# Different values of same physical quantity in time domain and space domain OR Physical explanation of Cauchy Schwarz result

I read this example in a popular math book I was browsing, and the author does not give any physical explanation.

The example: Consider a rock falling freely under gravity. There are no viscous or dissipative effects. We will find the average speed at the end of 5 seconds.

Time-domain. $v = 32 t \mbox{feet/sec}$ Average over $5$ seconds $$\langle v\rangle_\text{time} = \frac{1}{5} \int_0^5 32 t \mathrm{d}t = 80\text{ feet/sec}$$

Space Domain. At $t =5$ sec, $y = 400$ feet. $$\langle v\rangle_\text{space} = \frac{1}{400}\int_0^{400} 8\sqrt{y}\; \mathrm{d}y \approx 107\text{ feet/sec}$$

Sorry about the use of imperial units. But I just dont understand the physics of this. The source uses this only as a demonstration of Cauchy Schwarz inequality. Is there something elementary I am missing out here?

Essentially, he derives $$\frac{1}{T}\int_0^T v(t) \mathrm{d}t \leq \frac{1}{L}\int_0^L v(y)\mathrm{d}y$$

The math is ok to follow, but my physical intuition just cant get it somehow.

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I'm not sure I see what you're actually asking about. Are you confused about why the two different averages have different numerical values? (Also, no need to apologize for your units.) –  David Z Aug 22 '11 at 20:43
@David they are averaged over the same two events (classically), and represent the same physical quantity. Why are the values different, and in general, why is it an ineuqlaity? –  yayu Aug 22 '11 at 20:46

Let us denote by $\langle\cdot\rangle \equiv \int_0^T \cdot \frac{dt}{T}$ the averaging over time. Please notice that:

$$\frac{1}{L}\int_0^Lv\mathrm{d}y=\frac{1}{\langle v\rangle T}\int_0^Tv\frac{\mathrm{d}y}{\mathrm{d}t}\mathrm{d}t = \frac{1}{\langle v\rangle}\int_0^Tv^2\frac{\mathrm{d}t}{T} = \frac{\langle v^2\rangle}{\langle v\rangle}$$

Thus the inequality is given by: $\langle v\rangle \le \frac{\langle v^2\rangle}{\langle v\rangle}$, or $\langle v\rangle^2 \le \langle v^2\rangle$, which is indeed the the Cauchy-Schwarz inequality.

Thus averaging over time is not the same as averaging over the position. Averaging with respect to two variables is the same if and only if these two variables are linearly dependent (more generally related by an affine transformation). In our case, in general, the position is a nonlinear function of time, thus one obtains different results in averaging over time and position. The only case that these two averages are equal is the case of uniform motion with a constant velocity.

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Uniform motion is not the only case, it would still work with e.g. a superimposed harmonically periodic motion. It's just not equal for all start/end points anymore, then. –  leftaroundabout May 14 '12 at 18:30