## Saturday, July 27, 2013

### Compound Velocities in Relativity

There's one more class of relativity problems that you encounter in first-year physics that I haven't done yet with my pictorial method: that's the question of compound velocities. You know: the man in the train shoots a bullet inside the train...how fast does the observer on the ground see the bullet?

I already worked out the coordinate transformation for the train going at 80% of the speed of light, so it will be convenient to use that as our starting point. And let's have the man inside the train shoot a bullet at 80% of the speed of light. Obviously in Gallilean relativity the observer on the ground will see the bullet moving at 1.6 times the speed of light. But how does it work in relativity?

We already worked out that the coordinates of the point (0,1) moved to (4/3, 5/3) in the coordinate system of the moving train. And similarly for the point (1,0) which moves to (5/3, 4/3). (Remember, that any Lorentz transformation maps points along the trajectory of a hyperbola.) From these pieces of information, we can actually re-draw the entire coordinate system like so:

You should be able to verify that the equations I've written do indeed map points from the (x,t) coordinate system (the stationary system) to the (x', t') system (the moving system).

You should know that coordinate transformations of this kind can be written in matrix notation. We're going to do that, and the look at what happens to a point that is transformed twice: first, a stationary point that is transformed to the moving train, and then the same transformation applied within the moving train. On the graph, it looks like so:

Applying the transformation once moves the vector from a to b; and applying it again moves the vector from b to c. We can do this by matrix multiplication:
So the point (1,0) (that's "a") gets mapped to the point (41/9, 40/9) (that's "c"). If I know what I'm doing, that means the bullet inside the train is travelling at 40/41 times the speed of light. You might want to check me on that.

## Wednesday, July 24, 2013

### Why Does the Train Get Shorter?

I've been going through the basic steps of relativity over the last few days, and now I think I'm ready to do a typical first-year physics problem. They tell you that when you approach the speed of light, time gets slower and lengths get shorter. The other day I showed you how the world-line of the back of the train intersects hyperbolas passing through t = 1, 2, 3... etc; and how an observer on the ground, looking at those points, sees them as being longer than his normal seconds. Today I want to do the calculation of how the train gets shorter.

I'm going to use a speed of 0.8, or 4/5ths the speed of light; because then the numbers are going to work out. Let's start by sketching the world-lines of the train, front and back:
The green line represents the train as it's seen by a man riding inside. We talked the other day about how his x-axis gets tilted. You can see I didn't do a very good job of drawing it; I should have shown the angle between the green and blue being the same as the angle between the blue and brown. The first question is: when the back of the train passes through the origin, where does the man inside see the front of the train? That's our point (x, t).

We're going to assume the train is of unit length. Remember, the funny thing about that is you can't assume that it passes through the point (0,1). The world-line of the front of the train only passes through that point if the train is standing still. We can't assume anything if the train is moving, so we're going to call that point (0, h) for now. We'll find out that h is less than 1.

Now we're going to use the relativistic Law of Pythagoras (I think it's called the Minkowski metric) to measure the length of the train. We saw the other day how the hyperbolas in the lower half of the quadrant corresponded to equal time-like intervals, counted off in seconds. The same law tells us that hyperbolas in the upper half of the quadrant mark of points in space-time of equal distance from the origin. Since the train is of length 1, at the moment the back of the train passes through the origin, the front of the train must lie on the hyperbola x^2 - t^2 = 1. That's our first equation. (Actually, if I look back at my diagram, it's my second equation. But you get the idea.)

Then we just add in the fact that the train has a velocity of 0.8, and write the equation for the straight line passing through the point (0, h). Because we're trying to find out how long the speeding train appears to an observer standing on the ground.

It's not to hard to solve those two equations. I find that the front of the train is located at the position (4/3, 5/3) and consequently the world-line of the train passes through the point (0, 0.6). (EDIT: I had the coordinates reversed when I posted this morning, but I've fixed it now.) Here is another  drawing the train, this time showing the hyperbola which represents the locus of all points unit distance from the origin. No matter how fast the train is moving, at the moment the caboose is passing through the origin, a man inside the train will find that the engine is somewhere on this hyperbola:

I fixed my picture so that the green line and the brown line have the same angular separation from the light cone. I also drew in the hyperbola so that it's tangent to the world-line of the train. I'm not sure why it should be that way but I think it's right. You might want to check me on that.

