Natural Science Forum / Physics / General Physics / July 2008

Oh yes it would.

The ball leaves the cannon at speed 1.
The bat approaches the cannon at speed 1.
The ball meets the bat and rebounds/reflects.
How fast does the ball return to the cannon?

These two waves have the same amplitude, frequency and wavelength:

http://www.androcles01.pwp.blueyonder.co.uk/Wave/waves.htm

What is their speed?

Why did Einstein say
the speed of light from A to B is c-v,
the speed of light from B to A is c+v,
the "time" each way is the same?
  Ref:
http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img22.gif

Hint: Because he was a cretin.

'we establish by definition that the "time" required by
light to travel from A to B equals the "time" it requires
to travel from B to A' because I SAY SO and you have to
agree because I'm the great genius, STOOOPID, don't you
dare question it. -- Rabbi Albert "I'm a genius" Einstein.

You can have the complete answers when you ask complete questions.

 http://www.androcles01.pwp.blueyonder.co.uk/Doppler/Doppler.htm

The formula could be the same but because police cars radar
is connected to the speed of the car so it does not speed itself
and give a ticket to the "sitting still car".

Newtonian relativity does apply.
If not.. police cars could not move to use thier radar correctly

Doppler works with newton as well as it does with
any "newer" thoughts as long as you use it correctly.

Quentin, a proper question deserves a proper answer.

As you note, the doppler effect for sound is not symmetric between
emitter and receiver.  The two formulas differ according to whether it
is the emitter or the receiver that is moving w.r.t. the air, whle the
other is stationary.

But as your astute student noted, there is no medium for light waves,
according to current orthodox physics theory.  Since physics cannot
distinguish between moving emitter and moving receiver, motion being
an entirely relative concept, the relativistic formula for the doppler
shift has to be the same for both cases.

The answer can be found in any recent book on introductory physics, I
believe.  The answer turns out to be the geometric mean between the
two sound formulas.  You probably know these: Arithmetic mean: (x+y)/
2.  Geometric mean: sqrt(x*y).

The way that relativity comes in is through time dilation: there is a
factor of sqrt(1-v*v/c*c) in there somewhere.  You'll find it in the
books.

Cheers,

Uncle Ben

Quentin,

Beware of the answers you get here.  sci.physics is infested by those
obsessed with relativity and trying to prove it wrong.  I find this
interesting in that there is no similar group trying to disprove quantum
mechanics, which is (to me, anyway), even weirder than SR and GR.

Anyway, according to SR, there is no difference between the view of
a stationary police car and moving target car (frame of reference of
the police car), and a stationary target and moving police car (frame
of reference of the target), the signal received by the radar gun would be
the same.

I'm not going to waste time explaining the Doppler effect to you,
since you can find a simple expanation in any highschool level physics
textbook.  I'll only share with you that it has no connection to the
theory of relativity.

No offense intended, but we are not here to do your highschool
homework for you.

Harry C.

Okay, but you should be aware that that's an approximation assuming the
velocity of the car is much less than the velocity of the microwaves.
Totally sufficient for police work, but not good enough if you want to talk
about special relativity.

This would make for a good special relativity homework problem.  You'd
describe the setup with the approaching cars (according to relativity, it
doesn't matter which one is moving).  Then the first two parts would be:

(1) Calculate the (reflected wavelength)/(incident wavelength) ratio in the
frame of the police car.

(2) Calculate the (reflected wavelength)/(incident wavelength) ratio in the
frame of the civilian car.

(1) and (2) are just kinematics; you write the formula for the frequency
with which the waves strike the car and the frequency with which they are
reflected.  Those two frequencies have to be the same.

In part (1), the wavelength changes, but in part (2), it doesn't, even
though (1) and (2) both describe the same physical situation.  The key
point is that the wavelength of the microwaves depends on whose frame
you're measuring it in.

This would lead to part (3):

(3) Use this to derive the general transformation from the wavelength of
light in frame A to its wavelength in frame B, where A is moving w.r.t. B
at speed v in the same direction as the light.  The transformation is of
the form L_B = f(v) L_A.

Here you have to remember that in the police radar problem there are two
transformations of this type you have to apply -- the transformation from
police frame to civilian frame for the incident microwaves, and from the
civilian frame back to the police frame for the reflected ones.  But f(v)
is the same in both cases, so you can figure out what it is just by taking
a square root.

If you can come up with the same relativistic Doppler shift formula that
Einstein did in the reference Uncle Ben provided (remembering to convert
from wavelength to frequency, and that in Einstein's phi=0 case the
receiver is moving *away* from the source),

http://www.fourmilab.ch/etexts/einstein/specrel/www/#SECTION22

then you probably understand it.

Of course, for practical purposes, what matters is the wavelength as
measured by the radar gun, which will be doing measurements in the police
car's frame.  And so the answer to your original question is that the
measured wavelength shift will be the same in both cases.