Author Topic: Woodward's effect  (Read 285903 times)

Offline 93143

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Re: Woodward's effect
« Reply #660 on: 02/24/2016 10:34 PM »
The photon only ever transfers a small amount of its momentum to the solar sail due to its incredibly minuscule mass with respect to the sail.  This is the main reason photon propulsion is so inefficient.

No.  The photon transfers a very small amount of its energy to the solar sail due to its incredibly minuscule mass with respect to the sail.  Its momentum is, to first order, completely reversed (assuming a reflective sail) and thus it transfers twice its original momentum to the spacecraft.  That is as good as you can get.  Chasing the total amount of energy is a fool's errand; kinetic energy and collision mechanics simply don't work that way.

If you manage to get the solar sail to absorb 100% of the photon's energy, you've got a perfectly plastic collision and your momentum transfer is halved.  Virtually all of the absorbed energy is lost as heat.
« Last Edit: 02/24/2016 10:58 PM by 93143 »

Offline dustinthewind

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Re: Woodward's effect
« Reply #661 on: 02/25/2016 01:00 AM »
The photon only ever transfers a small amount of its momentum to the solar sail due to its incredibly minuscule mass with respect to the sail.  This is the main reason photon propulsion is so inefficient.

No.  The photon transfers a very small amount of its energy to the solar sail due to its incredibly minuscule mass with respect to the sail.  you say 'Tomato', I say 'Tomato'... Its momentum is, to first order, completely reversed (assuming a reflective sail) and thus it transfers twice its original momentum to the spacecraft.  That is as good as you can get.  Chasing the total amount of energy is a fool's errand; kinetic energy and collision mechanics simply don't work that way.  You don't think 2 ships with mirrors and light reflecting between them is a good idea? https://en.wikipedia.org/wiki/Photonic_laser_thruster

If you manage to get the solar sail to absorb 100% of the photon's energy, you've got a perfectly plastic collision and your momentum transfer is halved.  Virtually all of the absorbed energy is lost as heat.
We are again talking about a perfectly elastic collision. http://hyperphysics.phy-astr.gsu.edu/hbase/elacol.html

Offline 93143

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Re: Woodward's effect
« Reply #662 on: 02/26/2016 02:03 AM »
you say 'Tomato', I say 'Tomato'...

Momentum and kinetic energy are not interchangeable.  If you don't understand this, you need to go away and study mechanics until you do.

Please remember that momentum is a vector.  A very small object bouncing elastically off a very large object comes away with nearly the same absolute value of momentum in the centre-of-mass frame, but it's in the opposite direction.  Thus the momentum transferred to the large object is nearly twice what the small object originally had.  Its kinetic energy is of course a scalar, and doesn't change much.  (And of course all of this is out the window if you're using a different reference frame.)

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You don't think 2 ships with mirrors and light reflecting between them is a good idea?

That's a totally different idea; the one ship is using the other ship, not the light itself, as reaction mass.  The laser beam in that scheme is just a means of coupling the two massive objects, and the efficiency gain is entirely a function of photon recycling - they make multiple bounces, magnifying the thrust on both vehicles equally and in opposite directions.  It has nothing to do with "sucking the energy out of the light"; the energy is from the other spacecraft.  You can't reduce the scheme to just one massive object without having to rely on the photons themselves as reaction mass, and if you're relying on light as reaction mass you cannot exceed the efficiency of a photon thruster.

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If you manage to get the solar sail to absorb 100% of the photon's energy, you've got a perfectly plastic collision and your momentum transfer is halved.  Virtually all of the absorbed energy is lost as heat.
We are again talking about a perfectly elastic collision.

I already covered that.  In a perfectly elastic collision between a very small mass and a very large one, the energy transfer is very small.  Plasticizing the collision results in more energy transferred, but it is not bulk kinetic energy - it's heat, or strain energy, or acoustic energy, or something else frame-independent.  The bulk kinetic energy transfer is at a maximum in a perfectly elastic collision, and it's still minuscule in the centre-of-mass frame of reference.
« Last Edit: 02/26/2016 02:10 AM by 93143 »

Offline Paul451

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Re: Woodward's effect
« Reply #663 on: 02/26/2016 02:51 AM »
{Edit: While I was writing this, 93143 said much the same.)

