Special Theory of Relativity

1. Oct 31, 2007

BatmanACC

Thanks for taking a look. The following question is that is that from a grade 12 academic physics course.

1. The problem statement, all variables and given/known data

Scientist Ludwig von Drake, while in his laboratory, measures the half-life of some radioactive material which is in a bomb, approaching with speed v. Donald Duck, who is riding on the bomb, also measures the half-life. His answer is a factor of 2 smaller than Ludwig's. What is the value of v, expressed as a fraction of c?

2. Relevant equations

Special relativity equation: t=t0/square root (1-v2/c2) Note: 2 = squared
Other: I know there is at least one more I must use but for the life of me don't know what it is.

3. The attempt at a solution

First we set V=Drakes. If this is so than Donald's equation must be equal to:

t=[t0/square root (1-v2/c2)]/2

Therefore to= 2t[square root (1-v2/c2)]

The problem is in equating the equations. They end up cancelling out because one is a direct derivative of the other. This leads me to believe I need at least 1 more equation.

It must also be noted that while von Drake may use the special equation of relativity Donald duck cannot (at least this is what I figure). I say this because Donald duck is viewing the half-life from the bomb at rest, meaning he would be more in the realm of inertial frame of reference. What equation i now use knowing that I have no idea.

2. Oct 31, 2007

learningphysics

t=t0/square root (1-v2/c2) (this is the time dilation formula)

t is Drake's time

t0 is the proper time which is Donald's time.

we are given that t = 2t0.

so you have equations:

t=t0/square root (1-v2/c2)
t = 2t0

so solve these 2 equations.

3. Oct 31, 2007

dzogi

Donal Duck is measuring the proper time interval $$t_0$$, so the interval measured by Drake is $$t=t_o\lambda$$. We're given that $$t=2t_o$$. Dividing both equations we get that $$\lambda=(1-\frac{v^2}{c^2})^{-1/2}=2$$. Solving the equation yields the solution.

4. Nov 1, 2007

BatmanACC

Thanks guys. My solution set is similar to both of yours.

I used 2T= To/sqreroot(1-v2/c2) and 2T = To/x where x=.5

This yields sqreroot(1-v2/c2) = .5

Through squaring both sides and moving the variables/numbers around you obtain

.75c2=v2

Simply square root that to get the answer which is .866c=v

Once again thanks for your help!