Light Speed and Newton's Equation F=ma: Explanation

[Borek]

I am interested, and I wanted to see around how much U-235 you would have to fission (at 100% efficiency) to propel the space shuttle to light speed

Not sure what exactly 8th grader in US is, but let's try to take it to the level that should be understood by the 8th grader in Poland (AD 1977).

If you take a look at the relativitic mass calculated with Lorenz factor, you will see that the closer to the light speed you get, the larger the mass. For light speed denominator becomes 0 and mass becomes infinite (we can get bashed by nitpickers now, don't worry, just hold tight ). This in turn means that you need more and more energy to increase the speed - and to get to the exactly speed of light you will need infinite amount of energy. For that you need infinite amount of uranium.

 

[Dale]

You can also work this problem without using F=ma. Instead, I would recommend using conservation of energy. If you convert a rest mass m to energy with 100% efficiency you get an energy

E = mc²

Now, if you put that energy entirely into increasing the kinetic energy of a mass M, then by conservation of energy you have that the total relativistic energy of M (γ Mc²) is equal to the rest energy of M (Mc²) plus the rest energy of m (mc²):

γ Mc² = mc² + Mc²

where γ = 1/sqrt(1-v²/c²)

 

[Mettra]

The concept of what a derivative is will probably not come to you easily. There is a real importance and significance behind it which is very subtle (and did not reach me until I was out of high school altogether). I didn't understand it until doing numerous problems involving derivatives and using calculus for proofs. Everyone 'knows' what a derivative is intuitively, it's just difficult to make the connection with the EXPLICIT concept and what you already 'know'.

However, there are simple ways to get an idea of what a derivative is without learning high-level maths. Physically speaking (in a real physics problem, in other words) a derivative is the rate of change of something. This something might be, for example, velocity, or even just plain old position.

As an example:

Consider a ball rolling across the floor. An easy way to get measurements for this ball is to draw an x-y grid on the floor and just use a stop watch to examine it. You can also imagine, for the sake of this problem, that we can pause everything as if it was videotaped - this way we can think better about what is going on.

So let's say you draw an x-y grid with every 'space' being 1 meter.

So at the very beginning of the experiment (at time 'zero' in other words), we have our clock handy and we're ready to go. The ball moves 1 meter and you decide to pause everything. Your clock/stopwatch reads 1 second. So at time '1 second' (in other words t=1), we'll say that x=1 (if it's traveling along your x axis, of course)

But let's continue on. We unpause up until the stopwatch reads 2 seconds (at t=2). At this point, we notice the ball has moved 4 meters altogether (or another way to look at it, it has moved 3 meters in the last second). Interesting, as it seems the ball has somehow sped up. So at t=2, x=4.

Unpausing for another second, we find out that the ball has now moved a total of 9 meters (or 5 meters since the last pause). So at t=3, x=9.

--------
To summarize, we have:

t=0, x=0
t=1, x=1
t=2, x=4
t=3, x=9

So what is all this analysis telling us? Well we've got these numbers, so we can graph them and find the relationship between them. So let's say we make a graph where x is the x-axis and t is the y-axis.

It's hard to imagine what that would look like without actually graphing it, but I'm going to cheat a little bit and just link you here ( hyperphysics. phy-astr. gsu. edu/hbasees/acons.html sorry about the spaces, it won't let me post a url) to give you an idea. That's skipping ahead, but this would otherwise be a book-length post. Ignore the equations there for the moment and just look at the graphs.

The graph would look like the 'position' graph there at the top. That makes sense, because think about t and x. When t goes up a bit, x tends to go up even more. If you had [t=1,x=1] and [t=2,x=2] and [t=3,x=3] and so on, it would be a straight line (graph those points and see this for yourself if you can't imagine it).

There's a very useful mathematical analysis tool called the slope, analagous to the slope of a mountain or cliff. Normally, you can only take the slope of a straight line (rise over run, or y over x), but in this case, we can just look a bit and I'll cheat at the end to keep this from being too long (as if that isn't already the case ;) ).

You can see, looking at those graphs, the 'velocity' graph has a constant slope. In other words, it's just as steep at the top of the mountain as at the bottom. The angle with the ground doesn't change so it looks like a triangle.

The 'position' graph, on the other hand, gets steeper as you go up the mountain, the angle with the ground gets bigger as you climb up. So this graph doesn't have a single 'slope' value, the slope goes up as you move along the graph.

