# Is the brightness of light invariant?

Discussion in 'Physics & Math' started by Quantum Quack, Mar 14, 2006.

1. ### 2inquisitiveThe Devil is in the detailsRegistered Senior Member

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And which frame would both the detector AND the source be moving in? The only one I can think of would be an unspecified 'absolute rest frame'. Surely you don't mean a GLOBAL frame in which the CMB is at rest do you? Are you sure you understand frames of reference, Dale?

Where have I ever mentioned 'the detector's rest frame'? I have repeatedly stated the detector goes through a sequence of three frames. Are you stating a comoving inertial frame is a non-inertial frame? That was the first frame. The accelerating frame of the detector WAS a non-inertial frame, the second frame of reference. Are you stating that the last frame of reference in which the last readings were made was a non-inertial frame? The detector was not accelerating in that frame, but there was a constant relative velocity measured by Doppler shift.

You really don't know how frames work in General Relativity, do you? A non-inertial frame is NOT a 'rest frame'. The coordinates of the frame move through spacetime, spacetime does not 'move' while the coordinates of the detector are 'at rest'.

I ask again, where are the coordinates of this frame fixed? The coordinates are not the coordinates of the detector's 'rest frame' in that case. How can you possibly use 'a single inertial frame' unless it is an unspecified third frame of reference in which the detector feels no force of acceleration? I specified the detector DOES feel a momentary force, but not throughout the exercise.

3. ### DaleSpamTANSTAAFLRegistered Senior Member

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In the vast majority of possible inertial frames both the detector and the source will be moving. Why are you bringing up the CMB?

Why do you insist on using all of these frames when any single inertial frame is sufficient?

With 10 lightyears of distance between the star and the rocket I think the flat-spacetime assumption is good. There is no reason to bring in GR, the situation can be analyzed entirely with SR.

I really doubt that someone like you who obviously fails to grasp the basics of reference frames in SR is equipped to comment reasonably on GR reference frames.

You are so clueless. It is not necessary to have anything at rest in a reference frame.

Since a verbal description of how to solve the problem is obviously insufficient for you I will do it numerically.

Let's consider an inertial frame where the emitter and the detector are initially both moving at .33c in the positive x direction. We will place the origin such that the emitter is at (0,0) when the detector is at the middle of its acceleration. For convenience let's define a unit of time 1T=3.57ns=1/280MHz. We will also define a unit of distance 1D=1T*c. In these uints 10 years is approximately 87E15T=87PT.

The equation of motion for the emitter is s(t)=(t,t/3) and the equation of motion for a wave peak emitted at time τ is p<sub>τ</sub>(t)=(t,t-2τ/3). In order to give the emitter and the detector enough separation for 87PT to pass between emission and detection the equation of motion for the detector should be r(t)=(t,t/3+58PD) for t<-1PT and r(t)=(t,-t/3+58PD) for t>1PT. This gives a time of 2PT, or approximately 12 weeks, for the rocket to accelerate to from .33c to -.33c.

So, let's say that the emitter begins emitting with the first peak at s(-90PT) and continues emitting until the last peak at s(-84PT). The spacetime interval between the emission of any two peaks is 1T, so the coordinate time and location for the emission of the second peak can be found by solving |s(-90PT)-s(dt-90PT)|=1 to get dt=1.06T. The coordinate time and location for the detector detecting the first peak is found by solving p<sub>-90PT</sub>(t)=r(t) which gives (-3PT,57PD). The coordinate time and location for the detector detecting the second peak is found by solving p<sub>1.06T-90PT</sub>(t)=r(t) which gives (1.06T-3PT,.35D+57PD) the spacetime interval between the two is |(1.06T,.35D)|=1T so the detected frequency of the first wave is 280MHz.

Similarly, the coordinate time and location for the emission of the second to the last peak is found by solving |s(-84PT)-s(-dt-84PT)|=1 to again get dt=1.06T. The coordinate time and location for the detector detecting the last peak is found by solving p<sub>-84PT</sub>(t)=r(t) which gives (1.5PT,57.5PD). The coordinate time and location for the detector detecting the second to last peak is found by solving p<sub>-1.06T-84PT</sub>(t)=r(t) which gives (-.53T+1.5PT,.18D+57.5PD) the spacetime interval between the two is |(-.53T,.18D)|=1/2T so the detected frequency of the last wave is 560MHz.

Note that only a single frame was considered. Note also that neither the emitter nor the detector were ever at inertial rest in the frame of analysis. Note that no transformations between frames were even necessary in the analysis since spacetime intervals are frame-invariant. Note that the Doppler-shift formula was not used* during the working of the problem, but the Doppler effect was obtained directly from geometric first principles.

