Why does this ball go faster than c?

[Capiert]

In the animation’s right side (or half):

 

https://en.wikipedia.org/wiki/File:MichelsonMorleyAnimationDE.gif

 

Why does the red (ball) go faster than c?

 

Speed (v=d/t) is distance (d) per time (t).

 & the left animation

 (clearly) shows that the red ball moves

 (diagonally (up & then down))

 faster than the wave(_speed c).

 

(Simply compared:)

The ball is moving faster

 than the waves. E.g.

That (right_side’s red)

 ball travels more distance

 (diagonally);

 than the waves (can) do

 in the same amount of time.

I.e. longer (diagonal) path,

 compared to the horizontal path;

 & arrives

 (back at the gray half_mirror)

 “sooner” (=in less time)

 (than the blue ball).

 

Imagine (that), light (is) travelling faster than light.

=Is that suppose to be physics?

(It's a classic (joke or fake), but NOT classical.)

 

Edited December 16, 2017 by Capiert

[Capiert]

That's suppose to be a (classical=non_relativistic) demonstration,

 & it breaks the rules, both (classically & relativistically):

 of not obeying wave physics (of Michelson's time 1887),

 nor obeying relativity's light_speed c "limit" (Einstein 1905, 1915, 1916, 1920 .. til now).

[Strange]

3 hours ago, Capiert said:

Why does the red (ball) go faster than c?

It doesn’t. The right red ball reaches the screen after the left right ball  

 

[Capiert]

Thank's Strange it looks like you won

 & cleared that.

But why aren't both red balls going vertically up

 at the same speed?

That's suppose to be a comparison

 in the same medium.

Only the apparatus is suppose to be moving

 to the right til it (=the right red ball) hits the top mirror;

 or else travel diagonally.

I guess you mean,

 the right ball is moving slower vertically?

Ok, thanks.

It seems to make sense to me now.

Except the 90 degree angle is missing for the red incident ray (right side: diagonal up & right).

How do you explain that?

 

[swansont]

23 minutes ago, Capiert said:

Thank's Strange it looks like you won

 & cleared that.

But why aren't both red balls going vertically up

 at the same speed?

If the speed is limited to c and you have horizontal motion, you can't have the same vertical speed.

Look at fig 4 where this is used; there's a diagram

https://en.wikipedia.org/wiki/Michelson–Morley_experiment

 

23 minutes ago, Capiert said:

T It seems to make sense to me now.

Except the 90 degree angle is missing for the red incident ray (right side: diagonal up & right).

How do you explain that?

What 90 degree angle?

[Strange]

32 minutes ago, Capiert said:

But why aren't both red balls going vertically up at the same speed?

Because there is a horizontal component to its velocity. So the vertical component must be less to maintain the same total speed. 

[Capiert]

1 hour ago, swansont said:

If the speed is limited to c and you have horizontal motion, you can't have the same vertical speed.

Where does the (right red) ball's horizontal motion come from?

The right red ball has collided

 with the (horizontally) moving half mirror

 (& then moved up).

 But that red ball was bounced (=recoil, in the opposite direction).

(Naturally, the half mirror was only 45 degrees (instead of 90 degrees) orientation,

 but it's still a recoil (at least partially).)

So why should the right red ball have horizontal motion to the right

 after the 1st collision

 with the half mirror?

Quote

Look at fig 4 where this is used; there's a diagram

https://en.wikipedia.org/wiki/Michelson–Morley_experiment

Aren't the dotted (diagonal) lines' angles wishful thinking (from Michelson)?

He no longer has 90 degree incidence.

Quote

What 90 degree angle?

The 90 degrees

 that Michelson used

 to aim the light beam

 at the mirrors,

 for reflection (back to the half mirror).

1 hour ago, Strange said:

Because there is a horizontal component to its velocity. So the vertical component must be less to maintain the same total speed. 

Good answer. But (Michelson's) 90 degree angle?

The lab apparatus always had 90 degress,

 but his diagram did NOT.

What tells the photons to change their aimed angle (wrt v)? (Magic?)

=How do the photons know which angle they need

 to stay in sync?

(Michelson (thought he) knew,

 that's why he sketched the (other than 90 degree) angles.

But the photons didn't (know).

They only had 90 degree angled apparatus,

 I assume.)

=What (manually) re_aims the beam('s incident angle) wrt speed v? (Magic?)

Edited December 16, 2017 by Capiert

[swansont]

23 hours ago, Capiert said:

Where does the (right red) ball's horizontal motion come from?

