Rotating a laser beam faster than light?

[robbie7730]

Again I ask why do you think that information has passed faster than light? Saying "who knows?" is not an answer.


According to the frame that measures their speeds as being 0.60c, the two objects separate at a rate of 1.2c. So? Still nothing is moving faster than light and certainly no information has flowed faster than light.

well they would be moving away from each other faster than light, say they were launced toward each other, they`d expect to pass at 1.2c

but that would`t happen and they`d only pass each other at 1c when looking at each other

and also look where you highlighted my comment in red,,,, i have ``does`t`` but you obviously did`t see the T

 

[robbie7730]

People who have actually studied the subject, that's who.


No, that's clearly not true. if object A moves in one direction at 60% c (relative to the starting point) and object B moves in the opposite direction at 60%c (again relative to the starting point) then the speed of each, relative to the other, is about 88% the speed of light:
\frac{0.6c+ 0.6c}{1+ \frac{(0.6c)(0.6c)}{c^2}}= .88 c
nowhere near "faster than light". If you do not understand that calculation, then I recommend that you actually learn about physics before you talk about it.


Because, people have done experiments with things moving in space. And "relative to what" is the whole point! A single object will have many different speeds relative to different frames of reference. The basic concept, based on experiment, is that the "laws of physics" must be true in any frame of reference and that leads to relativity.

Now my question is, if you really believe that no one can "know" anything, and that science is meaningless, as you imply, why in the world are you looking at a board like this one?

Again i have to say about ``maybe i should learn about physics before i talk about it``
How much about physics should someone learn before they are qualified to talk about it?

Lucky people down through history did talk about stuff they did`d understand.


If one object does actually head in one direction at 0.6c, how long will it take to reach a light year distance, and will it be a shorter time for the occupants than the observer, and if so why, and can we be certain of the answer based on vessels traveling at 0.00005c with an atomic clock on it.

I think I am just as entitled to ask as a person who see`s himself as more qualified and see`s others as lesser entitled until they ``learn about physics`` . It did`t even say learn more about physics,,,

Anyway, it was interesting

 

[Doc Al]

well they would be moving away from each other faster than light, say they were launced toward each other, they`d expect to pass at 1.2c

Again, according to the frame that sees them traveling at 0.6c towards each other, they will close at a rate of 1.2c. They do not travel at 1.2c with respect to each other.

but that would`t happen and they`d only pass each other at 1c when looking at each other

Their relative speed is 0.88c, as explained by Halls, meaning that they each see the other passing by at 0.88c.

and also look where you highlighted my comment in red,,,, i have ``does`t`` but you obviously did`t see the T

Ah...You meant doesn't. (You might want to pay more attention to punctuation and spelling if you want to be understood.)

 

[robbie7730]

yes 0.6 pass each other at 0.88

i meant if they both at 1c they pass each other at 1c as observed from each other.

the spelling, well maybe your right there

im surprised at bein told i should learn about physics before commenting though, just because he thought i might not understand that formula, anyone has the right to
comment i would of thought. Is the title not physics help and maths help.

Maybe it should be changed to forum for physics experts and mathematicians only

 

[gonegahgah]

Thanks mgb.

Shuttle is in orbit at 300km with a period of 90mins gives a speed around 7.7Km/s
GPS satelites orbit at 20,000km but with a period of 12hours so around 4km/s

Cool. The following is true isn't it?
To get to a higher orbit when in orbit a ship or satellite accelerates in the direction it is orbiting. To get to a lower orbit a ship or satellite decelerates against the direction it is orbiting. So I would have thought that the higher up you were the higher your velocity.
Obviously not by the numbers you've worked out, but how does it work?
Could I also ask your forbearance and get you to kindly insert the math for me, if you would please?

A clock in orbit isn't weightless, it's simply falling just the same as if it was in a falling elevator. It's mass is the same.

