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Physics, Astronomy & Cosmology / Re: Are light and sound similar?

« on: 12/05/2020 17:09:33 »

Yeah BC, that's the one I was thinking of, no space suit.
=

But it is molecules particles 'vibrating', right? Gaining energy by me shouting, exiting them?

Imagine you had a lung full of air, You scream. The air is forced past your vocal chords which vibrate, modulating the air leaving your mouth and producing alternating waves of air compressions.  These compressions are further propagated by the air outside your body.
Now, what happens if there were no air outside of your body. 
The air inside your lungs is made up of molecules which are themselves traveling at something like 340 m/sec.  The air molecules are close enough together that they can't travel very far before bumping into another molecule (the mean free path).  This distance is very small compared to the wavelength of a sound wave. which is why air at standard pressure can carry sound waves for long distances.
So when the air which has been modulated by your vocal chords leaves your mouth, there are no air molecules out there to "contain" them.  Instead of bumping into other molecules and creating sound waves, they scatter and spread out a 340 m/sec.
So the "energy"  from the sound is rapidly being spread out thinner and thinner over this expanding volume.
Not only that, but the mean free path of the molecules increase;  they individually travel further and further before encountering another molecule. Thus the mean free path becomes long compared to the wavelength of the "sound", eventually surpassing it. 
The upshot is that the "sound energy" becomes more and more just a matter of the average kinetic energy of the molecules which is randomly distributed and no longer takes the form of discernible sound waves.

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Physics, Astronomy & Cosmology / Re: What do you know about supernovae?

« on: 11/04/2020 00:13:14 »

I'm not sure how to go about doing the math, but it should be pointed out that the majority of stars are not massive enough to create supernovae. About 75% of the stars in the Milky Way are red dwarfs, for example: https://www.bbc.com/news/science-environment-21350899

While this the state of the galaxy today, this not what we expect the early universe to be like. The main sequence stars you refer to are Population I stars, typically metal rich.    There are also Population II stars,  These are metal poor stars, they are also the oldest stars and were formed during the Early age of the universe.  Their lack of metal could be due to their being more likely the progeny of Type II supernovae, while Type IA supernovae became more prevalent later.

Population III stars (Or generation I stars) are the hypothetical earliest stars, they would have been massive, almost entirely Hydrogen and Helium, and due to their masses, short-lived, ending their lives as supernovae, which seeded the galaxy with elements used in the formation of later generations of stars

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Physics, Astronomy & Cosmology / Re: Can you help me understand "speed"?

« on: 04/04/2020 16:54:07 »

Both Alice and Bob agree that the reflected Laser hits Alice first.

Is there any reference frame which sees that reflected light hits Bob first due to relativity of simultaneity?

No.  Relativity of simultaneity cannot reverse the causality of events.  In this case, Alice is always between the mirror and Bob. and the the reflected light always leaves Bob, hits the mirror and returns to Bob, in that order. Ergo, the light must pass alice first on its way back to Bob.
Everyone in all  frames agree that the light leaves Bob as Alice passes,  both Bob's and Alice's clocks read 0, that Alice's clock reads 1 hr when the reflected light hits it and Bob's clock reads 2 hrs when the light reaches it.
All frames will also agree that Alice's clock reads 1 hr before Bob's clock reads 2 hr.

Below shows the Space-time diagrams for these events in as measured from different frame of reference.

Image11.gif (23.41 kB . 1536x1106 - viewed 3020 times)
Top left is "Bob(blue line) and mirror(red line) at rest", with Alice (green line) moving at 0.6c  The yellow lines are the light moving at c.
In these diagrams simultaneous events are directly horizontal to each other.
as we see, the returning light hits Alice before Bob.
Next over is "Alice at rest", with Bob and The mirror moving at 0.6 c .
Again, the reflected Laser hits Alice first on its way to Bob.

Next is a frame of reference has a relative velocity of 0.6c with respect to Bob and the mirror, but in the opposite direction from Alice.  Note while the vertical separation (representing time difference) is less for the period between the light hitting Alice and hitting Bob, it still hits Alice first.
Lastly, we have a frame with a relative velocity of ~0.882 c relative to Bob, in the same direction as the last frame.
The time gap between The light hitting Alice and Bob has shrunk even more, but it still hits Alice first.
we can keep increasing the velocity of this frame relative to Bob closer and closer and closer to c, and while the time gap between the light hitting Alice and hitting Bob will continue to decrease, it will reverse the order of these events (or even shrink the time gap to zero).

