Gravity effect on observed speed of light

[cos]

Depends what you mean exactly. If you mean that the position of the start and finish lines for both beams are seen to be the same (assumed to be far away from the gravitational influence of the two masses), then yes. But it's what happens in between that counts, and in between the distance markers and clocks have altered for one of the two beams - as referenced to the distant observer's clock and ruler. And that counts.

In fact we don't need any clocks or rods!

In compliance with the OP - the distant observer could see the beams emitted simultaneously (the sources are equidistant from him) and arrive at the finish line simultaneously.

As you point out - the position of the start and finish lines for both beams are seen, by the distant observer, to be the same ergo on the basis that the beams travel identical distances and, having been emitted simultaneously, arrive at the finish line simultaneously the observer is fully justified in concluding that, from HIS point of view, the beams took the same amount of time to make their trips.

So you agree time runs slower near a gravitating mass; but slower relative to what, if not to that distant gravity unaffected observer?

I freely admit that I am academically unqualified in the subject of physics but please stop treating me like a naive child.

As per above. Yes he is basing his findings on the gravity affected local values - all referenced to his own standards of measure!

As above. The situation as depicted is that there are NO gravity affected local values and he bases his conclusion on observation alone (simultaneous emission; the position of the start and finish lines for both beams are seen to be the same; simultaneous arrival at the finish line).

Alternately, as previously, he bases his determination of the identical distances using his meter stick held at arm's length and uses his clock in order to determine the time that it takes for the beams to make their trips. He cannot see the far distant clocks and rods.

Let us assume that the OP involves a semblance of reality and there are two black holes between which a beam of light travels. There are NO clocks or sticks dangling in space so the distant observer can only determine his measurements in accordance with his own equipment however all of a sudden clocks and sticks apear along the path of the beam.

Does that affect (alter) the distant observer's measuring rod or clock in any way?

Of course it does NOT!

Issue B: Can it ever be true that any object having some initial velocity v can first accelerate (or decelerate), then decelerate (or accelerate), in the direction of motion, finishing back at the initial velocity v, and maintain the same average velocity v?

It has been pointed out clearly in #7, #11 the reason that is not so. Let's use another example. Two cars are in a street drag race, only this race has a flying start where both cross the start line nose to nose and equal speed. Just like in 'Fast and Furious' one driver has a secret nitro booster that he activates and accelerates away, but not for too long - the booster is used up and he slows down to the same initial speed. From the vantage point of the other driver who had no such boost advantage, what will he have seen? Clearly the other car has gained ground during the boost phase, and despite both finishing at the same final speed = initial speed, the nitro driver wins. Conversely, if the other vehicle developed temporary engine trouble and slowed, then picked up to the original speed, he has lost ground and will finish behind. Do you not agree? Don't confuse equality of initial and final speed with equality of average speed!

You wrote "...the booster is used up and he slows down to the same initial speed." My depiction is that he does not slow down to the same initial speed but that his speed is then slower than its initial speed thereby enabling the other beam to catch up.

Tidal forces are proportional to the distance rate-of-change of 'g-force', which in turn is proportional to the distance rate of change of potential. It follows a very different power law and 'directionality' to change in light speed c which is linked more directly to the potential itself. So what is the verdict on how light speed (remember Issue A here!) varies with gravitational potential? Well here's one reference that backs up my quote in #6: http://en.wikipedia.org/wiki/Tests_of_general_relativity#Light_travel_time_delay_testing. I think you will search in vain for a reference that will back what I hope is now just your former position on this.

As I have previously pointed out - the Shapiro delay is relevant to a SINGLE gravitational field!

When somebody conducts a dual gravitational field experiment please let me know.

 

[cos]

Dear cos, contradictory claims are obviously not helpful!

Thanks for the response but there is nothing contradictory about the previous exchange.

A statement was made in relation to a beam of light traveling in a gravitational field and I simply asked in what direction it is traveling.

What almost certainly will be helpful is to read the last part of Einstein's paper as answer to your questions, which you can find here:

http://www.Alberteinstein.info/gallery/gtext3.html

It's a big document but worth waiting for it to download if you have a slow connection.