If I weren't specifically trying to make the point that the analysis can work in any frame then I would have worked in the emitter's rest frame for convenience. And if I weren't trying to demonstrate the principles I would have just used the Doppler formula for convenience. But this clearly demonstrates that those are only matters of convenience and not essential. The analysis can be performed in any single inertial frame as stipulated by the first postulate of SR.

-Dale

*the Doppler shift formula and the relativistic velocity addition formulas were used during the development of the problem to determine that .6c gives the desired Doppler factor of 2 and that .33c - -.33c gives .6c.

Last edited: Apr 7, 2006

5. ### 2inquisitiveThe Devil is in the detailsRegistered Senior Member

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OK, Dale, I didn't want you to go to this much trouble. I will read through your post and comment as best I can.
Well, you got me there! There can be an ALMOST infinite number of inertial frames throughout the universe to observe the motions from. In how many of those frames does the detector feel an acceleration between the two measurements of the frequency? I must really be ignorant as I can't figure out HOW the frequency can be MEASURED to be 280mHz in these other inertial frames for the first measurement taken. I can't figure out how the frequency can be measured to be 560mHz at the time of the second measurement in this infinity of inertial frames. I CAN understand how VELOCITY can be calculated using any inertial frame. Perhaps you can enlighten me concerning frequency measurements made by the detector?

No, but for it to be an inertial REST frame, the coordinates or SOMETHING has to be 'at rest', correct?
The detector has already made one measurement of 280mHz when the coordinates of this frame are plotted. What happened to those coordinates? Where is the observer plotting these coordinates located? I only know of one observer, the one on the ship along with the detector. But he can't 'feel' when he is in 'the middle' of his acceleration, correct?
Golly gee, and I thought a simple radio reciever could recieve a radio signal and display the recieved frequency. Was the frequency ever MEASURED in the middle of the acceleration of your beginning coordinates? From what location? Were any signals measured, or was the frequency of emission from the star already known? Were the relative velocities and locations of the emitter and detector known before the beginning of your exercise, and the frequencies CALCULATED from those known coordinates? How was the relative velocity difference (acceleration) from .33c to -.33c measured and from which frame of reference? In other words, who measured?

7. ### DaleSpamTANSTAAFLRegistered Senior Member

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It isn't just an ALMOST infinite number of inertial frames. There are an infinite number of possible translations, for every possible translation there are an infinite number of possible rotations, for every possible translation and rotation there are an infinite number of boosts. Thus there are as many possible inertial reference frames as there are points in a 9-dimensional real space.

In every single one of those inertial frames the detector feels an acceleration between the two measurements.

Certainly. What is a frequency measurement? A frequency measurement is nothing more nor less than the inverse of a period measurement. A period is the time between two successive peaks (or other features) of a repeating waveform. So a frequency measurement is essentially a time measurement. The proper time between any two events is a frame-invariant measurement and it is equal to the spacetime interval between those two events. So, since proper time is frame-invariant and proper frequency is the inverse of the proper time then proper frequency is also frame invariant. To get the frequency all you have to do is get the period, which you can do using a spacetime interval, which you can measure in any reference frame and all frames will agree.

Yes, obviously. The first postulate of SR, however, does not refer to "rest frames" nor to "inertial rest frames", but only "inertial frames". Hence my repeated comments about not needing anything to be at rest in a frame.

You do realize that coordinates can have negative numbers, right? The frame of analysis doesn't suddently come into existence at t=0. Also, why should anyone other than me be plotting these coordinates? I'm not trying to write a novel here, just solve a problem, so I don't understand your desire to introduce extra observers/characters to the problem.

No, I didn't calculate the measured frequency during the middle of the acceleration. This would be possible using the exact same technique I used previously. Specifically, solve for the intersection of the detector equation of motion with that of two successive wavefronts. Calculate the spacetime interval and take the reciprocal. I leave this as an exercise for the interested reader.

Alternatively you can just determine the relative velocity at that point and plug that into the Doppler formula. I leave that as an exercise for the less ambitious reader.

The relative velocity was calculated from your specified Doppler shift. If you had instead specified the relative velocities I could have calculated the Doppler shift. One uniquely determines the other. The frame of reference is completely arbitrary, both in terms of the origin and the boost. I just picked one where the equations of motion were non-zero but still relatively simple.

What do you mean "from which frame of reference"? I only used one frame! How can you possibly be confused about which frame when there is only one? How many times do I have to say that to get through to you? Which sun shines in the day? There's only one. One sun, one earth, one milky way, one God, and one frame!

