Notes on Special Relativity

 (updated 9/27/05)

NOTE: These are supplementary notes, you will still need to read your text book.
Caution, I've used symbol font in a few places- if your browser does not do symbol font these letters will show up as roman letters. I've tried to avoid this as much as I could.

What is a transformation?

Suppose you have two x-y coordinate systems lined up on the x-axis but not having the same origin (as shown) so the y-axises are separated by a distance d. The one on the left is the S system the one on the right the S' system (the names are not very creative but traditional).

Suppose both S and S' want to make a measurement of the object. notice that S gets the value x1 but S' gets the value x1'. They don't agree on the location. How can they resolve this? In order to compare values they need a transformation between the two coordinate systems: x1 = x1' + d     or for S' we have      x1' = x1 - d.

Now suppose S' is moving at a constant speed v. Now the value for d = vt where t is time (we assume the origins coincide when we start the clock at t = 0). And the transformation between coordinates are x1 = x1' + vt   and    x1' = x1 - vt. These are examples of Galilean transformations.

NOTE:

  • The y-locations don't change (so the  transformation is y = y'). But this is only true if the x-axis are parallel. If the x-axis are not parallel the transformations are more complicated.
  • Time is the same in both frames; t = t'.
  • Although the above seems pretty common sensical we now know it isn't right at high speeds.
  • We are only talking about reference frames that are moving at constant speed relative to each other (v is a constant value).
  • Notice that if S were moving to the left at speed v and S' were stationary the transformations would look exactly the same.

    • What is an invariant?

    Suppose the two observers in the two reference frames above wish to compare the length of an object? Notice that S will get a value of L = x2 - x1 and S' will have L' = x2' - x1' .

    How can they compare their values of L? Well we know  x1 = x1' + vt  and x2 = x2' + vt   so  L =  x2 - x1 = (x1' + vt ) - ( x2' + vt ) = x2' - x1' = L'. So for Galilean transformations we see that the length L is the same in both reference frames, L = L'. L is therefore an invariant under Galilean transformations, it has the same value in both frames. NOTE for further discussion: We assume here that we can make the measurements x1 and  x2  at the same time ie. simultaneous. See below why this assumption might be wrong.

    What about velocity?

    Well we know  x1' = x1 - vt  (according to Galileo anyway).
    1. If we take a derivative with respect to time of both sides we have  dx1'/dt  = dx1/dt - v. (Note that v is a constant - the reference frames move at constant velocity relative to each other.)
    2. Recall that dx1'/dt  = velocity of the object (definition of velocity) = u' (we use u instead of v for velocity because we already have used v) and dx1/dt = u, the velocity in the other reference frame. (Careful! v is the velocity of the reference frame S' compared to S. u is the velocity of the object as measured by S and u' is the velocity of the object as measured by S').
    3. So our conclusion is that velocity is NOT an invariant: u' = u - v.  In other words S' has to do something to (transform) the measurement that S has in order to compare the values.
    Example: S' is on a railway car traveling at 5 m/s (v = 5 m/s). She throws a ball forward at 2 m/s relative to her (u' = 2 m/s). How fast does S see the ball traveling?  u = u' + v so S has a value of u = 5 m/s + 2 m/s = 7 m/s for the ball relative to him (S).

    Why are invariants important?

    Taking a second derivative of  x1' = x1 - vt shows us that (since v is a constant) a' = a. In other words acceleration IS an invariant under Galilean transformations. This is good because now F = ma is the same as F' = m'a' (if we assume mass is also invariant). In other words Newton's laws are the same in the two reference frames if we use the Galilean transformations (the form of Newton's second law is invariant under Galilean transformations).

    We want the laws of physics to be the same in any reference frame ( i.e. invariant) otherwise we would have to learn different laws of physics for each new reference frame - different in our cars, planes, boats, on the moon etc. from the surface of the earth. So the laws ought to transform so as to be the same form in each frame.

    What is wrong with this picture?

    All of the above looks pretty reasonable but we now know the Galilean transformations aren't quite right. Here is why:
    1. Maxwell discovered the equations that explains that light is an electromagnetic wave (light includes radar, microwaves, x-rays, radio, gamma rays and much more). The equations predict that the speed of light is fixed in a vacuum (a constant for any observer). c = 3x108 m/s is an invariant.
    2. This would mean there is a problem with Galilean transformations. If S measures the speed of light to be c then S' should get u' = c - v (a number less than c). But Maxwell predicts c should be the SAME value for S and S' (c for both of them). So who is right? Galileo or Maxwell?
    3. Michelson and Morely tried to measure a change in the speed of light between two moving reference frames. They (and many others) found that the speed of light does NOT change from one frame to another, it is always c.
    4. NOTE: When light passes through something (like glass) it interacts with the material which causes a delay in the rate at which it goes through the material. So light can appear to slow down when it goes through something but that is not what we are talking about here.