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Physics, Astronomy & Cosmology / Re: Can you help me understand "speed"?

« on: 03/04/2020 17:38:41 »

A mirror is placed 1 light hour away from a powerful laser pointer. Right when the laser is turned on, Alice move from laser pointer towards the mirror at 1 m/s. Bob stays where the laser pointer is.
Who will see the reflected laser first?
What's the speed of light before and after reflection when measured by Alice? Is it still the same as distance divided by travel time?

To make this simple, we will assume Alice is already moving towards the mirror at 1 m/s relative to the pointer as she passes it and it is turned on.
Because she is moving at 1m/sec relative to both pointer and mirror, she would not measure the distance between them as being 1 light hr, but rather 0.99999999999999999443674971973191 light hr.  as measured relative to herself, the laser will travel at c, towards the the mirror, while the mirror moves towards her at 1m/sec.  It will take 0.99999999666435905358172976595532 hrs for the laser and mirror to meet.  During which time the distance between Alice and the mirror would decrease by 3599.9999879916925928942271574392 meters.  The laser will then return at c from this distance to Alice.  At this point, I'm going to quit giving numbers, because it is just too cumbersome to keep working with all these decimal places.
Instead, I'm going to change the scenario to use a velocity for Alice that is easier to work with.
Thus Alice now moves at 0.6c. relative to laser pointer and mirror.
Now:
As Alice passes the pointer, she will measure the distance to the mirror as being 0.8 light hrs, and the mirror will be approaching at 0.6c, while the light from the laser will be speeding towards it at c relative to Alice.  Thus, according to Alice, the light will take (0.8 lh/(1c+.6c) = 0.5 light hr to meet up with the mirror.  During which time, the distance between Alice and the Mirror will have decrease by 0.5 hr * 0.6c = 0.3 lh to 0.5 light hr.   The refelcted light, traveling at c will take 0.5 hr to get back to Alice.  Total time for Alice for round trip of laser  1 light hr.
Now, if we work it out from Bob's view:
The light travels at c towards the Mirror, arrives at the mirror in one hour and returns 1 hr later, total trip time 2 hr.
Alice, is moving towards the mirror at 0.6 c, so, in the time it take for the light to reach the Mirror, Alice will have moved to be 0.6 light hrs closer to the mirror to be 0.4 lh away from it, and will meet with the returning light in another (0.4 lh/(1c+0.c) = 0.25 hr,  total time between firing of laser and it meeting up again with Alice is 1.25 hrs.  However, since Alice is moving at 0.6 c relative to Bob, He will measure her clock as time dilated by a factor of 0.8 and only tick off 1.25hr * 0.8 = 1 hr.  The same amount of time that Alice says here clock ticked off.
If we go back to Alice's view, we can also work out how much time she would say ticks off on Bob's clock between Firing the laser and the light returning to Bob.
As noted above, Alice measures 1 hr for the light to meet up with her. During which time, she would measure Bob's clock as being time dilated and ticking off 0.8 hr.  Now in that hr, Bob, with a relative velocity of 0.6c, has moved 0.6 lh away from Alice.  The light passing Alice on it way to him has to "chase after" the receding Bob at c.  This takes 0.6 lh (1-0.6c) = 1.5 hours by Alice's clock. During which time Bob's clock accumulates 1.5 hr *.8 = 1.2 hr.  Added to the 0.8 hr already accumulated equals 2 hrs total time accumulated by Bob's clock according to Alice.  The same as what Bob's recorded.
Both Alice and Bob agree that the reflected Laser hits Alice first. ( though Alice would say that when the Light hits her, Bob's clock read 0.8 hr, and Bob would say that when the light hit Alice his clock read 1.25 hrs, And Alice would say that when the light returns to Bob, her clock reads 2.5 hrs, while Bob would say that her clock reads 1.6 hrs.
 
So the answer to what the speed of the light is according to Alice both before and after refection, it is c ( relative to Alice.) just like it is c relative to Bob as measured by Bob's.   It is also distance divided by travel time, keeping in mind that Bob and Alice do not measure either distance or time the same.

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General Science / Re: Manoeuvrability in space travel?

« on: 02/04/2020 18:12:11 »

It propels against the mass of it's own fuel! Brilliant!