Very surprisingly and astonishingly, if you work hard (really really hard) and figure out the slope by taking tiny increments in the 'position' graph, you will end up with points to plot the velocity graph! So what does that mean? Does it make sense?

If you think of the velocity graph as the slope of the position graph, it really does make sense. As you move to the right in the position graph, the slope steadily goes up higher and higher. That's exactly what the velocity graph is doing.

So what if we did it again? What if we graphed the slope of the velocity graph? Well the velocity graph has the same slope all the way, through right? So it would be just be some number, and that's what we see on the acceleration graph.

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So what does that all mean? Slopes and graphs, these are just vague concepts, just drawing stuff on paper - what is the meaning behind it?

Remember what I said originally way back up at the top: the derivative is the rate of change of something. Look at the acceleration graph. Is it changing? No, it's the same all the way through. What about the velocity graph? Is it changing? Yeah it's changing at a pretty constant pace. If you go right a bit, it goes up a bit. The position graph? It's changing, but the change isn't constant. It becomes bigger and bigger as you go on. It turns out that the slope IS the derivative!

So now we have a sort of definition to go with, the derivative is the slope/rate of change of something. This can be understood without any physical analysis of any real system, but I find it much easier to think of this stuff in terms of real things going on - baseballs, pots and pans, and boulders instead of just fuzzy concepts.

So let's think back to the ball rolling on the floor. The 'what's the rate of change of position' would translate into 'how fast is it going?'. That makes sense right? I mean if it's position is changing, it must be moving, and therefore it has a speed right? But looking at the velocity graph, it's not constant. If you measure the velocity at one instance, it will be different from another one. At t=0, it turns out, it wasn't moving at all!

'What's the rate of change of velocity' translates to 'how fast is it accelerating?' Well it accelerates at a constant rate, so at ANY TIME you can say the ball has an acceleration of 'blah' - whatever that acceleration is (in this case I chose an acceleration of 2 m/s/s).

-----
So do physicists sit around all day drawing graphs and taking slopes and whatnot? No way! That takes far too long, as I've demonstrated here. We use equations and their derivatives to help us speed this stuff up.

For example, for a case like this, where we can see that the acceleration is constant, we have a very special beloved equation that goes as follows:

[tex]x = x_0 + v_0t + \frac{1}{2}at^2


[/tex]
where x_0 is the beginning position, v_0 is the beginning velocity and a is the acceleration - all three of these are just numbers, so you can substitute any numbers you like in there.

In the example above, the beginning position is 0, and the beginning velocity is 0 - that makes it a bit simpler, but you can handle any problem with constant acceleration with this equation.

Taking the derivative directly of this equation (it's a simple mathematical process that you will learn in calculus) gives us an equation for velocity, since velocity is the derivative of position.

[tex]
v = v_0 + at
[/tex]

And finally, taking the derivative of that gives us an equation for acceleration since acceleration is the derivative of velocity.

[tex]
a = a
[/tex]
(just a number, remember)

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So the idea of a derivative is related to stuff you already understand pretty well. You understand that the rate of change of position is velocity and the rate of change of velocity is acceleration (to an extent, humans aren't so great at second derivatives intuitively). The strange thing is that it works out so well in mathematics in so many relationships, that it's difficult to bridge all the systems and rigor with your every-day experience.

I hope this was more helpful than short ;).

 

[malawi_glenn]

So by "scalar multiplication" hootenanny meant a multiplication of two scalars? That would explain the confusion. But the dot product is also called the scalar product.

dot product = inner product = scalar product

Those are different names for the same thing.

HAHA he OWNED us

 

[rocky14159]

Borek, I can't thank you enough! Now this all clicks!

 

[pmb_phy]

In special relativity it is also F = m₀A. And this forumla always works!

That expression only works when the object's mass is constant. It does not hold in general.

Pete

 

[rocky14159]

so in other words, a more accurate form of F=ma would simply be F=(ym)a where y is the Lorentz factor?

 

[Alfi]

Well I'm not 14 ( 53 is close ) and I've just started to look into math and science, again.
I'm very glad some of you can dummy down some answers because I've learned more here in the past few months than all of high school ever taught me.

Mettra - Thanks, nice build up explanation.

 

[pmb_phy]

so in other words, a more accurate form of F=ma would simply be F=(ym)a where y is the Lorentz factor?

No. It is more accurate to wrote F = dp/dt since it holds under all possible circumstances. Your expression holds only when the velocity is perpendicular to the acceleration.

