FREQUENTLY ASKED QUESTIONS

September 7, 2007

Content Questions

What's the definition of a rest frame?

Rest frames are defined with respect to objects. The rest frame of an object is the frame in which it's at rest. You can also have two objects at rest with respect to each other; the frame in which this is true is called their rest frame.

Is the rest frame arbitrary?

No, it's not arbitrary. For a given object, there is only one rest frame: the one at rest with respect to the object. (Note that there is no absolute rest frame: no object is special and can be said to have ``the'' rest frame of the universe.)

What's the equational proof that the speed of light is constant for all reference frames i.e. say one person is traveling at $v$, the other standing. Why is $c$ the same for both?

Einstein's statement that ``the speed of light is constant for all frames'' is not a proof- it's a postulate. It happens that this postulate is valid for our universe, within all the experimental precision we can muster. So you can consider Einstein's postulates to be experimentally true statements about our universe. As to ``why'' this is so: I don't think anyone really knows, although the ideas do satisfy some aesthetic principles of symmetry. In physics the game is to find the simplest, cleanest description of nature that works.

How did the observer in Einstein's train ``see'' the light from B first?

It is worth spending the time to think through this example carefully.

Both observers see flashes of light from each side of the train. Observer O on the ground sees them both arriving at the same time. Since he knows they come from A and B (due to the scorch marks on the ground), and he knows he is equidistant from A and B, and he has just read Einstein's paper so he knows that the speed of light is always $c$, he can deduce that the lightning flashes hit A and B simultaneously.

Observer O' in the center of the train boxcar also observes two flashes of light. Because the train has moved to the right while the light is traveling, the light flashes meet somewhere to the left of O'. In the S' frame, when the photons meet, the light from B' has already zipped by O', and the light from A' hasn't reached O' yet, due to the motion of the train. So what O' actually sees is a flash of light from B' first, then a flash of light from A'. O' knows that the lightning hit A' and B' (due to the scorch marks) and that she's equidistant from A' and B'. O' has also read Einstein's paper, so O' knows that the speed of light is $c$ in her frame, because it's $c$ in every inertial frame. So, since she saw the flash of light from B' first, she concludes that the lightning hit B' before it hit A'. In other words, the lightning strikes were not simultaneous. Who's right? Both of them! Something simultaneous in one frame may not be simultaneous in another. Weird? Yes! But this is the consequence of assuming that the speed of light is always $c$.

In Einstein's thought experiment, how does $O'$ see the flash from $B'$ first, then from $A'$? I was unclear about the position of the light beams meeting.

See above question. Because the train is moving to the right, the beams of light meet to the left of $O'$. That means that the beam from $B'$ sweeps by $O'$ before the beam from $A'$ has reached $O'$: so $O'$ sees the beam from $B'$ first.

For Einstein's train thought experiment, $O'$ sees the flash from $B'$ first because $O'$ is moving towards it?

See also above few questions. I guess you could say that it's because $O'$ is moving towards the source. If the train had been moving to the left, $O'$ would have seen the flash from $A'$ first.

In the first example with simultaneous lightning strikes, is it proved only by the movement of light rays?

I'm not sure what you mean exactly. See above few questions. The vital assumption in this example is that the speed of light is always $c$, no matter what frame. Normally we assume that time intervals which are the same in one frame are the same in another (and that light would have Galilean-ly added speeds in different frames): but if we assume the speed of light is always the same, we have to abandon the idea that time intervals between events are the same.

In the lightning bolt example, what if the the bolts are simultaneous in the train?

In this example we assumed that $O$ saw the equidistant flashes at the same time, and so the bolts were simultaneous on the ground. If we assume instead they are simultaneous in the train, they will not be simultaneous on the ground! It works symmetrically: to the person on the train, it looks as if the ground is moving. The analysis goes the same way: suppose $O'$ is at the center of the train and sees the flashes arrive at the same time. She's read Einstein's paper, so she concludes that because the ends of the train are equidistant, the flashes must have been simultaneous. The train moves to the right during the time the light is traveling. So, the flashes will meet to the right of $O$. The light from $A$ will arrive at $O$ before the light from $B$. So, observer $O$, also having read Einstein's paper and knowing that the speed of light is always $c$, concludes that in his frame, the lightning bolt hit $A$ first. The flashes are not simultaneous in $S$ if they are simultaneous in $S'$.

How did you derive the moving cart expression for $\gamma$?

First one imagines a light source, say a laser, sending light up to a mirror which reflects it back, and we consider the interval between when the light is emitted and when it returns. The time interval in the cart frame ($S'$) is $\Delta t'= \frac{2d}{c}$. The events - emission and return- happen at the same place in this frame. Now consider the ground frame, $S$. The ground observer sees light going up at an angle and down: see Fig. 2-5 in your text. By using the Pythagorean theorem, we can write $\left(\frac{c \Delta t}{2}\right)^2 = \left( \frac{v \Delta t}{2} \right)^2 + d^2$. Solving for $\Delta t$ in terms of the other quantities, we get $\Delta t = \frac{2d}{c \sqrt{1-v^2/c^2}}$, which according to our definition of $\gamma$ is $\Delta t = \gamma \Delta t'$.

What do you mean by events at the same place?

Events at the same place have the same space coordinates i.e. the events have the same $x$, $y$, $z$, although different times.
A proper time interval between two events is one measured when the events are at the same place.

How does it work if both observer and person in the train think that the other has slower time?

