Derivative ?

Question

Some of you may know this formula for the mass of an object near the speed of light (where m0 is the initial mass, v the speed and c the speed of light) :

m = m0 / (1 - v^2/c^2 )^½

I have to find dm/dv. I know it's something like dm/dt * dt/dv but I cant find the link...

Help me please !

right, chain rule...

By the way I know the answer but I need to know how to get there :

dm/dv = m0v / [ c^2 ( 1 - v^2/c^2 )^(3/2) ]

Answer 1

m0 is a constant, bring it outside. Taking the derivative does involve the chain rule. Consider it like this:

(1-v^2/c^2)^(-1/2) -> (x)^-1/2*dx/dv

=>-1/2*(1-v^2/c^2)*dx/dv
=>-1/2*(1-v^2/c^2)^(-3/2)*(-2v/c^2) (Derivative of the inside term)

Replacing m0

=>m0v/c^2*(1-v^2/c^2)^(-3/2)

Notice that if you call (1-v^2/c^2)^(-1/2)=gamma

and v/c=beta

then you can simplify:
d/dv(gamma)=beta/c*gamma^3

These are the terms commonly used in relativity.

Good luck!

PS: If your teacher tries to tell you your mass changes with velocity, you tell him/her that they're *wrong*. That's essentially that this derivative would "prove" to you, but it's simply not true. Your energy increases, but your mass is intrinsic.

Answer 2

I see two or three different opinions, so I'll try it myself. mo, the rest mass, is a constant, but m varies with v. Also, the way it's set up, I don't think you need to get dm/dt and dt/dv. Actually, dt/dv is really convoluted, since dv/dt is much more common. When used, t tends to be the independent variable and v is dependent.

[Edit: In relativity, t does vary with velocity, rather than the other way around. But you don't need that in this problem.]


To do this problem in a straightforward fashion, try this:

m = mo / (1 - v^2/c^2)^(1/2) = (mo/c)(c^2 - v^2)^(-1/2)

dm/dv = (mo/c)(-1/2)(c^2 - v^2)^(-3/2)(-2v)

dm/dv = (mo v/c) / (c^2 - v^2)^(3/2)

There's your derivative. You can tweak the form of it if you like ... the c can be brought back under the radical, and since you have a three-halves power in the denominator, you can rationalize it by multiplying top and bottom by sqrt(c^2 - v^2) if you wish.

But the basic derivative is right there.

Answer 3

dm/dv is notation for the derivativ of the equation, the derivative of m in terms of v and your m0 is a constant, so dont make it harder than it looks. Dont use the chain rule it is not the way. Use the quationt rule which is ((low times derivative of high)-)high times derivative of low)) divided by low, squared where the numerator is high and the denominator is low
p.s. no offense to the guy with a phd, hes write when talking about astropysics, but this is just basic calculus and you would get no credit for his answer

Answer 4

"dm/dt * dt/dv"

You mean something like the chain rule?