Pete

 

[rocky14159]

I understand that F=dp/dt is the true law from Newton, but what do the variables stand for?

 

[Ed Aboud]

Force is equal to the rate of change of momentum

 

[pmb_phy]

I understand that F=dp/dt is the true law from Newton, but what do the variables stand for?

F = 3-Force acting on object (as opposed to 4-force)

p = 3-momentum of object = mv

m = inertial mass of object

v = 3-velocity of object ( as opposed to 4-velocity)

Pete

 

[jlorda]

Certainly not. An object with rest mass m₀, moving at speed v, will have relativistic mass of
[tex]\frac{m_0}{\sqrt{1- \frac{v^2}{c^2}}}[/tex]

If you know about mass increasing I'm suprised you hadn't see that formula before.

Does this mean when the velocity = c then the answer is an imaginary number?

 

[rocky14159]

not necessarily, but if u plot it on a graph, it will end when v=c, so it is incalculatable (for now)

 

[jlorda]

incalculable is by definition an imaginary number ? square root of zero = i

 

[Dale]

I haven't been following the discussion, but the square root of zero is zero. It is calculable and well-defined.

 

[jlorda]

So M/0 = i imaginary? number

 

[Dale]

No, M/0 is undefined. To understand why, consider the following algebraic rearrangement.

M/x = N -> M = N*x

If M is not zero then there is no number N such that M = N*0 and if M is 0 then M = N*0 for all N. Therefore division by 0 is undefined.

 

[rocky14159]

in calcluable is not the definition of an imaginary number, to the extent of my knowledge, imaginary numbers are formed when you try to take the square root of a negative number. Is this true?

 

[Borek]

[tex]i = \sqrt -1[/tex]

 

[rocky14159]

Thanks

 

[aelbaruk]

The Sun Causes Blackouts !

According to my understanding of some of Einstein's theories , the speed of light is affected by gravity.Light being only a narrow band of electromagnetic waves; radio waves; X-rays and microwaves being the greater majority of them.

In explaining the bending of light waves, Einstein asks the reader to imagine the wave fronts striking the large mass - at right angles. The inner edge of the wave is slowed down due to the [sticky]gravitational force and the outer edge is not so greatly affected.

This is akin to watching a line of soldiers performing a "wheel" [in the trooping of the colour], where the "inner" soldier is marking time , marching on the spot and the outspreading soldiers are progressively moving using larger and larger paces in order to maintain a straight line.

The Earth is a relatively tiny mass in the general scope of things. Using the volume of the Earth as a unity value , with a radius of 6,378km
[For calculation purposes we assume the Earth to be a symmetrical sphere - instead of the poles flattened spheroid it is]
Our Sun has an estimated radius of 695,000km.
See 62229main_System.Lithograph.pdf at [www].nasa.gov

By dividing the Sun's radius by Earth's radius and cubing the result we get the figure of our Sun being 1.2967 million times the volume of Earth
[The 4/3pi factor canceling out top and bottom in the division]

And our Sun is a mere pipsqueak compared to other stars such as Aldeberan and Beteljuice (sic), they in turn being millions of times greater than our Sun.

So the gravitational effect by our Earth on the Sun's light waves will be very minor and any bending effect our Earth does have would be so minute as to be undetectable.

The effect of the Sun itself upon radio signals sent from a deep space probe back to Earth [ and vice versa ] when the Earth is in contradistinction will be noticeable.
Though how the affect is "measured" I have no idea, as for all intents the time a signal takes to arrive back to Earth from a probe in a contradistinct position - will be the time it takes !

Having no "none-gravitational" system to compare and contrast with - even though the scientists can claim the signal should [theoretically] take "x" seconds to arrive or effect a response at distance "y".
When the Earth , Sun and probe are in a direct line with the Sun intervening, I would expect there to be a communication black-out . The vast outpourings of radiation emanating from the sun totally drowning out any puny "six-volt" signal the hapless probe may be emitting.

All of which makes the recent successful landing of a probe on Mars , all the more incredible. For there must have been times during the probe's [8 year?] journey when there was absolutely no communication link with mission control on Earth ; when the intervening Sun quite literally blotted out and "blinded" any two-way signal.

[ I posit 8 years as it is my understanding that the team responsible sent the probe "around" the Sun - to use it's gravitational pull to "slingshot" it towards Mars' orbit.]