Correct on no net acceleration, however, the cavity has moved a bit having both accelerated and then decelerated.  This is still the first example where the photon recovers its original wavelength which is (blue shifted) with respect to its previous red shift but technically it just recovers its original wavelength.  We are dealing with a perfectly elastic photon so I should say we are dealing with a superconductive cavity.  The frame I am considering is a frame that is independent of the cavity or the photon.  Originally, it was stationary with the cavity.  Sort of a lab frame.  The wall that first welcomed the photon is moving away from that frame when re-emitting the photon so I consider it red-shifted as well as being red-shifted having imparted some of its energy into the cavity.
[...]
mmm, maybe I didn't clarify early on that the stipulation was that the photon was perfectly elastic.  A real photon of course will also generate waste heat [...] For being perfectly elastic all the energy goes into the ballistics and is converted into momentum (ideal assumption).

Sorry for the delayed reply, but reading your comments with fresh eyes made something click. I think I can see the fundamental error.

You've confused a couple of very different concepts. Relativistic red/blue shifting, and energy loss from photon re-transmission. They are not the same phenomenon and cannot be interchanged. But that is what you are doing.

When the reflected photon red-shifts because of the new motion of the reflective-wall, the photon is not losing energy. It appears to have less energy WRT the ship because the frame-of-reference for the ship changes. And so does the rest of the universe behind the ship, not just the photon that pushed the ship, all photons hitting the back of the ship are red-shifted by the same amount.

The reflected photon itself is unchanged (except that its vector of motion is reversed.)

From the point of view of a stationary external observer, the photon is not red-shifted, or blue-shifted. Or changed in any way but direction.

A reflective chamber cannot "increasingly" red-shift a photon. That's simply not a thing that happens. The photon would simply bounce around, unchanged, until it gets absorbed on one of its collision.

And that absorption brings us to...

The second phenomenon is where energy is partially lost. That's where a photon is not reflected. It is absorbed and another photon is re-emitted at a slightly longer wavelength; with the rest of the energy being converted to heat. This emission of a longer (redder) wavelength photon is what you've confused with relativistic red-shifting, but the "red-shifting" from re-emission has nothing to do with red-shift caused by relative velocity.

All photon reflections are "perfectly elastic". But different surfaces have different efficiencies of reflection, depending on how many photons are reflected (unchanged) and how many are absorbed and re-emitted.

Over multiple reflections between a (theoretical) perfectly reflective surface and a (more ordinary) very reflective surface, the energy of that original photon will gradually be transferred to the less perfectly reflective surface. However, absorbing the energy of a photon over multiple reflections inside a chamber is exactly the same as absorbing it once in a single "impact". The chamber does not and cannot increase the available energy or momentum from the photon.

The trick is finding some method of manipulating the effective mass of a photon.

A photon has no mass. It is a force vector exchanging momentum between two objects with mass; it does not have mass itself.

Which is the answer to the objection you probably thought of when reading the top section: "But where does the momentum come from to move a solar sail?" The answer is, "The momentum comes from the object which emitted the photon in the first place." That object moved in the opposite direction when it emitted the photon.

Offline dustinthewind

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Re: Woodward's effect
« Reply #664 on: 02/26/2016 05:02 AM »
you say 'Tomato', I say 'Tomato'...

Momentum and kinetic energy are not interchangeable.  If you don't understand this, you need to go away and study mechanics until you do.

Please remember that momentum is a vector.  A very small object bouncing elastically off a very large object comes away with nearly the same absolute value of momentum in the centre-of-mass frame, but it's in the opposite direction.  Thus the momentum transferred to the large object is nearly twice what the small object originally had.  Its kinetic energy is of course a scalar, and doesn't change much.  (And of course all of this is out the window if you're using a different reference frame.)
Dustinthewind: If the ratio of energy exchanged in a collision is very small for the large object and almost 100% for the photon then the absolute value of the photons momentum is also going to change very little hence my statement, "you say 'Tomato', I say 'Tomato'..."  A change in energy corresponds to a change in absolute momentum.  Momentum is conserved and so is energy. 

The photon only ever transfers a small amount of its momentum to the solar sail due to its incredibly minuscule mass with respect to the sail.  This is the main reason photon propulsion is so inefficient. ...


No.  The photon transfers a very small amount of its energy to the solar sail due to its incredibly minuscule mass with respect to the sail.  Its momentum is, to first order, completely reversed (assuming a reflective sail) and thus it transfers twice its original momentum to the spacecraft.  That is as good as you can get.  Chasing the total amount of energy is a fool's errand; kinetic energy and collision mechanics simply don't work that way.
...

Dustinthewind: It is not as good as you can get.  The photon has a very small relativistic mass so the energy/momentum exchange is almost nothing.  The photon escapes with most of its energy/momentum in a reflection.


