What is the exact change in frequency for an approaching object?

[lenfromkits]

When light is shone from an approaching object towards a stationary observer, the light is blue-shifted. What exactly is the amount of this change in frequency? So if the wave that leaves a radar gun is, say, 500 THz, when it is traveling 100km/h, what will it be when observed by the stationary observer?

This is NOT a homework question and I am not a student trying to do homework so please don't assume I'm being 'lazy' and not doing my homework myself. What I'm trying to figure out is if the change in frequency observed is linear compared to the speed, or if it is relativistic and a small change in frequency compared to the speed.

Thanks.

 

[bcrowell]

Look up the relativistic Doppler shift in any textbook.

 

[granpa]

set the object a certain distance from the gun.
how long before it reaches the gun.
in that time how many cycles of radiation will it reflect

 

[Passionflower]

When light is shone from an approaching object towards a stationary observer, the light is blue-shifted. What exactly is the amount of this change in frequency? So if the wave that leaves a radar gun is, say, 500 THz, when it is traveling 100km/h, what will it be when observed by the stationary observer?

Assuming a line of sight approach and a constant velocity.

The Doppler formula is for c=1:

$$\sqrt {{\frac {1+v}{1-v}}}$$

First we convert 100km/h in terms of c:
100 * 1000 /3600 / 299792458 = 9.2656693110597791549*10^(-8)

Thus we get a rate of frequency of:

1.0000000926566974032

What I'm trying to figure out is if the change in frequency observed is linear compared to the speed, or if it is relativistic and a small change in frequency compared to the speed.

Look at this graph that should give you the answer if it is linear or not:

 

[granpa]

I thought he wanted to know what a cop with a radar gun would see if a car was moving near light speed.

I didnt know that the radar gun was in the car (and was therefore time dilated)

 

[lenfromkits]

I thought he wanted to know what a cop with a radar gun would see if a car was moving near light speed.

I didnt know that the radar gun was in the car (and was therefore time dilated)

My understanding is that it doesn't matter if the radar gun is moving or stationary...it is only about the relative speed between the gun and the object it's pointing at? Just checking.

 

[lenfromkits]

Look up the relativistic Doppler shift in any textbook.

Funny...thanks. It didn't occur to me to search for 'relativistic' doppler shift - I had only looked into 'dopper shift' so I missed all the details on it. I get it now. And I see how at insignificant speeds there isn't much of a relativistic effect anyhow. This only leaves the effect for really fast speeds. I wonder if there has been any actual observed examples of that? I guess not yet?

 

[lenfromkits]

Assuming a line of sight approach and a constant velocity.

The Doppler formula is for c=1:

$$\sqrt {{\frac {1+v}{1-v}}}$$

First we convert 100km/h in terms of c:
100 * 1000 /3600 / 299792458 = 9.2656693110597791549*10^(-8)

Thus we get a rate of frequency of:

1.0000000926566974032


Look at this graph that should give you the answer if it is linear or not:

View attachment 28973

Thanks! You have no idea how much it helps to clear things up when someone spells it out like that :)

 

[lenfromkits]

Do we have yet any observational evidence of the relativistic aspect of the Doppler shift? (ie, for high velocities where 'v' is closer to 1 - are there any examples discovered yet?)

 

[Passionflower]

Here is a plot showing the relativistic and non-relativistic Doppler shift and the difference between them.

Did you know by the way that if the motion is not line of sight but instead transverse the Doppler effect is very different? In this case it is basically the relativistic time dilation, so for a 'train' approaching or departing the station it is:

$${1 \over \sqrt{1-v^2}}$$

My understanding is that it doesn't matter if the radar gun is moving or stationary...it is only about the relative speed between the gun and the object it's pointing at? Just checking.

That is exactly right!

What does matter is acceleration, if the gun or the target accelerates (or both) we need a different formula and in these cases even the distance of the radar gun matters.

If you keep accelerating hard enough you may even outrun an outer space speed check of your relativistic rocket.

