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michelson - morley experiment
I dont get this experiment at all. So, one ray of light is sent towards a mirror and reflected back again, parallel to the Earths surface. On its way to this mirror, it travels a farther distance then it does on the way back because the experiment is set up so that the light ray is emitted in the direction of Earths path through the ether.
Got that but then 
wtf? , if a light ray is sent perpendicular to the parallel one, how would it return to the source? unless it doesn't travel in a completely perpendicular direction to the ray parallel to Earth?
Got that but then wtf? , if a light ray is sent perpendicular to the parallel one, how would it return to the source? unless it doesn't travel in a completely perpendicular direction to the ray parallel to Earth?Scroll to see replies
I think you've got confused about the experimental set-up. It looks like this, in plan view if you are looking from above towards the floor:
It is assumed that the Earth passes through an Ether, which propagates light. Because the Earth is moving through the Ether, the light will travel at different speeds in the direction of motion of the Earth compared to perpendicular to that direction of motion. Therefore, 2 beams with equal path length should arrive at different times as one will propagate differently. You would then expect to see a shift in the fringe position, as the waves gain a shift in phase.
However, no shift was seen to within the accuracy of the experiment, suggesting that light travels the same speed in both directions and that there is no ether. The theory could though be retained if one assumed that the effect of the ether was exactly balanced by another factor which shrunk the length of the one arm. This was the Lorentz factor, but was only a fudge to maintain the ether. It just so happens to be the same factor that appears in special relativity, but it in no way implies relativity... it was just a fudge to save the then believed theory at the time.
The important part of the equipment is the partially silvered mirror in the middle, which allows part of the light through, and reflects the other half - therefore splitting the original beam into 2 perpendicular directions and later recombining them to be seen by a detector.
It is assumed that the Earth passes through an Ether, which propagates light. Because the Earth is moving through the Ether, the light will travel at different speeds in the direction of motion of the Earth compared to perpendicular to that direction of motion. Therefore, 2 beams with equal path length should arrive at different times as one will propagate differently. You would then expect to see a shift in the fringe position, as the waves gain a shift in phase.
However, no shift was seen to within the accuracy of the experiment, suggesting that light travels the same speed in both directions and that there is no ether. The theory could though be retained if one assumed that the effect of the ether was exactly balanced by another factor which shrunk the length of the one arm. This was the Lorentz factor, but was only a fudge to maintain the ether. It just so happens to be the same factor that appears in special relativity, but it in no way implies relativity... it was just a fudge to save the then believed theory at the time.
The important part of the equipment is the partially silvered mirror in the middle, which allows part of the light through, and reflects the other half - therefore splitting the original beam into 2 perpendicular directions and later recombining them to be seen by a detector.
oh, so the fact that both rays of light arrived at the same time means there is no ether, but doesn't the fact that the light ray perpendicular to Earths motion arrives back in the first place prove there is no ether, because if there was, by the time the light ray returned, the mirror would have moved.
Thanks, by the way
Thanks, by the way
PHKnows
oh, so the fact that both rays of light arrived at the same time means there is no ether, but doesn't the fact that the light ray perpendicular to Earths motion arrives back in the first place prove there is no ether, because if there was, by the time the light ray returned, the mirror would have moved.
Thanks, by the way
Thanks, by the way
That's the point. What it is actually showing, from the modern perspective, is that light travels at the speed of light whatever the relative speed of the objects around it. There is no preferential ether frame of reference, relative to which things should be measured. The laws of physics are the same in all inertial frames.
To rectify this, you need to introduce Lorentz contraction of the arm along the direction of motion. It wasn't seen in that way at the time though, it wasn't until later with Einstein etc that the ether theory could be proved to be wrong.
and no, it would still hit the mirror, since in a classical theory when the light was emitted it would have a forward component due to the motion of the mirror through the ether. Ie light would travel at v in the forward direction and c in the perpendicular direction giving it a resultant > c.
I get the implications of the fact that there is no interference of light, but only when i accept that the light ray perpendicular to earths motion has travelled farther, which is what i dont get.
I've done some really simple calculations and to me it seems like i've disproved the validity of the experiment for what it implies.lol.I think im definitely missing something.
Here are my calculations:
The Earth rotates around the Sun at a speed of roughly 30,000 m/s, which means it travels through the supposed ether at 30,000 m/s ????
A single light ray (for simplicity) is emitted perpendicular to earths motion, ie towards the sky, where it is reflected by a mirror and then detected when it returns.
Light travels at 3 x 10^8 m/s.
Assuming the mirror is 50m up, the light ray travels 100m before being detected. time = dist/speed = 100/3 x 10^8 = 0.3 microseconds
In this time, the earth has moved a distance of 30,000 x 0.3 microseconds = 0.01m , ie a cm. Therefore, in the time it takes for any given lightray to be emitted and then detected, the appuratus has only moved a cm through the ether, in which case it can still be detected.
