Relativity

[stevensfo]

As another example, consider a long jump rope held taut at both ends by two girls. If one girl shakes her end of the rope violently enough to send a wave down the rope to the other girl, the wave can jerk the other girl. The rope has not transported any mass, but it still carries momentum through its waving motion. In this way, waves can have no mass but still carry momentum.

That's a really good analogy. Thanks!


Steve

[odysseus2000]

As far as I know reflection & refraction at boundaries, whether conductors or not is normally done by considering wave behaviour. The calculations are reasonably complex, and rely on establishing the boundary conditions at the interface & applying Maxwell’s equations. There are many links on the net, often not entirely clear & these are two I came across that anyone seriously interested might find of interest, but they require some familiarity with mathematics, wave equations & Maxwell’s equations.

Reflection at at a dielectric boundary:

https://farside.ph.utexas.edu/teaching/ ... de104.html


Reflections from conductors (link seems to work despite not being all indicative of a link)

https://farside.ph.utexas.edu/teaching/ ... 0(i.e.%2C-,).,fraction%20of%20it%20is%20absorbed.

One important consideration is that in these derivations no consideration is given to light being composed of photons. The general belief is that in physics one sees light behaving as waves or as particles depending on the phenomena being studied; both indicative that out understanding of the phenomenon is limited & that we do not have a universal understanding or that no such universal understanding exists.

Regards,

Aside: Quite impressed by how well the chats did in explaining this subject.

[Watis]

Glad I'm not the only one confused by this stuff.

What if the clock is travelling away from me at three quarters the speed of light - and I start to move in the opposite direction at three quarters the speed of light?

Surely, relative to me, the clock is now travelling at one and a half times the speed of light?

Yet we're told that's not possible.

Watis

In relativity speeds don't add in that nice simple way.

See

https://galileo.phys.virginia.edu/classes/252/adding_vels.html

instead the addition of two speeds u and v is given by

( u + v ) / ( 1 + uv/c^2 )

If u = 0.75c and v = 0.75c then adding them gives. 1.5c / 1.5625 = 0.96c

Thank you for the reply to my question - and the link, which I've read but do not yet fully understand.

Watis

[odysseus2000]

In relativity speeds don't add in that nice simple way.

See

https://galileo.phys.virginia.edu/classes/252/adding_vels.html

instead the addition of two speeds u and v is given by

( u + v ) / ( 1 + uv/c^2 )

If u = 0.75c and v = 0.75c then adding them gives. 1.5c / 1.5625 = 0.96c

Thank you for the reply to my question - and the link, which I've read but do not yet fully understand.

Watis

Relativity is difficult & Einstein never received the Nobel Prize for it as it was too controversial. In many cases one has to be exact about where observers as otherwise this leads to many potential paradox that are only resolved by specifying where the observers are. This cartoon depiction video might help. One has to remember that there are never two different outcomes & this can help in resolving the paradox:

https://youtu.be/Xrqj88zQZJg?si=zzw69ZXNZiZCCqmz

Regards,

[stevensfo]

Relativity comes into play as well (QED incorporates relativity) in that it muddies the field with frames of reference. Because photons have no mass, they travel at the speed of light and so do not experience time.

Light not having mass has always puzzled me. So how do light sails work? i.e. the idea that probes can be propelled through space by light from the star. I know it's something to do with momentum, but how can you have momentum without mass? How is light bent by high gravity if it hasn't got mass?

Time for the 2nd aspirin!

Steve

Don't forget that the speed of light varies with the medium. That's how you get the effects of prisms and lenses. Solar wind is caused by the fact that space is not a void.

TJH

Thank goodness it's Friday!

How on earth can the speed of light slow down? If light hits a prism and slows down, then it must surely accelerate when it leaves the prism. How does it accelerate?

I liked the explanation of momentum vs mass. But surely even momentum loses energy over time. I once read about probes with solar sails that could sail with the sun's light, but after a while, would require lasers from the earth to keep it accelerating.

