# Why can light "push" but not be "pushed?"

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rigney: I suggest you look for some more reputable source, either an expert in the field or a respected science writer. You have found the ramblings of some random nut job who doesn't know what he's talking about. The internet is full of net jobs, crackpots, and charlatan, and you are very good at finding them.

You're right in one respect. I'm gullable enough to believe there may be a bit of truth in just about everything I read or hear. Tell me, is this guy on the level or just pounding salt? Since we both know that I'm not, may I ask if you are working at, or above his level of expertise?

http://www.vanderbilt.edu/AnS/physics/perakis/L9_00/

Edited by rigney

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My "guess" must have been wrong about flying into the projected beam absorbing the ships velocity... as a guess why there is no cumulative velocity. A case of "fuzzy" thinking. You are right. Once light is projected ahead, the ship will never catch up with it, let alone travel into it.

Which leaves the mystery as stated, i.e., why is the beam not boosted to 1&1/2 'c'? Clearly light can not be "pushed" faster than 'c', so what happens to the ship's velocity? I still don't know. But I'm working on it.

The "velocity relative to what?" question (in lieu of no "aether") can be answered by adding some props along the direction of travel. Say some space buoys had been placed along the way, brought to rest (zero velocity) relative to the approaching ship. Say they are placed every 186,000 miles, to make it easy. The ship's light beam passes one every second, while the ship, at 1/2 'c', passes one every two seconds.

After a minute the ship will have passed 30 buoys (having gone 5,580,000 miles) and its beam will have passed 60 of them (having gone 11,160,000 miles.) Then the distance between the ship and the far end of the beam is 5,580,000 miles, yet the light traveled 11,160,000 miles.

What am I missing here?

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Tell me, is this guy on the level or just pounding salt?

Too cutesy!!! And not historically correct. Read "rumor has it" as "I just made this story up".

What Einstein did do was try to envision what it would be like to "ride a beam of light" when he was 16. He didn't get all that far at this young age because he didn't have the necessary math or physics background. Ten years later he had the necessary tools and he had time to spare. He used this time to revisit his thoughts from a decade before. He found a novel and very simple way around one of the most vexing problems of the time, the inherent incompatibility of Newtonian mechanics and Maxwell's electrodynamics. Maxwell's equations strongly imply that the speed of light is the same to all observers. Most physicists were looking for ways around this implication (and around the conflict). Einstein simply took Maxwell's equations to heart. He accepted that the speed of light is the same to all observers and then determined the fall out from this assumption. This provided the answer to his thoughts from a decade before: You cannot ride a beam of light. That beam of light is speeding away from you at the speed of light. It doesn't matter how much faster you go. That beam of light is still going to go away from you at c. Always.

An object with non-zero mass must always move at subluminal speeds. Your mirror cannot go at the speed of light.

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The "velocity relative to what?" question (in lieu of no "aether") can be answered by adding some props along the direction of travel. Say some space buoys had been placed along the way, brought to rest (zero velocity) relative to the approaching ship. Say they are placed every 186,000 miles, to make it easy. The ship's light beam passes one every second, while the ship, at 1/2 'c', passes one every two seconds.

After a minute the ship will have passed 30 buoys (having gone 5,580,000 miles) and its beam will have passed 60 of them (having gone 11,160,000 miles.) Then the distance between the ship and the far end of the beam is 5,580,000 miles, yet the light traveled 11,160,000 miles.

What am I missing here?

If the buoys are at rest with respect to the ship, it doesn't pass by any of them.

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If the buoys are at rest with respect to the ship, it doesn't pass by any of them.

Right. I misspoke, which should have been obvious by the context of the ship and its light beam passing the buoys.

The buoys were meant to represent 'mileposts' on the 'road through space' traveled by the ship.

So, to be accurate, the ship which dropped the buoys, if it were going at half 'c' also would have programmed their retro-rockets to "brake", decreasing velocity relative to the ship until they each had lost 1/2 'c' of velocity.

OK? Now, again, "what am I missing" given that there are only 5,580,000 miles between the ship and the front end of the light beam after a minute of travel?

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Too cutesy!!! And not historically correct. Read "rumor has it" as "I just made this story up".