It's emitted from the source. Photons move at c.

Quote

The right red ball has collided

 with the (horizontally) moving half mirror

 (& then moved up).

 But that red ball was bounced (=recoil, in the opposite direction).

(Naturally, the half mirror was only 45 degrees (instead of 90 degrees) orientation,

 but it's still a recoil (at least partially).)

It's a reflection.

Quote

So why should the right red ball have horizontal motion to the right

 after the 1st collision

 with the half mirror?

Because the system is moving to the right. 

We know the light hits the mirror. If the mirror is moving to the right, the light must also be moving to the right, in order to hit the mirror.

Quote

Aren't the dotted (diagonal) lines' angles wishful thinking (from Michelson)?

He no longer has 90 degree incidence.

The 90 degrees

 that Michelson used

 to aim the light beam

 at the mirrors,

 for reflection (back to the half mirror).

Good answer. But (Michelson's) 90 degree angle?

The lab apparatus always had 90 degress,

 but his diagram did NOT.

What tells the photons to change their aimed angle (wrt v)? (Magic?)

Nothing "tells" the photons to do this. It's what happens when you view the system from another reference frame. You can ask the same thing about the mirrors in the picture. Why are they moving to the right? It's because we have chosen a frame where everything in the apparatus is moving.

You can ask the same question about a tree when you are driving in a car: how does it "know" to move past you?

 

[Strange]

On 16/12/2017 at 2:22 PM, Capiert said:

What tells the photons to change their aimed angle (wrt v)? (Magic?) =How do the photons know which angle they need to stay in sync?

Another way of looking at it (exactly equivalent to swansont's explanation) is that it is a choice made by the person drawing the diagram. There are multiple paths you could draw, not all of which are useful.  We need to trace the path of the light ray which will hit the upper mirror, then be reflected and hit the screen at the bottom. Because the apparatus is moving, you need to choose a path where the light goes at a slight forward angle in order to keep up with the apparatus. 

[swansont]

19 minutes ago, Strange said:

Another way of looking at it (exactly equivalent to swansont's explanation) is that it is a choice made by the person drawing the diagram. There are multiple paths you could draw, not all of which are useful.  We need to trace the path of the light ray which will hit the upper mirror, then be reflected and hit the screen at the bottom. Because the apparatus is moving, you need to choose a path where the light goes at a slight forward angle in order to keep up with the apparatus. 

To clarify this a bit: one must also note that there is no speed at which the photons will miss the mirror. We are not depending on a "spray" of photons (i.e. the drawer of the diagram is not just picking photons that hit; they always hit) The source is aimed at the mirror, which means that the photons hit the mirror. That must happen in all frames of reference. I cannot make the photons miss simply by being in motion.

[Capiert]

On ‎2017‎ ‎12‎ ‎18 at 12:55 PM, Strange said:

Another way of looking at it (exactly equivalent to swansont's explanation) is that it is a choice made by the person drawing the diagram. There are multiple paths you could draw, not all of which are useful.  We need to trace the path of the light ray which will hit the upper mirror, then be reflected and hit the screen at the bottom. Because the apparatus is moving,

So here it comes:

On ‎2017‎ ‎12‎ ‎18 at 12:55 PM, Strange said:

you

Not the photon!

On ‎2017‎ ‎12‎ ‎18 at 12:55 PM, Strange said:

need to choose a path where the light goes at a slight forward angle in order to keep up with the apparatus. 

Where did the photon

 get such intelligence? (I ask.)

That's a difficult task for any operator

 (to master, & calibrate).

How is a dumb, passive photon,

 supposed to be able

 to do that (calibration or re_aiming) task.

We normally have to train people

 to do that (aiming)

 to hit that (variable) target.

I doubt if you have recognized the situation.

Machines do NOT calibrate themselves.

If you are shooting, you must aim.

Direction (angles) is(/are) a very delicate principle;

 random will NOT do!

"Michelson aimed" the apparatus (only) once (~90°).

How does that account for a variable angle

 (with many possibilities, other than 90°)?

The sketch (Fig2, 1887)

 does NOT have 90°

 for the vertically_up incident_path.

The sketch, is thus,

 NOT an accurate display

 of the apparatus's behavior;

 but instead

 it is the fantasy (if I may say)

 of Michelson.

 

On ‎2017‎ ‎12‎ ‎18 at 1:18 PM, swansont said:

To clarify this a bit: one must also note that there is no speed at which the photons will miss the mirror.

That sounds like an acceptable statement.