If you weigh the clock on the shuttle it will show as zero weight therefore it is weightless, I'm pretty sure of that but correct me if not; but I certainly agree that it still has the same amount of mass no matter how much it weighs. Please assure me or correct me on the difference between weight and mass.

Because the front is always moving slower that will mean that during acceleration an object will continue to compress throughout its length getting shorter and shorter for the duration of the acceleration; doesn't it?

mgb, Is this correct also?

 

[mgb_phys]

The period of an orbit is given by Keplers third law (sorry can't get the tex to work out right - see wiki for nicer formatting)

(Period/2pi)^2 = radius^3 / G(M+m)
M is the mass of the planet and m is the mass of the orbiting body - which you can ignore for a tiny spaceship orbiting the Earth. Remember that radius is from the centre of the Earth - so you need to add 6400km to the altitude.
The speed is then simply the distance = 2 pi r, divided by the period.

The reason I made the point about weightless is that many people get confused about a body in orbit and believe that it is weightless because there is no gravity in space. If the shuttle was a large distance away from the solar system in inter-stellar space there would be no gravity and object would indeed be weightless. Objects that are in orbit (like the shuttle) are weightless because they are falling but they still receive (almost) the same force of gravity on them as they would on earth.

The accelrating rod getting shorter would depend on the speed of the accelrating force. If one end is being pushed faster than the speed of sound in the material then as long as it keeps up this speed then the rod will get shorter. You can see this in explosively deformed materials and things like shaped charges.

 

[russ_watters]

well they would be moving away from each other faster than light, say they were launced toward each other, they`d expect to pass at 1.2c

They might, but only if they don't know anything about Relativity. If they do know about Relativity, they would not expect to pass each other at 1.2C.

I know that some say the object in orbit is falling, causing it to be weightless, but is it not more true to say the outward force away from the planet being orbited due to centrifugal force from traveling in a circular path is equal to the objects gravitational mass and so appears weightless.

No. The "centrifugal force" is often considered to be "ficticious" because it is a secondary reaction force and isn't always felt. If you are driving in a car, you feel the car pushing you around a corner. In the case of an orbit, you don't feel the space station pushing your toward earh. The only force is the centripetal force of Earth pulling you toward it.

It's a chicken or egg thing: You are moving in a circular path because the Earth is pulling at you, not the other way around.

You also mixed the concepts of force and mass there.

 

[russ_watters]

The accelrating rod getting shorter would depend on the speed of the accelrating force. If one end is being pushed faster than the speed of sound in the material then as long as it keeps up this speed then the rod will get shorter. You can see this in explosively deformed materials and things like shaped charges.

Expansion: Again, think of the rod as a spring. If you accelerate the rod, it will shorten the way a spring does until the force of the coiled spring equals the force of the acceleration and the pressure wave of the start of the acceleration propagates through it. Then it will continue to accelerate as a shortened rod.

If you try to accelerate a rod by hitting it with something going faster than the speed of sound in the rod, it will push through its own shock wave and permanently deform/be destroyed and the rules change a bit. In other words, there is nothing preventing a rod from moving faster than its own speed of sound, but one part of a rod cannot be moving at the speed of sound while an adjacent piece is stationary, without damaging it.

 

[pallidin]

If you try to accelerate a rod by hitting it with something going faster than the speed of sound in the rod, it will push through its own shock wave and permanently deform/be destroyed and the rules change a bit. In other words, there is nothing preventing a rod from moving faster than its own speed of sound, but one part of a rod cannot be moving at the speed of sound while an adjacent piece is stationary, without damaging it.

Russ, that reminds me of what I read on some legitimate military web-site. Sorry, I don't recall the link.
Anyway, it involves the use of a flat, circular tungsten disk placed in some type of barrel that contains a high explosive.
When the explosive goes-off, the force is so great that the tungsten disk is deformed(conveniently into a conical shape) and expelled through the barrel at over 7,000 mph.

Not sure if this was a production or experimental weapon at the time of that writing but I think it was experimental.

