41

Physics, Astronomy & Cosmology / Re: What breaks the symmetry in this thought time dilation experiment?

« on: 05/05/2018 03:24:42 »

The fact that both an Earth observer and the ship observer locally measure 1g is a red herring.   Gravitational time dilation is due to potential difference. Think of potential difference as the energy per unit mass needed to lift something from a lower height to a higher height.    For the Earth observer, the difference in gravitational potential and thus the time dilation between the ship and himself is due to a gravitational field that decreases with strength as you move away from the center of the Earth.
 For the ship observer to claim that he is "at rest" he has to assume that the 1 g acceleration he feels is also due to a gravitational field, In his case, however it is a uniform field that does not diminish in magnitude with height ( if his ship was really long and he traveled from tail to nose, he would not measure and weakening in " gravity").  This means that "his" gravity field extends at a constant 1g all the way from him to the Earth.  Now it  would take an lot more energy to lift a mass that distance against 1g the whole way, than it does against a gravity field that weakens as you get higher.   Thus for the ship observer there is a much greater potential difference and thus time dilation between the Earth and himself than the Earth observer would measure between the ship and himself.    Therein is the break in symmetry between the two observers.

42

General Science / Re: How Can I Change the Mind Of A Flat-Earther ?

« on: 25/04/2018 14:56:03 »

To be honest, I don't think you can.  Some people are just immune to logical argument.

43

Physics, Astronomy & Cosmology / Re: Is there a planet the other side of the sun we cannot see?

« on: 20/04/2018 21:02:30 »

The point just opposite the Sun in Earth's orbit is Earth's  Lagrangian point #3
However unlike points 4 and 5, it is not fully stable.   An object put there would stay there only if it is left completely undisturbed. Think of it as a needle balanced on its point.  The slightest nudge of any kind and it will topple over.
Basically this would require the Solar system to be completely empty other than for the Sun, Earth and this object.   

In the real solar system, the object would be nudged out its L3 point, and once that happens, its tendency will be to continue drifting further and further from the L3 point.
One interesting possible outcome from this is that it ends up in a "horseshoe" orbit.
 Let's say that it gets nudged into a slightly lower orbit.  This will be faster orbit than the Earth's so it will start to catch up to the Earth.  As it gets closer to the Earth, the pull of Earth's gravity on it will get stronger and tend to pull forward on it.  This forward acceleration will have the effect of lifting the object to a higher orbit. A higher orbit is a slower orbit.  So oddly enough, the net effect of the Earth pulling forward on it is to slow it down.  It can even rise to an orbit that is further the Sun than the Earth and start to lose ground, with the Earth pulling further and further ahead of it.
Eventually the Earth gets so far ahead, that it is coming up from behind the object, threatening to "lap" it.  But now its gravity is pulling back on the object, causing it to drop into a lower faster orbit.  Now it picks up speed on the Earth and starts to pull away until it starts to catch up with the Earth once more and everything starts all over.

This is not just theoretical either.  The Earth already has a few asteroids in this type of orbit with respect to it and Saturn has a pair of moons with horseshoe orbits with respect to each other.

44

Physics, Astronomy & Cosmology / Re: Why do we see any stars at all?

« on: 11/04/2018 01:02:05 »

Over a million or billion year interval the photons coming from a star or galaxy have to miss every intervening particle or rock or comet. How do all those photons get through?

Let's look at some numbers.  The density of the local cloud of the interstellar medium has a density of ~0.1 particle per cc.
At sea level,  air has a density of ~1.225kg/cc.  Largely being made of nitrogen at ~2.34e-26 kg per atom, 1cc of sea level air contains ~5.24e19 atoms.  So how long would a column of interstellar space with a 1 square cm cross-section need to be to contain as many particles as 1 cc of air?  About 553.5 light years.    If our atmosphere was a uniform density from top to bottom, it would be 8.5 km thick. his means that for our column of interstellar medium to contain the same number of particles as there is in a 1 square cross-section column of air with the same thickness of of atmosphere, it would have to be 470 million light years long.   Light passes pretty well through the thickness of our atmosphere, and the individual stars we see are, at most, 1000's of light years away.   
When it comes to viewing distant galaxies, we are looking in directions non-parallel to the disk of the galaxy and most of the space the light is passing through is intergalactic space with a density  of ~0.000001 particle per cc. Now we are talking about distances that are large compared to the size of the observable universe before the light attenuation begins to even come close to being equal to that caused by passing through our atmosphere.  In other words, the air surrounding us is much more a detriment to the observation of distant stars and galaxies than all the intervening matter in space is.