    How can we fix the Galilean transformations?

    First we notice that we have two choices:
    1. Newton's laws are correct and look the same in any reference frame, the speed of light is NOT invariant . That implies the Galilean transformations are correct and Maxwell is wrong (Maxwell's equations are not invariant under Galilean transformations).
    2. Maxwell is right, the speed of light is an invariant, Maxwell's equations look the same in any reference frame and we need a new set of transformation equations (not Galilean). That means Newton's laws are wrong (are not the same in any reference frame).
    3. The laws of physics are not the same in different reference frames (in which case we have a whole lot more work to do or maybe we just can't even do physics).
    OUCH!?!? What to do?

    Well Lorentz worked out what the transformations would look like for Maxwell's equations (the transformations which leave Maxwell's equations invariant). These are called the Lorentz transformations. They are built on the assumption that the speed of light is an invariant. For the S and S' coordinate systems moving at speed v relative to each other and x axies parallel the transformations are:

    x' = (x - vt)/(1 -  v2/c2)1/2       t' = (t - vx/c2)/(1 -  v2/c2)1/2      y = y' and z = z'
    First notice a few things:

    What was Einstein's role in all this?

    Einstein started with two postulates:
    1. The laws of physics should have the same form in any reference frame moving at constant velocity (we need the laws themselves to be invariant).
    2. The speed of light is an invariant quantity (the same number in any constant velocity reference frame), just as Maxwell said.
    Some comments:

    What are some consequences of Einstein's postulates?

    As we will see applying the Lorenz transformations give different results than the Galilean transformations.
    For Lorenz transformations: NOTE: Remember what we mean by NOT invariant: S' has to do something to (transform) her measurement of length in order to compare with the measurement taken by S. An invariant has the same value in both frames.

    What is the difference between 1) transformations between frames and 2) measurements taken in each frame?

  • Here are the location transformations between two frames for the left and right sides of an object:

    •  x1' = (x1 - vt1)/(1 -  v2/c2)1/2 and  x2' = (x2 - vt2)/(1 -  v2/c2)1/2 .

    We want to compare the length of the object in the S frame (L = x2 - x1) with the length in the S' frame (Lp = x2'- x1') To make this easier we assume the time measurements are simultaneous in the S frame so t1 = t2. If we subtract these two and move things around (canceling the ts since they are the same) we get:

    L = Lp (1 -  v2/c2)1/2 .

  • Here is the time transformation between frames:


    •  t' = (t - vx/c2)/(1 -  v2/c2)1/2 .

    Now lets take a time interval (a time between two measurements) in S. So the time interval in the S frame (sometimes called the proper time) is t = t2 - t1 But look what happens when we use the Lorentz transformation to get the time interval in the S'  (a frame moving at speed v relative to S). In the special case where the location of the clock (x) is the same for both measurements (the clock does not moving in the S frame) (t2'- t1')  = (t2 - t1)/(1 -  v2/c2)1/2 (the vx/c2 part cancels out since x is the same location). Notice that 1/(1 -  v2/c2)1/2 is always > 1 so the the time measured in the other frame is always bigger than the proper time.

    'Weird' consequences and paradoxes of special relativity:

    Is any of this 'real'? Do we care?

    Yes! One example from everyday life is that the GPS (Global Position Satellite) systems used for hand held GPS and also in modern airplanes use a relativistic calculation to correct for the motion of the satellite. Without this correction a plane could miss the runway by as much as a km.

    A second example is the muon problem. Muons are subatomic particles with very short lifetimes (10-6 s) in the lab. They are created when cosmic rays hit the upper atmosphere. Even though they are traveling at 0.9c they do not live long enough to reach the earth's surface. But they do! We can measure lots of them at the earth's surface. So what is happening? Because they are in a moving frame we measure their clock to be running slow- they live longer than 10-6 s and so have time to reach the earth. From their reference frame they see their clock as normal but the distance they have to travel (the depth of the atmosphere) as much shorter because it is in a frame moving relative to them (they say they are stationary and the earth moves toward them). In both frames (earth frame and muon frame) the muon reaches the earth but our explanation for why is different (relative to which frame we are in).

    So how do we fix Newton's laws?

    If you take a derivative of the velocity transformation you see clearly that acceleration is NOT an invariant (note that in a = du/dt you have to transform not only u to u' but also t to t'). What then do we do with F = ma?