The answers to your question here above are in section 22, starting from on p.196 of the English version. Actually he describes clearer the effect on rods and clocks; but as you know that the ratio of distance/time must be c (normalised to 1 in that paper), you can easily figure it out (with added understanding!) from the predicted effect on rods and clocks.

Whilst I appreciate your having gone to the trouble of providing that information my comments regarding the speed of light relative to a gravitational field are in relation to what physically takes place i.e. to reality.

As regards "...the effect on rods and clocks;" please see my message pointing out that the far distant observer is making his determinations on the basis of his own measuring devices not on the basis of rods and clocks located close to those gravitational fields.

He may not be able to see those extremely far-distant rods and clocks thus can only go by his own determinations.

If those rods and clocks suddenly appeared (or disappeared) along the beam's trajectory his own measuring devices would remain unaffected!

 

[atyy]

The issue is how to interpret the redshift and time dilation. I personally like explanations that are in concordance with observation. For example, of all of the explanations of the twin paradox, I personally like the doppler explanation best. I'm not so arrogant as to foist that explanation on others. If you like some other interpretation better, so be it. There are a bunch of ways to resolve the twin paradox, all of which yield the same ultimate answer.

Distant observers cannot see photons when they are deep inside a gravity well; they have to wait for the photons to reach them before they can see them. The only observers who can see photons deep inside a gravity well are observers who themselves are deep inside a gravity well. Suppose some observer well outside a gravity well conducts local experiments to ascertain physical constants such as the speed of light, Planck's constant, and the fine structure constant. They then take the necessary equipment deep inside a gravity well and re-perform the experiments. They will get the same values as they obtained outside the gravity well. The physical constants are, well, constant.

So has the doppler shifted light in the twin paradox lost energy?

 

[harrylin]

Thanks for the response but there is nothing contradictory about the previous exchange.

A statement was made in relation to a beam of light traveling in a gravitational field and I simply asked in what direction it is traveling.

Whilst I appreciate your having gone to the trouble of providing that information my comments regarding the speed of light relative to a gravitational field are in relation to what physically takes place i.e. to reality.

As regards "...the effect on rods and clocks;" please see my message pointing out that the far distant observer is making his determinations on the basis of his own measuring devices not on the basis of rods and clocks located close to those gravitational fields.

He may not be able to see those extremely far-distant rods and clocks thus can only go by his own determinations.

If those rods and clocks suddenly appeared (or disappeared) along the beam's trajectory his own measuring devices would remain unaffected!

Einstein there explains how the new theory helps "to arrive at a closer approximation to reality"! (section 22), and he explains it roughly in the way you want it to be explained.

As a matter of fact, I recognise nothing at all of your issues in the text that you did not bother to read.

 

[PAllen]

So has the doppler shifted light in the twin paradox lost energy?

Isn't this obviously yes (?). KE and momentum are frame dependent, whether for light or bullets. In an accelerating frame, inertial objects appear to continuously change momentum and KE, and light appears to gain/lose energy. So, you can describe gravity well effects similarly.

 

[D H]

Let's get back to the original question.

An observer (in zero gravity) witnesses two parallel but widely separated beams of light that are transmited at the same time. One of the beams experiences zero gravity. The other beam travels between two massive objects with a very strong gravitational field, but arranged so that the beam direction is not changed.

I'm going to rephrase it slightly in a way that I hope eliminates some of the ambiguity and simultaneity issues. I'll use our universe and a near carbon copy of our universe.

Suppose an observer simultaneously sees two bursts of light in different parts of the night sky. The path from each source to the observer is more or less flat (no significant gravity wells along the way). Now imagine an alternate universe that is exactly the same as ours except that now one of the signals travels between two massive objects with very strong gravitational fields arranged so that the beam direction is not changed. Does our alternate observer still see the bursts as arriving simultaneously?

 

[PAllen]

Let's get back to the original question.I'm going to rephrase it slightly in a way that I hope eliminates some of the ambiguity and simultaneity issues. I'll use our universe and a near carbon copy of our universe.

Suppose an observer simultaneously sees two bursts of light in different parts of the night sky. The path from each source to the observer is more or less flat (no significant gravity wells along the way). Now imagine an alternate universe that is exactly the same as ours except that now one of the signals travels between two massive objects with very strong gravitational fields arranged so that the beam direction is not changed. Does our alternate observer still see the bursts as arriving simultaneously?

