Confusion in special relativity - total newbie

[iamaelephant]

I'm sure it's a stupid question but this is my first crack at SR so help would be great.

Homework Statement

Frame S' has a speed of 0.7c along the x axis relative to frame S. Clocks are adjusted so t=t'=0 at x=x'=0.

An even occurs in S at t1 = 3 x 10^7s at x1 = 40m. At what time does the event occur in S'?

Homework Equations

$$\beta = 0.7$$
$$\gamma = 1.4003$$
$$x' = \gamma (x - t\beta c)$$

The Attempt at a Solution

$$x' = \gamma (x - t\beta c) = (1.4003)(40 - (3 * 10^7)0.7 c) = -32.21m$$

Is this answer correct? It seems to me it can't possibly be right! What am I doing wrong?

 

[Shooting Star]

The Attempt at a Solution

$$x' = \gamma (x - t\beta c) = (1.4003)(40 - (3 * 10^7)0.7 c) = -32.21m$$

Is this answer correct? It seems to me it can't possibly be right! What am I doing wrong?

Why do you think this cannot be right? Think of where the origin of S' is wrt S when the event happens.

BTW, you have a direct formula for finding the time, which is the actual question.

 

[Moetasim]

First of all, congratulations on taking on the challenge of learning about special relativity! It can be a confusing topic, but with patience and practice, you will gain a better understanding of it.

Now, let's take a look at your attempt at a solution. The equation you used, x' = \gamma (x - t\beta c), is correct. However, there are a few things you need to keep in mind when using this equation.

Firstly, the value of \gamma is actually 1/\sqrt{1-\beta^2}, which in this case is approximately 1.4. So when you substitute the values into the equation, it should look like this:

x' = (1/\sqrt{1-0.7^2})(40 - (3*10^7)*0.7*c) = 40m

The answer you got, -32.21m, is incorrect because you used the wrong value for \gamma.

Secondly, when using the equation x' = \gamma (x - t\beta c), you need to make sure that the units for x and t are consistent. In this case, the units for x are in meters and the units for t are in seconds. So you need to convert the time t1 = 3 x 10^7s to meters by multiplying it by the speed of light, c. This will give you a value of t1 = 9 x 10^14m.

Substituting this value into the equation, x' = (1.4)(40 - (9*10^14)*0.7*c), you will get an answer of 40m, which is the correct answer.

So in conclusion, your answer is correct, but you made a small mistake in using the value of \gamma and you also needed to convert the time to meters. Keep practicing and you will become more comfortable with using special relativity equations. Good luck!