A sine wave traveling in +x direction can be represented by $\cos(\omega {t}- kx+\phi)$.

As shown in the drawing attached where $\phi=\frac{\pi}{2}$:

a) Is a plot that holds t=0 and two waves along +x direction.
$$t=0\Rightarrow\; \cos(\omega {t}- kx+\phi)\;=\;\cos(- kx+\frac{\pi}{2})$$
This implies maximum at $-kx+\frac{\pi}{2}\;=\;0$
$$k=\frac{2\pi}{\lambda}\;\Rightarrow\; \frac {2\pi}{\lambda}x=\frac {\pi}{2}\;\Rightarrow\;x=\frac{\lambda}{4}$$
This gives the wave form in RED that LAGs the $\cos(\omega{t}-kx)$.

b) Is a plot of the waveform at x=0 and the two waves vary with time t.
$$x=0\;\Rightarrow\; \cos(\omega {t}- kx+\phi)\;=\;\cos(\omega{t}+\frac{\pi}{2})$$
This implies maximum at $\omega{t}=-\frac{\pi}{2}$
$$\omega{t}\;=\;\frac{2\pi}{T}t\;=\;-\frac{\pi}{2}\;\Rightarrow\;t=-\frac{T}{4}$$
This means waveform in RED LEADs the original wave.

So the two cases give opposite result. Can anyone explain this to me?
Thanks

Alan
Attached Thumbnails

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 Lead means a given wave peaks before the referenced wave so in your case the red wave peaks before the blue wave in time. Lag means the blue reference wave peaks and then the red wave peaks at a later time.
 Thanks for the reply. But both cases comes from the exact same equation where $\phi=+\frac{\pi}{2}$. How do you justify one is lead and one is lag depends on what reference used? The other more basic question, Am I interpreting these correctly? It is very confusing.

I don't like that notation. Let's discuss this using $$\cos(kx-ct-\Phi)$$ It is exactly the same function because $\cos x = \cos -x$ ok?

Maybe we first find out why the function looks the way it does. Maybe you remember from school how to shift a function to the right. If you want to shift $f(x)$ by a value c. You subtract it from x. $f(x-c)$ is the same as $f(x)$ just shifted to the right by c units. Why minus and not plus? Functions are weird like that, and I believe it is a consequence of the higher mathematical magic of category theory, but then what do I know?

Anyhow you know what a cosine looks like $$\cos(x)$$ if you want it to wiggle faster you can multiply x by a factor. $$\cos(kx)$$ How weird! Did you notice that multiplying inside by k squeezes the function in the x direction, while multiplying on the outside stretches it in the y direction? That weird function behaviour again.

So now what is a plane wave? A plain wave is a cosine that shifts with time to the right. So it starts at zero, then at time 1 it has moved by c to the right at time 2 by 2c and so on. We can write this as a shift by ct.$$\cos\left(kx - ct \right)$$

Now we can shift the function an additional bit $\Phi$.

$$\cos\left(kx - ct - \Phi \right)$$

Now we understand why that thing looks the way it looks. And why it will shift the cosine to the right in the x direction. But what about the t direction? An additional shift can also be accomplished by letting a bit more time pass. Lets say we need a time of $\Delta$ to move the function by $\Phi$ then we could write instead
$$\cos\left(kx - c(t + \Delta) \right)$$
We need to add to t. Another way of looking at this is by looking at it as a function of the variable t. Remember we can multiply the inside by -1 and nothing changes. So we get
$$\cos(kx-ct-\Phi)=\cos(ct-kx+\Phi)$$
your formula. kx is now some fixed number shifting the function to the right in t direction, but $\Phi$ is positive so it will shift the formula to the left.

On the outside multiplying by -1 flips top and bottom, on the inside the function flips left and right. So in a way the reason why $\Phi$ shifts forward in x and backwards in t direction is because the t variable is "running backwards".

Maybe this helps, maybe not.

 Quote by yungman Thanks for the reply. But both cases comes from the exact same equation where $\phi=+\frac{\pi}{2}$. How do you justify one is lead and one is lag depends on what reference used? The other more basic question, Am I interpreting these correctly? It is very confusing.
I think they're basing it on the phase angle so a positive phase angle means the wave lags behind a wave of phase angle 0 and conversely a negative phase angle leads.

In the first example, they're looking at waves where t is fixed at time zero.

In the second example, they're looking at the same two waves where x position is fixed at 0 and the waves vary in time.

The reference wave is the one with phase angle 0.

