/c.

Therefore, I think the following question is more appropriate:

By the Michelson-Morley experiment, were we able to established a common time between the moving system S' and the stationary system S?

Yes.

That t'1 - t'0 is the time, with respect to the moving system S', the ray of light takes to move along its own x'-axis from x'0 = 0m to x' = x'1 while t1 - t0 the time, with respect to the stationary system S, the ray of light takes to move along the x'-axis of the moving system S' from x'0 = 0m to x' = x'1 implies that in order to establish a common time between the moving system S' and the stationary system S, an absolute frame of reference is not necessary.

 

[birulami]

The experiment confirmed that there was no absolute reference frame.Do you agree or disagree?

At best it showed that the experiment is not suited to prove the existence of an absolute reference frame or isolate it by measurement. This is not sufficient to prove that the absolute reference frame does not exist. Interestingly an absolute reference frame is indeed completely consistent with special relativity, but usually ruled out by occams razor.

See, for example, "How to teach relativity" from http://en.wikipedia.org/wiki/John_Stewart_Bell" .

 

[MikeLizzi]

Nice to read your statement, birulami. I thought I was alone in that opinion.

 

[russ_watters]

While that's true, it also says that such a reference frame could show no unique properties. So it really is completely pointless to consider the possibility. You may as well assume it doesn't exist, because it does nothing for you to assume it does. Just like the invisible/undetectable dragon living in my garage.

 

[MikeLizzi]

Russ, how could one experiment involving the behavior of light propagating within one reference frame show that a potential absolute reference frame would show no unique properties? Come on. No unique properties for light maybe. What about all the properties of matter?

 

[birulami]

While that's true, it also says that such a reference frame could show no unique properties. So it really is completely pointless to consider the possibility.

When you start learning geometry, you draw points and lines and triangles on paper despite the fact that those crude pictures have nearly nothing to do with the concepts introduced by the axioms about points and lines.

Similarly, the absolute reference frame would make people feel at home, for a start, when learning about special relativity. Basic geometry and algebra can then easily show why and how no observer is actually able to nail down the absolute frame with any measurement possible. Even more, all the perceived paradoxes just disappear. The only postulate necessary is the constant speed of light, and the absolute frame gets redundant without much fuss.

Harald.