imatfaal

I am going to have a go at this - even though I hope StudioT will provide a better answer and tell me where I have gone wrong

First line of equations

1. power equals voltage times current

2. as voltage equals Inductance times change in current over time (L di/dt) substitute this in for V. Giving power equals L di/dt i

3. rearrange to make the whole expression a change over time by changing i outside the influence of the di/dt to i^2/2 inside the influence

Second line of equations

1. Statement that Work is the integral of power over the time taken

2. Using the logic and expressions in the first line of equations you can rework this as integral of (L di/dt i) dt (not shown)

3. This collapse to integral with respect to di of L i. Remember i is a function of t

4. This integral equals 1/2L i^2

It would seem that part of what I want is a good explanation of the fundamental theorem(s) of calculus. I found something on berkeley dot edu, but it is actually a proof and not intuitive at all. https://math.berkeley.edu/~ogus/Math_1A/lectures/fundamental.pdf

I am confused by page two by the primes inside the functions: f(x') instead of f'(x).

Primed outside the brackets refers to the differential . The x being primed inside the brackets is saying this is a value of the variable which is different from x; in this case it is a difference less than delta which gives rise to a difference in the functions of less than epsilon

Look at the wiki on the definition of a limit of a function

https://en.wikipedia.org/wiki/Limit_(mathematics)#Limit_of_a_function

https://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit

Although I have just noticed wiki use different notation - typical.

In the spirit of the OP - and in that I am assuming that levitate means hold in the air without an immediately obvious means of support - I suppose one could make the kilo weight into a sphere and support it in an airflow like the pingpong ball at the end of a straw (which is used in demonstrations of Bernoulli).

But if the Universe is a set of events (in the spacetime sense , I suppose) is any event more central in any meaningful way than any other? Events "nearer" the BB?

The Big Bang happened everywhere. Remember everything is flying away from everything else - rather than everything flying away from a central explosion

I realise the "flying" bit is wrong - but ignore that bit - the important part is the fact that the big bang was/is a process which happened/is happening everywhere and that there was no core which expanded outwards just that everything moved away (fairly swiftly) from its neighbours

Why is it we always look at an object near c and never out from an object near c?

Firstly, the observer always views himself as stationary - there is no dilation nor contraction in the rest frame.

And secondly, it all works out the same; this is one of the basis of relativity that there are no privileged frames. Something flying past the solar system which we measure as .9c will see the solar system moving past them at .9c. And all the physics still works out.

For an example of something worked from both frames you can look up atmosphereic muon decay

http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/muon.html

A lateral 'hole' is a 'tunnel'.

Some of the wells in Sakhalin and in Qatar are very deep - but also have a a huge lateral component. If my memory serves me correctly the Sakhalin holes are 12000m deep with 7000 metres lateral - definitely not a tunnel; whereas the real biggies in Qatar are 12000m deep with 10000m lateral. But still an oilwell and not something I would call a tunnel

...

 

Apologies for this notation, but I am following Penrose (road to reality) where the idea is developed (pages 378 - 380) and applied to quantum computers (page 583)

Lol I love that book

I was distinctly underwhelmed - but it was a long time ago. So, StudioT/Mordred, is it worth a retry guys?

Sorry for the offtopicness.

Your in essence correct there are causally disconnected regions beyond the Cosmological event horizon. Those causally disconnected regions will never become causally connected on the future unless they were causally connected at one time in the past.

 

Inflation in solving the horizon problem connected previously disconnected regions but this is an extreme example that runs counter to the above.

 

So unless we have another inflationary scale event its unlikely to connect to further causal disconnected regions afiak.

 

Sometime in the future galaxies we see today will become causally disconnected never to be seen again.

 

The other problem is that we can now see the surface of last scattering which is the furthest we will ever see until we can measure the cosmic neutrino background. This then will be the new and furthest possible extent we will ever see. The neutrino background will allow us to see past the opaque fog prior to recombination.

Thanks Mordred - it was the inflationary period section that I was partially remembering. And yes regarding surface of last scattering - but I was thinking in terms of any possibility of connexion; neutrino, gravitational wave etc

my equation for KE is not wrong, it is the same as yours.

Yes it is - you are using the wrong value for v.

But I fear you are right about the acceleration.

As he has a further degree in physics and this is first year school physics I reckon he is right too.

I must ponder these equations and how a linear equation can return the same result as an exponential equation.