## Tuesday, July 23, 2013

### The Lorentz Transformation

The other day I showed you what a moving train looks like in relativity. Here is the graph I drew comparing the world-lines of a moving train and a stationary train:

It's important to note, among other things, that the lines for the front of the train do not intersect on the x-axis. I talked about this the other day. Today I'm going to try and explain why it has to work that way.

Suppose you are standing in the middle of the train and you turn on a light. The light travels to both ends of the train, bounces back (because we've put mirrors there) and you time how long it takes to return. Since you are standing in the middle of the train, it reaches both ends at the same time, and the beams come back simultaneously. Here is what the experiment looks like in a stationary train:

Relativity tells us the experiment has to look exactly the same from the point of view of an observer in a moving train. But whatever the train is doing, the trajectory of a light beam is always at 45 degrees to the x-t coordinate axes. So in the moving train, the experiment has to look like this:

For the stationary observer, events A and D occur at the same place, because the train is stationary; and the moving observer, who can't see outside the train, is entitled to make the same presumption about his points A and D. There is nothing strange about this; it's just what we call "Gallilean Relativity". We talked about how in the moving observer's coordinate system, the time axis is tilted up so it's parallel to AD.

Relativity comes into the picture when we look at simultaneity. Obviously for the stationary observer events B and C are simultaneous; relativity tells us that the moving observer also sees them as being simultaneous. And just the same way his time-axis tilted upwards so it became parallel to AD, so must his x-axis tilt sideways so it is parallel to CD. Simple geometry shows us that the two axes tilt by the same angle. (As always, we are using a system of units where c=1.)

But there's more. In the new coordinate system, time and distance are stretched. I'm not going to tell you why they get stretched today, but I'll tell you how they stretch. It's based on the relativistic Law of Pythagoras. You might think that the relativistic separation, or "distance" between two points on the x-t axis would just be given by the familiar equation:

But it doesn't work that way in relativity. Believe it or not, the relativistic separation between two points is calculated like so:
This has enormous implications. For example in the x-y coordinate system, (normal geometry) the distance from the origin is given by the normal Law of Pythagoras; so a circle drawn about the origin connects points that are all the same distance from the origin.

In the Relativistic Law of Pythagoras, instead of circles we get hyperbolas. (Because of the minus sign). So points along a hyperbola are the same relativistic separation from the origin. If that separation is along a line below the line x=t, then the separation is time-like; and so the intersection of the trains world-line with those hyperbolas marks off the time in seconds. It's hard to draw a good hyperbola in Paintbrush but it looks something like this:

For the observer in the moving train, those hyperbolas mark of the time in seconds: 1,2,3....  In this picture the man in the train has counted off three seconds while an observer standing beside the track has counted off something like four. Time is all screwed up in relativity.

## Sunday, July 21, 2013

### Relativity with Pictures

I don't like relativity. It's not my territory, and I'm not that comfortable with it. The last time I wrote about relativity was when I did e=mc-squared back here But some questions came up last week, and I've been working on it. Maybe you know I don't like working with equations; I need to be able to draw pictures. So here's what I've come up with.

We like to work in a one-dimensional universe...that is, just one spatial dimension and the time dimension. Here is a graph of our space, with units chosen so that the speed of light (shown as the blue arrow) is one:

The first thing we do is we put a train into the picture, and we start with a train standing still. It looks like this:

Now we ask the question: what if the train is moving? Let's give it something like one quarter the speed of light. Then obviously the world-line should look something like this:

The horrifying thing about relativity is that this picture is completely wrong. It's actually correct for what we call "Gallilean Relativity". But the universe we live in turns out to be different. Here is what the world-line looks like for an actual moving train:

Yes, this is it. There are a number of disturbing observations we need to make. First of all, notice that I haven't just drawn in the world-lines representing the front-end and the back-end of the train. I've drawn in the whole train to show it moving along. And I've tilted the moving train with respect to the stationary train. What does this mean?

The x-t axis is drawn from the perspective of a stationary observer, who could be inside or outside the stationary train. Points along the t-axis are stationary points: that is, they represent the same position at different times. But the other side of the coin is that points along the x-axis represent simultaneous points: that is, they pare points at the same time but at different locations. It goes both ways.

What Relativity says is that physics must have the same laws whether you are a moving observer or a stationary observer. So now we have to consider what the world looks like for the man inside the moving train. (Although I've drawn four cars, we should really think of it as one long tunnel-like car, where you can see clear from on end to another, but you can't see outside. Relativity asks what happens if the man inside doesn't know if he's stationary or if he's moving.