It seems to me that the light sail is a brilliant idea for sending out probes to other stars. Approach the other star and use the star to slow down?


Steve

[odysseus2000]

Don't forget that the speed of light varies with the medium. That's how you get the effects of prisms and lenses. Solar wind is caused by the fact that space is not a void.

TJH

Thank goodness it's Friday!

How on earth can the speed of light slow down? If light hits a prism and slows down, then it must surely accelerate when it leaves the prism. How does it accelerate?

I liked the explanation of momentum vs mass. But surely even momentum loses energy over time. I once read about probes with solar sails that could sail with the sun's light, but after a while, would require lasers from the earth to keep it accelerating.

It seems to me that the light sail is a brilliant idea for sending out probes to other stars. Approach the other star and use the star to slow down?


Steve

Light is, in the wave model, a set of perpendicular vectors, one is the Electric field, the other the Magnetic field.

When light goes through a medium, these two vectors interact with the medium with a macroscopic effect that light in a medium of refractive index n slows to (velocity of free space)/n and in some medium different wavelengths interact differently, so one gets a splitting of colour as in a prism. When light leaves the medium & goes into a vacuum it no longer has these interactions & travels at its full & only velocity that is believed to be a universal constant & which has never been shown experimentally to change within experimental uncertainties.

Under some conditions particles can travel faster in a medium than light & this leads to what is called Cherenkov radiation, when the faster than light in the medium particle interacts with the medium to create blue & shorter wavelength photons. Some talk of this as analogous to a sonic boom for the Cherenkov photons.

In a relativistic treatment a particle of zero mass has Energy = momentum; normally one puts the velocity of light, c, =1 since it simplifies calculations, from e=pc to e=p, if one has rest mass it has an additional term: E^2 = p^2 +.,m^2, against velocity of light =1 and different units are created. Most students hate this on first meeting & then having tried to show how easier it is to use units they know soon realise that units with c=1 are far easier to use.

So a photon of energy e has momentum e. It does not lose energy with distance or time, but for a sail, the photons from the sun spread out according to an inverse square; double the distance & the intensity is down by 4. To keep a propulsion going one has to increase the “wind” to compensate & this is done with a laser which is a directed source & not isotropic in to a 4xpi spherical emission as is the sun.

Regards,

[ursaminortaur]

Don't forget that the speed of light varies with the medium. That's how you get the effects of prisms and lenses. Solar wind is caused by the fact that space is not a void.

TJH

Thank goodness it's Friday!

How on earth can the speed of light slow down? If light hits a prism and slows down, then it must surely accelerate when it leaves the prism. How does it accelerate?

I liked the explanation of momentum vs mass. But surely even momentum loses energy over time. I once read about probes with solar sails that could sail with the sun's light, but after a while, would require lasers from the earth to keep it accelerating.

It seems to me that the light sail is a brilliant idea for sending out probes to other stars. Approach the other star and use the star to slow down?


Steve

The intensity of the light from the sun falls off as the square of the distance from the sun. Hence close to the sun the solar sails are receiving a large push but that declines as the solar sail gets further away. Note. That just means that the acceleration from solar photons eventually becomes almost zero it doesn't mean that the craft slows down.*

Lasers are therefore needed if you want to keep accelerating (since the lasers send out a concentrated beam of photons which spreads out very slowly such a beam can continue pushing the solar sail at far greater distances).

The same considerations would apply to using the destination star to slow down and hence you may need to get very close to the destination star to slow down sufficiently. Of course once you had setup a colony you could setup a laser in that system which could then be used both for sending ships back to our solar system and also to help in slowing ships coming into your new star system.


