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Is the speed of light constant relative to the observer? Hypothetically, I am standing at the front of a train. The train is traveling 100 mph. I shine a flashlight pointing forward in the direction the train is traveling.

Is the light from the flashlight traveling at 670,616,629 mph or (670,616,629 mph) +the speed of the train (100 mph) = 670,616,729 mph.?

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The light travels at 670,616,629 mph according to an observer on the train and also at 670,616,629 mph according to an observer on the ground.

The speed of light is the same for all inertial observers, despite their relative motion. Velocities don’t actually “add” the way you would think they should based on our everyday experience at low velocities. The formula for combining velocities in Special Relativity is more complicated than simple vector addition. At low velocities it reduces to the familiar addition but at high velocities $c$ acts as the “speed limit”.

If you are curious about the correct velocity-addition formula, see Wikipedia.

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  • $\begingroup$ I went to the link and tried to understand what was being said. Sadly, for me I did not find a direct answer to my question. I will answer my own question with this statement. The speed of light is not relative to the source nor the environment it is in. If I shine a flashlight out ahead of a moving train the light will travel at the speed of light irrespective of the source. If I am traveling at the speed of light and I shine a flashlight ahead of me in the direction of the light it will travel at the speed of light with no connection to or relative to the train. $\endgroup$ – farsideofourmoon May 15 '20 at 16:11
  • $\begingroup$ In the first equation, take $v’$ to be the speed of the light relative to the train, $u$ the speed of the train relative to the ground, and $v$ the speed of the light relative to the ground. When $v’$ is $c$, $v$ is also $c$. The formula is more interesting when you have a train moving at, say 2/3 the speed of light and someone on the train throws a ball at 2/3 the speed of light. See how fast the ball will move relative to the ground. It won’t be 4/3 of $c$; it will be less than $c$. $\endgroup$ – G. Smith May 15 '20 at 16:19
  • $\begingroup$ G. Smith, I thank you and others here for spending your time trying to educate me on this topic. I only asked the question because I have nothing better to do. At 70 and at home alone I find myself looking for reasons for being. This back and forth has given me food for thought even if its calory content is near zero, still thanks. Mr. Smith, I wish you, and all your fellow very bright friends an even brighter day tomorrow and beyond. bye $\endgroup$ – farsideofourmoon May 15 '20 at 16:33
  • $\begingroup$ @farsideofourmoon Best to you as well! $\endgroup$ – G. Smith May 15 '20 at 16:42

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