# Does this count as moving faster than light?

I'm not familiar with any complicated physics equation, however I do understand some basics. Suppose there is two objects, both of them are moving away from each other in a 3-dimensional space, which they both have the speed of half the speed of light ($c/2$). Relative to the reference frame of object A, object B would be moving away from itself equal to the speed of light.

   A          B
<-----      ----->
-c/2       c/2   to a 3rd observer
0          c    to A
-c          0    to B


This seems to break the law of "nothing can move faster than the speed of light". I believe that there is some other explanation for this case. I have done some research, and there are quite a few questions on this topic. Take this one and this one for example. However, they are about adding velocities together, which isn't quite the same as in the case I was describing.

This question is also similar to my case, however the two objects described are both moving toward the same direction, and I'm not shooting any beam or laser toward the other object.

Most probably in reality there are some extremely complex laws and equations which makes this question more complicated. However, the two objects are not interacting with each other and I'm not trying to "detect" the speed of the other object in the reference frame of A or B. Also none of the objects are moving at a significant fraction of the speed of light relative to the third observer, so special relativity doesn't apply(?) I just could not think of any reason why wouldn't B be moving away from A in the speed faster than light.

A simple explanation would be great.

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Just a comment about "However, they are about adding velocities together, which isn't quite the same as in the case I was describing.". They are exactly the same. You have situated "yourself" on the middle object (call it D) and asked the question from there, but the situation described is identical. D moves forward relative A and B moves forward relative D. Same thing. And you'll note that Floris gives the same answer. –  dmckee Aug 19 '14 at 13:11

In your frame of reference, it does indeed look as though the difference in speed between A and B is greater than $c$. But the question is - does A think that B is moving away at that speed? And the answer is "no".

There is a thing called the Lorentz transformation which describes how the observed speed of an object is a function of the speed of the observer; conveniently, this prevents the breaking of the speed limit of special relativity.

Without giving you the math (you asked for a "simple explanation"), there are two things that happen when you are in a moving inertial frame of reference: length contraction, and time dilation. Clocks moving relative to you seem to go slower, and distances become shorter. These changes are described in the Lorentz transform. Net result is that velocity is also changed - this is described in the Einstein velocity addition equation which states

$$u' = \frac{u+v}{1+\frac{uv}{c^2}}$$

When we put $u=v=c/2$, we get $u' = 0.8 c$, so no speed limit is broken.

The key to understand here (after I re-read your question I realized I needed to add this): it is OK for two things to appear to move faster than the speed of light relative to each other - for example, you can see two beams of light, one traveling to the left, and the other traveling to the right, and say "the difference in speed between these photons is 2c". However, there is no frame of reference in which you can observe anything moving faster than the speed of light - and THAT is the condition that special relativity imposes. Does that difference make sense?

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That light has a fixed velocity in vacuum comes from observations . In order to fit the data Lorenz transformation were imposed on the rigorous mathematical model for electromagnetism, Maxwell's equations.

It was the result of attempts by Lorentz and others to explain how the speed of light was observed to be independent of the reference frame, and to understand the symmetries of the laws of electromagnetism. The Lorentz transformation is in accordance with special relativity, but was derived before special relativity.

So your thought experiment is not in agreement with the data, measurements tell us that the speed of light in vacuum is the same no matter the frame of reference used. It is counter intuitive , you are using every day billiard ball intuition which is wrong for speeds compared to the speed of light. In mathematical terms you are using three dimensional vectors, whereas the behavior of nature is controlled at those speeds by the algebra of four dimensional vectors, where time is an imaginary number.

Einstein used the concept of Lorenz transformations in proposing special relativity which led to the famous E=m*c^2, and special relativity has never been falsified by experiment. It is the nature of reality for very fast moving matter.

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IN which way can this be seen as a simple change of dimensionality of the vector algebra? I am incompetent on this, and it is a naive question. –  babou Aug 19 '14 at 16:07
@babou well, read this en.wikipedia.org/wiki/Four-vector –  anna v Aug 19 '14 at 19:38

"Most probably in reality there are some extremely complex laws and equations which makes this question more complicated."

Not really. The equations are rather straightforward. Let's measure velocities in units of the speed of light and let's denote the velocity of $B$ as observed by $A$ as $v_{BA}$, the velocity of $A$ as observed by $C$ (the observer in the center) as $v_{AC}$, etc.

It seems natural to assume that $$v_{AB} + v_{BA}=0$$ $$v_{AC} + v_{CA}=0$$ $$v_{CB} + v_{BC}=0$$ and $$v_{AB} + v_{BC} + v_{CA} = 0$$

Indeed, till early 20th century these were considered universal truths. However, since 1905 we know that the last equation is only correct if all velocities are small compared to the speed of light (I.e. if all $v_{ij}$ in above equation are in absolute value much smaller than unity). If this is not the case, the equation $$v_{AB} + v_{BC} + v_{CA} + v_{AB} v_{BC} v_{CA} = 0$$ should be used.

You might want to check that for $v_{CA}=v_{BC}=1/2$, the inclusion of the triple product yields $v_{BA}=-v_{AB}=4/5$. Using the approximate (low speed) equation without the triple product term, you would derive the erroneous result $v_{BA}=-v_{AB}=1$.

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