I don't think I'm being very helpful here but I trying...
 Thanks 0xDEADBEEF, I understand this. My question is why the ∅ is lag ( to the right) in a) and lead ( to the left) in b)? I can even see why the phase is shifting to the left when holding x=0 and change t. I have no issue of understanding that either. Problem is the book called this as LEAD. That is where all the confusion started. In my understanding, a +ve ∅ ALWAYS imply lagging no matter what!!! Also thanks Jedishrfu for your help.
 Ok I think I see your point now. Sorry if I was too pedagogic. The name lead is indeed ****. The curve is located further to the left, this means it will arrive later at the point of observation. It is in no way leading. But I am just a physicist, maybe this naming makes sense in someone else's world.

 Quote by yungman A sine wave traveling in +x direction can be represented by $\cos(\omega {t}- kx+\phi)$. As shown in the drawing attached where $\phi=\frac{\pi}{2}$: a) Is a plot that holds t=0 and two waves along +x direction. $$t=0\Rightarrow\; \cos(\omega {t}- kx+\phi)\;=\;\cos(- kx+\frac{\pi}{2})$$ This implies maximum at $-kx+\frac{\pi}{2}\;=\;0$ $$k=\frac{2\pi}{\lambda}\;\Rightarrow\; \frac {2\pi}{\lambda}x=\frac {\pi}{2}\;\Rightarrow\;x=\frac{\lambda}{4}$$ This gives the wave form in RED that LAGs the $\cos(\omega{t}-kx)$.
How would you draw the conclusion that the red curve lags, from a plot versus distance?
The red is leading in both "cases". There are not two cases actually. You are just miss-interpreting the first plot.

By the way, leading and lagging are relative for continuous waves. The red is leading by T/4 or lagging by 3T/4.
Usually the least distance in time is considered. So here they would rather say is leading by T/4 rather than lagging by 3T/4.

 Quote by nasu How would you draw the conclusion that the red curve lags, from a plot versus distance? Think about this and you'll see (maybe) why are you confused. The red is leading in both "cases". There are not two cases actually. You are just miss-interpreting the first plot. By the way, leading and lagging are relative for continuous waves. The red is leading by T/4 or lagging by 3T/4. Usually the least distance in time is considered. So here they would rather say is leading by T/4 rather than lagging by 3T/4.
I saw it in the book, here's an article:

You can look at it this way: The cosine wave lead the sine wave by 90 deg because at x=0, cosine wave is at max while the sine is still going from 0 to the max. The sine wave won't reach the max until x=λ/4.

 Quote by 0xDEADBEEF Ok I think I see your point now. Sorry if I was too pedagogic. The name lead is indeed ****. The curve is located further to the left, this means it will arrive later at the point of observation. It is in no way leading. But I am just a physicist, maybe this naming makes sense in someone else's world.
This started out very simple and obvious, but the more I dig into it, the more confusing it gets. I am trying to look at the polarizing of a plain wave travel in +ve and -ve x direction and try to make sense how the phase constant play into it. This have effect on the LHC and RHC and direction of rotation.

 Quote by yungman I saw it in the book, here's an article: http://www.allaboutcircuits.com/vol_2/chpt_1/5.html You can look at it this way: The cosine wave lead the sine wave by 90 deg because at x=0, cosine wave is at max while the sine is still going from 0 to the max. The sine wave won't reach the max until x=λ/4.
You are still confused.
The first plot is for a given, fixed time. To say "until x=λ/4" does not make sense.
The point x=λ/4 exist at all times.

What you see there is like a snapshot. The leading wave is the one that is, well, ahead.
If you put a detector at some position along that line and let the time run, the red wave will reach the detector first*. The whole figure will shift to the right as the time passes with the red peaks remaining ahead (in space) of the blue ones by λ/4

*Unless you put it right between a blue and a red peak.

 Quote by nasu You are still confused. The first plot is for a given, fixed time. To say "until x=λ/4" does not make sense. The point x=λ/4 exist at all times. What you see there is like a snapshot. The leading wave is the one that is, well, ahead. If you put a detector at some position along that line and let the time run, the red wave will reach the detector first*. The whole figure will shift to the right as the time passes with the red peaks remaining ahead (in space) of the blue ones by λ/4 *Unless you put it right between a blue and a red peak.
At this point, the more important question is whether I am correct with my drawing that $\cos(-kx+\frac{\pi}{2})$ is the way I draw it in red in the top graph with x axis?