 

[mgb_phys]

It may have just been a high pressure experiment to create an unusual state of the metal.
You can make industrial parts by explosive forming, it's quite a good technique for small production runs where a high power press would be too expensive.
Basically you lay a piece of metal on top of a concrete or metal former shape, put a bit of C4 on top and surround the whole thing with water - then retire to a safe distance!

 

[madmike159]

I havn't read all of this post because its too long, so I'm sorry if some one has already asked this.
If you shon a laser on to the inside wall of a circle 1 light second away, it would have a circumference of about 60 light years. If your laser emmiter ratated at 1 rpm then would the laser light start to bend into a sprial because the point at the end would not be able to move as fast?

 

[Fredrik]

I havn't read all of this post because its too long, so I'm sorry if some one has already asked this.
If you shon a laser on to the inside wall of a circle 1 light year away, it would have a circumference of about 60 light years. If your laser emmiter ratated at 1 rpm then would the laser light start to bend into a sprial because the point at the end would not be able to move as fast?

The dot would move at about a light-year per second, but it wouldn't be at the location on the wall that the laser is aimed at. The shape of the beam would be curved and the dot would lag behind the laser. It would be a crazy spiral since the laser has turned over 500 thousand times when the dot starts moving.

 

[robbie7730]

i think the circumference would be 6.28 light seconds, or 1.8 million km`s roughly. I`d say the photons would form a spiral, they still travel in a straight line but i`d say they form a spiral as they radiate out from the spinning laser.

 

[robbie7730]

1 light year away would be just over 6 light years circumference

 

[mgb_phys]

I`d say the photons would form a spiral,

To be more precise - a line joining the position of all the photons at a point in time would form a spiral. It's important to remember that individual photons travel in a straight line (as robbie said)

 

[robbie7730]

To be more precise - a line joining the position of all the photons at a point in time would form a spiral. It's important to remember that individual photons travel in a straight line (as robbie said)

yes no doubt that's the more precise reality of it

 

[gonegahgah]

The reason I made the point about weightless is that many people get confused about a body in orbit and believe that it is weightless because there is no gravity in space. If the shuttle was a large distance away from the solar system in inter-stellar space there would be no gravity and object would indeed be weightless. Objects that are in orbit (like the shuttle) are weightless because they are falling but they still receive (almost) the same force of gravity on them as they would on earth.

Thanks mgb. I'm in a rush this morning so I'll have to study what you've written for me later. I just have a quick question.

I understand what you are saying that we can have weightless in a gravity field (falling) and weightless in the absense of gravity. That's ok. No problems with that.

Acceleration is supposed to be equivalent to gravity is it not? Does this apply in respect to time dilation as well? If it does then does the equivalent amount of acceleration produce the equivalent amount of time dilation as an equivalent amount of gravitation?

 

[Fredrik]

Acceleration is supposed to be equivalent to gravity is it not? Does this apply in respect to time dilation as well? If it does then does the equivalent amount of acceleration produce the equivalent amount of time dilation as an equivalent amount of gravitation?

Short answer: Yes.

Longer anwer: I think you should think of it the other way round. We can measure the acceleration by measuring the relative ticking rates of two clocks attached to opposite ends of a solid object. This would be a perfectly valid way to design an accelerometer.

Also, keep in mind that gravitational time dilation (which is equivalent to time dilation due to acceleration in flat spacetime) has an entirely different cause than the time dilation that's relevant when two clocks are moving with constant but different velocities in flat spacetime. In the latter case, which is the one that's easier to understand, both world lines are straight, but there's still time dilation due to the fact that different observers don't agree about which events are simultaneous. In the case of time dilation due to acceleration (either caused by gravity or by something else), the cause of the time dilation is that the world lines of the two clocks are curved by different amounts.