I'll throw in an ammendment to elimintate the alternate universe complication. Two pulsars in different parts of the sky are magically synchronized. The path from both initially encounter no significant graviy well. However, two neutron stars are heading for a collision directly on the path from one of the pulsars, perpendicular to its line of sight from us. As they get close to collision on the path from one pulsar, do we see the synchronization break down, and in what direction?

[Edit: Further thought: the image of the affected pulsar may become some complex shape due to graviational lensing, around the same time the graviational effects on the light become significant. This may cause the pulse to appear smeared in space and time. However, assuming perfect symmetry, there should remain some central point of the smeared image that has not appeared to shift in the sky. Will this part of the pulse image remain synchronized? ]

 

[atyy]

Isn't this obviously yes (?). KE and momentum are frame dependent, whether for light or bullets. In an accelerating frame, inertial objects appear to continuously change momentum and KE, and light appears to gain/lose energy. So, you can describe gravity well effects similarly.

How about the point of view of the inertial twin?

 

[PAllen]

How about the point of view of the inertial twin?

Again, seems inevitable. Red light has different energy than blue, per photon. If turnaround twin is sending pulses of constant energy from his/her point of view, inertial twin will receive red pulses with less energy and blue pulses with more energy. How could it be otherwise?

To make this better defined, assume the emitted pulses are directed, not spherical. Then, the turnaround twin changes momentum and KE a tiny bit after each pulse. Further, note that the angle of receiver needed to capture a complete directed pulse will change from the red shift case to the blue shift case. Assume receiver is big enough to capture whole pulse in both cases (relativistic beaming effects eliminated). Then, clearly the receiver will, as I said above, receive red pulses of lower total energy compared to the blue pulses. Small momentum and KE changes to the emitting, turnaround twin, preserve conservation of energy and momentum.

 

[atyy]

Again, seems inevitable. Red light has different energy than blue, per photon. If turnaround twin is sending pulses of constant energy from his/her point of view, inertial twin will receive red pulses with less energy and blue pulses with more energy. How could it be otherwise?

So from the inertial twin's point of view, the light lost energy?

 

[PAllen]

So from the inertial twin's point of view, the light lost energy?

See edited explanation above.

 

[Q-reeus]

As you point out - the position of the start and finish lines for both beams are seen, by the distant observer, to be the same ergo on the basis that the beams travel identical distances and, having been emitted simultaneously, arrive at the finish line simultaneously the observer is fully justified in concluding that, from HIS point of view, the beams took the same amount of time to make their trips...
..As above. The situation as depicted is that there are NO gravity affected local values and he bases his conclusion on observation alone (simultaneous emission; the position of the start and finish lines for both beams are seen to be the same; simultaneous arrival at the finish line)...

Again, you are not accepting that gravity alters distance and time scales. Identical start and finish lines does not mean identical path traversal in between. Two open ended rubber bands bridge across an equal width gap. One is stretched straight, the other is depressed down - 'saggy'. Two insects crawl across at equal local crawl speed, one on each rubber band. Do both arrive together because the gap is the same (your logic), or will the one on the saggy band arrive last (GTR = gravity warps spacetime logic)? Think before answering.

Alternately, as previously, he bases his determination of the identical distances using his meter stick held at arm's length and uses his clock in order to determine the time that it takes for the beams to make their trips. He cannot see the far distant clocks and rods...
..Let us assume that the OP involves a semblance of reality and there are two black holes between which a beam of light travels. There are NO clocks or sticks dangling in space so the distant observer can only determine his measurements in accordance with his own equipment however all of a sudden clocks and sticks appear along the path of the beam...
Does that affect (alter) the distant observer's measuring rod or clock in any way? Of course it does NOT!

So you genuinely believe that literal clocks and rulers somehow need to be present locally in order to effect things?! No - it is a very useful 'figure of speech' to refer to clocks and rulers as indicative that time and distance measure are effected by gravity in this case. You accept that clocks are used to measure time, and rulers to measure length, right? Get used to these terms as simply representative of the 'actual' quantities.

