A question on special relativity

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SUMMARY

The discussion centers on the relativistic addition of velocities in the context of special relativity. A user questioned how two cars, each traveling at 0.8c in opposite directions, could be perceived as exceeding the speed of light from one driver's frame of reference. The response clarified that using Galilean addition of velocities leads to incorrect conclusions, and instead, the relativistic formula should be applied: V_{a/c} = (V_{a/b} + V_{b/c}) / (1 + (V_{a/b} V_{b/c})/c^2), resulting in a relative speed of 0.975c, which adheres to the principles of special relativity.

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choon_min
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hi, everone I am new here, here is my 1st question, hope anyone can help me.

As the relative law, speed of any object can't exceed speed of light, c.
If got one car A move to right with speed 0.8c and another car B move to left 0.8c. (if assume car can move so fast la..)

So, if i am the driver of car A, for my frame of reference the car B speed relative to me is 1.6c, which is more than c, violate with theory of relativity, so any problem here?
 
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choon_min said:
So, if i am the driver of car A, for my frame of reference the car B speed relative to me is 1.6c, which is more than c, violate with theory of relativity, so any problem here?
That's only true if you use Galilean addition of velocities, which is only accurate for small speeds:

[tex]V_{a/c} = V_{a/b} + V_{b/c}[/tex]

The correct way to add velocities is to use the relativistic addition of velocities, which is correct for all speeds:

[tex]V_{a/c} = \frac{V_{a/b} + V_{b/c}}{1 + (V_{a/b} V_{b/c})/c^2}[/tex]

Thus the speed of car B with respect to you will be 0.975 c, not 1.6 c.

Note that for small speeds (small compared to light speed), the formula for relativistic addition of velocity is well approximated by the Galilean addition of velocity formula.
 

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