 

[gonegahgah]

Also, keep in mind that gravitational time dilation (which is equivalent to time dilation due to acceleration in flat spacetime) has an entirely different cause than the time dilation that's relevant when two clocks are moving with constant but different velocities in flat spacetime. In the latter case, which is the one that's easier to understand, both world lines are straight, but there's still time dilation due to the fact that different observers don't agree about which events are simultaneous. In the case of time dilation due to acceleration (either caused by gravity or by something else), the cause of the time dilation is that the world lines of the two clocks are curved by different amounts.

Thanks for jumping ahead on a question I was wondering Fredrik comparing relative speed time dilation (SR) and accelerated or gravitational time dilation (GR). So they are completely unrelated and can be treated separately?

SR time dilation produces a slowing of time as does GR time dilation. That is:
- if something is in greater gravity it will animate slower by our time.
- if something is under greater acceleration it will animate slower by our time.
- if something is going faster then it will animate slower by our time.
And this is accumulative isn't it?
ie. if you combine gravity with different relative velocity you sum their time dilation amounts.

 

[mgb_phys]

Thanks for jumping ahead on a question I was wondering Fredrik comparing relative speed time dilation (SR) and accelerated or gravitational time dilation (GR). So they are completely unrelated and can be treated separately?

Correct - see the above calcs for a shuttle and gps clock.

if something is in greater gravity it will animate slower by our time.
if something is under greater acceleration it will animate slower by our time.
if something is going faster then it will animate slower by our time.

Yes

ie. if you combine gravity with different relative velocity you sum their time dilation amounts.

Yes, that's why a shuttle clock runs slow and a gps clock runs fast overall.

 

[gonegahgah]

Thanks mgb.

So if you are weightless in deep space and someone is accelerating away from you they will have a component of animating slower than you added due to GR rules and anyone near a planet will have a component of animating slower than you added due to GR rules; ignoring the SR component for the moment.

That gives us an added GR component of:

in very deep space & steady (none) | very deep space & accelerating (slower)
----------------------------------------------------------------------------
near planet not accelerating (slower) |

I've left the fourth quadrant empty.
Let's say someone is in a car and they accelerate along the road will this add a GR component of slow down to how fast we see them animating on top of that due to the gravity?
Or are they mutually exclusive and only the greater of the two prevails?
Or does gravity prevail always?

 

[madmike159]

If you shon a laser on to the inside wall of a circle 1 light second away, it would have a circumference of about 60 light years.

Sorry no idea where 60 LY came from.

 

[gonegahgah]

mgb, if you don't know the answer are there any people you know that you find out for me.

There seems to be another thread on relativity and acceleration at https://www.physicsforums.com/showthread.php?t=273652".
Does this invalidate anything that we have here so far?

 

[JesseM]

So if you are weightless in deep space and someone is accelerating away from you they will have a component of animating slower than you added due to GR rules and anyone near a planet will have a component of animating slower than you added due to GR rules; ignoring the SR component for the moment.

GR is not needed to deal with acceleration in deep space, it's only needed for situations involving gravity. SR can deal just fine with an accelerating object of negligible mass in deep space.

 

[gonegahgah]

Thanks Jesse. That was what you said in the other thread. So are you saying that mgb was incorrect about acceleration being equivalent to gravity; specifically with respect to the effect upon time dilation. Are they non-equivalent in this specific matter?

 

[Jonathan Scott]

Thanks Jesse. That was what you said in the other thread. So are you saying that mgb was incorrect about acceleration being equivalent to gravity; specifically with respect to the effect upon time dilation. Are they non-equivalent in this specific matter?

I haven't been following this thread, but locally the effect of acceleration and the equivalent gravitational potential gradient on clocks is identical, so for example a clock at the front of an accelerating spaceship appears to run slightly faster than one at the back. This can be calculated by Special Relativity.

Also, if you consider something being accelerated in a rotating ring (like an Arthur C Clarke space-station) I think (if I remember correctly) that you will find that if you analyze the difference in time dilation between two different levels by Special Relativity given the velocity difference, you get the same result as if you analyze the difference based on the illusion of gravity seen in the frame of an observer in the ring.