You wrote "...the booster is used up and he slows down to the same initial speed." My depiction is that he does not slow down to the same initial speed but that his speed is then slower than its initial speed thereby enabling the other beam to catch up.

So here in #31, you finally let it be known entry and exit speeds are in your universe unequal, without giving any hint in either #3:
"I believe that the latter beam will accelerate as it approaches those objects then slow down as it travels away from same thereby arriving at the destination simultaneously with the uninterrupted beam. ", or #16:
"You may prefer to look at things from the point of view that light decelerates while falling into a gravity well and accelerates while climbing out however on the basis that this viewpoint leads to some apparently paradoxical results I prefer to look at things from the point of view that light accelerates while falling into a gravity well and decelerates whilst departing.", or #18:
"In my interpretation the train accelerates as it travels toward the objects then slows down as it departs same thus presenting an overall time equivalence as determined by the distant observer's gravity unaffected clock and rod. Will the two trains arrive together? Yes."

My strong suspicion is you meant equal in and out as suggested and inferred in the above three entries of yours, but when confronted with the inescapable logic that means necessarily unequal arrival times, you have shifted ground in order to preserve your assertion of equal arrival times. OK, you now want unequal in and out speed? Then realize this necessarily implies some kind of 'draggy' gravity theory at odds with both Newtonian theory and GTR. Recalling we are dealing with static, non-rotating masses as per OP, gravity is an entirely conservative effect - same in = same out, no if's, but's, or maybe's. You doubt that? Well argue it out with http://en.wikipedia.org/wiki/Conservative_force, and nothing changes qualitatively in GTR.

As I have previously pointed out - the Shapiro delay is relevant to a SINGLE gravitational field! When somebody conducts a dual gravitational field experiment please let me know.

No. The reference makes it plain curvature is insignificant, leaving only time delay. In the OP's setup (we have gone over this before), by symmetry of the twin masses curvature is canceled out, which only makes it worse from your pov - there is nothing but time delay as effect. Where do you go from there?

I freely admit that I am academically unqualified in the subject of physics but please stop treating me like a naive child.

Well so am I unqualified, but our difference is in the willingness to recognizing that 2+2 has to add up to 4, and if our pet idea disagrees, be man enough to acknowledge that, and move on a lit bit the wiser. That will make you a real adult. Cheers!

 

[PAllen]

I'll throw in an ammendment to elimintate the alternate universe complication. Two pulsars in different parts of the sky are magically synchronized. The path from both initially encounter no significant graviy well. However, two neutron stars are heading for a collision directly on the path from one of the pulsars, perpendicular to its line of sight from us. As they get close to collision on the path from one pulsar, do we see the synchronization break down, and in what direction?

[Edit: Further thought: the image of the affected pulsar may become some complex shape due to graviational lensing, around the same time the graviational effects on the light become significant. This may cause the pulse to appear smeared in space and time. However, assuming perfect symmetry, there should remain some central point of the smeared image that has not appeared to shift in the sky. Will this part of the pulse image remain synchronized? ]

Anyway, the upshot remains (based on the Shapiro effect analysis), that the pulsar whose signal path is affected by the converging neutron stars would fall out of synch, falling behind the other pulsar.

Is this the consensus of the experts (I'm not an expert)?

 

[Dale]

Using his meter stick (held at arm's length) he determines that the beams travel identical distances - from their source to the target. He does not measure 'a longer spacetime path' for one of those beams.

Remember, we are talking about curved geometry here. In curved geometry it is possible for a figure to have 4 straight sides with 4 right angles and still have one side which is longer than the opposite side. Of course the observer may use a coordinate system such that the coordinate distance is the same, but all distances based on the metric (measuring rods or light pulses) will be longer for the path through the gravity well.

 

[cos]

I recognise nothing at all of your issues in the text that you did not bother to read.

In that case I'm glad I didn't waste my time reading it.

 

[cos]

Let's get back to the original question.

I'm going to rephrase it slightly in a way that I hope eliminates some of the ambiguity and simultaneity issues. I'll use our universe and a near carbon copy of our universe.