And these are just the simple scenarios, for movement in three directions the problem becomes more difficult. As far as I know there is no literature on the exact Doppler formula for 3 dimensional acceleration of the target for a given velocity, or perhaps the "high priests" prefer to keep those formulas secret. Seriously though, if anyone knows please post or give a reference to the formulas, science should not be a secret.

 

[lenfromkits]

Here is a plot showing the relativistic and non-relativistic Doppler shift and the difference between them.

View attachment 29000

Did you know by the way that if the motion is not line of sight but instead transverse the Doppler effect is very different? In this case it is basically the relativistic time dilation, so for a 'train' approaching or departing the station it is:

$${1 \over \sqrt{1-v^2}}$$


That is exactly right!

What does matter is acceleration, if the gun or the target accelerates (or both) we need a different formula and in these cases even the distance of the radar gun matters.

If you keep accelerating hard enough you may even outrun an outer space speed check of your relativistic rocket.

And these are just the simple scenarios, for movement in three directions the problem becomes more difficult. As far as I know there is no literature on the exact Doppler formula for 3 dimensional acceleration of the target for a given velocity, or perhaps the "high priests" prefer to keep those formulas secret. Seriously though, if anyone knows please post or give a reference to the formulas, science should not be a secret.

thanks. yes, I was kind of avoiding the traverse situation.

3D movement? I would see that as always being just 2D (traverse) since the 3 axis are relative and you can always align one with the movment leaving only 2 of concern.

 

[Passionflower]

thanks. yes, I was kind of avoiding the traverse situation.

3D movement? I would see that as always being just 2D (traverse) since the 3 axis are relative and you can always align one with the movment leaving only 2 of concern.

Yes, a poor word choice from my side.

 

[lenfromkits]

Yes, a poor word choice from my side.

Hey, since you're so helpful, did you notice my question about what actual observations they've made of any sort of doppler shift than contained a relativistic effect? (ie, where v is closer to 1). (since as far as I can see, any doppler shift we measure is always with low values of v and so no real time dilation in there)

 

[lenfromkits]

or perhaps the "high priests" prefer to keep those formulas secret.

Hey, just curious, who are the "high priests?"

 

[Ich]

Hey, since you're so helpful, did you notice my question about what actual observations they've made of any sort of doppler shift than contained a relativistic effect?

You have to read the https://www.physicsforums.com/showthread.php?t=229034", section 4.

 

[granpa]

http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html#Tests_of_time_dilation

Tests of Time Dilation and Transverse Doppler Effect

The Doppler effect is the observed variation in frequency of a source[/color] when it is observed by a detector that is moving relative to the source. This effect is most pronounced when the source is moving directly toward or away from the detector, and in pre-relativity physics its value was zero for transverse motion (motion perpendicular to the source-detector line). In SR there is a non-zero Doppler effect for transverse motion, due to the relative time dilation of the source[/color] as seen by the detector.

I thought he wanted to know what a cop with a radar gun would see if a car was moving near light speed.

I didnt know that the radar gun (i.e. the source[/color]) was in the car (and was therefore time dilated[/color])

My understanding is that it doesn't matter if the radar gun is moving or stationary...it is only about the relative speed between the gun and the object it's pointing at? Just checking.

so if a laser beam of freq f from a stationary laser gun (the source) is reflected off a mirror moving at velocity v which is very near the speed of light then what is the freq of the reflected light as measured by the stationary observer? Note that the gun is the source.

 

[Passionflower]

Hey, just curious, who are the "high priests?"

I was just a figure of speech, it is an allusion to ancient history where the 'scientists' were actually members of the religious caste who kept their secrets from the common people.

 

[TrickyDicky]

What does matter is acceleration, if the gun or the target accelerates (or both) we need a different formula and in these cases even the distance of the radar gun matters.

And these are just the simple scenarios, for movement in three directions the problem becomes more difficult. As far as I know there is no literature on the exact Doppler formula for 3 dimensional acceleration of the target for a given velocity, or perhaps the "high priests" prefer to keep those formulas secret. Seriously though, if anyone knows please post or give a reference to the formulas, science should not be a secret.

