Einstein and "spooky actions"

[Reptillian]

You missed an important point:

http://www.religiousforums.com/forum/3333643-post367.html

True if the hand were somehow moving at the speed of light then there is no shadow, but in the setup the hand is moving at normal "everyday" speeds (small compared to light speed. The shadow is the thing that's length contracted since the shadow is moving close to (or supposedly faster than) the speed of light.

 

[zaybu]

True if the hand were somehow moving at the speed of light then there is no shadow, but in the setup the hand is moving at normal "everyday" speeds (small compared to light speed. The shadow is the thing that's length contracted since the shadow is moving close to (or supposedly faster than) the speed of light.

Fine, but now the calculations are different.

Sprinkles said:

For the sake of argument, say your hand is moving at velocity C, and thus it passes over the lamp in time T = W/C. A screen some distance away, with width L > W, is illuminated by the lamp. A shadow passes over the screen due to your hand passing over the lamp. How fast does the shadow move? The shadow's velocity across the screen is V = L/T = L/W x C, which is greater than C.

Now if the speed of the hand is no longer C, but say say v, much smaller than C then:

(1) T = W/v for the hand,
(2) Shadow's velocity V = L/T = (L/W) x v, and this is not greater than the speed of light.

SR comes into play only for v near the speed of light. In that scenario, the hand would contract by sqr ( 1 - v^2/c^2).

 

[Reptillian]

Fine, but now the calculations are different.

Sprinkles said:



Now if the speed of the hand is no longer C, but say say v, much smaller than C then:

(1) T = W/v for the hand,
(2) Shadow's velocity V = L/T = (L/W) x v, and this is not greater than the speed of light.

SR comes into play only for v near the speed of light. In that scenario, the hand would contract by sqr ( 1 - v^2/c^2).

The shadow is what I'm saying is length contracted not the hand. By assumption the shadow will be moving at relativistic speeds...the contention made was that the shadow will move faster than light for large enough screens, hand velocities, and distances.

 

[Mr Spinkles]

As for your calculation: the problem I see is that you've assumed that the shadow will travel across the screen in the same amount of time as the time it took for your hand to move across the source.

No no, I haven't assumed it. I calculated it explicitly. I'll clarify this below.

Time dilation is necessary since the shadow will be moving at relativistic speeds.

No. Lorentz transforms are not necessary because we are considering everything in one reference frame, S, at rest w.r.t. the lamp and the screen. Time dilation is something we would worry about if we wanted to re-do the problem in another reference frame, S'.

You are making this more complicated that it needs to be. (Why switch reference frames? Why draw the screen curved?) I understand why you are worried: usually, mistakes are made in relativity because the problem isn't as simple as we thought it was (e.g. we need to do Lorentz transforms). But this problem really is simple.

Let's go through the logic:

(1) Hand interrupts light beam at (t=0,y=0).
(2) Therefore, no light is registered on the screen at (t=d/c,Y=0).

With me so far?

 

[zaybu]

By assumption the shadow will be moving at relativistic speeds...the contention made was that the shadow will move faster than light for large enough screens, hand velocities, and distances.

Ok, but then your formula is wrong:

V (shadow) = (W/d) v(hand), where d is the distance between your hand and the lamp, w is the distance your hand has travelled across the lamp.

 

[Reptillian]

No no, I haven't assumed it. I calculated it explicitly. I'll clarify this below.

No. Lorentz transforms are not necessary because we are considering everything in one reference frame, S, at rest w.r.t. the lamp and the screen. Time dilation is something we would worry about if we wanted to re-do the problem in another reference frame, S'.

You are making this more complicated that it needs to be. (Why switch reference frames? Why draw the screen curved?) I understand why you are worried: usually, mistakes are made in relativity because the problem isn't as simple as we thought it was (e.g. we need to do Lorentz transforms). But this problem really is simple.

Let's go through the logic:

(1) Hand interrupts light beam at (t=0,y=0).
(2) Therefore, no light is registered on the screen at (t=d/c,Y=0).

With me so far?

Yeah, please continue....though don't you mean that the shadow isn't first noticed until time t=2d/c?

 

[Mr Spinkles]

Yeah, please continue....though don't you mean that the shadow isn't first noticed until time t=2d/c?

Why would it be 2d/c and not d/c? Recall the definition of d: "Additionally, imagine the light rays hitting the top and bottom of the screen forming an isosceles triangle, with sides of length "d" with the screen of length L as the base." **(Okay, a triangle with the tip cut off.)

