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Relative speed of two oppositely directed light beams


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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?

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7 minutes ago, Yanosh said:

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?

For the third observer, the mutual velocity of the two particles should not exceed 2C, since the velocity of each particle relative to this observer cannot exceed c. 

In the reference frame of the third observer, the velocities are added according to the rules of vector algebra without the use of the Lorentz factor.

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Of course there's an explanation.

The simplest is that, to measure the speed of the second oppositely directed light beam, you need to be in the FoR of the first light beam.
And since light has no valid FoR, you need to use SR addition rules ( Lorentz transformations )from a valid third frame.
Where you will always measure c to be the maximum speed.

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22 minutes ago, Yanosh said:

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?

 

An observer will see the beams separate at 2c. You can’t analyze this from the perspective of either light beam, since that does not represent a valid inertial frame if reference

If these were objects traveling at an attainable speed (i.e. less than c), then you can do this with the velocity addition formula. Let’s say they move at u and v relative to the central frame (of the person tossing the objects). They will see the total distance between them increase at (u+v) (using scalar speeds)

Each object will see the other recede at (u+v)/(1+uv/c^2)

http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/einvel2.html#c2

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This might give you more insight into SR rules:

Quote

Time is not frozen from light's perspective, because light does not have a perspective. There is no valid reference frame in which light is at rest. This statement is not a minor issue that can be approximated away or overcome by a different choice of words. This statement is fundamental to Einstein's theory of Special Relativity, which has been experimentally validated thousands of times over the last hundred years. The whole framework of Special Relativity is based on two fundamental postulates:

1. The laws of physics are the same in all inertial reference frames
2. The speed of light in vacuum is the same in all inertial reference frames.

If there were a valid reference frame in which light was at rest, then that would violate Postulate 2 because the speed of light would be different in various reference frames (i.e. the speed of light would be c in some frames and zero in its rest frame). And if Postulate 2 is discarded, then the entire theory of Special Relativity is discarded, because Special Relativity is derived from these two postulates. Asking the question, "If we just pretend that light has a reference frame, then what would happen?" will only lead to nonsense answers. Once you pretend that, you have thrown out all of Special Relativity, and you are just left with nonsense and science fiction.  Read more> https://wtamu.edu/~cbaird/sq/2014/11/03/why-is-time-frozen-from-lights-perspective/#:~:text=According to Special Relativity%2C as,exactly the speed of light.

 

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The speed of light is c in all inertial frames as you seem to know.  That means an observer would 'see' both beams moving at c.  That further means the closing speed would be 2c.  In other words, if the 2 beams were 2 Ls apart, after 1 second they would pass each other, but of course nothing is actually exceeding c.

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3 minutes ago, Bufofrog said:

The speed of light is c in all inertial frames as you seem to know.  That means an observer would 'see' both beams moving at c.  That further means the closing speed would be 2c.  In other words, if the 2 beams were 2 Ls apart, after 1 second they would pass each other, but of course nothing is actually exceeding c.

The OP seems to be conflating separation speed with light speed.

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  • 3 weeks later...

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. Which means that the relative speed of light beam is greater than C relative to the observer?

 

Am I wrong?

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Posted (edited)
24 minutes ago, Yanosh said:

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. Which means that the relative speed of light beam is greater than C relative to the observer?

Am I wrong?

 

Yes you are.  The light is blue shifted but the speed is still c.

Edited by Bufofrog
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8 hours ago, Yanosh said:

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. Which means that the relative speed of light beam is greater than C relative to the observer?

 

 

 

Am I wrong?

You are wrong in assuming Speed of the object is added to  the speed of light.

Here are two objects with a light source at rest between them. The light waves spread outward at c in all direction; each wave hitting both objects at the same time.

doppler1.gif.d8d0977e169b0b8479f807af5b56aa61.gif

Here's the scenario with the light initially being emitted when the source is halfway between the objects, but the source is moving towards the Blue observer.

doppler2.gif.d84ea13d6c6d06c9cbcc1def5fe1d5e3.gif

The waves still expand out from the source at c ( the first wave still reaches both observers at the same time.), however, each successive wave is emitted from a point that closer the blue observer and further from the red. Each wave emitted in the direction of Blue is emitted closer behind the previous wave and each wave emitted in the direction of Red trails further behind the previous wave.  Blue thus sees waves in a more rapid succession ( at a high frequency) than red does.  All without there being any change in the speed at which the light itself travels.

 

 

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