# Special Relativity basic concepts confusion

1. Oct 11, 2011

### math_maj0r

A rocket ship of length L (in the rocket's frame) leaves the earth at speed v. A light signal is sent after it which arrives at the rocket's tail at t'=0 and t=0 according to earth clocks.

My question is: Why is it that the total time (according to earth) for the signal to get to the head and back can be found by multiplying the rocket's time to get back by gamma, but the time to reach the head (according to earth) can't be found by multiplying the rocket's time to reach head by gamma?

I know it has to do with how you can multiply time by gamma only if the two events happen at the same place (same place in the proper frame? Is the Proper frame *defined* to be the frame in which the two events happen at the same place?), and that you can multiply distance by gamma only when the two events happen at the same time (in the proper frame? Which is *defined* to be the frame in which they happen at the same time)? So in the rocket problem we are allowed to multiply the length of the Rocket by gamma to get the length of the rocket according to earth people because it's possible for the earth people to find a ruler such that the ends of the ruler can touch the ends of the rocket at the same time? When we want to know the distance, the two events are the left end of the ruler on the left end of the rocket, and the right end of the ruler on the right end of the rocket? When we want to know the time, the two events are the light being at the tip of the rocket's tail before reaching the head and after coming back?

Can you give me an example of a rocket problem with a light signal in which to find the *length* of the rocket according to earth people I *can't* take the length of the rocket according to the rocket and multiply it by gamma, but in which to find the *time* according to earth people that it takes for the light signal to get from the tail to the head I *can* multiply the time that it takes in the rocket's frame for the light to go from the tail to the head by gamma?

I want to know a problem as similar to your rocket problem as possible where I can do the above.

2. Oct 12, 2011

### ghwellsjr

The term Proper Frame is defined as the rest frame of an object, so in your scenario, there are two Proper Frames, one for the earth and one for the rocket.

The correct way to calculate the times and locations of events from one frame to another is to use the Lorentz Transform but oftentimes, once people understand the general idea, they take a shortcut and just multiply or divide by gamma. If you want to know how the clocks run in one frame compared to the other, you multiply by gamma but if you want to know how lengths along the direction of motion transform, you need to divide by gamma.

So, for example, the earth will see the length of the rocket shorter by a factor of one over gamma (gamma is always greater than one) which is why it is called length contraction.

But getting back to the Lorentz Transform, when you want to see how the length of the rocket appears to the earth, you have to find two events, one located at the head of the rocket and the other located at its tail, that have the same time coordinate. I think this may be what you are asking about.

In a similar way, if you want to find out how the rocket determines its speed, you have to find two events located at the same place on earth with the different time coordinates.

Now you want an example. I'm going to use units of feet for distance and nanoseconds for time and I'm going to define the speed of light to be 1 ft/ns. Let's assume the rocket is 100 feet tall and it is launched from a tower that is also 100 feet tall. We'll have the rocket achieve a constant speed of 0.6c or 0.6 ft/ns by the time the tail has reached 1000 feet and we'll call that time 0. At a speed of 0.6c, the value of gamma is 1.25. For the sake of brevity, I will assume the rocket is being launched in the x direction and since y and z are always 0, I will leave them out of the nomenclature for the events. An event will be defined as [t,x].

Now in order to use the Lorentz Transform in its standard (easy) form, we have to define a common origin for both frames in which all the coordinates are zero. We'll do that for the tail of the rocket, 1000 feet of the ground at time zero.

So let's start with the earth frame. Here are some events in the earth frame at the start of our scenario (some time after launch):

[0,-1000] Bottom of launch tower.
[0,-900] Top of the launch tower.
[0,0] Tail of rocket.

Now we'll transform these events into the rocket frame:

[750,-1250] Bottom of launch tower.
[675,-1125] Top of the launch tower.
[0,0] Tail of rocket.

Notice that the times for the launch tower are not equal so we cannot use the distance parameters to see what the height of the tower is according to the rocket. What we have to do is adjust the time coordinate for either the event at the bottom of the tower or the event at the top of the tower to match the other one. For example, we could use this pair of events in the earth frame:

[0,-1000] Bottom of launch tower at time 0.
[60,-900] Top of the launch tower at time 60.

which transform to these events in the rocket frame:

[750,-1250] Bottom of launch tower at time 750.
[750,-1170] Top of the launch tower at time 750.

The difference between the distance coordinates between these two events (1250-1170=80) is the height of the tower as determined by the rocket.

We could have also adjusted the time at the bottom of the tower as follows:

[-60,-1000] Bottom of launch tower.
[0,-900] Top of the launch tower.

which transform to these events in the rocket frame:

[675,-1205] Bottom of launch tower.
[675,-1125] Top of the launch tower.

And again, we take the difference in the distance coordinates (1205-1125=80) and arrive at the same height (80 feet) for the launch tower as determined by the rocket frame. Note that we could have taken the shortcut and divided the earth height of the tower, 100 feet, by gamma, 1.25 and arrived at 80 feet much more quickly.

Now let's see how the earth views the rocket. Note that we have not defined an event for the head of the rocket because we can only do that in the rocket frame. So let's define two events for the tail and the head of the rocket in the rocket frame:

[0,0] Tail of the rocket.

When we transform these into the earth frame (using a speed of -0.6c) we get:

[0,0] Tail of the rocket.

Again, we need to find a time in the rocket frame for one of these events that will give the same times for both events in the earth frame. Here are two events in the rocket frame:

[0,0] Tail of the rocket.

that transform into these two events in the earth frame:

[0,0] Tail of the rocket.

And so we see that the earth determines that the rocket's length has been contracted to 80 feet.

Or, going back to the earth frame:

[0,0] Tail of the rocket.

This time, we need to find a time in the rocket frame for the other of these events that will give the same times for both events in the earth frame. Here are two events in the rocket frame:

[60,0] Tail of the rocket.

that transform into these two events in the earth frame:

[75,45] Tail of the rocket.

And the distance between the head and tail is 125-45=80 feet so we see that the earth again determines that the rocket's length has been contracted to 80 feet.

Now let's investigate the speed of the rocket. In the earth frame, we have to find two events separated in time that transform into the same location in the rocket frame (0). So in the earth frame:

[0,0] Tail of the rocket.
[100,60] Tail of the rocket 100 ns later.

which transforms into:

[0,0] Tail of the rocket.
[80,0] Tail of the rocket 80 ns later in the rocket frame.

So we see that earth determines that in 100 ns, the rocket will have traveled 60 feet for a speed of 0.6c.

In a similar manner, to see how the rocket determines its speed, we can use the event of the bottom of the launch tower in the rocket frame determined earlier:

[750,-1250] Bottom of launch tower.
[850,-1310] Bottom of launch tower 100 ns later.

Which transforms into the earth frame as:

[0,-1000] Bottom of launch tower.
[80,-1000] Bottom of launch tower 80 ns later in the earth frame.

The salient point here is that the bottom of the launch tower has moved -1310+1250=-60 feet in 100 ns for a speed of -0.6 ft/ns or -0.6c.

You also asked about a light signal sent after the rocket from earth and reaching the head and reflecting back to the tail but I'm not sure what you're asking about. Maybe understanding how to use the Lorentz Transform might clear that up for you but if not, ask again in a different way.

Last edited: Oct 12, 2011