What Einstein did do was try to envision what it would be like to "ride a beam of light" when he was 16. He didn't get all that far at this young age because he didn't have the necessary math or physics background. Ten years later he had the necessary tools and he had time to spare. He used this time to revisit his thoughts from a decade before. He found a novel and very simple way around one of the most vexing problems of the time, the inherent incompatibility of Newtonian mechanics and Maxwell's electrodynamics. Maxwell's equations strongly imply that the speed of light is the same to all observers. Most physicists were looking for ways around this implication (and around the conflict). Einstein simply took Maxwell's equations to heart. He accepted that the speed of light is the same to all observers and then determined the fall out from this assumption. This provided the answer to his thoughts from a decade before: You cannot ride a beam of light. That beam of light is speeding away from you at the speed of light. It doesn't matter how much faster you go. That beam of light is still going to go away from you at c. Always.

An object with non-zero mass must always move at subluminal speeds. Your mirror cannot go at the speed of light.

As ignorant as I am concerning both math and physics, I only invisioned the lazer traveling at c. and shooting out a beam also at c It stands to reason that a massless particle such as a photon would likely be the only thing that could travel at c. But now that a (particle/wave?) what ever, with a sub-mass (tactons) are staring us in the face, how do we address the issue? I know it's only theory that they exist, but what if? Edited by rigney
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What if? As DH said above... Basically you are asking "what do the laws of physics say would happen if we violated the laws of physics?"

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What if? As DH said above... Basically you are asking "what do the laws of physics say would happen if we violated the laws of physics?"

What will happen if they are violated?
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What will happen if they are violated?

Nothing.

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Nothing.

Right! We just pick up sticks and go on with it.
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What will happen if they are violated?

What happens in a game of chess when someone moves a rook along a diagonal?

To be a bit less flippant, there are two possible answers to your question.

1.
2. It can't happen, so stop asking.
This is equivalent to asking what happens when a rook moves along a diagonal. It's an illegal move. It can't happen. There is however a problem with this answer: It assumes the laws of physics perfectly describe the universe. They don't. The rules of chess are made by humankind, so we know exactly what they are. The rules of the universe are fumblingly ferreted out by humankind. Our job in doing so isn't perfect or complete. Our laws of physics are best guesses based on evidence, logic, and math. Not perfect. So what if they're wrong? That leads to answer #2:

3. We haven't the foggiest idea what would happen, so stop asking.
Physicists have occasionally performed experiments where the outcome was "Whoa! That can't happen!" These experiments, if confirmed, have the effect of turning physics upside down. The Michelson-Morley experiment is one of the most famous of such experiments. Isaac Asimov: The most exciting phrase to hear in science, the one that heralds the most discoveries, is not "Eureka!" (I found it!) but "That's funny..."

In the case of your mirror, we have very, very good reason to think that your question falls into category #1: "It can't happen, so stop asking." Suppose that it could happen. This means everything we know is wrong. Very, very wrong. This makes your question fall into category #2: "We haven't the foggiest idea what would happen, so stop asking."

Edited by D H
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So now my OP questions are ignored and the thread has gone into off-topic chatter. I would call it "thread hijacking" but I don't like the label, having been accused of it so much myself.

Thanks DH for making it personal and then derailing it onto a sidetrack.

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So now my OP questions are ignored and the thread has gone into off-topic chatter. I would call it "thread hijacking" but I don't like the label, having been accused of it so much myself.

Thanks DH for making it personal and then derailing it onto a sidetrack.

I apologise owl, it was my fault. Your questions were fine and on subject. Mine were simply unmindful questions on the issues. DHs answers were offensive, but hopefully he will eventually get over his arrogance. I wonder, does he even play chess? Edited by rigney
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Right! We just pick up sticks and go on with it.

Nothing that happens will violate physical laws. If something does, then they weren't laws.

Better?

What would happen if physical laws were violated?

If there's a law that says your answer isn't so or is impossible, just consider the law violated.

Though, a more reasonable answer comes from considering the limits of what is physically possible, for example not considering something moving at c but rather what happens in the limit as v approaches c. This is what Einstein did in some of the examples.

But to get back on track... owl, what is missing is a basic understanding of special relativity. I apologize, as having an incomplete understanding of things pointed out by someone seems to be considered offensive around here. Why not start with a simple example of SR found elsewhere on the web and ask about what doesn't make sense to you, rather than devising your own complicated thought experiment that doesn't help you see how it works?

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OK? Now, again, "what am I missing" given that there are only 5,580,000 miles between the ship and the front end of the light beam after a minute of travel?

You are once again assuming that the universe is Newtonian is by thinking that time and distance are immutable. They are not. What you are missing is that time is subject to time dilation, length to length contraction.

