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#1460 by meberbs on 10 Sep, 2019 19:18
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I would be careful and study the role of higher derivatives. I saw in textbooks that the exact formulas (in general relativity?) Are written as a polynomial.
GR has many complications including that there are few exact solutions. It is not however significant on the scale currently being discussed.I’m afraid the engineer’s intuition, suggests the possible important role of the “jerk”. For example, a jerk is used in the concept of laboratory generation of gravitational waves.
There is no such thing as laboratory generation of gravitational waves. The only gravitational waves that can be measured are from things like black hole mergers. You are just bringing up completely irrelevant things anyway here.The second / third derivative may produce a small effect, but in the resonator any force effect is automatically multiplied by the quality factor.
No, not necessarily true. Most of the time, it is more straightforward to write things in terms of the total energy in the cavity, in which case the Q factor is built in. Also, for some transient effects, the Q factor is actually irrelevant entirely. -
#1461 by Alex_O on 10 Sep, 2019 19:31
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The second stage is to prove that, taking into account a certain relativistic thing, for example, the small bottom in Emdrive looks like it is absolutely motionless at that moment in time when the Big bottom of the resonator has already begun to accelerate and the Doppler effect has ALREADY appeared on it. This will immediately show the difference between the ACTUAL pressure on the end walls versus the calculated one according to Maxwell.
What does absolutely motionless mean? Well, it’s somehow connected, that the speed of light is independent of the movement of the box with photons, I can’t explain it, intuition works.
Fixed - this probably means that the time intervals are multiples of fettoseconds, taking into account the time for photon reflection (this is re-emission, right?) And taking into account the time of flight of the photon through the resonator.
Type you can find such a moment in time - when on the big bottom - the Doppler effect has already happened, but on the small - did not have time. And when he succeeds, a photon with a different wavelength / energy will already fly in, slowly reflect, catch the buzz from the inverse Doppler effect on the small bottom, and fly back.
And, it seems, I understood - that’s what a fixed bottom is, in an already really accelerated resonator. This is when the expected The Doppler effect on it has not yet had time to happen, since the speed of light is finite! -
#1462 by Alex_O on 10 Sep, 2019 19:40
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No no has. One hundred years ago, Lebedev measured the pressure of light, and these were very measurable values. Any, even the weakest effect (micronewtons), and the photon pressure in the laboratory is whole micronewtons (large value!) Will be amplified in a high-Q cavity.I would be careful and study the role of higher derivatives. I saw in textbooks that the exact formulas (in general relativity?) Are written as a polynomial.
GR has many complications including that there are few exact solutions. It is not however significant on the scale currently being discussed. -
#1463 by Alex_O on 10 Sep, 2019 19:58
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Ok, let's go back to 1 point. Antenna on a cart, fixed wall. Wall pressure is easy to calculate. Taking into account the speed of the trolley is also simple. We will get a theoretical calculation and compare it with experiment, with acceleration, with a jerk of acceleration or in uniform motion. Let us immediately recall the Sagnac effect (I don’t know why, but to know that even weak derivatives / acceleration are well caught by laser gyroscopes, from the point of view of laboratory technology.)I would be careful and study the role of higher derivatives. I saw in textbooks that the exact formulas (in general relativity?) Are written as a polynomial.
No, not necessarily true. Most of the time, it is more straightforward to write things in terms of the total energy in the cavity, in which case the Q factor is built in. Also, for some transient effects, the Q factor is actually irrelevant entirely. -
#1464 by meberbs on 10 Sep, 2019 20:15
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Radiation pressure of light and GR are separate topics. GR effects are negligible for the experiments under discussion. Also microNewtons is a very small value and difficult to measure to begin with.
No no has. One hundred years ago, Lebedev measured the pressure of light, and these were very measurable values. Any, even the weakest effect (micronewtons), and the photon pressure in the laboratory is whole micronewtons (large value!) Will be amplified in a high-Q cavity.I would be careful and study the role of higher derivatives. I saw in textbooks that the exact formulas (in general relativity?) Are written as a polynomial.
GR has many complications including that there are few exact solutions. It is not however significant on the scale currently being discussed.
