Understanding electromagnetism

[lpetrich]

Good job, JimC. I'll give my more-general solution to the problem in ASCII form with the "super" tag as appropriate.

Start with the basic Newtonian equation of motion, where
X = {x,z} - x horizontal, z vertical
V = dX/dt
A = dV/dt
where
A = {0, -g}
where g is the acceleration of gravity.

Start the projectile at time t = 0, at position X = {0,0} and velocity V = {v*cos(a), v*sin(a)}
where v is the total starting velocity and a is the starting elevation angle.

Integrate A over t with the initial conditions to get V, and likewise on V to get X:
V = {v*cos(a), v*sin(a) - g*t}
X = {v*t*cos(a), v*t*sin(a) - (1/2)*g*t2}

X(vertical) reaches its maximum when V(vertical) = 0, or when t = v*sin(a)/g

This makes X(maximum height) = {v2*sin(2*a)/(2*g), v2*(sin(a))2/(2*g)}

The projectile reaches the surface when X(vertical) = 0 for some nonzero t, and that is for t = 2*v*sin(a)/g and
V = {v*cos(a), -v*sin(a)}
X = {v2*sin(2*a)/g, 0}

To maximize the horizontal distance, we find where d/da of it equals zero:
d(X(horizontal))/da = 2v2*cos(2*a)/g = 0

This has the nontrivial solution a = pi/4 = 45 d
It makes
x(maximum; end of flight) = v2/g
t(highest) = v/(g*sqrt(2))
z(maximum; middle of flight) = v2/(4*g)
t(total) = v*sqrt(2)/g

[Twiglet]

The proof that E=mc^2 reduces to the Newtonian KE=1/2mv^2 is quite fun, if anyone fancies a crack at it.

I'll give a couple of hints:

1) You need the full expression for what E=mc^2 means, because the m is KE=1/2mv^2 is just an identical symbol, but doesn't actually mean the same thing...
2) You will need to use the binomial expansion
3) After you have accounted both the above, you need to account for rest mass in the classical Newtonian expression and make a coherent argument about the terms being discarded from the binomial expansion.

btw.. the proof and question are pertinent to the OP for reasons I'll enter into once someone has solved the question, this isn't a derail. It's physics

[Tigger]

The proof that E=mc^2 reduces to the Newtonian KE=1/2mv^2 is quite fun, if anyone fancies a crack at it.

I'll give a couple of hints:

1) You need the full expression for what E=mc^2 means, because the m is KE=1/2mv^2 is just an identical symbol, but doesn't actually mean the same thing...
2) You will need to use the binomial expansion
3) After you have accounted both the above, you need to account for rest mass in the classical Newtonian expression and make a coherent argument about the terms being discarded from the binomial expansion.

btw.. the proof and question are pertinent to the OP for reasons I'll enter into once someone has solved the question, this isn't a derail. It's physics

Just as an aside, I did a QM module when I did my maths degree and I loved the derivation of E=MC2+ "something" iirc. With Hamiltonians and all that guff. It was a decade ago, and my maths has sadly totally deterirated. I can't even recall any of the trig identities in this thread! Is there a proof of what I'm after anywhere in anyone's portfolio?

[colubridae]

It's a smokescreen diversion of last resort. I've played the game before, spending hours on a careful mathematical presentation, only to have my detractors say that's wrong! when it isn't, or you cribbed it! when I didn't. Or both. Or anything else, or they do a runner. It's futile to do anything other than force the focus on the evidence. But this isn't easy when they won't read the OP or enter into the discussion, and are only interested in trashing it. mass, let's see how Twiglet responds to that before I deliver the coup de grace.

I picked the question I did for several reasons:

1) The solution requires a couple of easily written assumptions to be stated.
2) The entire solution is keyboard friendly, using x as an angle and v for velocity means no need for the greek alphabet
3) The solution (bar one small part) is entirely algebraic
4) The problem is set at a level where examiners could easily agree if the solution has been arrived at correctly
5) The time it would take anyone competent with the maths and physics to solve the problem is quite literally the time it takes to write it down. I could write the solution in less time than it has taken to draft this post.
6) A more complex problem would lead most of the readers of this thread unsure of whether you had answered it correctly. I wanted a problem where your failure or success would be glaring and apparant to a wide number of readers.

Also a cribbed solution would probably be needlessly overcomplicated, and most university level solutions would use "angle theta" and vectors, whereas this problem can easily be solved with scalars, algebraicly. The solution can be formulated in no more than 7-8 lines of text.