Suppose an observer simultaneously sees two bursts of light in different parts of the night sky. The path from each source to the observer is more or less flat (no significant gravity wells along the way). Now imagine an alternate universe that is exactly the same as ours except that now one of the signals travels between two massive objects with very strong gravitational fields arranged so that the beam direction is not changed. Does our alternate observer still see the bursts as arriving simultaneously?

Yes.

 

[cos]

I'll throw in an ammendment to elimintate the alternate universe complication. Two pulsars in different parts of the sky are magically synchronized. The path from both initially encounter no significant graviy well. However, two neutron stars are heading for a collision directly on the path from one of the pulsars, perpendicular to its line of sight from us. As they get close to collision on the path from one pulsar, do we see the synchronization break down, and in what direction?

The synchronization does not break down.

 

[harrylin]

In that case I'm glad I didn't waste my time reading it.

OK then, if you prefer to keep your issues, as you wish! I won't try anymore to help you get rid of them.

 

[cos]

Again, you are not accepting that gravity alters distance and time scales. Identical start and finish lines does not mean identical path traversal in between. Two open ended rubber bands bridge across an equal width gap. One is stretched straight, the other is depressed down - 'saggy'. Two insects crawl across at equal local crawl speed, one on each rubber band. Do both arrive together because the gap is the same (your logic), or will the one on the saggy band arrive last (GTR = gravity warps spacetime logic)? Think before answering.

Your having resorted to personal insults will be dealt with later!

Where, and in which direction, is the 'downward' (?) 'depression' of the path followed by that specific beam of light? Does that route 'sag' or bend in any direction?

Alternately, as previously, he bases his determination of the identical distances using his meter stick held at arm's length and uses his clock in order to determine the time that it takes for the beams to make their trips. He cannot see the far distant clocks and rods...
..Let us assume that the OP involves a semblance of reality and there are two black holes between which a beam of light travels. There are NO clocks or sticks dangling in space so the distant observer can only determine his measurements in accordance with his own equipment however all of a sudden clocks and sticks appear along the path of the beam...
Does that affect (alter) the distant observer's measuring rod or clock in any way? Of course it does NOT!

So you genuinely believe that literal clocks and rulers somehow need to be present locally in order to effect things?!

Typical ill-thought out response aimed specifically at obfuscation and deception!

I point out that the presence or absence of local clocks and rules can have NO effect on the distant observer's measuring devices and you snidely insinuate that I believe the opposite!

(<snip>Similar ill-thought out response aimed specifically at obfuscation and deception.)

...be man enough to acknowledge that, and move on a lit bit the wiser. That will make you a real adult.

It is inevitable that when a person realizes that they are losing an argument they start posting obfuscatory and deceptive material as well as personal insults to which I have, above, responded in kind.

Our discussion is accordingly terminated.

 

[cos]

Remember, we are talking about curved geometry here. In curved geometry it is possible for a figure to have 4 straight sides with 4 right angles and still have one side which is longer than the opposite side. Of course the observer may use a coordinate system such that the coordinate distance is the same, but all distances based on the metric (measuring rods or light pulses) will be longer for the path through the gravity well.

Trepidatiously anticipating eventual ad hominem attacks as have inevitably previously taken place - I submit that a person is looking out into the night sky; he sees, off to his left-hand side, one object that will eventually represent the start line of the beam that will pass between two (yet-to-appear) massive objects and, off to his right-hand side, another object that will eventually become the finish line.

He sees, initially, a beam of light cross the start line when his clock reads zero seconds and arrive at the finish line when his clock reads 5 seconds.

What is his estimation of the distance between the start line and the finish line? Is it, as I suspect it to be, (close to) 1·5 million Ks?

Having thereby determined the distance between the start and finish lines our observer then 'sees' two neutron stars appear on either side of the original beam's route which, as per my assumption below regarding the OP, generate no gravitational 'pull' on the start/finish lines.

(I assume that, in the OP, it is accepted that the intervening masses have no effect on the distance between the start and finish lines; that the start /finish lines are hypothetically immovable objects.)

According to responses - it will then take longer for the beam to traverse that distance as measured by his clock.

Does he conclude that the finish line and the starting line have physically moved further apart?

Prior to the appearance of the masses - another observer, located next to the start line, attaches a very long tape measure to it then travels over to the finish line feeding out the tape as he goes. No surprise - he determines a distance of 1·5 million Ks.