Also the force produced by light falling on a solar sail is pretty small so you would need really gigantic solar sails to accelerate any substantial payload - and those sails themselves need to be as light as possible which generally means they need to be extremely thin (which also means that they could easily be damaged by collisions with small particles). These considerations mean that at the moment the only potentially feasible project involving sending probes to other stars using solar sails envisages sending out a thousand extremely small probes (each probe being cubic centimetres in size and weighing a gram or so) on a fly-by mission to proxima centauri. Hopefully sending such a large number will mean that a few will reach the destination without having their solar sail and/or payload destroyed en route.

https://en.wikipedia.org/wiki/Breakthrough_Starshot



* I'm neglecting here any miniscule slowing due to the fact that space isn't a perfect vacuum.

[odysseus2000]

Most of the models for interstellar travel use one of the planets as a sling shot helper to gain velocity:

https://en.m.wikipedia.org/wiki/Gravity_assist

This can also be reversed by approaching a planet in the opposite direction to its rotation to slow a space craft.

Regards,

[doolally]

Most of the models for interstellar travel use one of the planets as a sling shot helper to gain velocity:

https://en.m.wikipedia.org/wiki/Gravity_assist

This can also be reversed by approaching a planet in the opposite direction to its rotation to slow a space craft.

Regards,

Rotation? You must mean orbit.
doolally

[odysseus2000]

Most of the models for interstellar travel use one of the planets as a sling shot helper to gain velocity:

https://en.m.wikipedia.org/wiki/Gravity_assist

This can also be reversed by approaching a planet in the opposite direction to its rotation to slow a space craft.

Regards,

Rotation? You must mean orbit.
doolally

For sling shots you most not get into orbit, as you would then be trapped.

For gravity assist, you approach parallel to the rotation so that the gravitational field of the planet pulls you in, but you don’t get close enough to be captured into an orbit.

For deceleration you go anti parallel to the rotation so that gravity pulls against you.

Voyager used gravity assist & there were concerns that extracting energy from a planet would perturb it, but the amount of energy extracted is low & although the planet does slows the effect is negligible. Carl Sagan did some video on the calculated perturbation of the planet during the Voyager mission.


Regards,

[ursaminortaur]

Most of the models for interstellar travel use one of the planets as a sling shot helper to gain velocity:

https://en.m.wikipedia.org/wiki/Gravity_assist

This can also be reversed by approaching a planet in the opposite direction to its rotation to slow a space craft.

Regards,

Gravity assist helps in getting around the Solar system , ie interplanetary travel, (though it also complicated missions since the planets have to be in right places in their orbits to make use of it). However for interstellar travel the speeds required to travel between the stars in any reasonable period of time are so much greater that the boosts provided by gravity assists are pretty much irrelevant.

[odysseus2000]

Most of the models for interstellar travel use one of the planets as a sling shot helper to gain velocity:

https://en.m.wikipedia.org/wiki/Gravity_assist

This can also be reversed by approaching a planet in the opposite direction to its rotation to slow a space craft.

Regards,

Gravity assist helps in getting around the Solar system , ie interplanetary travel, (though it also complicated missions since the planets have to be in right places in their orbits to make use of it). However for interstellar travel the speeds required to travel between the stars in any reasonable period of time are so much greater that the boosts provided by gravity assists are pretty much irrelevant.

Yes, but if you have no alternative, as in Voyager, then gravity assist is all that can be done & it’s mostly free not requiring much in payload to use. After a very long interstellar journey, gravity retardation might be all one has unless fuel is carried for retro rocket burn to retard velocity.


Regards,

[ReformedCharacter]

For gravity assist, you approach parallel to the rotation so that the gravitational field of the planet pulls you in, but you don’t get close enough to be captured into an orbit.

Regards,

I must admit to being confused by this explanation, what does the direction of planetary rotation (around its axis) have to do with the gravitational effect of the slingshot? I must be missing something because I assumed that the gravitational effect must be independent of the direction of rotation of the planet for a passing body. I note that your reference refers to:

A rotating black hole might provide additional assistance, if its spin axis is aligned the right way. General relativity predicts that a large spinning mass produces frame-dragging—close to the object, space itself is dragged around in the direction of the spin. Any ordinary rotating object produces this effect. Although attempts to measure frame dragging about the Sun have produced no clear evidence, experiments performed by Gravity Probe B have detected frame-dragging effects caused by Earth.