I would have been less sceptical of you if you had taken 3-5 minutes to answer after my original post. But even if you had answered it, it would only establish your ability to solve one Newtonian physics A level problem. A level of ability I am not at all convinced you possess.

If your shoes farsight, if I had been able to answer the problem easily off the top of my head, I'd have done so in a few lines and then gone on to complain about it, because that would have been by far the most effective way to make my request look a bit silly, and at least demonstrate that you know as much as a 16 year old taking an A level. As it stands,I just feel validated in the belief that you can't even solve a problem within the reach of a 16 year old child studying physics.

Just to add my two-pennies

This is from the a level maths mechanics paper 2006 It's not the exact twiglet question, if anything it's a bit tougher.

maths a level 2006 m2.jpg (59.47 KiB) Viewed 1901 times

[Twiglet]

Here you go col

(it was harder)

a) s (hor)=10
v=11
x=30

horizontally, s=ut+1/2at^2
a=0
u=11cos30=9.5263 m/s
10/9.5263=t

t=1.05 seconds.

b) vertically s=ut+1/2at^2
a=-9.8 (gravity)
t=1.05
u=11sin30=5.5

s=(1.05x5.5) - 0.5x9.8x1.05^2
s=0.373m above A


Target is 1 metre above A, so 1m-0.373=0.63m below the target, to 2 significant figures.

c)

Horizontally,

s=ut+1/2at^2, t is non zero, a=0
10=Vtcos30
t=10/(VCos30)
t=11.55/V

Vertically

a=-9.8
s=1
u=Vsin30=0.5V

s=ut+1/2at^2

1=0.5x11.55 -4.9x133.4/V^2
-4.78=-653.66/V^2
V^2=653.66/4.78=136.74
V=11.69 m/s

Crosschecking by back - substitution

Horizontally

10=11.69xCos30t gives t=0.988s

Vertically

?=11.69x0.5x0.988 - 4.9x0.988^2
=5.77 - 4.78
=0.99 which is fine.

d) Hitting the target is possible from a range of different angles and speeds, so for example. throwing the ball very fast directly at the target would work, as would throwing the ball at a high angle, more slowly

[JimC]

Twiglet wrote:

Hitting the target is possible from a range of different angles and speeds, so for example. throwing the ball very fast directly at the target would work, as would throwing the ball at a high angle, more slowly

It would be an interesting exercise to plot the correct combination of angles and speeds on a speed vs angle plot...

[Twiglet]

Twiglet wrote:

Hitting the target is possible from a range of different angles and speeds, so for example. throwing the ball very fast directly at the target would work, as would throwing the ball at a high angle, more slowly

It would be an interesting exercise to plot the correct combination of angles and speeds on a speed vs angle plot...

Actually not that hard. Just stick with Sin x and Cos x instead of plugging the numbers in, and you will end up with the general solution.

The specific case is harder because it's easier to fuck up piddling round with numbers.

[colubridae]

Shit dude. This is how I was taught it and I’m going way back

Free energy E is given by mc^2 / (sqrt (1 –(v/c)2)


Where1 / (sqrt (1 –(v/c)2) is the lorentz frame change transform.

This becomes on expansion


E = mc^2 times (1 + ((v/c)^2)/2 + terms with a higher exponent)


When v is much less than c this reduces to

E almost= mc^2 + m/2 * (v^2 * c^2 / c^2)

Giving

E almost= mc^2 + 1/2 m v^2

Taking the KE component of the energy of the free particle energy without the rest energy

Gives KE = 1/2 m v^2

Hope this is correct…

Sorry about the lack of equation capabilities it doesn’t transfer onto my explorer.

I have to do it in word otherwise my paper and pencil answer gets rewritten too many times.

[colubridae]

Twiglet wrote:

Hitting the target is possible from a range of different angles and speeds, so for example. throwing the ball very fast directly at the target would work, as would throwing the ball at a high angle, more slowly

It would be an interesting exercise to plot the correct combination of angles and speeds on a speed vs angle plot...

you need to get out more

[colubridae]

t+30 minutes.

Why am I not surprised.

if u want a challenge on basic physics ability then both parties need to agree terms and conditons first.

for farsight
How about this no maths, just understanding...
(You don't need an understanding of QM.)
3-4 sentences should do it.




This is the general form of the schrodinger equation for a particle in a potential well.

schrod.jpg (5.14 KiB) Viewed 1886 times

The term on the l.h.s. contains the symbol i.