The gravitational masses make their appearance (again without having any physical affect on the distance between the immovable start/finish lines) and he repeats his test.

As he enters progressively stronger gravitational tidal areas the tape measure, attached at the other end to an immovable object, proportionally stretches! It becomes spaghettified! The physical distances between the mm; cm; m; km divisions increase!

As he moves away from that midpoint the tape remains stretched albeit by progressively lesser amounts ergo, overall, he measures that the distance between the start/finish lines is shorter than it was before the masses appeared not longer as insisted upon in this thread.

I have no doubt whatsoever that it can be mathematically 'proven' that the physical results of his hands-on experiment are 'erroneous' - that the distance increases 'due to the Shapiro delay' however I know which version I would place my money on.

 

[Dale]

What is his estimation of the distance between the start line and the finish line?

The problem is that this question is ambiguous in curved spacetime. There are many different ways you could determine the distance, and in GR they can lead to different results. For example, you could simply subtract the coordinate positions, but since the coordinate positions are arbitrary so is this procedure. You could do a radar ranging experiment or you could use a tape measure, but those only work in a static spacetime. You could connect them by a spacelike geodesic and determine the spacetime interval, but there may be more than one geodesic. Etc.

In any case, you do not need to answer this question to answer the OP. It is clear that the pulse through the gravitational field will arrive later. If you prefer to attribute that to a changed (radar) distance or a changed (coordinate) speed of light is up to you. My preference is the former.

 

[atyy]

Again, seems inevitable. Red light has different energy than blue, per photon. If turnaround twin is sending pulses of constant energy from his/her point of view, inertial twin will receive red pulses with less energy and blue pulses with more energy. How could it be otherwise?

To make this better defined, assume the emitted pulses are directed, not spherical. Then, the turnaround twin changes momentum and KE a tiny bit after each pulse. Further, note that the angle of receiver needed to capture a complete directed pulse will change from the red shift case to the blue shift case. Assume receiver is big enough to capture whole pulse in both cases (relativistic beaming effects eliminated). Then, clearly the receiver will, as I said above, receive red pulses of lower total energy compared to the blue pulses. Small momentum and KE changes to the emitting, turnaround twin, preserve conservation of energy and momentum.

How is the equivalence of energy and frequency justified in classical physics?

 

[harrylin]

How is the equivalence of energy and frequency justified in classical physics?

Here we are discussing special relativity, according to which:

f'/f = E'/E

"It is remarkable that the energy and the frequency of a light complex vary with the state of motion of the observer in accordance with the same law."

- http://www.fourmilab.ch/etexts/einstein/specrel/www/ section 8 (Einstein, 1905).

 

[atyy]

Here we are discussing special relativity, according to which:

f'/f = E'/E

"It is remarkable that the energy and the frequency of a light complex vary with the state of motion of the observer in accordance with the same law."

- http://www.fourmilab.ch/etexts/einstein/specrel/www/ section 8 (Einstein, 1905).

But for anyone observer, there is no change in energy.

 

[PAllen]

But for anyone observer, there is no change in energy.

For any inertial observer there is no change in energy after the light is emitted. But two different observers disagree on energy of emitted light. Emitter thinks they emitted white light, total energy E, receding observer thinks they emitted red light of energy less than E.

On the other hand, for any observer whose state of motion is changing, they see a change in the energy of light they are passing through, as their state of motion changes.

I am really confused - I don't think any of this is controversial. If we just replace light with bullets, nobody disagrees; light is not fundamentally different. And for bullets you can certainly make an analogy between an accelerated observer seeing bullets lose KE, and bullets fired up gravity well lose KE. Specifically imagine a long rocket accelerating, bullets fired from its back to front. Someone at the front of the rocket will find tham less energetic than when they were fired at the bottom. You can explain this as accelerating toward the bullets or a fictitious gravity field. Now place same rocket on the surface of a planet. Bullets fired from the bottem will be less energetic at the top. You can say the lost energy going up a gravity well or that the rocket is accelerating upwards (to maintain static position in gravity well). Principle of equivalence says the situations are equivalent.

