Dimensional analysis

[ambros]

Sadly, no one managed to realize what are the units of this equation and point if there are any mistakes in this dimensional analysis:

 

[math]F_{12} = \frac {\mu_0}{4 \pi} \int \int \frac {I_1*dl_1 \times (I_2*dl_2 \times \hat{r}_{21} )}{|r|^2} = \frac {\mu_0*I_1* \int dl_1 \times (I_2* \int dl_2 \times \hat{r}_{21} )}{4 \pi*|r|^2}[/math]

 

 

 

[math]=> \frac {\mu_0*Q_1*v_1 \times (Q_2*v_2 \times \hat{r}_{21} )}{4 \pi*|r|^2}[/math]

 

 

[math]=> \frac {4 \pi*10^-7_{N/A^-2}*1_C*1_{m/s} \times (1_C*1_{m/s} \times 1)}{4 \pi*1_{m^2}}[/math]

 

 

[math]=> \frac {4 \pi*10^-7_{N/A^-2}*1_A*1_m \times (1_A*1_m)}{4 \pi*1_{m^2}} \ =OR= \ 10^-7_{N/A^-2}*1_{C/s} \times (1_{C/s})[/math]

 

 

[math]=> 10^-7_{N/A^-2}*1_A \times 1_A \ = \ 10^-7 \ N[/math]

 

 

 

 

This is not some kind of "taboo", is it now? This is *science* forum, right?

[Mr Skeptic]

Well for starters, re-opening closed topics is going to inevitably get you in trouble. Still, if it stays with this specific question you might be OK.

---

Merged post follows:

Consecutive posts merged

Yes, it should give units of force.

Edited April 7, 2010 by Mr Skeptic

[Cap'n Refsmmat]

[math]F_{12} = \frac {\mu_0}{4 \pi} \int \int \frac {I_1*dl_1 \times (I_2*dl_2 \times \hat{r}_{21} )}{|r|^2} = \frac {\mu_0*I_1* \int dl_1 \times (I_2* \int dl_2 \times \hat{r}_{21} )}{4 \pi*|r|^2}[/math]

 

You forgot that these are definite integrals along paths in the original formulas. You need to perform definite integration.

 

Now, let's not go around in the same circles again, shall we? Or this thread may have an extremely short life.

[swaha]

what are I1, I2. l! l2?

[darkenlighten]

You also have to remember that just because you have the correct dimensions, it does not necessarily mean you have the correct answer. It is just a step towards making sure you do. The next step is checking the math.

[ambros]

You forgot that these are definite integrals along paths in the original formulas. You need to perform definite integration.

 

Now, let's not go around in the same circles again, shall we?

 

Then answer the question or go away and stop trolling. This is not about solving integrals, it is simply about writing units down and performing the dimensional analysis. I'm not interested in your hallucinations, show your work and point exactly what term in what step do you imagine is incorrect, if you can - if you can not, then you are not qualified nor competent to be giving any advice. You only keep demonstrating how much you do not know, as no integrals need to be solved, for example this is how you do dimensional analysis of Biot-Savart law:

 

[math]B = \int \frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2} = \frac{\mu_0*I*\int dl \times \hat r}{4\pi*|r|^2}[/math]

 

[math]

=> \frac{4\pi*10^-7_{N/A^-2}*1_A* 1m \times 1}{4\pi*1_{m^2}} = \frac{10^-7_{N/A}}{1_m} = \ \ 10^-7 \ N/A*m \ = \ 10^-7 \ Tesla[/math]

 

 

 

Or this thread may have an extremely short life.

 

That scares me as much as if I was threatening you that I will never log in, or post on this forum again. Keep it up, and when all is left is three of you with bunch of teenagers parroting from the text books and unable to think, then you will surely enjoy every discussion, because then all will be saying exactly the same thing... without ever realizing or even being bothered if what they say is actually true or has any connection with reality. That is the only kind of place where you can fool yourself to know anything, but considering how my previous thread got closed and how many people was involved in that discussion and what they have said, I think you're doing pretty good job at it already, and once you get rid of me, you will again rule these barren lands, and 4+2 can go back to equal 71, or 55, as you see fit.

Edited April 8, 2010 by ambros

[Bignose]

Sadly, no one managed to realize what are the units of this equation and point if there are any mistakes in this dimensional analysis:

 

[math]F_{12} = \frac {\mu_0}{4 \pi} \int \int \frac {I_1*dl_1 \times (I_2*dl_2 \times \hat{r}_{21} )}{|r|^2} [/math]

 

A dimensional analysis doesn't have plugged-in numbers in it. All it tells you is what units the final results are in.