 

[Fredrik]

Thanks for jumping ahead on a question I was wondering Fredrik comparing relative speed time dilation (SR) and accelerated or gravitational time dilation (GR). So they are completely unrelated and can be treated separately?

I wouldn't say that they are unrelated. The standard SR time dilation describes the difference between clocks on straight (but not parallel) world lines. The time dilation due to acceleration describes the difference between clocks on different hyperbolic world lines. The different shapes of the hyperbolic world lines are often caused by the clocks being attached to opposite ends of the same solid object.

Also, I consider the time dilation due to acceleration to be a part of SR. (It's of course relevant in GR too, but not just in GR).

The scenarios you're describing are comparing a clock on a straight world line with a clock on a hyperbolic world line. This is just a small complication added to the standard scenario, because all we have to do is to compare the straight world line to the tangents of the hyperbolic world line, and those are just a bunch of straight lines. So we can still use the standard time dilation formula.

SR time dilation produces a slowing of time as does GR time dilation. That is:
- if something is in greater gravity it will animate slower by our time.
- if something is under greater acceleration it will animate slower by our time.
- if something is going faster then it will animate slower by our time.
And this is accumulative isn't it?
ie. if you combine gravity with different relative velocity you sum their time dilation amounts.

This looks correct to me, but it's possible that I have overlooked something.

So if you are weightless in deep space and someone is accelerating away from you they will have a component of animating slower than you added due to GR rules

GR has nothing to do with it, and if you're only interested in how much slower the other clock ticks relative to yours "right now", you can completely ignore the acceleration. Just compare your world line to the tangent of the other clock's world line by using the standard time dilation formula.

and anyone near a planet will have a component of animating slower than you added due to GR rules;

If you both use rockets to stay at a constant distance from the planet, then yes.

Let's say someone is in a car and they accelerate along the road will this add a GR component of slow down to how fast we see them animating on top of that due to the gravity?

Only if you're in the car too but behind the driver (and I still consider this to be SR, not GR). If you're standing on the side of the road, you can compare your world line to the tangents of the car's world line, and that means that you're just going to be using the standard formula for time dilation.

 

[gonegahgah]

So in other words, if you are steady in deep space and something accelerates relative to you it will appear to animate slower; but if you are steady in gravity and something accelerates relative to you - on the same world line - it will not animate slower than you?

So in deep space we apply the accelerating rule but in gravity we discard the accelerating rule completely and instead substitute it with the world line rule?

 

[gonegahgah]

Also, if you consider something being accelerated in a rotating ring (like an Arthur C Clarke space-station) I think (if I remember correctly) that you will find that if you analyze the difference in time dilation between two different levels by Special Relativity given the velocity difference, you get the same result as if you analyze the difference based on the illusion of gravity seen in the frame of an observer in the ring.

That's quite interesting Jonathan. Will have to look at that idea.

 

[Fredrik]

So in other words, if you are steady in deep space and something accelerates relative to you it will appear to animate slower; but if you are steady in gravity and something accelerates relative to you - on the same world line - it will not animate slower than you?

If you're "steady in gravity" (which I interpret as being held at a constant altitude above a fixed point on the surface of a non-rotating spherical mass) and something accelerates relative to you, then it's not on the same world line. Don't forget that a world line is a curve in spacetime, not in space, so if two objects move on the same paths in space, but with different velocities, their world lines are different.

If the other clock is accelerating relative to you but staying at your altitude the whole time, it should speed up, not slow down, because its world line is closer to a geodesic (free fall) than yours. (A higher "sideways" velocity will bring it closer to being in orbit. An orbit is a free fall that misses the Earth but fails to escape it. The world line of an object in free fall is a geodesic. A geodesic is a curve of maximum proper time. So a clock with a higher "sideways" velocity should tick faster).

