http://www.youtube.com/watch?v=aZrjMmMBa_8

Time dilation in the context of the theory of relativity does not seem to make sense to me. It is based on the premise that the speed of light is constant, which is said to be an exception when discussing relativity. For example, as demonstrated in the video above (5:00 to 8:00), the distance traversed by a particle of light that is emitted from a stationary car is said to be 'd=ct', whereas when such a particle is emitted by a car moving at velocity v, then the distance traversed by such a particle is represented by the formula d=(v+c)t, because the particle already possessed the velocity of the car when it was emitted. After this is explained logically in the video, it is said that in case 2, since the distance has increased, time must have increased as well. But clearly, the increase in time is due to the addition of 'v'. I understand that if the distance equation in the second case uses 'c' instead of 'v+c' as velocity, then it is reasonable that the time should increase, if this phenomena is observed by a stationary observer.

Why is it an exception that the speed of light emitted from a source does not depend on the speed of the source? If the same principle is applied to every other body, why is light exempted from it? It would be great if someone could explain this to me, or provide resources explaining this.

Edited October 27, 201312 yr by Ferreroire

Try to use relative simultaneity and relative length contraction.

The fact that the speed of light does not rely on the speed of the source is an empirical fact. It is something that we have found to be true through experiment.

 

As to why this is, the simple answer is that it is the way the universe is put together. It is due to the fact that there is something known as an invariant speed, a speed that does not change from observer to observer that also is finite. (In Newtonian physics, the invariant speed is infinity).

 

It turns out that light travels at this invariant speed since it massless, any anything massless must travel at this speed.

 

As to light not behaving to the same rules as the particle in your example. In fact it does, it is just that the rules for adding velocities even for particles is not as simple as the video presents.

 

The actual rule is

 

[math]V_t = \frac{v_1+v_2}{1+ \frac{v_1v_2}{c^2}}[/math]

 

where c is the speed of light.

 

As long as v1 and v2 are small, the answer comes out to very close to v1+v2

 

However, as one or both approach the speed of light, this is no longer the case.

 

If v1= 0.8c, and v2=0.19c, you don't get an answer of 0.99c, but an answer of 0.859c

 

and if either v1 or v2 = c, then you get c out as an answer.

 

Basically, as v1 and v2 approach c, the answer approaches c. It just seems like the light and everyday objects don't follow the same rules because everyday objects have speeds so much smaller than the natural limit.

Why is it an exception that the speed of light emitted from a source does not depend on the speed of the source? If the same principle is applied to every other body, why is light exempted from it? It would be great if someone could explain this to me, or provide resources explaining this.

 

Because experiment tells us that this is the case.

Thanks Janus! That was helpful. A few questions though - am I correct in understanding that in the formula you described, v1 (velocity of the car) is relative to the initial reference frame (in this case the road/earth), and v2 (velocity of light) is relative to v1? And Vt, as the function of V1 and V2, is relative to the initial reference frame?

Edited October 27, 201312 yr by Ferreroire

Thanks Janus! That was helpful. A few questions though - am I correct in understanding that in the formula you described, v1 (velocity of the car) is relative to the initial reference frame (in this case the road/earth), and v2 (velocity of light) is relative to v1? And Vt, as the function of V1 and V2, is relative to the initial reference frame?

1 and 2 refer to the respective frames. The speed of light is c.

Thanks Janus! That was helpful. A few questions though - am I correct in understanding that in the formula you described, v1 (velocity of the car) is relative to the initial reference frame (in this case the road/earth), and v2 (velocity of light) is relative to v1? And Vt, as the function of V1 and V2, is relative to the initial reference frame?

and v2=c(velocity of light) is relative to the car.And Vt is velocity of the light relative to the initial reference frame

 

Vt=(v1+c)/(1+v1c/c²)

Vt=(v1+c)/(1+v1/c)

Vt=(v1+c)/[(c+v1)/c]

Vt=(v1+c)c/(v1+c)

Vt=c

Edited October 28, 201312 yr by DimaMazin

Thanks Janus! That was helpful. A few questions though - am I correct in understanding that in the formula you described, v1 (velocity of the car) is relative to the initial reference frame (in this case the road/earth), and v2 (velocity of light) is relative to v1? And Vt, as the function of V1 and V2, is relative to the initial reference frame?

Here, let's use numbers.

 

If a pitcher riding on a train moving at 100 mph throws a 90 mph fastball in the direction of travel, someone watching on the ground using classical addition of velocities would say the ball was traveling at 190 mph.

 

Using the more accurate formula for adding velocities in relativity, we'd get this:

 

[math]V_t = \frac{v_1+v_2}{1+ \frac{v_1v_2}{c^2}}[/math]

 

[math]V_t = \frac{100+90}{1+ \frac{9000}{670616629^2}}[/math]

 

Which gives you a result of 189.9999999999961769 mph.

 

That's a difference of a quarter of a millionth of an inch per hour. So you can see why it looks like velocities add normally at speeds we usually encounter that are so far below lightspeed.

Thank you! I understand it now.

Edited October 28, 201312 yr by Ferreroire