 

[atyy]

For any inertial observer there is no change in energy after the light is emitted. But two different observers disagree on energy of emitted light. Emitter thinks they emitted white light, total energy E, receding observer thinks they emitted red light of energy less than E.

On the other hand, for any observer whose state of motion is changing, they see a change in the energy of light they are passing through, as their state of motion changes.

I am really confused - I don't think any of this is controversial. If we just replace light with bullets, nobody disagrees; light is not fundamentally different. And for bullets you can certainly make an analogy between an accelerated observer seeing bullets lose KE, and bullets fired up gravity well lose KE. Specifically imagine a long rocket accelerating, bullets fired from its back to front. Someone at the front of the rocket will find tham less energetic than when they were fired at the bottom. You can explain this as accelerating toward the bullets or a fictitious gravity field. Now place same rocket on the surface of a planet. Bullets fired from the bottem will be less energetic at the top. You can say the lost energy going up a gravity well or that the rocket is accelerating upwards (to maintain static position in gravity well). Principle of equivalence says the situations are equivalent.

No, I am confused. What's amazing to me is that the energy argument is not an equivalence principle argument. To me it is an argument from conservation of energy and semiclassical "old" quantum physics argument. From QM we have E=hf. From classical gravity, we have dE=g.dh, which implies df. No relativity even (all in one Galilean frame)! So somehow classical gravity knows about QM? (Yes, but how?)

The point of view from GR is just red shifts are due to light going along geodesics, and the local acceleration of the observer, and works even in cases where there is no conserved energy.

 

[PAllen]

No, I am confused. What's amazing to me is that the energy argument is not an equivalence principle argument. To me it is an argument from conservation of energy and semiclassical "old" quantum physics argument. From QM we have E=hf. From classical gravity, we have dE=g.dh, which implies df. No relativity even (all in one Galilean frame)! Some somehow classical gravity knows about QM? (Yes, but how?)

The point of view from GR is just red shifts are due to light going along geodesics, and the local acceleration of the observer, and works even in cases where there is no conserved energy.

Harrylin's link earlier show's how Einstein derived that different frames (similarly changing frames) see light energy and frequency proportionally changed, purely from SR and Maxwell. To be successful, QED had to incorporated SR + Maxwell in the classical limit; so it did, as a result of which E=hf can be said to follow.

My meta point is that there were features 'hidden' in classical theories that became clearer in the carry over to quanum theories. The classical theories didn't 'know about' quantum theories, instead the quantum theory clarified the classical theory.

An exampel is radiation in SR + Maxwell can be a very subtle issue, as radiation is non-local in this framework. In QED many of these tricky cases are very straightforward because radiation in QED is locally defined.

 

[harrylin]

No, I am confused. What's amazing to me is that the energy argument is not an equivalence principle argument. To me it is an argument from conservation of energy and semiclassical "old" quantum physics argument. From QM we have E=hf. From classical gravity, we have dE=g.dh, which implies df. No relativity even (all in one Galilean frame)! Some somehow classical gravity knows about QM?
[..]

E/E0=f/f0 => E ~ f.
Thus I would say, QM knows SR!

Edit: Oops I was delayed in replying, meanwhile PAllen said it all.

 

[atyy]

But the classical theories are complete in themselves. If we want to bring in QM, then we could just as well say QED is a low energy approximation to a non-relativistic theory.

Anyway, going back to Harrylin's point that purely clasically E and f transform the same way, does that necessarily mean that E~f, purely classically?

 

[PAllen]

But the classical theories are complete in themselves. If we want to bring in QM, then we could just as well say QED is a low energy approximation to a non-relativistic theory.

Anyway, going back to Harrylin's point that purely clasically E and f transform the same way, does that necessarily mean that E~f, purely classically?

It seems to. Assume E=g(f) for some arbitrary g, and that by SR, E0=g(f0). We also have E/E0 = f/ f0 = expression of v and c = constant relative to E and f. Call it k. Then E0=kE, f0=kf, then:

kE=g(k f)

k g(f) = g (k f)

This implies g'(f) = g'(kf) for all f (g' as derivative). Then g'(f)=g'(f/k)=g'(f/k^2)... If g continuous, the g'(f) = g'(0). Thus g' constant, thus g(f) = c f. QED (not quantum electrodynamics).