 

i.e. [math]\mathbf{F} = m\mathbf{a}[/math]

 

let the symbol [=] indicate "has units of"

 

m [=] m, mass (no big surprise there)

a [=] [math]\frac{l}{t^2}[/math] length per time squared

 

That means that F [=] [math]\frac{ml}{t^2}[/math]

F has units of mass*length per time squared.

 

That is all. The equation above boils to to having units also of mass*length per time squared. It says absolutely nothing about the correctness of the specific equation.

 

Think of it this way: A correct equation MUST have the correct units, but an equation with the correct units doesn't mean that the equation itself is correct.

 

Here is another example. Dimensionally, work and torque both have units of Newton*meter (or mass squared * length per time squared if you prefer). However, they are definitely not the same thing, even though dimensional analysis of both equations would show that they are dimensionally equivalent. Dimensional analysis is an excellent start, but it doesn't actually solve anything in the end.

Edited April 8, 2010 by Bignose
spelling

[Cap'n Refsmmat]

Then answer the question or go away and stop trolling. This is not about solving integrals, it is simply about writing units down and performing the dimensional analysis.

You'll note that if you, say, integrate velocity with respect to time, the result will be a distance, not a velocity. Integration matters.

 

I'm not interested in your hallucinations

Damn. Not even the fuzzy pink unicorns from last night? I swear they farted rainbows.

 

Keep it up, and when all is left is three of you with bunch of teenagers parroting from the text books and unable to think

I'm sure swansont's flattered to be lumped with the teenagers. Getting a PhD in physics while a teenager is a pretty good feat.

 

I think you're doing pretty good job at it already, and once you get rid of me, you will again rule these barren lands, and 4+2 can go back to equal 71, or 55, as you see fit.

Sixty-seven, I think. It's a far more pleasant number.

[Mr Skeptic]

You'll note that if you, say, integrate velocity with respect to time, the result will be a distance, not a velocity. Integration matters.

 

I'm pretty sure it doesn't matter for dimensional analysis. When it comes down to it, an integration is the sum of a whole bunch of very thin areas... the result would be the same as multiplying for the dimensional analysis, although of course the numbers would depend on the specifics.

 

---

 

To get the correct number out of that formula you really do need to do the integrals, and properly. Carrying numbers through the dimensional analysis doesn't cut it.

[Cap'n Refsmmat]

I'm pretty sure it doesn't matter for dimensional analysis. When it comes down to it, an integration is the sum of a whole bunch of very thin areas... the result would be the same as multiplying for the dimensional analysis, although of course the numbers would depend on the specifics.

Hmm. I suppose if you multiply by the unit of the infinitesimal, you'd be fine.

 

Hrm. Would cross products yield multiplied units? The magnitude of the cross product of two vectors is the same as the area of the parallelogram spanned by them, so I'd assume that cross products are like multiplications for units.

[Mr Skeptic]

Hrm. Would cross products yield multiplied units?

 

Yes, although as only one vector. Consider area vectors, for example -- they are a vector with units of area resulting from a cross product of two vectors with units of distance.

[Cap'n Refsmmat]

Yes, although as only one vector. Consider area vectors, for example -- they are a vector with units of area resulting from a cross product of two vectors with units of distance.

Sense is made!

 

[math]F_{12} = \frac {\mu_0}{4 \pi} \int \int \frac {I_1*dl_1 \times (I_2*dl_2 \times \hat{r}_{21} )}{|r|^2}[/math]

First, let's move that junk out of the integral.

 

[math]F_{12} = \frac {I_1 I_2 \mu_0}{4 \pi} \int \int \frac {dl_1 \times (dl_2 \times \hat{r}_{21} )}{|r|^2} = \frac {I_1 I_2 \mu_0}{4 \pi} \int dl_1 \times \int \frac {dl_2 \times \hat{r}_{21} }{|r|^2}[/math]

 

Now, let's look at that last integral specifically. Here's how its units work out:

 

[math]\int \frac {dl_2 \times \hat{r}_{21} }{|r|^2}= \int \frac{\mbox{m} \times \mbox{1}}{\mbox{m}^2} = \frac{1}{\mbox{m}}[/math]

 

(since r hat is a unit direction vector, it doesn't have units)

 

So.