It is simply the law of action and reaction.  In order to eject the fuel at high speed out of the nozzle, the rocket has to exert a force on it in that direction.  This, in turn, also results in an equal force pushing the rocket in the other direction.
There are a couple of misconceptions going around about rockets.  One is that they need something Air or something outside of the rocket to "Push against" in order work, which is not true.  If you were floating in the vacuum of space, holding a brick, and threw the brick away from yourself, you would move start moving in the opposite direction to the brick.  It's a matter of conservation of momentum.  Both you and the brick together start with zero momentum total.  You toss the brick to the right, giving it a non-zero momentum to the right.  In order for the combination of you and brick to maintain a total momentum of 0, you need to move at some speed to the left so your momentum to the left cancels out the brick's momentum to the right.
In reality, it is a disadvantage to have an atmosphere outside of the rocket. The outside air pushing in on the rocket nozzle decreases the speed at which the gasses can escape, and reduce the effective thrust of the rocket.

The other misconception is that a rocket's top speed is limited to the to its exhaust speed.   As long as a rocket is firing and expelling exhaust, it will continue to accelerate.  The only thing that limits its final velocity ( other than the Relativistic limit of the speed of light), is how much fuel it carries.

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

« on: 31/03/2020 17:17:53 »

One thing about this so-called paradox that is worth a little thought touches on the very basics of how we think about time travel. 

Dan and Emily are twins, and Dan makes the obligatory journey, after which he will be 20  yrs younger than his sister.  Fast forward to the point of return in (say) the year 2150.  Dan has aged 20 yrs less than Emily; he is at the stage she would have been at 20 yrs earlier.  It is easy to think of this as his having travelled 20 yrs into his sister’s past, but is that the wrong way of looking at it?  Both are at the same point in time – 2150.  In what sense has Dan travelled into the past? 

In no sense has he traveled into Emily's past.  He just experienced 20 yrs less time than she did between their separation and rejoining.  Relativity never claims that Dan travels into the past, only that the time interval he measured between the two events was shorter.  Time passed differently for the two.  If Dan left Emily when both their calendars read 2100, then when they met up again, Emily's calendar would read 2150, while Dan's would read 2130.
When They meet up again, it is 50 yrs into Emily's "future" and 30 yrs into Dan's future.
When you state that they are are the same point in time, 2150.  you are implying some type of absolute universal nature to time, and that 2150 is the "real" time.  It is only 2150 by the Earth clocks and this no more the "real time" than the  2130 according Dan's clocks.  It is just that when they meet up again, Dan agrees that by the Earth clock it is 2150 (twenty more years passed for Earth than did for him.)
There is the time as measured by Emily, and there is the time as measured by Dan. and there is no more meaning to time than that in this scenario. 

Consider another scenario.  This time there is no journey into space, but in 2150 Dan uses a time machine to travel back 20 yrs.  He meets Emily.  She is as she was in 2130.  He has travelled 20 yrs into her past and she is younger, not older, as is the case in the former scenario. 

Again, the SR twin scenario does not claim any travel "into the past".

Dan comes back from space younger than Emily.  Pop Sci books and on-line discussions provide the "hitch-hiker" with abundant explanations for that.  However, if, instead of the high-speed journey, Dan had been placed in some sort of “stasis chamber” in which his development and ageing had been halted for 20 yrs, the result in 2150 would be the same as in the first scenario.  He would appear to be 20 yrs younger than Emily.  So, would we claim that he had time-travelled?   

Once again.  SR does not imply "time travel" other than the time travel we all experience everyday from this moment to the next.  What SR does say is the there is no universal meaning to "the passage of time".  Every inertial reference frame measures time by its own independent standard, and that is the only meaning to "time" there is.
It's like the notions of left and right.  Everyone's "left" and "right" is unique to them.  If we are standing next to each and not facing the same direction, my "left" will not be the same as your "left'.   And there is no "universal" concept of "leftness".

Putting Dan in a stasis chamber, while he stays at rest with respect to Emily might, for Dan, "seem" the same personally, but only If Dan were not allowed to measure what was happening outside his chamber.  It might "mimic" the end result of Relativistic effects, but it wouldn't be the same. 

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

« on: 27/03/2020 16:14:54 »

With D moving at 0.5c towards the Earth and the Earth receding at 0.2c, the closing speed between D and the Earth, according to B is 0.3c,  at which rate, it will take 20 years for D to reach the Earth.