-Dale

8. ### 2inquisitiveThe Devil is in the detailsRegistered Senior Member

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No, the detector can only 'feel' an acceleration in its non-inertial frame. From all those other inertial frames, did the star accelerate or did the detector accelerate? If the frame is comoving with the detector, the star accelerates and it cannot be stated the detector 'feels' an acceleration. I am concerned you have a problem understanding what a 'comoving' frame refers to. It simply means that no relative velocity can be measured between two points that are comoving. It doesn't mean the HAVE to be moving, it is just that the notion of 'absolute rest' is discarded. Examples of comoving frames would be your house and the house next door to yours. They have no relative velocity when either is considered as a coordinate point, but that doesn't mean that the Earth is not rotating or moving through the universe, it is just irrelevant to the problem at hand. Another example would be you standing on the equator and considering a geostationary satellite overhead. There is '0' relative velocity between you and the satellite, but both coordinate points can be moving at different velocities relative to a different coordinate point, such as a heliocentric frame of reference.
No, that is not true. I am now thinking of a different frequency detected by the spaceship detector, two different frequency signals emitted from the star. What frequency is this?
I am not introducing additional observers. Only the observer located on the spaceship with the detector can measure a frequency, an observer located on the star can measure the emitted frequency and calculate what frequency the spaceship will measure, AS LONG AS NEITHER FRAME ACCELERATES during the 10 year travel time of the emitted signal. An observer located in any of these other 'infinite inertial frames' CANNOT determine what specific frequency is of interest to the problem as stated. Why do you think I specifed an 'indeterminate' radio frequency was emitted from the star? No one knows this frequency until it is measured by the detector on the spaceship.
I did not specify a Doppler shift. I stated the Doppler shift could be measured by the observer on the spaceship with the detector. The spaceship did not even exist when the signal was emitted from the star 10 years earlier. Don't you understand? I am the captain of a spaceship in the vacuum of space. I detect a radio signal comming from a distant source. Nothing is known about the source or the frequency of the signal until I detect the signal and begin my analysis. From this analysis, I can make all other statements about the source and my velocity relative to the source. I can determine my change in relative velocity wrt the source, and KNOW that it is me that changed velocity because I fired my engines, felt the acceleration, and measured a shift in the Doppler readings corresponding to my felt acceleration. No, I do not know my 'absolute' velocity relative to 'the universe', but I could get an estimate by measuring Doppler shifts in the CMB. That was not, and is not, the focus of this gedankin, however. The focus is that I CAN determine when my velocity has changed relative to another object, and know MY velocity has changed and not the other reference. I accelerated and not the universe.

9. ### DaleSpamTANSTAAFLRegistered Senior Member

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Why? You are going to have to justify that statement. I don't see why drawing and/or moving a bunch of coordinate system lines should have any impact on wether or not someone feels an acceleration.

Why not? You will have to justify that one too. I already showed mathematically how they could.

-Dale

10. ### 2inquisitiveThe Devil is in the detailsRegistered Senior Member

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3,181
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I will simply copy from wiki to educate you, Dale!

You cannot feel the force, or measure the frequency unless you do it from the accelerated frame. You CAN, however, describe the inertial frames from the accelerated frame, the frame of the detector.

11. ### Physics MonkeySnow Monkey and PhysicistRegistered Senior Member

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If I could clear up a confusion here, the detector feels an acceleration only if it is accelerated. It doesn't matter what reference frame you use to look at the detector. If I attach a mass to the detector via a spring, the mass will only deviate from its equilibrium position if the detector accelerates. Just looking at the mass-detector system in a noninertial frame won't make the mass move from it's equiliibrium position. That is nonsense.

12. ### DaleSpamTANSTAAFLRegistered Senior Member

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You obviously misunderstood the quotes you copied. They don't support your position.

I will continue to say "frame force" instead of "geeforce".

As a fan of GR I would think you would already be aware that frame forces are not "felt", only infered from observed motion in the accelerated frame. In other words, since they accelerate all parts of all objects equally they cannot be physically measured. For example, place a scale underneath a satellite in orbit and no force will be measured, despite the fact that the satellite is being accelerated by the gravitational frame force. Now, place the scale underneath a person on the ground and you measure a force. Which force? Not the gravitational frame force, the scale has already been shown to be insensitive to that. Instead you physically measure the action-reaction pair of contact forces between the person's feet and the ground.

The same analysis can be made in a rotating frame, like a rotating space station's rest frame. A scale placed under a person in the space station can measure a contact force between the person and the station. In the rotating frame he is not accelerating because the frame force is balanced by the contact force. Meanwhile, an object released from the space station will accelerate under the frame force in the rotating frame, but a scale placed under the object will measure nothing. Again, nothing measured with only the frame force, something measured with a frame and a holding force, so the frame force is not measurable only the holding force.

Note that frame forces do not come in action-reaction pairs (I am sure you remember that whole conversation with BillyT). And, as Wiki says, "An object accelerated to be stationary in the accelerated frame will 'feel' the presence of the field". So it is only through the application of some mechanical "holding force" that frame forces are felt. In other words, it is actually the mechanical force that is felt, not the frame force.

