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My limited understanding is: Photons travel at the speed of light (c). If we could theoretically (for the sake of argument) travel at the speed of light minus 1km/h (c-1km/h). And we turn on a flash light attached to our wing and it flashes both directions straight ahead and backward. Then the light in front of us would travel at 1km/h? How about the one in the back? would it go at the speed of 2c-1km/h relative to us?

Now in the above paragraph I mentioned speed of light minus 1km/h. I'm still unsure how our speed gets measured - relatively to which point in universe? Because from our perspective we would hold still (if we’re not accelerating) and the objects around us would be moving at approximately the speed of light.

If some of what I mentioned above is right than isn’t it just easy to shine flash light in both directions of XYZ axis and measure how quickly it returns back to us? Then the miniscule time differences would show us our movement relative to the center (or non-moving point(s) in universe[some people argue that there's no center) of the universe (or non-moving part of universe)?

Based on this video (time19:20): https://www.youtube.com/watch?v=IJhgZBn-LHg even our Milky Way is moving at 2.1 Million Km/h relative to the first detectable light in universe (approximately 0.2% of the speed of light).

thanks for your input. Paul

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Then the light in front of us would travel at 1km/h? How about the one in the back? would it go at the speed of 2c-1km/h relative to us?

We would see both beams of light traveling at c, relative to us, at every point in time, in every inertial frame that we measure from.

I'm still unsure how our speed gets measured - relatively to which point in universe?

Velocity, as you might or might not know, is a relative concept. That is, to determine velocity you must compare to another point of reference.

To answer your question, strictly from a conceptual/mathematical perspective, there is no such thing as a special frame. Even if we were to find an "ether" (the concept of a constant background point of reference that permeates the entire universe) that still wouldn't change the conceptual definition of velocity.

Hopefully you this answers your question.

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  • $\begingroup$ Sorry - this is a better description: Thank you - for responding point by point. Very helpful. Now mass cannot travel at the speed of light - so let's say I travel at 0.6c and you travel at the same speed but exactly the opposite direction against me. I turn on the flash light as described above. Would you observe my photons when they travel towards you and pass by you at the speed of 1.6c from your point of view (as it passes you (from your point of reference))? And what happens after the photons pass your point - they're so fast that you don't see them? Just curious. $\endgroup$ – Paul Jan 7 '17 at 19:01
  • $\begingroup$ Actually, I forgot to add your velocity against me. So I go 0.6c towards some center between us. You go 0.6c towards the same center. That means you travel 1.2c towards me from my point of view. You turn on the light and your photons then go 2.2c towards me. Just trying to understand this. Thanks for help. $\endgroup$ – Paul Jan 7 '17 at 19:13
  • $\begingroup$ @Paul You always observe photons traveling at c. This is a tricky aspect of relativity that takes some time to get used to. It helps to keep in mind that there's no frame of reference (OK, for the Milkyway there kind of is, the cosmic background radiation) but if you're in a car and your eyes are closed, you don't know how fast the car is moving, and there's no good way to test it unless the car makes a turn. Your ship moving a .6 c, if it's not accelerating, to you, it's standing still and everything is flying past it, from your ship's point of view. $\endgroup$ – userLTK Jan 7 '17 at 19:20
  • $\begingroup$ @Paul Hyperphysics has a cool calculator, .6 c in one direction + .6 c in the opposite direction = .88c not 1.2c. See here: hyperphysics.phy-astr.gsu.edu/hbase/Relativ/einvel2.html#c2 Distances and rate that time passes change which makes this kind of strange addition possible. $\endgroup$ – userLTK Jan 7 '17 at 19:23
  • $\begingroup$ thanks for the calculator - that's helpful in explaining the concept - a bit counterintuitive $\endgroup$ – Paul Jan 7 '17 at 19:32

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