Relative speed of two oppositely directed light beams

[Janos Meri]

TL;DR Summary

Relative speed of two oppositely directed light beams
1C or 2C

Can someone give a meaningful explanation that the relative speed of two oppositely directed light beams is why only one light speeds?
I understand that based on the Einstein relativity theory, the relative speed of two beams is C, because nothing can be quicker than light speed. However it is not an explanation, because based on Euclidean geometry, their relative velocities should be 2C.
It seems a paradox, because the two rules give different result.

Is there explanation that resolves this contradiction?

 

[Ibix]

Nothing can move faster than $c$ relative to any reference frame. Thus I can have two light beams traveling at $\pm c$ and the distance between them (as measured by me) falls at $2c$, and this is fine because nothing exceeds $c$ relative to my rest frame. However, I cannot transform into "the rest frame of one of the light beams" and measure the speed of the other, because it is not possible to define a frame where a light beam is at rest.

Euclidean geometry is the wrong tool to be using for this - you need to be using Minkowski geometry.

 

[Vanadium 50]

@Ibix is correct.

A question that does have an answer is that two massive bodies are moving away from each other (as seen by a third) at velocity v. What is the velocity of one as seen from the other as v →c? That velocity also approaches c.

 

[PeroK]

based on Euclidean geometry, their relative velocities should be 2C.
It seems a paradox, because the two rules give different result.

Based on the definition of velocity, their separation velocity is $\pm 2c$.

 

[FactChecker]

As @Ibix states, the separation speed as measured by an observer in the middle of two light beams going in the opposite direction would be $2c$. (You would have to be very cautious about using this speed. I don't know if there is any physically meaningful use for it.)

Although you can not talk about the speed of one light beam in the reference frame of the other, you can imagine two objects traveling in opposite directions, each at nearly light speed from an observer in the middle. In that case, you can talk about the speed of one object in the reference frame of the other. That will never be greater than $c$. An observer in the middle would see the total separation speed as nearly $2c$. The reason for the difference is the distortion of the Minkowski geometry. In the Minkowski geometry, there is no such thing as a relative speed greater than $c$

 

[DrGreg]

Edit: This post and the following posts that refer to Ilya B are about a post by Ilya B that has since been deleted.

Thus the question of the 'speed in the photon's coordinates' is irrelevant.

This is true, but not for the reasons given.

In the system of photon (moving at speed of light) the thickness of the universe is zero.

No, distance relative to a photon isn't "zero", it's "undefined".

in itself time there is zero time between start of journey and arrival.

No, time relative to a photon isn't "zero", it's "undefined".

 

[weirdoguy]

And, to state it explicitly, there is no such thing as "system (frame of reference) of a photon". I would recommend changing sources from which you are learning relativity.

It is defined.

No it is not. There is no reference frame in which photon is at rest. You are using formulas that I guess you don't understand.

 

[PeterDonis]

for system moving at v=1

There is no such system. The Lorentz transformation itself is undefined for $v = 1$.

 

[PeterDonis]

@Ilya B you are hijacking someone else's thread with uninformed and incorrect claims. If you make any further such posts in this thread, you will receive a warning and a thread ban. Please take heed.

 

[Janos Meri]

Thank you for all the answers.
I'm still trying to digest ...

 

[Janos Meri]

I know all the arguments are true, and most have become proven in practice in the meantime. I'm just thinking loudly, and something weird thing flashed in my mind.

Actually does anyone know the exact speed of light?
The speed of light was measured in our world, which is limited to 3 dimensions. However the universe is much wider space, all aspects.
Let's expand the study and consider the fourth dimension as well, the time axis. It is part of the modern physics, and this was raised by Einstein, firstly.
Let say there are two torches oppositely directed to each other. One light is red and other one light is green. (Color does not relevant) Both lamps emitted a beam of light at the same time.
While one beam of light arrives to the other point, it moved in the space (x, y, and z - axles),
BUT! - In the same time it moves on the 4th axis, on the time axis as well. (t - Axis)

The measured and accepted value of the speed of light is 300,000 km/s. But what if, if this is the just apparent speed of light? The real speed is in the 4 dimensional inertial system is greater. Let say 2C.

Rename the light of speed (C) to apparent light of speed: C_A=300,000 km/s
Let's install a new concept: real speed of light: C_R=600,000 km/s

A2 + B2 = C2

Replace variables by our variables:
C_A² + T² = C_R²
C_A² + T² = (2C_A)²
C_A² + T² = 4C_A²
T² = 3C_A²
T = √3 C_A
T = √3 * 300,000
T = 1.73 * 300,000
T = 519,000 km/s

So, the speed of anything on time axis is nearly 520,000 km/s.

The real speed of the light is 600,000 km/s and the 2 light beams relative speed is 600,000 km/s, what is equal by real light speed, which is not greater than maximum speed of the universe.

In this form, Euclidean geometry is also true and Einstein's rule is true as well.

Plus, we also calculated a hitherto unknown constant.

 

[weirdoguy]

Actually does anyone know the exact speed of light?

Yes, it's defined to be $299792458\frac{m}{s}$

 

[PeterDonis]

Actually does anyone know the exact speed of light?

Yes, because it is now defined, in SI units, to be exactly $299,792,458$ meters per second.

The reason this is possible is that what you are calling "the speed of light" is really an inherent property of spacetime. It's not something we "measure". In "natural" units, it is just $1$; the only reason we use a weird number like $299,792,458$ instead of $1$ is that we humans have had a weird history of developing units for things.

what if, if this is the just apparent speed of light?