On ‎2017‎ ‎12‎ ‎18 at 1:18 PM, swansont said:

We are not depending on a "spray" of photons (i.e. the drawer of the diagram is not just picking photons that hit; they always hit).

Here's where things get murky.

Whoever drew the diagram

 (I don't want to mention (any) names

 but I will assume Michelson did.)

 chose 1 specific angle

 (for the beam travelling diagonally_up to the right);

 but it was NOT 90°!

 (..which the experiment had!).

But that doesn't matter,

 I mean,

 do scientists always

 have to do things right?

A little experimental error

 never hurt the funding.

Especially when dealing with famous (telephone) people

 like A. G. Bell.

The point is, Michelson

 had very sensitive apparatus (+1 point for precision);

 & it was locked_on=calibrated

 in only 1 position (angle, approx. 90°);

 & stayed so (calibrated, for producing fringes),

 so he could (freely) rotate the stone 360°.

But that 90° angle

 is NOT demonstrated

 in his sketch's (Fig 2, 1887) vertical path.

Any child can see that the sketch's (vertical) angle is NOT 90°;

 & yet the experiment('s setup) was.

The experiment

 versus the diagram (Fig 2, 1887)

 do NOT correlate

 on that point

 of dealing with

 the sketch's_vertical incident_angle.

On ‎2017‎ ‎12‎ ‎18 at 1:18 PM, swansont said:

 The source is aimed at the mirror,

i.e. half mirror

On ‎2017‎ ‎12‎ ‎18 at 1:18 PM, swansont said:

which means that the photons hit the

(=that 45°, half)

On ‎2017‎ ‎12‎ ‎18 at 1:18 PM, swansont said:

mirror. That must happen in all frames of reference.

Meaning any speed v.

But here's the catch.

On ‎2017‎ ‎12‎ ‎18 at 1:18 PM, swansont said:

I cannot make the photons miss simply by being in motion.

That's true:

 once the beam is aimed correctly

 (then) it stays correctly_aimed,

 for any angle

 the apparatus

 is rotated (into).

But for all practical purposes,

 it is aimed in "only 1" direction

 wrt the apparatus.

(&) Mechanically we can "measure" (that angle was) approximately 90° (for the lab apparatus).

Why (then) is that (~90° angle) NOT in Michelson's (Fig 2, 1887) sketch?

Why does his "sketch" use a "smaller" angle than (the experimental) 90° (incidence)?

 

Why should I bother with a diagram

 that does NOT accurately describe the experiment?

[Capiert]

On 17 December 2017 at 1:59 PM, swansont said:

We know the light hits the mirror. If the mirror is moving to the right, the light must also be moving to the right, in order to hit the mirror.

In which case, wasn't that angle decided by the draftsperson (that was) sketching?

 (because in that case a new angle would be needed for every different speed v ?);

 or else if the upper mirror is long enough to be hit at 90 degrees,

 & reflection occurs in a wavefront,

 wouldn't the diagonal beam, where it's twice Michealson's (drafted) reflected angle,

 hit the semi_mirror in perfect sync with the horizontal path? i.e. identical delays for both (vertical & horizontal) paths?

 

(I mean the (reflected) angle in question is <1 degree (difference, wrt  incidence at 90 degrees) for the intended speed v (~c/1000).

Surely a slur distortion, or front might account for that interception (fringe intensity) at the semi_mirror.?)

Edited September 11, 2018 by Capiert

[swansont]

45 minutes ago, Capiert said:

In which case, wasn't that angle decided by the draftsperson (that was) sketching?

 (because in that case a new angle would be needed for every different speed v ?);

The angle depends on the speed. That's not up to the person making the drawing, it's up to the conditions of the example.

Quote

 or else if the upper mirror is long enough to be hit at 90 degrees,

 & reflection occurs in a wavefront,

 wouldn't the diagonal beam, where it's twice Michealson's (drafted) reflected angle,

 hit the semi_mirror in perfect sync with the horizontal path? i.e. identical delays for both (vertical & horizontal) paths?

 

(I mean the (reflected) angle in question is <1 degree (difference, wrt  incidence at 90 degrees) for the intended speed v (~c/1000).

Surely a slur distortion, or front might account for that interception (fringe intensity) at the semi_mirror.?)

We're trying to capture what is actually happening, instead of trying to find a loophole that obscures the physics.

[Capiert]

20 hours ago, swansont said:

The angle depends on the speed. That's not up to the person making the drawing, it's up to the conditions of the example.

We're trying to capture what is actually happening, instead of trying to find a loophole that obscures the physics.