The Sagnac effect involves the entire device, emitter, detector, and mirrors rotating. It measures that different parts of the device are moving in different directions due to the nature of rotations. It is not in any way applicable to the experiment you described.
Ok, let's go back to 1 point. Antenna on a cart, fixed wall. Wall pressure is easy to calculate. Taking into account the speed of the trolley is also simple. We will get a theoretical calculation and compare it with experiment, with acceleration, with a jerk of acceleration or in uniform motion. Let us immediately recall the Sagnac effect (I don’t know why, but to know that even weak derivatives / acceleration are well caught by laser gyroscopes, from the point of view of laboratory technology.)I would be careful and study the role of higher derivatives. I saw in textbooks that the exact formulas (in general relativity?) Are written as a polynomial.
No, not necessarily true. Most of the time, it is more straightforward to write things in terms of the total energy in the cavity, in which case the Q factor is built in. Also, for some transient effects, the Q factor is actually irrelevant entirely.
For your experiment, you would simply get a Doppler shift proportional to the relative velocity of the screen and the antenna at any given time. -
#1465 by Alex_O on 10 Sep, 2019 20:32
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OK, will the pressure on the fixed screen depend on the Doppler shift of the wave frequency? agree?
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#1466 by dustinthewind on 10 Sep, 2019 21:08
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Ok, let's go back to 1 point. Antenna on a cart, fixed wall. Wall pressure is easy to calculate. Taking into account the speed of the trolley is also simple. We will get a theoretical calculation and compare it with experiment, with acceleration, with a jerk of acceleration or in uniform motion. Let us immediately recall the Sagnac effect (I don’t know why, but to know that even weak derivatives / acceleration are well caught by laser gyroscopes, from the point of view of laboratory technology.)I would be careful and study the role of higher derivatives. I saw in textbooks that the exact formulas (in general relativity?) Are written as a polynomial.
No, not necessarily true. Most of the time, it is more straightforward to write things in terms of the total energy in the cavity, in which case the Q factor is built in. Also, for some transient effects, the Q factor is actually irrelevant entirely.
I think what TT was talking about was Doppler shifting but not in the sense of the Doppler shift but rather more like the Compton effect where light loses energy to accelerating an electron. because of the claim that radiation is accelerating the cavity. Problem is how much pressure do you even need to break the devices static friction before it even begins to accelerate. If it never actually accelerates with just a pulse of radiation via breaking static friction, I don't think one could even claim accelerating the cavity has an effect like the Compton effect. The Compton effect would also be much smaller because of the immense size of the cavity compared to that of an electron. Unless the photons are accelerating something else like electrons in the cavity. Then the question is why are photons losing energy to electrons in the cavity and how would that provide a propulsive effect. -
#1467 by meberbs on 10 Sep, 2019 22:00
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OK, will the pressure on the fixed screen depend on the Doppler shift of the wave frequency? agree?
Yes. -
#1468 by TheTraveller on 11 Sep, 2019 03:30
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Even if you had a magic device to work around this, you still couldn't change the fact that you wouldn't be measuring the net momentum inside the cavity, since photons are travelling in different directions.
During the 5 x TC cavity stored energy decay time, sidewall and coupler losses will not cause photon wavelength change. Just reducing cavity stored energy.
We are not trying to measure directional momentum change. Only measuring net lengthening of photon wavelengths vs injected wavelength. From the wavelength difference, the amount of photon energy & momentum that was transferred to cavity energy & momentum gain can be calculated & compared to that gained by the accelerating cavity.
Maybe think about how net photon wavelength could lengthen inside an accelerating cavity and then NOT lengthen inside a cavity that is not accelerating? -
#1469 by meberbs on 11 Sep, 2019 03:46
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This statement is a complete non-sequiter, it has nothing to do with anything I said.Even if you had a magic device to work around this, you still couldn't change the fact that you wouldn't be measuring the net momentum inside the cavity, since photons are travelling in different directions.
During the 5 x TC cavity stored energy decay time, sidewall and coupler losses will not cause photon wavelength change. Just reducing cavity stored energy.We are not trying to measure directional momentum change. Only measuring net lengthening of photon wavelengths vs injected wavelength. From the wavelength difference, the amount of photon energy & momentum that was transferred to cavity energy & momentum gain can be calculated & compared to that gained by the accelerating cavity.
