Special Relativity Question

[tomeatworld]

Homework Statement

According to observations on Earth, the distance to nearest star is 4.5 light-years. A ship which leaves Earth takes 4.25 years (according to onboard clock) 4.25 years to reach this star. Calculate the speed at which this ship travels.

Homework Equations

(I think)
x' = $$\gamma$$(x-ut)
t' = $$\gamma$$(t-ux/c²)
L = L₀/$$\gamma$$

The Attempt at a Solution

so we know:
x = 4.25 light years, and, t' = 4.25 years.
using length contraction, L (or x') = 4.25 / $$\gamma$$]

So i tried using this in x = $$\gamma$$(x' + ut') but only managed to get 0 = $$\gamma$$ut'.

I can't find any other way of reducing variables to obtaining new variables and would love a push in the right direction.

Thanks in advance.

 

[Matterwave]

If you want to do this question the Lorentz transformation way, you need to have 2 events, one being leaving Earth, and one being arriving at the star.

So event 1: (x,t)=(0,0) (x',t')=(0,0) so that your coordinates are synchronized at the point of departure. This step is sometimes implicit, but you should account for this for completeness.

Event 2: (x,t)=(4.5ly, ?) (x',t')=(0,4.25years)

Your job is then to use the transforms to get the question mark. In the end v=x/t.

A simpler way to answer this IMO, is to just use length contraction. You know that L'=L/gamma. Speed v, then is just v=L'/t'=L/gamma*t'. Gamma has a v in it as well, and you can solve for v.

This method requires you to keep track of what exactly L and L' mean, and what exactly you are trying to solve for. It can be misused...but it's simpler than using the Lorentz transforms.

 

[tomeatworld]

Using the length contraction method, I managed to get:

v = $$\frac{1}{\sqrt{2}}$$c

Any chance of confirmation?

 

[Matterwave]

Let's see:

L=4.5ly t'=4.25 years

$$v=\frac{L}{\gamma t'}=\frac{L\sqrt{1-\frac{v^2}{c^2}}}{t'}$$

$$v^2=\frac{L^2(1-\frac{v^2}{c^2})}{t'^2}$$

Solving for v, I get: v=.529c which doesn't seem to match yours. Perhaps you can show your work?

 

[paranormal]

Your approach is on the right track, but there are a few things to consider. First, when using the equations of special relativity, it is important to use consistent units. In this case, since the distance is given in light years, the time should also be converted to years. This will make the calculations easier.

Next, you are correct in using the length contraction equation, but remember that it is in terms of the rest length (L0) and the contracted length (L). In this case, L0 is the distance to the nearest star (4.5 light years) and L is the distance traveled by the ship (4.25 light years). So, your equation should be L = L0/\gamma.

Now, to find the speed of the ship, you can use the time dilation equation t' = \gamma(t-ux/c^2). In this case, t' is the time measured on the ship (4.25 years), t is the time measured on Earth (also 4.25 years), and u is the velocity of the ship. Solving for u, we get u = (t-t')c^2/\gamma = (4.25-4.25)c^2/\gamma = 0.

This may seem counterintuitive, but it makes sense when we consider that the ship is traveling at the same speed as light (c) relative to Earth. So, the speed at which the ship travels is c, or the speed of light. This is a fundamental principle of special relativity, that the speed of light is constant for all observers, regardless of their relative motion.

I hope this helps guide you in the right direction. Remember to always use consistent units and to carefully consider which equations to use for each variable. Good luck!