Johnny5

sorry to bore you guys with my simplicity but could someone please tell me if i'm getting this. the E^2 = (mc^2)^2 part is derived from the formula E=mc^2 and is thus the energy of the particle if it was converted to energy at the speed of light and the (pv)^2 is the kinetic energy? if so why cant E=hf be the energy of the photon that comes out of the converted particle. or am i just stupid................HIGH SCHOOL IS FUN!

No, you are confused about E=Mc^2 in the first place. Let's see your derivation of the formula, if you don't have one, I can provide you with one.

Regards

Ok' date=' so how does Newton discover force and momentum and their proper definition??? through discovery and experimenting?? back then there was no spaceship, how did he discover such things that would not or rarely happen on Earth??

 

Albert[/quote']

 

Albert, I asked myself the same question long ago, and I can tell you what I think.

 

First of all, he used different words than we do today. He did not use the word 'momentum', he used the word 'motion. He would refer to the "quantity of motion" in a thing. And he also referred to the "vis insita" which usually gets interpreted as inertia, and he also says somewhere in Principia, that the exact formula for gravitation remains for the mathematicians to figure out, so I am not sure even if he got that right. Feynman couldn't understand Newton's proof for gravity being inverse square, if that tells you anything, and he tried very hard to.

 

But really, if you pay attention to history, you will see that all Newton did, was try to restate what Galileo had already figured out. In other words, Newton really didn't figure out Newton's laws, and this is my stance.

 

Instead, I hold that Galileo figured out "Newton's laws," only Galileo stated them in Latin and Italian, Newton re-stated them in English.

 

The reason I say this is obvious. Of the two men, Galileo was the one who performed experiments. It is obvious to me, that Galileo was trying very hard to figure something out, and wasn't going to stop until he did. He performed experiments with pendulums trying to figure something out. He performed experiments in gravity, trying to figure something out, and he performed experiments with 'friction' trying to figure something out. His greatest experiment of all, is the one which helped him figure out his Law of Inertia, which is exactly what Newton's first law is.

 

And more to the point, in Galileo's free fall experiments, it was absolutely essential that he figure out how do define speed, and acceleration. And in one of his books (i think it was Dialogue concerning two new sciences) he meticulously explains the modern definition of speed and acceleration.

 

It is right there in black and white, Galileo invented the differential calculus. You can see galileo focusing on tiny changes in time, in order to figure out how to define acceleration, a concept he initially struggled with. But of course once he figured it out, he knew how to explain free fall.

 

I might add, that Leonardo Da Vinci also tried to explain free fall mathematically, although his explanation was incorrect. Galileo realized that in order to figure out the correct mathematical formula, he had to perform experiments, so that is exactly what he did, and he was successful.

 

Since free fall happens so fast, Galileo needed a way to slow things down, so he decided to use gently sloping inclined planes. This way, the time it takes something to reach the ground from a height of three feet, is about a minute or so. Then, Galileo needed a way to measure time, accurate to the second. At first he used his pulse, but eventually figured out a way to supposedly get an accuracy to the hundredth of a second, by developing a water clock. You make a small hole in a container, fill it up to a certain point with water, then when you take your finger off the hole the clock is ticking, and then the draining water fills a bucket. Later you weigh the bucket, and deduce the time.

 

But my whole point is, the only way to figure out "Netwons laws" is to do the experiments that Galileo did (most importantly the free fall ones, since you need them in order to figure out the meaning of acceleration, in fact you need them to even have a reason to figure out the meaning of acceleration).

 

Also note that galileo was at some point concerned with the fact that different objects fall at the same rate of speed regardless of their weights. He performed many many experiments, designed to figure out whether or not this was true. Eventually he was able to convince himself that it is indeed true that "all objects fall to earth with the same acceleration regardless of their weights, when dropped from the same height"

 

And lastly, Galileo thought quite a bit about the motion of projectiles. He figured out that the path is a parabola, and realized that horizontal and vertical motion could be treated separately. It is Galileo, not Newton, who figured out that the parabolic motion is caused by the horizontal part following Galileo's law of inertia, and the vertical part having the acceleration caused by earths gravity.

 

But not to detract from Newton, I think Newton's third law was due purely to Newton, not Galileo, but then I could be wrong about this.