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You don't think 2 ships with mirrors and light reflecting between them is a good idea?

That's a totally different idea; the one ship is using the other ship, not the light itself, as reaction mass.  The laser beam in that scheme is just a means of coupling the two massive objects, and the efficiency gain is entirely a function of photon recycling - they make multiple bounces, magnifying the thrust on both vehicles equally and in opposite directions.  It has nothing to do with "sucking the energy out of the light"; the energy is from the other spacecraft.  You can't reduce the scheme to just one massive object without having to rely on the photons themselves as reaction mass, and if you're relying on light as reaction mass you cannot exceed the efficiency of a photon thruster.
Dustinthewind: Lets talk about the one ship using the other ship.  The mirrors have infinite Q and the light is perfectly collimated.  The light is between the 2 mirrors and the ships aren't moving as they are attached.  There is a force but no red-shifting of light.  As soon as the ships are released and they starting moving away from each other, we get red shifting.  We get red-shifting because F.dx = energy and that energy is coming from the light (energy conserved).  Now lets take our ships and push them back together.  They are now working against the force and light should be being blue shifted.  F.dx=E again we are storing energy in the spring.  This form of propulsion is efficient because the photons are not escaping with all their momentum.  It is draining the energy out of the photons efficiently.  They are getting better thrust than a photon rocket by draining the energy out of the photon by elastic collisions and not letting that energy escape, like a solar sail does.  My method also uses elastic collisions to do the same but it's still to be seen if there is any real way to modify the relativistic mass of a photon.

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If you manage to get the solar sail to absorb 100% of the photon's energy, you've got a perfectly plastic collision and your momentum transfer is halved.  Virtually all of the absorbed energy is lost as heat.
We are again talking about a perfectly elastic collision.

I already covered that.  In a perfectly elastic collision between a very small mass and a very large one, the energy transfer is very small.  Plasticizing the collision results in more energy transferred, but it is not bulk kinetic energy - it's heat, or strain energy, or acoustic energy, or something else frame-independent.  The bulk kinetic energy transfer is at a maximum in a perfectly elastic collision, and it's still minuscule in the centre-of-mass frame of reference.
« Last Edit: 02/26/2016 05:23 AM by dustinthewind »

Offline Paul451

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Re: Woodward's effect
« Reply #665 on: 02/26/2016 03:27 PM »
As soon as the ships are released and they starting moving away from each other, we get red shifting.  We get red-shifting because F.dx = energy and that energy is coming from the light (energy conserved).

Incorrect. The energy is coming from the two ships. F = -F

The photons are unchanged by their reflection, with respect to a stationary observer. The photons are not red-shifted, they are not blue-shifted. They are just changing direction.

That's the thing you need to understand. The photons themselves have not been changed, they have not lost energy, only their direction has been changed. And it's the change in direction that is moving the ships (F = -F) not a loss of energy by the photon themselves.

You are confusing the mathematical short-hand of vector addition/subtraction of moving energy, with actual energy. The energy to move the ship is not coming from the energy of the photon, it's coming from the change-of-direction of the energy of the photon. It's a vector change, not a scalar change. You're then jumbling this up with red/blue-shift and changing mass, and getting yourself tied in knots.

A photon has 100 scalar units of energy. It is moving from right to left, so it has 100 vector units of energy to the left.

It reflects off a mirror/sail/wall/ship. The photon is now moving to the right. It still has 100 scalar units of energy, so it now has 100 vector units of energy to the right. It therefore has taken 200 vector units of rightward energy from the mirror/ship; even though its scalar energy hasn't changed. The ship, in order to balance the equation, now has 200 vector units of leftward energy, so if it was originally stationary, it starts moving to the left.

The photon has 100 scalar units of energy. It is moving from left to right, so it has 100 vector units of energy to the right.

The photon has 100 scalar units of energy.

Offline Paul451

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Re: Woodward's effect
« Reply #666 on: 02/26/2016 05:16 PM »
Four scenarios, each adding one effect.

---

A laser is firing a beam of photons at a perfectly reflective, free-flying surface.

Each photon has the same wavelength, which gives each photon 100 completely-arbitrary-units of energy.

A photon strikes the perfectly reflective surface in a perfectly elastic collision. The photon bounces and reverses direction back towards the laser. The free-flying reflective surface is pushed away from the laser (F = -F) and the photon has its direction reversed.

Actually 2F = -2F because it's a reflection.