In the frame of the ship, those buoys are not 186,000 miles apart. They are only 161080.725 miles apart. At the time the ship emits the beam, the ships captain will compute that the beam will travel 4,832,421.75 miles. (Note well: It's not 60*161080.725 miles = 9,664,843.5 miles. It's half that. In the captain's frame, that last buoy is moving toward the captain at 1/2 c.) The clocks on the ship and the clocks in the space harbor master's frame (the frame in which the buoys are at rest) aren't running at the same rates, either. At the point that the ship has passed 30 buoys it's clock will not show 30 seconds. It will instead show 25.9807 seconds. From the captain's perspective, the beam of light is moving at 4,832,421.75 miles / 25.9807 seconds = 186,000 miles/second. (I've used your value for the speed of light here.)

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Right. I misspoke, which should have been obvious by the context of the ship and its light beam passing the buoys.

The buoys were meant to represent 'mileposts' on the 'road through space' traveled by the ship.

So, to be accurate, the ship which dropped the buoys, if it were going at half 'c' also would have programmed their retro-rockets to "brake", decreasing velocity relative to the ship until they each had lost 1/2 'c' of velocity.

OK? Now, again, "what am I missing" given that there are only 5,580,000 miles between the ship and the front end of the light beam after a minute of travel?

What frame is used to measure the distance between the buoys? Relative to the rest frame of the buoys the distance will be contracted in the spaceship frame by a factor of 0.866 from traveling at c/2. And that minute of travel — is it a minute as measured in the buoy frame? From your setup it's not clear.

(edit: xpost with D H)

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What frame is used to measure the distance between the buoys? Relative to the rest frame of the buoys the distance will be contracted in the spaceship frame by a factor of 0.866 from traveling at c/2. And that minute of travel — is it a minute as measured in the buoy frame? From your setup it's not clear.

(edit: xpost with D H)

I was obviously using the earth standard mile as derived from earth's frame. That is what "a mile" means. So the ship that dropped the buoys would have used that standard for the distances between buoys, once 'fixed' in place.

Of course the ship would have compensated for its 1/2 'c' velocity and the resulting .866 apparent contraction of distance between drops.

I was striving to demonstrate how the whole example would 'pan out' in old fashioned earth miles marked out over the course of the travel through space in question.

Another part of the photon "pushing" issue raised in the OP was raised again in post 4.

In all possible cases of light pushing or being pushed, the question for me still remains, "How is it that mass-less light acts like mass? Or, "how does mass-less momentum "push" at all with no 'substance' (so to speak) to impact whatever it pushes on?"

I redirect attention to the above and ask again.

Edited by owl
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You seem to be asking about a change in the rest mass of light, even though light can never be at rest by definition. Perhaps that's the source of your confusion?

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In all possible cases of light pushing or being pushed, the question for me still remains, "How is it that mass-less light acts like mass? Or, "how does mass-less momentum "push" at all with no 'substance' (so to speak) to impact whatever it pushes on?"

Because a massless particle, i.e. in this case photon, still has momentum. Classical physics is not completely correct and its failure is obvious in some circumstances but these are situations not encountered in everyday experience. The effects from moving fast, strong gravity and/or having massless particles are not explained by pre-1900 physics. The fact that EM radiation would have momentum, though, is part of classical theory. If you have a transfer of energy you are going to transfer momentum as well.

I was obviously using the earth standard mile as derived from earth's frame. That is what "a mile" means. So the ship that dropped the buoys would have used that standard for the distances between buoys, once 'fixed' in place.

Of course the ship would have compensated for its 1/2 'c' velocity and the resulting .866 apparent contraction of distance between drops.

I was striving to demonstrate how the whole example would 'pan out' in old fashioned earth miles marked out over the course of the travel through space in question.

The definition of the mile is not the issue, it was the frame in which the mile was measured. But this is now moot, as D H has explained the details by choosing one of the frames.

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Which leaves the mystery as stated, i.e., why is the beam not boosted to 1&1/2 'c'? Clearly light can not be "pushed" faster than 'c', so what happens to the ship's velocity? I still don't know. But I'm working on it.

The "velocity relative to what?" question (in lieu of no "aether") can be answered by adding some props along the direction of travel. Say some space buoys had been placed along the way, brought to rest (zero velocity) relative to the approaching ship. Say they are placed every 186,000 miles, to make it easy. The ship's light beam passes one every second, while the ship, at 1/2 'c', passes one every two seconds.