45

Physics, Astronomy & Cosmology / Re: Which is the Closest Star to Planet Earth ?

« on: 22/03/2018 00:21:21 »

@CliffordK I have to echo @Bill S comment.
That chart is fascinating. I knew some stars are getting closer, but didn’t realise they had comet like tracks. The dynamics of this system would be interesting to know.

They don't.   The graph shows distance vs. time. The shape shown for the the curves is an artifact of this.

Assume that the asterisks in the first line below is a star passing our solar system ( the asterisk in the second line) at some fixed speed in a straight line.  Thus each asterisk is the star's position relative to our sun at fixed time period apart.
**********************************
----------------*-----------------
A line joining any given asterisk on the top line with the asterisk on the bottom line gives the star's distance for that moment in time.
If you were to plot the individual lengths of these lines on a graph, you'd get a curve like the ones shown in the provided graph. You would be basically be plotting the graph the the function  sqrt(x^2+(vt)^2), where x is the distance of closest approach, v is the speed of the star relative to the Sun, and t is the time as measured from the moment of closest approach

46

Physics, Astronomy & Cosmology / Re: What have I missed regarding This Einstein thought experiment?

« on: 21/03/2018 16:12:47 »

If two object's are mutually receding at .9  from a third observer then that observer can say they are separating at up to 1.8c  but each object will only see the other moving at .9c (if that is relevant)

I'm unclear about the set-up, here.  Are the two objects receding from each other; both travelling at 0.9c? 
Where is the observer relative to the two objects?
Are the two objects receding from the observer?

Assume observer 1 is at point A.  Object 2 is receding from point A to the East at 0.9c relative to point A and object 2 is receding from point A to the West at 0.9c relative to point A.
Observer 1 and object A will measure their respective relative velocity as being 0.9c
Observer 1 and object B will measure their respective relative velocity as being 0.9c
Observer 1 will measure the respective velocity difference between A and B's velocities as being 1.8c
Objects A and B will measure their relative velocity to each other as being ~0.994475c ( addition of velocities theorem)

Acknowledging that no craft could travel at c, but bowing to Einstein's thought experiment:  what would be the external observer's measurement of the speed of light within the craft?  My instinctive thought would be that she would measure both as c. which would mean that the light was perceived as stationary relative to the craft.

That depends on the direction the light is traveling.
If the ship is moving at 0.999c, relative to our observer, then he will measure the difference between light traveling in the same direction as the ship as moving at 0.001c with respect to the ship, but light traveling in the other direction will be moving at 1.999c with respect to the ship. (The light traveling from the face of the ship passenger to mirror will take longer to reach the mirror than the light reflecting off the mirror will take to make the round trip.)

47

Physics, Astronomy & Cosmology / Re: Can we see space expanding?

« on: 13/03/2018 15:46:31 »

The expansion of the Universe is not the applicable over all scales.   It only applies at scales large enough that local forces aren't strong enough to overcome it.  Your 2 year old son is not expanding because the bonds between the molecules holding him together prevent it.  The solar system is not expanding because the gravity of the Sun prevents its.  Our galaxy isn't expanding because its own mutual gravity holds it together.   Even our local group of galaxies is gravitational bound together against the expansion of the universe (Evidence of this is the fact that our neighboring galaxy of Andromeda is actually decreasing its distance from us).  It isn't until you get past the scale of galaxy clusters that these clusters are far enough apart that their gravitational attraction to each other is not enough to stop them from being pulled along with the expansion of the universe as a whole. 

48

Physics, Astronomy & Cosmology / Re: Do I weigh as much as the Earth?

« on: 10/03/2018 20:22:53 »

The gravitational field strength of a person is far less than that of the earth at equal radial distances from the centre of mass of each of them. If the mass of the person were all within its own Schwarzscild radius then there would be a radial distance from its event horizon where the gravitational field strength would equal that at the surface of the earth. If the surface of the earth were at this radial distance from the centre of the black hole then the person would weigh the same as the planet. However, the definition of person is now lost.