Momentum is a vector quantity. That means if you don't know its direction, you have no way of comparing it to anything. With half of the momentum of the photons in one direction, and half in the other, the total momentum would be zero, and no amount of measuring frequency would tell you that.Maybe think about how net photon wavelength could lengthen inside an accelerating cavity and then NOT lengthen inside a cavity that is not accelerating?
Maybe you should actually read the post that I have linked you to repeatedly, and answer the simple questions I asked you. At this point it is extremely rude of you to keep using the quote button on my posts, but refuse to read what I am writing, or actually engage in a conversation. You are just repeating the same nonsensical statements that I already addressed. -
#1470 by TheTraveller on 11 Sep, 2019 09:43
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Momentum is a vector quantity. That means if you don't know its direction, you have no way of comparing it to anything. With half of the momentum of the photons in one direction, and half in the other, the total momentum would be zero, and no amount of measuring frequency would tell you that.
Will try to explain this one more time.
Do understand photon momentum has a vector.
This discussion has nothing to do with that.
Do you also understand the amount of a photons momentum is related to its wavelength? Lower momentum = longer wavelength?
To measure photon momentum loss.
Inject photons into the cavity for 5 x TC seconds or a shorter period.
If the photons do transfer momentum to the accelerating cavity, their net wavelength will be increasing, indicating that overall the trapped cavity photons momentum has decreased from that present during injection.
Trying to measure increasing wavelength during photon injection is really not doable, because as you stated they mix. However once injection is completed & as the photons naturally decay, it is possible to measure their wavelength increase or not.
Will admit that doing this will not indicate the direction of momentum transfer. Would suggest the only way for the photons to lose some momentum is via transfer to the accelerating cavity. Especially when the measured photon momentum loss, via net wavelength increase, closely matches the momentum gain of the accelerating cavity. -
#1471 by meberbs on 11 Sep, 2019 13:56
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This discussion has everything to do with that, because you ignoring the direction of momentum or flipping it around randomly is a common reason for your incorrect statements. You cannot show conservation of momentum (the goal you stated) without accounting for direction. If you think otherwise, you do not actually understand anything about momentum and need to go take an introductory physics class.Momentum is a vector quantity. That means if you don't know its direction, you have no way of comparing it to anything. With half of the momentum of the photons in one direction, and half in the other, the total momentum would be zero, and no amount of measuring frequency would tell you that.
Will try to explain this one more time.
Do understand photon momentum has a vector.
This discussion has nothing to do with that.Do you also understand the amount of a photons momentum is related to its wavelength? Lower momentum = longer wavelength?
That is only true when you have a single photon, or all photons are travelling in the same direction, which is not the case here.Will admit that doing this will not indicate the direction of momentum transfer.
It won't tell you total momentum either, because photons will be travelling in opposite directions. You will also be measuring them relative to the moving cavity rather than the lab frame, as you admitted, so that is another level of measuring the wrong thing.Would suggest the only way for the photons to lose some momentum is via transfer to the accelerating cavity.
Doing an equal amount in opposite directions transfers nothing though. All you are proving here is that you will stubbornly ignore anything you don't like such as the simple questions I keep asking you to respond to.Especially when the measured photon momentum loss, via net wavelength increase, closely matches the momentum gain of the accelerating cavity.
There literally is no possible outcome to your experiment that would show both a working emDrive and conservation of momentum. This is obvious to anyone willing to actually address the simple questions I asked you. -
#1472 by Chris Bergin on 11 Sep, 2019 15:11
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Seeing a few "Someone disagreed with me on the internet" style report to mods. That's allowed.
What isn't allowed is where people are uncivil. So none of that please.
"He didn't answer my question" is not a report to mod. He disagreed with me is not a report to mod. Frustration with someone who sounds like a scratched record and won't debate said points is annoying, but not a report to mod.
Report to mod is where there is a breach of forum rules. Don't ask a mod to step in because someone's equation didn't pass your sniff test.
