Relativity of 2 cars' velocities

[kyle1320]

Suppose 2 cars are on a highway going toward each other, each at half the speed of light. Can the driver of one car see the other car as it is going towards him, past him, and away from him? I imagine as the other car was coming towards him, it would get blue shifted out of the visible spectrum, and oppositely as the other car was going away it would get red shifted out of the visible spectrum. But as they passed each other, would the other car appear to be right next to him, and stay that way? Or would he only see the other car for the exact moment they were next to each other? Would he never see the other car? Thanks in advance :)

 

[tiny-tim]

welcome to pf!

hi kyle1320! welcome to pf!

the can always see each other …

to add velocities, see http://en.wikipedia.org/wiki/Velocity-addition_formula#Special_theory_of_relativity"

 

[kyle1320]

-REMOVED- weird. I only took general physics, didn't even know about that. I just still have a lot of questions from that class. :P

 

[kyle1320]

Sorry for the double post, but:

s = v+u / 1+(vu/c^2)

s = actual velocity of fly (car 1)
v = actual velocity of ship (car 2)
u = velocity of fly (car 1) as calculated on ship (car 2)
c = speed of light

so:

.5 = .5+u / 1+(.5u/1^2)
.5 = .5+u / 1+.5u
u = 0

So to each other, the cars would appear to be not moving. Is this what you meant by they would always see each other? What would the surrounding landscape look like then? Would the other car appear to be moving backwards?

 

[tiny-tim]

no, u and v are the actual speeds of the two cars, = 0.5,

so their relative speed = (0.5 + 0.5)/(1 + 0.5*0.5) = 1/(5/4) = 4/5

 

[kyle1320]

"As Galileo observed, if a ship is moving relative to the shore at velocity v, and a fly is moving with velocity u as measured on the ship"
"where s is the velocity of the fly relative to the shore."

 

[tiny-tim]

Galileo's ship is your highway

 

[DaveC426913]

As tt points out, the cars observe each other moving toward them at 4/5c.

I'm not sure what your Galileo reference is intended to portray.

 

[tiny-tim]

follow the link, dave!

 

[kyle1320]

The ship and the fly are the cars, and the highway is the shore, as it is a stationary point of reference as to the velocity of the 2 cars.

 

[mrspeedybob]

The OP made specific reference to the effects of redshift and blue shift of the light on the visibility of the cars, I don't think this was a question about velocity addition but rather about the extent of frequency shifting moving things out of the visible spectrum.

Light that is outside the visible spectrum is simply light that your eye does not register. The fact that the light bouncing off of a car is undetectable to your eye will not make the car invisible since you still will not be able to see the light from what is behind the car, since the car is in the way. Your eye simply will not detect the light reflected off of the car. Light not detected looks just like no light, which looks black. So if a red car is approaching at a high enough speed it will look blue, If there is a blue car right beside it approaching you at the same speed it will look black. As the 2 cars pass you the blue one will shift to red and the red one will shift to black.

 

[DaveC426913]

follow the link, dave!

Yeah yeah. Quoting without context is meaningless.

 

[tiny-tim]

hi kyle1320!

(just got up :zzz: …)

The ship and the fly are the cars, and the highway is the shore, as it is a stationary point of reference as to the velocity of the 2 cars.

but the shore (in http://en.wikipedia.org/wiki/Velocity-addition_formula" ) is not the stationary point of reference as to the given velocity v of the fly