 

[math]F_{12} = \frac {I_1 I_2 \mu_0}{4 \pi} \int dl_1 \times \frac{1}{\mbox{m}} = \frac {\mbox{A A N}}{\mbox{A}^2} \mbox{m} \frac{1}{\mbox{m}} = \mbox{N}[/math]

 

I think that's right. So the equation at least has the right units.

[ambros]

A dimensional analysis doesn't have plugged-in numbers in it. All it tells you is what units the final results are in.

 

Numbers tell you what is the constant' date=' and in what units that constant ends up.

 

 

i.e. [math]\mathbf{F} = m\mathbf{a}[/math]

 

let the symbol [=] indicate "has units of"

 

m [=] m, mass (no big surprise there)

a [=] [math]\frac{l}{t^2}[/math] length per time squared

 

That means that F [=] [math]\frac{ml}{t^2}[/math]

F has units of mass*length per time squared.

 

How about you first learn this:

 

unit for time = "s" - second

unit for length = "m" - meter

unit for mass = "kg" - kilogram

 

http://en.wikipedia.org/wiki/Newton_(unit)

 

Is that supposed to be a news to someone here? I said equation is in Newtons, didn't I?

 

 

[math]

=> 10^-7_{N/A^-2}*1_A \times 1_A \ = \ 10^-7 \ N \ = \ 10^-7 \ kg*m/s^2

[/math]

 

Is this better?

 

 

That is all. The equation above boils to to having units also of mass*length per time squared. It says absolutely nothing about the correctness of the specific equation.

 

Think of it this way: A correct equation MUST have the correct units, but an equation with the correct units doesn't mean that the equation itself is correct.

 

And your conclusion is then, what? Are you saying that equation given by the International Bureau of Weights and Measures (BIPM), that defines the ampere unit, and therefore all the electronics and electrics on this planet, is not correct, or that I made some mistake in my dimensional analysis?

[Cap'n Refsmmat]

How about you first learn this:

 

unit for time = "s" - second

unit for length = "m" - meter

unit for mass = "kg" - kilogram

 

http://en.wikipedia.org/wiki/Newton_(unit)

 

Is that supposed to be a news to someone here? I said equation is in Newtons, didn't I?

It doesn't matter what symbols you use, as long as you know what they mean.

 

And your conclusion is then, what? Are you saying that equation given by the International Bureau of Weights and Measures (BIPM), that defines the ampere unit, and therefore all the electronics and electrics on this planet, is not correct, or that I made some mistake in my dimensional analysis?

 

I believe Bignose is implying that the units can be correct but the math can still be wrong. But I won't put words in his mouth.

[Mr Skeptic]

OK, now how about we do a dimensional analysis of this equation:

[math]KE = \int_{v = 0 m/s}^{v = 10 m/s}1 kg * a * dr = \int_{v = 0 m/s}^{v = 10 m/s}1 kg * v * dv[/math]

[ambros]

Sense is made!

 

 

First' date=' let's move that junk out of the integral.

 

[math']F_{12} = \frac {I_1 I_2 \mu_0}{4 \pi} \int \int \frac {dl_1 \times (dl_2 \times \hat{r}_{21} )}{|r|^2} = \frac {I_1 I_2 \mu_0}{4 \pi} \int dl_1 \times \int \frac {dl_2 \times \hat{r}_{21} }{|r|^2}[/math]

 

Now, let's look at that last integral specifically. Here's how its units work out:

 

[math]\int \frac {dl_2 \times \hat{r}_{21} }{|r|^2}= \int \frac{\mbox{m} \times \mbox{1}}{\mbox{m}^2} = \frac{1}{\mbox{m}}[/math]

 

(since r hat is a unit direction vector, it doesn't have units)

 

So.

 

[math]F_{12} = \frac {I_1 I_2 \mu_0}{4 \pi} \int dl_1 \times \frac{1}{\mbox{m}} = \frac {\mbox{A A N}}{\mbox{A}^2} \mbox{m} \frac{1}{\mbox{m}} = \mbox{N}[/math]

 

I think that's right. So the equation at least has the right units.

 

At least? Do you not see that is THE EQUATION, the one that defines the ampere unit, and with that all the rest of equations that have anything to do with any el. currents in classical electromagnetism. You do not need to split equation to parts, you did the same thing as I did only messier and without showing all the steps properly, but at least you arrived at correct conclusion, finally. Look again, all you did is to write those two "dl" as "1m".