No.  If our hypothetical person were compressed to an sphere with a radius of ~1.38e-5 m, then his surface gravity would be ~ 1g.    However, this does not mean that the force between himself and the Earth as a whole would increase by any significant amount.  If he were resting on the surface of the Earth, only that part of the Earth's surface touching him would be subject to that 1 g pull.  His gravitational pull is still subject to the inverse square law.  Any part of the Earth 2.76e-5 m away from his center would only experience 1/4 g.  At 5.52e-5m  this falls to 1/8g etc.  at the distance of the center of the Earth it will have fallen to ~4.6e-24g
The total "weight" felt by the "person" would be the result of the integration of the gravitational pull on all the different parts of the Earth, and would still add up to ~150 lbs.  The only small increase seen would be due to the fact that the center of mass for the compressed sphere would be just the slightest bit closer to the center of the Earth then the full-sized person would be.

49

Physics, Astronomy & Cosmology / Re: Which twin is older when they meet again?

« on: 05/03/2018 18:05:07 »

Instantaneous aging? Is this because the mathematics are too hard?

"Near" instantaneous would be better.  There is nothing that prevents us with dealing with low accelerations over extended periods, it's just that for the purposes of illustration of the principles involved, it jut not worth the trouble.  For example, let's consider A during that part of their outbound trip where they come to a rest with respect to B some ten light years distant from B.
If he were to do this by maintaining a constant 1g acceleration( as measured by himself), then he would have to start his deceleration when he is 0.15 ly (as measured by B) from where he wants to stop.   If you want to calculate how much time passes on B's clock according to A during this period, you have to factor in the fact that his relative velocity with respect to A is constantly changing, and thus the time dilation due to that.  You also have to figure in the effects caused by his being in an accelerated frame.   Clocks in the direction of the Acceleration (towards B) will run fast, by a factor determined by the magnitude of his acceleration and the distance to those clocks.  In the case of B's clock this distance is increasing as A goes travels between the point where they first start the deceleration and where they come to a stop with respect to B.
Not an impossible task, but an extra complication that doesn't really aid in the example.   It is much simpler to assume that the acceleration is very high and over a very short distance. 
So, for example, I gave the respective ages for A and B as being 67.32 and ~69 years.   But these answers are a result of rounding.  As long as we assume that the acceleration phases for either A and B occurred over a very short distance any difference in the answer made by accounting for them are too small to effect the rounding.

The problem with using truly Instantaneous changes in velocity is that this equates to an infinite acceleration over zero time, and anytime you bring infinities (especially coupled with a division by zero) in, you are inviting headaches.

50

Physics, Astronomy & Cosmology / Re: What is the mean time dilation at the surface of the earth

« on: 04/03/2018 15:50:45 »

Can we determine, with respect to the centre of our galaxy, what the time dilation is on the surface of the earth. This would mean determining a value for the mass of the galaxy with the dark matter element included. Then calculating the potential that the earth is at. I was considering this with respect to external galaxies. From our perspective do we view the universe as running faster than it actually is due to our time being time dilated. This would skew our measurements of expansion.

For your intended purpose ( how it effects observations of distant galaxies), you don't want the Earth's gravitational potential with respect to the our galaxy's center but with respect to some far distant observer.   For that, you can get a rough estimate from our distance from the center of the galaxy and the mass of that part of the galaxy closer to the center than we are.  This comes out to a time dilation of 1 part in 624,000 due to gravitational potential.   However, this would be mitigated some by the fact that the Sun has a orbital velocity with respect to the Galaxy's center,  which accounts for about 1 part in 3.5 million. ( if we are viewing some distant point in the universe, we would see its clock run fast due to the relative gravitational potential, but run slow due to our  relative velocity difference.
So how would this effect our observation of the Universe?  Consider our nearest large neighbor, the Andromeda galaxy. We measure its radial velocity as being 301 ±1 km/sec.  Which means we can only measure it to ab accuracy of 1 part in 301.    In other words, the excepted error bar in our observation far exceed any correction required to account for time dilation.

51

Physics, Astronomy & Cosmology / Re: Which twin is older when they meet again?

« on: 03/03/2018 22:18:48 »

If we assume that B starts after A after A reaches his destination and come to a halt, then
A will age 17.32 yrs during its outbound trip and then 50 yrs waiting for B to arrive for a total of 67.32 years
B will age 20 years during A's outbound trip and then age ~49 years during his trip for a total of ~69 years, and thus will be older than A when he arrives.