The lightray parallel to earths motion has travelled exactly the same distance because when it returns to the detector, the earth moving in the opposite direction to it through the ether compensates for when the lightray is emitted and has to "catch up" the mirror.
Therefore, both lightrays have travelled the same distance, and interference shouldn't be observed anyway, assuming the detector can "detect" within the range of a cm.
And thats when the mirrors are 50m away. When they are 5m away, the distance travelled through the ether is only a mm.
As you,ve probably guessed, im imagining the light moving in the same line through the ether, ie like if you were to throw a ball directly up while running, the ball would not have any horizontal movement, and this makes sense to me because electromagnetic field dynamics are different to human-ball dynamics in my eyes.
I've done some really simple calculations and to me it seems like i've disproved the validity of the experiment for what it implies.lol.I think im definitely missing something.
Here are my calculations:
The Earth rotates around the Sun at a speed of roughly 30,000 m/s, which means it travels through the supposed ether at 30,000 m/s ????
A single light ray (for simplicity) is emitted perpendicular to earths motion, ie towards the sky, where it is reflected by a mirror and then detected when it returns.
Light travels at 3 x 10^8 m/s.
Assuming the mirror is 50m up, the light ray travels 100m before being detected. time = dist/speed = 100/3 x 10^8 = 0.3 microseconds
In this time, the earth has moved a distance of 30,000 x 0.3 microseconds = 0.01m , ie a cm. Therefore, in the time it takes for any given lightray to be emitted and then detected, the appuratus has only moved a cm through the ether, in which case it can still be detected.
The lightray parallel to earths motion has travelled exactly the same distance because when it returns to the detector, the earth moving in the opposite direction to it through the ether compensates for when the lightray is emitted and has to "catch up" the mirror.
Therefore, both lightrays have travelled the same distance, and interference shouldn't be observed anyway, assuming the detector can "detect" within the range of a cm.
And thats when the mirrors are 50m away. When they are 5m away, the distance travelled through the ether is only a mm.
As you,ve probably guessed, im imagining the light moving in the same line through the ether, ie like if you were to throw a ball directly up while running, the ball would not have any horizontal movement, and this makes sense to me because electromagnetic field dynamics are different to human-ball dynamics in my eyes.
See if you can make sense of this, which is calculating the length travelled by the 2 beams:
http://www.relativitycalculator.com/Albert_Michelson_Part_II.shtml
http://www.relativitycalculator.com/Albert_Michelson_Part_II.shtml
how does L1/(c+v) + L1/(c-v) = 2L1/(c^2 - v^2)??
PHKnows
how does L1/(c+v) + L1/(c-v) = 2L1/(c^2 - v^2)??
combine fractions and multiply out the 2 factors in the denominator. Difference of 2 squares.
F1 fanatic
See if you can make sense of this, which is calculating the length travelled by the 2 beams:
http://www.relativitycalculator.com/Albert_Michelson_Part_II.shtml
http://www.relativitycalculator.com/Albert_Michelson_Part_II.shtml
I cant really follow it, the maths doesnt make sense, and i asked on the maths thread and according to a guy on there :
L1/(c+v) + L1/(c-v) = 2L1c/(c^2 - v^2). The c's missing on the webpage
I think i get the jist of whats going on but it relies on the fact that the perpendicular to earths motion lightray has a horizontal component. And my calculations show that it doesnt have to have a horizontal component to still be detected.
PHKnows
I cant really follow it, the maths doesnt make sense, and i asked on the maths thread and according to a guy on there :
L1/(c+v) + L1/(c-v) = 2L1c/(c^2 - v^2). The c's missing on the webpage
I think i get the jist of whats going on but it relies on the fact that the perpendicular to earths motion lightray has a horizontal component. And my calculations show that it doesnt have to have a horizontal component to still be detected.
L1/(c+v) + L1/(c-v) = 2L1c/(c^2 - v^2). The c's missing on the webpage
I think i get the jist of whats going on but it relies on the fact that the perpendicular to earths motion lightray has a horizontal component. And my calculations show that it doesnt have to have a horizontal component to still be detected.
yeah you're right, the c comes back in the next line on the webpage (implicitly).
What you have above is wrong though. Firstly, you don't normally consider it as going towards the sky, but the other perpendicular direction (ie still parallel to the floor, but away from the motion of the earth. You don't need to know the direction of Earth's motion, the fact they are perpendicular means it doesn't matter.
Secondly, your saying that 1cm makes no difference makes no sense. It may still be detected, it depends on the size of your mirror, but the point is that the mirror has moved on a distance and so in order to hit it and bounce back the light must travel at an angle. The size of the mirror isn't really important.
Similarly, when the light comes back, it won't "catch up" because the times are clearly different.