That doesn't seem like relativistic velocity addition.
To make the scenario perfectly symmetrical using those numbers, in B frame D should aim to planet Y which is 10 lightyears away. At speed of 0.5c, it will take 20 years to reach planet Y.

As point out by Halc, it isn't needed here.
Example, You have Alice, Bob, and Charlie.
Alice "stays home", while as measured by her, Bob and Charlie both head off in opposite directions at 0.5c
According to Alice,  With Bob going in one direction and Charlie in the other, the distance between them is increasing at a rate of 0.5c + 0.5c = 1c . After 1 hr they will be 1 light hr apart from each other ( Bob will be 0.5 light hr from Alice in one direction, and Charlie 0.5 light hr from Alice in the the Other.) etc.
For Bob to get the relative speed between himself and Charlie, he adds the 0.5c he measures relative to Alice, to the 0.5c Alice measures as the relative velocity between herself and Charlie, using relativistic velocity addition.
(.05c+..5c)/1(1+0.5c(0.5c)/c^2) = 0.8c
The difference between Charlie's velocity and Alice's velocity as measured by Bob is 0.8c-0.5c = 0.3c
After 1 hr ( by Bob's clock) Bob measures Alice to be 0.5 light hr away, and Charlie as being 0.8 light hr way. A difference of 0.3 light hr.

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

« on: 26/03/2020 15:46:38 »

Let's make the situation more symmetrical by adding planet Y, which is stationary in B frame and located 10 light years behind, measured in B frame. How far is the distance between B and planet Y when measured by earth observer?
Let's say the earth observer also has a twin, called D. When B is passing by and A jump to earth, D jump into B's space ship. After 30 years in the ship, D return to earth at 0.5c measured in B frame. Will he arrive on earth at the same time as planet Y passing by earth?

From the Earth frame:
Y is 9.798 ly away as B passes, and is approaching at 0.2c.  It will take 48.99 yrs to reach Earth.
D has jumped on to B's ship, an stays for 30 yrs (B's time)., This takes 30.62 years by Earth's clock.  This means that at that point B and D will be 0.2c x 30.62 = 6.124 ly from Earth.
D heads back towards Earth at 0.5c relative to B as measured by B,  this equates to (0.2c-0.5c)/(1-(0.2c)(0.5c)) = -0.333...c
(minus sign represents direction back towards Earth.)
6.124 ly/ 0.333c = 18.372 y
Added to the 30.62 yrs equals 48.99 yrs.  The same amount of time that passed between B passing and Planet Y arriving.

In B's frame:
Y is 10 light years away when he passes Earth and D jumps aboard. After 30 yrs of flight, B and D are 6 ly from Earth. And Y is 4 ly from the Earth.  With D moving at 0.5c towards the Earth and the Earth receding at 0.2c, the closing speed between D and the Earth, according to B is 0.3c,  at which rate, it will take 20 years for D to reach the Earth. It also will take 20 yrs for planet Y to cover 4 ly at 0.2c.  So according to B, it takes 50 yrs between his passing Earth and both D and Y to meet up at the Earth.   Meanwhile, according to B, the clock on Earth has been time dilated and accumulated 48.99 yrs. The exact same amount of time as determined by the Earth.

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

« on: 24/03/2020 19:41:38 »

Quote from: hamdani yusuf on Yesterday at 08:47:43
Here is the same scenario, but the twins start from earth simultaneously.
And here is the same scenario, but the twins finish on planet X simultaneously.

We can see that in B's frame, B doesn't change velocity since beginning of the trip until the end, although the diagrams don't show B's position before the journey start and after it ends.
Would it make a difference if B never changes velocity at all?
Let's say that the twin were born in a space ship moving at 0.2c on a line connecting earth and Planet X. When the ship is passing earth, twin A stops by on earth for 30 years in earth time, and then catching up B with faster space ship at 0.5c relative to earth, which makes them meets again after 20 years journey from earth to planet X (measured in earth time). The question is, what's the symmetry breaker between B and earth observer, giving that they never really change their frame of reference?

If B never changes velocity, then the symmetry between B and the Earth observer is never broken. 

Space-time diagram from the Earth observer frame.
 