52

Physics, Astronomy & Cosmology / Re: Is this challenge to the equivalence principle valid?

« on: 02/03/2018 19:50:02 »

The quickest thing was to find the original notes from which I quoted in the other thread.  What I posted then was probably an extract from this.

In closed windowless box we should be unable to tell the difference between acceleration and gravity, there should be no experiment we could perform that would give us that information.

Suppose you have with you in your box two marbles.  Can you use these in an experiment to discover if you are being mechanically accelerated, or are stationary on the surface of the Earth?  Surely you can, if you have a sufficiently sensitive measuring instrument.

Release your marbles simultaneously from the top of the box.  They will fall to the bottom.  If you are being accelerated, their trajectories will be parallel, but if you are on the surface of a planet their trajectories will converge on the centre of the planet, because gravity operates as though the entire mass of the gravitating object were at its centre; so the marbles will move towards the centre of the planet; thus they will converge as they fall.

Another experiment you could try would be to release one from the top of the box and one from waist height: under acceleration, they will maintain that separation until the first one hits the floor.  Under gravity, however, the lower marble will fall faster than the upper one, so the separation will increase.  This happens because the lower marble is closer to the centre of gravity than is the upper one, and as gravity varies according to the inverse square law the force on the marbles is different.

The problem with these types of tests is that they are being perform over a region (such as between the top and bottom of the box),  But the equivalence principle only requires that they be indistinguishable locally (over a small enough region.)

Or let's put it another way.  You have your box sitting on the Earth.  As you pointed out, there will be differences when dropping objects from top bottom in it.     But let's move this box to a planet with 4 times the mass and twice the radius of the Earth.  You still have 1g of gravity at the bottom of the box.  But now the difference between the results you get vs. an accelerating box are less apparent. (for example, the angle between objects falling on different sides of the box will be smaller.)   We can keep moving our box further and further from our gravity source while increases the source mass to compensate. As we do so, the falling objects in the box behave more and more like those in an accelerating box. 

This is the fundamental idea of the equivalence principle, that as you move your box further away from the gravity source, the behavior of the falling object approach the a "limit" in their behavior equivalent to that of an accelerating box. 
Gravity and Acceleration are equivalent  "in principle".

As far as the article goes,  it fully depends upon when we can actually perform the experiments that will confirm the predictions one way or another.   The exciting part is that it will finally gives us an idea of what needs to "tweaked" in order to get GR and GM into agreement.

53

Physics, Astronomy & Cosmology / Re: Would a Hohmann Transfer be more fuel efficient to attain lunar orbit?

« on: 20/02/2018 21:27:44 »

To achieve lunar orbit would it be more fuel efficient to utilize a Hohmann Transfer, rather than a "direct" approach, as the Apollo program used?

Yes, the Hohmann transfer orbit is the most fuel-efficient, if all other things were unchanged - but they aren't.

The Hohmann transfer orbit uses just two (quite intense) engine burns.
But it takes about half an orbit between the two engine burns - in the case of an Earth to Moon transfer, that is 2 weeks, compared to about 3 days for the Apollo approach.

That means 4 times as much food, oxygen and muscle cramps, which means more mass, and more fuel.

In the case of Apollo 13, when their oxygen tank exploded, depriving them of electrical power and heating, they continued on their orbit around the Moon, and back to Earth after about 5 days. If they had been on a Hohmann transfer orbit, it would have taken about 4 weeks to return to Earth orbit - and they would have died.

A Hohmann transfer orbit from LEO to Moon orbit only takes 10 days total.  What Apollo 13 used was a  circumlunar free return trajectory.   This basically means that you can whip around the Moon and use its gravity to  return you to LEO.  However, in this case it required a corrective burn using the LEM to accomplish.  Unlike Apollos 8, 10 and 11 which were put directly into circumlunar free return trajectories , Apollo 13 was initially put into a highly elliptical orbit  which fell well short of the Moon, which would return them to the Earth. Then after a docking with the LEM and a system check out, this was changed to  a non-return trans-lunar trajectory.  The accident occurred after this course correction, which required the LEM burn to put them into that circumlunar return trajectory (they could have done a direct abort by using the Service Module engines, but were worried that the explosion may have compromised the SM's structural integrity).