For this observer, the back end of the train (and the front end) are both stationary points. But we already figured out for a stationary observer that the world-line of a stationary point runs parallel to the t-axis. So for the man inside the train, the time axis must follow the world-line from the back of the train. It is tilted.

There is nothing so unusual about this: it is exactly the same thing in traditional Gallilean relativity. It doesn't mean time is distorted...it means points along the time-axis are stationary points. It's very straightforward.

But I've drawn the train as being tilted with respect to the x-axis. This is something entirely different and it is brand-new with relativity. It says that in addition to the time-axis being tilted, the space axis is also tilted. But we already figured out that for a stationary observer, points along a line parallel to the space axis are simultaneous points. So if the space axis really is tilted, then it tells us that the way I have drawn, from the point of view of the man inside the train, simultaneous "snapshots" of the location of the whole train.

We're soon going to see why it has to be this way; but before we do, there's just a couple more important observations to make about my drawing. If you look carefully, you'll see that the world-lines of the front end of the train do not intersect along the x-axis, but in fact just a little bit to the right. This is not an accident: I've drawn it that way on purpose. And furthermore, if you look at the intersection of those world-lines, neither is it simultaneous with the back of the train passing through the origin. Although you can't see it from my drawing, the location of the front of the train at the moment the back end passes through the origin is actually above the horizontal world line by approxamately the same amount as it is below the world line when it crosses the x-axis - something like this:

The time axis has shifted...that already happened in Gallilean relativity, and it just meant that position is relative. But here the space-axis also shifts...and it means that simultaneity is relative. In fact, we will eventually see that the space axis tilts exactly the same amount as a the time axis - at least, it does when we choose our units so that the speed of light is unity (c=1).

I've told you as best I can what the world looks like in relativity. When we return, I'll try to explain why it looks that way.

## Thursday, July 18, 2013

### Derivatives of the Delta Function

Last time I left off with a group of integrals involving delta functions, and today I'm going to show you how they work. The basic idea is that the dot product of the delta function with any other function simply evaluates that function at the origin. Because the area under the delta function is just one.

What about the dot product of the derivative of the delta function with a target function? It turns out that this operation simply evaluates the derivative of the target function. You can "prove" this using traditional integration by parts and the properties of the delta function, but you can actually see it pretty convincingly from  the picture I left off yesterday:

You can see the derivative of the delta function is like two oversized delta functions just to the left and right of the origin. So the first spike evaluates the function just to the left, and the second spike evaluates it just to the right, except negative. If the target function is flat, they cancel out. If the target function is changing, then you get a measure of how fast it's changing. What else is this but the derivative? (Except that you can see the polarity is wrong. Delta functions do that...you have to reverse the polarity on all the odd orders to get the correct sign for the derivatives.)

With this we can evaluate our original integral, which was of course

It's not hard to see that this function is zero at the origin, so the middle term of our "dot products" goes away. And the first derivative is also zero (you can see this if you sketch the graph...it's a local minimum) so the first term also goes away. All that's left is the third term, which you know evaluates to 2. So the definite integral is 2.

You might think all that business with the delta functions was a waste of time because the terms ended up disappearing. But they don't always. The easiest way to see this is if you try to take this integral from one to infinity instead of zero to infinity. That means you have to move the delta functions over to the right by one unit, and now you're evaluating the function and its derivative for x=1. I'm getting -1/e + 4/e + 2/e = 5/e for an answer (don't forget you've got to reverse the sign on the first derivative!). You can check me on that but I think it's right.

## Sunday, July 14, 2013

### More Fun With Delta Functions

I told you yesterday that I was going to integrate by parts using delta functions. We're going to see how it looks for the familiar problem:

I told you I wanted to treat this as the dot product of x-squared and e-to-the-minus-x; but the "dot product" idea only works when you integrate over the whole range of the function from minus to plus infinity. So I figured out that you can "fix" this problem by using the integral of the Dirac Delta function, also known as the unit step or the Heaviside Step function. I like to draw it graphically like so:

And since I know what a dot product is in terms of functions, I don't need to keep writing the integral sign. I can just write the functions, like so:
Oh yeah...I slipped in one more thing. The D-squared operator is the differentiation operator. I can throw it in with impunity because when it operates on the exponential, it just returns the same function. So I can throw it in for free.