The energy/momentum to create the photon comes from the energy/momentum of the laser.

The energy/momentum to change the direction of the photon comes from the energy/momentum of the reflective surface.

The photon acts as a force carrier between the two.

So we look at the universe from the frame-of-reference of the reflective surface. Because the reflective surface changes velocity after the collision, it alters its own frame-of-reference. It now sees the reflected photon as being red-shifted, therefore the photon now has only 99 completely-arbitrary units of energy.

Aha!, you say, 1 completely-arbitrary unit of energy has been removed from the photon.

Wrong.

The laser emits another photon at exactly the same wavelength from the laser's POV, and it has 100 completely-arbitrary-units of energy from the laser's POV.

But the reflective surface sees the laser emitting a new photon with just 99 completely arbitrary units of energy, exactly the same amount as the reflected photon. Indeed every photon from that direction has been red-shifted and lost 1 arbitrary-unit of energy, as has that half of the universe. Where has that extra energy gone?

Nowhere. Because the apparent change in energy of the reflected photon did not come from an actual change in energy of the photon, it came from a change in the frame-of-reference of the reflective surface. And you can't compare linearly across a changing frame-of-reference without correcting for that change.

---

So lets use a reflective cavity. The cavity begins at rest WRT a stationary observer. Its left and right walls are perfectly, equally reflective. All photon collisions are perfectly elastic.

Inside the cavity is a random photon, moving to the left. It has 100 completely-arbitrary units of energy, and 1 different-but-also-arbitrary unit of momentum

The photon hits the left wall and is perfectly reflected backwards.

After the first collision, the cavity is now moving to the left WRT to the stationary observer. The photon is now moving towards the right wall.

From the frame-of-reference of the stationary observer, the wavelength of the photon is exactly the same. The photon still has 100 units of energy, and 1 unit of momentum. Only the direction has changed.

The cavity is now moving to the left with 2 units of momentum. The photon is moving to the right with 1 unit of momentum.

(Initial state: 0 + 1 units to the left, ie, 1 unit to the left. Final state: 2 + (-1) units to the left, ie, 1 unit to the left. No change.)

Now we look at the frame-of-reference of the left wall. Before the collision it is stationary. It experiences a burst of acceleration during the collision, and then is stationary again. But after the collision, the photon is red-shifted and has just 99 units of energy. The photon also has slightly less than 1 units of momentum.

Now we look at the frame-of-reference of the right wall. Before the collision it is stationary. It experiences a burst of acceleration during the collision, and then is stationary again. But after the first collision, the photon has been BLUE-shifted and has 101 units of energy and slightly more than 1 unit of momentum.

When the photon strikes the right wall, it has slightly more energy and momentum than it did when it struck the left wall.

The photon hits the right wall and is reflected to the left. From the frame-of-reference of the stationary observer, the cavity is pushed to the right by 2 units of momentum, therefore coming to a rest. The photon is reflected to the left and has 1 unit of momentum.

(Initial state: 2 + (-1) = 1 unit to the left. Final state: 0 + 1 = 1 unit to the left. No change.)

The cavity has moved slightly to the left, correctly accounting for the energy and momentum of the photon.

Over time, the stationary observer sees the cavity moving to the left with 1 unit of momentum, gained from containing the photon. It is as if the cavity was simply a perfectly black wall that absorbed the photon in a single collision. The only difference the reflection back-and-forth within the cavity makes is that it causes the cavity vibrate slightly back-and-forth. The vibration is symmetrical. The cavity does not "amplify" the momentum of the photon in any way, it does not extract more momentum from the photon that a single collision with a black wall.

WRT the left/right walls, their view of the photon has perfectly reversed after the second collision. The photon is now red-shifted according to the frame-of-reference of the right wall, and blue-shifted according to the frame-of-reference of the left wall. And by exactly the same amount as the first collision. When it reflects off the left wall again, the situation reverses perfectly again. The two states are perfectly symmetrical. The photon does not increasingly red-shift, it does not lose energy; from the frame-of-reference of either wall, it merely changes from red-shift to blue-shift after each collision. And between collisions, one wall always sees the photon red-shifted and one wall sees it blue-shifted, then collision, and they reverse. Red and blue, collision, blue and red, collision, red and blue, collision, blue and red; with the amounts always the same.

---

This time we again have a reflective cavity with two walls, left and right. The left wall is still a perfectly reflective surface, all collisions are perfectly elastic. But the right wall is slightly imperfectly reflective, slightly inelastic.

