Travelling at the speed of light?

In summary, the equation v = (u + v')/(1 + uv'/c2) allows for the calculation of relative velocities in special relativity. For velocities significantly smaller than the speed of light, the equation simplifies to v ≈ u + v'. This can also be applied to photons moving towards each other, but reference frames at the speed of light are forbidden. The relative velocities between the two photons will be equal in magnitude but opposite in direction. However, it is important to note that this equation is only valid for relative velocities with magnitudes less than the speed of light.
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If I was traveling at half the speed of light from A to B and another person traveling at half the speed of light from B to A (not directly towards each other, side-by-side), would i be traveling at the speed of light relative to the other person. If so, what would i see when i look at this other person?
 
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  • #2
v = (u + v')/(1 + uv'/c2)

This is how to sum two velocites, u and v', in SR where c is the speed of light in the vacuum.

If we input the velocities into the equation as a fraction of c, then all the c's oin the bottom of the equation cancel and we're left with:

v = (0.5c + 0.5c)/(1 + 0.25) = 0.8c

Therefore you will observe the other person traveling at 0.8c or 4/5 of the speed of light relative to you.
 
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  • #3
I'll just add for speeds a lot lower that c the two velocities do add algerbraically, that is :

v ≈ u + v'

for values of u and v' significantly smaller than c.

For example, taking two objects that both appear to be traveling ~650 mph (10-6 c) in oppoiste directions to a 'stationary' observer the difference in observed velocities will only be about 1 part in a trillion.
 
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  • #4
So, according to your theory:

If 2 photons were moving towards each other, both moving at the speed of light:

v = (c + c)/(1 + c^2/c^2) = 2c/2 = c

The photons would observe each other moving at the speed of light? If that's the case, how can it tell whether the other photon is moving at all of not?

v = (c + 0)/(1 + 0/c^2) = c/1 = c
 
  • #5
It's not my theory, it's Lorentz's/Einstein's (the equation is derived from the Lorentz transformations), it's called the special theory of relativity.

According to the transformation, both photons would observe each other traveling at the speed of light, but reference frames at this speed are forbidden anyway.
 
  • #6
Originally posted by jcsd
I'll just add for speeds a lot lower that c the two velocities do add algerbraically, that is :

v ≈ u + v'

for values of u and v' significantly smaller than c.

For example, taking two objects that both appear to be traveling ~650 mph (10-6 c) in oppoiste directions to a 'stationary' observer the difference in observed velocities will only be about 1 part in a trillion.

Actually the same rule holds for low speeds - it holds for all speeds, but if you try it for speeds around a billionth of c, which are the speeds we encounter (feet per second) then the Lorentz formula becomes normal addition to a high degree of accuracy. But there is no cutoff where relativity addition stops and the familiar addition begins.
 
  • #7
One photon has velocity +c and the other photon has velocity -c, both relative to some initially-given reference frame.
The relative velocities ought to be calculated by a difference formula:

u' = (u - v)/(1 - uv/c2)

.

So, for the velocity of the +c photon relative to the -c photon,

u' = ((+c) - (-c))/(1-(+c)(-c)/c2) = (+2c)/(+2) = +c

For the velocity of the -c photon relative to the +c photon,

u' = ((-c) - (+c))/(1-(-c)(+c)/c2) = (-2c)/(+2) = -c

. The relative velocities would be equal in magnitude, but oppositely directed.

There is a problem with all this. In order to derive composition theorems like this, guaranteeing that the postulates of special relativity are enforced, it is necessary to assume implicitly that all relative velocities between observers have magnitudes less than c. So, it is not a foregone conclusion that the two photons define reference frames for which the composition formula is valid. But it does appear to give "appropriate" answers. One photon gets a relative velocity of +c and the other gets a relative velocity of -c.
 

1. What is the speed of light?

The speed of light, denoted by the symbol c, is approximately 299,792,458 meters per second. This is the speed at which light travels in a vacuum and is considered to be the fastest possible speed in the universe.

2. Is it possible for an object to travel at the speed of light?

According to Einstein's theory of relativity, it is not possible for an object with mass to travel at the speed of light. As an object's speed increases, its mass also increases, making it impossible for an object with mass to reach the speed of light.

3. What are the implications of travelling at the speed of light?

Travelling at the speed of light would have many implications, including time dilation, length contraction, and an increase in mass. It would also require an infinite amount of energy to reach this speed, making it impossible for our current technology to achieve.

4. Can we observe objects travelling at the speed of light?

No, we cannot observe objects travelling at the speed of light. As an object approaches the speed of light, its mass and energy increase, causing it to become infinitely heavy and requiring an infinite amount of energy to accelerate further. Therefore, we can only observe objects that are travelling significantly slower than the speed of light.

5. How does travelling at the speed of light affect time?

Travelling at the speed of light would cause time to slow down for the traveller relative to an observer. This is known as time dilation and is a consequence of Einstein's theory of relativity. As an object approaches the speed of light, time would slow down significantly for the object, while time would continue at its regular pace for an observer.

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