Relative speed of two oppositely directed light beams

[PeterDonis]

the relative speed of the light beam compared to us, should be greater than C otherwise would not be blue-shift effect.

No. The speed of a light beam relative to anyone is always c.

"Each wave crest has a bit less distance to travel to reach you than the previous one so spends a bit less time in flight and arrives a bit sooner than if the source were not moving."
This condition is only met if speed increased.

No. It is true as long as the source is traveling towards the observer. There is no need for the source's speed relative to the observer to change.

 

[PeroK]

What about the blue-shift effect? For me, those 2 things are somewhat contradictory.

If I interpret it correctly, the blue-shift effect is created because the speed at which a given object approaching the observer is added to the speed of light beam. Which means that the relative speed of light beam is greater than C relative to the observer?

Am I wrong?

Yes, you are wrong.

 

[Nugatory]

This condition is only met if speed increased.

It is not. If the time between the emission of successive crests is $\Delta{t}$, the source is approaching us at speed $v$ (assume for simplicity that $v$ is small compared with $c$ so that we don't have to mess with time dilation and other relativistic corrections):

A crest leaves the source at time $T$ when the source is at distance $D$. This crest arrives at the receiver at time $T+D/c$. The source is moving towards the receiver so at time $T+\Delta{t}$ the distance to the receiver is $D-v\Delta{t}$ andthe next crest, emitted at time $T+\Delta{t}$ arrives at $T+\Delta{t}+(D-v\Delta{t})/c$; the time between crest arrivals is less than $\Delta{t}$ even though the crests are all traveling at speed $c$.

 

[Sagittarius A-Star]

Of course the speed of the light is C still, but the relative speed of the light beam compared to us, should be greater than C otherwise would not be blue-shift effect.

At first you should say, what is your reference frame.

A) Description in the frame of the sender:

The sender is at rest. The light is moving with $c$ towards the receiver. The receiver is moving with $v$ towards the light. You get (includes an additional blue-shift by factor $\gamma$, because the moving receiver is time-dilated):
$\frac{f_R}{f_S} = \gamma \frac{c+v}{c} => \frac{f_R}{f_S} = \gamma (1+v/c)\ \ \ \ \$(1)

B) Description in the frame of the receiver:

The sender is moving with $v$ towards the receiver. The light is moving with $c$ towards the receiver. The receiver is at rest. You get:
$\frac{f_S}{f_R} = \gamma \frac{c-v}{c} => \frac{f_R}{f_S} = 1/\gamma * \frac{1}{(1-v/c)}\ \ \ \ \$(2)

With the time dilation factor $\gamma = 1/\sqrt{(1-v/c) * (1+v/c)}$ you can see, that the above equations (1) and (2) are equivalent and can also be written as:
$\frac{f_R}{f_S} =\sqrt{\frac{1+v/c}{1-v/c}}$

 

[David Lewis]

[T]he relative speed of two beams is C, because nothing can be quicker than light speed. However it is not an explanation ... because their relative velocities should be 2C.

It's not necessary to use light beams. If you observe two spaceships moving towards each other at +/- 0.6c then closing speed (rate at which the distance between them is decreasing) is 1.2c in your frame of reference. No physical law is violated.

 

[Ibix]

It's not necessary to use light beams. If you observe two spaceships moving towards each other at +/- 0.6c then closing speed (rate at which the distance between them is decreasing) is 1.2c in your frame of reference. No physical law is violated.

A point of clarification on this: in Newtonian physics closing speed and relative velocity are equal. Not so in relativistic physics, where the closing speed is limited to $2c$ and relative velocity to $c$.

 

[Grasshopper]

And, as has been pointed out repeatedly, you are analysing this completely wrongly. You need to use Minkowski geometry, not Euclidean geometry.
No. In fact, the 4d analog of distance along a light path is zero (hence the alternative name "null worldline" for lightlike worldlines). And you can't define "speed" through 4d spacetime (Brian Greene notwithstanding) because time isn't a separate thing.

Hey real quick hijack question, then I’ll leave:

If you try to do the “Pythagorean theorem” in Minkowski space for light, it would look like this, right?
$s^2 = (ct)^2 - x_{1}^2 - x_{2}^2 - x_{3}^2 = 0$Or did I miss something?

 

[Nugatory]

Or did I miss something?

Nope, that’s pretty much got it. But the implications of subtracting instead of adding as we do with Euclidean geometry are huge - most of special relativity flows from that one difference.

 

[Ibix]

Or did I miss something?

Looks fine. Pedantically, you want some deltas in there ($\Delta s$, $\Delta t$ etc) because those are differences you are working with. The distinction doesn't matter much here, but becomes important in non-inertial frames and GR.