 

 

 

BIPM: -"The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 metre apart in vacuum, would produce between these conductors a force equal to 2 x 10^-7 newton per metre of length."

 

 

What you just proved is that the force on ONE wire, as defined by the BIPM equation for magnetic force that you just dimensionally solved, gives the result that is equal to 10^7 N, (the constant). And so to get the correct result using this equation both F12 and F21 need to be taken into account.

 

F(between) = F12(magnitude)+F21(magnitude) = 2*10^-7 N

 

F(m) = F/1m = 2*10^-7 N/m

[Cap'n Refsmmat]

At least? Do you not see that is THE EQUATION, the one that defines the ampere unit, and with that all the rest of equations that have anything to do with any el. currents in classical electromagnetism. You do not need to split equation to parts, you did the same thing as I did only messier and without showing all the steps properly, but at least you arrived at correct conclusion, finally. Look again, all you did is to write those two "dl" as "1m".

I suppose. But there's no sense doing something the easy way when you can make it more complicated.

 

What you just proved is that the force on ONE wire, as defined by the BIPM equation for magnetic force that you just dimensionally solved, gives the result that is equal to 10^7 N, (the constant). And so to get the correct result using this equation both F12 and F21 need to be taken into account.

 

F(between) = F12(magnitude)+F21(magnitude) = 2*10^-7 N

 

F(m) = F/1m = 2*10^-7 N/m

 

We've been over this before, haven't we?

[ambros]

OK, now how about we do a dimensional analysis of this equation:

[math]KE = \int_{v = 0 m/s}^{v = 10 m/s}1 kg * a * dr = \int_{v = 0 m/s}^{v = 10 m/s}1 kg * v * dv[/math]

 

No, we are integrating over distance. That is extremely different from your example. No other examples are like this, we are talking about the real world here and there are very specific things that I want to point out, which I can not do on anything else but on this very particular unit and the definition of ampere. This is the integral we are talking about here:

 

[math]I\int dl \ => \ 1 A = 1\tfrac C s \ => \ \mathrm{C} = 1 \mathrm{A} \cdot 1 \mathrm{s}[/math]

 

You think there is some time integral or spatial derivation involved here? Tell me then, what speed is that of 1C/s? How many meters per second is that? AMPERE: -"6.242 × 10^18 electrons passing a given point each second constitutes one ampere." -- Passing a given POINT? What speed is that? How many meters per second is that? What is the LENGTH of this "point"?

Edited April 8, 2010 by ambros

[Mr Skeptic]

How about:

[math]KE = \int_{v = 0 m/s}^{v = 10 m/s}1 kg * \frac{1}{t^2} * r * dr[/math]

 

That's a nice little integral over distance.

[ambros]

We've been over this before' date=' haven't we?[/quote']

 

Yes, but this is the first time you actually solved it, and hence you proved my point. -- By doing dimensional analysis what you did is to actually solve that equation for the force F12 exerted on wire '1' , where there are two parallel wires and with given unit values of: I1=1A, I2=1A, distance r=1m. -- You got result which is 1 N, and that leaves us with the NUMERICAL value of the magnetic constant, in Newtons, of course:

 

[math]F_{12} = \frac {\mu_0}{4 \pi} = 10^-7 N[/math]

 

[math]F_{21} = \frac {\mu_0}{4 \pi} = 10^-7 N[/math]

 

[math]F_{between \ two} = F_{12 magnitude} + F_{21 magnitude}= \ 2 *10^-7 N[/math]

 

 

I'm glad we agree now, and thanks for the help.

Edited April 8, 2010 by ambros

[Cap'n Refsmmat]

Sigh.

 

I did dimensional analysis. I did not compute the value of the function. I have shown how to compute its value before.

[ambros]

How about:

[math]KE = \int_{v = 0 m/s}^{v = 10 m/s}1 kg * \frac{1}{t^2} * r * dr[/math]

 

That's a nice little integral over distance.

 

You closed my last thread by saying it is going nowhere and now instead of talking about what this thread is about you are trying to get off topic?!

 

 

CAN YOU RESPOND TO THIS:

 

[math]

I\int dl \ => \ 1 A = 1\tfrac C s \ => \ \mathrm{C} = 1 \mathrm{A} \cdot 1 \mathrm{s}

[/math]

 

What speed is that of 1C/s?

 

How many meters per second is that?