Image1.gif (6.81 kB . 360x566 - viewed 4556 times)

Dark blue line - Earth
Green line - distant planet
blue line - B
Red line - A (if he leaves Earth at the same time as B)
magenta line -A ( if he waits at Earth and then leaves)
Times for each clock shown.  It is assumed that in the Earth rest frame clocks at the distant planet are synchronized to the Earth clock.
The two times separated by the / represents Distant planet time/ A's time

If you shift to B's frame of reference with no changes in velocity, you get this:


Image2.gif (7.43 kB . 277x562 - viewed 4544 times)
When B passes Earth the Earth clock reads zero, B sets his clock to zero also (or B is  born, if you like.)  At this moment the clock at the distant planet already reads 2 years past 0.
B takes ~ 49 years by his clock to reach the distant planet, during which time, Both the Earth clock and distant planet clock advances ~48 years.  Thus the planet's clock reads 2+48 = 50 years upon his arrival, while according to him the Earth clock reads 0+48 = 48 years at that same moment.

The cyan line represents the "checkpoint" that you mentioned earlier and where B would be at when A(red line ) reaches the planet (as measured in the Earth-planet frame.)  In B's frame he doesn't reach this point until after A has reached the distant planet.

The point is that if B never changes velocity with respect to the Earth, B and and the Earth will disagree as to who aged more between B passing Earth and passing the planet.

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Physics, Astronomy & Cosmology / Re: What would I find in a square centimetre of space?

« on: 07/03/2020 15:58:28 »

As the title says, just what would I find in a square centimetre of space? If that bit of space was in the vacuum of outer space?

It depends on where in "outer space" you are.  The interstellar medium can vary from a density of 1e-4 atoms per cm^3 to 10^6 atoms per cm^3. While 1 million atoms may seem like a lot, it is still 1/10,000 as many as you would find per cubic centimeter in a man-made high-vacuum chamber.

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Physics, Astronomy & Cosmology / Re: What Do You Call A Moon Of A Moon ?

« on: 14/12/2019 16:27:36 »

One  specific answer is that we call it the Moon.
The Earth is a "moon" of the Sun.

Thank Ewe BC. So, our "moon" is called "The Moon" is that it's name ? or does our Moon not have one ?

Does this then mean that the Sun is a moon of Sagittarius A ?

To be accurate, there is only one "Moon". It is the natural satellite of the Earth that takes roughly a month to complete one synodic period (the word moon comes from the word meaning "month".

The tendency to call other natural satellites "moons" is similar to calling all open-end adjustable wrenches "Crescent" wrenches.  Crescent is the brand name, and only the wrenches made by that company are truly "Crescent wrenches".  The company name just got closely associated with that type of wrench.

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Physics, Astronomy & Cosmology / Re: Is everything in the universe in motion?

« on: 12/12/2019 21:23:23 »

It is said that our solar system is orbiting the galaxy and the galaxies are orbiting each other.. When we look out into space it seems as if everything is stationary but that is clearly not the case.

The Solar system has a motion relative to the center of the galaxy. The galaxy has a motion relative to the barycenter of the local group.     If we look out at a distant galaxy, it has a motion relative to us.   The key word in all of these statements is "relative".   Motion can only be measured with respect to a reference, and that reference can be arbitrary.  When we look at that distant galaxy, and claim that it moves relative to our galaxy, we could just as easily claim that is our galaxy moving relative to it.   We generally say the former because it is the most convenient.   Thus is the case for any reference frame, we simply choose the one that is convenient for our needs.    There is no absolute reference against which we can measure motion.

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Physics, Astronomy & Cosmology / Re: A reported object travelling faster than the speed of light?

« on: 26/11/2019 18:13:31 »

 A key is understanding the difference between "apparently" moving faster than light and actually moving faster than light.
Simple example:  And object starts 8 light mins away traveling towards you at 0.8c.
The light from the moment it starts at the distance of 8 light min, arrives 8 min after it starts.  The object itself arrives in 10 min, or only 2 min after the light of it being 8 light min away does.  Thus it would appear to you that the object crossed the 8 light min distance in just 2 min and thus traveled at 4c.   But this is just an visual effect caused by the fact that the object is following closely behind its own light. The object was always traveling at 0.8c You merely saw its 10 min trip compressed into 2 min.

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New Theories / Re: Does Einstein’s Equivalence Principle have any physical standing?

« on: 17/11/2019 16:54:02 »

Let’s just suppose (as a special case) that in interstellar space there is a planet exactly like earth about 50 billion miles away from the space ship, which weighs 2 tons or 2000 Kg. Then the acceleration due to that  planet would be 9.8m/s2 and the gravitational  attraction between the planet and the space ship would be  G x (5.972 x 1024 x 2x 103)/((5X1010)2) = 0.000318860846 m3 kg-1 s-2 .