Curiously enough after a brief search on the web, I find many doppler formulas for velocity but none for acceleration. I guess the derivation is easy and they don't bother to show it.
Once you have worked out the Doppler shift of frequencies if the shift is constant there is no acceleration and if varies with time you calculate this variation per unit time and convert it to an acceleration:

$$a=\frac{c \Delta fd}{2tfr}$$
a= acceleration
c=light speed constant
t=time it takes for the Doppler shift variation
$$\Delta fd$$=Doppler shift variation
fr= frequency echoed by the source

 

[jtbell]

Do we have yet any observational evidence of the relativistic aspect of the Doppler shift?

See section 4 of the following FAQ:

Experimental Basis of Special Relativity

 

[Passionflower]

so if a laser beam of freq f from a stationary laser gun (the source) is reflected off a mirror moving at velocity v which is very near the speed of light then what is the freq of the reflected light as measured by the stationary observer? Note that the gun is the source.

Relativity teaches us that when a light source approaches we can still have it being redshifted.

Here is a 3D graph that shows the Doppler of an approaching source for different angles. It is hard to see but if you look good you can see the shift is maximum if the angle is 0 and that for very wide angles there is actually redshift instead of blueshift (e.g. the flat parts on the sides go below 1 for high velocities).

Curiously enough after a brief search on the web, I find many doppler formulas for velocity but none for acceleration. I guess the derivation is easy and they don't bother to show it.

The formula for an accelerating observer of an inertial light source in the line of sight is (c=1):

$$-g \left( 1+gx \right) ^{-1} \left( \ln \left( 1-{\frac {g}{v}}<br /> \right) \right) ^{-1}$$

 

[TrickyDicky]

The formula for an accelerating observer of an inertial light source in the line of sight is (c=1):

$$-g \left( 1+gx \right) ^{-1} \left( \ln \left( 1-{\frac {g}{v}}<br /> \right) \right) ^{-1}$$

Huh? That might work if you already have the acceleration and you then just plug it in the formula, I thought what you asked for was an equation to obtain the acceleration not the frequency. Your formula has nothing to do with this.

 

[Passionflower]

I thought what you asked for was an equation to obtain the acceleration not the frequency. Your formula has nothing to do with this.

No, I was asking for the above formula but then generic so that it also works with angles.

I do not believe obtaining the acceleration from a given Doppler shift and a velocity can be expressed analytically in a formula.

 

[TrickyDicky]

No, I was asking for the above formula but then generic so that it also works with angles.

But why do you call it a formula for acceleration, maybe that's why you can't find it, I don't think there is any conspiration to keep that in secret.

I do not believe obtaining the acceleration from a given Doppler shift and a velocity can be expressed analytically in a formula.

Why not?, I just did. It is used all the time for doppler radar applications (weather radars etc)

 

[Passionflower]

But why do you call it a formula for acceleration, maybe that's why you can't find it, I don't think there is any conspiration to keep that in secret.

What are you talking about? Please read again what I wrote, it seems you take things a bit too literally.

Why not?, I just did. It is used all the time for doppler radar applications (weather radars etc)

$$a=\frac{c \Delta fd}{2tfr}$$
a= acceleration
c=light speed constant
t=time it takes for the Doppler shift variation
$$\Delta fd$$=Doppler shift variation
fr= frequency echoed by the source

So you say this formula obtains the proper acceleration of an observer based on the Doppler shift from an inertial source? You do not think the issue is time dilation and a change in velocity during the proper acceleration?

 

[TrickyDicky]

What are you talking about? Please read again what I wrote, it seems you take things a bit too literally.

So you say this formula obtains the proper acceleration of an observer based on the Doppler shift from an inertial source? You do not think the issue is time dilation and a change in velocity during the proper acceleration?

Nothing of the kind, that formula obtains the acceleration of the source, relative to the observer, no time dilation involved, it's all in tne same reference frame. By now I get this has nothing to do with what you looked for. Hope someone else can help you.