None of this makes any sense.

 

[PeroK]

Thank you for all the answers.
I'm still trying to digest ...

It looks like you didn't manage to digest what we tried to explain to you!

 

[weirdoguy]

Euclidean geometry

Specetime geometry is not Euclidean, so you can't use Pythagorean theorem. I think it would be better to spent some time reading textbooks on relativity rather than wasting your time on things that are known to be wrong

 

[Ibix]

It is part of the modern physics, and this was raised by Einstein, firstly.

It was actually Minkowski who pointed out that Einstein's maths could be interpreted as implying a 4d structure to the universe.

The rest of what you've written, charitably, looks like you've seen but failed to understand a Minkowski diagram. You appear to be trying to apply Euclidean geometry in a situation where you need to use Minkowski geometry. I'd strongly recommend spending some time with a textbook.

 

[Janos Meri]

I would respectfully note that I wrote essentially the same thing. The difference is that I rounded the measured light speed to 300,000 km/s. Furthermore, I did not refer to SI as it is obvious.

Yes, according to wikipedia, the accepted last measurement is:
“After centuries of increasingly precise measurements, in 1975 the speed of light was known to be 299792458 m/s (983571056 ft/s; 186282.397 mi/s) with a measurement uncertainty of 4 parts per billion. In 1983, the metre was redefined in the International System of Units (SI) as the distance traveled by light in vacuum in 1 / 299792458 of a second.”

Maybe I didn't express myself well. (Sorry, English is not my native language.)
The point of what I wanted to describe is that this measurement is limited to the 3-dimensional world, but Einstein says there is a fourth dimension, the time axis.
Everything is moving forward in time on the time axis. And from this perspective, all spatial motion is more complex, as it moves not only in space (x, y, z axles) but also on the time axis (t axis).

So in a four-dimensional coordinate system, the speed of light must be greater than 300,000 km/s, because the light also moves along on time axis as well.

 

[Ibix]

but Einstein says there is a fourth dimension, the time axis.
Everything is moving forward in time on the time axis. And from this perspective, all spatial motion is more complex, as it moves not only in space (x, y, z axles) but also on the time axis (t axis).

And, as has been pointed out repeatedly, you are analysing this completely wrongly. You need to use Minkowski geometry, not Euclidean geometry.

So in a four-dimensional coordinate system, the speed of light must be greater than 300,000 km/s, because the light also moves along on time axis as well.

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.

 

[weirdoguy]

The point of what I wanted to describe is (...) in a four-dimensional coordinate system, the speed of light must be greater than 300,000 km/s, because the light also moves along on time axis as well.

And we've already addresed it - it's wrong. Spacetime is not Euclidean, there is no meaningfull way to define "speed along time axis", and other things that you can find in most of the textbooks on relativity.

 

[Janos Meri]

Ok, I'm got it. I'm absolutely wrong.

Sorry that all of you wasted your time for it!

 

[Ibix]

Ok, I'm got it. I'm absolutely wrong.

Sorry that all of you wasted your time for it!

If you want to find out the correct model I would recommend a textbook. Spacetime Physics by Taylor and Wheeler is an excellent text and now free to download from Taylor's website.

 

[Janos Meri]

Thank you very much!

 

[David Lewis]

Summary:: Relative speed of two oppositely directed light beams
1C or 2C... based on Euclidean geometry, their relative velocities should be 2C.

I think you mean Galilean Transform, not Euclidean geometry.

Einstein says there is a fourth dimension, the time axis.

The idea of time as a 4th dimension was proposed, even by laypeople, long before Einstein came on the scene. Einstein and Minkowski were the first to couch spacetime in a formal geometry that agreed with experiment, simplified laws of nature, and was compatible with Maxwell Equations.

 

[Janos Meri]

In the Minkowski geometry, there is no such thing as a relative speed greater than C

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?

 

[weirdoguy]

Am I wrong?

Yes: https://en.wikipedia.org/wiki/Velocity-addition_formula#Special_relativity

 

[Ibix]

Am I wrong?

Yes. The speed of light is always $c$. Speeds do not add linearly - if I see two objects with velocities $u$ and $v$ in the same direction then the one doing $v$ will see the other having velocity $w$, where$$w=\frac{u-v}{1-uv/c^2}$$Note that at speeds that are low compared to the speed of light $1-uv/c^2\approx 1$ and hence $w\approx u-v$, which is why we can get away with pretending that speeds add in every day usage. Note also that if $u=c$ then $w=c$ whatever $v$ is, so if I see an object doing $c$ so will everyone else, as per my second sentence.

 

[Janos Meri]

Yes: https://en.wikipedia.org/wiki/Velocity-addition_formula#Special_relativity

Thanks.
Nice big formula set ...
The next year will be enough for me to digest and understand all of this.

 

[Nugatory]

Am I wrong?

Yes, and googling for “Doppler effect” will find some good explanations.

When the source is approaching you, 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. Thus the frequency at the receiver is greater even though the speed of the light is unchanged.

 

[FactChecker]

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?

No. The light is still coming to the observer at a speed of $c$ but the time between peaks in the frequency changes because the distance traveled is changing as the object travels toward or away. There is no addition of velocities that is relevant.

 

[Janos Meri]

When the source is approaching you, 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. Thus the frequency at the receiver is greater even though the speed of the light is unchanged.

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.
As you wrote:
"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.

In common sense, this should be the case that is the reason why I can't digest, but I'm working on it.

Thanks all answers!

 

[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.