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#1473 by Alex_O on 12 Sep, 2019 11:59
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Thank. But I don’t understand what to do nextOK, will the pressure on the fixed screen depend on the Doppler shift of the wave frequency? agree?
Yes.
To ALL. Is there anyone with experience in calculating radiation pressure taking into account the doppler effect?
Hello TT. I read a huge amount of your messages and really appreciate you for your contribution to studying the concepts of emdrive. I also remember that you had minor health difficulties, I sincerely hope that everything is fine now?
Can you ask Roger Shawyer good data about acceleration diagrams in the Emdrive system? Good modeling requires analysis at time intervals (possibly) less than a hundredth of a second.
And the second question - how did Roger Shawyer even guess about the idea of Doppler? Where did he see it? Did he measure the floating frequency in the resonator while moving the assembly on a linear bench? And he measured the quality factor loss in the resonator during the movement on long rails?
I have one unexpected version of how emdrive works, it can be said in a nutshell.
- Emdrive - "digs a hole."
- The hole does "oops"!
- And the emdrive dives into the hole.
- And so in the loop.
Here the walls of the hole and this "oops" are somehow built from electromagnetic forces, this thing is connected from - I don’t know how to say it better - we see a standing wave in the resonator, we see that the EM field is locked "in the cell", and we see how the EM field is trying "break out of the cage." Therefore, the slightest push, vibration - creates a small (ordinary) physics, which allows the EM field, which is generally unique, unusual! distributed in the "cell" in the "wedge shape" create a small push. Which works only on condition that Emdrive is helped a little by external force? But only once, in the beginning. Remembering the microphone effect in TESLA superconducting resonators, it seems to me that this may be like the truth.
Upd.
I’ll add another verification option. Emdrive of a different geometry, for example, built from two cones (mirror image through a small bottom), if installed on a stand and turned on the RF power, can offer resistance - an attempt to move it from its initial position - if a small force (push) is applied along the axis.
By the way, it will be cool, it looks as if the inertial mass was added to the drive.
By the way, any electromagnetic cavity emits gravitational waves well, in the EM cavity there are several places - where gravitational waves are “born”, and it seems I read that processes in the skin layer have not yet been studied as a place where gravitational waves are generated . And I read about the waves of Unro ...
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#1474 by meberbs on 12 Sep, 2019 15:09
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First calculate the frequency shift as a function of velocity. This is simply a factor that multiplies the original frequency. It is easy enough to google the formula.
Thank. But I don’t understand what to do nextOK, will the pressure on the fixed screen depend on the Doppler shift of the wave frequency? agree?
Yes.
To ALL. Is there anyone with experience in calculating radiation pressure taking into account the doppler effect?
The energy and momentum in a photon are both proportional to the frequency, so they increase by the same factor.
The radiation pressure is proportional to the power hitting the surface, where the power can basically be considered as (photons/second)*(energy/photon). Therefore the radiation pressure also increases by the same factor.
There really isn't anywhere to go from there, because you aren't breaking any conservation laws with this, the antenna emitting the EM waves feels an equal amount of force in the opposite direction.Therefore, the slightest push, vibration - creates a small (ordinary) physics, which allows the EM field, which is generally unique, unusual! distributed in the "cell" in the "wedge shape" create a small push. Which works only on condition that Emdrive is helped a little by external force? But only once, in the beginning.
Please see my previous explanation, the net force from the photons inside the cavity is in the opposite direction of the externally applied acceleration. This differential radiation pressure goes away as soon as the externally applied acceleration does, it has no mechanism to be self-sustaining, because it always opposes applied acceleration.By the way, any electromagnetic cavity emits gravitational waves well, in the EM cavity there are several places - where gravitational waves are “born”, and it seems I read that processes in the skin layer have not yet been studied as a place where gravitational waves are generated . And I read about the waves of Unro ...
No, that isn't how gravitational waves work, and no significant or detectable gravitational waves could be generated by a small mass like that. -
#1475 by Alex_O on 12 Sep, 2019 16:24
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First calculate the frequency shift as a function of velocity. This is simply a factor that multiplies the original frequency. It is easy enough to google the formula.
The energy and momentum in a photon are both proportional to the frequency, so they increase by the same factor.
