Hard physics Relativity homework

[Hybird]

Two space ships (F and M) start at point x=x'=0 and t=t'=0. F stays put while M takes off in the x direction at 0.6c. After 350 min on F's clock, M turns around and heads back (assuming instantly). 350 more min on F's clock and M is back at start. T = 700 min.

calculate time t' that has elapsed on M's clock for the first half of the trip.

Now I would assume this is just where you use the Lorenz factor isn't it? So that M's time should measure shorter than 350 min for the first half.. but in the question it has the hint: Calculate M's time coordinate t', for the point tr=350 min, xr=v*tr

shouldnt it just be... Gamma = 1.25, so t' = 280??

 

[anathema]

Yes, you are correct. The Lorentz factor, represented by the symbol gamma (γ), is used to calculate the time dilation effect between two reference frames in special relativity. In this scenario, the reference frame of F is at rest, while the reference frame of M is moving with a velocity of 0.6c.

To calculate the time dilation effect, we use the formula t' = γt, where t is the time measured in the rest frame and t' is the time measured in the moving frame. In this case, t = 350 min and γ = 1.25. Therefore, t' = 1.25 * 350 = 437.5 min.

However, the hint given in the question suggests that we should also consider the position of M at the given time t = 350 min. At this point, M has traveled a distance of xr = v*t = 0.6c * 350 min = 210 ly (light-years) in the x direction. This means that M is now at a different position in the x direction compared to the start point, x' = 210 ly.

To calculate the time coordinate t' for this new position, we use the formula t' = γ(t - vx'/c^2), where v is the velocity of M and c is the speed of light. Plugging in the values, we get t' = 1.25(350 min - (0.6c * 210 ly)/c^2) = 280 min.

So, the time elapsed on M's clock for the first half of the trip is 280 min. This is shorter than the time measured in the rest frame, as expected due to the time dilation effect.

 

[quicknote]

Yes, you are correct in using the Lorentz factor to calculate the time elapsed on M's clock for the first half of the trip. The Lorentz factor (represented by the symbol γ) is given by:

γ = 1 / √(1 - v^2/c^2)

where v is the velocity of M (0.6c in this case) and c is the speed of light.

Using this formula, we can calculate the Lorentz factor to be 1.25. This means that time on M's clock will pass 1.25 times slower than time on F's clock. Therefore, for 350 minutes on F's clock, only 280 minutes will have elapsed on M's clock.

However, the hint in the question asks you to calculate M's time coordinate (t') at a specific point in time (tr=350 min) and space (xr=v*tr). This is where the concept of time dilation comes into play. Time dilation is the phenomenon in which time appears to pass slower for an object moving at high speeds relative to an observer.

In this case, we can use the time dilation formula to calculate M's time coordinate at tr=350 min:

t' = γt - vγx

where t is the time on F's clock (350 min), v is the velocity of M (0.6c), and x is the distance traveled by M (v*tr). Plugging in the values, we get:

t' = (1.25)(350 min) - (0.6c)(1.25)(350 min)

t' = 437.5 min - 262.5 min

t' = 175 min

This means that at tr=350 min, M's clock will show a time of 175 minutes, while F's clock will show 350 minutes. This is because M is moving at a high speed, causing time to appear to pass slower for M relative to F.

Overall, the calculation of t' = 280 minutes for the first half of the trip is correct, but the hint in the question is asking for the specific time coordinate at a specific point in time and space, which takes into account the effects of time dilation.