For example, suppose that you're standing on the ground of a non-rotating planet right next to a train track that's been built along a great circle around the planet, and that you have two synchronized clocks. You put one of them on the train and send it off to go around the planet. When it comes back it will be ahead of yours, unless of course its speed around the planet was much faster than the speed needed for a low altitude orbit. If it goes fast enough, it will behind yours when it gets back.

So in deep space we apply the accelerating rule but in gravity we discard the accelerating rule completely and instead substitute it with the world line rule?

Actually there's only one rule: What a clock measures is the integral of \sqrt{-g_{\mu\nu}dx^\mu dx^\nu} along the curve in spacetime that represents its motion. In other words, it measures the proper time of its world line. (In an inertial frame in 1+1-dimensional SR, the expression above can be simplified to \sqrt{dt^2-dx^2}).

The "standard time dilation in SR", "time dilation due to acceleration in flat spacetime" and "time dilation due to curvature of spacetime" are all special cases. The standard time dilation formula compares the proper times of two straight world lines. What I've been calling "time dilation due to acceleration" compares the proper times of two hyperbolic world lines associated with opposite ends of a solid object. Gravitational time dilation is also usually about comparing the world lines on opposite ends of a solid object (e.g. two different floors of the same building), but in curved spacetime.

In complicated situations, it's pointless to try to use the rules that apply to special cases separately. You're just going to have to calculate the proper time along each clock's world line.

 

[MeJennifer]

For example, suppose that you're standing on the ground of a non-rotating planet right next to a train track that's been built along a great circle around the planet, and that you have two synchronized clocks. You put one of them on the train and send it off to go around the planet. When it comes back it will be ahead of yours, unless of course its speed around the planet was much faster than the speed needed for a low altitude orbit. If it goes fast enough, it will behind yours when it gets back.

I question the correctness of the phrase I marked in red.

 

[Austin0]

I haven't been following this thread, but locally the effect of acceleration and the equivalent gravitational potential gradient on clocks is identical, so for example a clock at the front of an accelerating spaceship appears to run slightly faster than one at the back. This can be calculated by Special Relativity.
___________________________________________________________________-

Hi I am a little confused. I had thought that the desynchronization of clocks was due to clocks running slightly faster in the BACK due to acceleration?. Is this incorrect?

Thanks

 

[JesseM]

Thanks Jesse. That was what you said in the other thread. So are you saying that mgb was incorrect about acceleration being equivalent to gravity; specifically with respect to the effect upon time dilation. Are they non-equivalent in this specific matter?

According to the equivalence principle, if you're in a room in a very small region of spacetime that's sitting in a gravitational field, physics will look exactly the same as it does in an accelerating room in empty space. If both the floor and the ceiling of the room are at a fixed height in the gravitational field, I believe this would be equivalent to the floor and the ceiling in empty space undergoing Born rigid acceleration, in which the distance from one to the other in each one's instantaneous inertial rest frame at any given moment remains constant from one moment to another. This means that in a given inertial frame, it will appear that the floor and ceiling are accelerating at different rates (each is experiencing a constant G-force, but the G-force experienced by the floor is slightly larger)--their worldlines will look like the hyperbolae shown in the second diagram on this page (and note the paragraph immediately below that diagram which says 'We can imagine a flotilla of spaceships, each remaining at a fixed value of s by accelerating at 1/s. In principle, these ships could be physically connected together by ladders, allowing passengers to move between them. Although each ship would have a different proper acceleration, the spacing between them would remain constant as far as each of them was concerned.'). Because the velocity of the floor and ceiling are different at any given moment in this inertial frame, the time dilation factor of clocks at the floor and ceiling will be different, and this will cause the one on the floor to run slower by the same amount as the clock on the floor in the room that's at a fixed height in a gravitational field (this is my understanding anyway, I haven't actually done the math, but see pervect's post #75 on this thread).