As before, the photon hits the left wall and is perfectly reflected.

As before, from the frame-of-reference of a stationary observer, the photon's energy is unchanged, only its direction.

As before, from the frames-of-reference of the left and right walls, the photon is red and blue shifted by 1 unit of energy, respectively.

As before, the cavity starts moving to the left with 2 units of momentum, the photon to the right with 1 unit.

Now the photon strikes the right wall. It is an imperfect collision. The photon has less energy.

In reality, the photon is absorbed and a new photon is emitted at a slight longer (redder) wavelength. The lost energy is transferred to the vibrational-energy (heat) of the atom that absorbed/re-emitted the photon.

From the frame-of-reference of the stationary observer, the photon has lost energy in addition to changing direction. It went from having 100 units of energy to the right to now having 99 units of energy to the left. And the right wall is 1 unit of energy hotter.

WRT the stationary observer, the cavity has received a net transfer of 1.99 units of momentum to the right. Not 2 units. The cavity has a momentum after collision of 0.01 units of momentum to the left. (Initial state: 2 + (-1) units to the left. Final state: 0.01 + .99 units to the left.)

Therefore unlike the previous example, WRT the stationary observer, the cavity has not come to a rest. It is moving very slowly to the left.

As the photon bounces back and forth inside the cavity, more and more of its energy is absorbed by the right wall. But the cavity eventually moves to the left with 1 unit of momentum. The situation is exactly the same as if the left wall was a mirror and the right wall perfectly black and the energy and momentum of the photon was reflected once and then 100% absorbed. Which is exactly the same as if the cavity was a single black wall. Even though the walls and their collisions are no longer symmetrical, the net outcome is exactly the same.

The interesting and potentially confusing thing here is that the energy is being absorbed by the right wall, but the transferred momentum is to the left.

That's because scalar units are not the same as vector units.

----

So now we invent a new technology, it is able to change the "mass" of the photon, to increase energy, by {shrug} "superconductors". The extra energy given to the photon comes from an electrical source powering the wall, so total energy is conserved.

We create a cavity in which the left wall is made of the new technology and the right wall is merely a perfect mirror. The left wall's magic technology is powered from a battery attached to the cavity.

When a photon bounces off the left wall, it gains energy. In reaction, the left wall (and hence the cavity) move to the left with more than 1 unit of momentum.

But when that photon bounces off the right wall, perfectly, it has that greater-than-1-unit of momentum. Therefore the right wall (and hence the cavity) moves to the right with exactly the same momentum as the left wall moved left.

WRT a stationary observer, the two sides balance out. All that's left is the original 1 unit of momentum of the original photon.

Over time, all the energy in the battery will be transferred to the photon. Instead of 1 unit of momentum, it will have 1001 units of momentum.

And the cavity will still only move to the left by 1 unit of momentum.

Because the extra 1000 scalar units of extra momentum have no net vector units of momentum. Zero.

1 + (+1000) + (-1000) units to the left

= 1 + 0 units to the left

= 1 unit to the left

Because you cannot treat scalar units and vector units the same, unless you are very very careful to remember where the vector is.

This scenario ends up being exactly the same as scenario 2. Except that instead of a symmetrical 1 unit vibration, you have a symmetrical 1001 unit vibration.

If we instead use one magic super-wall and one imperfectly reflective inelastic wall, we end up with scenario 3. Again the cavity ends up moving left with 1 unit of momentum. The only difference is that instead of slowly absorbing the photon's energy and becoming hotter, the right wall will absorb the photon's original energy and energy from the battery, becoming much hotter. It will still only move to the left with 1 unit of momentum.
« Last Edit: 02/26/2016 05:19 PM by Paul451 »

Offline dustinthewind

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Re: Woodward's effect
« Reply #667 on: 02/26/2016 06:36 PM »
As soon as the ships are released and they starting moving away from each other, we get red shifting.  We get red-shifting because F.dx = energy and that energy is coming from the light (energy conserved).

Incorrect. The energy is coming from the two ships. F = -F

The photons are unchanged by their reflection, with respect to a stationary observer. The photons are not red-shifted, they are not blue-shifted. They are just changing direction.

That's the thing you need to understand. The photons themselves have not been changed, they have not lost energy, only their direction has been changed. And it's the change in direction that is moving the ships (F = -F) not a loss of energy by the photon themselves.
...
The energy to move the ship is not coming from the energy of the photon, it's coming from the change-of-direction of the energy of the photon. ...

The photon has 100 scalar units of energy. It is moving from left to right, so it has 100 vector units of energy to the right.