For one, you screwed up your equation.  You put the (2 x 10^3), which I assume was a rough attempt to convert miles to meters*,  in the wrong half of the equation   It needed to be multiplied by the 5 x 10 so that the you had
 G x (5.972 x 1024 )/((5X1010x 2x 103)2) . In other words, the numerical value of your number was off by a factor of at ~ 8e10. in addition to being expressed in the wrong units.  Correctly used, the equation gives an answer in m/s^2 or acceleration.  For example, the equation used for the surface of the Earth (a distance of ~6378000 meters from the Earth center gives
G(5.972e24/6378000^2 = 9.8 m/s^2 or the acceleration due to gravity at the surface.)
A more accurate value for your scenario is 6.23e-14 m/s^2.    Which is the acceleration due to gravity at this distance from the Earth. 

It is now possible to see that the gravity exerted on the space ship is very small, therefore if the mass of the spaceship were to be multiplied by the force of gravity a total force of 0.00031 N which is an extremely small force would be exerted by gravity. If the ship were being accelerated at 9.8m/s2 then the force exerted on the space ship would be equal to F = 2000 Kg x 0.00003 m/s1 = 0.62 N which is a ridiculously small force and certainly would not have the effect of pushing anyone in spaceship back against their seats.

The above arrived at value 6.23e-14 m/s^2 multiplied by the 2000 kg of the rocket gives the amount of thrust or force the rocket would have to generate in order to "hold station" at this distance and not slowly begin to drift towards the planet (1.25e-10 N).
A 80 kg man standing on the floor of said rocket would feel a force of ~5e-12 N Small, yes, but no more or less than what he would feel if standing on the surface of a planet with a mass equal to that of the Earth and 50 billion miles in radius.  It is also exactly the same force he would feel pressing him against the floor if the rocket were accelerating at 6.23e10-14 m/s^2 in empty space with no gravitational forces present.  The point of the equivalence principle is that our passenger, would not be able to distinguish between "hovering" 50 billion miles from an Earth massed planet and the rocket accelerating at 6.23e10-14 m/s^2 through free space with there being no gravity.  The fact that the force he feels is in all practical effect too small to be measured doesn't mean that it is the same in both cases.

Where am I going with this? I am trying to show that the equivalence principle of Einstein’s has absolutely no real meaning, it is just using gravitation as an explanation of itself. Here we have an object in space being accelerated to an equivalent acceleration of the acceleration due to the gravity on earth (i.e., 9.8 m/s2 ) yet it has no effect, some gravitational mass is needed close by to make it work.

No. What you have shown here is that you don't actually grasp the meaning behind the equivalence principle. The flaw is in your understanding, not the principle itself.

*more accurate would have been 8/5 x 10^3. And if you are going to rough out this by this degree (2 instead of 1.6, or only to to 1 significant digit), it really doesn't make sense to carry out the Earth's mass to 5 significant digits.

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Marine Science / Re: How do tides work?

« on: 15/10/2019 16:28:42 »

Tides are the result of the difference in gravitational pull across the Earth caused by the Sun and Moon.
So for example, the Moon pulls on the near side of the Earth more than it pulls on its center, and pulls on the center harder than it pulls on the far side.  Also, since the Moon's gravity pulls towards the center of the Moon, it also tends to slightly squeeze in on the sides of the Earth. This results in the two tidal bulges we get.  So that when you have a high tide on one side of the Earth you will have a high tide on the other side. ( this is why we get 2 high tides and 2 low tides in roughly 24 hrs.)
The Earth is tilted with respect to the both its orbit around the Sun and the Moon's orbit around it.  This results is the tidal bulge peaks to be at latitudes away from the equator most of the time.  Sometimes, like during full and new Moons, the Sun Earth and Moon are more or less in a line and the pulls of the Sun and Moon align.  This results in their tidal effects adding up and we get higher than average high tides and lower than average low tides, known as Spring tides.  During quarter Moons, the tidal effects tend to cancel each other out and we get the least variation between low and high tides, these are known as Neap tides.
Inland lakes can have tides, but the lake has to be fairly large for them to be measurable. Lake Superior has a tide, but it's Spring tides are only 5 cm in height.

Another effect is that open ocean tides are only about 1 meter in height.  The much larger tides we get at the coasts of continents is due to a "sloshing" effect.   If you are on an Island in the tropics, you will only see the open ocean tidal difference.