And now comes the good part. Instead of operating to the right, I operate to the left. I differentiate x-squared, because I'm allowed to operate in either direction. But there's a catch...it's not just x-squared: it's x-squared multiplied by the unit step!

We can easily do this differentiation using the product rule compounded twice. You know:

(fg)'' = f''g + 2f'g' + fg''

The easiest way to differentiate those step functions is pictorially. I hope the following picture is self-explanatory:

Are you still with me? This is what we get when we differentiate to the left. Now we just have to multiply all these terms by the right-hand side, namely exp(-x). This is what we get:
Remember, these are my "dot products", which means they are really just integrals between plus and minus infinity. And they're easy to evaluate! The one on the right hand side is just the exponential multiplied by twice the unit step, which kills everything to the left of zero. We know this integral evaluates to 2*1 = 2.

What about those crazy integrals with the delta function and its derivatives? Those are especially nice, because the delta functions kill everything except what's going on right at the origin. You can actually prove this with integration by parts (the "traditional" way), but you should be able to convince yourself of it pictorially.

Why don't  you think about that and I'll finish it off when we return...

### Integration by Parts using Delta Functions

Today I'm going to talk about something I figured out maybe twenty-some years ago. Everyone knows about integration by parts: you do this thing with uv -vdu, and you can do certain classes of definite integrals. Well, I figured out my own way of doing it, and I think it's pretty cool. I don't really like explaining it on paper; I'd rather demonstrate it for people on a blackboard. But it came up a couple of times last week, so I thought maybe I'd try and write it up.

It's really a mathematical technique which draws on physics for its inspiration. You know in quantum mechanics you have the bras and kets, and you always calculate the expecation value of an operator by using this type of expression:

<phi*|H|phi>

The phi's are "state vectors", and the H is an "operator". (The asterik represents complex conjugation, which is a bit of a bookkeeping detail we don't need to worry about too much here.) This is the abstract form; but often the state vectors can be represented as functions in space, and the operator is an action performed on those functions. For example, the operator which evaluates the momentum of a function is just differentiation (actually, the del operator in three dimensions).

Sometimes the operator is just the "do-nothing" operator; in which case, the calculation simply evaluates the dot product of the two state vectors, which of course is the square of the total amplitude. Or, in the functional representation, it evaluates the integral between plus-and-minus infinity of the product of the function with itself. Which we can also interpret as the amplitude-squared of the function.

Sometimes the bra and ket are two different state vectors, in which case you have the dot product of two vectors. Or in the case of two functions multiplied by each other and integrated over all space, it's what you can consider as the "dot product" of those two function. Like the vector dot product, when you divide by the total amplitude, it gives you a measure of the extent to which those functions are pointing in the same direction. If the dot product is zero, we can say that those functions are "orthogonal".

So what does this have to do with integration by parts?  Well, you know that there are two functions involved, which we call u(x) and v(x). What I like to do is say that we can think of them as bras and kets, with differentiation being the operator:

<u(x)|D|v(x)>

The "D" represents differentiation so what I've written here is just the plain dot product of u and dv/dx, after you've done the differentiation. This is exactly what integration by parts is...well, almost but not quite.

I didn't exactly emphasize this, but this "dot-product"interpretation only really works if the limits of your integration are plus/minus infinity. That's not exactly the usual case for integration by parts, where you're normally integrating between specific limits, often for example from zero to infinity. For example, you'll be doing the product of x-squared and exp(-x), from zero to infinity. You can't integrate it all the way from negative infinity because it diverges. And an integral of this kind is not a dot product.

What I figured out is you can make it into a dot product using the unit step function. The unit step function H(x) is defined as zero to the left of the y-axis and 1 to the right. Or alternately it's the integral of the Dirac delta function. (The H is for Heaviside, an electrical engineer who was a few years ahead of Dirac in these things. So my dot product looks like this:

<H(x)u(x)|D|v(x)>

Now we're ready to do integration by parts. So I'm going to do uv-vdu, right? No. It's a lot nicer than that. What I say is this: the D operator can work in either direction. It looks like you're operating to the right, on v(x), in which case you have the "dot product:

H(x)u(x) *dv/dx

Now you know in quantum mechanics they're always reminding you that things aren't commutative. But that doesn't mean you can't operate in the opposite direction! You can. And when you do, it looks like this:

d/dx{H(x)u(x)}*v(x)

That's integration by parts, and when I come back I'll show you how it works.