After a minute the ship will have passed 30 buoys (having gone 5,580,000 miles) and its beam will have passed 60 of them (having gone 11,160,000 miles.) Then the distance between the ship and the far end of the beam is 5,580,000 miles, yet the light traveled 11,160,000 miles.

What am I missing here?

Start view from Earth:

> Spaceship

- Lightray

o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o bouys

Finish view from Earth:

________________________ 30 buoys __________________________> Spaceship

______________________________________________________________________________________ 60 buoys ________________________- Lightray

o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o bouys

From an Earth view nothing happens to the ships velocity, it is moving in the trail behind the photons it emitts, getting further and further behind the front peak of the lightray. The mystery is how the crew on the ship can measure the lightray to move away from the ship with c when a measurement from Earth shows a speed difference of 1/2 c between the ship and the lightray. The answer from relativity is that the crew is in *distorted* space and time compared to observers on Earth, rulers and clocks on the spaceship are no longer calibrated to an equal scale as if they where on Earth.

(Of course from the crews point of view it is the observers on Earth that are in *distorted* space and time.)

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You seem to be asking about a change in the rest mass of light, even though light can never be at rest by definition. Perhaps that's the source of your confusion?

I know that light does not "rest" to be measured as resting mass and that its momentum acts like mass. My question is, how does the energy/momentum of photons "impact" or apply force to objects without a "substantial" agent or carrier of that force, i.e., without mass.

I really don't understand how momentum sans mass acts as a force "pushing" on things as in the OP examples.

Because a massless particle, i.e. in this case photon, still has momentum.

I know and have been referring to that fact.

Classical physics is not completely correct and its failure is obvious in some circumstances but these are situations not encountered in everyday experience. The effects from moving fast, strong gravity and/or having massless particles are not explained by pre-1900 physics. The fact that EM radiation would have momentum, though, is part of classical theory. If you have a transfer of energy you are going to transfer momentum as well.

Check. Also already known. I am not arguing for classical physics. See above reply to iNow for my continuing inquiry about how momentum transfers energy and force ("push") as in my examples.

I also already know that clocks slow down in rate of "timekeeping" as they move faster or experience stronger gravity and that lengths/distances appear contracted from higher velocity frames.

We need not go over all that again.

The definition of the mile is not the issue, it was the frame in which the mile was measured. But this is now moot, as D H has explained the details by choosing one of the frames.

It's my inquiry and I'll tell you what my issue is.

I am not interested in how the string of buoys *appears* to the moving ship. The ship would have used its SR manual and computers to adjust for its clock running slower than earth clocks and for the apparent length contraction (to .866/1 if you figure is correct) it experiences because of that while dropping the buoys.

But its mission was to place the buoys 186,000 earth-based miles apart... not length-contracted "miles" as "seen" from the ship's high velocity.

I have no argument with what DH said about that as per "in the frame of the ship.":

In the frame of the ship, those buoys are not 186,000 miles apart. They are only 161080.725 miles apart.

Also no argument with the rest of that post and all the math.

My thought experiment specified earth miles as the units between buoys. The question then remains regarding the distance between the ship and the end of the light beam (5,580,000 earth-frame miles) after a minute of travel.

I agree with Spyman here:

The answer from relativity is that the crew is in *distorted* space and time compared to observers on Earth, rulers and clocks on the spaceship are no longer calibrated to an equal scale as if they where on Earth.

(Of course from the crews point of view it is the observers on Earth that are in *distorted* space and time.)

My setup was meant to illustrate the example as seen from earth's frame of reference, assuming that the whole operation (both ships) were earth-based at launch and that their 1/2 'c' velocity was relative to earth... not about creating an alternative reality "as seen from the ships' frame of reference."... again, not disputing the observations as seen from the ships.

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I know that light does not "rest" to be measured as resting mass and that its momentum acts like mass. My question is, how does the energy/momentum of photons "impact" or apply force to objects without a "substantial" agent or carrier of that force, i.e., without mass.

I really don't understand how momentum sans mass acts as a force "pushing" on things as in the OP examples.

I know and have been referring to that fact.

Check. Also already known. I am not arguing for classical physics. See above reply to iNow for my continuing inquiry about how momentum transfers energy and force ("push") as in my examples.

A change in momentum is the same as a push. You can't have one without the other. The amount of push additionally depends on the time interval over which the change in momentum happened. If this is from a reflection, then the "how" can be viewed as that the photon is absorbed and re-emitted in the opposite direction. There's an E&M discussion as well, which involves applying the boundary conditions of how the fields have to behave when you go from one medium to the next. In either case, you have redirected the light, so there is a change in momentum. Thus, a force.