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Physics, Astronomy & Cosmology / Re: What is the cosmic microwave background?

« on: 15/10/2019 01:38:04 »

Well, I just don't understand it... If it is infinite then why all the to-do and palava about "We've calculated the mass of the entire universe and it's been found wanting, and in need of a shed-load more mass, which just happens to be invisible" (Dark Matter). It's this statement that inspired me to question what CMBR actually is and why somebody thought it suggested there wasn't enough of it (mass that is).

Is dark-matter another myth that's been blown out of proportion?

We have looked at the observable universe how it behaves.  It behaves as if it contains a lot more mass than the mass we can see or detect via the electromagnetic spectrum.  So we have calculated the mass of the entire "observable universe", and found that visible mass only makes up a part of it.
On smaller scales, we look at galaxies. We can measure how fast the stars in them orbit.  But if we calculate how fast they they should orbit based on the matter we can see, they are orbiting too fast.  Not only that, but the way their orbital speeds change as you move out from the center of the galaxy isn't consistent with the way the matter we see is distributed. So for example with a typical spiral galaxy, instead of behaving like the matter is constrained to the visible disk shape, they behave as if the majority of the matter is spread out in a sphere which extends above and below the disk of the galaxy.
There are two possible explanations for this: Either there is something with mass there that we can't see, or our understanding of gravity is wrong.   Both possibilities have been examined and considered. But as more and more data comes in, the dark matter keeps gaining evidence and the other model loses it. For example, we've recently found some galaxies that behave as if they have little to no dark matter.  Now while is is perfectly reasonable for this to occur in the dark matter model, it is less reasonable that gravity itself behave differently in these galaxies then other.  If the rules governing gravity are different than those we think they are, they would still need to be consistent from galaxy to galaxy.
And It's not as if we don't already know of particles that behave similar to the way dark matter would. The neutrino is a massive particle, with gravity, which does not interact via the electromagnetic interaction. It is effectively invisible.  It is a WIMP (Weakly Interacting Massive Particle)*, which is one of the possible candidates for dark matter, In fact, one hypothesized type of neutrino( the Sterile Neutrino) has been suggested as a suspect.

*The other candidate is MACHOs or MAssive Compact Halo Objects

17

Physics, Astronomy & Cosmology / Re: Is light an independent entity, or is it an effect of a larger mechanism ?

« on: 14/10/2019 18:13:16 »

Is this a good time to ask about something I came across recently in a popular science video somewhere....

Concerning a particular type of particle which, when its 'state' was changed then a matching particle's state also changed - instantaneously, no matter how widely seperated in distance (thus appearing to break the lightspeed limit).

Sorry I forget the details and the video, because I didn't pay too much attention to it. But just wondering if this rings any bells here?

Like as already been alluded to such "quantum entanglement" can't be used to send actual information faster than c (speed of light in a vacuum.)  Using the glove example.  In QM, until you open an envelope, not only do you not know whether it contains a LH or RH glove, but neither does the "glove" know* which it is. It is in some undetermined state.  When you open your envelope and your glove "decides" what it is, the other entangled glove "decides" its the other.
The problem is that when you open your envelope to ans see a LH glove, you don't know if that is because you opened the envelope and set off the decision making process, or if someone else had already opened the other envelope, and that event decided what glove you would see. It would take some slower than light communication between you and where the other envelope is to determine this.  The same goes with the changing of state (turning the glove inside out).  You don't know if the inside-out glove was the original unchanged state or the "changed" state without communicating with the other end by some slower than light means.

* This is not to say that you could actually do this with an actual pair of gloves, the glove here is just stand-in for a subatomic particle. One of the reasons quantum behavior seems so strange to us is that by the time you get to the macroscopic scales these effects "average out" to being immeasurable, and how objects appear to behave at this scale is what we accept as "natural".

18

Physics, Astronomy & Cosmology / Re: Do half lives of radioactive materials affected by gravity potential?

« on: 09/10/2019 23:00:29 »

Two samples of identical radioactive material are placed separately, one on earth surface at sea level, while the other are in a satellite orbiting the earth. Will they have different half lives ?

Well no.  If some element has a half life of a million years, it will still be a million years anywhere.
But if I measure that million years using a clock that's not present with the element decaying, the dilation between the clock and the element being measured will not be the same, so the result can vary one way or another.