 

Read Galileo's book and decide for yourself whether or not I am right, and think about the experiments Galileo was doing, and ask yourself this, "What was this man (Galileo) trying to figure out?"

 

Regards

The proof is quite nice; you have to use the fact that:

 

[math]\sum_{k=1}^{n} \left[(k+1)^3-k^3\right] = (n+1)^3-1[/math]

 

which is fairly obvious; just start writing it out and you'll see that all the terms cancel apart from the ones given.

 

Now also note that:

 

[math]\sum_{k=1}^{n} \left[(k+1)^3-k^3\right] = \sum_{k=1}^{n} (3k^2+3k+1)[/math]

 

By swapping things around and equating bits and pieces' date=' you get:

 

[math'](n+1)^3-1-n-\tfrac{3}{2}n(n+1) = 3\sum_{k=1}^{n} k^2[/math].

 

So by simplifying everything down:

 

[math]\sum_{k=1}^{n} k^2 = \frac{n(2n-1)(n+1)}{6}[/math].

This was terrific, thank you very much. I couldn't remember how to do it. There was some sort of procedure to start you off in the general case, which resembles the middle part of your proof. I'm still trying to recall it, I think it involved synthetic division. Hmm I'm sure it will come to me.

Thanks again

The forces can always be broken down into their components. The perpendicular component does no work.

Why do you say that the perpendicular component does no work?

A proton is flying through a linear accelerator, something is applying a force to it, to cause it to linearly accelerate.

Now, someone way down the tube is waiting for it to pass them. As the proton flies by, they "punch" the proton, in a direction perpendicular to its original motion.

In that scenario, work is done by the "punch" force. So I am not following you. I do better with equations.

I have a question.

What standard things do you think the final 'model' of physics should explain.

Right now there are thousands of experiments discovering the tiniest minutia, but no mathematical theory is going to be able to explain one billion bits of minute experimental data.

I can sort of start this off.

1. magnetic moment of proton

2. magnetic moment of neutron

3. magnetic moment of electron

4. g factors

5. Why the nuclear force range is what it is.

6. Fine structure constant.

7. Formula for gravity

8. Show how the fundamental constants of nature are related.

9. Equation of motion for particles

10. Explain masses of elementary particles

11. Explain superconductivity

etc

You get the idea.

Here is the question again...

What things must the final supertheory of physics explain?

Does anyone know how to use the finite discrete difference calculus to find a formula for the following finite sum?

[math] \sum_{n=1}^{n=m} n^2 [/math]

Thank you

an identical magnitude force perpendicular to the motion does no work' date=' and so does not change the speed, so dm/dt = 0

[/quote']

 

This is what I don't understand. If whilst the object was being accelerated linearly, say along an x axis, in the direction of increasing x coordinates, you instantaneously applied any force whatsover f, in the j^ direction (as the particle passed by you), then you would alter the trajectory of the particle. It would no longer have been purely linearly accelerated.

 

At the moment the force in the j^ direction was applied, the particle would move in that direction, and so dR in that direction would be nonzero, and therefore an identical magnitude force F, perpendicular to the original velocity vector V, would have done work, since F*dR would be nonzero.

 

So I don't understand what you are trying to say I guess, but I'm trying to.

 

Thank you though

From your"photonic frame" perspective are all massive objects at "c" or some other constant?

From the perspective of a photon at rest, the speed of other things would not be constant, nor would they all be c, but to be sure, large things would be zipping by. (I am using the Galilean transformations by the way)

Regards

There are no vector symbols - that's the problem.

 

F=dP/dt = mdv/dt + vdm/dt

 

if the force is in the direction of motion' date=' the mass must change: dm/dt is not zero [/quote']

 

Lets do an actual problem please.

 

Proton being linearly accelerated

 

Initially, a proton is at rest inside a linear accelerator.

Let the rest mass m0 of the proton be exactly 1.67 x 10^-27 kilograms.

 

So the proton is just floating there, it is at rest and remaining at rest. There are currently no external forces acting upon this proton. This frame is an inertial reference frame.

 

Then at some moment in time, an external force acts upon this proton, to accelerate it linearly.