The photon has 100 scalar units of energy.
How then do you rectify that a radar gun measures a cars velocity by the change in wavelength of the emitted light.  https://en.wikipedia.org/wiki/Radar_gun  The gun exist in the stationary frame that the photon was emitted in, and if it is as you claim, it appears the photon when returning, will have no change in wavelength to indicate if the car is moving either away or towards the radar gun. 

Now that I think of it, I am not sure how related the shift in wavelength of a radar gun is compared to the car imparting energy into the photon.  I'm going to have to work through some problems before I can say this.
« Last Edit: 02/28/2016 04:33 AM by dustinthewind »

Offline Paul451

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Re: Woodward's effect
« Reply #668 on: 02/26/2016 07:03 PM »
The photons are unchanged by their reflection, with respect to a stationary observer. The photons are not red-shifted, they are not blue-shifted. They are just changing direction.
How then do you rectify that a radar gun measures a cars velocity by the change in wavelength of the emitted light.  The gun exist in the stationary frame that the photon was emitted in, and if it is as you claim, it appears the photon when returning, will have no change in wavelength to indicate if the car is moving either away or towards the radar gun.

The reflective sail is initially stationary WRT the observer. There is no doppler shift of the photon seen by the observer, but the sail is pushed in the opposite direction.

So where does the energy come from, if not by red/blue shifting the photon?

That's what I was trying to get across.
« Last Edit: 02/26/2016 07:05 PM by Paul451 »

Offline dustinthewind

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Re: Woodward's effect
« Reply #669 on: 02/28/2016 04:29 AM »
I think you are right.  There is a lot more going on than I expected that is shifting the wavelength of light.  There appears to be Doppler shifting due to constant velocity, possibly with energy exchange, and along with relativity.  I think I really need to work through some of the details.  I appreciate all your comments Paul and 93143 and making me think about some of these issues. 
« Last Edit: 02/29/2016 02:46 AM by dustinthewind »

Offline dustinthewind

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Re: Woodward's effect
« Reply #670 on: 02/29/2016 02:20 AM »
Here is a classical derivation of the Doppler shift of a reflected photon that is pushing on a mirror.  All I used was conservation of Momentum, h*f/c = photon momentum, photon energy = h*f and then Conservation of energy.  The program used is wxMaxima.  I appear to get a classical shift in frequency that is twice that of the Doppler effect listed here: http://imagine.gsfc.nasa.gov/features/yba/M31_velocity/spectrum/doppler_more.html .  See for instance: https://en.wikipedia.org/wiki/Radar_gun I believe it may be because I substituted for the change in velocity of the mirror a photon that is reflected.  A reflected photon imparts twice the kick of an emitted photon.  Its curious that this classical prediction appears to give a limit of speed c.  Was it the energy/momentum I used of the photon?

This is in regards to my suggesting that when the photon hit one side of the cavity (accelerating it), that it was red-shifted, having given some of its energy to the cavity.  I'm still not sure it's possible to modify the relativistic mass of light, conserving momentum.  If it were possible we could possibly pull off a Woodward effect using light.  One thing bothers me about the cavity I suggested is that while it might conserve the absolute value of momentum it appears to re-direct momentum.  On the other hand, I suspect this is what any Woodward effect device would do.  Take for instance the idea of energy moving between a capacitor-inductor circuit with pezio-electric actuator that re-directs the expanding/contracting momentum (change in mass via energy) into a single direction.  Is it possible momentum is conserved in absolute but not necessarily direction?
« Last Edit: 10/26/2016 02:27 PM by dustinthewind »

Offline ppnl

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Re: Woodward's effect
« Reply #671 on: 02/29/2016 04:31 AM »
Is it possible momentum is conserved in absolute but not necessarily direction?

No, momentum is a vector. It inherently has both magnitude and direction. It is not like kinetic energy which is only a scalar.

But even though Kinetic energy is a scalar it is frame dependent which means its value depends on who is looking.

Offline dustinthewind

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Re: Woodward's effect
« Reply #672 on: 02/29/2016 04:53 AM »
Is it possible momentum is conserved in absolute but not necessarily direction?

No, momentum is a vector. It inherently has both magnitude and direction. It is not like kinetic energy which is only a scalar.

But even though Kinetic energy is a scalar it is frame dependent which means its value depends on who is looking.