It's my inquiry and I'll tell you what my issue is.

I am not interested in how the string of buoys *appears* to the moving ship. The ship would have used its SR manual and computers to adjust for its clock running slower than earth clocks and for the apparent length contraction (to .866/1 if you figure is correct) it experiences because of that while dropping the buoys.

But its mission was to place the buoys 186,000 earth-based miles apart... not length-contracted "miles" as "seen" from the ship's high velocity.

I have no argument with what DH said about that as per "in the frame of the ship.":

Also no argument with the rest of that post and all the math.

My thought experiment specified earth miles as the units between buoys. The question then remains regarding the distance between the ship and the end of the light beam (5,580,000 earth-frame miles) after a minute of travel.

The point I was making is that any time you reference a distance, you must say in what frame it has been measured. You have specified the earth frame, so you see a difference of 5,580,000 miles between the light and the ship, because one moved at c and one moved at c/2.

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I have no argument with what DH said about that as per "in the frame of the ship.":

Also no argument with the rest of that post and all the math.

My thought experiment specified earth miles as the units between buoys. The question then remains regarding the distance between the ship and the end of the light beam (5,580,000 earth-frame miles) after a minute of travel.

What question? I laid it all out. What part did you not understand?

My setup was meant to illustrate the example as seen from earth's frame of reference, assuming that the whole operation (both ships) were earth-based at launch and that their 1/2 'c' velocity was relative to earth... not about creating an alternative reality "as seen from the ships' frame of reference."... again, not disputing the observations as seen from the ships.

Alternative reality? The spaceship POV is just as real as the Earth-based POV, and it is part and parcel of the same reality. It just isn't your Newtonian reality, which you apparently just cannot give up. You don't even know that you are making assumptions of a Newtonian universe in every post you make. You say you accept the math, but it's obvious that deep down, you don't.

You will not be able to understand these concepts until you can see that (1) you are assuming a Newtonian universe, and (2) you are willing to give up that point of view. It's not easy. Relativity is weird and can be very counterintuitive to our intuition.

Our physical intuition is at best Newtonian, and more often than not its even more primitive than that. Our physical intuition is based on what we see around us. For example, the natural state of a rock is obviously at rest. A rock might roll down a hill if someone or something gives it a shove, but it inevitably comes to a stop. This makes it hard for many to accept Newtonian mechanics, let alone something as foreign as relativity.

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The point I was making is that any time you reference a distance, you must say in what frame it has been measured. You have specified the earth frame, so you see a difference of 5,580,000 miles between the light and the ship, because one moved at c and one moved at c/2.

First, on the mechanics of light pushing on things...

say on a solar sail just to keep it simple: How does the energy of its momentum transfer force to the sail with no mass to push against the matter of the sail? The sail absorbs the light, no doubt, but how does the "push" manifest?

As to "distance":

I said that the distance between buoys must be 186,000 in earth miles, not contracted "miles" as measured from the speeding ship... requiring the adjustment to compensate as already discussed. Also I already specified that there is, as you said "a difference of 5,580,000 miles between the light and the ship, because one moved at c and one moved at c/2." I laid that out already. Do you have a point in repeating it all?

The question remains, why are there only 5,580,000 earth-frame miles between the ship and the far end of the light beam if the light traveled at 'c' ahead relative to the ship as well as relative to the buoys during that minute?

Edited by owl
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First, on the mechanics of light pushing on things...

say on a solar sail just to keep it simple: How does the energy of its momentum transfer force to the sail with no mass to push against the matter of the sail? The sail absorbs the light, no doubt, but how does the "push" manifest?

Newtonian mechanics implies that massless objects can't have momentum. You are once again showing your Newtonian biases. You need to drop this. If you cannot do that you will not understand.

The answer is simple: Massless objects can and do have momentum. Even though photons are massless, they do have momentum and energy.

As to "distance":

I said that the distance between buoys must be 186,000 in earth miles, not contracted "miles" as measured from the speeding ship... requiring the adjustment to compensate as already discussed. Also I already specified that there is, as you said "a difference of 5,580,000 miles between the light and the ship, because one moved at c and one moved at c/2." I laid that out already. Do you have a point in repeating it all?

The question remains, why are there only 5,580,000 earth-frame miles between the ship and the far end of the light beam if the light traveled at 'c' ahead relative to the ship as well as relative to the buoys during that minute?

I. Give. Up.

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