Continuing:

(3) The hand also interrupts the light beam at y=W. When does the hand get there? At t=W/V.

(4) This sends a shadow to position Y=L on the screen. When does the shadow get there? It takes a time d/c after the hand interrupted the beam. Since the hand interrupted the beam at t=W/V, the shadow arrives at t = (W/V) + d/c.

The logic here is almost identical to what I said in my previous post. Still with me?

 

[Reptillian]

Why would it be 2d/c and not d/c? Recall the definition of d: "Additionally, imagine the light rays hitting the top and bottom of the screen forming an isosceles triangle, with sides of length "d" with the screen of length L as the base."

Here I'm struggling. Who's the observer here? The guy waving his hand? (t=2d/c) A dude sitting on the screen? (t=d/c) Am I taking some data from the guy waving his hand and some data from the guy on the screen to calculate this superluminal propagation?

the 2 in the 2d/c comes from the fact that the light has to travel from the screen back to the observer.

 

[Mr Spinkles]

Here I'm struggling. Who's the observer here? The guy waving his hand? (t=2d/c) A dude sitting on the screen? (t=d/c) Am I taking some data from the guy waving his hand and some data from the guy on the screen to calculate this superluminal propagation?

Oh okay, yes, I see what's confusing you.

The "observer" can be anyone at rest w.r.t. reference frame S. We have to go back to the definition of a reference frame S, which involves an array of observers and clocks at rest w.r.t. each other, at every location in S.

It may help you to imagine that you, personally, are located at the hand initial position. Your colleagues are located at the other 3 positions (hand final position; shadow initial position; shadow final position). You and your colleagues have all paced out the distances and synchronized your watches beforehand. If a hand (or shadow) arrives at someone's location, they record the time and immediately send a light signal to the other 3 colleagues. Therefore, you personally will observe the hand at its initial position, and sometime later, you will receive light signals from your colleagues reporting their observations. From those signals, you personally can reconstruct all the events.

So for example, you receive a light signal from your colleague reporting the observation of the shadow initial position at time 2d/c. But you know (because you are a competent observer) that meant the shadow actually arrived at the screen at time d/c. In relativity, we aren't talking about what people in frame S "see", but rather, what actually happened by the agreement of all competent observers in frame S. So we don't worry about what you vs. your colleagues personally witnessed. Rather, we consider what everyone in frame S agrees actually happened: the shadow actually arrived at time d/c at the screen, even though you personally witnessed only the light signal from your colleague at time 2d/c to report this event.

Does that make sense?

 

[Reptillian]

Oh okay, yes, I see what's confusing you.

The "observer" can be anyone at rest w.r.t. reference frame S. We have to go back to the definition of a reference frame S, which involves an array of observers and clocks at rest w.r.t. each other, at every location in S.

It may help you to imagine that you, personally, are located at the hand initial position. Your colleagues are located at the other 3 positions (hand final position; shadow initial position; shadow final position). You and your colleagues have all paced out the distances and synchronized your watches beforehand. If a hand (or shadow) arrives at someone's location, they record the time and immediately send a light signal to the other 3 colleagues. Therefore, you personally will observe the hand at its initial position, and sometime later, you will receive light signals from your colleagues reporting their observations. From those signals, you personally can reconstruct all the events.

So for example, you receive a light signal from your colleague reporting the observation of the shadow initial position at time 2d/c. But you know (because you are a competent observer) that meant the shadow actually arrived at the screen at time d/c. In relativity, we aren't talking about what people in frame S "see", but rather, what actually happened by the agreement of all competent observers in frame S. So we don't worry about what you vs. your colleagues personally witnessed. Rather, we consider what everyone in frame S agrees actually happened: the shadow actually arrived at time d/c at the screen, even though you personally witnessed only the light signal from your colleague at time 2d/c to report this event.

Does that make sense?

Ok I think I get what you're saying...here's what I'm saying, suppose I have a laser pointer and a round spherical screen centered at me with a radius of one light year. I point my laser pointer at theta=0 radians and wait to see the dot. 2 years later, there's the dot. Now I move my arm up to theta=pi/2 radians in a time of 5 seconds. Two years later, there's the dot. So the laser pointer dot moved a little over one and a half light years in about 4 years time.

 

[PolyHedral]

Why draw the screen curved?

Doing the whole problem with a point source projected onto a curved screen makes the calculation easier, because all the points of the screen are equally close to the source. Doing it with a planar light source and a flat screen doesn't produce a shadow with the require properties.