 

Assume the relativistic mass formula is correct. Therefore, the mass of the proton in this frame at all moments in time is given by:

 

[math] M = \frac{m_0}{\sqrt{1-v^2/c^2}} [/math]

 

Since this is an inertial reference frame the following statement is true in this frame:

 

[math] \vec F = \frac{d\vec P}{dt} [/math]

 

The momentum P of this proton in this reference frame is given by

 

[math] \vec P = M \vec v [/math]

 

If you carry out the differentiation you get:

 

[math] \vec F = \frac{m_0}{(1-v^2/c^2)^{3/2}} \vec a [/math]

 

The formula above is valid for a proton starting at rest, and having the velocity and acceleration be in the same direction, which is the case with linear acceleration, so the formula is valid for this exact problem.

 

Now you say:

 

but an identical magnitude force perpendicular to the motion does no work' date=' and so does not change the speed, so dm/dt = 0

 

[/quote']

 

Would you be kind enough to prove this for me? (Thanks in advance)

 

Next you say:

 

The only terms that can be different is dv/dt' date=' which is the acceleration, or m

 

[/quote']

 

I don't understand this, can you explain it more please?

 

Lastly you say:

 

If you want to use the inertial idea that m = F/a' date=' the only way to do that is if the mass is no longer a scalar[/quote']

 

After you answer my other questions, lets focus on this question, which I think is the most important one of all. I have had discussions with several others on this exact question. Some say that division of a vector by a vector is undefined, but I myself realize that if F and a have the same direction, then you can just cancel out the directions, and are left with a scalar.

 

Others (electrical engineers) have mentioned something about this as being similar to how to define an electrical impedance (R=V/i), and insist that this is the way to define m to represent inertia, in the Galilean sense. Then they talk about drag force formulas, and lose me very fast. One of them mentioned something about a phenomenon that happens when you reach a certain speed, (sonic boom), and talked about some kind of mathematical miracle that would happen if you ever got something to reach the speed of light. (The effect they were talking about was something like what happens when you break the sound barrier, but they weren't very clear). And lastly, I recall one of the EE's saying something about the mathematics needed for this kind of analysis, but I cannot remember the name. I think it was vector geometry, but I'm not sure.

 

Anyway, I have many questions about m=F/a, but I will wait for you to answer my first few questions.

 

Thank you

 

J5

I'm saying that if you use that as your definition of mass, then you get a given force causing different accelerations in different directions. You have no velocity in a direction perpendicular to your motion, thus you get a different answer. Your term for mass that is direction dependent, i.e. it's no longer a scalar. That's the problem. Using rest mass and relativistic formulae alleviates this problem.

Exactly why do you get a given force causing different accelerations in different directions? I understand what you are saying, I just want to know why you are saying it.

Suppose that:

[math] M = \frac{m_0}{\sqrt{1-v^2/c^2}} [/math]

I don't see any vector symbols there, so could you explain yourself more?

Regards

I don't want this post to end here.

 

Suppose that before a photon is emitted' date=' it is whirling in a circle in the rest frame of the emitter. Just like Newton's pail. Then we let it go, and it flies off in a straight line.

 

There were parameters to the motion of the photon before emission. The photon had a period T, and a circumference [math'] 2 \pi R [/math]

 

Its tangential speed would be:

 

[math] v_t = \frac{2 \pi R}{T} [/math]

 

f = frequency = 1/T, hence

 

[math] v_t = 2 \pi f R [/math]

 

and

 

[math] \omega = 2 \pi f [/math]

 

Which gets us here:

 

[math] v_t = \omega R [/math]

 

Which is correct for circular motion. Setting the tangential speed vt equal to the speed of light relative to the emitter gives:

 

[math] c = \omega R [/math]

 

Now I guess use the galilean transformations from here.

 

Has anyone seen this before?

 

Regards

To continue a bit, we have:

[math] \frac{dc}{dt} = \frac{d}{dt}(\omega R) = R\frac{d\omega}{dt} + \omega \frac{dR}{dt} [/math]

Now, invoke the postulate that the speed of light relative to the emitter is a fundamental constant of nature. Hence dc/dt=0 so that we have:

[math] 0 = R\frac{d\omega}{dt} + \omega \frac{dR}{dt} [/math]

From which it follows that:

[math] R\frac{d\omega}{dt} = - \omega \frac{dR}{dt} [/math]

Has anyone seen this before?

I have to say that from a strictly mathematical point of view, we don't really care all that much about what the variables in the equation actually represent. There's nothing in any definition of the derivative of a function to say something completely different for time-based functions.