 

 

AMPERE: -"6.242 × 10^18 electrons passing a given point each second constitutes one ampere."

 

Passing a given POINT? What speed is that?

[Cap'n Refsmmat]

I think this should settle things nicely... (click to see large version)

 

 

Here's what I did in Mathematica:

 

I defined C1 and C2, the paths representing the wires, as in the original equation:

[math]\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_1} \oint_{C_2} \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2}[/math]

 

You can see that they are 1 meter apart, since C2 is one meter higher in the y dimension.

 

Next, I specified their derivatives, which gives us [math]d \mathbf{s_1}[/math] and [math]d \mathbf{s_2}[/math].

 

Next, I specified [math]\mu_0[/math].

 

Then I wrote out the Ampere force law as an integral. You'll note the limits on the integral. Ampere's force law gives us the force on a piece of wire -- its units are Newtons, not Newtons per meter. So one wire is infinite, with limits from negative infinity to positive infinity. For the other wire, we will consider a one-meter segment, so we establish the force on one meter of one wire.

 

Regarding the Norm[]s: Norm[] computes the magnitude of a vector. C1 - C2 gives us the vector pointing from one to the other, as you know from basic vector analysis. The fraction (C1 - C2 divided by its norm) gives us a vector pointing from one to the other, of unit length -- [math]\hat{\mathbf{r}}_{12}[/math].

 

I evaluated it. You can see that the result is [math]2 \times 10^{-7} \mbox{ N}[/math] in the y direction. Not [math]1 \times 10^{-7} \mbox{ N}[/math]. The one meter segment of wire experiences a force of [math]2 \times 10^{-7} \mbox{ N}[/math] from its interaction with the infinite wire.

 

If I extended the one meter wire to infinity, the force it experiences would become infinitely large, of course, so we have to calculate the force per meter. That's why I limit the integral to one meter.

Edited April 8, 2010 by Cap'n Refsmmat

[ambros]

I think this should settle things nicely... (click to see large version)

 

[attach]2456[/attach]

 

Here's what I did in Mathematica:

 

Please copy/paste that here in text format so I can easier refer to it.

 

* int[54] is wrong, move y to 1

 

* why is the x coordinate marked as "t", some relation to time?

 

* where did you get idea that 2nd integral is from - to + infinity?

 

* "t" is not initialized and it does not appear in the final equation?

 

* write down final equation, so I can see better and point to rest.

 

 

Will you please plug in the equation EXACTLY as it is written, which means that both of those two integrals integrate over the same distance - IN PARALLEL... and see what Mathematica will tell you.

 

 

Ampere's force law gives us the force on a piece of wire -- its units are Newtons, not Newtons per meter.

 

Close, but you are still mixing two very different equations. Magnetic force IS NOT DEFINED with Ampere's force law, it is defined with Biot-Savart law and Lorentz force equations. This is Ampere's force law, it is some OLD equation that specifically relates only to parallel wires and force per unit length:

[math]F_m = F/1_m = 2 k_A \frac {I_1 I_2 } {r}[/math]

 

 

This below is NOT Ampere's force law, and these two equations ARE NOT EQUAL:

[math]\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_1} \oint_{C_2} \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2}

[/math]

 

...this is Biot-Savart and Lorentz force combined, and the difference should be obvious, beside that the units are different, as you finally have realized now. We are getting closer to truth. Very good, I'm proud of you.

 

 

 

...so we have to calculate the force per meter.

 

Yes, and if that doesn't work just use a bigger hammer.

Edited April 8, 2010 by ambros

[Mr Skeptic]

You closed my last thread by saying it is going nowhere

 

Wasn't it?

 

and now instead of talking about what this thread is about you are trying to get off topic?!

 

My question is a dimensional analysis question, which seems to be the topic. But hey, if you can't answer it that's cool.

 

CAN YOU RESPOND TO THIS:

 

[math]

I\int dl \ => \ 1 A = 1\tfrac C s \ => \ \mathrm{C} = 1 \mathrm{A} \cdot 1 \mathrm{s}

[/math]

 

What speed is that of 1C/s?

 

Speed has units of distance over time. This isn't a speed, it's a flow.

 

How many meters per second is that?

 

 

AMPERE: -"6.242 × 10^18 electrons passing a given point each second constitutes one ampere."

 

Passing a given POINT? What speed is that?

 

An area if you prefer. You need to take into account the free electron density and the wire's cross sectional area before you can find the average drift velocity of an electron.