54

Physics, Astronomy & Cosmology / Re: Why must c be an absolute "speed limit"?

« on: 20/02/2018 17:43:13 »

I have a simple, and yet nagging question...

Why must the speed of light, c, be an absolute "speed limit"?

Why can't something, anything exceed 299,792,458 m/s?  Why can something go 299,792,459 m/s?

It boils down to the nature of time and space and their interrelationship to each other.   The speed of c plays an important part of that relationship.      It is the invariant speed for the Universe (the speed everyone measures as having the same value with respect to themselves) Newtonian physics also can be said to have an invariant speed, but it is infinite in value.   c is finite.
Once you have a finite invariant speed, it automatically falls out that it becomes the universal speed limit. 

If you are asking why the universe has an finite invariant speed, then the best I can say is that you should be glad that it does.  Many of the fundamental interactions that allow complex structures, from subatomic particles on up, to even exist rely on Relativity and the fact that this finite invariant speed exists.    In other words, it might not even be possible for a universe without it to have the type of complexity needed for beings capable of asking the question to form in the first place.

55

Question of the Week / Re: QotW - 18.02.11 - How does size affect death from falling?

« on: 13/02/2018 16:49:49 »

A key point here is the square-cube law.   Assuming that density remains constant, As an object increases in size, it's cross-sectional size grows slower than its mass.    With a living being, the cross-sectional size determines things like muscle and bone strength etc.    Terminal velocity is also determined by mass and cross-sectional area.  The terminal velocity for a skydiver is ~53 m/s.    If we were to evenly reduce his size to 10% of its normal value, then his cross-sectional area would decrease to 1/100 its original amount and his mass to 1/1000th. If all else remains constant, this reduces his terminal velocity by a factor of 10 to 5.3 m/s. This is the equivalent of how fast you would be moving after a fall of 1.43 m ( 4.7 ft) with no air resistance.  So, in other words, no matter from how high you fell, you wouldn't hit the ground any faster than you would if you had fell from 1.43 meter.

56

Physics, Astronomy & Cosmology / Re: What happens to cepheid variable stars as they age?

« on: 12/02/2018 17:05:06 »

It depends on what type of Cepheid you are considering.  Classical, or Type I Cepheid stars are much more massive than our Sun and Are likely to supernova ( the variable brightness is due to conditions in the outer layers and not the fusion at the core*).
Type II Cepheid stars are generally thought to be about the mass of the Sun or smaller, so they would likely end up as white dwarf stars.

* Helium in the outer layers is ionized which makes it more opaque, causing it to trap light and dim the star.  The trapped light heats the Helium causing the outer layers to expand, which allows it to cool, de-ionize, become more transparent and let more light escape,  Gravity then pulls the now cooled expanded layers back, where they reheat and the process starts all over again.

57

Physics, Astronomy & Cosmology / Re: Is this calculation of simultaneity right?

« on: 10/02/2018 17:18:40 »

Janus, thank you.  If I may be allowed to expound, and I mean not impertinence.

As I read the first part of your explanation,  "You use the Einstein Synchronization method.   You put a light source halfway between the to clocks, It sends a synchronizing signal to both clocks which tells them to start."  I immediately thought of Zeno's Paradoxes, specifically Dichotomy.  To refresh your memory, Zeno correctly asserts that in order for anything that is in motion from point A to point B it must first achieve the halfway point.  In order for it to make it to the halfway point it must first get to the halfway point of that, or the one-quarter (overall) point, and in order to reach that it must cross the half-way of the halfway of the half-way, or the one-eighth point, and so on.

In a similar fashion, this is the same as folding, or tearing, a piece of paper in half, and then in half again, and so on.  Can you tear the paper in half enough times to make the paper cease to exist?  Obviously, no.

So, back to the Einstein Synchronization method...

Each iteration of the halving of the distance from Earth to AC would cause a time lag.  Not in the halving, but in the fact that the signal still has to travel from one point to another.

Thank you for indulging me. - Pete

Zeno's paradox has no relevance here (besides, it fail to take into account the concept of a "limit".)