Surely any frame-dragging would be negligible with regard to a slingshot maneuver?

RC

[hiriskpaul]

Most of the models for interstellar travel use one of the planets as a sling shot helper to gain velocity:

https://en.m.wikipedia.org/wiki/Gravity_assist

This can also be reversed by approaching a planet in the opposite direction to its rotation to slow a space craft.

Regards,

Rotation? You must mean orbit.
doolally

Yes orbit. Gravity assist has nothing to do with a planets rotation. To slow down (lose kinetic energy) space probes pass on the inside of the orbit. ie between the planet and the sun. It is not necessary to approach in the opposite direction to the planet's orbit.

[odysseus2000]

For gravity assist, you approach parallel to the rotation so that the gravitational field of the planet pulls you in, but you don’t get close enough to be captured into an orbit.

Regards,

I must admit to being confused by this explanation, what does the direction of planetary rotation (around its axis) have to do with the gravitational effect of the slingshot? I must be missing something because I assumed that the gravitational effect must be independent of the direction of rotation of the planet for a passing body. I note that your reference refers to:

A rotating black hole might provide additional assistance, if its spin axis is aligned the right way. General relativity predicts that a large spinning mass produces frame-dragging—close to the object, space itself is dragged around in the direction of the spin. Any ordinary rotating object produces this effect. Although attempts to measure frame dragging about the Sun have produced no clear evidence, experiments performed by Gravity Probe B have detected frame-dragging effects caused by Earth.

Surely any frame-dragging would be negligible with regard to a slingshot maneuver?

RC

As I think of it, the space ship begins to feel the gravity of the large mass & as the space ship moves, the force towards the centre of the large mass now acts on the new position of the space ship & during each of the infinitesimal times (using a calculus analogy), the force moves in the direction of the rotation of the large mass & alternatively if the large mass is rotating in the opposite direction.


Alternatively one can think of the large mass deforming space time & the deformation of space time rotating with the large mass (unless it is entirely symmetric) & this dragging the space ship. The effect of mass on space time is enough to bend light, so must have an effect on finite mass. Is this what you mean by frame dragging?

There was once a code, maybe now an App, that would allow calculation of gravity boosts etc & permit the user to determine optimum times for travel within the solar system. If anyone has this it would be interesting to see if it calculates any difference between approaching parallel or anti parallel to a large mass rotation direction. This was all worked out for the voyager mission, determining launch dates & course as I remember it, but perhaps there were adjustments during the early stages of the mission.

It’s an interesting field so it would be nice to know if there is or is not a rotation component with gravity assist. There is I believe a rotation dependence for earth based launches, using the velocity of the launch pad as an aid to lift off, such that the ideal place for a rocket launch pad is on the equator.

Please correct if wrong.

Regards,

[ursaminortaur]

I must admit to being confused by this explanation, what does the direction of planetary rotation (around its axis) have to do with the gravitational effect of the slingshot? I must be missing something because I assumed that the gravitational effect must be independent of the direction of rotation of the planet for a passing body. I note that your reference refers to:


Surely any frame-dragging would be negligible with regard to a slingshot maneuver?

RC

As I think of it, the space ship begins to feel the gravity of the large mass & as the space ship moves, the force towards the centre of the large mass now acts on the new position of the space ship & during each of the infinitesimal times (using a calculus analogy), the force moves in the direction of the rotation of the large mass & alternatively if the large mass is rotating in the opposite direction.


Alternatively one can think of the large mass deforming space time & the deformation of space time rotating with the large mass (unless it is entirely symmetric) & this dragging the space ship. The effect of mass on space time is enough to bend light, so must have an effect on finite mass. Is this what you mean by frame dragging?