    Einstein requires momentum conservation (one of our fundamental laws) be invariant from one frame to another. For momentum to be conserved all the external forces are zero and the internal forces all cancel exactly because of the third law. This is convenient since we can find out what happens to m in F = ma without worrying about F.

    One way to write the conservation of momentum is to say sum of all initial momentum = sum of all final momentum. Another way is to say sum pf - sum pi = 0. What Einstein wants is that when sum pf - sum pi = 0 is true then sum pf' - sum pi' = 0 is also true (where p' is measured in the S' frame and p is measured in the S frame).

    CAUTION! Conservation is not the same thing as invariance. Conservation says some quantity (in this case momentum) is the same before as it is after something happens (no change of reference frames). Invariance says that when S gets a case of conservation then S' does too (a change of reference frames). But notice that the initial (=  final)  momentum in the S' frame (pi') does not have to be the same number as the initial (=  final) momentum in the S frame (pi). pi' and pi  are NOT equal. So S might say the total momentum is 10kgm/s before and after the collision while S' says the momentum is 15 kgm/s before and after. They agree on conservation but momentum is not an invariant.

    Ok so what happens if you require the momentum conservation law to be invariant?

    (i.e. If momentum is conserved in one from it will be conserved in any frame.) The result is that we have to re-define momentum:

    p = mu/(1 - u2/c2)1/2. CAUTION! The u here is the speed of the mass m, NOT the speed v of the reference frame. Your book uses the Greek letter gamma (g) for both 1/(1 - u2/c2)1/2 and 1/(1 - v2/c2)1/2 .

    If we now define Newton's law to be F = dp/dt we do have an invariant form of Newton's laws (i.e. F' = dp'/dt' is also true).

    What about energy?

    E = total energy = Ek +   mc2 = mc2/(1 - u2/c2)1/2 which gives us our relativistic definition of total energy.

    What is the difference between 1) transformations between frames and 2) measurements taken in each frame (part 2)?

    Look very carefully at the definition of energy and momentum in your book.

    E = mc2/(1 - u2/c2)1/2.
    px = mux/(1 - u2/c2)1/2.
    etc.
    The above are measurements in the the S frame.

    For measurements in the S' frame we have:

    E' = mc2/(1 - u'2/c2)1/2.
    px' = mux'/(1 - u'2/c2)1/2.
    etc.

    Notice that the speed of the object is different in each frame.

    Transformations between frames look like:

    E' = (E - vpx)/(1 - u2/c2)1/2 where v is the speed of the reference frame and u is the speed of the object.

    The mass = energy connection.

    From the above definition of total energy (and the invariant mass/energy four vector (mc2)2 = E2 - (pxc)2 + (pyc)2 + (pzc)2) we see mass and energy are combined in special relativity and that total mass/energy (mc2) is invariant. For units of energy we can use the electron volts (eV and MeV = 106 eV). Then since mass appears in the same equation but multiplied by c2 it becomes useful to use MeV/c2 for the units of mass. From these considerations it becomes clear that mass and energy are pretty much the same thing (both are components of the same four vector for example). What are the implications?
     
  • If we add up the rest mass of a proton and a neutron we get 938.28 MeV/c2 + 939.57 MeV/c2 = 1877.85 MeVc2. But if we measure the rest mass of a deuteron (a proton and a neutron bound together) we only get 1875.63 MeV/c2. What happened to the "missing" 2.22MeV/c2? Well if we add exactly 2.22 MeV of energy we find we can separate the deuteron into proton and neutron (each with the correct mass). So it is useful to think of the 2.22 MeV as being converted into the "missing mass". This "missing mass" is called the binding energy.
  • Suppose we have two objects of the same mass. Obviously the have more energy if they are moving towards each other at the same speed than if they are at rest relative to each other. But if they move at the same velocity towards each other the speed of the system of two particles considered together is zero. This means the rest mass of the system of the two moving particles is greater than the rest mass of a system of two non moving particles. So, for example, we expect a container of gas to increase in mass as the temperature of the gas is increased (the gas molecules have more kinetic energy which is equivalent to more mass). This also means a spinning object has more mass than the same object at rest. Weird huh?
  • We can have particles that are all energy (zero rest mass) but only if they travel at speed c. These would be photons (light particles) or gravitons (gravity particles). Note that, although the rest mass is zero the mass measured in a frame moving relative to the photon is not zero (light has an 'effective mass').
  • We can have creation of particles from energy and annihilation of particles into energy. The total mass/energy has to be conserved but we can have it all be energy (light) or all be mass or part mass and part energy.


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