This stand for the square root of -1.

This number does not exist.

Can you explain why it is there?

Looking for a bit more than ‘it makes the equation work’

[Twiglet]

Shit dude. This is how I was taught it and I’m going way back

Free energy E is given by mc^2 / (sqrt (1 –(v/c)2)


Where1 / (sqrt (1 –(v/c)2) is the lorentz frame change transform.

This becomes on expansion


E = mc^2 times (1 + ((v/c)^2)/2 + terms with a higher exponent)


When v is much less than c this reduces to

E almost= mc^2 + m/2 * (v^2 * c^2 / c^2)

Giving

E almost= mc^2 + 1/2 m v^2

Taking the KE component of the energy of the free particle energy without the rest energy

Gives KE = 1/2 m v^2

Hope this is correct…

Sorry about the lack of equation capabilities it doesn’t transfer onto my explorer.

I have to do it in word otherwise my paper and pencil answer gets rewritten too many times.

Spot on Col, the approximation v<<c makes the higher order terms in the expansion vanishingly small, leaving the classical KE by itself with the rest mass energy.

This link shows the expansion in full http://www.relativitycalculator.com/bin ... ries.shtml

[Farsight]

Believe it or not we are studying it. We actually use these insights to decide how to approach other scientists. i.e. We psychologically profile a scientist before interacting with them, and have preset methods as a guide. i.e. Certain universities enforce and condition standards of rigid conformity, such as cambridge, oxford, while others such as UCLA are more freewheeling, open minded and creative...

Interesting stuff Brain Man.

yeh ive seen that one. They reason they accuse you of smoke and mirrors is because its a projection of their own processes.

And how, just look at the thread, totally derailed with schoolboy maths that's totally off-topic, and absolutely unrelated to Twiglet's challenge as to why c is a limit. I do find this kind of thing pretty amazing. I've had dingdongs with Young Earth Creationists and Islamic Fundamentalists, and have seen Morton's Demon in action in a religious context, dismissing all evidence that threatens conviction. But it's still rather odd to be reminded that this is a human trait rather than something unique to religion. For me a prime example was seeing Stephen Hawking on Discovery talking earnestly about time travel, the latter being something of a personal bugbear, and the reason why I started posting on this forum. It does give me pause for thought about the nature of human consciousness, along the lines of this:

Q: Am I conscious and self aware?
A: No. you only think you are.

Edit: come on Twiglet, that arithmetical exercise is nothing to do with why c is the limit. Now I've taken time out to post up threads explaining energy and mass, whereafter I'll explain the speed limit. Do be so good as to read them and give comment, otherwise people will think you're dodging your own challenge.

[colubridae]

Shit dude. This is how I was taught it and I’m going way back

Free energy E is given by mc^2 / (sqrt (1 –(v/c)2)


Where1 / (sqrt (1 –(v/c)2) is the lorentz frame change transform.

This becomes on expansion


E = mc^2 times (1 + ((v/c)^2)/2 + terms with a higher exponent)


When v is much less than c this reduces to

E almost= mc^2 + m/2 * (v^2 * c^2 / c^2)

Giving

E almost= mc^2 + 1/2 m v^2

Taking the KE component of the energy of the free particle energy without the rest energy

Gives KE = 1/2 m v^2

Hope this is correct…

Sorry about the lack of equation capabilities it doesn’t transfer onto my explorer.

I have to do it in word otherwise my paper and pencil answer gets rewritten too many times.

Spot on Col, the approximation v<<c makes the higher order terms in the expansion vanishingly small, leaving the classical KE by itself with the rest mass energy.

This link shows the expansion in full http://www.relativitycalculator.com/bin ... ries.shtml

Thanks. Need a lie down after that though.

[colubridae]

made it to 1000. without a ban.

[Twiglet]

Edit: come on Twiglet, that arithmetical exercise is nothing to do with why c is the limit. Now I've taken time out to post up threads explaining energy and mass, whereafter I'll explain the speed limit. Do be so good as to read them and give comment, otherwise people will think you're dodging your own challenge.

It could hardly be simpler. E=mc^2=m0c^2/(1-v^2/c^2)^1/2
In the limit as v approaches c, E approaches infinity, hence infinite energy is required to propel any particle with inertial mass to the speed of light.
There is another, non trivial proof that any particle which does not possess inertial mass must by definition be travelling at the speed of light, but I can't remember what it's called. Someone can probably enlighten me.