L/(c+v) =/= L/(c-v)
The times are different. Don't worry if you don't understand this, it's really not something you need to worry about until degree level when you will be shown how to do it properly.
Oh i see. So the lightrays are both emitted along the surface of the earth and perpendicular in that plane. But whats really troubling me is that, it seems as though, according to that webpage, the lightray returns to the exact position it left from. How does light "know how to do that"?
PHKnows
Oh i see. So the lightrays are both emitted along the surface of the earth and perpendicular in that plane. But whats really troubling me is that, it seems as though, according to that webpage, the lightray returns to the exact position it left from. How does light "know how to do that"?
position? you mean position as in relative position on the instrument, rather than absolute position in space? Ahh, now that's the crucial thing. That's why the light perpendicular is at an angle. It's not that the light knows, the light is emitted in all directions, but you only choose to look at the ray that starts and ends at the same point since that's what a perpendicular light beam would do between 2 mirrors were they not moving.
i mean position in the ether. if the earth is moving and emits a lightray, the lightray would not return to the exact same position on the appuratus coz the earth is moving. thats what my calculations were for. the earth would have moved 1mm through the ether by the time it took the light ray to be emitted and returned. So how does the lightray just randomly travel angularly to the pure perpendicular direction to compensate for this so that it returns to the same position it was emitted?
I'm confused, it doesn't return to the same place in the ether. Looking at it from above you would see it do this:
_____
/\
/ \
/ \
Thats whats troubling me, i just think its a huge coincidence that the lightray follows a path through the ether which allows it to return to its original position, no matter what the speed of the earth and orientation of solar system etc. If it did return a mm or so away from where it was emitted, would it still be detectable?
PHKnows
Thats whats troubling me, i just think its a huge coincidence that the lightray follows a path through the ether which allows it to return to its original position, no matter what the speed of the earth and orientation of solar system etc. If it did return a mm or so away from where it was emitted, would it still be detectable?
but it clearly hasn't returned to it's previous position, seeing as it's in a different place in space before and after. If it were in the same place it would be on top of the outgoing ray.
but then there shouldn't be interference anyway, because both lightrays have travelled the same distance.
PHKnows
but then there shouldn't be interference anyway, because both lightrays have travelled the same distance.
huh? that's one light ray, it goes from the start, to the first mirror, and the bounces back to the same plane from which is started but displaced along the x axis. The other ray isn't even considered here.
displaced along the x axis? so it has moved conveniantly?
Ok. I'll try and be more clear on what i dont get.
Prior to the michelson-morley experiment, newtonian physics suggested that light could be "caught up" if you chased it at a speed above c.
Light was thought, (and still is?), to travel through the ether. Theoretically, as the earth orbits around the sun, and the solar system around the centre of the milkyway, it must be travelling through the ether.
If you use an inferometer, you are shooting light out in a 360 degree fashion. At different points in this rotation, the earths motion relative to the ether is the same but the light is being shot at different angles across the ether.
What i dont get is how the lightray returns when the earth is moving through the ether in the first place. An analogy: you are on a train. you also have a baseball. the train is the earth and the baseball is the lightray. the atmosphere surrounding the train is the ether.
according to newtonian physics, when you throw the baseball out the window, it will not travel in a complete perpendicular direction to the trains motion. it will also have a horizontal component, assuming the trains motion is horizontal. now imagine a mirror attatched to the outside of the train which you throw the ball against. The ball(a lightray) would have to travel horizontally on both the out and return journey, to return to your hand.
So light returns to its original position in this experiment only by not travelling completely "straight" through the ether. It randomly moves in a specific pathway to be detected again. Is this what is happening?
Prior to the michelson-morley experiment, newtonian physics suggested that light could be "caught up" if you chased it at a speed above c.
Light was thought, (and still is?), to travel through the ether. Theoretically, as the earth orbits around the sun, and the solar system around the centre of the milkyway, it must be travelling through the ether.
If you use an inferometer, you are shooting light out in a 360 degree fashion. At different points in this rotation, the earths motion relative to the ether is the same but the light is being shot at different angles across the ether.
What i dont get is how the lightray returns when the earth is moving through the ether in the first place. An analogy: you are on a train. you also have a baseball. the train is the earth and the baseball is the lightray. the atmosphere surrounding the train is the ether.
according to newtonian physics, when you throw the baseball out the window, it will not travel in a complete perpendicular direction to the trains motion. it will also have a horizontal component, assuming the trains motion is horizontal. now imagine a mirror attatched to the outside of the train which you throw the ball against. The ball(a lightray) would have to travel horizontally on both the out and return journey, to return to your hand.
So light returns to its original position in this experiment only by not travelling completely "straight" through the ether. It randomly moves in a specific pathway to be detected again. Is this what is happening?
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