The radiation pressure is proportional to the power hitting the surface, where the power can basically be considered as (photons/second)*(energy/photon). Therefore the radiation pressure also increases by the same factor.
Thank you, I saw these formulas, it seems that there is a simple proportion - if the frequency increases by 10%, then the pressure also increases by 10%. But it seems to me that it is too simple.
For example:.
1. Not movable perfect screen. The energy of the incident photon is equal to the energy of the reflected one. The screen receives a momentum from the photon = 2p. (where p is the photon momentum)
2. Movable perfect screen. The frequency of the incident photon is not equal to the frequency of the reflected one. The screen will receive an impulse = 2p + dp when a photon is reflected, where dp is the contribution of the Doppler effect
All Right? for the first stage? -
#1476 by meberbs on 12 Sep, 2019 17:52
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The complications you bring up here don't change this part of the calculation.
First calculate the frequency shift as a function of velocity. This is simply a factor that multiplies the original frequency. It is easy enough to google the formula.
The energy and momentum in a photon are both proportional to the frequency, so they increase by the same factor.
The radiation pressure is proportional to the power hitting the surface, where the power can basically be considered as (photons/second)*(energy/photon). Therefore the radiation pressure also increases by the same factor.
Thank you, I saw these formulas, it seems that there is a simple proportion - if the frequency increases by 10%, then the pressure also increases by 10%. But it seems to me that it is too simple.1. Not movable perfect screen. The energy of the incident photon is equal to the energy of the reflected one. The screen receives a momentum from the photon = 2p. (where p is the photon momentum)
By "perfect screen" I assume you meant to write "perfect mirror" in which case this statement is correct. A perfrectly black screen would of course be half of this.2. Movable perfect screen. The frequency of the incident photon is not equal to the frequency of the reflected one. The screen will receive an impulse = 2p + dp when a photon is reflected, where dp is the contribution of the Doppler effect
The reflected photon would have less momentum after reflection so dp would be negative, unless you are working in a frame where the mirror has initial velocity towards the signal source. For a moving mirror, you simply calculate the incident and reflected momentum to get the total force on the mirror, where the reflected momentum is just calculated by the Doppler shift due to reflection.
For when the mirror is initially starting out from 0 velocity, but is allowed to move, there is still a small Doppler shift due to energy transferred to the mirror as it starts moving. I did the calculations for this here. This is generally a negligible effect, as the mass of the mirror is going to be very large compared to the energy in a photon.
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#1477 by Alex_O on 12 Sep, 2019 18:06
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For when the mirror is initially starting out from 0 velocity, but is allowed to move, there is still a small Doppler shift due to energy transferred to the mirror as it starts moving. I did the calculations for this here. This is generally a negligible effect, as the mass of the mirror is going to be very large compared to the energy in a photon.
OK, I understand your calculation, for one photon a very small value. But in the radiation pressure of the photon flux with a power of 1 kW, we will already have some deviation from the magnitude of the order of 3.3 μN, for each reflection, and this small deviation should be multiplied by the value of the quality factor (of the order of 5 * 104-109). It seems to me that you did not take into account the Q factor in your calculation.
I will go ahead.
Now consider two designs of an exemplary photon rocket. With one and two mirrors. Two mirrors are an emdrive without side walls. We will study three options.
1. The rocket stands on a movable trolley with zero friction in the supports. After turning on the RF power, the dinanometer attached to the trolley will show the thrust of the rocket = 3.33 μN / kW.
2. The rocket is on, hit the cart with a hammer.
a) The blow was struck from the side of emitter
b) The blow was struck from the side of a large mirror.
After the impact, the rocket’s design was deformed, the deformation propagates with the speed of sound in the metal.
3. The rocket is on, its parts have experienced thermal stress. In this case, different parts of the rocket could make relative movements at different speeds.
Now we need to very carefully formulate the difficult question in the next post.
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#1478 by meberbs on 12 Sep, 2019 18:24
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OK, I understand your calculation, for one photon a very small value. But in the radiation pressure of the photon flux with a power of 1 kW, we will already have some deviation from the magnitude of the order of 3.3 μN, for each reflection, and this small deviation should be multiplied by the value of the quality factor (of the order of 5 * 104-109). It seems to me that you did not take into account the Q factor in your calculation.