 

[Reptillian]

Doing the whole problem with a point source projected onto a curved screen makes the calculation easier, because all the points of the screen are equally close to the source. Doing it with a planar light source and a flat screen doesn't produce a shadow with the require properties.

Exactly Spherical polar coordinates rock! lol

 

[Mr Spinkles]

Ok I think I get what you're saying...here's what I'm saying, suppose I have a laser pointer and a round spherical screen centered at me with a radius of one light year. I point my laser pointer at theta=0 radians and wait to see the dot. 2 years later, there's the dot. Now I move my arm up to theta=pi/2 radians in a time of 5 seconds. Two years later, there's the dot. So the laser pointer dot moved a little over one and a half light years in about 4 years time.

Yes but for the first (2 years - 5 seconds) after you move your hand, the dot does not move at all. Then the dot moves the entire distance in 5 seconds.

Superficially this seems to violate special relativity, but it actually doesn't. It's understandable if this seems confusing. I'm happy to discuss it with you.

 

[Mr Spinkles]

Doing the whole problem with a point source projected onto a curved screen makes the calculation easier, because all the points of the screen are equally close to the source. Doing it with a planar light source and a flat screen doesn't produce a shadow with the require properties.

It's not a planar light source. Just the motion of the hand, and the screen, are planar. With a curved screen, the initial and final times and positions of the shadow are the same as with a flat screen. The only difference is that with a curved screen, the shadow traveled a farther total distance along the screen. This can only increase the velocity of the shadow, and thus it does not affect our conclusions.

 

[Reptillian]

Yes but for the first (2 years - 5 seconds) after you move your hand, the dot does not move at all. Then the dot moves the entire distance in 5 seconds.

Superficially this seems to violate special relativity, but it actually doesn't. It's understandable if this seems confusing. I'm happy to discuss it with you.

Sure I was using average velocity, but are you certain it moves the entire distance in 5 seconds? We'll need to use calculus and special relativity to decide. I'd suggest that the dot 'lags' and that the one and a half light years is indeed traversed over the course of 2 years. Well under the speed of light.

 

[idav]

Yes but for the first (2 years - 5 seconds) after you move your hand, the dot does not move at all. Then the dot moves the entire distance in 5 seconds.

Superficially this seems to violate special relativity, but it actually doesn't. It's understandable if this seems confusing. I'm happy to discuss it with you.

The dot is moving but the light is still going the normal speed. A photon from the laser pointer will hit point A while a completely different photon will hit point B both of which going normal speed.

 

[Mr Spinkles]

Sure I was using average velocity, but are you certain it moves the entire distance in 5 seconds? We'll need to use calculus and special relativity to decide. I'd suggest that the dot 'lags' and that the one and a half light years is indeed traversed over the course of 2 years. Well under the speed of light.

No no ... ! Hold your horses. You are making things way more complicated than necessary. Trust me.

Break the problem into tiny, manageable pieces. You turn on your laser pointer and aim it at a screen 1 light year away. For 2 years, you continue to see no spot. Then, 2 years later, you see the spot. (**Note: it's also correct to say that 1 year later, the laser dot hits the screen. But we'll stick to your convention of adopting the POV of the guy pointing the laser.)

You turn off your laser pointer. For 2 years, the spot remains. Then after 2 year, the spot disappears.

You move your laser pointer. For 2 years, the spot does not move. Then after 2 years, it moves.

With me so far?

 

[Mr Spinkles]

The dot is moving but the light is still going the normal speed. A photon from the laser pointer will hit point A while a completely different photon will hit point B both of which going normal speed.

Exactly, well said.

 

[PolyHedral]

It's not a planar light source. Just the motion of the hand, and the screen, are planar. With a curved screen, the initial and final times and positions of the shadow are the same as with a flat screen. The only difference is that with a curved screen, the shadow traveled a farther total distance along the screen. This can only increase the velocity of the shadow, and thus it does not affect our conclusions.

Doing it with a point source and a planar screen means that the ends of the screen are further away from the source than the middle, which makes the calculations more difficult.

 

[Mr Spinkles]

Doing it with a point source and a planar screen means that the ends of the screen are further away from the source than the middle, which makes the calculations more difficult.

But we only need to calculate the position and time of the shadow at two points: at the bottom, and top of the screen. If the average speed between these points is greater than c, then the shadow must have traveled faster than c, if not during the entire journey, then at least during part of the journey. In other words, the exact motion of the shadow in between the initial and final positions does not affect our conclusions.