My question was, does the difference calculus take this into account.

A - B

is totally different from

B - A

In order to define the difference operator [math] \Delta [/math] the values of the physical quantity which changed are necessarily values at different moments in time. So a convention must be used, in order to specify which is the latter and which is the earlier value. So I am thinking that there is already a convention in place, for the unidirectionality of the "flow of time".

To make this totally clear, let Q denote an arbitary physical quantity. The difference in Q is defined as:

[math] \Delta Q = Q2 - Q1 [/math]

So if I don't choose a convention right now, a student will come along and ask, which came first the Q2 or the Q1. Am I right in saying this?

Regards

Johnny5, I did not present an argument in favor of expanding space. I presented an argument that expanding space does not imply 2=1 (which was, after all, your original point). Your response, which is an opinion against expanding space, does not address that.

Here was your model for space expansion:

T: R³xR-->R³xR, T(x,y,z,t)=((x+|a|Dt,y,z,t+Dt)

You have the x axis stretching.

Use this model of space expansion to analyze a rigid bar for me, if you don't mind.

At some moment in time, let one end of a steel bar have x coordinate 0, and let the other end of the bar have x coordinate 3. So the bar is three meters long presently, at moment in time t=0.

Then exactly one moment in time later, let space have expanded in that model. Tell me where the center of inertia is now, and where the ends of the bar are now.

One more thing... at moment in time t=0, the temperature of the bar is absolute zero, the density of the bar is uniform, and the bar is at rest in the frame. You can introduce any other physical concepts you need.

Regards

What you are saying is that in a fixed cartesian coordinate system the points cannot move with respect to each other.

 

But is that the way space "is"?

Yes, that's what is being said. That the points in space have a 1-1 mapping to the points in the coordinate system.

If space is indeed created (or I should say' date=' if the [b']spatial separation of 2 objects is increasing as a result of cosmic inflation[/b]), then it is possible to construct a map from a coordinate axis at one time t to the same axis at another time t such that all points from the first map continusously on to the second.

In order to mathematically define the "spatial location of an object" one needs to have already chosen some reference frame in which to define that location. The location of the object is to be taken as the location of the objects center of inertia.

The spatial separation of two objects would then be measured by a rigid ruler (ruler at absolute zero kelvin) at rest in the measurement frame.

Now the location of a center of inertia is just a point in the reference frame.

So to say that two objects got further apart we do not mean that two points in the reference frame moved. Instead we mean that there was relative motion of material, and the center of inertia of at least one of the objects, now has a new location in the original frame.

Or to put this another way.

Take a rigid steel bar, and locate the center of inertia. Now bend the bar into a V shape. The center of inertia no longer lies inside the material, it lies exterior to the material. You did not bend space in order to change the coordinates of the center of inertia... no sir... you bent the bar.

Regards

Assuming [/b'] no two points can coincide you are correct.

Its a conclusion, not an assumption.

Set up a three dimensional rectangular coordinate system whose origin is the center of mass of the universe. Let the center of mass of the universe be permanently at rest in this frame.

Make this frame a minimum energy frame. That means you don't want the axes spinning wildly relative to the fixed stars.

Mathematically this means we want Galileo's law of inertia to be true in this coordinate system.

Ok all that being done, consider two locations in space A,B relative to the center of mass of the universe, which is at point C.

I say that the distance between A and B cannot change. It is a temporal constant. For suppose not. Let it be the case that A can move relative to B. Let a force act upon A, to give it an impulse in the direction of B, with just the right final speed so that at some point in the future, A coincides with B.

At that moment in time, the point A is spatially equivalent to the point B.

But the points are different by stipulation.

Therefore, point A cannot move through the coordinate system.

Regards

(If the argument seems corny, that's not my fault. It constantly boils down to not (2=1), because if at any moment in time the point A moved it would be on top of and in (coinciding) with some other distinct point D in the frame. And in a rigid frame such as this, the Pythagorean theorem is a true statement)

I don't want this post to end here.

Suppose that before a photon is emitted, it is whirling in a circle in the rest frame of the emitter. Just like Newton's pail. Then we let it go, and it flies off in a straight line.