What in the world are you talking about? The setup that you have been asking about and what my calculation applies to is not a resonator, there is no Q factor.1. The rocket stands on a movable trolley with zero friction in the supports. After turning on the RF power, the dinanometer attached to the trolley will show the thrust of the rocket = 3.33 μN / kW.
Yes for your first picture where you are just radiating away (pretending perfect directionality), though lets just ignore the details of how you measure this result, and just assume that we know the power, the mass, and can observe the rate of change of velocity.2. The rocket is on, hit the cart with a hammer.
The answer to all of these is very simple, conservation of momentum holds, and you don't get any different results. If you think you get different results, you made a mistake. Plausible common mistakes include things like missing that the for the case in your first picture, the initial emission of the photons transfers momentum to the structure to the right exactly equal magnitude to the momentum in the photons as they go to the left.
a) The blow was struck from the side of emitter
b) The blow was struck from the side of a large mirror.
After the impact, the rocket’s design was deformed, the deformation propagates with the speed of sound in the metal.
3. The rocket is on, its parts have experienced thermal stress. In this case, different parts of the rocket could make relative movements at different speeds.
Now we need to very carefully formulate the difficult question in the next post.
For your picture with 2 mirrors, it will not go anywhere, ignoring any losses that are radiated away (which will total to no more than 3.33 μN / kW) -
#1479 by Alex_O on 12 Sep, 2019 18:38
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OK, I understand your calculation, for one photon a very small value. But in the radiation pressure of the photon flux with a power of 1 kW, we will already have some deviation from the magnitude of the order of 3.3 μN, for each reflection, and this small deviation should be multiplied by the value of the quality factor (of the order of 5 * 104-109). It seems to me that you did not take into account the Q factor in your calculation.
What in the world are you talking about? The setup that you have been asking about and what my calculation applies to is not a resonator, there is no Q factor.1. The rocket stands on a movable trolley with zero friction in the supports. After turning on the RF power, the dinanometer attached to the trolley will show the thrust of the rocket = 3.33 μN / kW.
Yes for your first picture where you are just radiating away (pretending perfect directionality), though lets just ignore the details of how you measure this result, and just assume that we know the power, the mass, and can observe the rate of change of velocity.2. The rocket is on, hit the cart with a hammer.
The answer to all of these is very simple, conservation of momentum holds, and you don't get any different results. If you think you get different results, you made a mistake. Plausible common mistakes include things like missing that the for the case in your first picture, the initial emission of the photons transfers momentum to the structure to the right exactly equal magnitude to the momentum in the photons as they go to the left.
a) The blow was struck from the side of emitter
b) The blow was struck from the side of a large mirror.
After the impact, the rocket’s design was deformed, the deformation propagates with the speed of sound in the metal.
3. The rocket is on, its parts have experienced thermal stress. In this case, different parts of the rocket could make relative movements at different speeds.
Now we need to very carefully formulate the difficult question in the next post.
For your picture with 2 mirrors, it will not go anywhere, ignoring any losses that are radiated away (which will total to no more than 3.33 μN / kW)
No no, we do not want to violate the law of conservation of momentum. We just want to calculate the manifestations of the Doppler effect at different points of the rocket, and then, already think about what to do next.
When parts of a rocket move, and different parts of a rocket make microscopic movements with the speed of sound in a solid, then if we want to take into account the Doppler effect.
Since the speed of the photon >> is greater than the speed of sound, options are possible.
a) The pressure on the large mirror already has an additive from the Doppler effect. And the pressure on the small mirror for a "long, long time" will have an initial value.
During this "long, long time", photons will have time to reflect many, many times between the mirrors, and in each transaction, create a pressure difference on the mirrors that will accumulate.
Here, "long, long time" is obvious = Length of the rocket / speed of sound minus Length of the rocket / speed of light.
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What is it? I hit Emdrive with a hammer or thermally processed it, and is there a very real moment in time when at its different ends, for a very short time, but the photon pressure changed very, very much? With the accumulation of the effect due to the quality factor of the resonator?