 

[Alfi]

To be more precise - a line joining the position of all the photons at a point in time would form a spiral. It's important to remember that individual photons travel in a straight line (as robbie said)

does this imply that there is a limit to the rotational speed of the laser shooting it's photons one at a time based on the frequency. Even with our fastest ejection of individual photons,
If the laser was spun fast enough, would there be blank spots on the screen light years away?
( photons would hit further than one wavelength apart)

 

[Fredrik]

Hi I am a little confused. I had thought that the desynchronization of clocks was due to clocks running slightly faster in the BACK due to acceleration?. Is this incorrect?

Thanks

Yes, it's incorrect. In the original rest frame, the rocket keeps getting shorter due to Lorentz contraction. This means that the front and rear can't have the same acceleration profile. The rear must accelerate faster than the front for the rocket to get shorter. What a clock measures is the integral of \sqrt{dt^2-dx^2} along its world line, and since the rear is accelerating faster, the integral along its world line gets a bigger contribution from the dx² term. Because of the minus sign in front of it, that makes the integral smalller. That's why the clock in the rear will be behind the clock in the front.

 

[mgb_phys]

does this imply that there is a limit to the rotational speed of the laser shooting it's photons one at a time based on the frequency. Even with our fastest ejection of individual photons,
If the laser was spun fast enough, would there be blank spots on the screen light years away?
( photons would hit further than one wavelength apart)

There are always gaps since the photons come out individually, it's just that there are so many of them that it appears as a constant beam. As you sweep the laser over larger distances the gaps between them become larger
You can easily work out how many photons a laser emits per second from the power (the number of J/s ) and the energy of a photon of that wavelenght ( E = hc/wavelength)

 

[gonegahgah]

mgb, I wasn't aware there was a relationship between wavelength and the number of photons. Don't get me wrong; I think this whole sweeping laser notion in no way represents faster than speed motion; but I was just curious about the inference I got from what you posted about a relationship. Are you saying one wavelength is equivalent to one photon? Could you explain this further for me, please?

 

[JesseM]

mgb, I wasn't aware there was a relationship between wavelength and the number of photons.

A single photon can have any wavelength, but there is a relationship between a photon's energy and its wavelength (for an individual photon, relation[/url] says E=hf where f is the frequency and h is Planck's constant, and f is related to the wavelength \lambda by f = c/\lambda so E = hc/\lambda), so if you know the power of a light wave hitting a surface (energy/time) and you know its wavelength, that tells you the number of photons hitting the surface per unit time.

 

[gonegahgah]

So ergo one frequency (& wavelength) is implied by that to correspond to a single photon.

Someone is supposed to have taken a photo of light.
The photo of a 'photon' that I saw comprised multiple wavelengths in a single 'photon' not just a single frequency (or wavelength).

 

[JesseM]

So ergo one frequency (& wavelength) is implied by that to correspond to a single photon.

I don't understand what you mean by this, or how it's supposed to follow from what I said. A single photon can travel a distance that corresponds to multiple wavelengths, do you disagree? Likewise, if multiple photons hit a surface, the distance between the positions where they hit can be much larger or smaller than each photon's wavelength.

 

[gonegahgah]

My apologies. I should have worded that better. My question isn't about the intensity, nor arrival timing of photons, nor how far photons can travel.

You said that for an "individual photon ... E=h/f"
I assumed you meant E(individual photon) = h / f. Is that correct?
If that is correct it would imply that one cycle of light through a point (or a wavelength) corresponds to a self-contained single photon.

 

[jtbell]

I assumed you meant E(individual photon) = h / f. Is that correct?

That is correct.

If that is correct it would imply that one cycle of light through a point (or a wavelength) corresponds to a self-contained single photon.

No, it doesn't. Why would you think it does?