Right, that is what I would think too and I can't think of any natural systems that would violate it.  If the Woodward concept works then it almost implies it may be possible to redirect momentum.  Unless I am mistaken but at the moment, it sure seems that way.  I mean the simple device they have at the top of this link seems to imply a re-direction of momentum: https://en.wikipedia.org/wiki/Woodward_effect .   

I have noticed the cavity I proposed that modifies the mass of light also re-directs the momentum.  I think it is a property that is at the heart of the Mach-effect/Woodward effect concept. 
« Last Edit: 02/29/2016 05:10 AM by dustinthewind »

Offline ppnl

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Re: Woodward's effect
« Reply #673 on: 02/29/2016 07:38 AM »
Is it possible momentum is conserved in absolute but not necessarily direction?

No, momentum is a vector. It inherently has both magnitude and direction. It is not like kinetic energy which is only a scalar.

But even though Kinetic energy is a scalar it is frame dependent which means its value depends on who is looking.

Right, that is what I would think too and I can't think of any natural systems that would violate it.

No system can violate it. And that fact isn't a physical fact so much as a mathematical fact. You would have to rewrite physics.

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If the Woodward concept works then it almost implies it may be possible to redirect momentum.


Another way to say "redirect momentum" is "violate conservation of momentum." As long as you understand that then explore the possibilities all you want.

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Unless I am mistaken but at the moment, it sure seems that way.  I mean the simple device they have at the top of this link seems to imply a re-direction of momentum: https://en.wikipedia.org/wiki/Woodward_effect .

I really can't make much sense of anything on that page.
   
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I have noticed the cavity I proposed that modifies the mass of light also re-directs the momentum.  I think it is a property that is at the heart of the Mach-effect/Woodward effect concept.

As I said above you cannot change the mass of a photon. If you could it would be an immediate and obvious violation of conservation of momentum. You may as well be invoking invisible blue fairies.

Or you could try this:

http://star.psy.ohio-state.edu/coglab/Miracle.html

Offline dustinthewind

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Re: Woodward's effect
« Reply #674 on: 03/01/2016 04:58 AM »
Is it possible momentum is conserved in absolute but not necessarily direction?

No, momentum is a vector. It inherently has both magnitude and direction. It is not like kinetic energy which is only a scalar.

But even though Kinetic energy is a scalar it is frame dependent which means its value depends on who is looking.

Right, that is what I would think too and I can't think of any natural systems that would violate it.

No system can violate it. And that fact isn't a physical fact so much as a mathematical fact. You would have to rewrite physics.

Quote
If the Woodward concept works then it almost implies it may be possible to redirect momentum.


Another way to say "redirect momentum" is "violate conservation of momentum." As long as you understand that then explore the possibilities all you want.

Dustinthewind: By re-direct momentum I am saying not to change the absolute value of the momentum but to change the direction which in effect is a violation but re-direct implies not changing the value.

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Unless I am mistaken but at the moment, it sure seems that way.  I mean the simple device they have at the top of this link seems to imply a re-direction of momentum: https://en.wikipedia.org/wiki/Woodward_effect .

I really can't make much sense of anything on that page.
Dustinthewind: The device at the top, right hand side, that changes mass and the pulls it self together like an accordion.  It then transfers energy to the other side, changing its mass, and then expands each time pushing in one direction.  How does a device that vibrates and has zero momentum, (if it doesn't transfer mass), push it self in such a manner with out propellant?  It implies a re-direction of momentum during the expanding/contracting phase.
   
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I have noticed the cavity I proposed that modifies the mass of light also re-directs the momentum.  I think it is a property that is at the heart of the Mach-effect/Woodward effect concept.

As I said above you cannot change the mass of a photon. If you could it would be an immediate and obvious violation of conservation of momentum. You may as well be invoking invisible blue fairies.

Or you could try this:

http://star.psy.ohio-state.edu/coglab/Miracle.html

Dustinthewind: If changing the mass of a photon/object could give us such an ability as to re-direct momentum eliminating the need for a propellant, I think it would at least be worth considering if it were a possibility, considering the rewards.  Some things in the past that were considered blue fairies/impossible were later found to be possible so why not keep an open mind?  I read some where, superconductors make photons massive objects, and when light enters a dielectric it slows down.  What if light slows down in a dielectric because its effective relativistic mass increases and momentum is conserved? 
URL: Mass change of dielectric media induced by propagation of electromagnetic waves by C.Z. Tan
« Last Edit: 03/01/2016 05:19 AM by dustinthewind »

Offline ChrisWilson68

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Re: Woodward's effect
« Reply #675 on: 03/01/2016 05:57 AM »
Is it possible momentum is conserved in absolute but not necessarily direction?

