# Addition of Velocities: A photon emitted backwards

• wainker1
In summary, the conversation discusses the concept of deriving the addition of velocities in the context of relativity and a specific scenario involving a rocket traveling at the speed of light and emitting a photon. The conversation also touches on the possibility of using L'Hospital's rule to find the velocity of the photon relative to the lab, but ultimately realizes that the correct mathematics was not applied.
wainker1
So I just finished learning about the derivation of the addition of velocities...this should imply that I am a newcomer to relativity. So here's my question:

We have a rocket that is traveling away from a lab at the relative velocity of c. (I know this is impossible because it would take infinite energy, but hear me out...). Within the rocket frame, it emits a photon that travels towards the rear of the ship (towards the lab) at velocity -c relative to the rocket. What is the velocity of this photon relative to the lab? My prediction was either it is -c or it is 0...leaning towards the former.

I was looking at this formula: v = (vrel + v') / (1 + vrel*v'). v is the photon's speed relative to the lab. vrel is the rocket's speed relative to the lab (c). v' is the photon's speed relative to the rocket (-c).

Here's where I got:

Let's have the rocket approach the speed of light and let's just have c = 1 and -c = -1. Then the lim as vrel → 1 = (1 + -1) / (1 + 1*-1) = 0 / 0...indeterminate form.

So I tried applying L'hospital's rule (which I don't know if that's even legal). I assumed v' is a constant in the formula since it is a photon that has a constant velocity so the variable to take a derivation is vrel. After taking a derivation of the top and bottom we get: (1 + 0) / (1 + v') = 1 / (1 + v'). Applying the lim vrel → 1, we get 1 / (1 + -1) = 1 / 0 = ∞?

I know my logic is flawed somewhere so any pointers would help.

Thanks.

wainker1 said:
We have a rocket that is traveling away from a lab at the relative velocity of c. (I know this is impossible because it would take infinite energy, but hear me out...)

Mentz114 said:

Ok, but let's then look at what happens when the rocket gets really really really really close to c...to do this we take a limit as vrel → 1. Then if you follow through the steps of my post, I eventually get an answer of ∞. So even if we remove the impossibility, my logic still has a flaw somewhere.

wainker1 said:
Let's have the rocket approach the speed of light and let's just have c = 1 and -c = -1. Then the lim as vrel → 1 = (1 + -1) / (1 + 1*-1) = 0 / 0...indeterminate form.
Not too surprising to get an indeterminate result when you are calculating something impossible physically.

wainker1 said:
So I tried applying L'hospital's rule (which I don't know if that's even legal). I assumed v' is a constant in the formula since it is a photon that has a constant velocity so the variable to take a derivation is vrel. After taking a derivation of the top and bottom we get: (1 + 0) / (1 + v') = 1 / (1 + v'). Applying the lim vrel → 1, we get 1 / (1 + -1) = 1 / 0 = ∞?

I know my logic is flawed somewhere so any pointers would help.

Thanks.

Using v' = -c = -1 :

The derivative of the numerator (vrel + v') = (vrel -1) wrt vrel is 1.

The derivative of the denominator (1 + vrel*v') = (1 - vrel) wrt vrel is -1.

The result is 1/-1 = -1

So surprisingly we get the right result, probably for all the wrong reasons.

Last edited:
yuiop said:
The derivative of the numerator (vrel + v') = (vrel -1) wrt vrel is 1.

The derivative of the denominator (1 + vrel*v') = (1 - vrel) wrt vrel is -1.

The result is 1/-1 = -1

Ah yes, correct mathematics would solve my problem. Thank you for pointing out my mistake and for the help.

## 1. What is the concept of "Addition of Velocities"?

The concept of "Addition of Velocities" refers to the mathematical method used to calculate the combined velocity of two objects that are moving in relation to each other. It takes into account the initial velocities of both objects and their relative motion to determine the final velocity.

## 2. How does the addition of velocities apply to a photon emitted backwards?

The addition of velocities still applies to a photon emitted backwards, as it is still a moving object. However, the speed of light is constant, so the added velocity will not change the speed of the photon, only the direction in which it is traveling.

## 3. Can the addition of velocities change the speed of light?

No, the speed of light is a fundamental constant and cannot be changed by the addition of velocities. The addition of velocities can only change the direction in which the light is traveling.

## 4. How does the addition of velocities affect the wavelength and frequency of light?

The addition of velocities does not affect the wavelength and frequency of light, as these properties are intrinsic to the light itself and do not change with relative motion. However, the observed wavelength and frequency may appear to change due to the Doppler effect, which is a separate phenomenon.

## 5. Is the addition of velocities applicable to all types of light, or just photons?

The addition of velocities is applicable to all types of light, not just photons. It can be used to calculate the combined velocity of any moving object, including particles with mass and other forms of electromagnetic radiation such as radio waves, microwaves, and x-rays.

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