If both observers go along forever in their respective IRFs, then each will, forever, see the other's clock's time as slower. That's all there is to it. However I suspect the source of your discomfort is along the lines of what's known as ``the twin paradox'', which we will likely be getting to soon. The ``paradox'' is this: suppose one twin goes on a long relativistic trip and comes back. Another twin stays home on Earth. If each one sees the other's clock as slow, symmetrically, will they both be the same age when they meet? If not, which one will be older?

The resolution to this ``paradox'' lies in realizing that the symmetry holds only so long as the two reference frames are always moving at speed $v$ with respect to each other; in other words so long as they are inertial frames. In the twin astronaut example, if one twin travels away at some speed and comes back, she must necessarily change reference frames, and to do that she has to accelerate. In the simplest case, she goes off in a straight line and comes back. Then, even if she spends most of her time in an inertial frame, at some point she has to turn around and change frames from one moving with $v$ to one moving with $-v$: she must accelerate at some time, invalidating the assumptions of special relativity. This means that the symmetry need no longer apply. (The answer is that the one that traveled will be younger.)

BTW, this can all be understood more naturally in the context of general relativity (which treats non-inertial reference frames). In GR, the shortest path in ``space-time'' is the one with the longest time interval. We probably won't get to this material in this class, though.

If both clocks are moving slow from the opposite viewpoint, what happens when they stop? Do the clocks match?

See above question. When one clock stops moving with respect to the other, it's no longer in an inertial reference frame, and the symmetry between the clocks is no longer valid. The difference between the clock readings will depend on the particular path one clock has taken with respect to the other.

Can you go over the length contraction again?

Consider a ruler at rest in $S$, with its ends at $x_1$ and $x_2$. Its proper length (the length in the frame in which it's at rest) is $L=x_2-x_1$. Now imagine a train rolling by at speed $v$, in frame $S'$. You can imagine two events: first, a person in the train touches one side of the ruler, and next, that person touches the other side of the ruler; the time interval between these events, in the ruler frame $S$, is $\Delta t$, and the proper length is $L=v \Delta t$. By our time dilation result, the interval between these same events in frame $S'$ is $\Delta t'= \Delta t/\gamma$. (In this case, $\Delta t'$ is the proper time, since the two events - the two touchings of the ruler ends - happen in the same place in $S'$.) So the length $L'$ of the ruler in $S'$ is $L' = v \Delta t'= v \Delta t/\gamma$, which is $L' = L/\gamma$. This is length contraction: the length of a ruler moving with respect to you is shorter than the length of the ruler in its rest frame.

Moving objects can't actually be shorter, given the barn yard paradox. What's the difference between perceived and real length?

They are indeed really shorter. The perceived and real lengths are the same. We will see more later about how this can be consistent with the ``barn yard paradox''; this is actually in your homework set, cast in the form of a pole vaulter problem.

The scheme for measuring proper length that you presented is somewhat confusing because the two events (moving observer passing one end of ruler, moving observer passing right end of ruler) are not simultaneous in either observer's frame. Generally, when I want to measure the proper length between two events, I go to a frame in which those two events are simultaneous. A better scheme would be to have the stationary observer place synchronized time-bombs at each end of the rule, then measure the time difference between arrival of the wavefronts from the two explosions.

There is no need to use simultaneous events in order to determine the relation between measured and proper length. (Note that ``proper length between two events'' does not have exactly the same meaning as the ``proper length of an object'': what I think you mean by ``proper length between two events'' is the ``proper distance between two space-time events'', which is indeed the space-time interval for which those events are simultaneous.) Yes, if you explode simultaneous time bombs at the ends of a rigid body, the interval between those events will be the proper length of that object. I don't see any simple way to derive the $L=L_0/\gamma$ relation with a setup like this (comparing time interval between when light leaves the end, and when two beams meet in the two frames). Note that the time dilation equation does not work for events which are simultaneous in some frame: they can not happen in the same place in any frame, so they do not have a proper time interval.

How has the length contraction been confirmed for the matter itself beyond just our interaction with them?

I'm not sure what you mean by ``beyond just our interaction with them'' (we have to interact with matter to make measurements). In particle and nuclear physics, relativistic effects are part of everyday life, and effects are very measurable, as for the study of squished relativistic nuclei.

Can you post an example of when proper and prime are the same and when they are not the same?

In today's time dilation and length contraction examples, the proper time was in the cart (prime) frame, and the proper length was in the ground (unprime) frame. Prime and unprime are just arbitrary labels; you can label frames as you like. The point I wanted to make with the warning was this: never use prime and unprime as a guide to determining whether some time interval is a proper time, or whether some length is a proper length. In each case, consider the meaning of proper time and proper length. The proper time is the interval for the frame in which events happen in the same place, and the proper length is the length in the frame in which the object is at rest.

What is the way to use $\gamma$/relativity in the Lagrangian/ Hamiltonian formulation?

We won't be getting to this in this course, but in general one requires in a relativistic theory that a Lagrangian be ``invariant under Lorentz transformation'', i.e. that it leads to the same physics in any frame. (We will be defining Lorentz transformations in a bit).

In what depth does this course cover relativity?

We have a few more lectures in which we will cover a few more consequences of Einstein's postulates, plus treatment of momentum and energy. We will get to the Lorentz transform, but not much further (we won't do much in 4-vector notation). We will (probably) not cover anything in general relativity in this course.

Speaking of muons, will we hear more about your research in lecture?

Yes, probably. I work in neutrino physics. Also you will hear about research from other professors in the department over the course of the semester.