 

[Austin0]

Yes, it's incorrect. In the original rest frame, the rocket keeps getting shorter due to Lorentz contraction. This means that the front and rear can't have the same acceleration profile. The rear must accelerate faster than the front for the rocket to get shorter. What a clock measures is the integral of \sqrt{dt^2-dx^2} along its world line, and since the rear is accelerating faster, the integral along its world line gets a bigger contribution from the dx² term. Because of the minus sign in front of it, that makes the integral smalller. That's why the clock in the rear will be behind the clock in the front.

Fredrik --thanks for your responce but I am still confused.
As I understand it, the desynchronuzation of clocks is responcible for the invariance of the measured speed of light? The only way this seems to make sense is if the clocks at the rear are running ahead of the clocks in front. Ie: showing a later time.
This seems to mean that if you have distributed , synchronized clocks in a rest frame and then accelerate the frame , the clock in the rear would end up running ahead of one in the front. Does this make sense or am I seriously missing something??

 

[JesseM]

You said that for an "individual photon ... E=h/f"
I assumed you meant E(individual photon) = h / f. Is that correct?
If that is correct it would imply that one cycle of light through a point (or a wavelength) corresponds to a self-contained single photon.

Why would it imply that? Can you explain your reasoning?

 

[Fredrik]

As I understand it, the desynchronuzation of clocks is responcible for the invariance of the measured speed of light?

I don't know if it works that way, i.e. if you can prove the invariance using only what we know about desynchronization of clocks attached to the same solid object. Maybe you can, but it seems to be an unnecessarily complicated way to think about it. I think of SR as a physical theory that consists of a mathematical model (Minkowski space) and a set of postulates that identifies things in the model with things in the real world. The metric can be used to derive anything about the model, so I'd rather say that the Minkowski metric is the cause of both desynchronization and invariance.

Anyway, Einstein's second "postulate" says that the speed of light is the same in all inertial frames. It's not the same in arbitrary coordinate systems.

The only way this seems to make sense is if the clocks at the rear are running ahead of the clocks in front. Ie: showing a later time.

I don't see why.

This seems to mean that if you have distributed , synchronized clocks in a rest frame and then accelerate the frame , the clock in the rear would end up running ahead of one in the front. Does this make sense or am I seriously missing something??

If the red "this" refers to what I said, then the red "ahead" should be "behind". Anyway, if you take a grid of rulers and synchronized clocks and accelerate them by starting a tiny rocket attached to each clock when they all show the same time, and then shut them off when they have reached velocity v in the original rest frame, then they will be doing inertial motion again, but they won't be synchronized, so the numbers you can read off the grid won't agree with the coordinates assigned by an inertial frame.

 

[gonegahgah]

Why would it imply that? Can you explain your reasoning?

Well, you said that E=hc/λ.
I assume you were transfering that through (by the progression of the paragraph) as meaning E(individual photon) = hc/λ. So constant divided by wavelength.

I read your words and this to say that one photon = one wavelength. If the photon were multiple wavelengths joined end to end then the total energy of an individual photon would be (number of wavelengths) * hc / λ as the equation implies that it is per wavelength.

Maybe you could clarify it for me by showing how you would work out how many photons arrived given the energy and time.

 

[Austin0]

Expansion: Again, think of the rod as a spring. If you accelerate the rod, it will shorten the way a spring does until the force of the coiled spring equals the force of the acceleration and the pressure wave of the start of the acceleration propagates through it. Then it will continue to accelerate as a shortened rod.
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Hi One question.
Given that the rate of propagation [speed of sound] is very slow compared to intermolecular elastic reactions , wouldn't both the force and the compression move as a reciprocal occilation with no net compaction or actual motion at the point of acceleration until the whole system attained sufficient energy to begin moving as a body?

In the spring context; Wouldn't continuous acceleration result in continued expansion /compression of the spring with the force of acceleration never reaching equilibrium with the intermolecular tensile force in a state of maximum compression?

This is assuming that the magnitude of the force does not exceed the systems ability to distribute it fast enough.

This is a question I have mulled over before in other contexts so would appreciate any clarification.