No, momentum is a vector. It inherently has both magnitude and direction. It is not like kinetic energy which is only a scalar.

But even though Kinetic energy is a scalar it is frame dependent which means its value depends on who is looking.

Right, that is what I would think too and I can't think of any natural systems that would violate it.

No system can violate it. And that fact isn't a physical fact so much as a mathematical fact. You would have to rewrite physics.

Quote
If the Woodward concept works then it almost implies it may be possible to redirect momentum.

Another way to say "redirect momentum" is "violate conservation of momentum." As long as you understand that then explore the possibilities all you want.

By re-direct momentum I am saying not to change the absolute value of the momentum but to change the direction which in effect is a violation but re-direct implies not changing the value.

Everyone understands what you're saying, and everyone has been trying to tell you that changing the direction is a violation of conservation of momentum.  Keeping the magnitude the same but changing the direction is a violation of conservation of momentum.  It's that simple.

Also, please learn to follow the standard quoting conventions for this forum, like 99% of other posters do.  It's more difficult for readers when not everyone follows the conventions.

Offline 93143

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Re: Woodward's effect
« Reply #676 on: 03/01/2016 06:05 AM »
I mean the simple device they have at the top of this link seems to imply a re-direction of momentum: https://en.wikipedia.org/wiki/Woodward_effect .

No, you haven't understood it properly.  It's a device that changes the mass of its components in order to produce a (vector) momentum imbalance.  And even that would be impossible in isolation - Woodward is proposing that the mass changes are due to a Wheeler-Feynman-type gravity-wave interaction with the rest of the universe, which is where the momentum and energy conservation would come from.  Aside from the usual complications re: cosmological redshift and so on, the whole system should make sense in Newtonian terms if the principle of operation is valid.

Please consider carefully the fact that momentum is frame-dependent.  If you understand this, you will see that scalar conservation of momentum doesn't even make sense.
« Last Edit: 03/01/2016 06:20 AM by 93143 »

Offline ppnl

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Re: Woodward's effect
« Reply #677 on: 03/01/2016 06:23 AM »

Dustinthewind,

No you cannot "redirect" the momentum. That is a momentum conservation violation and leads directly to an energy conservation violation. Study some vector math and understand that it is the vector that must be conserved not the absolute value of its magnitude.

No, you cannot change the mass of anything. If you could that would be a momentum conservation violation and an energy conservation violation. If you could do this then an reactionless drive would be the very least of your powers. You could call on infinite energy to do anything you wanted.

Keeping an open mind is good but what you are proposing contradicts itself. You say you don't intend to violate conservation of momentum and then you talk about redirecting momentum.

You. Can't. Do. That.

The problem isn't that you are violating some supposed law of physics. You are but that is the least of your problems. Mathematically momentum is a vector quantity. Mathematically that vector quantity is what is conserved. Mathematically you cannot "redirect" that vector quantity and also conserve it. You cannot talk about "redirecting" the momentum vector while conserving momentum without abandoning all math, logic and reason.


Offline dustinthewind

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Re: Woodward's effect
« Reply #678 on: 03/01/2016 06:44 AM »
I don't think anyone is telling me anything that I already don't know.  That momentum is a vector and has direction and to conserve it you need to keep it as a vector. 

All I am saying is if a projectile with momentum gets trapped in an osculating system and hits one wall giving up 20% of is energy but the impact with the other wall only gives back 10% of its energy that the projectile will end up giving its energy to the composite walls in a way that drains its energy.  This transferring of energy to the composite walls is completely determined by the ratio of energy given upon impact per wall and completely disregards the original momentum of the object that gets trapped.  If its offensive to consider such, i'll just have to keep it to myself. 
« Last Edit: 03/01/2016 07:00 AM by dustinthewind »

Offline Paul451

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Re: Woodward's effect
« Reply #679 on: 03/01/2016 04:53 PM »
All I am saying is if a projectile with momentum gets trapped in an osculating system and hits one wall giving up 20% of [its] energy but the impact with the other wall only gives back 10% of its energy that the projectile will end up giving its energy to the composite walls in a way that drains its energy.

You seem to be confusing a non-vector form of energy with a vector form.

If the energy of the photon is transferred to an object, the object gets hotter, not faster.

You can convert between vector/scalar energy, but there needs to be a second system in play to balance the force, otherwise you are violating CoM. You have to pay for the conversion by adding negative (in a vector sense) momentum somewhere else.