There was once a code, maybe now an App, that would allow calculation of gravity boosts etc & permit the user to determine optimum times for travel within the solar system. If anyone has this it would be interesting to see if it calculates any difference between approaching parallel or anti parallel to a large mass rotation direction. This was all worked out for the voyager mission, determining launch dates & course as I remember it, but perhaps there were adjustments during the early stages of the mission.

It’s an interesting field so it would be nice to know if there is or is not a rotation component with gravity assist. There is I believe a rotation dependence for earth based launches, using the velocity of the launch pad as an aid to lift off, such that the ideal place for a rocket launch pad is on the equator.

Please correct if wrong.

Regards,

Launches into geostationary orbit or other equatorial orbits generally take place near the equator with the launch being directed eastward this is because the Earth rotates eastward and is rotating fastest at the equator this boosts the speed of the rocket by 460 m/s which reduces fuel usage.

https://en.wikipedia.org/wiki/Near-equatorial_orbit

Equatorial orbits can be advantageous for several reasons. For launches of human technology to space, sites near the Equator, such as the Guiana Space Centre in Kourou, French Guiana, or Alcantara Launch Centre in Brazil, can be good locations for spaceports as they provide some additional orbital speed to the launch vehicle by imparting the rotational speed of the Earth, 460 m/s, to the spacecraft at launch.[1] The added velocity reduces the fuel needed to launch spacecraft to orbit. Since Earth rotates eastward, only launches eastward take advantage of this boost of speed. Westward launches, in fact, are especially difficult from the Equator because of the need to counteract the extra rotational speed.

If you want to put your satellite into a polar orbit it is better to launch from a higher latitude since the craft will need to countract the rotational speed of the Earth and that will be lower the further away you are from the equator.

[hiriskpaul]

I must admit to being confused by this explanation, what does the direction of planetary rotation (around its axis) have to do with the gravitational effect of the slingshot? I must be missing something because I assumed that the gravitational effect must be independent of the direction of rotation of the planet for a passing body. I note that your reference refers to:


Surely any frame-dragging would be negligible with regard to a slingshot maneuver?

RC

As I think of it, the space ship begins to feel the gravity of the large mass & as the space ship moves, the force towards the centre of the large mass now acts on the new position of the space ship & during each of the infinitesimal times (using a calculus analogy), the force moves in the direction of the rotation of the large mass & alternatively if the large mass is rotating in the opposite direction.


Alternatively one can think of the large mass deforming space time & the deformation of space time rotating with the large mass (unless it is entirely symmetric) & this dragging the space ship. The effect of mass on space time is enough to bend light, so must have an effect on finite mass. Is this what you mean by frame dragging?

There was once a code, maybe now an App, that would allow calculation of gravity boosts etc & permit the user to determine optimum times for travel within the solar system. If anyone has this it would be interesting to see if it calculates any difference between approaching parallel or anti parallel to a large mass rotation direction. This was all worked out for the voyager mission, determining launch dates & course as I remember it, but perhaps there were adjustments during the early stages of the mission.

It’s an interesting field so it would be nice to know if there is or is not a rotation component with gravity assist. There is I believe a rotation dependence for earth based launches, using the velocity of the launch pad as an aid to lift off, such that the ideal place for a rocket launch pad is on the equator.

Please correct if wrong.

Regards,

You seem to be barking up the wrong tree here and overcomplicating. Just consider what is happening from the frame of reference of the planet. Someone sitting on the planet sees the probe fly in and back out on a hyperbolic trajectory, as a result of a 2 body inverse square law solution. The probe leaves at the same speed it arrived, but in a different direction. This is just an approximation, as the gravitational field of the Sun perturbs this a little, but for the flyby the planet is the dominant gravitational field.

From the point of view of the Sun's reference frame the velocity of the planet gets added in to the trajectory. Some simple vector addition will get you there.

[ReformedCharacter]

There is I believe a rotation dependence for earth based launches, using the velocity of the launch pad as an aid to lift off, such that the ideal place for a rocket launch pad is on the equator.

Please correct if wrong.

