Frequency, Velocity and Wavelength of Light

[Nugso]

Hello all. When light goes through one medium to another, its velocity and wavelength change but frequency remains the same. If the wavelength of light decreases, the length between two crests decreases, doesn't that mean there will be more wave hence frequency will have to increase? I tried to find an animation that explains it, but couldn't find one.

The other question is, when light is affected by gravity, this time its frequency and wavelength change, but its velocity remains the same. Well, light is not directly by gravity but its spacetime path is curved. Since we know E=hf, and in that case f changes thus light's energy also changes. How can its energy change when there's no force acting upon it?

Why does sometime frequency change and sometimes velocity?

Edit: Forgot to ask my last question. When measuring the frequency of light when it's under the effect of gravity, how do we know the device measuring the frequency is not affected by gravity?

 

[Nugatory]

Hello all. When light goes through one medium to another, its velocity and wavelength change but frequency remains the same. If the wavelength of light decreases, the length between two crests decreases, doesn't that mean there will be more wave hence frequency will have to increase? I tried to find an animation that explains it, but couldn't find one.

The speed is equal to the frequency times the wavelength (number of peaks passing by in a given period of time, times the distance between peaks). So it's possible to change any two while keeping the third constant.

The other question is, when light is affected by gravity, this time its frequency and wavelength change, but its velocity remains the same. Well, light is not directly by gravity but its spacetime path is curved. Since we know E=hf, and in that case f changes thus light's energy also changes. How can its energy change when there's no force acting upon it?

in general relativity all gravitational effects on everything, regardless of composition, are due to curvature not a force acting between things. Thus, GR predicts that gravity will affect everything, including light, the same way - and in fact that's where the prediction of gravitational redshift came from. As for the obvious follow-on question "how does following a curved path in free-fall lead to an energy change?"... You'll find plenty of previous threads discussing this.

Why does sometime frequency change and sometimes velocity?

it depends on the physics of the particular situation you're considering. In the gravitational redshift case, the speed of light in a vacuum must always be $c$, so the loss of energy when light travels "uphill" requires that the frequency fall and the wavelength increase (speed equals frequency times wavelength, remember). In a medium in which light travels at a speed less than c, the frequency can remain constant while the speed and the wavelength both go down.

Edit: Forgot to ask my last question. When measuring the frequency of light when it's under the effect of gravity, how do we know the device measuring the frequency is not affected by gravity?

We can lower the light source into the gravity well, watching the frequency decrease as the depth increases. Or (which comes down to the same thing) we set up multiple identically constructed light sources at different depths, compare the measured frequency of each.

 

[Nugso]

Thanks Nugatory,

in general relativity all gravitational effects on everything, regardless of composition, are due to curvature not a force acting between things. Thus, GR predicts that gravity will affect everything, including light, the same way - and in fact that's where the prediction of gravitational redshift came from. As for the obvious follow-on question "how does following a curved path in free-fall lead to an energy change?"... You'll find plenty of previous threads discussing this.

"how does following a curved path in free-fall lead to an energy change?" Thanks! I just couldn't articulate myself. Searching for the threads.

it depends on the physics of the particular situation you're considering. In the gravitational redshift case, the speed of light in a vacuum must always be $c$, so the loss of energy when light travels "uphill" requires that the frequency fall and the wavelength increase (speed equals frequency times wavelength, remember). In a medium in which light travels at a speed less than c, the frequency can remain constant while the speed and the wavelength both go down.

Do you know of a situation where wavelength remains the same whereas speed and frequency change?

According to the gravitational redshift case, there's also "Gravitational Time Dilation", which says that a clock that's under the effect of gravitational field runs more slowly.

And according to the picture above, the distance light travels extends due to gravitational field. We know that x = v.t, and in that case x increases while t decreases(?). How's that possible?

 

[yuiop]

Look closely at the green dot representing the GR case. It arrive at the observer after the black and red dots so both x and t increase relative to the other cases.

 

[Nugso]

Look closely at the green dot representing the GR case. It arrive at the observer after the black and red dots so both x and t increase relative to the other cases.

You're right but the gravitational time dilation says that a clock runs more slowly. I don't seem to get the relation between that sentence and the picture.

 

[yuiop]

You're right but the gravitational time dilation says that a clock runs more slowly. I don't seem to get the relation between that sentence and the picture.

The time t is the time according to the time at the source and at the observer far away from the gravitational body. Photons do not record time, but if the photon was replaced by a particle moving close to c, then the travel time recorded by the particle would be less than the travel time according to the far away observer.

 

[WannabeNewton]

...but its velocity remains the same.

Keep in mind that this is only true locally.

Anyways, you can use a simple conservation of energy argument to show that gravitational redshift is necessary. Also, being affected by non-trivial space-time geometry is the exact same thing as being affected by gravity!

 

[Nugso]

@yuiop, So it's kind of all about relativity of time, right?
@WannaBeNewton, Can you please elaborate? Both my English and Physic levels are low so I usually have a hard time grasping and understanding things.

 

[WannabeNewton]

@WannaBeNewton, Can you please elaborate?

Elaborate on what specifically, sorry? I'd be happy to elaborate on whichever point you wish!

 

[Nugso]

Elaborate on what specifically, sorry? I'd be happy to elaborate on whichever point you wish!

Ah, I should have eleborated first! What did you mean by "locally"?

 

[WannabeNewton]

Ah, I should have eleborated first! What did you mean by "locally"?

Locally in this context means that if the worldline of an observer intersects the worldline of a light ray at some event and the observer measures the speed of the light ray at that coincident event then he will find that $v = 1$. When an observer measures the speed of a light ray or the relative speed of another observer, the measurement is always done at a single event that lies on both their worldlines; this is what is meant by locally.

 

[Nugso]

http://www.sketchtoy.com/53576707

You mean in the yellow area, the observer(blue circle) will find that the red light's speed is 1? How do we get the "1"?

 

[WannabeNewton]

Haha that drawing reminds me of the old Nickelodeon show Chalk Zone :p

But yes that is exactly what happens. $v = 1$ is in natural units (so that in particular $c = 1$) meaning it's the same thing as $v = c$.

 

[Nugso]

Haha that drawing reminds me of the old Nickelodeon show Chalk Zone :p

But yes that is exactly what happens. $v = 1$ is in natural units (so that in particular $c = 1$) meaning it's the same thing as $v = c$.

Couldn't have expressed myself better!

I was thinking how could it be 1. Turns out it's in natural units. Many thanks!