Question

relation between energy and amplitude of a light wave, if amplitudes a factor in energy why isn't it in E=hf

Answer 1

The "amplitude" of light isn't a readily measurable quantity. That's why we stick with frequency.

Answer 2

so does amplitude of light even effect energy

Answer 3

It does, since the energy of a wave can always be expressed in some form of the equation:\[E_{wave} = {1\over2} \kappa A^2\], where kappa is some constant, and A is some amplitude. It's just not really useful to write it in such a way.

Answer 4

cool! can you derive this or show a link to more information

Answer 5

This is a slightly different case, but it illustrates the point (uses classical mechanics. Sorry if I jump over a few steps):
However, in simple harmonic motion, you CAN measure the amplitudes of many things:\[E_{spring} = {1\over2}mv^2 + {1\over2} kx^2\]Due to conservation of energy, we see that the amplitude of x and the amplitude of v are related by \[mv_{\max}^2 = kx_{\max}^2\]Or if you want to simplify, you can just make a relation\[E={1\over2}m(\omega r)^2 + {1\over2}m \omega^2 x^2 = {1\over2}m \omega^2 (r^2+x^2)\]Thus we can see that\[\kappa = m \omega^2; A = \sqrt{r^2 + x^2}\]

Answer 6

very nice. so the actual equation for the energy of a light wave is:
E=hf+.5kA^2

Answer 7

It's helpful to note that\[v = \omega r; k = m \omega^2; \omega = \sqrt{k \over m}\]
Don't worry if this doesn't make sense... I hope it helps but if you haven't gotten this far in physics, then you can look forward to these sweet (and elegant!) relations.

Answer 8

No, it's actually just \[E=hf \]The amplitude energy is basically encoded into the frequency.

Answer 9

@jwuphysics, your formulation isn't quite correct. Light DOES have amplitude, and corresponds to the amplitudes of the electric and magnetic oscillations in the light wave. The reason the formula given isn't dependent on amplitude is because that is the amount of energy that is in EACH carrier (i.e., each photon). When you make a light brighter, you increase the amplitude, and thus you have more photons. Thus, you have more overall energy, though you have the same amount of energy in each photon. Changing the color of the light changes the amount of energy per carrier.

Answer 10

@Orbitalshawn, no that isn't correct. You're adding two different types of energies...one is the energy per photon, and the other is the supposed energy for the whole light, which doesn't even make sense, since it's not dependent on how much light there is! (Two bulbs = twice as much energy released as one)

Answer 11

@yakeyglee Yes, I did say that light has an amplitude, but (correct me if I'm wrong), it doesn't have a physical interpretation. A closer analog might be intensity, but that also hinges on the frequency of light.

Answer 12

The physical interpretation is the magnitude of the electric field in the oscillation. Intensity comes from electromagnetic energy, which is related to the square of the electric field, which is why we care about squares of amplitudes in Quantum Mechanics when we consider probabilities.

Answer 13

Use the Poynting vector to determine the energy flow. Let us consider the following simple plane wave:\[\vec E = E_0 \cos(\omega t - kx) \hat y\]\[\vec B = B_0 \cos(\omega t - kx) \hat z\]The Poynting vector, which is the magnitude of the energy density in the direction of its flow, is given as follows:\[\vec S \overset{\text{def}}{=} \frac{\vec E \times \vec B}{\mu_0}=\frac{E_0B_0}{\mu_0}\cos^2(\omega t - kx)\hat x\]Time-averaging \(\cos^2\) is simply \(\frac{1}{2}\).\[\vec S = \frac{E_0B_0}{2\mu_0}\]We also know that for electromagnetic waves, \(B_0=\frac{E_0}{c}=E_0 \sqrt{\mu_0 \epsilon_0}\).\[\vec S = \frac{E_0^2}{2}\sqrt{\frac{\epsilon_0}{\mu_0}}\propto E_0^2\]Note how \(E_0\) is the amplitude of the electric field in the electromagnetic wave.

Answer 14

I forgot to indicate the direction in the last two equations, but \(\vec S\) is parallel to \(\hat x\).

Answer 15

so that means that the amplitude of one light wave is dependent on frequency. The only way to change amplitude is by adding photons?

Answer 16

Amplitude is not dependent on frequency. The magnitude of the time-averaged Poynting vector gives you time-averaged energy per unit volume, while \(E=hf\) gives you energy per carrier. Increasing the amplitude thus increases the amount of carriers per unit volume, which is the same as making the light brighter.

Answer 17

In fact, interestingly enough, using this, you could calculate the number of photons per unit volume in light with electric amplitude \(E_0\) and frequency \(f\) by the following scheme:\[\frac{\text{Carriers}}{\text{Unit Volume}}=\frac{\displaystyle \left( \frac{\text{Energy}}{\text{Unit Volume}} \right)}{\displaystyle \left( \frac{\text{Energy}}{\text{Carrier}} \right)}=\frac{\dfrac{E_0^2}{2}\sqrt{\dfrac{\epsilon_0}{\mu_0}}}{hf} = \boxed{\displaystyle \frac{E_0^2}{2hf}\sqrt{\frac{\epsilon_0}{\mu_0}}}\]

Answer 18

That last part wasn't really relevant...I just thought it was interesting. Did I manage to answer your question to your satisfaction?

Answer 19

@yakeyglee: Small change to what you wrote:

Energy per unit volume is \(\Large \frac{\epsilon_0 E_0^2}{2}\)

and \(\Large \frac{ E_0^2}{2} \sqrt \frac{\epsilon_0}{\mu_0}\) is power per unit area.

ratio of those is \(c\), as energy is carried along by the wave at the speed of light.

Answer 20

Ah...thanks for the correction...you are correct!

Answer 21

Even so, @Orbitalshawn, the proportionality is still the same.