Relativity Question: Light Moving at 2 Speeds at Once?

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This discussion centers on the relativistic effects of light speed as perceived by different observers. It clarifies that while Person A on a spaceship moving at the speed of light perceives light from a flashlight as moving at light speed, Person B on the ground also measures the light at the same speed due to the principles of special relativity. The Galilean addition of velocities is invalid at relativistic speeds, and the correct formula for velocity addition is v(total) = (v1 + v2) / (1 + (v1 * v2)/c²). Understanding time dilation and length contraction is crucial for grasping these concepts.

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  • Understanding of special relativity principles
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Chiko
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Hey, i asked my physics teacher this, but he wasent interested in explaining it to me :mad:

Person A is on a boat moving at 10km/h
Person B is on the ground standing still
Person C is walking foward on the boat at 2km/h

Person C in relation to PA is moving at 2km/h
and PC in relation to PB is moving at 12km/h

I understand this concept, but my book said that it does not apply at speeds near the speed of light. So here's my question:

Person A is on a spaceship (pretend, for simplicity reasons) cruizing at the Speed of light.
Person B in standing still on the ground.

PA turns on a flashlight pointing foward.
The light in relation to PA moves foward at the speed of light.
But in relation to PB the light would move at a speed of 2 times the speed of light, right?

However, my teacher told me that in relation to both people the light would be moving at the same rate. Which would mean that the light is traveling at 2 speeds at once (SoL in relation to PA and 2SoL in relation to PB - because the spaceship is cruizing at SoL). This highly confuses me, please explain?

Thanks in Advance
Chiko
 
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However, my teacher told me that in relation to both people the light would be moving at the same rate. Which would mean that the light is traveling at 2 speeds at once (SoL in relation to PA and 2SoL in relation to PB - because the spaceship is cruizing at SoL). This highly confuses me, please explain?
This is a basic problem in relativity. To understand it with no math, you have to remember that relativity also "generates" (the word isn't correct because the effect would also exist if we hadnt discovered SR !) time dilatation and length contraction.

Where you think there should be different speeds, there is only one, and this is easily explained when you remeber those effects : the distance light has to travel is in fact a bit shorter and time is dillated.

Moreover, the well-knnown formula v(total) = v1 + v2 to add velocities is false in SR. vtotal = (v1 + v2) / (1 + (v1 * v2)/c²)

Sorry if my message is a bit confusing but it's 1.30am in France and i am quite tired :(

PS : Be careful, you cannot travel at the speed of light, and you cannot use SR in frameworks going at the speed of light.
 
Last edited:
Chiko said:
Person A is on a spaceship (pretend, for simplicity reasons) cruizing at the Speed of light.
Excuse me?

The light in relation to PA moves foward at the speed of light.
But in relation to PB the light would move at a speed of 2 times the speed of light, right?

Let's say the spaceship is going past the person at .99C, than the light from the flashlight is passing the spaceship at the speed of light. The light from the flashlight is moving away from the "stationary" person at the speed of light also. That's why there is time dialation.
 
Time dilation huh? I am thinking this question is over the head of a just starting physics one student, hehe. Cause I am still totally lost. Because I am relating the speed the light is traveling to the speed person C traveled in the first problem - just they have different speeds. But time dilation and such thrown into the mixture would change stuff i assume.

Ill read over this a few times and see if i can make sense - Wish me luck! ^.~

Chiko
 
If you shine a flashlight on earth, you can measure it to be going exactly what we think it would go. If you do the same on a the moon, you would see it is the exact same velocity as it was on earth, yet the moon is not stationariy relative to the earth. If you understand the concept that the speed of light will always be going the same exact velocity relative to anything, you may be closer at understanding a lot of things.

Or, if you are drifting in space, you may feel as if you were stationary. You shine a flashlight and you measure it to be exactly what it is on earth. Yet, relative to certain stars, you could be going extremely fast.
 
Chiko said:
Person A is on a boat moving at 10km/h
Person B is on the ground standing still
Person C is walking foward on the boat at 2km/h

Person C in relation to PA is moving at 2km/h
and PC in relation to PB is moving at 12km/h
Think for a second about the reason this is true. It does seem intuitively obvious because the low speeds we constantly deal with in every day life do behave this way, but is there any real reason to believe this is the way the universe should behave? Is it a priori (something you know independent of experience) or a posteriori (something you know because of experience)? It may be hard to accept, but it turns out that the Galilean addition of velocities law (v_{AC}=v_{AB}+v_{BC}) is not independent from experience. We know we can add velocities only because it works so well in every day activities.

However, Einstein showed us that the Galilean equation for the addition of velocities is only an approximation. The way relative velocities truly compare in our universe is similar but not quite Galilean:

v_{AC}=\frac{v_{AB}+v_{BC}}{1+\frac{v_{AB}v_{BC}}{c^2}}

Using this equation we can see that if observer C is moving 2 km/h relative to observer B (v_{BC}=2 km/h) and observer B is moving 10 km/h relative to observer A (v_{AB}=10 km/h) then observer C is moving extremely close to 12 km/h relative to observer A (v_{AC}{\approx}12 km/h) since the speed of light is so large (c=1,079,252,848.8 km/h), but as v_{AB} or v_{BC} gets closer and closer to c, the more the Galilean answer starts to deviate from the relativistic addition of velocities. And as you can see, if you fill in either v_{AB}=c or v_{BC}=c then v_{AC} will always equal c.
 
Thats quite interesting! I am sooo glad i finally got into the world of physics. I am closer to understanding this, but not quite there yet (maybe because the equations i don't know yet... ill look up those and read about them sometime soon.) Thanks for all the imput!

-Chiko
 
just curious, why was this moved to the homework section? it didnt sound like homework to me.. :confused:
 

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