The bottom line is that the "felt" force is part of the physics of the problem. It is an observable fact. If you make any description where that observable fact disappears then you are not correctly describing the situation. Simply drawing lines and saying "this point is (20,7)" or "this point is (-5,41)" cannot make measurable forces appear or disappear. Regardless of wether the lines are straight or curved, accelerating or inertial, all physically measurable forces will still be felt. If you actually understood reference frames you would realize that the fact that the frame forces are not measurable is actually the only reason that it is possible to use accelerating frames in an analysis. You can add as many frame forces (or as few, even 0) as you wish and you will not change any measurement results.

So, in our rocket scenario what is the measurable force? The rocket exhaust produces a measurable force on the rocket and the rocket produces a measurable force on the pilot etc. No other forces are measurable. Any analysis which is correct must maintain that fact: the force between the rocket and the pilot must be measurable during the engine burn and must be zero otherwise. So let's examine the situation in an inertial and an accelerating frame and let's consider the pilot and scale as a single free body. The only forces acting on the body are any frame forces and the contact force from the ship.

Accelerating frame without engine burn: Since we are in the accelerating frame there is a frame force pushing down on the body. The engine is off so there is no opposing contact force. The unbalanced frame force leads both to the observed downward acceleration and the observed measurement of no contact force on the scale.

Inertial frame without engine burn: Since we are in the inertial frame there is no frame force. The engine is off so there is no contact force. This lack of forces leads both to the observed lack of acceleration and the observed measurement of no contact force on the scale.

Accelerating frame with engine burn: Since we are in the accelerating frame there is a frame force pushing down on the body. Since the engine is on there is a balancing contact force pushing up. The balanced pair of forces leads both to the observed lack of acceleration and the measurement of the contact force on the scale.

Inertial frame with engine burn: Since we are in the inertial frame there is no frame force. Since the engine is on there is an unbalanced contact force pushing up. This unbalanced contact force leads both to the observed upwards acceleration and the measurement of the contact force on the scale.

The take home message is that the rocket's force is an observable force. As such it is observed in all frames, inertial and non-inertial. Although frames disagree about the motion, all frames agree that the motion is consistent with the observed forces.

I can hardly believe that you are still so ignorant about reference frames with as much time as you and I have spent discussing them. I am at a loss to figure out how to get these ideas through to you. I have tried description after description, math, logic, examples, and still you don't understand. I don't know if you are being deliberately ignorant or if this is just a difficult concept for you to grasp. All I know is we have been discussing these ideas for literally months now and you are still completely unable to use even the most basic concepts. I am truly sorry for my failure to get through to you.

-Dale

Last edited: Apr 8, 2006
13. ### 2inquisitiveThe Devil is in the detailsRegistered Senior Member

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3,181
Sorry about the late response, but I just turned on my computer a few minutes ago.

To begin with, how can you 'continue' to say "frame force" instead of "geeforce" when there are no frame forces evident in this exercise? Frame forces are centrifugal forces and the coriolis force, NOT acceleration forces. The satellite in orbit is in an INERTIAL frame in General Relativity, but since rotating frames do not give reciprocial results in special relativity, the satellite is generally considered to be in a non-inertial frame.
Why are you bring up all these diversions? To show you have a greater knowledge of physics than I? I will concede that fact without challenge, I have repeatedly stated I AM NOT a physicist. I do know the phrase 'gravitational frame force' is poppycock. Gravity is not a 'force' in General Relativity, it is an acceleration caused by the curvature of spacetime. And, yes, an acceleration due to the rocket's firing CAN be measured, and felt, in the frame. It is just not described as a 'force', let alone a 'frame force'.

You do understand WHY gravity is not described as a 'force' in relativity, don't you? Do you believe Newton's inverse square rule for gravitational force is essentially correct? How can it be based on an inverse square of DISTANCE in relativity, when distances are variable? Special Relativity CANNOT deal with gravity. According to STR, if you are in an inertial frame and determine the distance between two gravitating objects in the direction of your travel (vector), the distance will vary according to your velocity relative to them. Their true gravitational attraction remains an inverse square of a single, true distance measured between them, not 'contracted distances'.
HaHa, thanks for your obsfucation of a simple gedankin in this thread, Dale! It is a perfect illustration of the effort the 'faithful' will employ to support a flawed theory.

14. ### Quantum QuackLife's a tease...Valued Senior Member

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Now thats food for thought....

15. ### DaleSpamTANSTAAFLRegistered Senior Member

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I was refering to our previous conversations on other threads where we have discussed frame forces at length.

Frame forces are present in any and all non-inertial frames, not just rotating frames.

I didn't bring it up. You brought it up. You are the one asked "In how many of those frames does the detector feel an acceleration?", and stated, "the detector can only 'feel' an acceleration in its non-inertial frame." All these "diversions" are in direct refutation of your claim.