Relativity: Mass with respect to velocity

[ek=1/2mv^2]

Homework Statement

In the theory of relativity, mass of a particle is

m= m_{0}/ \sqrt{1-v^{2}/c^{2}}

m is the mass when the particle moves at v relative to the observer, and m_{0} is the rest mass.

Sketch the graph of m with respect to v.

Homework Equations

E = mc^{2}

The Attempt at a Solution

No idea where to start :(

 

[Curious3141]

Don't you know how to sketch a curve? Think of what the vertical intercept is, and whether there's an asymptote.

 

[LCKurtz]

Homework Statement

In the theory of relativity, mass of a particle is

m= m_{0}/ \sqrt{1-v^{2}/c^{2}}

m is the mass when the particle moves at v relative to the observer, and m_{0} is the rest mass.

Sketch the graph of m with respect to v.

Homework Equations

E = mc^{2}

The Attempt at a Solution

No idea where to start :(

Would you know where to start if the problem was to graph$$
y = \frac 1 {\sqrt{1-\frac{x^2} 9}}$$?

 

[ek=1/2mv^2]

Theory of special relativity graph

Homework Statement

In the theory of special relativity, mass of a particle is

m = \frac{m_{0}}{\sqrt{1-v^{2}/c^{2}}}

m is the mass when the particle moves at v relative to the observer, and m0 is the rest mass.

Sketch the graph of m with respect to v. 2. Homework Equations

m = \frac{m_{0}}{\sqrt{1-v^{2}/c^{2}}}

The Attempt at a Solution

In order to sketch the graph, I had to first (before finding the derivative and second derivative):

- Define the domain
- Find the x and y intercepts
- Find where there would be asymptotes (both vertical and horizontal)
- Check for symmetry

DOMAIN

When the x-axis is the velocity of a particle, the domain is 0 \leq x\leq c

INTERCEPTS

f(x) = y = m

x = v

therefore,

f(x) = \frac{m_{0}}{\sqrt{1-x^{2}/c^{2}}}
I didn't find any x-intercepts

Found a y-intercept at y = m_{0}

ASYMPTOTES
since domain has been restricted to be between 0 and C, there is no horizontal asymptote.

Vertical asymptote:
Took the just the denominator and made it = to zero, got v=c. Since v is x. It is x = c, the vertical asymptote. SYMMETRY

There is Even symmetry, f(x) = f(-x)

Ok, so now the 1st and 2nd derivatives

1st derivative

m^{\prime} = \frac{-mv}{c^{2}-v^{2}}

x-intercept at 0, therefore there is an interval of increase between 0 and c.

2nd derivative

m^{\prime \prime} = \frac{-(m + v) + 2m^{\prime}v}{c^{2} - v^{2}}

Here is when i am really stuck, how do we find the zeros for the second derivative?

BTW, pic of question and answer is attached. (number 53)

 

[Simon Bridge]

You have an equation for m and m' to substitute in, m''=0, so solve for v.
Shortcut - what would m''=0 tell you?
What would that mean for relativity?
Is that reasonable physically?

http://www.weburbia.com/physics/mass.html

 

[ek=1/2mv^2]

Thanks, that helped alot. Have a great day.