There is no exponential equation there. x^2 is not exponential. 5^x is exponential

Learn some basic maths and basic physics - it is well worth it. Many people here will help - but fewer will help if you pretend great knowledge and then demonstrate great ignorance

That's true, but machinery would all be electrical, not generating huge amounts of heat. They coped with the problem 45 years ago, so it shouldn't be too much of a problem now. Astronauts produce heat, but the suits coped with that, and the direct sunlight.

Cooling by radiation will be much faster with no clouds or atmosphere. Even at mid Lunar Day, a radiant surface that is shaded would shed heat very fast.

Hmm.

[latex]\frac{q}{A}=\sigma \cdot T^4 [/latex]

where q is the Watts given off , A is the Area (so LHS is Power per area, ie the Joules per second per Square metre), Sigma is the Stefan-Boltzman Constant,and T is the absolute Temperature. If you want to be keeping stuff at superconductor temperature then lets say that your ratdiators are at 200K

[latex]\frac{q}{A}=\sigma \cdot T^4 = 5.67*10^{-8} \cdot 200^4 = 5.67*16 = 91 Watts/m^2[/latex]

That's pretty damn poor. And that is basis a perfect blackbody radiator at abs zero. Your surface will be a grey body and will not be at abs zero

[latex]\frac{q}{A}= \epsilon \cdot \sigma \cdot (T_{radiator}^4 - T_{surroundings}^4)[/latex]

T_surroundings will be greater than zero so will reduce your emission and epsilon will be less than 1 so will further reduce them

No I wasn't questioning that . I was wondering if it might have its own intrinsic"speed limit" (or "rate limit" ) entirely separate to the speed limit of GR.

 

Naively it seems more fundamental than the GR speed limit. (maybe "fundamental" implies a hierarchy that does not exist)

There will be a distance / sphere beyond which we presume lie stars/galaxies/clusters etc which we have never been causally connected to and which we have never/will never see any light from; I think this is correct*. There will be an associated rate of expansion between us and something at that distance/on that sphere - whilst it is not a physical limit it is a point at which expansion becomes meaningless as nothing we will ever have any inkling of will ever have a higher rate.

* Hopefully Mordred can tell me if this is true - in the meantime I will do a little more reading. Galaxies which we will one day see because of expansion is a - to me at least - a counter intuitive subject. I am pretty sure this is the particle horizon at 14.4Gpc - although I had a feeling that expansion for some odd reason meant that eventually we might see more although this sounds like rubbish

...

 

Answer: Sort of. You will get a square but it will be rotated.

 

...

 

1) We had a long thread about a video of the companion 'train in the tunnel' version. I will try to find it unless someone else can.

 

2) I believe you are referring to 'Thomas precession'

 

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

Or is it Terrell-Penrose rotation that Mike is referrng to (which is the combination of actual SR contraction and visual distortion due to finite/invariant speed of light transmission to observer). There was a superb animation explaining that posted a while back too. I will look for both

https://faraday.physics.utoronto.ca/PVB/Harrison/SpecRel/Flash/ContractInvisible.html Terrell rotation and the inability to actually observe length contraction but we actually see rotation

 

....

We MUST add the impact of the expansion to our formula.

No, we MUST NOT. In the situations in which it is used it is great. You should try some actual study before making pronouncements. At the beginning of this thread, it was clear that you had no idea of the concepts of Special relativity (you claimed the absolute nature of time and distance remember?) - by post#53 you are demanding the equations be re-written. I have counted at least three knowledgeable and helpful members explaining to you the difference between a metric expansion and relative motion through space; you seem to hurry through the carefully prepared posts in your rush to find yet more things you can say must change about modern physics.

Oh I thought mass was mass, and energy was mass and everything was the same. How much energy would it take to sheer apart a massive black hole like a planet's gravity sheers apart a micro blackhole?

To the best of our knowledge, things outside blackhole only interact gravitationally (and to a minuscule extent electromagnetically in the case of a charged blackhole). Once inside the event horizon, the only interaction is swallowing by the blackhole giving a change in mass (and charge and angular momentum). I have never heard of anything which shears apart a blackhole.

A microblackhole in the immediate vicinity of earth would be far more concern to us - those things are hot - very hot, and only getting hotter. But I do not believe they are altered by external gravity

Official KE equation is 1/2mv2

I wrote 0.5*m*sqr(v)

 

imatfaal's equations seem to imply that both m and sqr(v) should be multiplied by 0.5.

this is the descrepancy.

1/2m

looks to me like

.5m.

But apparently it means that it applies to the whole thing, not just m.

