The star Delta goes supernova. One year later and 2.0 ly away,...

Question

The star Delta goes supernova. One year later and 2.0 ly away, as measured by astronomers in the galaxy, star Epsilon explodes. Let the explosion of Delta be at x_D = 0 and t_D = 0. The explosions are observed by three spaceships cruising through the galaxy in the direction from Delta to Epsilon at velocities v₁ = 0.30c, v₂ = 0.50c, and v₃ = 0.70c. All three spaceships, each at the origin of its reference frame, happen to pass Delta as it explodes. What are the times of the two explosions as measured by scientists on each of the three spaceships?

Accepted Answer

Hey, everyone in this problem, we're told that as measured by observers on earth, two stars A and B undergo supernova first star A explodes and two years later, a second explosion is detected from star B. A distance of five light years separates the two stars. We're told to assume that the explosion of A occurs at X A equals zero and T A equals zero. And there are two scientific spacecraft SC one and SC two traveling from A to B at a constant velocity of 0.25 C and 0.4 C respectively. When star A explodes SC one and SC two are positioned at the origin of their respective reference frames and coincidentally pass by A. So we're asked to determine the time difference between the two explosions detected by those two spacecraft crews. KSC one and SC two, we have four answer choices here A through D, each of them containing a different time difference for those two crews. And we're gonna come back to those as we work through this problem. Now, we're told a lot of information in this problem. OK. But we're gonna work through it bit by bit. Now, the main thing here is this is a relativity problem. OK. So we want to start by thinking about the frame of references that we're dealing with here, right? So we have the earth, OK. So let's think about what we've been told from our earth frame of reference and as measured by earth, OK, these stars explode two years apart. So t the time between explosions is going to be two years and the stars are separated by a distance of five light years. And so X is going to be five light years, the distance between those two stars. So we've written all the information we know about the earth. And let's go ahead and label the earth frame of reference. We're gonna call this frame. All right, then we have spacecraft one. Hey, what is spacecraft one? What do we know about it? Well, it's moving at this velocity of 0.25 C. So we have that V one is gonna be 0.25 C. OK. And we want to know the time difference between the explosions in T um relative to spacecraft one. OK? Or according to their crew. And so we're interested in T one prime. Hm. All right. Yeah, SC one. OK. We called the earth frame S, we'll call SC one frame S prime and then we have SC two. So the second spacecraft now we're told it has a speed of 0.4 C or velocity So V two is going to be 0.4 C and then we have T two double prime and we wanna know that value as well. OK. So again, that's the time difference. Now, in this case, according to the spacecraft two crew. All right. So just like we had these other frames of references for the earth in spacecraft one, we're gonna create another one for spacecraft two and that's gonna be frame s double prime. All right. Now, one really important thing that we've been told in this problem is that the explosion of A occurs at X A equals zero and T A equals zero. Then we've also been told that when star A explodes sc one and SC two are positioned at the origin of their reference frames and coincidentally pass by A. OK. So what this means is that explosion of A occurs at the origin of all of these reference frames. OK. So the origins coincide because the origins coincide, we can use our Laurents transformations. OK. All right. So let's go ahead and work this out for each of our spacecrafts. Now, spacecraft one we're gonna do in blue spacecraft two, we're gonna do in green just so we can see the difference and make sure that we're not mixing them up. So for spacecraft one, what do we have or recall that our Lorentz transformation for time we have T one prime is going to be equal to gamma multiplied by T minus V one X divided by C squared. OK. We have T we have V one, we have XC is the speed of light. So we know that that's constant. What about this gamma value? I'll recall that gamma is equal to one divided by the square root of one minus beta squared. OK. In this case, we have, we're gonna call it beta one because we're dealing with spacecraft one and beta is equal to V divided by C. So again, in this case, V one divided by CC, this is a 0.25 C divided by C. And so we get a beta value of 0.25. All right. So we know beta which allows us to find gamma. And then the other values we've mentioned, we already know them all. So we can substitute in and solve for T one prime. So we have that T one prime is going to be equal to T which is two years minus V 10.25 C multiplied by X five, what years divided by the squad? And all of this is gonna be divided by the square root of one minus beta squared, which is one minus 0.25 squared. All right. So let's go ahead now. OK. We have C divided by C squared. OK. So we can divide out one of those C values. Then we have light years divided by C. Now recall that the speed of light C is equivalent to a light year per year. So if we have a light year divided by a light year per year, that's gonna leave us with a unit of years. And then we just have those two numbers multiplied together. So we get that T one prime is going to be equal to two years minus 1.25 years divided by the square root 0.93 75. And we can work this out on our calculator to get a final value of T one for the spacecraft. One of 0.77 for six years. Hm. All right. So event A occurred at the origin and according to the spacecraft, one crew, the event B or the explosion of star B happens 0.7746 years or later. OK. So let's take a look at our answer traces. See if we can narrow anything down. Option A and option B have the explosion of B happening before the explosion of A. So that's not what we found. So we can eliminate those options. C and D have spacecraft one crew finds the explosion of B happening 0.77 years after the explosion of A. So that's what we just found in blue. And so either of those options could be correct, we can move on to spacecraft two to figure out what answer choice is going to be correct. So for spacecraft two, we're going to do the exact same thing. OK, we just have some different values to use. So using our Lawrence transformation again, for time T two double prime is going to be equal to gamma multiplied by T minus V two multiplied by X divided by C squared. Now gamma is the same in this case, in terms of its equation one divided by the square root of one minus beta squared. But because we have a different V value, that means we have a different beta value and therefore different gamma value. So beta in this case is going to be V two divided by C which is 0.4 C divided by C. And this gives us a beta value of 0.4. So we can substitute it to our T two equation T two double prime is going to be equal to that time again of two years. Hey, that's the time between our explosions in the earth frame of reference. OK. Then we're gonna subtract the speed 0.4 C multiplied by the distance in that original frame of reference, five light years divided by C squared. And all of this is gonna be divided by that square root. And in this case, that square root is one minus 0.4 squared. All right. So the same thing is gonna happen in the numerator with these CS and one of them is gonna divide out the other light year per year is going to give us a unit of years in that numerator. And so we get that T two double prime is going to be equal to two years minus two years divided by the square root of one minus 0.4 squared. OK? Two years minus two years. Well, that's just gonna give us zero. And so what we get is that T two double prime is equal to zero. What this tells us is that the time between the explosions is zero. OK. This means that the explosions occur simultaneously, they occur at the same time according to spacecraft crew. Two. All right. So let's get back to our answer choices. We've answered both parts of the problem. Now, we had to narrow it down to C or D and we now know that according to spacecraft two crew, the explosions are simultaneous. And so the correct answer here is going to be option D spacecraft one crew finds that the explosion of B happen 0.77 years after a in spacecraft two crew finds that the two explosions are simultaneous. Thanks everyone for watching. I hope this video helped see you in the next one.

Key Concepts

Relativity of Simultaneity

Lorentz Transformation

Time Dilation

Watch next