Regards,

That's true if the desired orbit is close to the equatorial plane. IIRC the orbit of the ISS is a compromise for Russian and US launches because of the differences in latitude of the launch sites. For polar orbits the closer that the launch site is to either of the poles the more efficient the trajectory. AFAIK 'change of plane' maneuvers are expensive in terms of the energy required to make such a change. I presume this is why SpaceX, for example, uses Vandenburg for some of its Starlink launches because it is more efficient for reaching certain orbital planes. But I'm not a rocket scientist

RC

[ursaminortaur]

Gravity assist helps in getting around the Solar system , ie interplanetary travel, (though it also complicated missions since the planets have to be in right places in their orbits to make use of it). However for interstellar travel the speeds required to travel between the stars in any reasonable period of time are so much greater that the boosts provided by gravity assists are pretty much irrelevant.

Yes, but if you have no alternative, as in Voyager, then gravity assist is all that can be done & it’s mostly free not requiring much in payload to use. After a very long interstellar journey, gravity retardation might be all one has unless fuel is carried for retro rocket burn to retard velocity.


Regards,

A spacecraft travelling at Voyager speeds would take 73,000 years to reach Proxima Centauri. Noone is going to even send a probe that takes that long to reach its destination let alone a ship with colonists. The Voyager craft themselves will probably run out of power in about five years and after that just continue drifting through space pretty much forever as they were never aimed at any particular target outside the solar system.

https://imagine.gsfc.nasa.gov/features/cosmic/nearest_star_info.html

The Voyager 1 spacecraft is on an interstellar mission. It is traveling away from the Sun at a rate of 17.3 km/s. If Voyager were to travel to Proxima Centauri, at this rate, it would take over 73,000 years to arrive.

[odysseus2000]

As I think of it, the space ship begins to feel the gravity of the large mass & as the space ship moves, the force towards the centre of the large mass now acts on the new position of the space ship & during each of the infinitesimal times (using a calculus analogy), the force moves in the direction of the rotation of the large mass & alternatively if the large mass is rotating in the opposite direction.


Alternatively one can think of the large mass deforming space time & the deformation of space time rotating with the large mass (unless it is entirely symmetric) & this dragging the space ship. The effect of mass on space time is enough to bend light, so must have an effect on finite mass. Is this what you mean by frame dragging?

There was once a code, maybe now an App, that would allow calculation of gravity boosts etc & permit the user to determine optimum times for travel within the solar system. If anyone has this it would be interesting to see if it calculates any difference between approaching parallel or anti parallel to a large mass rotation direction. This was all worked out for the voyager mission, determining launch dates & course as I remember it, but perhaps there were adjustments during the early stages of the mission.

It’s an interesting field so it would be nice to know if there is or is not a rotation component with gravity assist. There is I believe a rotation dependence for earth based launches, using the velocity of the launch pad as an aid to lift off, such that the ideal place for a rocket launch pad is on the equator.

Please correct if wrong.

Regards,

You seem to be barking up the wrong tree here and overcomplicating. Just consider what is happening from the frame of reference of the planet. Someone sitting on the planet sees the probe fly in and back out on a hyperbolic trajectory, as a result of a 2 body inverse square law solution. The probe leaves at the same speed it arrived, but in a different direction. This is just an approximation, as the gravitational field of the Sun perturbs this a little, but for the flyby the planet is the dominant gravitational field.

From the point of view of the Sun's reference frame the velocity of the planet gets added in to the trajectory. Some simple vector addition will get you there.

I am good at over complicating & getting it wrong!

Looking at this 5 minute YouTube, certainly shows how simple it can be made, with a spacecraft using the hyperbolic trajectory entering at 3x(velocity of Saturn) & leaving at 5x(velocity of Saturn) taking into account nothing but the orbital velocity of Saturn, so my approach was wrong:

https://youtu.be/1s6_4qX-u2o?si=mUG2DRUf7SKRj6PP

Thank you for the corrections.

Regards,