I'm sorry that did not answer my questions, e.g. y/n's.

[swansont]

20 minutes ago, Capiert said:

I'm sorry that did not answer my questions, e.g. y/n's.

I don't really understand your questions, but it seems like you are trying to find a way for your idea to work if you make the mirrors big enough, or the speed small enough, and that's missing the point. The apparatus in the thought experiment can't rely on such factors. The concept has to work for any speed and an arbitrarily small mirror size.

So anything along the lines of "if the mirror is long enough" means you are trying to get around the physics being discussed.  

[Capiert]

On 12 September 2018 at 12:37 PM, swansont said:

I don't really understand your questions,

I'm sorry, I did not make myself clear enough.

I hope you do not make random answers when you answer.

Quote

but it seems like you are trying to find a way for your idea to work if you make the mirrors big

 (=wide)

Quote

enough

 (which they were)

Quote

, or the speed small enough,

'any' speed except c,

 but its true Michelson's sketch is an exageration. Was it v=c/10?

Quote

and that's missing the point.

I thought the point was (for) the M&M experiment of 1888, fig 2. 

Quote

The apparatus in the thought experiment can't rely on such factors.

Yes, so why aren't we (still) talking about the real experiment?

Quote

The concept has to work for any speed

Yes, any reasonable speed.

Quote

and an arbitrarily

Does that mean trial & error? (E.g. decide (=arbitrate) from random values.)

E.g. It's (=the mirror has) got to be wide (=big) enough.?

Quote

small mirror size.

They were 5 cm wide (& made of metal).

Quote

So anything along the lines of "if the mirror is long enough" means you are trying to get around the physics being discussed.  

Quite the contrary, (I'm interested in specific details). If the mirror is made too small, wouldn't a small (& coherent enough) light beam get cutoff?

E.g. Wouldn't a (vertically) 90 degree incident ray (either) partially (or never) hit a too small mirror b that (mirror b) is moving away (to the right) at v?

How can you change 90 degree x,y,z orthogonality, by changing reference systems? E.g. Formula?

Edited September 13, 2018 by Capiert

[swansont]

12 hours ago, Capiert said:

E.g. It's (=the mirror has) got to be wide (=big) enough.?

No. From a conceptual point, the experiment working does not depend on the side of the mirror. That's an experimental detail, because the light beam is  of a certain size, and you want to capture the beam. But you are NOT relying on the mirror size to capture the beam because of the motion. The light remains at the same position in the mirror. It doesn't shift because of the motion.

12 hours ago, Capiert said:

They were 5 cm wide (& made of metal).

Because that's a physical mirror and you have to manufacture it.

12 hours ago, Capiert said:

Quite the contrary, (I'm interested in specific details). If the mirror is made too small, wouldn't a small (& coherent enough) light beam get cutoff?

That's a practical consideration, not one that is affected by the theory.

12 hours ago, Capiert said:

E.g. Wouldn't a (vertically) 90 degree incident ray (either) partially (or never) hit a too small mirror b that (mirror b) is moving away (to the right) at v?

How can you change 90 degree x,y,z orthogonality, by changing reference systems? E.g. Formula?

If it hits it while stationary, it will hit it when it's moving. The beam does not shift relative to the mirror.

[Janus]

18 hours ago, Capiert said:

They were 5 cm wide (& made of metal).

Quite the contrary, (I'm interested in specific details). If the mirror is made too small, wouldn't a small (& coherent enough) light beam get cutoff?

E.g. Wouldn't a (vertically) 90 degree incident ray (either) partially (or never) hit a too small mirror b that (mirror b) is moving away (to the right) at v?

How can you change 90 degree x,y,z orthogonality, by changing reference systems? E.g. Formula?

if the ether the M&M experiment was trying to measure the Earth's drift with respect to existed, then, yes there would have been a slight offset in the perpendicular light. (with the 11 m path length, and a "drift velocity of 30 km/sec, it would have only amounted to a bit over 1 mm.)  This would not however have effected the time it took for the perpendicular light to travel it path, and it was the timing or phase difference the experiment was designed to measure.

However, the M&M experiment, and those following it, failed to detect said ether.  No ether, no drift.   Thus "moving" or "stationary" the beam travels at a 90 degree angle to the apparatus.

If I'm playing catch with someone across the width of  a railway car traveling at 60 mph relative to the tracks, I don't have to "lead" him with my toss. I toss the ball directly at him.  Light in a vacuum behaves the same way, as long as the target has no velocity relative to me, I just aim my light at the object.

Edited September 14, 2018 by Janus