OK you need to learn very basic maths. You also need to learn to use notation in the way the rest of the world does

1. multiplication - it doesn't matter if you multiply the m by 1/2 or multiply the m by v^2 and then by 1/2; they are both exactly the same. This is basic junior school maths

2. sqr(v) does not normally mean v² and it is very close to be confused with sqrt(v) which stands for the square-root

When you are getting confused DO THE ACTUAL MATHS - don't guess!

What do you think the three main branches of philosophy are? To be honest I think there are four but then I know most people limit it at three.

This is homework help - not a place for just asking and getting a ready-made answer. What are your initial thoughts?

[latex]

PE_{at 9.8m} = mgh = 4 * 9.8 * 9.8 = 384.16J[/latex]

[latex]KE_{groundlevel}= \frac{mv^2}{2} = \frac{4 * v^2}{2}[/latex]

but what is v? look at suvat

[latex]v^2 = u^2 +2as[/latex]

u=0m/s (starts from a standstill) a=9.8m/s^2 s=9.8m

[latex]v^2 = 0+2 * 9.8 * 9.8 = 192.08[/latex]

so

[latex]KE_{groundlevel} = \frac{mv^2}{2} = \frac{4 * 192.08}{2} = 384.16J[/latex]

So your 384.16 Joules of PE at 9.8 metres is converted entirely to 384.16J of KE by the time it reaches the ground

Sorry Guys are you all forgetting all the green stuff on our planet. That green stuff takes energy in the form of radiation and uses it to make complex molecules out of a selection of less complex molecules.

It is long wavelength - ie very cold - radiation that we cannot do much with; although Blackholes will be converting it to mass (or whatever is inside a blackhole) for many many times the current lifetime of the universe yet to come

When you add energy to a particle, doesnt that make it denser? Because dense just means more mass inside a small volume.

the particle Im talking about would probably require a collider that circles the galaxy to make.

Difficult to say really. Volume is a quantity which is normally measured in the rest frame - but in the rest frame the mass-energy is equal to the rest mass; in the moving frame viewed from a relative frame what is the volume?

To make a black hole the entire mass must be inside the schwarzchild radius.

[latex]r_{schwarzchild}=\frac{2GM}{c^2}[/latex]

That would give a proton size black hole a mass in the order of 10^11 kg. So a speed that would be many many orders of magnitude faster than LHC

BUT ALL THAT IS IMMATERIAL - a blackhole mass must be measured in its rest frame; ie the rest mass

Horizon is on BBC at the moment, about telescopes. The James Webb space telescope has a plastic shade, to effectively block out the Sun's rays.

It's thin plastic sheeting, about the thickness of a human hair. About the size of a tennis court, and several layers, with a gap between each layer of a few inches. This is designed to keep the temperature of the telescope steady at minus 220c. so that it can pick up infra-red from the universe without swamping it with it's own infra-red.

 

Going by that, preventing things from overheating on the Moon would be pretty easy.

Not a huge number of moving parts generating their own heat through friction, nor engines operating inefficiently, and the rest of the telescope (ie everything that may generate heat) is cooled to much lower by active systems so no bleed through from the rest. Any machinery on moon would be generating heat and that can only be radiated away or actively cooled by refrigerants

Well someone needs to figure it out. This is going to bug me for the rest of my life.

lets see if I remember how to math, that expression reads

the imaginary value of velocity squared divided by the speed of light squared miltiplied by zero point five. And once that is simplified you have the resulting relative passage of time.

You cannot use special relativity for an inertial frame which is travelling at light speed - the physics doesn't work and you get a nonsense answer. Nothing with a mass is able to travel at the speed of light - so we cannot really envisage the "perspective" or "time flow" for something (massless) which does travel at c

Why dont super fast objects, despite having potentially more mass then a black hole not exert black hole levels of gravity?

You need high rest mass for a singularity and event horizon to form. Just realised that is not what you are asking.

start again. The superfast objects we actually know of are things like neutrinos (within a few billionths of the speed of light) - but these thing are insanely small too. As a massive over estimate the maximum rest mass of a neutrino is an electronVolt. 1 eV boosted up to 0.999,999,997c would give a total mass-energy of about 13keV - which is still pretty tiny

The protons in the LHC have a much higher mass-energy - they are travelling with a speed of the same magnitude but are about a billion times heavier; they have a mass-energy of around 10^-23 kg (from memory this should be around 4TeV so I think I am a little high). Which is huge for a proton but still tiny. Just gonna work out what the lead ions pack - a little less than a 100 times more (about 300TeV).