Is this true?

1) c + 25 = c

Also how was the speed of light measured? Did they use a black body to capture the excessive light back then?


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    $\begingroup$ I assume you're trying to add two velocities with one being the speed of light? $\endgroup$ – Kyle Kanos Nov 28 '18 at 20:58
  • $\begingroup$ en.wikipedia.org/wiki/Speed_of_light#History $\endgroup$ – BowlOfRed Nov 28 '18 at 21:00
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    $\begingroup$ c+25 =c+25, otherwise 25=0 which is a contradiction $\endgroup$ – Wolphram jonny Nov 28 '18 at 21:21
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    $\begingroup$ You should've posted two separate questions instead of asking both in the same post. $\endgroup$ – João Vítor G. Lima Nov 28 '18 at 21:31
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    $\begingroup$ you might want to read about how velocity addition works in special relativity: en.wikipedia.org/wiki/… $\endgroup$ – Paul T. Nov 28 '18 at 21:32

$$c + 25 = c$$

This is false. The value of $c$ is:

$$c = 299\,792\,458\, \mathrm{m/s}$$

(Of course, using SI units). The speed of light $c$ is a universal constant. According to Einstein's postulates for Special Relativity, all reference frames see light moving with the same speed, $c$, regardless of their motion relative to the light beam.

The reason you might believe that $c + 25 = c$ is merely because of a misconception about velocity composition.

See, if I am inside a car moving at speed $v = 25\, \mathrm{m/s}$ and I fire a laser (speed $c$) in the direction of my velocity, the Classical formula for velocity composition tells us that a person standing on the ground would see light move with speed $c + 25$ (SI units). But the person should see light moving at $c$, so that implies that $c + 25 = c$.

Except the classical formula for velocity composition is wrong!

In the theory of relativity, it can be shown that the correct formula for velocity addition is different.

If I am inside the car, moving with speed $v$, and fire a bullet with speed $u$ towards the direction of my motion, then the classical formula says that a person standing on the ground measures the bullet's speed ($u'$) to be

$$u' = u + v$$

But the relativistic formula, the formula for composition of velocities in Relativity, says that the speed will be

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

If the speeds $u$ and $v$ are very small compared to $c$, then $uv/c^2$ becomes approximately 0 and we get the classical formula back. This is the case for speeds of bullets, cars, planes etc., the classical formula will work for objects moving with speeds less than 10% of $c$.

In the case of the car with the laser, we would get the velocity of light measured by the standing man to be:

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


$$u' = \frac{c + 25}{1 + \frac{25}{c}}$$

Therefore $u' = c$. See? We get $c$ as the speed of light in both frames (the car and the standing man).

The reason you got the (false) result that $c + 25 = c$ is merely because you tried to apply the classical formula when you shouldn't. Light moves at (of course) the speed of light. In any case where the speeds of objects are comparable to the speed of light (more than 10% of $c$), you have to use the relativistic formula. So when dealing with a light beam, the relativistic formula must be used.

As for measuring $c$, as far as I'm aware the first experiment that accurately predicted the speed of light without using astronomical data was Fizeau's experiment. (https://en.m.wikipedia.org/wiki/Fizeau_experiment).


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