Yes, time dilation effects everything.  Below 3200 km, the orbital speed is high enough that time would run slower.  The ISS is a case of this.  Above that altitude, the speed is lower and the gravitational potential is higher, so time runs faster.  GPS satellites are an example of this.

Let's say the half life of the sample when measured on earth surface is 1 year. Let one sample orbits the earth for 10 years, and then sent back to earth. Will those samples have the same mass?

It depends on the altitude of the orbit.   A clock sitting on the surface of the Earth, as measured by a distant observer shows a time time dilation due to it gravitational potential relative to the Earth. For an orbiting clock, both the potential and orbital speed come into play.   
Now according to our far off observer, the higher orbiting clock run faster than the lower surface clock if they just consider gravity, but the orbital speed adds a factor which adds an additional amount of slowing.  Now the lower the orbit, the less the difference in gravitational potential, but the higher the orbital speed.
So for example, it the orbiting clock were able to orbit the Earth at the same height as the ground clock, then there would be no difference in tick rate due to gravitational potential, any time dilation difference would be just due to orbital velocity, and the orbiting clock would be ticking slower.
However, if you move the orbiting clock to some really high altitude ( say moon orbit distance), the gravitational potential difference grows and the orbital speed goes down. Now the orbiting clock is runs fast compared to the ground clock by an amount that is greater than what it is slowed by orbital speed.  The orbiting clock runs fast compared to the ground clock.
Somewhere between these two cases is an orbit where the two effects cancel each other out and the orbiting clock and ground clock tick at the same rate.
If R is the radius of the Earth, then r the radius of the orbit  when this occurs is found by
r = 3R/2
Ergo, if the orbiting clock orbits at an altitude 1/2 an Earth radius above the ground, it ticks at the same rate as the ground clock.  Orbiting lower means it ticks slower than the ground clock and orbiting higher means it ticks faster.

So for example, a GPS clock would be expected to run slow by ~38 microseconds per day.  Over ten years, that would accumulate to a difference of ~0.13 sec.  So an isotope with a half-life of one year would decay imperceptibly less in orbit than on the Earth.  Since radioactive decay is statistical in nature, the difference may even be within that statistical variation. (in other words, two samples side by side may vary in result by more than the difference expected from time dilation in this example for any reasonable size of sample.)

19

The Environment / Re: Why do clouds stay high up when they are heavier than air?

« on: 08/10/2019 17:47:50 »

Is a cloud gaseous water or condensed water?

yes


The boundary of the cloud is not a true edge: water is going in and out of the cloud all the time. But only within the region of the cloud is the aerosol stable. Any water outside of the cloud is either in the gaseous state, or is precipitations (solid or liquid).

A good example of this is the lenticular cloud that can form on the leeward side of a mountain.  The wind pushes humid air up the windward slope which condenses on the leeward side.  The cloud formed remains in position with respect to the mountain, but the moisture forming it is constantly being removed by the wind, as the same wind brings in new moisture.

20

Physics, Astronomy & Cosmology / Re: What is the cosmic microwave background?

« on: 08/10/2019 17:28:34 »

And just to add to What Kryptid already said. While the photons we detect today are different ones than the ones we detected yesterday or will detect tomorrow. The one we detected yesterday where a bit "younger" and and the one we will detect tomorrow will be a bit "older".   In other words, the CMB is still decreasing in temp as the universe continues to expand.  The decrease is very small, so it would take a long time before it became enough to be discernible by our measurement standards, but it exists.


 

So the radiation we saw last week must now be gone forever. It sounds like this radiation must have been tavelling past us and 'disappearing forever' for the last umpteen billion years. why is it STIIL travelling past us in a way that we can register with instruments?

This is like saying that because the average water molecule speed is something is in the 100's of meters per second, that a fish in a lake should not detect any water around him because the water molecules that he felt yesterday has traveled off to some other part of the Lake. New water molecules have replaced them.*

*Now it is true that with the lake being a confined body of fixed size, he can encounter the same water molecule more than once.  This is note completely inconceivable for the universe if it were of a finite fixed size.  a Photon could pass the Earth, "circumnavigate" the universe and pass it again. How long it takes between passes would depend on the size of the universe. ( in a really small "closed" universe you could conceivably look off into the distance with a telescope and see the back of your own head)
 But with an expanding universe, even if it is finite, the size and rate of expansion could be such that the "circumference" grows faster than light can traverse it, and light passing you at any given moment can never return. ( but this still doesn't mean that it won't be replaced by a different photon.)