Does Light Reach Objects Moving at High Speeds?

[themaster1j]

As I just registered as a member, I don't know if this question has been asked before on this forum, and I'm sorry if I'm asking it again.

From what I understand, even if an object is moving away from light with a significant speed, the light would still approach that object as if it would be standing still.
If an object travels at 50% of the speed of light and a light beam is emitted at ~300,000km behind that object, how can the light reach the object in 1 second if the object is moving away from the light? Doesn't it mean that the distance the light would have to travel would be more than 300,000km?

I apologize for my basic understanding of Physics.

 

[1994Bhaskar]

Well for the object with speed 50% of speed of light time will increase will increase by the formula t=t0/(sqrt(1-(v/c)^2)), where t0 is original time.Similar way is for distance.The distance from your frame will have increased so that light cover's only 300000 km in one second.

 

[ZealScience]

Read about Lorentz trans formation. Speed of light is postulated to be constant by relativity because of causality. Lorentz transformation is what makes speed of light constant.

Last but not least, welcome to PF!

 

[pervect]

As I just registered as a member, I don't know if this question has been asked before on this forum, and I'm sorry if I'm asking it again.

From what I understand, even if an object is moving away from light with a significant speed, the light would still approach that object as if it would be standing still.
If an object travels at 50% of the speed of light and a light beam is emitted at ~300,000km behind that object, how can the light reach the object in 1 second if the object is moving away from the light? Doesn't it mean that the distance the light would have to travel would be more than 300,000km?

I apologize for my basic understanding of Physics.

In the frame where the object is moving away, it does takes more than one second. In fact you can write

x1 = -3e8 m + 3e8(m/s) * t
x2 = 1.5e8 (m/s) *t

and solve for the value of time at which x1 = x2 as t=2

However, it is still true that if you transform to the frame of the moving object, so that it is standing still, the light beam still approaches it at 'c'. You have to be careful how you analzye this, becuase the notion of simultaneity is the moving frame is different - specifically, the coordinates of the object (t=0, x=-3e8 meters) transform to something like (t=.5*gamma seconds, x = -3e8*gamma meters), where gamma 2/sqrt(3).

This is done via the Lorentz transform
$$x' = \gamma \left( x - v\,t\right)$$
$$t' = \gamma \left( t - \frac{v}{c^2} \, x\right)$$

 

[themaster1j]

So the distance the light has to travel to reach the object doesn't increase, even though the object is moving away from the light?