There were parameters to the motion of the photon before emission. The photon had a period T, and a circumference [math] 2 \pi R [/math]

Its tangential speed would be:

[math] v_t = \frac{2 \pi R}{T} [/math]

f = frequency = 1/T, hence

[math] v_t = 2 \pi f R [/math]

and

[math] \omega = 2 \pi f [/math]

Which gets us here:

[math] v_t = \omega R [/math]

Which is correct for circular motion. Setting the tangential speed vt equal to the speed of light relative to the emitter gives:

[math] c = \omega R [/math]

Now I guess use the galilean transformations from here.

Has anyone seen this before?

Regards

2 nickels =1 dime, it's all[/b'] about assumptions

That's cute, but in the example I provided, the argument goes something like this...

2 points = 1 point

Therefore, 2=1.

And 2 does not equal 1, hence....

Regards

What I meant was, that in a function of a variable, for example [math] f(x) = x^{2} + 3x - 2 [/math'], the value of the function is dependant upon the value of the variable (in this example, x), which continuously moves down the x-axis, so that the change in the variable (again, in this example, x), and thus the differential, is always positive.

You say that the differential is always positive, do you mean df, or do you mean dx?

The following is basic calculus:

[math] y = f(x) = x^{2} + 3x - 2 [/math]

The picture is of a parabola. And we can discover its vertex.

Taking the first derivative we have:

[math] dy/dx = \frac{df}{dx} = 2x + 3 [/math]

Setting that equal to zero gives:

[math] 0 = 2x + 3 [/math]

So if dy/dx=0 then x=-3/2

dy/dx is the slope of the tangent line as a function of x.

Setting dy/dx=0 is the criterion for the tangent to the curve to be a horizontal line. This is the place where the function assumes a local max, or local min.

The second derivative is given by:

[math] d^2y/dx^2 = 2 [/math]

Positive so concave up.

So when x=-3/2, the function has a local minimum.

[math] y = (-3/2)^{2} + 3(-3/2) - 2 = -4 [/math]

So the vertex of the parabola is at (x,y)=(-3/2,-4).

Picking up from here, what do you mean that dx is necessarily positive?

Regards

Didn't read it' date=' eh?

[/quote']

 

I started to. In article 3.5 he is already discussing boosted frames, but I have a problem with his worldline argument in 3.3.

 

Regards

I think your conclusion is wrong and one can[/b'] equal two using the assumptions you proposed.

No, one cannot equal two, this has nothing to do with assumptions.

My conclusions about the vacuum are reached by using what I might as well call spatio-temporal logic. Logicians are still working out the details of temporal logic, physicists have actually worked out spatio-temporal logic. In a way, that's what reasoning about motion leads to.

Regards

Fine with me' date=' because there's no debate to be had.

 

Section 3.3 of the following document explains quite clearly why you are wrong.

 

http://www.physics.nyu.edu/hogg/sr/sr.pdf[/quote']

 

I did look over section 3.3, but it had no effect.

 

You can't fit a twenty foot ladder inside a ten foot barn no matter how fast you run.

 

I would think something is wrong with the math, before I would think something is wrong with reality.

 

Regards

You seem to hold a double standard. Your "reason" does not logically necessitate your position, and yet you seem to insist on a proof from anyone who disagrees with it. Why are your bald assertions more acceptable than anyone else's?

You are right, that reason doesn't logically necessitate that position. You are absolutely right.

Ok I guess I will have to show you my real argument, though I don't want it ridiculed because I like it.

Suppose that space could be created. Therefore, points in a coordinate system could move relative to one another. If a point in a coordinate system could move in a coordinate system, then two points in the coordinate system would simultaneously be located at one point in the coordinate system, whence 1=2. Therefore, locations in the vaccum cannot be created.

That was my secret argument for maintaining my position.

Regards

I agree that space isn't a thing' date=' but your conclusion simply does not follow from that fact. Your view seems to stem from treating space as a container in which physical objects move around in. This view is rooted in the assumption that, if you took the objects away, the containing space would remain.

 

Is that your basic position?

[/quote']

 

It most certainly is.

 

Regards

I haven't. I'm saying it's not, because of the contradiction I gave. You get the right answer, mathematically, for things like energy and momentum. But you run into the problem I described.

Ok, help me out here.

You are saying the relativistic mass formula isn't true?

What is the problem you described?

Regards