There is no series of "dividing in half" involved.  You simply have a light source that is mid way between the clocks you want to synchronize.  It does not matter how you determined the midpoint, You could have stretched a measuring tape between the light source and one clock and then between the light source and the other clock, or used some other method. 
The light then emits one flash which then starts the two clocks, The delay time between the emission of the light and its reception at the clocks is of no matter (you don't even need to know the speed of light for it to work), because it's the fact that the flashes reach both clocks at the same time that is used to synchronize the clocks.  The two clocks involved do not have to communicate with each other at all.

58

Physics, Astronomy & Cosmology / Re: Is this calculation of simultaneity right?

« on: 08/02/2018 17:00:27 »

I'm stuck with the base supposition.  How could a clock 4-light years distant, be it atomic or anything else, be synchronised with a clock on Earth?  What would "carry" the synchronizing information to the clock in the AC system?  And do so instantly, without any loss of time in the transfer?

When clocks are synchronized there is always a variation in the "time", in most cases imperceivable, but it exists.  So, again, the two clocks would have to be out of sync by however long it takes the synchronizing information to reach the second clock.

However, if a clock could be constructed that utilizes quantum "teleportation", then I suppose two clocks separated by great distances could be synchronized.

You use the Einstein Synchronization method.   You put a light source halfway between the to clocks, It sends a synchronizing signal to both clocks which tells them to start.  This only has to be done once, as it is assumed that both clocks from then on will remain in sync in their rest frame (in thought experiments, we assume ideal clocks, in real life we would use clocks who's known limits of drift are much smaller than the effect we are attempting to measure. 
Another method is to send a signal from one clock to the other and just account for the distance the signal has to travel. If Clock 1 sends a signal to clock 2 and they are 4 light years apart, then when Clock 2 gets the signal it knows that the light left 4 years ago and can set itself accordingly to sync itself with Clock 1.  The point is that in principle it is possible to sync the clocks exactly in their rest frame and in practice it is possible to sync them close enough that the difference is insignificant to the experiment.

59

Physics, Astronomy & Cosmology / Re: How does a geostationary orbit work?

« on: 08/02/2018 16:47:02 »

As long as the satellite has a small mass with respect to the Earth (as are any of the ones we have put up), then the non-rotating Earth centered frame is what one measures the sattelite's velocity with respect to.   In fact, in the formula V= sqrt(GM/R) for orbital velocity, R is the distance between the centers of the orbiting mass and the primary body.

This changes as the mass of the satellite increases. At some point(when this is depends on just how accurate of an answer you require), you have to start including the mass of the satellite into the calculation.  At this point you no longer use the Earth's center as the focus of the orbit, but rather the "barycenter",  which is the center of  mass of the two bodies combined.
You also consider both bodies as orbiting this point. (this is not to say that small satellites don't orbit a barycenter, but that is is so close to the center of the Earth as to make no practical difference.) 
An example is the Earth-Moon pair. the Moon is massive enough and far enough away that the Earth-Moon barycenter is some 4267 km from the Earth's center and 2111 km below it surface.  The Earth's center orbits around this point with the same period as the Moon does.   (With many astronomical resources, it is common to use the E-M barycenter as the reference upon which our orbit around the Sun is based rather than the Earth's center.)
The Sun-Jupiter barycenter is just above the Sun's surface.(The earliest detection of extra-solar planets were based on this. The planets detected this way were so massive that they caused a large enough shift in the position of its star as it orbited the barycenter that we were able to measure it. This is also how the first black hole candidate was discovered. )

60

Physics, Astronomy & Cosmology / Re: Is it possible to build a gravity shield?

« on: 02/02/2018 21:49:49 »

Let's shift gears a bit.  Consider gravitational waves.  Would it be theoretically possible to emit a cancelling gravitation wave, thereby cancelling gravity out in the region where they collide?

No, because gravitational waves are not responsible for gravity.  They carry information about changes in the gravitational field.   Thus if you were to induce a change in the gravitational field at some point, then the information of this change would propagate at c to other points of the field as a gravitational wave.  To create a second cancelling wave, you would have to induce another change in the field. For this to cancel out the first wave entirely, it would have to cancel out the initial change in the field.    You might be able to arrange to gravitational waves to intersect at a point in such a way that they cancel out at that point.  But that just means that that point doesn't measure or detect the distant changes in the field that created the waves.  Thus its local field strength simply remains the same and it is not subject to any change in gravity.