Root problem.

[Gareth56]

In a book I have the following step isn't clear to me:-

 

r = m x sqrt(kT/m) ------> r = sqrt(mkT)

 

I just can't see the steps that go from the first equation to the second. Could anyone show me the missing step(s)?

[timo]

Here's a possible explanation. Feel free to skip any step if you find it unnecessary and also free to ask if one of the steps is not clear to you: [math] m \sqrt{ kT/m } = \sqrt{m^2} \sqrt{ kT/m } = \sqrt{ m^2kT/m } = \sqrt{mkT} [/math].

Edit: I was assuming that m is a mass, k the Boltzman constant and T the temperature and that hence all of these three variables are positive. You might have to be careful if some of them can be negative, e.g. m=sqrt(m^2) does not hold when m<0.

[Gareth56]

Hi can I just ask how and why [math]m[/math] can become [math]\sqrt{m^2}[/math] ?

 

 

I've just realised that [math]m[/math] is the same as [math]\sqrt{m^2}[/math]

 

So was it really necessary to simplify [math]m \sqrt{ kT/m }[/math] to [math]\sqrt{mkT}[/math] or could you just have left it as it was?

Edited November 6, 2009 by Gareth56
Penny dropped!!!

[timo]

It's not absolutely needed and depending on the context either version could be preferable. But usually, the shorter an expression and the less frequent symbols appear, the better.

 

OLD POST, REDUNDANT:

Sure, I am just not sure how to explain it. Seems pretty obvious that the square root of a number squared gives the number. The square root is something like the anti-operation (the technical term is "inverse") of squaring a number. Would you agree that [math]\sqrt{x}^2 = x[/math]? In that case you can rearrange it as [math] \sqrt{m^2} = \sqrt{m\cdot m} = \sqrt m \sqrt m = \sqrt{m}^2 = m[/math]. Perhaps you should also just do some reading on what the square root is and how you rearrange expressions involving it.

 

Or a somewhat more intuitive explanation:

The square root of X tells me which value Y=sqrt(X) I'd have to square to get X. So which value do I have to square to get m-squared? m. Hence m = sqrt(m*m).

[Gareth56]

Many thanks. 'O' level maths was so long ago

[Bignose]

m is almost certainly the symbol for mass, which will always be a positive number. In that case, the square root of the square is always positive as well - there is no ambiguity in taking the square root.

 

I think that [math]\sqrt{mkT}[/math] looks better than [math]m\sqrt{\frac{kT}{m}}[/math], but in the end it is just aesthetics.

[the tree]

So was it really necessary to simplify [math]m \sqrt{ kT/m }[/math] to [math]\sqrt{mkT}[/math] or could you just have left it as it was?

When you're trying to find a value for this, two multiplications and a square root is preferable to two multiplications, a square root and a division. Partially because any operation is going to take up computation time and cause a loss of accuracy and also because division in particular is very sensitive to small inaccuracies.

[magicalprimate]

The first equation was m^1 x (kT)^1/2/(m)^1/2, which is (kT)^1/2(m^1)/[m^1/2].

 

m^1 divided by m^1/2 is m^1/2, and (kT)^1/2 is multiplied by that in the above to equal (kT^1/2)(m^1/2), and since the square root of a a dimensionless quantity multiplied by the square root of another dimensionless quantity is the same as the square root of both of those things multiplied together, the previous equation is (kT x m)^1/2 which is of course the same as sqrt(mkT)