### Quantum Mechanics Midterm Problem #1

You are gliding along in a revolutionary high-speed vehicle at a constant 0.8c. On the ground are two stations, X and Y, that according to ground observers have clocks synchronized and are 9.0×10

Question by Randy Harris.

The first two parts of this problem are relatively simpler than the third, since they're both at the location of the event (glider passing station Y) so both parties will agree on what each clock says, since the clocks are in the same place in both frames. Classically, we would expect all the clocks to agree all the time, but since we are going ridiculously fast, it's not the case.

Part A - Clock on the Glider

This requires using the simple equation speed × time = distance. We want to know howw long it takes to travel between the stations, so solve for time:

t = d ⁄ v

The problem is that it won't appear to be 9.0×10

L = L

Where L

We use this length contraction formula to calculate the length of the path between the two stations in the frame of the glider. Once we know how far apart the two stations are, it's trivial to calculate how long it takes to travel between them.

L = 9.0×10

γ

γ

L = 5.4×10

Now that we know how far the glider has to travel, we plug that in as the distance in the speed equation:

t = 5.4×10

t = 22.5ms

So both parties, the glider crew and the observer at station Y will agree that the clock in the glider reads 22.5ms.

Part B - Clock in Station Y

This part is even simpler than part A, since the path between X and Y isn't moving in this frame. The only thing moving in this frame is the glider, which isn't a problem since the clock is only one place in the glider, so it's length contraction can be ignored.

t = 9.0×10

t = 37.5ms

Again, since this clock is at the location of the event, both parties will agree that the clock says 37.5ms.

Part C - Clock in Station X

This is where the calculations start getting gnarly. This clock is not at the location of the event (glider passing Y), but is 9 million meters away from it, meaning that there is no requirement for the two parties to agree on what time it displays (don't you love relativity?).

In the frame of the stations, there isn't any problem. The clocks at the two stations are synchronized, and since we already calculated the time on the clock at Y, the clock at station X in the same frame will be the same: 37.5ms.

The problem is in the frame of the glider. From the glider's perspective, station X is moving at 0.8c away from it, so time dilation dictates that the clock in station X should show a time less than the 22.5ms that passed on the glider, from the gliders frame of reference.

This can be calculated using the Lorentz transformation for time:

t = γ

Calling the position of the glider the origin in its frame, the x location of station X would be -5.4×10

t = γ

t = 13.5ms

Now this is pretty crazy. You on the glider and another person on station Y's platform both look at each other, and agree what each other's clocks say, but then look back at station X, and disagree by 24ms. This also means that from the glider's perspective, every place along the path from X to Y is currently at a different time.

But that's relativity for you... and this has just been three weeks...

^{6}m apart. Just as you pass station X, your clock and the station X clock read zero. Your passing station Y is an event. When this event occurs, what does (a) your clock read according to you, and according to someone in station Y, (b) the station Y clock read according to you, and according to someone in station Y, (c) the station X clock read according to you, and according to someone in station X?Question by Randy Harris.

The first two parts of this problem are relatively simpler than the third, since they're both at the location of the event (glider passing station Y) so both parties will agree on what each clock says, since the clocks are in the same place in both frames. Classically, we would expect all the clocks to agree all the time, but since we are going ridiculously fast, it's not the case.

Part A - Clock on the Glider

This requires using the simple equation speed × time = distance. We want to know howw long it takes to travel between the stations, so solve for time:

t = d ⁄ v

The problem is that it won't appear to be 9.0×10

^{6}m between the two stations, but some value less than that. This is due to the effect of length contraction, which causes moving objects to be shorter. The equation for length contraction is:L = L

_{0}⁄ γ_{v}Where L

_{0}is the length of the object in a stationary frame, and γ is the Lorentz factor found throughout special relativity, as a function of v.We use this length contraction formula to calculate the length of the path between the two stations in the frame of the glider. Once we know how far apart the two stations are, it's trivial to calculate how long it takes to travel between them.

L = 9.0×10

^{6}m ⁄ γ_{0.8c}γ

_{v}= (1 − v² ⁄ c²)^{-0.5}γ

_{0.8c}= (1 − 0.8²)^{-0.5}L = 5.4×10

^{6}mNow that we know how far the glider has to travel, we plug that in as the distance in the speed equation:

t = 5.4×10

^{6}m ⁄ 0.8ct = 22.5ms

So both parties, the glider crew and the observer at station Y will agree that the clock in the glider reads 22.5ms.

Part B - Clock in Station Y

This part is even simpler than part A, since the path between X and Y isn't moving in this frame. The only thing moving in this frame is the glider, which isn't a problem since the clock is only one place in the glider, so it's length contraction can be ignored.

t = 9.0×10

^{6}m ⁄ 0.8ct = 37.5ms

Again, since this clock is at the location of the event, both parties will agree that the clock says 37.5ms.

Part C - Clock in Station X

This is where the calculations start getting gnarly. This clock is not at the location of the event (glider passing Y), but is 9 million meters away from it, meaning that there is no requirement for the two parties to agree on what time it displays (don't you love relativity?).

In the frame of the stations, there isn't any problem. The clocks at the two stations are synchronized, and since we already calculated the time on the clock at Y, the clock at station X in the same frame will be the same: 37.5ms.

The problem is in the frame of the glider. From the glider's perspective, station X is moving at 0.8c away from it, so time dilation dictates that the clock in station X should show a time less than the 22.5ms that passed on the glider, from the gliders frame of reference.

This can be calculated using the Lorentz transformation for time:

t = γ

_{v}(vx′ ⁄ c² + t′)Calling the position of the glider the origin in its frame, the x location of station X would be -5.4×10

^{6}, and the time in the glider's frame would be the time shown on the glider's clock: 22.5ms. Plug it all in, and we get:t = γ

_{0.8c}(0.8 × -5.4×10^{6}⁄ c + 22.5ms)t = 13.5ms

Now this is pretty crazy. You on the glider and another person on station Y's platform both look at each other, and agree what each other's clocks say, but then look back at station X, and disagree by 24ms. This also means that from the glider's perspective, every place along the path from X to Y is currently at a different time.

But that's relativity for you... and this has just been three weeks...