You’re in good company if you find this impossible to believe. Suppose I shot a ball forward at $50\text{m}/\text{s}$ while driving past you at $30\text{m}/\text{s}$. You would certainly see the ball traveling at $80\text{m}/\text{s}$ relative to you and the ground. What we’re saying with regard to light is equivalent to saying that the ball travels at $50\text{m}/\text{s}$ relative to my car and at the same time travels at $50\text{m}/\text{s}$ relative to the ground, even though the car is moving across the ground at $30\text{m}/\text{s}$. It seems logically impossible.

You might think that this is merely a matter of semantics. If we can just get our definitions and use of words straight, then the mystery and confusion will disappear. Or perhaps the difficulty is a confusion between what we “see” versus what “really happens.” In other words, a better analysis, one that focuses on what really happens, would find that light “really” travels at different speeds in different reference frames.

Alas, what “really happens” is that light travels at $3.00\times {10}^8\text{m}/\text{s}$ in every inertial reference frame, regardless of how the reference frames are moving with respect to each other. It’s not a trick. There remains only one way to escape the logical contradictions.

The definition of velocity is $u=\text{Δ}x/\text{Δ}t$, the ratio of a distance traveled to the time interval in which the travel occurs. Suppose you and I both make measurements on an object as it moves, but you happen to be moving relative to me. Perhaps I’m standing on the corner, you’re driving past in your car, and we’re both trying to measure the velocity of a bicycle. Further, suppose we have agreed in advance to measure the position of the bicycle first as it passes the tree in FIGURE 36.10, then later as it passes the lamppost. Your $\text{Δ}x\prime$, the bicycle’s displacement, differs from my $\text{Δ}x$ because of your motion relative to me, causing you to calculate a bicycle velocity $u\prime$ in your reference frame that differs from its velocity u in my reference frame. This is just the Galilean transformations showing up again.

FIGURE 36.10 Measuring the velocity of an object by appealing to the basic definition $u=\text{Δ}x/\text{Δ}t$.

Now let’s repeat the measurements, but this time let’s measure the velocity of a light wave as it travels from the tree to the lamppost. Once again, your $\text{Δ}x\prime$ differs from my $\text{Δ}x$, and the obvious conclusion is that your light speed $u\prime$ differs from my light speed u. The difference will be very small if you’re driving past in your car, very large if you’re flying past in a rocket traveling at nearly the speed of light. Although this conclusion seems obvious, it is wrong. Experiments show that, for a light wave, we’ll get the same values: $u\prime =u$.

The only way this can be true is if your $\text{Δ}t$ is not the same as my $\text{Δ}t$. If the time it takes the light to move from the tree to the lamppost in your reference frame, a time we’ll now call $\text{Δ}t\prime$, differs from the time $\text{Δ}t$ it takes the light to move from the tree to the lamppost in my reference frame, then we might find that $\text{Δ}x\prime /\text{Δ}t\prime =\text{Δ}x/\text{Δ}t$. That is, $u\prime =u$ even though you are moving with respect to me.

We’ve assumed, since the beginning of this textbook, that time is simply time. It flows along like a river, and all experimenters in all reference frames simply use it. For example, suppose the tree and the lamppost both have big clocks that we both can see. Shouldn’t we be able to agree on the time interval $\text{Δ}t$ the light needs to move from the tree to the lamppost?

Perhaps not. It’s demonstrably true that $\text{Δ}x\prime \ne \text{Δ}x$. It’s experimentally verified that $u\prime =u$ for light waves. Something must be wrong with assumptions that we’ve made about the nature of time. The principle of relativity has painted us into a corner, and our only way out is to reexamine our understanding of time.